Chapter 10 Network functions and s-domain analysis

電路學講義第10章10-1

Chapter 10 Network functions and s-domain analysis-option-

10.1 Complex frequency and generalized impedancecomplex frequency, generalized impedance and admittance

10.2 Network functionnetwork functions and impedance analysis

10.4 s-domain analysisPoles and zeros, forced response, natural response, stability

10.5 Network scalingimpedance scaling, frequency scaling


10.1 Complex frequency and generalized impedance Basics1. Complex-frequency waveform representation of a real signal

x

x

jm

sttjwjmx

tm

eXXjws

eXeeXwteXtxφ

σφσ

σφ

=+≡

≡=+= +

frequency,complex :

]Re[]Re[)cos()( )(

3. admittance generlized:)(impedance, generlized:)(then

]Re[)cos()(],Re[)cos()(

VIsY

IVsZ

eVwteVtveIwteIti stv

tm

sti

tm

≡≡

=+==+= φφ σσ

2. For a two-terminal element, if

1/sLsC1/RY(s)sL1/sCRZ(s)

inductorcapacitorresistor


4. s-domain diagram

5. Phasor representation is a special form as s=jw, σ=0.Frequency-diagram is a special case of s-domain diagram with s=jw,σ=0.

6. Application: more general form to analyze the circuit forced response. Chapter 13 “Laplace transform” will be intoduced to analyze the complete response (zero-input response and forced response).


Discussion1. Ex. 10.2 find i(t), L=1H, R=5Ω, C=0.1F )904cos(20)( , 2 ot tetv += −

)1.534cos(8)(,1.5381.1435.2

9020

1.1435.2)42(,42,9020V ,(s)

1

1RCs1

1(s)

2

2

°−=°−∠=°∠

∠=

°∠=+−+−=∠==

+++=

++=

++=

− tetiI

jZjsZVI

RCsRLsRLCsRsL

sCR

sLZ

to

o


2. Ex.10.3 Miller-effect capacitance

=+=−

−=

=+=→

+=+=→

−=

−=−=

212

11

1

1

)11(

)1(

)1()(

)(1)(

sCA

sCVAI

VI

sCAsCVI

VAsCVAVsCIVAV

VVsC

sC

VVI

inout

in

ininin

inout

outinoutin


3. Ex. 10.4 Generalized impedance converter, find Zin(s)

resistor negativedependent -frequency

)(1)(,1,1inductor capacitor elements, resistive ,,,

,

)()3(,)4(

)5...(

)4...(

)3...(

)2...(

op 2nd for the

)1...( op1st for the

222242153

25314

42

531

531

42

5

31

42

31

42)5(),2(

34

31

2

3

1

21)1(

33

43

5

21

1

=−==→=====

→=⇒

=−−

=→−=−=→

=−+

=−=

−=

−=

=

+=

wRRCw

jwZRCs

ZRZZZsC

ZZ

ZZZZZZ

ZZZZZZ

VZZZ

ZZZ

VZZZZVV

IVZZZZVIZVV

VZZ

ZZ

VVZ

ZZ

ZVVI

ZVVI

ZVVI

VV

VZZ

ZV

eqinin

in

in

in

ininin

inininbc

in

ininba

cb

cinin

bin

ain


10.2 Network functionsBasics1. t-domain approach: for a second-order circuit, its input signal x(t) and

response y(t) are related by

functiontransfer

function,point -drivingfunction,network :)(

)()('"'"]Re[)(],Re[)(

12

2

12

2

12

212

21212

sHasasabsbsb

XY

XbsbsbYasasaxbxbxbyayayaeYtyeXtx

o

o

oooo

stst

≡++++=⇒

++=++→++=++==

2. s-domain approach: much easy!!• use s-domain impedance representation• use methods of series-parallel reduction, source conversion,

proportionality, node and mesh equation, voltage and current dividers…to obtain response

•XYsH =)(


Discussion1. Ex.10.5 use t-domain approach to find

LCs

LRs

sVVsHVsV

LCs

LRs

vvLC

vLRv

LCs

LRs

sL

VIsHVs

LI

LCs

LRs

vL

iLC

iLRi

vdtiC

RiLiKVL

s

LsL

sLLL

dtd

Liv

s

LsL

sLLL

dtd

sLLL

LL

1)(,]1[

1)1(

1

1

)(,1]1[

11)1(

)1...(1

2

2

222

"'"

