logics of the laplace transform

1

LOGICS OF The Laplace Transform BY TARUN GEHLOT

Introduction

(1) System analysis

static system: y(t) = ax(t) => easy (simple processing)

dynamic system:

)(...)()(...)()( )(

01

)1(

1

)(

txbdt

tdxbtyadt

tdyadt

tdymm

m

nn

n

n

n

Can we determine y(t) for given u(t) easily?

Easier solution method

))(()()()()(

1 fYFtyfXfHfY

(1) we have systematic way to obtain H(f) based on the differential equation(2) we can obtain X(f)

Fourier transform: an easier way

(2) ProblemFourier transform of the input signal:

DynamicSystem

x(t) input y(t) output

A processor whichprocesses the input signalto produce the output

DynamicSystem

X(f) Y(f )

H(f )Algebraic equation, notdifferential equation

2

1||

)()(

2

2

ftj

ftj

e

dtetxfX

if x(t) does not go to zero when t and t X(f) typically does not exist! (the existence of X(f) if not guaranteed)

A very strong condition, can not be satisfied by many signals!

(3) Solution:

Why should we care about t<0 for system analysis? We do not care!x(t) does not go to zero (when t ), but tetx )( may!

(Will be much easier. ) use “single-sided” Fourier transform of tetx )( , instead of “double-

sided” Fourier transform of x(t).

0 0

)()())(( dtetxdteetx tjtjt

very useful, let’s use a new name for it: Laplace transform.

(4) Laplace Transform

Definition: Laplace transform of x(t)

0

)()()()]([ jsdtetxsXtxL st

What is a Laplace transform of x(t)?A time function? No, t has been eliminated by the integral with respect to t!A function of s ( s is complex variable)

(5) System analysis using Laplace transform

3

))(()()()()(

1 sYLtysXsGsY

(6) How is the inverse transform defined?

j

j

stdsesXj

tx

)(

21)(

(7) Will we often use the definition of the inverse transform to find timefunction?

No!What will we do?

Express X(s) as sum of terms for which we know the inverse transforms!

5-2 Examples of Evaluating Laplace Transforms using the definition

(1) x(t)=1 and step function x(t)=u(t)

0(Re(s))1)]([)1(

1)0sin0(cos1)0(,)0(,01|(|

)(1)()]()([

00

00

000

stuLL

sj

s

ifeifee

es

ees

ees

es

e

stdes

dtedtetxtutxL

j

jjt

t

tjtt

t

st

t

t

ststst

(2) )()( tuetx t

0

)(

0

)]([ dtedteetueL tssttt

Define a new complex variable ss~

DynamicSystem

X(s) Y(s )

G(s )

Inverse Laplace transform

4

0

~

dte ts

we know 0)Re(1

0

ss

dte st

seL

sorss

tueL

ss

dte

ss

dte

t

t

ts

ts

1][

)Re()Re(0)Re(1)]([

0)Re(1

0)Re(1

0

)(

~

~0

~

(3) )()( ttx

1

)sin(cos

)()]([

0

00

0

tt

ttjt

tst

st

tjte

eee

dtettL

No constraint on s.

5-2B Discussion: Convergence of the Laplace Transform

(1) To assure

00

)()( dteetxdtetx tjtst converge, )Re(s must be psotive

enough such that tetx )( goes to zero when t goes to positive infinite

(2) Region of absolute convergence and pole

-5

(3) How to obtain Fourier transform form Laplace transform:

))(()()()]([ txFjXsXtxLjs

Important: why introduce Laplace transform; definition of Laplace transform as amodification of Fourier transform; find the Laplace transforms of the three basicfunctions based on the (mathematical) definition of Laplace transform.

Chapter structure

Part one: Definition (5.1)Part two: Direct evaluation of Laplace transforms of simple time-

domain function.Easy? Not at all!

complex (not simple) functions: even harder tools for application of Laplace transform in system analysis

Part three: Rest of the chapter

What tools: tools for easier Laplace transform evaluationtools for easy inverse transform

Page 5-5




Chapter structure






Page 5-5




Chapter structure






6

5-3 Some Laplace Transform theorems(Tools for evaluating Laplace transform based on the Laplace

transforms of the basic functions)

5.3.1 LinearityAssume )()()( 2211 txatxatx ( 1a and 2a are time independent)

)]([)()],([)( 2211 txLsXtxLsX then )()()]([)( 2211 sXasXatxLsX

HW#2-1: Assume )()()()()( 2211 txtatxtatx , )]([)( 11 txLsX and )]([)( 22 txLsX .

