CHAPTER 6

TWO PORT NETWORKS

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 OBJECTIVES
To understand about two port networks
and its functions.
To understand the different between z-
parameter, y-parameter, ABCD- parameter
and terminated two port networks.
To investigate and analysis the behavior of
two port networks.

2
 SUB - TOPICS
6-1 Z PARAMETER
6-2 Y PARAMETER
6-3 ABCD PARAMETER
6-4 TERMINATED TWO PORT
NETWORKS

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 TWO PORT NETWORKS
A pair of terminals through which a current may enter
or leave a network is known as a port.
Two terminal devices or elements (such as resistors,
capacitors, and inductors) results in one port
network.
Most of the circuits we have dealt with so far are two
terminal or one port circuits. (Fig. a)
A two port network is an electrical network with two
separate ports for input and output.
It has two terminal pairs acting as access points. The
current entering one terminal of a pair leaves the
other terminal in the pair. (Fig. b)

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 One port or two

terminal circuit

Two port or four

terminal circuit
It is an electrical
network with two
separate ports for
input and output.

No independent
sources.

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 6-1 Z PARAMETER
Z parameter also called as impedance parameter and
the units is ohm ()
The black box is replace with Z-parameter is as
shown below. I
I1 2

Z11 Z12
+ +
V1 V2
- -
Z21 Z22

V1 z11I1 z12I 2 V1 z11 z12 I1 I1

V2 z 21I1 z 22I 2 V z z I z I
2 21 22 2 2
where the z terms are called the impedance parameters, or
simply z parameters, and have units of ohms.
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 V1 V2
z11 and z 21
I1 I 2 0
I1 I 2 0

z11 = Open-circuit input impedance

z21 = Open-circuit transfer
impedance from port 1 to port 2

V1 V2
z12 and z 22
I2 I1 0
I2 I1 0

z12 = Open-circuit transfer

impedance from port 2 to port 1
z22 = Open-circuit output
impedance

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 Example 1

Find the Z parameter of the circuit below.

I1 I2
+ +

V1 240 V2
120
_ _

40

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 Solution
When I2 = 0(open circuit port 2). Redraw the circuit.
I1 Ia
V1 120 I b .......(1)
+ +
240 280
Ib I1......(2)
V Ib 120 V2
400
1 sub (1) (2)
_ _ V
Z11 1 84
40 I1
V2 240 I a .......(3)
120
Ia I1.......(4)
400
sub (4) (3)
V2
Z 21 72
I1 9
When I1 = 0 (open circuit port 1). Redraw the circuit.
Iy I2
V2 240 I x .......(1)
+ +
160
Ix I 2 .......(2)
V1 240 V2 400
120 Ix
sub (1) (2)
_ _
V2
Z 22 96
40 I2
V1 120 I y .......(3)
240
Iy I 2 .......( 4)
400
sub (4) (3) 84 72
In matrix form: Z
V1 72
Z12 72 96
I2
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 Example 2

Find the Z parameter of the circuit below

I1 2 j4 10 I2

+ +
+
10I2
V1 V2
_
_ -j20 _

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 Solution
i) I2 = 0 (open circuit port 2). Redraw the circuit.

I1 2 j4 I2 = 0 V1 I1 (2 j4)
V1
+ + Z11 (2 j4)
I1
V2
V1 V2 0 (short circuit)
_ _
Z 21 0

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ii) I1 = 0 (open circuit port 1). Redraw the circuit.

I1 = 0 10 I2
V1 10I 2
+ V1
+ Z12 10
+ I2
V1 10I2
V2
_ _ V2 V2 - 10I 2
-j20 _ I2
j20 10
j 1
In matrix form; 2I 2 V2
(2 j4) 0 20 10
Z V2
10 (16 - j8) Z 22 (16 - j8)
I2

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 6-2 Y - PARAMETER
Y parameter also called admittance parameter and
the units is siemens (S).
The black box that we want to replace with the Y-
parameter is shown below.
I1 I2

Y11 Y12
+ +
V1 V2
- -
Y21 Y22

I1 y11V1 y12V2 I1 y11 y12 V1 V1

I 2 y 21V1 y 22V2 I y y V y V
2 21 22 2 2
where the y terms are called the admittance parameters, or
simply y parameters, and they have units of Siemens.
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 I1 I2
y11 and y 21
V1 V2 0
V1 V2 0

y11 = Short-circuit input

admittance
y21 = Short-circuit transfer
admittance from port 1 to port 2

I1 I2
y12 and y 22
V2 V1 0
V2 V1 0

y12 = Short-circuit transfer

admittance from port 2 to port 1
y22 = Short-circuit output
admittance
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 Example 3

Find the Y parameter of the circuit shown

below.
5
I1 I2
+ +

V1 20 15 V2

_ _

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 Solution
i) V2 = 0
V1 20 I a .......(1)
5 I2 5
I1 Ia I1.......(2)
+
25
sub (1) (2)
V1 20
Ia I1 1
Y11 S
_ V1 4
V1 5 I 2
I2 1
Y21 S
V1 5

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ii) V1 = 0

I1 5
I2 V2 15I x .......(3)
+ 5
Ix I 2 .......(4)
15 Ix V2
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sub (3) (4)
_ I 4
Y22 2 S
V2 15
In matrix form;
V2 5I1
1 1
4 I1 1
Y 1 5S Y12 S
4 V2 5

5 15

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 Example 4

Find the Y parameters of the circuit

shown.
I1 2 j4 10 I2

+ +
+
10I2
V1 _ V2
_ -j20 _

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 Solution
i) V2 = 0 (short circuit port 2). Redraw the circuit.
I1 2 j4 10 I2

