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