Chapter 13 Laplace transform analysis

電路學講義第13章13-1

Chapter 13 Laplace transform analysis13.1 Laplace transform

definition, properties, Laplace transform pairs13.2 Transform inversion

partial-fraction expansion, complex poles, repeated poles, time delay, initial and final values

13.3 Transform circuit analysiszero-state response, natural response, forced response, zero-input response, complete response

13.4 Transform analysis with mutual inductance (optional) 13.5 Impulses and convolution (optional)

impulses, transforms with impluses, convolution and impulse response


13.1 Laplace transforms Basics1. Purpose of using Laplace transform analysis for linear circuits

• Equivalent to time-domain analysis (differential equations)• Use algebraic equation for simpler operation

2. Laplace transformDiscussion1. Properties

∫∞ −−

≡=0

)()]([)( dtetftfLsF st

)()( tBgtAf +

ssFfsfsFs

fssFsFe

dssdFasF

sBGsAF

ost

/)()0(')0()(

)0()()(/)()(

)()(

2 −−

−

−

−−

−

−+

+

∫ −

−−

−

t

oo

at

df

tftf

ttuttfttf

tfe

0)(

)(")('

)()()(

)(

λλ


2. Proof of properties

)(1)(1))((

)()()()]([])([,)()()7(

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

)]('[)0(')(')(')(')]("[)6(

)()0()()()()]('[)5(

)( )()()]()([)4(

)()()()()()()]([)3(

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

(s)(s))()()]()([)]()([)1(

000

0000

2

000

000

0

)(

0000

0

)(

0

0 00

sFs

dttfes

dgs

es

etg

sedtgdtetgtgLdfLdftg

sFssffssFfsf

tfsLfdtdt

detfetfdtedt

tdftfL

ssFfdtdt

detfetfdtedt

tdftfL

sFedefdtettfttuttfL

sFdsddtetf

dsddt

dsdetfdtetftdtettfttfL

asFdtetfdtetfetfeL

BGAFdtetgBdtetfAdtetBgtAftBgtAfL

ststst

stst

ststst

ststst

sttstt

t

stooo

stst

stst

tasstatat

ststst

ooo

o

==−

−−

=

−====

+−−=+−+−=

+−=−==

+−=−==

==−=−−

===−−==

+===

+=+=+=+

∫∫

∫∫∫∫

∫∫

∫∫

∫∫

∫∫∫∫

∫∫

∫ ∫∫

∞ −∞ −∞−

∞ −∞ −∞∞

−−−−

−∞ −∞−∞ −

−∞ −∞−∞ −

−∞ +−−=∞ −

∞ −∞ −∞ −∞ −

∞ +−∞ −−−

∞ ∞ −−∞ −

−−−

−−−−

−−−

−−−

++

−−−−

−−

− −−

λλλλ

λλ λλ


3. Laplace transform pairs

2222

2222

12

12

)(sincos)()cos( ,sincos)cos(

os ,sin

)(1)1(1 ,

)(! ,

)(1 ,1

! ,1

1)()( ,

βφβφφβ

βφβφφβ

ββ

βββ

++−+↔+

+−↔+

+↔

+↔

+↔−−

+↔

+↔

+↔

↔↔

−↔−−↔

−

−+

−−−

+

−

asaste

sst

sstc

st

asse

aasret

aset

ase

srt

st

seDtutu

sAA

at

atr

atratat

rr

sD


4. Proof of Laplace transform pair

220

2

1

433

322

2

000

]11[2

12

][sin)4(

)(!

)()!1(][

)(11][,1][),()()3(

!)!1(][...

