Chapter 8 Two Port Network & Network Functions

ECE -2103: Network Analysis

Department of Electronics and Communication Engineering, M.I.T. Manipal.

Chapter 8 Two Port Network & Network Functions

A pair of terminals through which a current may enter or leave a network is known as a port. A

port is an access to the network and consists of a pair of terminals; the current entering one

terminal leaves through the other terminal so that the net current entering the port equals zero.

For example, most circuits have two ports. We may apply an input signal in one port and obtain

an output signal from the other port. The parameters of a two-port network completely describes

its behaviour in terms of the voltage and current at each port. Thus, knowing the parameters of a

two port network permits us to describe its operation when it is connected into a larger network.

Two-port networks are also important in modeling electronic devices and system components.

For example, in electronics, two-port networks are employed to model transistors and Op-amps.

Other examples of electrical components modeled by two-ports are transformers and

transmission lines.

Four popular types of two-port parameters are examined here: impedance, admittance, hybrid,

and transmission. We show the usefulness of each set of parameters, demonstrate how they are

related to each other.

Learning Outcomes:

At the end of this module, students will be able to:

1.Differentiate one port and two port network devices.

2.Calculate two port network parameters such as z, y, ABCD and

h parameters for given electrical network.

3.Relate different two port network parameters.

4.Simplify the complex network such as cascade, parallel networks using

fundamental two port network parameters.

5.Find the various driving point & transfer functions of two port network.



A Typical one port or two terminal network is shown in figure 1.1. For example resistor,

capacitor and inductor are one port network.

Fig.1.1

Fig. 1.2 represents a two-port network.A four terminal network is called a two-port network

when the current entering one terminal of a pair exits the other terminal in the pair. For example,

I1 enters terminal ‘a’ and exit terminal ‘b’ of the input terminal pair ‘a-b’. Example for four-

terminal or two-port circuits are op amps, transistors, and transformers.

Fig.1.2

To characterize a two-port network requires that we relate the terminal quantities

2121 ,, IandIVV .The various terms that relate these voltages and currents are called parameters.

Our goal is to derive four sets of these parameters.

1.3 Open circuit Impedance Parameter (z Parameter):

Let us assume the two port network shown in figure is a linear network then using superposition

theorem, we can write the input and output voltages as the sum of two components, one due to I1

and other due to I2:



2221212

2121111

IzIzV

IzIzV

Putting the above equations in matrix form, we get

2

1

2

1

2221

1211

2

1][

I

Iz

I

I

zz

zz

V

V

the z terms are called the z parameters, and have units of ohms. The values of the parameters can

be evaluated by setting 01 I or 02 I .

The z parameters are defined as follows:

Thus

01

111

2

II

Vz

01

221

2

II

Vz

02

112

1

II

Vz

02

222

1

II

Vz

In the preceding equations, letting 01 I or 02 I is equivalent to open-circuiting the input or

output port. Hence, the z parameters are called open-circuit impedance parameters.

Here 11z is defined as the open-circuit input impedance, 22z is called the open-circuit output

impedance, and 12z and 21z are called the open-circuit transfer impedances.

If 12z = 21z , the network is said to be reciprocal network. Also, if 11z = 22z then the network is

called a symmetrical network.

We obtain 11z and 21z by connecting a voltage 1V (or a current source 1I ) to port 1 with port 2

open-circuited as in fig.

Similarly 12z and 22z by connecting a voltage 2V (or a current source 2I ) to port 2 with port 1

open-circuited as in fig.

A two-port is reciprocal if interchanging an ideal voltage source at one port with an ideal

ammeter at the other port gives the same ammeter reading.



Example 8.1

Determine the z parameters for the circuit in the following figure and then compute the

current in a 4Ω load if a 0024 V source is connected at the input port.

To find 11z and 21z , the output terminals are open circuited. Also connect a voltage source 1V to

the input terminals. This gives a circuit diagram as shown in Fig

Applying KVL to the left-mesh, we get

111 612 VII

11 18IV

Hence

01

111

2II

Vz 18Ω

Applying KVL to the right-mesh, we get

0603 12 IV

12 6IV

Hence

01

221

2

II

Vz = 6 Ω

To find 12z and 22z , the input terminals are open circuited. Also connect a voltage source V2 to

the output terminals. This gives a circuit diagram as shown in Fig.



