Page 183 - Schaum's Outline of Theory and Problems of Electric Circuits

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
172
p ffiffiffiffiffi [CHAP. 8
The numerator of HðsÞ in Example 8.8 is zero when s ¼ j 12. Consequently, a voltage function
at this frequency results in a current of zero. In Chapter 12 where series and parallel resonance are
p ffiffiffiffiffiffiffi
5
discussed, it will be found that the parallel LC circuit is resonant at ! ