SECTION 6: TIME-DOMAIN ANALYSIS - College of...
Transcript of SECTION 6: TIME-DOMAIN ANALYSIS - College of...
ESE 330 ndash Modeling amp Analysis of Dynamic Systems
SECTION 6 TIME-DOMAIN ANALYSIS
K Webb ESE 330
This first sub-section of notes continues where the previous section left off and will explore the difference between the forced and natural responses of a dynamic system
Natural and Forced Responses2
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3
Natural and Forced Responses
In the previous section we used Laplace transforms to determine the response of a system to a step input given zero initial conditions The driven response
Now consider the response of the same system to non-zero initial conditions only The natural response
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Natural Response
119910 + 119887119887119898119898119910 + 119896119896
119898119898119910119910 = 0 (1)
Use the derivative property to Laplace transform (1) Allow for non-zero initial-conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119904119904119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 0 (2)
Same springmassdamper system Set the input to zero Second-order ODE for displacement
of the mass
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Natural Response
Solving (2) for 119884119884 119904119904 gives the Laplace transform of the output due solely to initial conditions
Laplace transform of the natural response
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(3)
Consider the under-damped system with the following initial conditions
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01119898119898119904119904
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Natural Response
Substituting component parameters and initial conditions into (3)
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
(4)
Remember itrsquos the roots of the denominator polynomial that dictate the form of the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
For the under-damped case roots are complex Complete the square Partial fraction expansion has the form
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
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Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
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Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
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Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
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Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
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Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
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Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
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Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
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14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
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Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
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Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
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Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
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Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
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Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
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Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
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Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
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Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
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Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
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Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
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Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
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Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
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Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
This first sub-section of notes continues where the previous section left off and will explore the difference between the forced and natural responses of a dynamic system
Natural and Forced Responses2
K Webb ESE 330
3
Natural and Forced Responses
In the previous section we used Laplace transforms to determine the response of a system to a step input given zero initial conditions The driven response
Now consider the response of the same system to non-zero initial conditions only The natural response
K Webb ESE 330
4
Natural Response
119910 + 119887119887119898119898119910 + 119896119896
119898119898119910119910 = 0 (1)
Use the derivative property to Laplace transform (1) Allow for non-zero initial-conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119904119904119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 0 (2)
Same springmassdamper system Set the input to zero Second-order ODE for displacement
of the mass
K Webb ESE 330
5
Natural Response
Solving (2) for 119884119884 119904119904 gives the Laplace transform of the output due solely to initial conditions
Laplace transform of the natural response
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(3)
Consider the under-damped system with the following initial conditions
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01119898119898119904119904
K Webb ESE 330
6
Natural Response
Substituting component parameters and initial conditions into (3)
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
(4)
Remember itrsquos the roots of the denominator polynomial that dictate the form of the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
For the under-damped case roots are complex Complete the square Partial fraction expansion has the form
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
K Webb ESE 330
7
Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
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8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
3
Natural and Forced Responses
In the previous section we used Laplace transforms to determine the response of a system to a step input given zero initial conditions The driven response
Now consider the response of the same system to non-zero initial conditions only The natural response
K Webb ESE 330
4
Natural Response
119910 + 119887119887119898119898119910 + 119896119896
119898119898119910119910 = 0 (1)
Use the derivative property to Laplace transform (1) Allow for non-zero initial-conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119904119904119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 0 (2)
Same springmassdamper system Set the input to zero Second-order ODE for displacement
of the mass
K Webb ESE 330
5
Natural Response
Solving (2) for 119884119884 119904119904 gives the Laplace transform of the output due solely to initial conditions
Laplace transform of the natural response
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(3)
Consider the under-damped system with the following initial conditions
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01119898119898119904119904
K Webb ESE 330
6
Natural Response
Substituting component parameters and initial conditions into (3)
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
(4)
Remember itrsquos the roots of the denominator polynomial that dictate the form of the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
For the under-damped case roots are complex Complete the square Partial fraction expansion has the form
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
K Webb ESE 330
7
Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
K Webb ESE 330
8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
4
Natural Response
119910 + 119887119887119898119898119910 + 119896119896
119898119898119910119910 = 0 (1)
Use the derivative property to Laplace transform (1) Allow for non-zero initial-conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119904119904119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 0 (2)
Same springmassdamper system Set the input to zero Second-order ODE for displacement
of the mass
K Webb ESE 330
5
Natural Response
Solving (2) for 119884119884 119904119904 gives the Laplace transform of the output due solely to initial conditions
Laplace transform of the natural response
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(3)
Consider the under-damped system with the following initial conditions
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01119898119898119904119904
K Webb ESE 330
6
Natural Response
Substituting component parameters and initial conditions into (3)
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
(4)
