EE 401 Control Systems Analysis and Design A Review of The Laplace Transform Wednesday 27 Aug 2014...

EE 401 Control Systems Analysis and Design

A Review of The Laplace Transform

Wednesday 27 Aug 2014 EE 401: Control Systems Analysis and Design Slide 1 of 18

http://www.erau.edu/

The Laplace Transform

Wednesday 27 Aug 2014

• Question: What is the utility of this mathematical tool (the T)?

• Answer: 1. It greatly simplifies the process of solving Linear Time-

Invariant (LTI), homogeneous, Ordinary Differential Equations (ODEs)

2. Provides the basis for a “qualitative” evaluation of linear systems.

The T converts differential equations into algebraic equations

EE 401: Control Systems Analysis and Design Slide 2 of 19


The Laplace Transform


• Definition: The (unilateral) Laplace Transform is defined as

(1)

where is the complex frequency variable with units sec–1. Assumes that (i.e., causal signal)

• Note: Equation (1) is meaningful only if the integral converges, i.e.,

This may be true for only a region of convergence within the imaginary plane.

0

( ) ( ) ( )stf t f t e dt F s

L

0

( ) stf t e dt



Some LT Examples


Example 1: T of an exponential (a > 0);

Similarly,

0

at at ste e e dt

L

For now we will ignore this Region of Convergence.

1ates a

L

( )

0

s a te dt

( )

0

1( )

s a tes a

( ) ( )

0

1( )

s a t s a t

t te e

s a

1s a

Region of Convergence

0 iff Re{s}>a 1



Some LT Examples


Example 2: LT of a constant; . Often denoted by and referred to as the unit step function or Heaviside function

• Note:

0

( ) ( ) stu t u t e dt

L

( ) ( ) aau t a u ts

L L

0

1 stes

1s



Some LT Examples


Example 3: LT of a sinusoidal;

• Also,

0

cos( )2

jbt jbtste ebt e dt

L

2 2sin( ) bbts b

L

Recall Euler’s Formula

cos( )2 cos( ) sin( )

sin( )2

j j

jj j

e e

e je e

j

( ) ( )0 0

1 12( ) 2( )

s jb t s jb te es jb s jb

1 1

2( ) 2( )s jb s jb

( ) ( )2( )( )s jb s jbs jb s jb

2 22

2( )s

s b

2 2s

s b



Some LT Examples


Example 4: LT of a powers of t;

i.e., (LT of a unit ramp)

0

n n stt t e dt

L

21ts

L

u dv

Integration by Parts

( )d uv udv vdudt

( )d uv udv vdudt

21 nn ts

L

1

0 0

1 stn st n et e nt dt

s s

1

0

n stn t e dts

1nn ts

L

1!nns

udv uv vdu



Using MATLAB Directly

Symbolic Toolbox

Using MATLAB MuPAD notepad

MATLAB Symbolic EditorSometimes different syntax

Some LT ExamplesThe MATLAB Symbolic Math Toolbox

Wednesday 27 Aug 2014 EE 401: Control Systems Analysis and Design Slide 8 of 19


Some LT ExamplesThe MATLAB Symbolic Math Toolbox


• Using Mathematica



Some LT Examples


Example 5: LT of a unit impulse;

0

( ) ( ) stt t e dt

L Note: We need to include t = 0

in the integral.

See Handout “Laplace_Transform_Table.pdf” for a table of Laplace Transforms.

