1. CHAPTER 6
TWO PORT NETWORKS

2. 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.

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

4. 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)

5. 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.

6. 6-1 Z – PARAMETER
Z – parameter also called as impedance parameter and the units is ohm (Ω)
The “black box” is replace with Z-parameter is as shown below. + V1 - I1 I2 V2
Z11
Z21
Z12
Z22
where the z terms are called the impedance parameters, or simply z parameters, and have units of ohms.

7. z11 = Open-circuit input impedance
z21 = Open-circuit transfer impedance from port 1 to port 2
z12 = Open-circuit transfer impedance from port 2 to port 1
z22 = Open-circuit output impedance

8. Example 1 Find the Z – parameter of the circuit below. 40Ω 240Ω 120Ω +V1 _ V2 I1 I2

9. Solution When I2 = 0(open circuit port 2). Redraw the circuit. Ia I1 +
40Ω
240Ω
120Ω + V1 _ V2 I1 Ia Ib

10. When I1 = 0 (open circuit port 1). Redraw the circuit.
40Ω
240Ω
120Ω + V1 _ V2 Iy I2 Ix
In matrix form:

11. Example 2 Find the Z – parameter of the circuit below + _ V1 V2 -j20Ω
10Ω
j4Ω 2Ω
10I2 I2 I1

12. Solution i) I2 = 0 (open circuit port 2). Redraw the circuit. + V1 _
j4Ω 2Ω I1 V2
I2 = 0

13. ii) I1 = 0 (open circuit port 1). Redraw the circuit.+ _ V1 V2
-j20Ω
10Ω
10I2 I2
I1 = 0

14. 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. + V1 - I1 I2 V2
Y11
Y21
Y12
Y22
where the y terms are called the admittance parameters, or simply y parameters, and they have units of Siemens.

15. y11 = Short-circuit input admittance
y21 = Short-circuit transfer admittance from port 1 to port 2
y12 = Short-circuit transfer admittance from port 2 to port 1
y22 = Short-circuit output
admittance

16. Example 3 Find the Y – parameter of the circuit shown below. 5Ω 15Ω
20Ω + V1 _ V2 I1 I2

17. Solution
V2 = 0 5Ω
20Ω + V1 _ I1 I2 Ia

18. ii) V1 = 0
In matrix form; 5Ω
15Ω + V2 _ I1 I2 Ix

19. Example 4 Find the Y – parameters of the circuit shown. V1 V2 -j20Ω+ _ V1 V2
-j20Ω
10Ω
j4Ω 2Ω
10I2 I2 I1

20. Solution i) V2 = 0 (short – circuit port 2). Redraw the circuit. V1+ _ V1
10Ω
j4Ω 2Ω
10I2 I2 I1

21. ii) V1 = 0 (short – circuit port 1). Redraw the circuit.+ _ V2
-j20Ω
10Ω
j4Ω 2Ω
10I2 I2 I1

22. 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. + V1 - I1 I2 V2
A11
C21
B12
D22

23. where the T terms are called the transmission parameters, or simply T or ABCD parameters, and each parameter has
different units.
B= negative short-circuit transfer impedance ()
A=open-circuit voltage ratio
C= open-circuit transfer admittance (S)
D=negative short-circuit current ratio

24. Example 5 Find the ABCD – parameter of the circuit shown below. 2Ω 10Ω+ V2 _ I1 I2 V1 4Ω

25. Solution
i) I2 = 0, 2Ω
10Ω + V2 _ I1 V1

26. ii) V2 = 0, 2Ω
10Ω I1 I2 + V1 _ 4Ω
I1 + I2
In matrix form;

27. 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. + V1 - I1 I2 V2  Zg ZL Vg
Two – port network
Terminated two-port parameter can be implement to z-parameter,
Y-parameter and ABCD-parameter.

28. 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;

29. 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,
2) KVL at input,
3) KVL at the output,
From these equations, all the characteristic can be obtained.

30. 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
With Rs = 5Ω,what would be the value of + V1 - I1 I2 V2  Rs 4Ω Vs Z

31. Solution Z-parameter equation becomes; KVL at the output;
Subs. (3) into (2);
Subs. (4) into (1);
Hence, the input impedance;
For the circuit to have maximum power,

32. To find at max. power transfer, voltage drop at Z1 is half ofVs
From equations (3), (4), (5) & (6),
Overall voltage gain,

33. Example 7 The ABCD parameter of two – port network shown below are.
The output port is connected to a variable load for a maximum power transfer. Find RL and the maximum power transferred.

34. Solution ABCD parameter equation becomes V1 = 4V2 – 20I2
I1 = 0.1V2 – 2I2
At the input port, V1 = -10I1
(1)
(2)
(3)

35. But from Figure (b), we know that V1 = 50 – 10I1 and I2 =0
(3) Into (1)
-10I1 = 4V2 – 20I2
I1 = -0.4V2 + 2I2
(4) Into (2)
-0.4V2 + 2I2 = 0.1V2 – 2I2
0.5V2 = 4I2
(4)
(5)
From (5);
ZTH = V2/I2 = 8Ω
(6)
But from Figure (b), we know that
V1 = 50 – 10I1 and I2 =0
Sub. these into (1) and (2)
50 – 10I1 = 4V2
I1 = 0.1V2
(7)
(8)
Sub (8) into (7); V2 = 10
Thus, VTH = V2 = 10V
RL for maximum power transfer, RL = ZTH = 8Ω
The maximum power;
P = I2RL = (VTH/2RL)2 x RL = V2TH/4RL = 3.125W