3EC1 MATHEMATICS-III UNIT 1: LAPLACE TRANSFORM - Laplace ...
MESB 374 System Modeling and Analysis Laplace Transform and Its Applications.
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Transcript of MESB 374 System Modeling and Analysis Laplace Transform and Its Applications.
MESB 374 System Modeling and Analysis
Laplace Transform and Its Applications
Laplace Transform
• Motivation• Laplace Transform
– Review of Complex Numbers
– Definition
– Time Domain vs s-Domain
– Important Properties
• Inverse Laplace Transform• Solving ODEs with Laplace Transform
Motivation
A quick way tosolve for the solution of a linear-time-invariant (LTI) ODE
for various inputs with either zero or non-zero ICs
Time domainmodel
Frequency domainmodel
L
Time domainsolutionClassic calculus techniques
Integration & Convolution
0
t
td
Frequency domainsolutionAlgebraic techniques
Free ResponseForced Responsedue to . .due to input ( )
( ) ( ) ( )F N
I C sU s
Y s Y s Y s 1-L
d
dt
s
( )y t
( )Y s
( ) ( ) ( )FY s G s U s
Review of Complex Numbers
• Common Forms of a Complex Number:
– Coordinate Form:
– Phasor (Euler) Form:
• Moving Between Representations– Phasor (Euler) Form Coordinate Form
– Coordinate Form Phasor (Euler) Form
Az x j y
z Ae A jj (cos sin )
1
2 21
1
tan ( ) when is in the 1st or 4th quadrant
atan2( , ) tan ( ) when is in the 2nd quadrantatan2( , )
tan ( ) when is in the 3rd quadrant
yx
yx
yx
zA x y
y x zy x
z
Real
Img.
z
y
x
x A
y Ae jj
cos
sincos sin
Definition of Laplace Transform
• Laplace Transform– One Sided Laplace Transform
where s is a complex variable that can be represented by s=j f (t) is a function of time that equals to 0 when t < 0.
• Inverse Laplace Transform
0( ) ( ) ( )£ stF s f t f t e dt
11
( ) ( ) ( )2£
c j
c j
stf t F s F s e dsj
A function ofcomplex variable s
A function oftime t
A function oftime t
A function ofcomplex variable s
Laplace Transforms of Common Functions
Some simple examplesStepsExponentialsRampsTrigonometricImpulses
t
y 0
00
( ) ( )
1
1
st
st st
Y s y t e dt
e dt es
s
t
y0
( )
0
( )
1 1
t st
s t
Y s e e dt
es s
t
y
0
2
( )
1
stY s te dt
s
Unit Step
Unit Ramp
Exponential
Laplace Transforms of Common Functions
0 0
00
2 2
( ) sin2
1 1
2 2
y y
y y
y y
j t j tst st
y
j s t j s tj s t j s t
y y
y
y
e eY s t e dt e dt
j
e ee e dt
j j j s j s
s
0
0 0 00
0
( ) ( ) ( ) 0
( ) 1
st st stY s t e dt t e dt e dt
t dt
t
y
t
y
Trigonometric
Unit Impulse
( ) sin yy t t
( ) ( )y t t
Important Properties• Linearity
Given
a and b are arbitrary constants,then
Q: If u(t) = u1(t) + 4 u2(t) what is the Laplace transform of u(t) ?
• DifferentiationGiven
The Laplace transform of the derivative of f (t) is:
– For zero initial condition:
1 1
2 2
( ) ( )
( ) ( )
££
F s f t
F s f t
1 2
1 2
( ) ( )
( ) ( )
£ a f t b f t
aF s bF s
0
0 0
0
2
( ) ( )
(0) ( ) (0)
( ) ( ) ( ) (0)
( ) (0) (0)( ) (0) (0)
( ) ( ) ( ) (
£
£ £ £
£ £ £
st
st st
st
dfddt dt
ddt
ddt
f t e dt
e f f de
f s fe dt sF s f
f t f t s f t f
s sF s f fs F s sf f
f t f t s f t f
2
3 2
0)
( ) (0) (0) (0)
( ) (0) (0) (0)
s s F s sf f f
s F s s f sf f
1 Multiplication by
Differentiation
££
£ s
ds
dt
( ) ( )£F s f t
1 2
1 2
( ) 4 ( )
( ) 4 ( )
£ u t u t
U s U s
Important Properties• Integration
Given
The Laplace transform of the definite integral of f (t) is:
– Conclusion:
Q : Given that the Laplace transforms of a unit step function u(t) = 1 and f(t) = sin(2t) are
What is the Laplace transform of 0 0 0
0 00
0
( ) ( )
1 1( ) ( )
10 ( )
1( )
£ t tst
tst st
st
f d f d e dt
e f d e f t dts s
f t e dts
F ss
1
0
Integration Division by
£1
££
t
s
ds
2
1( ) 1
2( ) sin(2 )
4
£
£
U ss
F s ts
e t t t t( ) cos( ) 2 3 5 4 22
( ) ( )£F s f t
2
( ) 2 3 10 20 0
22 3
42 3 4
£ ( ) £ 1 £ 1 £ £ sin(2 )
22 3 10 2 sin 2 04
2 3 10
t tE s
s
s
de t dt d t
dt
sss s s
s s s
Obtaining Time Information from Frequency Domain
• Initial Value Theorem
Ex:
• Final Value Theorem
Ex:
f sF ss
( ) lim ( )0
f f t sF st s
( ) lim ( ) lim ( ) 0
1
lim)0(
)(lim)0(
22
2
s
sy
ssYy
s
s
t
y
22)(
s
ssY
y(t)£
1
lim)(
)(lim)(
0
0
ass
say
ssYy
s
s
Note: FVT applies only when f (∞) exists !
t
y ass
asY
)(
y(t)£
Inverse Laplace TransformGiven an s-domain function F(s), the inverse Laplace transform is used to obtain the corresponding time domain function f (t). Procedure:
– Write F(s) as a rational function of s.
