1

Part III Network Analysis

Topics to be covered

a) Conventions for Describing Networks

b) Network Equations c) Impedance Functions and Network Theorems

d) Two-port Parameters e) Sinusoidal Steady-state Analysis

III.1) Conventions for Describing Networks III.1.1 Reference directions for current and voltage

i

v+ _

Remarks: Above figure shows the reference directions for voltage and current for all passive elements. One terminal of the sources is marked plus, the other minus. When the polarity of the source coincides with the reference marks, the voltage is described by a positive number. When the polarity of the source is opposite to the reference marks, the voltage is designated by a negative number. Example 3.1 The reference directions applied to resistor, the inductor and the capacitor.

a) )()( tRitV RR

b) )()( tidt

dLtV LL

c)

t

CC dttiC

tV )(1

)(

2

III.1.2 Topological Description of networks

Graph Oriented Graph Remarks: we replace all elements of the network with lines, constructing a skeleton of the network, which is called the graph. If we also indicate a reference direction by an arrow for each line of the graph, then it is known as an oriented graph.

i) The line in the graph are identified as branches ii) The junction of two or more branches is known as a node ( and also as a

vertex)

Example 3.2 special topological structures:

T-network

3

-network

ladder-network

Bridged T-network

4

Bridge network

The lattice network

Remarks: Planar graph maybe drawn on a sheet of paper without crossing lines. Node pair: two nodes which are identified for specifying a voltage variable. Loop (mesh) is a closed path in graph formed by a number of connected branches.

5

Sub-graph is formed by removing branches from the original graph. A tree: a tree of connected graph has following properties: 1) it contains all of the nodes of the graph; nodes are not left in isolated positions 2) it contains n-1 branches 3) there are no closed path. Chords or links: branches removed from the graph in forming a tree are called chords or links Number of branches in a tree= number of nodes-1 Number of the node is exactly the same as the number of nodes in the corresponding graph. For the graph with b branches, c chords, and n nodes, their relationship:

C =b-(n-1) = b-n+1 Example 3.3 Three graphs are shown in figure. Classify each of the graph as planar or non-planar. Identify the nodes, branches, and possible trees.

a

b

c

d

6

III.2) Network Equations III.2.1) Kirchhoff’s laws

a) Kirchhoff’s voltage law: the algebraic sum of all branch voltages around any closed loop of a network is zero at all instant times.

b) Kirchhoff’s current law: the algebraic sum of all branch currents leaving a node is zero at all instant times.

III.2.2) The number of network equations Consider a network composed of b branches. The unknown quantities of interest are the branch voltages and the branch currents, making total 2b unknowns for the b branches. Since the voltage current relationship is known for each branch, the unknown quantities may be reduced from 2b to b. We will then use Kirchhoff’s laws to write b equations in b unknowns. III.2.3) Source transformations Example 3.4: Combining sources of voltage and current.

7

III.2.4) Formulation of network equations

a) Loop variable analysis Example 3.5: Using Kirchhoff’s voltage law

V1

R1

C

L1

R2

tvdtiiC

iR 2111

1

8

01

222

112 iRdt

diLdtii

C

b) Node variable analysis

Example 3.6: Using Kirchhoff’s current law

ivvdt

dCv

R 2111

1

01

32222

121 vvdt

dCv

Rvv

dt

dC

01

23233

vvdt

dCv

R

c) State variable analysis

C

Vc

R

L

Vs

9

Kirchhoff’s current law: Lc i

dt

dvC

Kirchhoff voltage law: cLsL vRiv

dt

diL

We rearrange these two equations to the form

Lcc i

Cv

dt

dv 1*0

L

vi

L

Rv

Ldt

di sLc

L

1

We are said to be in state form. In general :

112121111 yxaxaxa

dt

dxnn

222221212 yxaxaxa

dt

dxnn

nnnnnnn yxaxaxa

dt

dx 2211

III.3) Impedance Functions and Network Theorems

III.3.1) Introduction The solution of the differential equations for networks has given rise to time-domain functions of the form

tsn

neK

Where nnn jws

tjwt

ntjw

nts

nnnnnn eeKeKeK

1) When 0nw tn

tjwn

tsn

nnnn eKeKeK 2) When 0n twjtwKeKeKeK nnn

tjwn

tjwn

tsn

nnnn sincos

III.3.2) Transform Impedance and transform of circuits

10

1） Resistance

)()( tRitV RR or )()( tGVti RR , R

G1

The corresponding transform equations are

)()( sRisV RR or )()( sGVsi RR

2） Inductance

dt

tdiLtV L

L

)()( or

t

LL dttvL

ti )(1

)(

The equivalent transform equation for the voltage expression is )0()()( LLL issILsV