Applying KVL to the left-mesh, we get

21

21

6

6012

IV

IV

02

112

1

II

Vz =6 Ω

Applying KVL to the right-mesh, we get

063 222 IIV

22 9IV

02

222

1

II

Vz =9 Ω

The equations for the two-port network are, therefore

211 618 IIV (1)

212 96 IIV (2)

The terminal voltages for the network shown in Fig.8.2 are 0

1 024V (3)

22 4IV (4)

Fig.8.2

Combining equations (1) and (2) with equations (3) and (4) yields

21

0 618024 II

21 1360 II

On Solving, we get 0

2 073.0 I A



1.4 Admittance Parameter( y Parameter):

The terminal currents can be expressed in terms of the terminal voltages: The y terms are known

as the admittance parameters (or, simply, y parameters) and have units of siemens.

2221212

2121111

VyVyI

VyVyI


2

1

2

1

2221

1211

2

1][

V

Vy

V

V

yy

yy

I

I

The y terms are called the y parameters, and have units of seimens. The values of the parameters

can be evaluated by setting 01 V or 02 V .

The y parameters are defined as follows:

Thus

01

111

2

VV

Iy

01

221

2

VV

Iy

02

112

1

VV

Iy

02

222

1

VV

Iy

In the preceding equations, letting 01 V or 02 V is equivalent to short-circuiting the input or

output port. Hence, the y parameters are called short-circuit admittance parameters.

If 12y = 21y , the network is said to be reciprocal network. Also, if 11y = 22y then it is called a

symmetrical network.



A reciprocal network ( 12y = 21y ) can be modeled by the equivalent circuit in Fig.8.3

Fig.8.3

Example 8.4

Determine the admittance parameters of the T network shown in Fig.8.4

Fig.8.4

To find 11y and 21y , we have to short the output terminals and connect a current source 1I to the

input terminals. The circuit so obtained is shown in Fig.

5

22

224

111

VVI

Hence SV

Iy

V5

1

01

111

2

Using the principle of current division,

222

2 112

III

Appliying KVL to first loop => 211 105 IIV

Hence SV

Iy

V10

1

01

221

2



To find 12y and 22y , we have to short-circuit the input terminals and connect a current source

2I to the output terminals. The circuit so obtained is shown in Fig.

10

3

24

242

222

VVI

Hence SV

Iy

V10

3

02

222

1

3

42

2

21

21

II

II

Applying KVL to loop 2

1122 1024 IIIV

Hence SV

Iy

V10

1

02

112

1

It may be noted that, 12y = 21y therefore the network is reciprocal .

Thus, in matrix form we have

YVI

2

1

2

1

10

3

10

110

1

5

1

V

V

I

I



Example 8.5

For the network shown in Fig.8,5 determine the y parameters.

Fig.8.5 Fig.8.6

To find 11y and 21y short the output terminals and connect a current source 1I to the input

terminals. The resulting circuit diagram is as shown in Fig. 8.6

KCL at node 1V :

2

11

111

aVVVI

11 23 IVV a

KCL at node aV :

021

0

2

1 11

V

VVV aa

0aV

Substituting this in above equation => 113 IV

SV

Iy

V

3

01

111

2

SV

Iy

V

0

01

221

2

To find 22y and 12y short-circuit the input terminals and connect a current source I2 to the

output terminals. The resulting circuit diagram is shown in Fig. 8.6



Fig.8.6

KCL at node V2: 222

1

2

1I

VVV a

223 IVV a …(5)

KCL at node Va:

001

2

1

0 2

VVV aa

03 2 VVa

23

1VVa ………(6)

Substituting equation (6) in equation (5) yields

22

23

3 IV

V

223

8IV

Hence

SV

Iy

V3

8

02

222

2

We have 23

1VVa

Also 031 II

31 II

aa V

V2

2

1 ………(7)