Remember itrsquos the roots of the denominator polynomial that dictate the form of the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
For the under-damped case roots are complex Complete the square Partial fraction expansion has the form
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
K Webb ESE 330
7
Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
K Webb ESE 330
8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
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22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
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23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
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25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
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27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
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33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
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34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
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Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
5
Natural Response
Solving (2) for 119884119884 119904119904 gives the Laplace transform of the output due solely to initial conditions
Laplace transform of the natural response
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(3)
Consider the under-damped system with the following initial conditions
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01119898119898119904119904
K Webb ESE 330
6
Natural Response
Substituting component parameters and initial conditions into (3)
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
(4)
Remember itrsquos the roots of the denominator polynomial that dictate the form of the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
For the under-damped case roots are complex Complete the square Partial fraction expansion has the form
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
K Webb ESE 330
7
Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
K Webb ESE 330
8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
6
Natural Response
Substituting component parameters and initial conditions into (3)
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
(4)
Remember itrsquos the roots of the denominator polynomial that dictate the form of the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
For the under-damped case roots are complex Complete the square Partial fraction expansion has the form
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
K Webb ESE 330
7
Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
K Webb ESE 330
8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
7
Natural Response
119884119884 119904119904 = 015119904119904 +071199041199042+4119904119904+16
= 1199031199031 119904119904+2 +1199031199032 3464119904119904+2 2+ 3464 2 (5)
Multiply both sides of (5) by the denominator of the left-hand side
015119904119904 + 07 = 1199031199031119904119904 + 21199031199031 + 34641199031199032
Equating coefficients and solving for 1199031199031 and 1199031199032 gives
1199031199031 = 015 1199031199032 = 0115
The Laplace transform of the natural response
119884119884 119904119904 = 015 119904119904+2119904119904+2 2+ 3464 2 + 0115 3464
119904119904+2 2+ 3464 2 (6)
K Webb ESE 330
8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
8
Natural Response
Inverse Laplace transform is the natural response119910119910 119905119905 = 015119890119890minus2119905119905 cos 3464 119905119905 + 0115119890119890minus2119905119905 sin 3464 119905119905 (7)
Under-damped response is the sum of decaying sine and cosine terms
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
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22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
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23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
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25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
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27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
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33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
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Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
9
Driven Response with Non-Zero ICrsquos
Now Laplace transform allowing for both non-zero input and initial conditions
1199041199042119884119884 119904119904 minus 119904119904119910119910 0 minus 119910 0 + 119887119887119898119898119884119884 119904119904 minus 119887119887
119898119898119910119910 0 + 119896119896
119898119898119884119884 119904119904 = 1
119898119898119865119865119894119894119894119894 119904119904
Solving for 119884119884 119904119904 gives the Laplace transform of the response to both the input and the initial conditions
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0 +1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(8)
119910 +119887119887119898119898119910 +
119896119896119898119898119910119910 =
1119898119898119865119865119894119894119894119894 119905119905
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
10
Driven Response with Non-Zero ICrsquos
Laplace transform of the response has two components
119884119884 119904119904 =119904119904 119910119910 0 + 119910 0 + 119887119887
119898119898119910119910 0
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
+1119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
Total response is a superposition of the initial condition response and the driven response
Both have the same denominator polynomial Same roots same type of response
Over- under- critically-damped
Natural response - initial conditions Driven response - input
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
11
Driven Response with Non-Zero ICrsquos
Laplace transform of the total response
119884119884 119904119904 =015119904119904 + 07 + 1
1199041199041199041199042 + 119887119887
119898119898 119904119904 + 119896119896119898119898
=0151199041199042 + 07119904119904 + 1
119904119904 1199041199042 + 119887119887119898119898 119904119904 + 119896119896
119898119898
Transform back to time domain via partial fraction expansion
119884119884 119904119904 =1199031199031119904119904 +
1199031199032 119904119904 + 2119904119904 + 2 2 + 3464 2 +
1199031199033 3464119904119904 + 2 2 + 3464 2
Solving for the residues gives
1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119910119910 0 = 015 119898119898
119910 0 = 01 119898119898119904119904
119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
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23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
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25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
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27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
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33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
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Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
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Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
12
Driven Response with Non-Zero ICrsquos
Total response119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464 119905119905 + 00794119890119890minus2119905119905 sin 3464 119905119905
Superposition of two components Natural response
due to initial conditions
Driven response due to the input
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
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59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
Next wersquoll apply the Laplace transform to the entire state-space model in matrix form just as we did for single differential equations
Solving the State-Space Model13
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
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15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
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21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
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22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
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35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
14
Solving the State-Space Model
Wersquove seen how to use the Laplace transform to solve individual differential equations
Now wersquoll apply the Laplace transform to the full state-space system model
First wersquoll look at the same simple example Later wersquoll take a more generalized approach
State-space model is
119901119909 =
minus119887119887119898119898 minus119896119896
1119898119898 0
119901119901119909119909 + 1
0 119865119865119894119894119894119894 119905119905
(1)119910119910 = 0 1
119901119901119909119909
Note that because this model was derived from a bond graph model the state variables are now momentum and displacement