0

st

te

1


http://mercury.pr.erau.edu/~bruders/teaching/2014_fall/EE401/handouts/Laplace_Transform_Table.pdf

http://mercury.pr.erau.edu/~bruders/teaching/2014_fall/EE401/handouts/Laplace_Transform_Table.pdf


Some Properties of the LT


• Linearity:

Example:

0

( ) ( ) sta f t a f t e dt

L

0

( ) sta f t e dt

( )a f t L

0

( ) ( ) ( ) ( ) stf t g t f t g t e dt

L

0 0

( ) ( )st stf t e dt g t e dt

( ) ( )f t g t L L

3 ( ) 2 3 ( ) 2t tt e t e L L L

13 21s

3 5

1ss





• Differentiation:

'

0

( ) ( ) std f t f t e dtdt

L

00

( ) ( )st stf t e s f t e dt

( ) (0)sF s f

du v

F(s)


𝑠Input Output

+-

( )F s ( ) ( ) (0)Y s sF s f

𝑑𝑑𝑡

Input Output

( )f t ( ) '( )y t f t

(0)f




• Integration:

0 0 0

( ) ( )t t

stf d f d e dt

L

0 00

1 1( ) ( )t

st stf d e f t e dts s

1 ( )F ss

u dv

(.)Input Output

( )f t


1𝑠

Input Output

( ) ( )dy t f ( )F s1( ) ( )Y s F ss




• Convolution:

0

( ) ( ) ( ) ( )t

f g t d F s G s

L

𝑔∗Input Output

( )f t ( )y t f g


𝐺 (𝑠 )Input Output

( ) ( ) ( )Y s F s G s( )F s

(t)* ( ) ( ) ( )g f t F s G sL


Inverting the Laplace Transform


• Inverting the Laplace Transform:

• Use the tables instead!!

1 1( ) ( ) ( )2π

jst

j

f t F s F s e dsj

L

DO NOT USE THIS FORMULA



Inverting the Laplace TransformExample #1:


Solve the first order ODE (ordinary differential equation)

Take the LT of the equation.

5 ( ) 10 ( ) 3 ( ), with (0) 1y t y t u t y

35 ( ) (0) 10 ( )sY s y Y ss

35( 2) ( ) 5s Y ss

1 0.6( )2 ( 2)

Y ss s s

1 2( ) 0.32 ( 2)

Y ss s s

2 2( ) 0.3(1 ), 0t ty t e e t

1s a

( )a

s s a

= 1

Unforced Response(due to initial conditions)

Forced Response(due to input 3u(t))





Solve the second order ODE (ordinary differential equation)

Take the LT of the equation.

( ) 3 ( ) 2 ( ) 5 ( ), with (0) 1, (0) 2y t y t y t u t y y

2 5( ) (0) (0) 3 ( ) (0) 2 ( )s Y s sy y sY s y Y ss

2 5( 3 2) ( ) (0) (0) 3 (0)s s Y s sy y ys

2( 1)( 2) ( ) 5 2 3s s s Y s s s s 2 5( )

( 1)( 2)s sY s

s s s

1 2

A B Cs s s

5 / 2 5 3 / 21 2s s s

25 3( ) 5 , 0

2 2t ty t e e t





Partial fraction expansion for the case of complex roots

2

3( )2 5

Y ss s s

2 2 5A Bs Cs s s

3 3 3( ) cos(2 ) sin(2 ), 05 5 10

t ty t e t e t t

3( 1 2 )( 1 2 )s s j s j

2

3 63 / 5 5 5

2 5

s

s s s

20

3 352 5 s

As s

233 2 5

5s s Bs C s

2 23 65 5s s Bs Cs

2 2

3 / 5 3 25 1 2

ss s

2 22 2

3 / 5 3 1 1 25 21 2 1 2

ss s s

cos(2 )te t L

sin(2 )te t L





Partial fraction for the case of complex rootsThis result can be further simplified:

3 3 3( ) cos(2 ) sin(2 )5 5 10

t ty t e t e t

3 3 1cos(2 ) sin(2 )5 5 2

te t t

cos( )cos( ) sin( )sin( )r a b r a b

cos( )r a b

221 1 / 2r

1/ 2arctan1

b

22 13 3 1 1/ 2 cos(2 tan (0.5))5 5

te t

0.6 0.6708 cos(2 26.6 )te t



EE 401 Control Systems Analysis and Design A Review of The Laplace Transform Wednesday 27 Aug 2014...

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