)()0()( sLsILisV LLL designating the transform voltage, which is the sum of the applied voltage and initial-

current voltage, as )(1 sV We see that the transform impedance for the indictor becomes

LssZsI

sVL

L

)()(

)(1

3） Capacitance The time-domain relationship between voltage and current for a capacitance is given as

dt

tdvCti C

C

)( , and

t

CC dttiC

tv1

The equivalent transform equation for the voltage expression is

s

q

s

sI

CsV C

C

0)(1

Cs

q

Cs

sIsV C

C

0)(

Where C

q 0is the initial voltage of the capacitor which, due to the reference

direction for voltage is C

qV

00 , thus

s

VsV

Cs

sI

Cs

q

Cs

sIsV C

CCC

0)(0)(

11

Designating the transform voltage of )(sZC as s

VsVsV C

01 )(

The ratio of the transform voltage to the transform current is

Cs

sZsI

sVC

C

1

)(

)(1

III.3.3) Series and parallel combinations of elements

1) For series combination of elements, the impedance will be the sum of individual elements.

n

kk sZsZ

1

)(

2) For parallel combination of elements, the admittance will be the sum of

individual admittance.

n

kk sYsY

1

)(

III.4) Two-port Parameters III.4.1) Relationship of two-port variables

port1 Two portNet work port2

+

_ _

+

V1 V2

I1 I2

In above two port network, we see four variables identified—two voltages and two currents.

12

Now only two of the four variables are independent, and the specification of any two of them determines the remaining two. There are total six combinations of two port parameters.

Functions Name Express In terms of

Equations

Open circuit impedance

21,VV 21 , II

2221212

2121111

IZIZV

IZIZV

Short circuit admittance

21, II 21 ,VV

2221212

2121111

VYVYI

VYVYI

Transmission 11, IV 22 , IV

221

221

DICVI

BIAVV

Inverse transmission

22 , IV 11 , IV

112

112

''

''

IDVCI

IBVAV

Hybrid 21, IV 21 ,VI

2221212

2121111

VhIhI

VhIhV

Inverse Hybrid 21,VI 21 , IV

2221212

2121111

IgVgV

IgVgI

III.4.2) Short circuit admittance parameters

2221212

2121111

VYVYI

VYVYI

observe that if either 1V or 2V is zero, the four parameters may be defined in

terms of a voltage and current; thus

01

111

2

V

V

Iy

01

221

2

V

V

Iy

02

112

1

V

V

Iy

02

222

1

V

V

Iy

Since short circuit conditions are specified for each of the functions in above equations, the parameters are known as short-circuit admittance parameter.

13

If the network being studied is reciprocal, as we have

01

221

02

112

21

VV

V

Iy

V

Iy

Network+-V1

I1

I2

Network +-

I2I1

V2

Example 3.7 Consider the network in which AY , BY , CY are the admittance of subnetworks.

Determine the short circuit admittance.

YB

YC

YA

Port 1 Port 2

Solution:

CA

V

YYV

Iy

01

111

2

14

C

V

YV

Iy

01

221

2

C

V

YV

Iy

02

112

1

BC

V

YYV

Iy

02

222

1

III.4.3) Open circuit impedance parameters

2221212

2121111

IZIZV

IZIZV

Observe that if either 1I or 2I is zero, the four parameters may be defined in terms of a voltage and current; thus

01

111

2

I

I

Vz

01

221

2

I

I

Vz

02

112

1

I

I

Vz

02

222

1

I

I

Vz

The condition 01 I or 02 I implies an open circuit at port 1 or port 2, accounting for the designation of the parameters as open-circuit impedance parameters.

