Page 193 - Schaum's Outline of Theory and Problems of Electric Circuits

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
182

Fig. 8-25 [CHAP. 8

8.17 Obtain the natural frequencies of the network shown in Fig. 8-26 by driving it with a conveniently
located current source.

Fig. 8-26
0
The response to a current source connected at xx is a voltage across these same terminals; hence the
network function HðsÞ¼ VðsÞ=IðsÞ¼ ZðsÞ. Then,

2
1 1 1 1 1 s þ 2:5s þ 1:5
¼ þ þ ¼
ZðsÞ 1 2=s 2 þ 4s 2 s þ 0:5
s þ 0:5 s þ 0:5
Thus, ZðsÞ¼ð2Þ ¼ð2Þ
2
s þ 2:5s þ 1:5 ðs þ 1Þðs þ 1:5Þ
The natural frequencies are the poles of the network function, s ¼ 1:0Np=s ¼ 2 and s ¼ 1:5 Np/s.

8.18 Repeat Problem 8.17, now driving the network with a conveniently located voltage source.

The conductor at yy 0 in Fig. 8-26 can be opened and a voltage source inserted. Then,
HðsÞ¼ IðsÞ=VðsÞ¼ 1=ZðsÞ:
0
The impedance of the netework at terminals yy is
2
1ð2=sÞ s þ 2:5s þ 1:5
ZðsÞ¼ 2 þ 4s þ ¼ð4Þ
1 þ 2=s s þ 2

1 1 s þ 2
Then; HðsÞ¼ ¼
2
ZðsÞ 4 s þ 2:5s þ 1:5