lecture 25 laplace transform hung-yi lee. reference textbook 13.1, 13.2

Lecture 25Laplace Transform

Hung-yi Lee

Reference

• Textbook 13.1, 13.2

Laplace TransformMotivation and Introduction

Laplace Transform

dtetf st

0)(sF

tf sF

dsesFj

tf stj

j

)(

2

1

c

c

se

R

Laplace Transform ( L[f(t)] )

Inverse Laplace Transform (L-1[F(s)])

Time domain s-domain

Note

dtetftf st

0)(L

dtetfdtetf stst

0

0

0)()(

Always 0?When f(0)=∞, it may not be zero

Note

tf

sF

Laplace Transform (L)

Inverse Laplace Transform (L-1)

Time domain

s-domain

tf̂=

When t≥0

1

,,1

tutu

tu

Domain

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25°1'9"N   121°32'31"E

Different domains means view the same thing in different perspectives

Position:

Domain

x

y 1,1 xy 0,2

Different domains means view the same thing in different perspectives

Linear Algebra:

Domain

1tfTime domain

ss /1F s-domain

SignalMuggle

Engineer LaplaceTransform

Inverse LaplaceTransform

Transform: switch between different domains

Fourier Series

Fourier Transform

TimeDomain

TimeDomain

FrequencyDomain

FrequencyDomain

https://www.youtube.com/watch?v=r4c9ojz6hJg

Fourier Transform

dtetf tj

)(F

deFt tj

)(

2

1f

Laplace Transform

dtetfs st

0)(F

dsesFj

tf stj

j

)(

2

1

Laplace Transform

v.s. Fourier

Transform

dtetfs st

0)(F

dtetf tj

)(F

Fourier Transform:

Laplace Transform:

dteetf tjt

0)(

dtetuetf tjt

)(

Laplace Transform of f(t)

Fourier Transform of f(t)e-σtu(t)

=

Multiply e-σt

Do Fourier Transform

Laplace Transform

v.s. Fourier

TransformMultiply u(t)

dtetfs st

0)(F

dtetuetf tjt

)(

Transformable Function

Fourier Series

Fourier Transform

Laplace Transform

All Functions2te

te

1

0 D

Periodic Functions

Why we do Laplace transform?

Transfer Function

H(s)(circuit, filter …)

)(tx

)(ty

Laplace transform can help us find y(t) easily

Transfer Function ttx 0

t cosAe)( 0

00t

0 cosAe||)( 0 sHtsHty

H(s)(circuit, filter …)

The signal with complex frequency s0 = σ0 + ω0 t0

t sinjAe 0

00t

0 sinAe|| 0 sHtsHj

tj 0eAe s

tjHj0

00 eAee|H| sss

Transfer Function

H(s) H(s)

(z is complex)

tj 0eAe stx

tjHj

000 eAee|H| sss

ty

(H(s0) is complex)

t0Zestx

t0

0ZessHty

Transfer Function

H(s)

dsesFj

tx stj

j

)(

2

1

dsesFsHj

ty stj

j

)(

2

1

)(L sFtx

)(HL sFsty We do not know y(t), but we know its Laplace transform

Transfer Function

H(s)

ty

tx )(sF

)(H sFs

L

-1L

Laplace TransformLaplace Transform Pairs

Laplace Transform Pairs (1/4)

1)( tf

01)(L dtetf st

101

s s

1

t

t

stes 0

1

If Re[s]=σ>0

Time domain: 1 s-domain: 1/s

ROC: σ > 0

ROC

Time domain: u(t)(Only consider t>0)

Laplace Transform Pairs (2/4)

00

)(L dtedteetf tsastat

atetf )(

t

t

tsaesa 0

1 10

1

sa as

1

If Re[a+s]=a+σ>0

Time domain: e-at s-domain: 1/(s+a)ROC: σ > -a

ROC-a

Laplace Transform Pairs (3/4)

0 2dte

j

ee sttjtj

sincos

sincos

je

jej

j

ttf sin)(

0

sin)(L dtettf st

002

1 dtedtej

tjstjs

222

sj

jsjs

jsjsj

11

2

122

s

j

ee

ee

jj

jj

2sin

2cos

If Re[s]=σ>0

Laplace Transform Pairs (4/4)