Making use of equation (7) in (6) yields

21

3

1

2V

I

Hence SV

Iy

V3

2

02

112

1

1.6 Transmission Parameters:

The transmission parameters are defined by the equations:

221

221

DICVI

BIAVV


2

2

2

2

1

1][

I

VT

I

V

DC

BA

I

V

02

1

2

IV

VA

02

1

2

IV

IC

02

1

2

VI

VB

02

1

2

VI

ID

A, B, C and D parameters represent the open-circuit voltage ratio, the negative short-circuit

transfer impedance, the open-circuit transfer admittance, and the negative short-circuit current

ratio, respectively.

Example 8.7

Determine the transmission parameters in the s domain for the network shown in Fig.8.7

Fig.8.7 Fig.8.8

The s domain equivalent circuit with the assumption that all the initial conditions are zero is

shown in Fig. 8.8



To find the parameters A and C, open-circuit the output port and connect a voltage source 1V at

the input port as shown in Fig. below (Left side)

1

1

111

112

111

s

VI

sV

s

sV

s

VI

Therefore 1

02

1

2

sV

VA

I

And sV

IC

I

02

1

2

To find the parameters B and D, short-circuit the output port and connect a voltage source 1V to

the input port as shown in Fig. above (Right)

Applying the current division formula

1

112

sII

Hence )1(

02

1

2

sI

ID

V

The total impedance as seen by the source 1V is

)2(

)1()1(

1

2

11

11

1

121

1

1

s

sVsII

I

V

s

s

s

sZ

Hence )2(

02

1

2

sI

VB

V



Example 8.9

Find the transmission parameters for the network shown in Fig.8.9

Fig.8.9

To find the parameters A and C, open the output port and connect a voltage source V1 to the

input port as shown in Fig.8.10

Fig.8.10

Applying KVL to the input loop, we get

2

3

1

3

1 10105.1 VIV

Also KCL at node a gives

01040

403

21

VI

2

6

3

21 1025.6

10160V

VI

Substitute the value of 1I in the preceding loop equation we get

2

3

2

63

1 10)1025.6(105.1 VVV

2

3

1 10375.8 VV

Hence 3

02

1 10375.8

2

IV

VA



6

02

1 1025.6

2

IV

IC

To find the parameters B and D, refer the circuit shown in Fig.8.11

Fig.8.11

Applying KCL at node b, we find

21 040 II

12 40II

40

1

02

1

2

VI

ID

Applying KVL to the input loop, we get

40105.1

105.1

23

1

1

3

1

IV

IV

5.3740

105.1 3

02

1

2VI

VB

1.7 Hybrid Parameters ( h parameters):

2221212

2121111

VhIhI

VhIhV


2

1

2

1

2221

1211

2

1][

V

Ih

V

I

hh

hh

I

V

01

111

2

VI

Vh

01

221

2

VI

Ih

02

112

1

IV

Vh

02

222

1

IV

Ih



The parameters 11h , 12h , 21h and 22h represent the short circuit input impedance, the open circuit

reverse voltage gain, the short-circuit forward current gain, and the open-circuit output

admittance respectively. Because of this mix of parameters, they are called hybrid parameters.

Example 8.12

For the network shown in Fig.8.12 determine the h parameters.

Fig.8.12

To find 11h and 21h short-circuit the output terminals so that 02 V .Also connect a current

Source 1I to the input port as in Fig.

Applying KCL at node x

00

11

IR

V

R

VI

C

x

B

x

CB

xRR

VI11

11

CB

CBx

RR

RRIV

1)1(

Hence,

01

111

2

VI

Vh



A

CB

CB

A

CB

CB

V

Ax

RRR

RR

RIRR

RRI

I

RIV

)(

)1(

)(

)1(

1

1

01

1

2

KCL at node y

00

0

21

321

C

x

R

VII

III

Substitute for xV in the above equation and simplifying results in

)(

)(

01

221

2BC

BC

VRR

RR

I

Ih

To find 12h and 22h open-circuit the input port so that 1I . Also, connect a voltage source 2V

between the output terminals as shown in Fig.