21

2

''"

'

2

2

'

++===++→

=++→

++===++→

=++→

=++→

=

∫

s

L

s

L

VVsH

VIsH == )(,)( 21


2. Repeat Ex.10.5 using s-domain impedance analysis

LCs

LRs

ssRCLCs

LCsVVsHV

sCRsL

sLV

LCs

LRs

sL

sRCLCssC

VIsH

sCRsL

VI

s

LsL

s

LsL

11)(1

1

1

1)(1

2

2

2

2

2

221

++=

++==→

++=

++=

++==→

++=


3. Ex.10.6 obtain the network functions and differential equations torelate iL, vC to vS

ssLLL

S

L

SSL

vRCvRiLiRLCiRLsRLCs

RCsVIsH

sRCRsL

V

sCR

sL

VI

+=++→

+++==→

++

=

++

=

''"

211)(

111

sCCC

S

C

SSC

RvRvLvRLCvRLsRLCs

RVVsH

sRCRsL

sRCR

V

sCR

sL

sCRVV

=++→

++==→

++

+=

++

+=

'"

22 )(

1

1

11

11


4. Ex.10.7 obtain the network functions to relate v1, v2, v3 to vin

[ ]

[ ]

[ ] [ ][ ]

[ ]

ininin

in

in

in

in

in

in

S

SSSout

s

Fout

VV

VV

VV

VGVsC

VVV

GsCGsCGGsC

GsCGsC

VVVG

VGVsC

VGVGVsC

I

IGIIVV

GsCGsCGGsCsCGsC

Y

IVYY-

vvR

RRv

RRR

v

321

3

2

1

3

2

13

~

~~

333

,,

0)(20

)(0)(2

000000

200

00

2

,equation constraint

220022

analysismatrix node

)1(

⇒

=

+−−−+

+−+

+

=

+=

+

==

+−−−+−+

=

=

=+−

=+

=

µ

µµ

µ

µµ

µ

µµ

µ

µ


10.4 s-domain analysisBasics1.Properties of network function H(s)

• It’s much easier to obtain in the s-domain than in t-domain• It simplifies the calculation of forced response• One can determine the natural response

2. It is equivalent to• impulse response or system response in the courses of linear

system, communication theory, signal processing• Green’s function in electromagnetics

3.

))(( eg. pairs cojugatecomplex in or R,, R,K

factorgain :zero,:,pole:,,elements storage-energy ofnumber the toequals ,

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

......)(

21

21

11

1

11

1

∗

−−

−−

−−∈∈

∈≤

−−−−−−=

++++++==

iiii

iiii

n

m

on

nn

n

om

mm

m

pspszpKzpRba

nnmpspspszszszsK

asasbsabsbsbsb

XYsH


4. A network function is completely defined by gain factor, zeros and poles.

5. Pole and zero are displayed in s-plane s=σ+jw as pole-zero pattern

)]74()][74([)10()(

jsjsssKsH

−−−+−−+=

poles are in the LHP(left half plane)


6. Forced response calculation• given H(s)• if • then

]Re[)( tsoeXtx =

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

......

)(

)()()(

])...)(()...)((Re[])(Re[]Re[)(

21

21

21

21

21

21

pspszszsKsH

pspszszs

KsH

sHsHsH

eXpspszszsKeXsHeYty

oo

ooo

oo

ooo

ooo

ts

oo

ootso

ts ooo

−∠+−∠−−∠+−∠+∠=∠

−−−−

=

∠=−−−−===


7. Natural response calculation• given

• then )()(

)()(

))(())(()(

21212

21212

21

21

sPsZK

ppsppszzszzsK

pspszszsK

XYsH =

++−++−=

−−−−==

mode. TEM e.g. netics,electromagin used widely is mode termTheresponse natural theof mode"" one:)cos(2

...)cos(2...)(

,then ,0, e.g.response natural scircuit' of valuessticcharacterior es,eigen valu are poles

)()( polynomial sticcharacteri

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

1

21212

21'

21"

21212121

iit

i

iit

itjwt

itjwt

iN

iiiiiii

NNN

AtweA

AtweAeAeAtyjwpwjwp

ppsppssPyppyppy

xzzxzzxKyppyppy

i

iiiii

∠+

+∠+=++=

−=≠+=→→

++−=→

=++−→

++−=++−

−∗+

+

σ

σσσ

σσ


8. Modes associated with distinct poles

• For repeated poles, the mode waveform is multiplied by t, or power of t.

ptrrN

r

etAtAAtyppp

)...()(then ,... e.g.