(a) Is )]([)( txLsX equal to )()()()( 2211 sXtasXta ?(Answer: Yes or No)

(b) If )]([)( 11 taLsA , is it true)()()()()]([ 2211 sXsAsXsAtxL ?

(Answer: Yes or No)

Example 5.1(1) Find )(cos 0tL

Key to solution : express )(cos 0t as linear combination of )(t , )(tu ,and/or te :

seL

stuL

tL

t 1][

1)]([

1)]([

let 0 j

0

1][ 0

jseL tj

let 0 j

0

)( 1][][ 00

jseLeL tjtj

7

Can we use tje 0 and tje 0 to express )cos( 0t ?

)sin()cos(

)sin()cos()sin()cos(

00

00

00

0

0

tjte

tjttjte

tj

tj

20

2

00

00

00

0

0

0

))(()()(

21

1121

)]()([21)][cos(

2)cos(

)cos(2

00

00

00

ss

jsjsjsjs

jsjs

eLeLtL

eet

tee

tjtj

tjtj

tjtj

(2) Find ][sin 0tL

5-3-2 Transforms of Derivatives

Assume )]([)( txLsX

Then )0()()(

xssX

dttdxL

Proof:

(1) Definition

00

)()()()( tdxedte

dttdx

tdtdxL stst

(2) Integration by parts:General equation:

8

b

a

bt

at

b

a

tdutvtvtutdvtu )()()()()()(

(3) Use the above equation

Why? If we assumestetutxtv )(,)()(

dtetxsdetxtdutv stst

000

)()()()(

)()( txLsX

](1)from)([)()()(

)()(,)(

00

dttdxLtdxetdvtu

txtvetu

st

st

)]([)(

)()(

)()()()(

0

00

0

txsLtxe

dtetxstxe

tdutvtvtu

t

tst

stt

tst

b

a

t

t

)(lim txe st

t

must go to zero. Otherwise,

0

)()]([ dtetxtxL st does not exist !

)0()0()( 00 xxetxe st

tst

use0 as lower limit => )0( x

)0()()()0(

)()()()(0

0

xssXssXx

tdutvtvtu t

t

9

HW#2-2: Assume )]([)( txLsX . Prove

)0()0()()( )1(22

xsxsXs

dttxdL

HW#2-3: Express

n

n

dttxdL )()(

using )]([ txL

Example 5-2

Find i(t) using Laplace transform methodfor t>0

Solution:(1) Before switched from 1 to 2 at t=0

AiAi 2)0(224

(2) System equation (t>0)

KVL:0)(2)(

)2()1(0)()(

tidt

tdi

ohmRHLtRidt

tdiL

(3) Solve system equation using Laplace transform

Atuetis

sI

sIssIissI

tiLdt

tdiLtidt

tdiL

t )(2)(2

2)(

02)()2()(2)0()(

)]([2)()(2)(

2

10

5-3-3 Laplace Transform of an integral

Assume )]([)(,)()( txLsXdxtyt

Thens

yssXdxL

t )0()()(

where

0

)()0( dxy

Proof :

00

0

0

0

)()(

vduuvudv

dtedxdxL

tt

stt

(1)

sy

dxs

edxse

sedxuv

stst

t

t

t

sttt

t

)0(

)()(lim

)(

00

00

(2)

)(1)(1)(000

sXs

dtetxs

dttxs

evdu stst

(3)

sy

ssX

ssX

sydxL

t )0()()()0()(

Proved!

0

1

)0( y

11

Example 5.3:

Find I(s) = L(i(t))

Solution:(1) Differential equationKVL :