!32][,21][,11][,)()()2(

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

)]([)10(][)1(

ββ

βββ

ββ

+=

+−

−=−=

+=

+−−=

+=

+−=

+=+↔

=−−=

=−==−==−=−↔

−=−=−−

==−

−=−

===

−∞ −

−

−−−

+

−−

∞−∞ −∞ −

∫

∫∫

−

−−−

sjsjsjdte

jeetL

asr

asr

dsdetL

asasdsdteL

aseLasFtfe

sr

sr

dsdtL

ssdsdtL

ssdsdtL

ssdsdtL

dssdFttf

setuLe

sDtutuL

tAuLsA

sA

seAdteAdtAeAL

sttjtj

rratr

atatat

rrr

sDsD

ststst

p

p


22

222222

2222

00

1

0

)(sincos)()]cos([

sincossincossinsincoscos)][cos(

]sin[1][cos),0()()]('[,][sin)6(

)1(11,])(

1[

1,)()(1)5(

βφβφφβ

βφβφφ

ββφ

βφβφβφβ

ββ

ββ

βββ

λ

λλ

λλ

++−+=+

+−=

+−

+=−=+

+==−=

+=

−−

=−

==+

+↔↔

−

−

−−−−

−

−−

−

�

�

asasteL

ss

ssstttL

sst

dtdLtLfssFtfL

stL

ea

ea

deass

L

asedfsF

s

at

attat a

att

�

�


5. Switched first-order circuit, find vc(t) for t>0

0),159()]1(51[756][)(

)5(75

56

)5(756

12.0/152.1

15)]0([2.015'12

6812

815)0(

5551 >+−=−−+==

++

+=

++=

++=

=+−→=+×→

=+

×=

−−−−

−

−

teeeVLtv

ssssss

ssV

VvsVLvCvKVL

v

ttt

6. Second-order circuit, find iL(t) for t>0, iin(t)=I1, t<0, I2, t>0

]/1)/([/)/(

1/

/][][

''

)0(,)0(

221

21

2211

21

1

2

11

LCsLRssLCIsILRsI

RCsLCssIRCILCsII

sIIRIsVCVRIIsIL

IiCvvRiLi

RIvIi

++++=

++++=→

=+−=+−

→

=+=+

== −−


13.2 Transform inversionBasics1. Partial-fractional expansion

0,...)(

)()(,...

1,...