121

21+++=

===oscillationα=0stable

LHPunstableRHP

overdamped

−−−=−+−=

222

221

o

o

wpwp

αααα

underdamped

−−−=−+−=

222

221

αααα

o

o

wjpwjp

• 2nd-order circuit

0'2" 2 =++ ywyy oα

critically dampedα−== 21 pp


9. A network is stable when all of its poles are in the left half of the s-plane, irrespective of pole-zero cancellation.

0 as ,0)(0LHP in the →→=⇒<→+= + teAtyjwp tjwtiNiiii

iiσσσ

10. Any network that contains resistance and no controlled sources is stable.


Discussion1. Ex. 10.8 find poles and zeros

)]20()][20()][6()][6()][32([)]108()][108([5

)40040)(36)(32(164165)(

2

22

234

−−−−−−−−−−−−+−−−=

++++++−=

ssjsjssjsjss

ssssssssH


2. Ex. 10.9 given H(s) and x(t), find forced response

°−∠==→

°−∠=−−−+−−

−=

===−−−+−−

−=++

−=

+−−

5.1087.23)(

5.10837.2)]36()][36([

6)(

]10Re[]Re[3cos10)()]36()][36([

64512

6)(

)34(4

2

XsHYjsjs

ssH

eeXtetxjsjs

sssssH

o

oo

oo

tjtst o


3. Ex. 10.10 analyze H2(s) in Ex.10.7

response unstable2responsen oscillatio2

response dunderdampe2125)510()510(,,5

]25)510(2)[5()(

5 if,1])2(2)[(

])2([)(

2321

2

22

222

2

→>→=

→<<−−±−−=−=

+−++=

==

+−+++−+==

µµ

µµµ

µ

µµ

ppp

ssssP

aRC

a

aassasaassa

VVsH

in


10.5 Network scalingBasics1. Network scaling

fm

fmmm

f

m

f

mmm

mm

mnf

qmm

ifiifin

m

nff

mff

kkCC

CsCskk

sCkZkZ

kLkLLs

kLsksLkZkZ

RkRZkZ

VIsHIVsH

qKkkKZkZ

pkpzkzpspszszsK

pksp

ks

zksz

ks

KsH

=→≡===

=→≡===

=→=

=−=

==→=

==−−−−=

−−

−−=

−

ˆˆˆ

1ˆ

ˆ

ˆˆˆˆˆ

ˆˆotherwise,0

/)(,1/)(,1

,ˆˆ

ˆ,ˆ,)ˆˆ)...(ˆˆ()ˆˆ)...(ˆˆ(ˆ

)ˆ

)...(ˆ

(

)ˆ

)...(ˆ

()(ˆ

1

1

1

1

H(s)R, L, C, w wCLR

sH

ˆ,ˆ,ˆ,ˆ)(ˆ

0,ˆ scalingfrequency >= ff ksks

0,ˆ scaling impedance >= mm kZkZ

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

1

1

n

m

pspszszsKsH

−−−−=


Discussion1. Ex. 10.11 find H(s) using network scaling

))((

1025105.0104.0

1)(

11

21

1

93

s-ps-ps-zK

ssV

IsH

sCR

sL

VI

s

L

sL

=

×+×+

==

++

=

−−

25,10)31(,,102

ˆ,ˆ,ˆ

1ˆ,62ˆ,ˆ,4ˆ

)ˆˆ)(ˆˆ(ˆˆ

40ˆ4ˆ4ˆ

101

40ˆ

1ˆ

1)(ˆ

025.0ˆ,1ˆ,10ˆ

102,105let

321

4

211

21

12

43

=×±−=×−=→

===

=±−=−=

−−−=

+++=

++

=

=====×=

×=×= −−

Kjppz

Kkk

Kpkpzkz

Kjppz

pspszs

sss

sssH

kkCCL

kkLRkR

kk

i

m

fifiifi

fmf

mm

fm

Chapter 10 Network functions and s-domain analysis

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Transcript of Chapter 10 Network functions and s-domain analysis