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
15
Laplace Transform of the State-Space Model
For now focus on the state equation Output is a linear combination of states and inputs Determining the state trajectory is the important thing
Use the derivative property to Laplace transform the state equation
119904119904 119875119875 119904119904119883119883 119904119904 minus 119901119901 0
119909119909 0 =minus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 + 1
0 119865119865119894119894119894119894 119904119904
Rearranging to put all transformed state vectors on the left-hand side
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
16
Laplace Transform of the State-Space Model
119904119904 119875119875 119904119904119883119883 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (2)
Can factor out the transformed state vector from the left-hand side Must multiply 119904119904 by a 2 times 2 identity matrix
119904119904119868119868 minusminus 119887119887
119898119898minus119896119896
1119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 00 119904119904 minus
minus 119887119887119898119898
minus1198961198961119898119898
0119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
17
Laplace Transform of the State-Space Model
119904119904 + 119887119887119898119898
119896119896
minus 1119898119898
119904119904119875119875 119904119904119883119883 119904119904 = 119901119901 0
119909119909 0 + 10 119865119865119894119894119894119894 119904119904 (3)
Note the form of (3) The LHS is 119904119904119868119868 minus 119860119860 119831119831 119904119904 where 119860119860 is the system matrix Everything on the RHS reduces to a 2 times 1 vector A known matrix times a vector of unknowns equals a
known vector If we can solve for 119875119875 119904119904 andor 119883119883 119904119904 we can inverse
transform to get 119901119901 119905119905 andor 119909119909 119905119905 Use Cramerrsquos Rule
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
18
Cramerrsquos Rule
Given a matrix equation
119808119808119808119808 = 119858119858
We can solve for elements of 119808119808 as follows
119909119909119894119894 =det 119808119808119894119894det 119808119808
=119808119808119894119894119808119808
The matrix 119808119808119894119894 is formed by replacing the 119894119894119905119905119905 column of 119808119808 with the vector 119858119858
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
19
Laplace Transform of the State-Space Model
According to Cramerrsquos Rule
119883119883 119904119904 =
119904119904+119887119887119898119898 119901119901 0 +119865119865119894119894119894119894 119904119904
minus 1119898119898 119909119909 0
119904119904+119887119887119898119898 119896119896
minus1119898119898 119904119904
119883119883 119904119904 =119904119904 119909119909 0 +119887119887
119898119898 119909119909 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(4)
According to the output equation from (1)119884119884 119904119904 = 119883119883 119904119904
Equation (4) is identical to (8) from the previous subsection of notes which we arrived at differently
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
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25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
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27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
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33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
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35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
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Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
20
Laplace Transform of the State-Space Model
119884119884 119904119904 =119904119904 119910119910 0 +119887119887
119898119898 119910119910 0 minus minus1119898119898119901119901 0 minus 1
119898119898119865119865119894119894119894119894 119904119904
1199041199042+119887119887119898119898119904119904+
119896119896119898119898
(5)
Next sub in parameter values ICrsquos and an input Use PFE to inverse transform to 119910119910(119905119905) Again consider the under-damped system
Let the input be a 1119873119873 step 119865119865119894119894119894119894 119905119905 = 1119873119873 1 119905119905
119898119898 = 1 119896119896119896119896
119896119896 = 16 119873119873119898119898
119887119887 = 4 119873119873119904119904119898119898
119909119909 0 = 015 119898119898
119901119901 0 = 01 119873119873 119904119904
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
21
Laplace Transform of the State-Space Model
The Laplace transform of the output becomes
119884119884 119904119904 =015119904119904 +06 minus minus01minus1119904119904
1199041199042+4119904119904+16
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
(6)
Inverse transform via partial fraction expansion
119884119884 119904119904 = 0151199041199042 +07119904119904+1119904119904 1199041199042+4119904119904+16
= 1199031199031119904119904
+ 1199031199032 119904119904+2 +1199031199033 3464119904119904+2 2+ 3464 2 (7)
Multiply both sides by left-hand-side denominator0151199041199042 + 07119904119904 + 1 = 11990311990311199041199042 + 41199031199031119904119904 + 161199031199031 + 11990311990321199041199042 + 21199031199032119904119904 + 34641199031199033119904119904
Equating coefficients and solving yields1199031199031 = 00625 1199031199032 = 00875 1199031199033 = 00794
K Webb ESE 330
22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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2nd-Order Pole Locations and Damping
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59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
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In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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Convolution
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80
Convolution
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22
Laplace Transform of the State-Space Model
The Laplace transform of the system response is
119884119884 119904119904 = 00625119904119904
+ 00875 119904119904+2119904119904+2 2+ 3464 2 + 00794 3464
119904119904+2 2+ 3464 2 (8)
The time-domain response is119910119910 119905119905 = 00625 + 00875119890119890minus2119905119905 cos 3464119905119905 + 00794119890119890minus2119905119905 sin 3464119905119905 (9)
Transient portion Due to initial conditions
and input step Decays to zero
Steady-State portion Due to constant input Does not decay
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23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
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24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
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Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
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26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
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27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
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33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
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Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
23
Laplace Transform of the State-Space Model
Now wersquoll apply the Laplace transform to the solution of the state-space model in general form
119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
For now focus on the state equation only Output is derived from states and inputs
Laplace transform of the state equation
119904119904119831119831 119904119904 minus 119808119808 0 = 119808119808119831119831 119904119904 + 119809119809119809119809 119904119904
Rearranging
119904119904119831119831 119904119904 minus 119808119808119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0
Factoring out the transformed state from the left-hand side
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
24
Laplace Transform of the State-Space Model
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119809119809 119904119904 + 119808119808 0 (10)
Remember the dimensions of each term in (10) 119904119904119816119816 minus 119808119808 119899119899 times 119899119899 119831119831 119904119904 119899119899 times 1 119904119904119816119816 minus 119808119808 119831119831 119904119904 119899119899 times 1
119809119809119809119809 119904119904 119899119899 times 1 119808119808 0 119899119899 times 1
Apply Cramerrsquos rule to solve for the Laplace transform of the 119894119894119905119905119905 state variable
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
The matrix 119904119904119816119816 minus 119808119808 119894119894 is formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with 119809119809119809119809 119904119904 + 119808119808 0 An 119899119899 times 1 vector of known values
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
25
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894119904119904119816119816minus119808119808
(11)