)()()()( tvtvtvtx RCL

t

CC

t

CC

tt

C

C

C

diC

tvv

diC

vtv

diC

dv

dttiC

tdv

dttdvCti

)(1)(0)(

)(1)()(

)(1)(

)(1)(

)()(

)()(

)(1)(

)()(

tRitv

diC

tv

dttdiLtv

R

t

C

L

(2) Laplace transform

)()(

)0(1)(1

)(11)(1

)(1)(1)(

)0()()]0()([)(

)()()()(

0

0

sRIsV

vs

sICs

dics

sICs

diss

sIC

sV

zeroisLsIissILsV

sVsVsVsX

R

c

C

L

RCL

12

LCs

LRsL

vssX

sRC

Ls

vssX

RCs

Ls

vs

sXsI

sRIvs

sICs

sLsIsX

C

C

C

C

1)0()(

1)0()(

1

)0(1)()(

)()0(1)(1)()(

2

2

5-3-4. Complex Frequency shift (s-shift) Theorem

Assume tetxty )()()]([)()]([)( tyLsYtxLsX

Then )()( sXsY

setuL

stuL t 1])([,1)]([

20

20

020

20 ][sin][cos

s

tLs

stL

=>2

020

020

20 )(][sin

)(][cos

×

×

setL

ssetL tt

Example 5-4 Find136

8)]([)( 211

sssLsXLtx

Solution:

2222

22

2)3(2)2/5(

2)3(3

4965)3(

1368)(

´

sss

sss

ssssX

13

)0(2sin252cos)]([)( 331 ttetesXLtx tt

5-3-4 Delay Theorem question: How to express delayed function?

Assume )()]()([)]([ sXtutxLtxL

Then )0()()]()([ 0000 tsXettuttxL st

(If )0( 0 t , it will not be a delay!)Proof :

)()()(

)()(

)()(

)()()()(

)()(

)]()([

0000

0

00

0

00

0

0

0

00

0)(

0

)(00

000

00

000

00

sXedtetxedexe

ttdettxe

dtettxdtettx

dtettuttxdtettuttx

dtettuttx

ttuttxL

stststssttt

t

ttsst

t

sttts

t

st

t

stt

st

st

14

Question: will )()]()([ 000 sXettuttxL st be true if ?00 t

No! (it will not be a delay)

Example 5-5: Square wave beginning at t = 0

sT

sT

e

sTsTsTsT

sq

e

essss

ss

sssss

es

es

es

ess

txL

sT

2

2

32

432

223

2

0

0

021

00

00

1

11111

)1(21

...)(21

...121212121

121212121)]([

5-3-5 ConvolutionSignal 1: )(1 tx Signal 2 : )(2 tx

dtxxtxtxty

)()()(*)()( 2121

if 00)(1 ttx

dtxxty

0

21 )()()(

15

if )0)((00)( 22 ttxttx

dtxxtyt

0

21 )()()(

Therefore, if 00)(,0)( 21 ttxtx

dtxxdtxxdtxxt

)()()()()()( 210

210

21

])()([])()([])()([ 210

210

21 dtxxLdtxxLdtxxLt

0 0

210 0

21 ])()[(])()([ ddtetxxdtedtxx stst

Look at

)()(

)(

)()(

20

2

00)(

2

0

)(2

02

2

0

sXedexe

dexe

dteetxdtetx

sssx

ss

dtd

t

stsst

tt

Then

)()(

)()(

)()()(

21

012

021

sXsX

dexsX

dsXexsY

s

s

t

16

5-3-7 Product5-3-8 Initial Value Theorem

)(lim)0(

)]([)( 1

ssXx

sXLtx

s

Example: A demonstration where x(0) is obvious)(cos)( 0 ttuetx t

It is evident: 10cos)0( 00 ex

Using Laplace transform

20

2)()]([)(

sstxLsX

122lim

/)22(/)2(lim

222lim

/)2(/)(lim

2lim

)()(lim)(lim)0(

20

22

2

20

22

2

20

2

sss

s

s

ss

dssddssd

ss

dsssddsssd

ssss

sssssXx

5-3-9 Final Value Theorem: if )(tx and dttdx /)( are Laplace transformable, then)(lim)(lim

0ssXtx

st

(condition: )(ssX has no poles on axisj or in the right-half s-planor )(lim txt exists)

5-3-10 Scalinga>0: x(at) a times fast (if a>1)

or slow (if a<1) as x(t)

)]([)( txLsX

What do we expect on )]([ atxL ?

17

?)()]([asXatxL

asX

adex

a

atdeatxa

dteatxatxL

as

ta

at

atasa

st

1)(1

)()(1)()]([

00

0

)(0

0

5-4 Inversion of Rational Functions(1) Ways to find )(tx from )(sX

(1)

dtesX

jtx st)(

21)( (Contour Integral)

(2) Transform pair

)(1

)(1

tues

tus

t

Therefore )()(1)( tuetxs

sX t

(2) All kinds of Laplace Transform ? No!