NetworkI1 V2

+

_

+

_

V1

15

Network I2V2

+

_

+

_

V1

Example 3.8 Consider the T network in which AZ , BZ , CZ are the impedance of

subnetworks. Determine the open circuit impedance

ZB ZC

ZA

Port 1 Port 2

Solution:

BA

I

ZZI

Vz

01

111

2

A

I

ZI

Vz

01

221

2

A

I

ZI

Vz

02

112

1

CA

I

ZZI

Vz

02

222

1

We can determine the relationship between Z and Y parameters.

16

From 2221212

2121111

IZIZV

IZIZV

, we can determine 21, II in terms of 21 ,VV

211

121

2

212

122

1

VZ

VZ

I

VZ

VZ

I

1121

1222

2221

1211ZZ

ZZ

yy

yy

III.4.4)Transmission parameters The transmission parameters serve to relate the voltage and current at one port to voltage and current at the other port. In equation form,

221

221

DICVI

BIAVV

Remarks： 1）A,B,C, and D are the transmission parameters. 2) The negative sign for the second term in the equation arises from two different conventions in assigning a positive direction to 2I . In power transmission problems, it is conventional for the current to be assigned a reference direction which is opposite to that in the block diagram. Thus the minus signs are for 2I , not for B and D.

01

2

2

1

I

V

V

A

01

2

2

1

V

V

I

B

01

2

2

1

I

I

V

C

01

2

2

1

V

I

I

D

17

We observe that A

1 is an open-circuit voltage gain, that B

1 is a short-circuit

transfer admittance, C

1 is an open circuit transfer impedance, D

1 is a short circuit

current gain. The transmission parameters are useful in describing two-port networks which are connected in cascade.

Aa Ba

Ca Da

Ab Bb

Cb Db

Na Nb

+ + + + + +

__ _ _ _ _

V1 V1a V2a V1b V2b V2

I1 I1a I2a I1b I2b I2

N

Consider the above network, the two cascaded networks are Na and Nb . For these two networks, corresponding transmission equations are

a

a

aa

aa

a

a

aaaaa

aaaaa

I

V

DC

BA

I

V

IDVCIIBVAV

2

2

1

1

221

221

And

b

b

bb

bb

b

b

I

V

DC

BA

I

V

2

2

1

1

For the composite network, N, we have

2

2

1

1

I

V

DC

BA

I

V

18

Our objective is to determine A,B,C,D in this equation in terms of the transmission parameters of previous equations. By observing that aVV 11 , bVV 22 , aII 11 , ba VV 12 , ba II 12 , 22 II b

2

2

2

2

1

1

2

2

1

1

1

1

I

V

DC

BA

DC

BA

I

V

DC

BA

DC

BA

I

V

DC

BA

I

V

DC

BA

I

V

I

V

bb

bb

aa

aa

b

b

bb

bb

aa

aa

b

b

aa

aa

a

a

aa

aa

a

a

Thus

bb

bb

aa

aa

DC

BA

DC

BA

DC

BA

We expressed 1V and 1I in terms of 2V and 2I , if we expressed 2V and 2I in terms of 1V and 1I , then the equations are written

112

112

''

''

IDVCI

IBVAV

And the inverse transmission parameters are A’,B’,C’, and D’. These equations apply for transmission in the opposite direction to that in following equation:

2

2

1

1

I

V

DC

BA

I

V

Just as we have shown that z function can be expressed in terms of y functions, so we may show that

2111 / zzA , 21/ zzB , 21/1 zC , 2122 / zzD Observe that 2112 / zzBCAD , for reciprocal network, 1/ 2112 zz Thus 1 BCAD for reciprocal network. Similarly we may show that the inverse transmission parameter, 1'''' CBDA for reciprocal network.

19

III.4.5) The Hybrid Parameters The hybrid parameters have wide usage in electronic circuits, especially in constructing models for transistors.

2221212

2121111

VhIhI

VhIhV

The h parameters are defined in terms of two of the variables by letting 01 I or

02 V .

01

111

2

V

I

Vh

01

221

2

V

I

Ih

02

112

1

I

V

Vh

02

222

1

I

V

Ih

Remarks:

11h is the short circuit input impedance. 21h is the short circuit current gain.