Denominator of (11) is the determinant of 119904119904119816119816 minus 119808119808 119904119904119816119816 minus 119808119808 is an 119899119899 times 119899119899 matrix
Each diagonal term is a first-order polynomial in 119904119904 One term in the determinant is the trace of the matrix the product terms along
the diagonal 119904119904119816119816 minus 119808119808 is an 119951119951119957119957119957119957-order polynomial in 119956119956
The characteristic polynomial
Δ 119904119904 = 119904119904119816119816 minus 119808119808 (12)
Roots of Δ 119904119904 are values of 119904119904 that satisfy the characteristic equation
Δ 119904119904 = 0 (13) Poles of (11) Eigenvalues of system matrix 119808119808
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
26
Laplace Transform of the State-Space Model
Denominator of every state variablersquos Laplace transform contains the characteristic polynomial
119883119883119894119894 119904119904 = 119904119904119816119816minus119808119808 119894119894Δ 119904119904
(14)
A characteristic of the system Remember denominator roots (ie poles) determine the nature of
the response Real roots ndash decaying exponentials Complex roots ndash decaying sinusoids
Responses of all state variables have same components Numerators of transforms determine the differences
Output transform has the same denominator Δ 119904119904 Linear combination of states and input Response includes the same sinusoidal andor exponential components
K Webb ESE 330
27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
K Webb ESE 330
29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
K Webb ESE 330
32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
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44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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2nd-Order Pole Locations and Damping
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59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
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In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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Convolution
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80
Convolution
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27
Laplace Transform of the State-Space Model
Assume zero initial conditions 119808119808 0 = 120782120782 SISO system single input ndash 119880119880 119904119904 is a scalar transform
Can factor out the input from the numerator of (14)
119904119904119816119816 minus 119808119808 119894119894 = 119880119880 119904119904 119904119904119816119816 minus 119808119808 119894119894lowast
where 119904119904119816119816 minus 119808119808 119894119894lowast is the 119899119899 times 119899119899 matrix formed by replacing the 119894119894119905119905119905 column of 119904119904119816119816 minus 119808119808 with the 119899119899 times 1vector 119809119809 119880119880 119904119904 appears in every term of one column of 119904119904119816119816 minus 119808119808 119894119894 119880119880 119904119904 appears in every term of the determinant
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Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
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Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
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54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
28
Laplace Transform of the State-Space Model
Can now write the Laplace transform of the state variable response as
119883119883119894119894 119904119904 = 119880119880 119904119904 119904119904119816119816minus119808119808 119894119894lowast
119904119904119904119904minus119860119860= 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904
Δ 119904119904(15)
119873119873119873119873119898119898119894119894 119904119904 is in general different for each state variable At most an 119899119899 minus 1 119904119904119905119905 order polynomial in 119904119904
Components of every state variable (and output) response determined by The characteristic polynomial Δ 119904119904 The input 119880119880 119904119904
Numerator 119873119873119873119873119898119898119894119894 119904119904 determines exact response Weighting of each sinusoidal andor exponential component
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29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
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80
Convolution
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29
Laplace Transform of the State-Space Model
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
(15)
Laplace transform of each state variable response 119883119883119894119894 119904119904 is the Laplace transform of the input scaled by
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
119880119880 119904119904 119883119883119894119894 119904119904
In the next sub-section wersquoll explore a related concept ndashtransfer functions
119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
K Webb ESE 330
Transfer Functions30
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
Transfer Functions30
K Webb ESE 330
31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
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31
Transfer Functions
Now come back to the full state-space model including the output equation ndash (SISO case assumed here - 119873119873 and 119910119910 are scalars)
119808 = 119808119808119808119808 + 119809119809119873119873119910119910 = 119810119810119808119808 + D119873119873
Assume zero initial conditions and Laplace transform the whole model
119904119904119831119831 119904119904 = 119808119808119831119831 119904119904 + 119809119809119880119880 119904119904 (1)
119884119884 119904119904 = 119810119810119831119831 119904119904 + D119880119880 119904119904 (2)
Simplify the state equation as before
119904119904119816119816 minus 119808119808 119831119831 119904119904 = 119809119809119880119880 119904119904
Solving for the state vector
119831119831 119904119904 = 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) (3)
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
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32
Transfer Functions
Substituting (3) into (2) gives the Laplace transform of the output
119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809119880119880(119904119904) + D119880119880 119904119904
Factoring out the input119884119884 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D 119880119880 119904119904 (4)
Transform of the output is the input scaled by the stuff in the square brackets
Dividing through by the input gives the transfer function
119866119866 119904119904 = 119884119884 119904119904119880119880 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
Ratio of systemrsquos output to input in the Laplace domain assuming zero initial conditions
An alternative to the state-space (time-domain) model for mathematically representing a system
K Webb ESE 330
33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
K Webb ESE 330
34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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Convolution
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Convolution
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33
Transfer Matrix ndash MIMO Systems
For MIMO systems119808 = 119808119808119808119808 + 119809119809119809119809119858119858 = 119810119810119808119808 + 119811119811119809119809
119898119898 inputs 119809119809 is 119898119898 times 1 119809119809 is 119899119899 times 119898119898 119901119901 outputs 119858119858 is 119901119901 times 1 119810119810 is 119901119901 times 119899119899
Transfer function becomes a 119953119953 times 119950119950 matrix
119814119814 119904119904 = 119832119832 119904119904119809119809 119904119904
= 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + 119811119811
Transfer function 119866119866119894119894119894119894 119904119904 relates the 119894119894119905119905119905 output to the 119895119895119905119905119905input
119866119866119894119894119894119894 119904119904 = 119884119884119894119894 119904119904119880119880119895119895 119904119904
Wersquoll continue to assume SISO systems in this course
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Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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2nd-Order Pole Locations and Damping
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59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
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In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
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34
Transfer Functions
System output in the Laplace domain is the input multiplied by the transfer function
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904
We saw earlier that state variables are given by
119883119883119894119894 119904119904 = 119880119880 119904119904 119873119873119873119873119898119898119894119894 119904119904Δ 119904119904
