Network Analysis and Synthesis

Chapter 2Network transform representation

and analysis

2.1 The transformed circuit

• When analyzing a network in time domain we will be dealing with– Derivation and – Integration

• However, when transformed to complex frequency domain these become– Derivation -> multiplication by ‘s’– Integration -> division by ‘s’

• Hence, it is easier to do network analysis in complex frequency domain.

• The voltage current relationships of network elements in time domain and complex frequency domain are given as:

• Resistor

)()( tRitv )()( sRIsV

• Inductor– The time domain relation ships are

– In frequency domain they become

)0()(1

)(

)()(

0

idvL

ti

dt

tdiLtv

t

s

i

sL

sVsI

LissLIsV

)0()()(

)0()()(

• An inductor is represented in frequency domain as– An impedance sL in series with a voltage source

Used in mesh analysis.or

– An admittance 1/sL in parallel with a current source

Used in nodal analysis.

• Capacitor– The time domain relation ships are

– In frequency domain they becomedt

tdvCti

vdiC

tvt

)()(

)0()(1

)(0

)0()()(

)0()()(

CvssCVsI

s

v

sC

sIsV

• A capacitor is represented in frequency domain as– An impedance 1/sC in series with a voltage

sourceUsed in mesh analysis.or

– An admittance sC in parallel with a current sourceUsed in nodal analysis.

Example 1

• In the figure below, the switch is switched from postion 1 to 2 at t=0. Draw its transformed circuit and write the transformed equations using mesh analysis.

• The transformed circuit is

• The transformed equations become

Example 2

• The switch is thrown to position 2 at t=0. Find i(t). Vv

ampi

C

L

2)0(

2)0(

• The transformed circuit is

• Writing the transformed equation

• Solving for I(s)• Inverse transforming

)(2

32

25

sIs

sss

1

1

2

1)(

)1)(2(

32)(

sssI

ss

ssI

tt eeti 2)(

Example 3

• At t=0, the switch is opened. Find the node voltages v1 and v2

v1 mho 1

f1 2

1

VG

ChL

• The transformed circuit becomes

• The transformed equations become

• Solving these 2 equations

2.2 System function

• The excitation , e(t), and response, r(t), of a linear system are related by a linear differential equation.

• When transformed to complex frequency domain the relationship between excitation and response is algebraic one.

• When the system is initially inert, the excitation and response are related by the system function H(s) given by

)()()( sEsHsR

• The system function may have many different forms and may have special names. Such as:– Driving point admittance– Transfer impedance– Voltage or current ratio transfer function

• This is because the excitation and response may be taken from the same port or different ports and the excitation and response can be either voltage or current.

Impedance

• Transfer impedance is when the excitation is a current source and the response is a voltage.

• When both the excitation and response is at the same port it is called driving point impedance.

)(

)()( 0

sI

sVsH

g

sLsC

sLsCRsH

1

1

)(

Admittance

• Transfer admittance is when the excitation is a voltage source and the response is a current.

)(

)()( 0

sV

sIsH

g

RsC

sLsH

11

)(

Voltage ratio transfer function

• When the excitation is a voltage source and the response is a voltage.

)(

)()( 0

sV

sVsH

g

)()(

)()(

21

2

sZsZ

sZsH

Current ratio transfer function

• When the excitation is a current source and the response is a current.

)(

)()( 0

sI

sIsH

g

sCsLR

sCsH

RsLsC

RsLsH

1

1

)(

1

1

)(

• Note that, the system function is a function of the system elements only.

• It is obtained from the network by using the standard circuit laws. Such as:– Kirchhoffs law– Nodal analysis– Mesh analysis

Example 4

• Obtain the driving point impedance of the network. Then using the following excitations determine the response.1. 2. The square pulse on figure b3. The waveform on figure c

)()( tutSinwti og

a b c

• First lets find the driving point impedance• Note that it is the equivalent impedance of

the 3 elements

CLs

CG

sC

s

GsL

sCsH

111

)(2

1. Its transform is

Hence, the response is

)()( tutSinwti og

220)(

ows

wSI

LCCGssC

s

ws

wsHsIsV

o

ogo 1

.)()()(2

22

2. The excitation is given as

Hence, the response is

asess

sI

atututi

11

)(

)()()(

LCCGssC

s

s

esHsIsV

as

go 1.

1)()()(

2

3. The excitation is given as

)()()()( atua

attu

a

ttutig

22

11)(

as

e

asssI

as

• Consider the partial fraction expansion of R(s)where si are the poles of H(s) and sj are the poles of E(s).

• Taking the inverse Laplace transform of R(s)

• The terms are associated with the system H(s) and are called the free response terms.

tsi

ieA

• The terms are due to the excitation E(s) and are called the forced response terms.

• The frequencies si are the natural frequency of the system, while the frequencies sj are the frequencies of the excitation.

tsj

jeB

Problem

• Find the free response and the forced response for the circuit below. The system is inert before applying the source.

)()(cos2

1)( tuttvg

2.3 Poles and zeros of system

• We will discuss the relationship between the poles and zeros of a system function and its steady state sinusoidal response.

• In other words, we will investigate the effect of positions of poles and zeros upon H(s) on the jw axis.

• To find the steady-state sinusoidal response of a system function we replace ‘s’ by ‘jw’.

• Hence, the system function becomes

Where M(w) is the amplitude or magnitude response φ(w) is the phase response

)()()(

|)()(wj

jws

ewMjwH

sHjwH

• The amplitude and phase response of a system provide valuable information in the analysis and design of transmission circuits.

• Consider the low pass filter• Observe that– It passes only frequency

below wc– The phase response is

almost linear till wc

• Hence, if all the significant harmonic terms are less than wc , then the system will produce minimum phase distortion.

• In the rest of this section, we will concentrate on methods to obtain amplitude and phase response curves.

R-C network

•

• To obtain H(jw) we substitute s by jw.RCs

RC

sCR

sCsV

sVsH

1

1

1

1

)(

)()(

1

2

RCjw

RCjwH1

1

)(

• In polar form H(jw) becomes

)(tan

2

1

222

)(1

1

)(1 wjwRCj ewMe

CRw

RCjwH

wRCw

CRw

RCwM

1

2

1

222

tan)(

1

1

)(

• The amplitude is unity and the phase is zero degrees at w=0.

• The amplitude and phase decrease monotonically as we increase w.

• When w=1/RC, the amplitude is 0.707 and phase is -450.

• As w increases to infinity M(w) goes to zero and the phase approaches -900.

Half power point

Amplitude and phase from pole-zero diagram

• For the system function

• H(jw) can be written as

• Each one of the or represent a vector from zi or pj to the jw axis at w.

))()((

))(()(

210

100

pspsps

zszsAsH

))()((

))(()(

210

100

pjwpjwpjw

zjwzjwAjwH

)( izjw )( jpjw

• If we express

• Then H(jw) can be given as

jij

jjj

ii eMpjweNzjw ,

210210

210

210)( jeMMM

NNAjwH

• In general,

Example

• For find the magnitude and phase for w=2.

• Solution– First let us find the zeros and poles

– Zero at jw=0– Poles at

22

4)(

2

ss

ssF

)1)(1(

4)(

jjwjjw

jwjwF

)1( )1( jjwandjjw

• Magnitude

• Phase

5

4

10*2

2*4)2( jM

0000 8.26458.7190)2( j

Exercise

• Examine the property of F(s) around the poles and zeroes.

Bode plots

• In this section we turn our attention to semi logarithmic plots of system function, called Bode plots.

• In these plots we take the logarithm of the amplitude and plot it on linear frequency scale.

• For amplitude M(jw), if we express in terms of decibel it becomes )(log20 jwM

• For system function

• If we express the amplitude in terms of decibels we have

|)(|

|)(||)(|)(

)(

)()(

jwD

jwNjwHjwM

sD

sNsH

|)(|log20|)(|log20)(log20 jwDjwNjwM

• In factored from both N(s) and D(s) are made up of 4 kinds of terms1. Constant K2. A root at origin, s3. A simple real root, s-a4. A complex set of roots,

• To understand the nature of log-amplitude plots, we only need to discuss the amplitude response of these 4 terms.

• If the term is on the numerator it carries positive sign, if on denominator negative sign.

222 2 ss

1. Constant K

• The dB gain or loss is

• K2 is either positive |K|>1 or negative |K|<1.• The phase is either 00 for K>0, or 1800 for K<0.

2log20 KK

Single root at origin, s

• The loss or gain of a single root at origin is

• Thus the plot of magnitude in dB vs frequency is a straight line with slope of 20 or -20.

• 20 when s is in the numerator.• -20 when s is in the denominator.

• The phase is either 900 or -900.• 900 when s is in the numerator.• -900 when s is in the denominator.

wjw log20||log20

The factor s+α

• For convenience lets set α=1. Then the magnitude is

• The phase is

• A straight line approximation can be obtained by examining the asymptotic behavior of the factor jw+1.

2

12 1log20|1|log20 wjw

wjw 1tan)1arg(

• For w<<1, the low frequency asymptote is

• For w>>1, the high frequency asymptote is

Which has a slope of • These 2 asymptotic approximations meet at w=1.

dBw 01log201log20 2

12

ww log201log20 2

12

cadedecibel/de log20 w

• Note that the maximum error is for w=1 or for the non normalized one w=α.

• For the general case α different from 1, we normalize the term by dividing by α.

• The low frequency asymptote is

• The high frequency asymptote isdB

w01log201log20

2

1

2

2

log20log201log202

1

2

2

w

w

For complex conjugates

• For complex conjugates it is convenient to adopt a standard symbol.

• We describe the pole (zero) in terms of magnitude ω0 and angle θ measured from the negative real axis.

• These parameters that describe the pole (zero) are ω0, the undamped frequency of oscillation, and ζ, the damping factor.

• If the pole (zero) pair is given as

• α and β are related to ω0 and ζ with

• Substituting these terms in the conjugate equation

jp 2,1

200

00

1sin

cos

))(( 21 psps

2

002

200

200

2

11))((

jww

jjwjjwjjwjjw

• For ω0=1 (for convenience), the magnitude of conjugate pairs can be expressed as

• The phase is

2122222 41log2021log20 wwwjw

21

1

22tan)(

ww

• The asymptotic behavior is – For low frequency, w<<1

– For high frequency, w>>1

which is a straight line with slope of 40dB/decade.• These 2 asymptotes meet at w=1.

dBww 01log2041log20 2

12222

www log4041log20 2

12222

Example

• Using Bode plot asymptotes, draw the magnitude vs. frequency for the following system function

1

1010*161

50

1.0)(

34

2 sss

ssG

Actual plot