– Use long division to write F(s) as the sum of a strictly proper rational function and a quotient part.
– Use Partial-Fraction Expansion (PFE) to break up the strictly proper rational function as a series of components, whose inverse Laplace transforms are known.
– Apply inverse Laplace transform to individual components.
One Example
• Find inverse Laplace transform of2
3
6 8( )
4
sF s
s s
2 2
1 2 33
6 8 6 8( )
4 2 2 2 2
s s a a aF s
s s s s j s j s s j s j
1 0 0( ) lim ( ) 2
s sa sF s sF s
2 22 ( ) 2
s ja s j F s
3 2
2 ( ) 2s j
a s j F s
2
2 2 2 2( ) 4
2 2 4
sF s
s s j s j s s
1 1 12 4 2
1£ ( ) £ £
42 4cos(2 )
f ts
F ss s
t
• Residual Formula
Use Laplace Transform to Solve ODEs
Y(s): Solution in Laplace Domain
y(t): Solution in Time Domain
Differential Equations (ODEs)
+Initial Conditions (ICs)
(Time Domain)
Algebraic Equations
( s-domain )
L [ ] L -1 [ ]
Solve ODE
Solve Algebraic Equation
ExamplesQ: Use LT to solve the free response of a
1st Order System.
Q: Use LT to find the step response of a 1st Order System.
( )y y y y 0 0 0
( )y y K y 1 0 0
£ £ 0y y
( ) (0) ( ) 0sY s y Y s
01 ( )s Y s y
0( )1
yY s
s
01 1( )
1
0
1£ ( ) £
1y t y
t
Y ss
y e
£ £y y K
( ) (0) ( )K
sY s y Y ss
1 ( )K
s Y ss
1
1 1( )
1 1( 1) ( )
KY s K K
s s ss s s
1( ) 1£ ( )t
y t K eY s
( ) 5 ( )K
sY s Y ss
5
( )1 1
KY s
s s s
1( ) 1 5£ ( )t t
y t K e eY s
Q: What is the step response when the initial condition is not zero, say y(0) = 5.
Use LT and ILT to Solve for Responses
Find the free response of a 2nd order system with two distinct real characteristic roots:
2
£££
( ) (0) (0) 9 ( ) (0) 18 ( ) 0yyy
s Y s sy y sY s y Y s
( ) ( )y y y y y 9 18 0 0 0 0 3where and
1 2
2
3 3( )
9 18 3 6 3 6
a aY s
s s s s s s
1 33 ( ) 1
sa s Y s
2 6
6 ( ) 1s
a s Y s
1 21 1 1( )
3 6
1 1£ ( ) £ £
3 6
1 ( 1)
y t a a
t t
Y ss s
e e
Use LT and ILT to Solve for Responses
Find the free response of a 2nd order system with two identical real characteristic roots:
2
25
5 ( ) 10s
a s Y s
( ) ( )y y y y y 10 25 0 0 1 0 5where and
2
£££
( ) (0) (0) 10 ( ) (0) 25 ( ) 0yyy
s Y s sy y sY s y Y s
1 2
2 22
15 15( )
10 25 55 5
s s a aY s
s s ss s
2
1 2 15 5
15 ( ) 15 1
2 1 ! s s
d da s Y s s
ds ds
1 2
1 1 1( ) 2
5 5
1 1£ ( ) £ £
5 5
10
y t a a
t t
Y ss s
e te
Use LT and ILT to Solve for Responses
Find the free response of a 2nd order system with complex characteristic roots:
2
£££
( ) (0) (0) 6 ( ) (0) 25 ( ) 0yyy
s Y s sy y sY s y Y s
( ) ( )y y y y y 6 25 0 0 1 0 3where and
1 2
221 2
3 3( )
6 25 3 16
s s a aY s
s s s p s ps
1,2 3 4p j
1
1 1
1( )
2s pa s p Y s
22 2 1
1( )
2s pa s p Y s a
1 2
1 1
1 1 1( )
1 2
1 1
1 1
11 111 1
3 4 3
1 1£ ( ) £ £
1 1£ £
2Re
12Re cos(4 )
2
y t a a
a a
p tp t p t
j t t
Y ss p s p
s p s p
a e a e a e
e e t
Use LT and ILT to Solve for Responses
Find the unit step response of a 2nd order system:
2
££ £ ££
( ) (0) (0) 6 ( ) (0) 8 ( ) ( ) (0) 3 ( )yy u uy
s Y s sy y sY s y Y s sU s u U s
6 8 3 where (0) 0 and (0) 0y y y u u y y
1 2 3
2 2
3 3 1( ) ( )
6 8 6 8 2 4
s s a a aY s U s
s s s s s s s s
1 0
3( )
8sa sY s
2 2
12 ( )
4sa s Y s
1 2 31 1 1 1( )
2 4
1 1 1£ ( ) £ £ £
2 4
3 1 18 4 8
y t a a a
t t
Y ss s s
e e
3 4
14 ( )
8sa s Y s