where Δ 119904119904 = 119904119904119816119816 minus 119808119808 is the characteristic polynomial
Output is linear combination of states and input so wersquod expect the denominator of 119866119866 119904119904 to be Δ 119904119904 as well Is it What is the denominator of 119866119866 119904119904
K Webb ESE 330
35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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79
Convolution
K Webb ESE 330
80
Convolution
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35
Transfer Functions
119866119866 119904119904 = 119810119810 119904119904119816119816 minus 119808119808 minus1119809119809 + D (5)
The matrix inverse term in (5) is given by
119904119904119816119816 minus 119808119808 minus1 =119886119886119886119886119895119895 119904119904119816119816 minus 119808119808119904119904119816119816 minus 119808119808
where the numerator is the adjoint of 119904119904119816119816 minus 119808119808 Equation (5) can be rewritten as
119866119866 119904119904 = 119810119810 119886119886119886119886119894119894 119904119904119816119816minus119808119808 119809119809+D 119904119904119816119816minus119808119808119904119904119816119816minus119808119808
(6)
Transfer function denominator is the characteristic polynomial Poles of the transfer function are roots of Δ 119904119904
System poles or eigenvalues Eigenvalues of the system matrix 119808119808 Along with the input system poles determine the nature of the time-domain
response
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
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50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
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52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
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53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
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56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
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Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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2nd-Order Pole Locations and Damping
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59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
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Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
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Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
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Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
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Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
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75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
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In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
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Convolution
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80
Convolution
K Webb ESE 330
This sub-section of notes takes a bit of a tangent to explain the use of the term eigenvalues when referring to system poles
Eigenvalues36
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
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38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
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39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
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40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
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41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
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Using the Transfer Function to Determine System Response42
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43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
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45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
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46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
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49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
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58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
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65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
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67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
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68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
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69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
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71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
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Step Response Characteristics72
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73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
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74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
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77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
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78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
37
Eigenvalues Wersquove been using the term eigenvalue when referring to system poles ndash why Recall from linear algebra the eigenvalue problem
119808119808119808119808 = 120582120582119808119808 (1)
where 119808119808 is an 119899119899 times 119899119899 matrix119808119808 is an 119899119899 times 1 vector ndash an eigenvector120582120582 is a scalar ndash an eigenvalue
Eigenvalue problem involves finding both the eigenvalues and the eigenvectorsthat satisfy (1)
Eigenvalues and eigenvectors are specific to (characteristics of) the matrix 119808119808 An 119899119899 times 119899119899 matrix will have at most 119899119899 eigenvalues and 119899119899 corresponding eigenvectors
Equation (1) says An 119899119899 times 1 eigenvector 119808119808 left-multiplied by an 119899119899 times 119899119899 matrix 119808119808 results in an 119899119899 times 1 vector The resulting vector is the eigenvector scaled by the eigenvalue 120582120582 Result is in the same direction as 119808119808 ndash ie not rotated
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
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48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
38
Eigenvalues and Eigenvectors
Geometrically multiplication of a vector by a matrix results in two things Scaling and rotation
Consider the matrix
119808119808 = 1 34 2
And the vectors
1198081198081 = 11 1198081198082 = 1
0 Compute the product
119858119858 = 119808119808119808119808
In both cases results have different magnitudes and different directions
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
39
Eigenvalues and Eigenvectors
Multiplication of a matrix and one of its eigenvectors results in scaling only No rotation
The 2 times 2 matrix
119808119808 = 1 34 2
has two eigenvectors (normalized)
1198081198081 = minus07070707 and 1198081198082 = minus06
minus08and two corresponding eigenvalues
λ1 = minus2 and λ2 = 5
such that
119808119808119808119808120783120783 = 120582120582120783120783119808119808120783120783 and 119808119808119808119808120784120784 = 120582120582120784120784119808119808120784120784
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
40
Eigenvalues and Eigenvectors
A full-rank 119899119899 times 119899119899 matrix will have 119899119899 pairs of eigenvalues and eigenvectors
To find all eigenvalues and eigenvectors that satisfy (1)
119808119808119808119808 = 120582120582119808119808 (1)rearrange
120582120582119808119808 minus 119808119808119808119808 = 120782120782
and factor out the eigenvector term
120582120582119816119816 minus 119808119808 119808119808 = 0 (2)
If 120582120582119816119816 minus 119808119808 minus1 exists then 119808119808 = 120782120782 which is the trivial solution and of no interest
Wersquore interested in values of 120582120582 and 119808119808 that satisfy (2) when 120582120582119816119816 minus 119808119808 is not invertible ndash when it is singular
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
41
Eigenvalues and Eigenvectors
Want to find values of 120582120582 for which (120582120582119816119816 minus 119808119808) is singular A matrix is singular if its determinant is zero
120582120582119816119816 minus 119808119808 = 0 (3)
Equation (3) is the characteristic equation for 119808119808 120582120582119816119816 minus 119808119808 is the characteristic polynomial Δ 120582120582 An 119899119899119905119905119905-order polynomial in 120582120582
Eigenvalues of matrix 119808119808 are all 119899119899 values of 120582120582 that satisfy (3) Roots of the characteristic polynomial Find the corresponding eigenvectors by substituting 120582120582 into (2) and
solving for 119808119808