0 2dte

ee sttjtj

ttf cos)(

0

cos)(L dtettf st

002

1 dtedte tjstjs

222

s

jsjs

jsjs

11

2

122

s

sIf Re[s]=σ>0

sincos

sincos

je

jej

j

j

ee

ee

jj

jj

2sin

2cos

Summary of Transform Pairs

• More complete Transform Pairs: http://www.vibrationdata.com/math/Laplace_Transforms.pdf

The 4 transform pairs are sufficient to imply all transform pairs in Table 13.2.

as 1ate

s

1122

s

tsin

22 ss tcos

s-domainTime domain

Note: Impulse function

• What is L-1[1]? L-1[1]=δ(t) (impulse function, Dirac delta function)

0""

00

t

tt

0""

00

0

00 t

ttt

1dtt

dt/1

dt

tdut 0

1

0t

1

Laplace TransformProperties

The six properties in Table 13.1 (P585)

Property 1: Linear Combination• Let L[f(t)]=F(s) and L[g(t)]=G(s)

)()()()(0

dtetgtftgtfL st

00

)()( dtetgdtetf stst

)()( sGsF

)()(0

dtetAftAfL st )(sAF

Property 2: Multiplication by e-at

• Let L[f(t)]=F(s)

0

)()(L dtetfetfe statat

0

)( dtetf tas )( asF

as 1ate

s

11

ROC

ROC-a

Multiplication by e-at

Property 3: Multiplication by t• Let L[f(t)]=F(s)

)()(L sFds

dttf

0

)( dtetfds

dsF

ds

d st

0dt

ds

detf

st

0

dtettf st

0

dtettf st

Property 4: Time Delay

00

0

00

)(

0

)( )()(

ttttf

tt

ttuttftg

0tt

Delay by t0 and zero-padding up to t0

t

tt

stdtettftg0

)()(L 0

0tt 0tt

0

0)( def ts

ddt

0

)(0 defe sst sFe st0

Property 5: Differentiation

• Let L[f(t)]=F(s) b

a

b

a

b

adtuvuvudtv

Integration by parts:

00

)()()( dtestfetf stst

0

)()(L dte

dt

tdftf st

0

)()0( dtetfsf st )0()( fssF

v’

v v

u

u u’

Property 5: Differentiation

• Let L[f(t)]=F(s)

0)(L fssFtf

1

1LsF

s

tf tetf

te tf 1

1L

s

tf

Example

0)(L fssFtf 11

1

ss

1

1

s

Property 5: Differentiation

0)(L fssFtf

sFtf )(L

00)(L ffssFstf

002 fsfsFs

……

000)(L )1(21 nnnnn ffsfssFstf

Property 6: Integration

tdftg

0)()(

• Let L[f(t)]=F(s)

0 0)()(L dtedftg stt

0 0)( dtdef

t st

0)(

ddtef st

(You can use integration by parts as in P584)

t

t

dtdef st)(

tdftg

0)()(

• Let L[f(t)]=F(s)

Property 6: Integration

0)(L

ddteftg st

0)(

ddtef st

0

1)( de

sf st

0

1)( des

f s

0

)(1 defs

s

ss

F1

Laplace Transform Properties

)(1

)( n Integratio

)0()( )( ation Differenti

)( )()( delay Time

)( )( by tion Multiplica

)( )( by tion Multiplica

)()( )()( n CombinatioLinear

-

0

0

00

sFs

df

fssFtf

sFettuttf

dssdFttft

asFtfee

sBGsAFtBgtAf

t

st

atat

Table 13.1 Laplace Transform Properties (P585)Operation Time Function Laplace Transform

More properties (in Homework)

Time-scaling property

Periodic function

Integral of F(s)

sFtf /L

tftdssFs

1L

tf tgf(t)=0 outside 0<t<T

TtuTtfTtuTtftftg 22

…..

sTe

sFtg

1

L

22

22

22

1

2

1

2

sincos )cos(

sincos )cos(

sin

!

1

1

!

1

1 )()(

as

aste

s

st

st

as

ret

aste

ase

s

rt

st

s

eDtutu

s

AA

at

ratr

at

at

rr

sD

Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)

1

0 D

D-ttuL u

D-tLtuL u sD-e11

ss

22

22

22

1

2

1

2

sincos )cos(

sincos )cos(

sin

!

1

1

!