...)()()(

2121

2

2

1

1

11

1

11

1

≥+++=↔

−=−

++−

+−

=

−≤++++

++++==

=

−−

−−

teAeAeAtf

sFssAss

Ass

Ass

A

nmasasas

bsbsbsbsDsNsF

tsn

tsts

ssiin

n

on

nn

om

mm

m

n

i

2. Complex poles

0),cos(

][21

21

21)(

,2/2/

,)()(2

)(

)()()()(

222222222

≥+=

+=+=↔

∠=++

+−+

=

−=++

+=−++

+=++

+=

−

+−+−+−∗−−

∗

tteK

eeeKeKKetf

KKjs

Ksjs

K

ws

CBswsCBs

wssCBssF

tm

tjtjtm

tjtj

m

ooo

φβ

φβαβα

αββαααα

α

φβφβαβαβα


3. Repeated poles

iii

iii

ssiissiissii

tsi

tsi

tsi

i

i

i

i

i

i

sFssdsdAsFss

dsdAsFssA

tetAteAeAtfss

Ass

Ass

AsF

===−=−=−=

≥++=↔−

+−

+−

=

)]()[(!2

1,)]()[(,)()(

0,21)(

)()()(

32

2

33

23

1

23213

32

21

4. Time-delay termost

oo esFsFsFttuttftutftf −+=↔−−+= )()()()()()()()( 2121

5. Initial-value theorem final-value theorem

)()(lim )0()(lim

)0()0(

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

)(')0()(lim 0)0()(lim

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

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

0

00

0

∞=→=→

=

−∞=−−=

=−=−

=−=

−−=

→

+

∞→

−−

−−+

∞ −−

→

−

∞→

∞ −−

−+

∫

∫

−

−

fssFfssFff

ffffssFssF

dtetffssFfssF

dtetffssFtfL

fftftf

ss

c

c

st

sccs

stcccc

c

p

電路學講義第13章

Discussion1. Ex.13.5

13-10

)( find,]/1)/([

/)/(2

212

1 tiLCsLRss

LCIsILRsII++++=

0,52)(10

12

52

1)2(

40242)()10(

5)10(

40242)()2(

2)10)(2(40242)(

102)10)(2(40242

)2012(40242

2,2,20/1,1,12

102

10

2

03

2

2

02

0

2

01

3212

2

221

≥+−=→+

++−+=

=+

+−+−=+=

−=+

+−+−=+=

=++

+−+−==

++

++=

+++−+−=

+++−+−=→

=−====

−−

−==

−==

==

teetisss

I

sssssIsA

sssssIsA

ssssssIA

sA

sA

sA

sssss

sssssI

IICLR

tt

ss

ss

ss


2. Ex.13.6 )( find,)256)(2(

71615)( 2

2

tfsss

sssF+++

−−=

0),4.674cos(265)(

4.672624108

40962)43(

66102

)43(2/

)43(2/

2566610,66,10

2125)230()5(71615)2)(()256(571615

5256

71615

2562)256)(2(71615)(

32

43

2

22

222

2

2

1

21

2

2

≥°++=

°∠=+=+−=++−=

+++

−+=

++−−==→

++++++=−−

+++++=−−→

=++

−−=

++++

+=

+++−−=

−−

+−=

∗

−=

tteetf

jj

jjs

sK

jsK

jsK

sssCB

CsBCsBsssCBsssss

ssssA

ssCBs

sA

ssssssF

tt

js

s


3. Ex.13.7 )( find,)5()2(

142)( 3

2

tfsssssF

+++−−=

0,3)(,0,1142)64302080(

)486956()1213()1(142)4()5(6)5)(4()5()4(

6)5(

142)()4(

1)4(

142)()5(

5)4()4(4)(

54241211

21211

12112

12113

11

2312

211

4

2

5

313

53

2

54

43

132

1211

≥++−==−=→

+−−=++++

++++++++

+−−=+++++++++→

=+

+−−=+=

=+

+−−=+=

++

++

++

+=

−−−

−=−=

−=−=

teetetfAAssAA

sAAsAAsAssssssAssA

ssssFsA

ssssFsA

sA

sA

sA

sAsF

ttt

ss

ss


4. Ex.13.8 input signal x(t)=20u(t)-40u(t-3), response y(t) satisfying y’(t)-5y(t)=-x(t)=-20u(t)+40u(t-3), find zero-state response

)3(]88[44)3()3(2)()(,44)(

4,4)5(205)5(

20

2)5(

402040205)0(

)3(5511

51

212121

1

311

33

−−−−=−−−=−=

−==→+−=−→−

+=−

−=

−=−

+−=→+−=−−

−

−−−

−

tueetutftftyetf

AAsAsAsA

sA

ssF

eFFss

eYs

eYfsY

ttt

ttt


13.3 Transform circuit analysisBasics1. Zero-state response y(0-)=0, Y(s)=H(s)X(s) →y(t)

zero response = natural response + forced response2. Zero-input response y(0-) as sources → Y(s), y(t)

CC

C

CCC

CC

IsCs

vV

vsVCICvi

1)0(

)]0([

'

+=

−=→

=

−

−

LL

L

LLL

LL

VsLs

iI

isILVLiv

1)0(

)]0([

'

+=

−=→

=

−

−


3. Complete response = zero-state response + zero-input responseDiscussion1. Ex. 13.10 find step response i(t)

5.0)(lim)(,0)(lim)0(0,5.05.0)(

5.08)144()5.0(8)4()4(5.08

18,21

)4(8,

)4(4)4(8

321621621

)()(

16232

1621

1)(

0

442121

221

212

422

0212

222112

2

==∞==≥−−=

=→+−+++=+→

−+++=+

−=+==++=

++

++=

++=

+++==

=+

+=+

+=

→∞→

+

−−

−==

ssIissIitteeti

AsAsAssssAss

ssA

ssA

sA

sA

sA

sss

sss

ssZVsI

IV

sssssZ

ss

tt

ss

s

s


2. Ex. 13.11 vs(t)=50cos8t, t≧0, find i(t)

0),()()(),9.818cos(07.7)(

9.8107.7718

56282/

82/

6456

10)(,)4(

104

11,56,1

01664025640081664508104

0)16640256()81664()8104()(40050

)4)(()64(10)64)(4(40050

106440050,

64)4(4)4)(64(40050

32162162

850

)()(

850

82

44211

11

11

11

11

11112

113

112

22211

24

2

2

12221211

22

2

222

22

≥+=°−=

°−∠=−=++=→

++

−=

++=

−−=+−+

+−=→−===→

=+−=++=++−

=++−++++++−++=+→

++++−++=+

−=++=

+++

++

+=

+++=

+++

+==

+=

=

∗

−−

−=

ttitititti

jjs

sKjs

Kjs

KssI

teetiss

IABA

BABAABAA

AABAsBAAsBAAsAAss

sBAssssAss

sssA

sBAs

sA

sA

ssss

sss

ss

sZVsI

ssV

FNF

jsF

ttNN

s

s

s


3. Given tt etxetxtyss

sH 23 10)((2)10)((1)for )( find,)2)(1(

1)( −− ==++

=

ttt

t

FNttt

t

teeetysssss

sHsXsY

Hs

Xetx

tytyeeetyssssss

sHsXsY

sXetx

22

22

2

32

3

101010)()2(

102

101

10)2)(1(

10)()()(

(-2) invalid is methodphasor 2

10(s)10)( )2(

)()(5105)(3

52

101

5)3)(2)(1(

10)()()(

310(s)10)( )1(

−−−

−

−−−

−

−−=+−+

+−+

+=

++==

∞=+

=→=

+=+−=+

++

−++

=+++

==

+=→=

p


4. Given is(0-)=6, is(0+) =0, find zero-input response iL

0),455cos(26)(

4526335)5(

6062

5)5(2/

5)5(2/

501060612120)220100(

)0(100

)0(

120)0(20)0(122)0(

,6)0(

5

55

2

'