12h is the open circuit reverse voltage gain, 22h is open circuit output admittance. The inverse hybrid parameters, or the g parameters, are those in the equations:

2221212

2121111

IgVgV

IgVgI

From these, we see that

01

111

2

I

V

Ig

20

01

221

2

I

V

Vg

02

112

1

V

I

Ig

02

222

1

V

I

Vg

Remarks:

11g is the open circuit input admittance. 21g is the open circuit voltage ratio.

12g is the short circuit current ratio, 22g is short circuit input impedance at port 2. Figure for the h and g parameters:

Example 3.9 for transistors:

21

III.4.6) Relationship between parameter sets z y T 'T h g

z

2221

1211

zz

zz

y

y

y

yy

y

y

y

1121

1222

C

D

C

CC

A T

1

'

'

'

''

1

'

'

C

A

C

TCC

D

2222

21

22

12

22

1

hh

hh

h

hh

1111

21

11

12

11

1

gg

gg

g

g

g

y

z

z

z

zz

z

z

z

1121

1222

2221

1211

yy

yy

B

A

B

BB

D T

1

'

'

'

''

1

'

'

B

D

B

TBB

A

1111

21

11

12

11

1

hh

hh

h

h

h

2222

21

22

12

22

1

gg

gg

g

gg

T

21

22

21

2121

11

1

z

z

z

z

z

z

z

21

11

21

2121

22 1

y

y

y

yyy

y

DC

BA

'

'

'

''

'

'

'

T

A

T

CT

B

T

D

2121

22

21

11

21

1

hh

hh

h

hh

2121

11

21

22

21

1

gg

gg

g

g

g

'T

12

11

12

1212

22

1

z

z

z

z

z

z

z

12

22

12

1212

11 1

y

y

y

yyy

y

TT

TT

AC

BD

''

''

DC

BA

1212

22

12

11

12

1

hh

hh

h

h

h

1212

11

12

22

12

1

gg

gg

g

gg

h

2222

21

22

12

22

1

zz

zz

z

z

z

1111

21

11

12

11

1

y

y

y

yy

y

y

D

C

D

DD

B T

1

'

'

'

'

1

'

'

A

C

A

TAA

B

2221

1211

hh

hh

gg

gg

gg

gg

1121

1222

g

1111

21

11

12

11

1

z

z

z

zz

z

z

2222

21

22

12

22

1

yy

yy

y

y

y

A

B

A

AA

C T

1

'

'

'

'

1

'

'

D

B

D

TDD

C

hh

hh

hh

hh

1121

1222

2221

1211

gg

gg

22

Example 3.10 conversion from h to T

2221212

2121111VhIhIVhIhV

221

221DICVIBIAVV

Solution:

212221

221

22112121111

2121111

221

221

222

21

222

2112221212

1

11

VhIh

Vh

hhVhIhV

VhIhV

Ih

Vh

hV

h

hI

hIVhIhI

221

112

212

21

112

21

122122112122

212

21

22111

1I

h

hV

h

hI

h

hV

h

hhhhVhI

hV

h

hhV

Thus:

221

112

211 I

h

hV

h

hV

221

221

221

1I

hV

h

hI

2

2

2

2

2121

22

21

11

21

1

1

1 I

V

DC

BA

I

V

hh

hh

h

h

h

I

V

Some parameter simplification for passive, reciprocal networks Parameter Condition for passive

networks Condition for electrical symmetry

Z 2112 zz 2211 zz

Y 2112 yy 2211 yy

ABCD 1 BCAD DA A’B’C’D’ 1'''' CBDA '' DA H

2112 hh 1 h

G 2112 gg 1 g

23

III.4.7) Parallel connection of two port networks

Na

Nb

+

+ +

+ +

+

_

_

_

_ _

_

V1a V2a

V1b V2b

V2

I1

I1a I2a

I1b I2b

I2

V1

We have

a

a

aa

aa

a

a

aaaaa

aaaaa

V

V

yy

yy

I

I

VyVyIVyVyI

2

1

2212

1211

2

1

2221122

2121111

And

b

b

bb

bb

b

b

V

V

yy

yy

I

I

2

1

2212

1211

2

1

Assume the parallel connection can be made, then the connection requires that

ba VVV 111 and ba VVV 222

Further

ba III 111 and ba III 222

From kirchhoff’s current law, combining these equations gives:

2121211111111 VyyVyyIII bababa

2222211212222 VyyVyyIII bababa

24

Example 3.11 The network of the figure is of the type used for the so called notch filter. For the element values that are given, determine the y parameters

Solution :

III.5) Sinusoidal Steady-state Analysis

III.5.1) The sinusoidal steady state If a sinusoidal source is connected to a network of linear passive elements, then every voltage and current in that network will be sinusoidal in the steady state, differing from the source waveform only in amplitude and phase angle. Remarks: 1) The sinusoidal may be repeatedly differentiated or integrated and still be a sinusoid of the same frequency.