1

1 )()(

as

aste

s

st

st

as

ret

aste

ase

s

rt

st

s

eDtutu

s

AA

at

ratr

at

at

rr

sD

Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)

1

0 1

1-ttuL us

-se1

1

0 1

1-ttu u

1-ttut u

t

1-ttutL udss

d-se1

2

-s-s e1e

s

s

2

-s-s e-e-1

s

s

Example for Periodic function

1

0 1

2

-s-s e-e-1

s

sL

1

0 1 2 3 4 5

L

TtuTtfTtuTtftftg 22

sTe

sFtg

1

L

2

-s-s e-e-1

s

sse 21

1

22

22

22

1

2

1

2

sincos )cos(

sincos )cos(

sin

!

1

1

!

1

1 )()(

as

aste

s

st

st

as

ret

aste

ase

s

rt

st

s

eDtutu

s

AA

at

ratr

at

at

rr

sD

Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)

1 1s

tds

ds 1

2s

2tds

ds 2

32 s

3tds

sd 32

432 s

rt 132 rrs 1! rsr

L

L

L

L……

L

22

22

22

1

2

1

2

sincos )cos(

sincos )cos(

sin

!

1

1

!

1

1 )()(

as

aste

s

st

st

as

ret

aste

ase

s

rt

st

s

eDtutu

s

AA

at

ratr

at

at

rr

sD

Table 13.2 Laplace Transform Pairs (P585) f(t) F(s)

tcos

sinsincoscos tt

tcosL

2222

sincos

ss

s

Note: Euler’s formula

tjte tj00 sincos0

0

1

js 20

2 ss

20

20

s

j

0

0

0

1

js

js

js

20

20

s

js

Note: Multiplication

)()( sGsFtgtfL

tsinL22

s

t2sinL

2

22

s

2

2cos1sin 2 t

t

t2sinL 22 22

1

2

1

s

s

s

Laplace TransformInverse Laplace Transform

Partial-Fraction Expansions

• Rational Function

)(

)()(

011

1

011

1

sD

sN

asasas

bsbsbsbsF

nn

n

mm

mm

n

n

ss

A

ss

A

ss

A

2

2

1

1

nssssss

sN

21

)( s1, s2, ……, sn are the roots of D(s) ( poles of F(s) )

We only consider the case that m < n.(strictly proper rational function)

One pole, one term

Partial-Fraction Expansions

• Rational Function

)(

)()(

011

1

011

1

sD

sN

asasas

bsbsbsbsF

nn

n

mm

mm

sDsR

bs m

)(F

If m = n

δ(t)

If m = n + 1

sDsR

csbs m

)(F

1

0fssFtf

sdδ(t)/dt ……

multiply sdifferentiate

Partial-Fraction Expansions

• Rational Function

n

n

ss

A

ss

A

ss

AsF

2

2

1

1)(

ateas

L

11

tsn

tsts neAeAeAtfsFL 2121

1 )(

tsi

i

i ieAss

AL

1

0t

There are three tips you should know.

Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An

•Example 13.5

102102

40242)(F 321

2

s

A

s

A

s

A

sss

sss

Panacea: reduce to a common denominator, and then compare the coefficients

21010240242 3212 ssAssAssAss

ssAssAssA 2102012 23

22

21

2321 AAA

2421012 321 AAA

4020 1 A

Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An

•cover-up rule

212

22

)(then , setting A

ss

sNss

2

2

1

1

21

)()(

ss

A

ss

A

ssss

sNsF

2

211

2

)(

ss

AssA

ss

sN

1

21

11

)(then , setting A

ss

sNss

2

1

21

1

)(A

ss

ssA

ss

sN

Tip 1: How to find A1,A2,∙∙∙∙∙∙∙,An

• Example 13.5: cover-up rule

102102

40242)(F 321

2

s

A

s

A

s

A

sss

sss

Take A2 as example. Multiplying s+2 at both sides

Set s=-2:

102

2102

402422 321

2

s

A

s

A

s

As

sss

sss

10

22

10

40242 32

12

s

AsA

s

As

ss

ss

2

2

1022

4022422A

52 A

Tip 2: For Complex Poles

• Example 13.6: 2562

71615)(

2

2

sss

sssF Find f(t) = L-1[F(s)]

43 ,2 js

43432)( 22211

js

A

js

A

s

AsF

It is not easy to

find A21 and A22.