'

≥°−=

°−∠=−=++

+=

+++

−+=

+++=→+=++→

+=→

−=→=

=×=

−=−=→

==

−

+−=

∗

−

−

−−

−

−

tteti

jjs

sK

jsK

jsK

sssI

sIs

sKVL

svI

sV

CvsCVICviiv

sILisLIVLivi

tL

js

LL

CCC

CCCCC

LC

LLLL

LLL


5. Given vs=20 for t<0, vs =-20 for t≧0, find vC

0),6.268cos(52020)(

6.26520)2(20]8)4[()808(202,20

808)808(20

8)5(2/

8)4(2/

)808()808(20

5.02)202()540

(

)0(

20)0(

22

)0(

,4)0(

4

84

2

02

2

2

2

'

'

≥°−+−=

°−∠=−=++−+=−=

++−+=

+++

−++=

++−+=→

=+−++→

−=→=

=

−=−=→

==

−

+−==

∗

−

−

−

−

ttetv

jjss

ssKssssA

jsK

jsK

sA

sssssV

ssVVsKCL

CvsCVICviv

IsLisLIV

Livi

tC

jss

C

CC

CCCCC

C

LLLL

LLL


13.4 (Optional)13.5 Impulses and convolution (optional)Basics1. Unit impulse

1)()]([

)()()()(

1)()(

0,0)( properties

)()(

0 0

0

0

===•

==−•

==•

≠=•

=

∫

∫

∫ ∫

∞

=

−−

∞

∞− =

∞

∞−

−

+

−

t

stst

otto

edtettL

tftftttf

dttdtt

tt

tudtdt

o

δδ

δ

δδ

δ

δ


2. Convolution ∫∞

∞−−≡∗ λλλ dtgftgtf )()()()(


3. )]([)]([)]()([,0for 0)()( tgLtfLtgtfLttgtf ×=∗<==

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

])([)(

\])([)(

])()[(

])()[(

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

0 0

0 0

0

0

)(

0 0

0 0

tgLtfLdegdef

ddegef

ddegef

ddegf

ddtetgf

dtedtgftgtfL

ss

ss

ss

st

st

st

×=×=

=

=

=

−=

−=∗

� �

� �

� �

� �

� �

� �

∞ ∞ −−

∞ ∞ −−

∞ ∞

−

−−

∞ ∞

−

+−−=

∞ ∞ −

−∞ ∞

− −

− −

−

−

− −

− −

µµλλ

λµµλ

λµµλ

λµµλ

λλλ

λλλ

µλ

µλ

λµλ

λλµ

λµ


4. Linear circuit )()()()()()( txthtysXsHsY ∗=⇔=

)]([)(),()()]()([)]([)( 111 sHLthtxthsXsHLsYLty −−− =∗===

5. Impulse response h(t) )()(),()()()(1)(),()( thtysHsXsHsYsXttx ===→==δ


Discussion1. v(t)=80u(t), find i(t)

)7.333cos(8.4)(2)(

7.338.431

2042,31

2/31

2/2

1022042

1024082

4008040)204(80)(

)204(4008040

151

20

120)(

31

22

2

2

2

2

2

°−+=

°−∠=++

+=++

+−+

+=

++++=

++++=

++++=

++++=

+++=

−

+−=

∗

tetti

jssK

jsK

jsK

sss

ssss

sssss

ssI

sssss

sss

sZ

t

js

δ


2. v(t)=[u(t)-u(t-D)]/D, find zero-state response i(t)

)()51()(

)5(5

51

25102510

1)(

1)()()(

5

22

tuetth

sssss

sssZsV

sIsH

t−−=

+−+

+=

++=

++===


)(555

5

0

5

0

5

1)51()(,)3(

1)51()(,0)2(

0)(,0)1(

)()()()()(

0)],())[1)(,)51()(

Dttt

Dt

tt

t

t

eD

DteDtd

DetiDt

eDtd

DetiDt

tit

dtvhtvthti

tDtutuD

tvetth

−−−

−

−

−−

−

−−=−=>

=−=<<

=<

−=∗=

≥−−=−=

�

�

�

λλ

λλ

λλλ

λ

λ

Chapter 13 Laplace transform analysis

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