KCL at node y:

01211

I

R

VV

R

V

CB

Since 01 I we get

CCB

CCB

R

V

RRV

R

V

R

V

R

V

21

211

11

0

CB

B

IRR

R

V

Vh

02

112

1

Applying KVL to the output mesh, we get

0)( 2212 IRIIRV BC

Since 01 I we get



222 VIRIR BC

Hence, BCI

RRV

Ih

1

02

222

1

Example 8.13

Determine the h parameters of the circuit shown in Fig.8.13

Fig.8.13

Performing Δ to Y transformation, the network shown in Fig.8.13 takes the form as shown in

Fig.8.14

Since all the resistors are of same value RRY

3

1

Fig.8.14

To find h11 and h21, short-circuit the output port and connect a current source I1 to the input

port as in Fig. 8.15

Fig.8.15



)]44(4[11 IV

11 6IV

Hence

6

01

111

2VI

Vh

Using the principle of current division

444

12

II

2

12

II

2

1

01

221

2

VI

Ih

To find h12 and h22, open-circuit the input port and connect a voltage sourceV2 to the output port

as shown in Fig. 8.16

Fig.8.16

Using the principle of voltage division, we get

444

21

VV

2

1

02

112

1

IV

Vh

222 8)44( IIV

Hence, SV

Ih

I8

1

02

222

1



1.8 Relationship between the two port parameters:

If all the two-port parameters for a network exist, it is possible to relate one set of parameters to

another, since these parameters interrelate the variables 1V , 1I , 2V and 2I .

1.9 Relation between the z parameters and y parameters:

Or

2

11

2

1][

V

Vz

I

I

And y Parameters are related by

2

1

2

1

2221

1211

2

1][

V

Vy

V

V

yy

yy

I

I

Equating above two equations we see that

1 zy

To find 1z

z

zadjoz

)int(1 where 21122211

2221

1211zzzz

zz

zzz

And

1121

1222)int(

zz

zzzadjo

Hence,

z

z

z

zz

z

z

z

yy

yy

1121

1222

2221

1211

1.10 Relation between z and h parameters:

2221212

2121111

IzIzV

IzIzV

From second equation

2

22

1

22

212

1V

zI

z

zI

Substitute this in 1st equation

2

22

121

22

211222111 V

z

zI

z

zzzzV

Writing above equation in matrix form

2

1

2

1

2221

1211

2

1][

I

Iz

I

I

zz

zz

V

V



2

1

2222

21

22

12

22

2

1

1 V

I

zz

z

z

z

z

z

I

V

Compare this with h parameter equation

2

1

2221

1211

2

1

V

I

hh

hh

I

V

2222

21

22

12

22

2221

1211

1

zz

z

z

z

z

z

hh

hh

Similarly all parameters can be related as shown in Table 1.11

Table 1.11



Example 8.17

Determine the y parameters for a two-port network if the z parameters are:

95

510z

6555910 z

65

10

65

565

5

65

9

1121

1222

2221

1211

z

z

z

zz

z

z

z

yy

yy

Example 8.18

Following are the hybrid parameters for a network:

63

25h Determine the y parameters for the network.

Solution

242365 h

5

24

5

35

6

5

11

1111

21

11

22

11

2221

1211

h

h

h

h

h

h

h

yy

yy

Example 8.19

Determine the y and z parameters for a two-port network.

Applying KCL to the node 1, 121 5.05.1 IVV

Applying KCL to the node 2, 1221 35.0 IIVV

Writing above equations in the matrix form

2

1

2

1

13

01

15.0

5.05.1

I

I

V

V

2

1

1

2

1

13

01

15.0

5.05.1

I

I

V

V



=

2

1

2.12.3

4.04.0

I

I

Therefore

2.12.3

4.04.0][z

Hence

5.04

5.05.1][ 1zy

1.12 Interconnection of Two port network:

A large, complex network may be divided into subnetworks for the purposes of analysis and

design. The subnetworks are modeled as two-port networks, interconnected to form the original

network. The two-port networks may therefore be regarded as building blocks that can be

interconnected to form a complex network. The interconnection can be in series, in parallel, or in

cascade. Although the interconnected network can be described by any of the six parameter sets,

a certain set of parameters may have a definite advantage. For example, when the networks are in

series, their individual z parameters add up to give the z parameters of the larger network. When

they are in parallel, their individual y parameters add up to give the y parameters of the larger

network. When they are cascaded, their individual transmission parameters can be multiplied

together to get the transmission parameters of the larger network.