2) The sum of a number of sinusoids of one frequency, but of arbitrary amplitude and phase, is the sinusoid of the same frequency.

25

III.5.2) The sinusoidal and jwte The sinusoidal waveform is described by the equation

wtVv sin Where V is the amplitude of the sine wave, w is the frequency in radians per sec, and is the

phase angle of v with respect to the reference wtV sin . The period of v is the time interval between 0wt and 2wt radians. Let the second value of time be Tt so that

wTwT

22

wtjwte jwt sincos and wtjwte jwt sincos

Thus

j

eewt

jwtjwt

2sin

and

2cos

jwtjwt eewt

III.5.3) Solution using jwte Consider the network in which the source is sinusoidal and the response )(ti is required for steady-state operation.

For this network, wtVv cos , which may be written

2cos

jwtjwt eeVwtV

26

The generator wtVv cos is seen to be equivalent to two generators, one generating 2

jwteV ,

and the other generating 2

jwteV

.

Using the principle of superposition, we may consider the driving forces separately and then combine the resulting current to obtain the solution. For the first generator, the differential equation becomes:

2

jwteVRi

dt

diL

To solve this equation, we let jwtss Aei 1 , where A is the coefficient to be determined.

Thus we will have

jwLR

VA

VRAjwLA

2/

2

Similarly, we may let jwtss Bei 2 be the steady state solution of differential equation.

jwLR

VB

VRBjwLB

2/

2

The solution for the steady state becomes

R

wLwt

LwR

VwtwLwtR

LwR

V

jwLR

e

jwLR

eVBeAei

jwtjwtjwtjwt

ss

1

222222tancossincos

2

Remarks:

1) The steady state response to the excitation wtVv cos may be found by

determining the response to the excitation jwte , multiplying this response by jVeV , and then determining the real part of this product.

2) The steady state response to the excitation wtVv sin may be found by

determining the response to the excitation jwte , multiplying this response by jVeV , and then determining the imaginary part of this product.

III.5.4) Phasors and phasor diagram Consider a network of L independent loops excited by sinusoidal sources, each operating at the same frequency. We are required to determine the steady-state response in this network.

27

Suppose the L loops are written in transform form. jws The impedance function become jwzik , the loop voltage is jwt

ieV , the current is in phasor form jwt

k eI

We have

LLLLLL

LL

LL

VIjwzIjwzIjwz

VIjwzIjwzIjwz

VIjwzIjwzIjwz

2211

22222121

11212111

Where the exponential jwte has been cancelled in each of the L equation. When the required response phasor kI is found through routine algebraic manipulation of

complex numbers, it will be a phase of the form kiKk eII

From which the response in the time domain may be written by using the magnitude and phase information kKk wtIti cos

Following are the steps:

1) Express the network equations in transform and let jws in all network functions. This w is the frequency of the sinusoidal sources

2) All initial-condition sources are set equal to zero, because initial conditions are not relevant to the determination of a steady-state solution

3) All source voltages are described in terms of the cosine function iii wtVtv cos)( From which the phasor is ij

ii eVV . Useful equations in changing sine functions to

cosines are

2cossin

wtwt

4) The system of equations results can be solved by usual method to find kI 5) From which the response in the time domain may be written by using the magnitude and

phase information kKk wtIti cos

28

L

V

R

C

Example 3.12: The network is described by the equation,

11 VIjwZVVV CLR

Where

wCwLjRjwZ

1

Since wtv cos1 , we have 11 V And

wCwLjR

Z

VeII j

111

111

From this we see,

22

1

1

1

wCwLR

I

And

RwC

wL1

tan 11

Finally the time-domain expression is )cos()( 111 wtIti

Ex. 2-12, 2-16,3-23, 9-7, 9-9, 11-3, 11-21,12-4,12-10,