+
+
10I2
V1 _
_

I0
V1 (2 j4)I1
I1 1
Y11 (0.1 - j0.2) S
V1 2 j4
I2
Y21 0S
V1 20
 ii) V1 = 0 (short circuit port 1). Redraw the circuit.
I1 2 j4 10 I2

+
+
10I2
_ V2
-j20 _

I2
Y22 (0.05 j0.025) S
- 10I 2 V2
I1 ........(1)
2 j4 sub (2) (1)
V2 V2 - 10I 2 I
I2 Y12 1 (-0.1 j0.075) S
- j20 10 V2
In matrix form;
1 1
2I 2 V2 .......(2) 0.1 j0.2 0.1 j0.075
10 - j20 Y S
0 0.05 j0.025
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 6-3 T (ABCD) PARAMETER
T parameter or also ABCD parameter is a another
set of parameters relates the variables at the input
port to those at the output port.
T parameter also called transmission parameters
because this parameter are useful in the analysis of
transmission lines because they express sending end
variables (V1 and I1) in terms of the receiving end
variables (V2 and -I2).
The black box that we want to replace with T
parameter is as shown below.
I1 I2
A11 B12
+ +
V1 V2
- C21 D22 -
22
 V1 AV2 BI 2
I1 CV2 DI 2

V1 A B V2 V2
I C D I T I
1 2 2
where the T terms are called the transmission parameters,
or simply T or ABCD parameters, and each parameter has
different units.

V A=open-circuit V B= negative short-

A 1 voltage ratio B 1 circuit transfer
V2 I2 0
I2 V2 0 impedance ()

I1 C= open-circuit I1
C transfer admittance D D=negative short-
V2 (S)
I2 V2 0
circuit current ratio
I2 0
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 Example 5

Find the ABCD parameter of the circuit

shown below.
I1 2 4 I2

+ +

V1 10 V2

_ _

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 Solution
i) I2 = 0,
V2 10 I1
I1 2
I1
C 0.1S
+ + V2
V1 10 V2 V1 2 I1 V2
V2 6
_ _ V1 2 V2 V2
10 5
V1
A 1.2
V2

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ii) V2 = 0,

I1 2 4 I2 10
I2 I1
+ 14
I1
V1 10 D 1.4
I1 + I2 I2
_ V1 2 I1 10I1 I 2
V1 12 I1 10 I 2
In matrix form; 14
V1 12 I 2 10 I 2
1.2 6.8 10
T
0.1S 1.4 B
V1
6.8
I2

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 6-4 TERMINATED TWO
PORT NETWORKS
In typical application of two port network, the circuit
is driven at port 1 and loaded at port 2.
Figure below shows the typical terminated 2 port
model.
Zg I1 I2

+ +
+ Two port
Vg V1 V2 ZL
network
- -

Terminated two-port parameter can be implement to z-parameter,

Y-parameter and ABCD-parameter.

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 Zg represents the internal impedance of the source and
Vg is the internal voltage of the source and ZL is the
load impedance.
There are a few characteristics of the terminated two-
port network and some of them are;
V1
i) input impedance, Zi
I1
V2
ii) output impedance, Zo
I2
I2
iii) current gain, A i
I1
V2
iv) voltage gain, A v
V1
V2
v) overall voltage gain, A g
Vg
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The derivation of any one of the desired expression
involves the algebraic manipulation of the two port
equation. The equation are:
1) the two-port parameter equation either Z or Y or ABCD.

For example, Z-parameter, V1 Z11I1 Z12I 2 .......(1)

V2 Z21I1 Z22I 2 .......(2)
2) KVL at input, V1 Vg I1Zg .......(3)

3) KVL at the output, V I Z .......(4)

2 2 L

From these equations, all the characteristic can be obtained.

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 Example 6
For the two-port shown below, obtain the suitable value of
Rs such that maximum power is available at the input
terminal. The Z-parameter of the two-port network is given
as
Z11 Z12 6 2
Z
21 Z 22 4 4

V2
With Rs = 5,what would be the value of
Vs
Rs I1 I2

+ +
+ Z
Vs V1 V2 4
- -

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 Solution
Z-parameter equation becomes; V1 6 I1 2 I 2 .......(1)
V2 4 I1 4 I 2 .......(2)

KVL at the output;

V2 4I 2 .......(3)
Subs. (3) into (2); I1
I2 .......( 4)
2
Subs. (4) into (1); V1 5I1.......(5)

Hence, the input impedance; Z V1 5

1
I1

For the circuit to have maximum power, Rs Z1 5

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 V
To find 2 at max. power transfer, voltage drop at Z1 is half of
Vs
Vs
Vs
V1 .......(6)
2

From equations (3), (4), (5) & (6),

Overall voltage gain, V2 1

Ag
Vs 5

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 Example 7
The ABCD parameter of two port network shown below
are.
4 20
0.1S 2

The output port is connected to a variable load for a
maximum power transfer. Find RL and the maximum power
transferred.

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 Solution

ABCD parameter equation becomes

V1 = 4V2 20I2 (1)
I1 = 0.1V2 2I2 (2)

At the input port, V1 = -10I1 (3)

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(3) Into (1) (4) = (2)
-10I1 = 4V2 20I2 -0.4V2 + 2I2 =0.1V2 2I2
I1 = -0.4V2 + 2I2 (4) 0.5V2 = 4I1 (5)

From (5);
ZTH = V2/I2 = 8 (6)
But from Figure (b), we know that
V1 = 50 10I1 and I2 =0
Sub. these into (1) and (2)
50 10I1 = 4V2 (7)
I1 = 0.1V2 (8)

Sub (8) into (7); V2 = 10 Thus, VTH = V2 = 10V

RL for maximum power transfer, RL = ZTH = 8

The maximum power;
P = I2RL = (VTH/2RL)2 x RL = V2TH/4RL = 3.125W
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