We almost only see

s

nnnn

mmmm

easasas

bsbsbsb

11

1

11

10

...

...

rational function X(s) Delay

)()(])([)(

)]([)(1

1

tutxesXLty

sXLtxs

18

Consider Rational Functions only!

(3) Non proper Rational Functionproper Rational Function

Non proper m>=n )0( 0 bproper m<n

Non proper => proper + Polynomial (using long division)

2351

23764

22

23

ss

ssss

sss

53

74764

1

2

2

23 232

ssss

sssss

sss

How to find inverse Laplace transform for polynomials?

)()()1()()1()(

)1(111

)(

ttLsLsLts nn

consider proper rational functions only!

(4) Proper Rational Functions: Partial Fraction Expansion

sum of

t

t

t

tn

n

es

tus

t

ttues

ttues

s

tunet

s

1

)(1)(1

)(sin)(

)(cos)(

)(!)(

1

020

20

020

2

1

19

Let’s look at examples, and then summarize!

Techniques:Common Denominator Factorize first!Specific value of s Expand second!Heaviside’s Expansion Find coefficients third!Matlab

Example 5-9: Simple Factors

161010)( 2

ss

sY

Solution:

(1) Factorize and expand28)2)(8(

10)(

s

Bs

Ass

sY

(2) Common Denominator Methods

3/53/56/1010821082

010)8()2(

)2)(8()8()2(

)2)(8(10

ABBBBA

BABAsBsA

sssBsA

ss

)()35

35()(

21

35

81

35)(

28 tueety

sssY

tt

specific values of s

41041010

282810

28)2)(8(10

2

0

BA

BAs

Bs

Ass

s

s

Can you solve for A and B?

20

Heaviside Expansion

3/528

102

)8(2

1028)2)(8(

10

8

AssBA

s

sB

sA

sss

and 3/582

108

)2(8

10 2

BBssA

s

s

Example 5-10 Imaginary Roots

)8)(2)(1(202515)( 2

2

sss

sssY

Solution: what do we have:

82)( 4321

sA

sA

jsA

jsAsY

821)()(

821)()(

)(

432

1221

432

21

sA

sA

sjAAsAA

sA

sA

sjsAjsAsY

A1+A2 must be real number(-A1+A2)j must be real number

821)( 43

221

sA

sA

scscsY

Heaviside Expansion => A3=1 and A4= - 2.

82

21

1)8)(2)(1(202515)( 2

212

2

ssscsc

ssssssY

s=1 =>92

31

2932202515 21

´́ cc

s=2 => 102

41

52

104520225415 21

cc

Same as real roots!

21

Can we solve for c1 and c2?c1=1 c2=1

82

21

11

1

82

21

11)(

22

2

sssss

sssssY

=> )(]2sin[cos 82 tueett tt Too complex: use MATLAB

Example 5-11 Repeated linear Factors

2321

2

)2(28

)8()2(10)(

sA

sA

sA

ssssY

Example 5-12

34

2321

3

)2()2(28

)8()2(10)(

sA

sA

sA

sA

ssssY

Example 5-13 Complex - Conjugate Factors

jsA

jsA

sA

ssss

ssssssY

112]1)1)[(2(662

)22)(2(662)(

3212

2

2

2

Example 5-14 Repeated Quadratic Factors

22

2222

2111

22

234

)1(12

)1)(2(157125)(

scsB

scsB

sA

sssssssY

* Summary of Partial–Fraction Expansion(1) Expansion Structure:

Simple Roots (including complex conjugate)

=>j

j

sA

could be complex.

Repeated Roots: m multiplicity

=> mj

m

jj sB

sB

sB

)(...

)( 221

real number or complex number

(2) Avoid complex numberFor complex conjugates: jbaj

kkj

kk

j

k

j

j

j

j

basDcs

s

Bs

B

basDcs

sA

sA

])[()(

)(

22*

*

22*

*

(3) Inverse Laplace transform

2)()!1()(

)(

1

ktuk

etAs

A

tueAs

A

tk

kkj

j

tj

j

j

j

j

23

)(]sin(cos)sin(cos[)!1(

)(][)!1()()(

1

1

*

*

*

tubtjbtBbtjbtAekt

tueBeAkt

sB

sA

jjat

kjba

jba

tj

tj

k

kj

jk

j

j

j

j

jj

logics of the laplace transform

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