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

電路學講義第13章13-2

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

電路學講義第13章13-3

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

電路學講義第13章13-4

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

電路學講義第13章13-5

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

電路學講義第13章13-6

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

�

�

電路學講義第13章13-7

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

++++=

++++=→

=+−=+−

→

=+=+

== −−

電路學講義第13章13-8

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

電路學講義第13章13-9

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

電路學講義第13章13-11

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

電路學講義第13章13-12

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

電路學講義第13章13-13

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

電路學講義第13章13-14

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([

'

+=

−=→

=

−

−

電路學講義第13章13-15

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

電路學講義第13章13-16

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

電路學講義第13章13-17

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

電路學講義第13章13-18

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

電路學講義第13章13-19

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

電路學講義第13章13-20

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

電路學講義第13章13-21

2. Convolution ∫∞

∞−−≡∗ λλλ dtgftgtf )()()()(

電路學講義第13章13-22

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

電路學講義第13章13-23

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

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

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

電路學講義第13章13-24

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

電路學講義第13章13-25

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−−=

+−+

+=

++=

++===

電路學講義第13章13-26

)(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