Letting 120582120582 = 119904119904 (3) becomes the denominator of the system transfer function 119866119866 119904119904
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
Using the Transfer Function to Determine System Response42
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
43
Using 119866119866 119904119904 to determine System Response
System output in the Laplace domain can be expressed in terms of the transfer function as
119884119884 119904119904 = 119880119880 119904119904 119866119866 119904119904 (1)
Laplace-domain output is the product of the Laplace-domain input and the transfer function
Response to two specific types of inputs often used to characterize dynamic systems Impulse response Step response
Wersquoll use the approach of (1) to determine these responses
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
44
Impulse response
Impulse function
120575120575 119905119905 = 0 119905119905 ne 0
intminusinfininfin 120575120575 119905119905 119886119886119905119905 = 1
Laplace transform of the impulse function is
ℒ 120575120575 119905119905 = 1
Impulse response in the Laplace domain is
119884119884 119904119904 = 1 119866119866 119904119904 = 119866119866 119904119904
The transfer function is the Laplace transform of the impulse response Impulse response in the time domain is the inverse transform of the
transfer function
119910119910 119905119905 = 119896119896 119905119905 = ℒminus1 119866119866 119904119904
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
45
Step Response
Step function 1 119905119905 = 0 119905119905 lt 01 119905119905 ge 0
Laplace transform of the step function
ℒ 1 119905119905 = 1119904119904
Laplace-domain step response
119884119884 119904119904 = 1119904119904 119866119866 119904119904
Time-domain step response
119910119910 119905119905 = ℒminus1 1119904119904 119866119866 119904119904
Recall the integral property of the Laplace transform
ℒ int0119905119905 119896119896 120591120591 119886119886120591120591 = 1
119904119904 119866119866 119904119904 and ℒminus1 1
119904119904 119866119866 119904119904 = int0
119905119905 119896119896 120591120591 119886119886120591120591
The step response is the integral of the impulse response
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
46
First- and Second-Order Systems
All transfer functions can be decomposed into 1st- and 2nd-order terms by factoring Δ 119904119904 Real poles ndash 1st-order terms Complex-conjugate poles ndash 2nd-order terms
These terms and therefore the poles determine the nature of the time-domain response Real poles ndash decaying exponentials Complex-conjugate poles - decaying sinusoids
All time-domain responses will be a superposition of decaying exponentials and decaying sinusoids These are the natural modes or eigenmodes of the system
Instructive to examine the responses of 1st- and 2nd-order systems Gain insight into relationships between pole location and response
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
Response of First-Order Systems47
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
48
First-Order System ndash Impulse Response
First-order transfer function
119866119866 119904119904 = 119860119860119904119904+120590120590
Single real pole at
119904119904 = minus120590120590 = minus1120591120591
where 120591120591 is the system time constant Impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904 = 119860119860119890119890minus120590120590119905119905 = 119860119860119890119890minus119905119905120591120591
119896119896 119905119905 = 119860119860119890119890minus119905119905120591120591
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
49
First-Order System ndash Impulse Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Decays to zero as long as the pole is negative
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
50
Impulse Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
51
First-Order System ndash Step Response Step response in the Laplace domain
119884119884 119904119904 = 1119904119904 119866119866 119904119904 = 119860119860
119904119904 119904119904+120590120590
Inverse transform back to time domain via partial fraction expansion
119884119884 119904119904 = 119860119860119904119904 119904119904+120590120590
= 1199031199031119904119904
+ 1199031199032119904119904+120590120590
119860119860 = 1199031199031 + 1199031199032 119904119904 + 1205901205901199031199031
1199041199040 1205901205901199031199031 = 119860119860 rarr 1199031199031 = 119860119860120590120590
1199041199041 1199031199031 + 1199031199032 = 0 rarr 1199031199032 = minus119860119860120590120590
119884119884 119904119904 = 119860119860120590120590119904119904minus 119860119860120590120590
119904119904+120590120590
Time-domain step response
119910119910 119905119905 =119860119860120590120590minus119860119860120590120590119890119890minus120590120590119905119905 = 119861119861 minus 119861119861119890119890minus
119905119905120591120591
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
52
First-Order System ndash Step Response
Initial slope is inversely proportional to time constant
Response completes 63 of transition after one time constant
Almost completely settled after 7120591120591
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
53
Step Response vs Pole Location
Increasing 120590120590 corresponds to decreasing 120591120591 and a faster response
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
54
Pole Location and Stability
First-order transfer function
119866119866 119904119904 =119860119860
119904119904 minus 119901119901
where 119901119901 is the system pole Impulse response is
119896119896 119905119905 = 119860119860119890119890119901119901119905119905
If 119901119901 lt 0 119896119896 119905119905 decays to zero Pole in the left half-plane System is stable
If 119901119901 gt 0 119896119896 119905119905 grows without bound Pole in the right half-plane System is unstable
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
Response of Second-Order Systems55
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
56
Second-Order Systems
Second-order transfer function
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+1198861198861119904119904+1198861198860
= 119873119873119873119873119898119898 119904119904119904119904+120590120590 2+120596120596119889119889
2 (1)
where 120596120596119886119886 is the damped natural frequency
Can also express the 2nd-order transfer function as
119866119866 119904119904 = 119873119873119873119873119898119898 1199041199041199041199042+2120577120577120596120596119894119894119904119904+120596120596119894119894
2 (2)
where 120596120596119894119894 is the un-damped natural frequency and 120577120577 is the damping ratio
120596120596119886119886 = 120596120596119894119894 1 minus 1205771205772
120577120577 = 120590120590120596120596119894119894
Two poles at 11990411990412 = minus120590120590 plusmn 1205901205902 minus 1205961205961198941198942 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
57
Categories of Second-Order Systems The 2nd-order system poles are
11990411990412 = minus120577120577120596120596119894119894 plusmn 120596120596119894119894 1205771205772 minus 1
Value of 120577120577 determines the nature of the poles and therefore the response
120635120635 gt 120783120783 Over-damped Two distinct real poles ndash sum of decaying exponentials ndash treat as two first-order terms 1199041199041 = minus1205901205901 1199041199042 = minus1205901205902
120635120635 = 120783120783 Critically-damped Two identical real poles ndash time-scaled decaying exponentials 11990411990412 = minus120590120590 = minus120577120577120596120596119894119894 = minus120596120596119894119894
120782120782 lt 120635120635 lt 120783120783 Under-damped Complex-conjugate pair of poles ndash sum of decaying sinusoids 11990411990412 = minus120590120590 plusmn 119895119895120596120596119886119886 = minus120577120577120596120596119894119894 plusmn 119895119895120596120596119894119894 1 minus 1205771205772
120635120635 = 120782120782 Un-damped Purely-imaginary conjugate pair of poles ndash sum of non-decaying sinusoids 11990411990412 = plusmn119895119895120596120596119894119894
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
58
2nd-Order Pole Locations and Damping
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
59
Second-Order Poles - 0 le 120577120577 le 1
Can relate 120590120590 120596120596119886119886 120596120596119894119894 and 120577120577to pole location geometry
120596120596119894119894 is the magnitude of the poles
120577120577 is a measure of system damping
120577120577 = 120590120590120596120596119894119894
= sin 120579120579
120577120577 = 0 Two purely imaginary poles
120577120577 = 1 Two identical real poles
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
60
Impulse Response ndash Critically-Damped
For 120577120577 = 1 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 2120596120596119894119894119904119904 + 1205961205961198941198942=
119860119860119904119904 + 120596120596119894119894 2 =
119860119860119904119904 + 120590120590 2