The basic interconnections to be considered are: parallel, series and cascade.

PARALLEL: Voltages are the same.Current of interconnection is the sum of currents.

SERIES: Currents are the same. Voltage of interconnection is the sum of voltages.

CASCADE: Output of first subsystem acts as input for the second.



1.13 Networks in parallel:

Two two-port networks are in parallel when their port voltages are equal and the port currents of

the larger network are the sums of the individual port currents. In addition, each circuit must

have a common reference and when the networks are connected together, they must all have their

common references tied together.



Fig.8.20

YVI

V

V

yy

yy

I

I

2

1

2221

1211

2

1

aaa

aa

aa

a

a

a

a

a

a

a

VYI

yy

yyY

V

VV

I

II

2221

1211

2

1

2

1,,

bbb VYI

mannersimilar a In

baba

baba

VVVVVV

IIIIII

222111

222111

,

,

ba

ba

VVV

III

VYYVYVYI babbaa )(

ba YYY

Hence if two port sub networks are connected in parallel then the resultant network y parameters

can be calculated by adding individual y parameters of sub networks.

1.14 Neworks in series

The networks are regarded as being in series because their input currents are the same and their

voltages add. In addition, each network has a common reference, and when the circuits are

placed in series, the common reference points of each circuit are connected together.



Fig.8.21

Description of subnetworks

bbbaaa IZVIZV ,

Where

bb

bb

b

aa

aa

azz

zzZ

zz

zzZ

2221

1211

2221

1211,

Since by definition III ba and ba VVV 111

IZZV ba )(1

ba ZZZ

1.15 Networks in cascade:

Output of first subsystem acts as input for the second

Fig.8.22

Interconnection constraints

baba IIVV 212

aa VVVV 1111 and

ba IIII 2211

a

a

aa

aa

a

a

I

V

DC

BA

I

V

I

V

2

2

1

1

1

1

b

b

bb

bb

b

b

a

a

I

V

DC

BA

I

V

I

V

2

2

1

1

2

2

Hence

2

2

1

1

I

V

DC

BA

DC

BA

I

V

bb

bb

aa

aa



It is this property that makes the transmission parameters so useful. Keep in mind that the

multiplication of the matrices must be in the order in which the networks Na and Nb are

cascaded.

Example 8.23:Comptute the y parameter of the network shown in Fig.Q.8.23

Fig.Q.8.23

Let us refer to the upper network as Na and the lower one as Nb. The two networks are

connected in parallel. Comparing Na and Nb with the circuit shown below.

we obtain

324 122212112112 aaaaaa yyyyyjy

Or

43424 22112112 jyjyyjy aaaa

i.e. Sjj

jjYa

434

442

And

624 122212112112 jyyjyyyy bbbbbb

or

64244 22112112 jyjyyy bbbb

i.e. Sj

jYb

644

424

Therefore, overall y parameters are



Sjj

jjYYY ba

2744

4426

1.16 Network Functions for One Port and Two Port Network

1.17 Driving Point Functions:

The impedance or admittance found at a given port is called a driving point impedance(or

admittance).

1.Driving Point Impedance:

Z11(s) = V1(s)/I1(s)

2. Driving Point Admittance:

Y11(s) = I1(s)/V1(s) = 1/Z11(s)

1.18 Transfer Function:

The transfer function relates the transform of a quantity at one port to the transform of another

quantity at another port.Thus transfer functions which relate voltages and currents have

following possible forms:

The ratio of one voltage to another voltage, or the voltage transfer ratio.