43

43

2

43)(43 22

211

js

AjsA

s

AjssFjs

432

71615 2

jss

ss

Set s=-3+j4 to find A21…….

Tip 2: For Complex Poles

• Example 13.6: 2562

71615)(

2

2

sss

sssF Find f(t) = L-1[F(s)]

2562

5)(

2

ss

CBs

ssF Do not split the complex poles

Find B and C

2562

71615

2562

221253052

2

2

22

sss

ss

sss

CCsBsBsss

66 72125

10 155

CC

BB

256

6610

2

5)(

2

ss

s

ssF

Tip 2: For Complex Poles

• Example 13.6: 2562

71615)(

2

2

sss

sssF Find f(t) = L-1[F(s)]

2562

5)(

2

ss

CBs

ssF Do not split the complex poles

Find B and C (another approach)

2562

515)(

2

2

3

3

ss

CsBs

s

s

s

sssF

BssFs

515lim

252

5

252

7)0(

CF

10B

66C

ssFs lim

)0(F

Tip 2: For Complex Poles

• Example 13.6: 2562

71615)(

2

2

sss

sssF Find f(t) = L-1[F(s)]

256

6610L

21

ss

sFind

221

21

43

96310L

256

6610L

s

s

ss

s

(Refer to P593)

te

s

s t

cosL22

1

te

st

sinL

221

22

1

221

43

4L24

43

3L10

ss

s

256

6610

2

5)(

2

ss

s

ssF

Tip 3: Repeated Poles

• Exercise 13.31

22

34

10124)(

ss

sssF

34321

s

AA

s

A

221

2

2

3434

10124)(

s

A

s

A

ss

sssF

order=2order=3

212

2

43

10124

AsAs

ss

212

2

496

10124

AsAss

ss

41 A

126 21 AA

1049 21 AA

?

334321

s

A

s

A

s

A

Tip 3: Repeated Poles

• Exercise 13.31

2321

2

2

33434

10124)(

s

A

s

A

s

A

ss

sssF

(Refer to P596 – 597)

232

12

2

3

4

3

4

3

10124

s

As

s

AsA

s

ss61 A

32

122

34

3

4

10124AAs

s

As

s

ss

103 A

We cannot find A2 by multiplying (s+3)

34

3

34

10124 32

12

s

AA

s

As

ss

ss

Tip 3: Repeated Poles

• Exercise 13.31

22

2

2

3

10

34

6

34

10124)(

ss

A

sss

sssF

22

2

2

3

10

34

6

34

10124)(

s

s

s

sA

s

s

ss

sssssF

s 4 6 02A 2A2

Exercise 13.10

2

8

3

4235F

s

esss

s

ses

s

s8

23

42

3

5

te 35

L-1

)(0 sFe st)()( 00 ttuttf Time delay

Delay by 8 and zero-padding up to 8

?

L-1

23

42F

s

ss

221

3

A

3

A

ss

21 AA342 ss

10A2 3s

Exercise 13.10

2

8

3

4235F

s

esss

s

ses

s

s8

23

42

3

5

te 35

L-1

)(0 sFe st)()( 00 ttuttf Time delay

Delay by 8 and zero-padding up to 8

?

L-1

23

42F

s

ss

21

3

10

3

A

ss

23

42Fs

s

sss

21

3

10

3

A

s

sss

s 1A2

Exercise 13.10

2

8

3

4235F

s

esss

s

ses

s

s8

23

42

3

5

te 35

L-1

)(0 sFe st)()( 00 ttuttf Time delay

Delay by 8 and zero-padding up to 8

?

L-1

23

42F

s

ss

23

10

3

2

ss

te 32

L-1

tte 310 L-1

88

5tf 3

tutf

e t

tt teetf 33 102

Initial and Final Values

• We can find the value f(0+) and f(∞) from F(s)

Initial-value Theorem

sFtf LWe know

)(lim)0( ssFfs

Because F(s) is strictly proper, )(lim ssF

s is defined.

