Introduction to Electric Circuits

226 Two-port networks

With the output port short circuited so that V2 - 0,

VI=BI2 and 11=DI2

The short circuit input impedance is Zsc = (V~/I~)Jv2=o = (BI2)/(DI2), so

Zsc- B/D (9.65)

From Equations (9.64) and (9.65) we have that ZocZsc- AB/CD and, for a
symmetrical network, A = D, so

ZocZsc = B//C

Now from Equation (9.63), Zo - ~/(B/C), so

Zo = V/(ZocZsc) (9.66)

Example 9.13

Determine (1) the ABCD-parameters and (2) the characteristic impedance of
the network shown in Fig. 9.19.

O I1

V1 Y

Z

I I

(25 + j0)~

j0.2S j0.2S

i-1

I2

Y V2

O

Figure 9.19

Solution

From matrix Equation (9.57) we have for the zr-network:

A = D = l + ZY= l + (25 • jO.2) = l + j5

= V'(12 + 52)z_tan -~ (5//1) = 5.09/_78.69 ~

B=Z=25fl

C = 2Y + ZY 2 = j0.4 + (25)00.2) 2 = (-1 + j0.4) S

This means that C is in the third quadrant so that

C = V'(12 + 0.4)2/[180 - tan -1 (0.4//1)] = 1.07/158.22 ~ S

From Equation (9.63)

Zo - ~/(B//C) = V/[25/(1.07/-158.22~ = 4.83/-158.22 ~ 1~

9.10 IMAGE IMPEDANCES

Suppose that for a non-symmetrical two-port network, terminating the output

port with an impedance Z~2 results in an impedance looking into the input