218 Two-port networksTherefore
I,/(1 + ZY)= V1/Z
Also, we saw above that 12
B = V,/I2 = Z
In matrix form= I,/(1 + ZY) so that (^12) = V~/Z. Therefore
[v,] [1 zj[ ]
I~ Y I+ZY 12
Note that in this case A 4= D. This two-port network is not symmetrical because
the input and output ports are not interchangeable.
Example 9.7
Obtain the ABCD-parameters for the network shown in Fig. 9.10.
]1
V1
-0~
~ Y
I2 9 ~ O+
V2
O-
Figure 9.10
Solution
This is called a shunt admittance network.
A = V1/VzII~=O._ With I2 - O, V1 - V2 so that A - 1.
From Equation (9.38), C = I~/Vz],~=o. With I2
V1 = V2. Therefore
From Equation (9.36),
= 0, 11- VIY- VzY since
C- I,/V2- Y
From Equation (9.39), D - l~/I21v~=o. With V2 - 0, 11 -- I 2 since Y is short
circuited, so that D- 1. As expected then for this symmetrical network,
A=D.
From Equation (9.37), B - V~/12lv~=o. With V2 - 0, V1 - 0 and so B = 0.
In matrix form