The ratio of one current to another current ,or the current transfer ratio.

The ratio of one current to another voltage or one voltage to another voltage.

Transfer function for the two port network:

Denominator Numerator

)(2 sV )(2 sI

)(1 sV )(12 sG )(12 sY

)(1 sI )(12 sZ )(12 s



Example 8.24

Find driving point impedance Z(s) for the network shown in Fig.Q.8.24

Solution

The driving point impedance for RLC series circuit is

Fig.Q.8.24

sC

sRCLCs

sCLsRsZ

11)(

2

Example 8.25

Find driving point impedance Z(s) for the network shown in Fig.Q.8.25

Solution

The driving point impedance for series RL network shunted by a capacitor is

Fig.8.25

LCs

LRs

LRs

CsLR

sCsZ

1)(

1

)(1

1)(

2

and driving point admittance function )(

1)(

sZsY



Example 8.26

For the network shown find the voltage ratio )(

)()(

1

212

sV

sVsG and

)(

)()(

1

11sV

sIsY by

considering no current in the output terminals.

Fig.8.26

Solution

The network acts as a voltage divider .

)()(1

)()(1

)(

2

1

sVsIsC

and

sVsIsC

sRI

Therefore the ratio of these equations is

RCs

RC

sIsCR

sIsC

sV

sVsG

/1

/1

)()/1(

)()/1(

)(

)()(

1

212

RCs

s

RsCRsV

sIsY

/1

1

/1

1

)(

)()(

1

11



Example 8.27

For the network shown in Fig.8.27 find G12(s)

Fig.8.27

Solution

LCs

LC

LCssCsL

sC

VV

V

sV

sVsG

CL

C

/1

/1

1

1

/1

/1

)(

)()(

22

1

212

Where LV and CV are the voltages across Inductor and capacitor respectively.

Example 8.28

For the network shown in Fig. 8.28 find G12(s)

Solution Fig.8.28

1R and sC/1 can be combined into an equivalent impedance having the value

1/1

1)(

1

1

1

CsR

R

RsCsZ eq

)()(

)()(

2

2

1

212

sZR

R

sV

sVsG

eq

CRRRRs

CRs

RRCRsR

RCRsRsG

2121

1

2121

22112

/)(

/1)(

Exercise: Q.101

Obtain the ABCD parameter representation of circuit shown in Fig.Q.101



Fig.Q.101

Ans:

32425.0

220025.29

ST

Q.102 for the bridge circuit shown in Fig.Q.102. Find the transmission parameters.

Fig.Q.102

Q.103 The network of Fig.Q.103 contains both a dependent current source and a

dependent voltage source. Determine y and z parameters.

Fig.Q.103



Q.104 Find the h parameters for the network shown in Fig.Q.104. Keep the result in s

domain.

Fig.Q.104

Ans : )2(2)2(2

)4(

)2(2

4

)2(2

4522211211

s

sh

s

sh

s

sh

s

sh

105. For the network shown in Fig.Q.105, determine the h parameters

Fig.Q.105

Ans:

CBCB

B

CB

BCA

CB

CB

RRh

RR

Rh

RR

RRhR

RR

RRh

1)()1(22122111



Q.106 For what value of ‘a’ is the circuit shown in Fig.Q.106 is reciprocal? Also find h

parameters.

Fig.Q.106

Ans: a=2 and

5.02

22h

Q107. What is the value of n for the network shown in Fig.Q.107 to be reciprocal? Also find

y12 and y21 for that n

Fig.Q.107

Ans: n= -0.01 and y12 = y21=-0.002



Summary

1. Various electronic devices such as transistors, transformers etc can be modelled

using various two port network parameters such as impedance, admittance,

transmission and hybrid parameters.

2. Relation between various two port network parameters studied and verified for

different two port parameters.

3. Complex two port network in cascade, parallel or any other form can be modeled

using simple two port network parameters.

4. Various driving point & transfer functions are defined and determined for the two

port networks.

Chapter 8 Two Port Network & Network Functions

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Transcript of Chapter 8 Two Port Network & Network Functions