Impulse response119896119896 119905119905 = ℒminus1 119866119866 119904119904
119896119896 119905119905 = 119860119860119905119905119890119890minus120590120590119905119905
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
61
Impulse Response ndash Critically-Damped
Speed of response is proportional to 120590120590
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
62
Impulse Response ndash Under-Damped For 0 lt 120577120577 lt 1 the transfer function is
119866119866 119904119904 =119860119860
1199041199042 + 2120577120577120596120596119894119894119904119904 + 1205961205961198941198942
Complete the square on the denominator
119866119866 119904119904 =119860119860
119904119904 + 120577120577120596120596119894119894 2 + 120596120596119894119894 1 minus 12057712057722 =
119860119860119904119904 + 120577120577120596120596119894119894 2 + 120596120596119886119886
2
Rewrite in the form of a damped sinusoid
119866119866 119904119904 =119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120577120577120596120596119894119894 2 + 1205961205961198861198862 =
119860119860120596120596119886119886
120596120596119886119886
119904119904 + 120590120590 2 + 1205961205961198861198862
Inverse Laplace transform for the time-domain impulse response
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905sin(120596120596119886119886119905119905)
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
63
Under-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
64
Under-Damped Impulse Response vs 120577120577
119896119896 119905119905 =119860119860120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905 = 119861119861119890119890minus120577120577120596120596119894119894119905119905 sin 120596120596119894119894 1 minus 1205771205772119905119905
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
65
Impulse Response ndash Un-Damped
For 120577120577 = 0 the transfer function reduces to
119866119866 119904119904 =119860119860
1199041199042 + 1205961205961198941198942
Putting into the form of a sinusoid
119866119866 119904119904 =119860119860120596120596119894119894
1205961205961198941198941199041199042 + 1205961205961198941198942
Inverse transform to get the time-domain impulse response
119896119896 119905119905 = ℒminus1 119866119866 119904119904
An un-damped sinusoid
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
66
Un-Damped Impulse Response vs 120596120596119894119894
119896119896 119905119905 =119860119860120596120596119894119894
sin 120596120596119894119894119905119905
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
67
Second-Order Step Response
The Laplace transform of the step response is
119884119884 119904119904 =1119904119904119866119866 119904119904
The time-domain step response for each damping case can be derived as the the inverse transform of 119884119884 119904119904
119910119910 119905119905 = ℒminus1 119884119884 119904119904
or as the integral of the corresponding impulse response
119910119910 119905119905 = 0
119905119905119896119896 120591120591 119886119886120591120591
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
68
Critically-Damped Step Response vs 120590120590
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205901205902
1 minus 119890119890minus120590120590119905119905 minus 120590120590119905119905119890119890minus120590120590119905119905
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
69
Under-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
70
Under-Damped Step Response vs 120577120577
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus 119890119890minus120590120590119905119905 cos 120596120596119886119886119905119905 minus120590120590120596120596119886119886
119890119890minus120590120590119905119905 sin 120596120596119886119886119905119905
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
71
Un-Damped Step Response vs 120596120596119894119894
119910119910 119905119905 = ℒminus11119904119904119866119866 119904119904 =
1198601198601205961205961198941198942
1 minus cos 120596120596119894119894119905119905
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
Step Response Characteristics72
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
73
Step Response ndash Risetime
Risetime is the time it takes a signal to transition between two set levels Typically 10 to 90
of full transition Sometimes 20 to
80
A measure of the speed of a response
Very rough approximation
119905119905119903119903 asymp18120596120596119894119894
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
74
Step Response ndash Overshoot
Overshoot is the magnitude of a signalrsquos excursion beyond its final value Expressed as a
percentage of full-scale swing
Overshoot increases as 120577120577 decreases
120635120635 OS
045 20
05 16
06 10
07 5
119874119874119874119874 = 119890119890minus 120577120577120577120577
1minus1205771205772 sdot 100
120577120577 =minus ln 119874119874119874119874
100
1205871205872 + ln2 119874119874119874119874100
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
75
Step Response ndashSettling Time
Settling time is the time it takes a signal to settle finally to within some percentage of its final value Typically plusmn1 or plusmn5
Inversely proportional to the real part of the poles 120590120590
For plusmn1 settling
119905119905119904119904 asymp46120590120590
= 46120577120577120596120596119894119894
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
In this sub-section wersquoll see that the time-domain output of a system is given by the convolution of its time-domain input and its impulse response
The Convolution Integral76
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
77
Convolution Integral
Laplace transform of a system output is given by the product of the transform of the input signal and the transfer function
119884119884 119904119904 = 119866119866 119904119904 119880119880 119904119904
Recall that multiplication in the Laplace domain corresponds to convolution in the time domain
119910119910 119905119905 = ℒminus1 119866119866 119904119904 119880119880 119904119904 = 119896119896 119905119905 lowast 119873119873 119905119905
Time-domain output given by the convolution of the input signal and the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = int0119905119905 119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
78
Convolution
Time-domain output is the input convolved with the impulse response
119910119910 119905119905 = 119896119896 119905119905 lowast 119873119873 119905119905 = 0
119905119905119896119896 120591120591 119873119873 119905119905 minus 120591120591 119886119886120591120591
Input is flipped in time and shifted by 119905119905
Multiply impulse response and flippedshifted input
Integrate over 120591120591 = 0 hellip 119905119905
Each time point of 119910119910 119905119905 given by result of integral with 119873119873 minus120591120591 shifted by 119905119905
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
79
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
80
Convolution
K Webb ESE 330
A few of MATLABrsquos many built-in functions that are useful for simulating linear systems are listed in the following sub-section
Time-Domain Analysis in MATLAB81
K Webb ESE 330
82
System Objects
MATLAB has data types dedicated to linear system models
Two primary system model objects State-space model Transfer function model
Objects created by calling MATLAB functions ssm ndash creates a state-space model tfm ndash creates a transfer function model
K Webb ESE 330
83
State-Space Model ndash ss(hellip)
sys = ss(ABCD)
A system matrix - 119899119899 times 119899119899 B input matrix - 119899119899 times 119898119898 C output matrix - 119901119901 times 119899119899 D feed-through matrix - 119901119901 times 119898119898 sys state-space model object
State-space model object will be used as an input to other MATLAB functions
K Webb ESE 330
84
Transfer Function Model ndash tf(hellip)
sys = tf(NumDen)
Num vector of numerator polynomial coefficients Den vector of denominator polynomial coefficients sys transfer function model object
Transfer function is assumed to be of the form
119866119866 119904119904 =1198871198871119904119904119903119903 + 1198871198872119904119904119903119903minus1 + ⋯+ 119887119887119903119903119904119904 + 119887119887119903119903+11198861198861119904119904119894119894 + 1198861198862119904119904119894119894minus1 + ⋯+ 119886119886119894119894119904119904 + 119886119886119894119894+1
Inputs to tf(hellip) are Num = [b1b2hellipbr+1] Den = [a1a2hellipan+1]
K Webb ESE 330
85
Step Response Simulation ndash step(hellip)
[yt] = step(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output step response t output time vector
If no outputs are specified step response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
86
Impulse Response Simulation ndash impulse(hellip)
[yt] = impulse(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output impulse response t output time vector
If no outputs are specified impulse response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
82
System Objects
MATLAB has data types dedicated to linear system models