Example 13.9

• Find the initial value f(0+)

8009018

16005)(F

23

3

ssss

ss 50 f

)(lim)0( ssFfs

8009018

16005lim

23

3

sss

ss

5

Example 13.9

• Find the initial slope f’(0+)

8009018

16005)(F

23

3

ssss

ss 50 f

)(lim)0( sFsfs

)(FsfL s

0)(FsfL fss 58009018

1600523

3

sss

s

s

sss

ssf

s5

8009018

16005lim)0(

23

3

8009018

8009018516005lim

23

233

sss

sssssss

90

∞ - ∞ ?

Initial and Final Values

• We can find the value f(0+) and f(∞) from F(s)

)(lim)(0

ssFfs

Initial-value Theorem

Final-value Theorem

sFtf LWe know

)(lim)0( ssFfs

If the final value exists(Can be known from the poles)

Because F(s) is strictly proper, )(lim ssF

s is defined.

Final Values

4 regions

Region A

Region D

Region C

Region B

Final Values

Region A

as

As

)(F

rr

as

A

as

A

as

As

2

21)(F

cbss

Css

2

B)(F

a>0

a>0

b<0

atet )(f

No final value

No final value

attet )(f

22

s

CsB

α<0 teC

teBtt

t

sin

cos)(f

No final value

Final Values

Region B

as

As

)(F

rr

as

A

as

A

as

As

2

21)(F

cbss

Css

2

B)(F

a<0

a<0

b>0

atet )(fFinal value = 0

Final value = 0

attet )(f

22

s

CsB

α>0 teC

teBtt

t

sin

cos)(f

Final value = 0

Region C

Final Values

22

B)(F

s

Css

tCtBt sincos)(f

No final value

2222CB

ss

s

Final Values

Region D

s

As )(F A)(f t

final value = constant

rr

s

A

s

A

s

As

221)(F tt 2A)(f No final value

Summary for Final Values

Final value exists (1) Poles on the left plane, or (2) single pole at the origin

(P601)

Non zero final valuesingle pole at the origin

Final Values )(lim)(0

ssFfs

Final-value Theorem

22

F

s

ss

The final value exists iff the poles are in this region

Only one pole 0)(lim0

ssFs

ts cosf The final value not exists

The final-value theorem gives the wrong answer when the final value does not exist.

as

s

1

F

0a0)(lim

0

ssF

s

ates f The final value not exists

Final Values )(lim)(0

ssFfs

Final-value Theorem

The final value exists iff the poles are in this region

Only one pole

The final value is not zero iff there is only one pole at the origin

2

2

1

1)(Fas

A

as

A

s

As

The final value is clearly A

The final value is 0

The final value is A

)(lim)(A0

ssFfs

Example 13.9

• Find the final value

8009018

16005)(F

23

3

ssss

ss 50 f

Four poles: 0, -10, -4+8j, -4-8j The final value exists.

The final value is not zero.

)(lim)(0

ssFfs

8009018

16005lim

23

3

0

sss

ss

2

Laplace TransformApplication

Differential Equation

Find v(t)

6VV15812

8)0(

v

BVtvtRi )()( )()( tvCti

BVtvtvRC )()(

15)()(60

112 tvtv

15)()(2.0 tvtv

15L)(L)(L2.0 tvtv

)(L)( tvsV

6)(

)0()(L)(L

ssV

vtvstv

s

1515L

s

sVssV15

)(6)(2.0

Differential Equation

0t

s

sVssV15

)(6)(2.0

55

756

12.0

152.1

)(

s

B

s

A

ss

s

sssV

0s55

756)(

s

sBA

s

sssV 15A

B5756

)(5

s

As

s

ssVs 5s 9B

5

915)(

ss

sV

tetv

sV5

1

915)(

)(L

Homework

• 13.6, 13.9, 13.10, 13.16, 13.25, 13.28, 13.35, 13.38. 13.46

Thank you!

Answer

• 13.6: derive by yourself• 13.9: proof by yourself• 13.10: proof by yourself• 13.16: F(s)=2(1-3se-2s-e-3s)/s2

• 13.25: f(t)=-2+5e-2t-3e-4t-e-6t

• 13.28: f(t)=5-5e-4t+10e-3tcos(t-36.9 。 )• 13.35: f(t)=2te-tcos(2t-180 。 )• 13.46: f(0+)=2, f’(0+)=-5, f(∞) not exist

Appendix

Fourier Series

• Periodic Function: f(t) = f(t+nT)

Fourier Series:

Laplace Transform Pairs (1/4)

s

11L ste11 0s

σ=0

0 ω

?