Two primary system model objects State-space model Transfer function model
Objects created by calling MATLAB functions ssm ndash creates a state-space model tfm ndash creates a transfer function model
K Webb ESE 330
83
State-Space Model ndash ss(hellip)
sys = ss(ABCD)
A system matrix - 119899119899 times 119899119899 B input matrix - 119899119899 times 119898119898 C output matrix - 119901119901 times 119899119899 D feed-through matrix - 119901119901 times 119898119898 sys state-space model object
State-space model object will be used as an input to other MATLAB functions
K Webb ESE 330
84
Transfer Function Model ndash tf(hellip)
sys = tf(NumDen)
Num vector of numerator polynomial coefficients Den vector of denominator polynomial coefficients sys transfer function model object
Transfer function is assumed to be of the form
119866119866 119904119904 =1198871198871119904119904119903119903 + 1198871198872119904119904119903119903minus1 + ⋯+ 119887119887119903119903119904119904 + 119887119887119903119903+11198861198861119904119904119894119894 + 1198861198862119904119904119894119894minus1 + ⋯+ 119886119886119894119894119904119904 + 119886119886119894119894+1
Inputs to tf(hellip) are Num = [b1b2hellipbr+1] Den = [a1a2hellipan+1]
K Webb ESE 330
85
Step Response Simulation ndash step(hellip)
[yt] = step(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output step response t output time vector
If no outputs are specified step response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
86
Impulse Response Simulation ndash impulse(hellip)
[yt] = impulse(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output impulse response t output time vector
If no outputs are specified impulse response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
83
State-Space Model ndash ss(hellip)
sys = ss(ABCD)
A system matrix - 119899119899 times 119899119899 B input matrix - 119899119899 times 119898119898 C output matrix - 119901119901 times 119899119899 D feed-through matrix - 119901119901 times 119898119898 sys state-space model object
State-space model object will be used as an input to other MATLAB functions
K Webb ESE 330
84
Transfer Function Model ndash tf(hellip)
sys = tf(NumDen)
Num vector of numerator polynomial coefficients Den vector of denominator polynomial coefficients sys transfer function model object
Transfer function is assumed to be of the form
119866119866 119904119904 =1198871198871119904119904119903119903 + 1198871198872119904119904119903119903minus1 + ⋯+ 119887119887119903119903119904119904 + 119887119887119903119903+11198861198861119904119904119894119894 + 1198861198862119904119904119894119894minus1 + ⋯+ 119886119886119894119894119904119904 + 119886119886119894119894+1
Inputs to tf(hellip) are Num = [b1b2hellipbr+1] Den = [a1a2hellipan+1]
K Webb ESE 330
85
Step Response Simulation ndash step(hellip)
[yt] = step(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output step response t output time vector
If no outputs are specified step response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
86
Impulse Response Simulation ndash impulse(hellip)
[yt] = impulse(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output impulse response t output time vector
If no outputs are specified impulse response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
84
Transfer Function Model ndash tf(hellip)
sys = tf(NumDen)
Num vector of numerator polynomial coefficients Den vector of denominator polynomial coefficients sys transfer function model object
Transfer function is assumed to be of the form
119866119866 119904119904 =1198871198871119904119904119903119903 + 1198871198872119904119904119903119903minus1 + ⋯+ 119887119887119903119903119904119904 + 119887119887119903119903+11198861198861119904119904119894119894 + 1198861198862119904119904119894119894minus1 + ⋯+ 119886119886119894119894119904119904 + 119886119886119894119894+1
Inputs to tf(hellip) are Num = [b1b2hellipbr+1] Den = [a1a2hellipan+1]
K Webb ESE 330
85
Step Response Simulation ndash step(hellip)
[yt] = step(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output step response t output time vector
If no outputs are specified step response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
86
Impulse Response Simulation ndash impulse(hellip)
[yt] = impulse(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output impulse response t output time vector
If no outputs are specified impulse response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
85
Step Response Simulation ndash step(hellip)
[yt] = step(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output step response t output time vector
If no outputs are specified step response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
86
Impulse Response Simulation ndash impulse(hellip)
[yt] = impulse(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output impulse response t output time vector
If no outputs are specified impulse response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
86
Impulse Response Simulation ndash impulse(hellip)
[yt] = impulse(syst)
sys system model ndash state-space or transfer function t optional time vector or final time value y output impulse response t output time vector
If no outputs are specified impulse response is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
87
Natural Response ndash initial(hellip)
[ytx] = initial(sysx0t)
sys state-space system model function x0 initial value of the state - 119899119899 times 1 vector t optional time vector or final time value y response to initial conditions - length(t)times 1 vector t output time vector x trajectory of all states - length(t)times 119899119899 matrix
If no outputs are specified response to initial conditions is automatically plotted
Time vector (or final value) input is optional If not specified MATLAB will generate automatically
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
88
Linear System Simulation ndash lsim(hellip)
[ytx] = lsim(sysutx0)
sys system model ndash state-space or transfer function u input signal vector t time vector corresponding to the input signal x0 optional initial conditions ndash (for ss model only) y output response t output time vector x optional trajectory of all states ndash (for ss model only)
If no outputs are specified response is automatically plotted
Input can be any arbitrary signal
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-
K Webb ESE 330
89
More MATLAB Functions
A few more useful MATLAB functions
Polezero analysis pzmap(hellip) pole(hellip) zero(hellip) eig(hellip)
Input signal generation gensig(hellip)
Refer to MATLAB help documentation for more information
- Section 6 Time-Domain Analysis
- Natural and Forced Responses
- Natural and Forced Responses
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Natural Response
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Driven Response with Non-Zero ICrsquos
- Solving the State-Space Model
- Solving the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Cramerrsquos Rule
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Laplace Transform of the State-Space Model
- Transfer Functions
- Transfer Functions
- Transfer Functions
- Transfer Matrix ndash MIMO Systems
- Transfer Functions
- Transfer Functions
- Eigenvalues
- Eigenvalues
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Eigenvalues and Eigenvectors
- Using the Transfer Function to Determine System Response
- Using 119866 119904 to determine System Response
- Impulse response
- Step Response
- First- and Second-Order Systems
- Response of First-Order Systems
- First-Order System ndash Impulse Response
- First-Order System ndash Impulse Response
- Impulse Response vs Pole Location
- First-Order System ndash Step Response
- First-Order System ndash Step Response
- Step Response vs Pole Location
- Pole Location and Stability
- Response of Second-Order Systems
- Second-Order Systems
- Categories of Second-Order Systems
- 2nd-Order Pole Locations and Damping
- Second-Order Poles - 0le120577le1
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Critically-Damped
- Impulse Response ndash Under-Damped
- Under-Damped Impulse Response vs 120596 119899
- Under-Damped Impulse Response vs 120577
- Impulse Response ndash Un-Damped
- Un-Damped Impulse Response vs 120596 119899
- Second-Order Step Response
- Critically-Damped Step Response vs 120590
- Under-Damped Step Response vs 120596 119899
- Under-Damped Step Response vs 120577
- Un-Damped Step Response vs 120596 119899
- Step Response Characteristics
- Step Response ndash Risetime
- Step Response ndash Overshoot
- Step Response ndashSettling Time
- The Convolution Integral
- Convolution Integral
- Convolution
- Convolution
- Convolution
- Time-Domain Analysis in MATLAB
- System Objects
- State-Space Model ndash ss(hellip)
- Transfer Function Model ndash tf(hellip)
- Step Response Simulation ndash step(hellip)
- Impulse Response Simulation ndash impulse(hellip)
- Natural Response ndash initial(hellip)
- Linear System Simulation ndash lsim(hellip)
- More MATLAB Functions
-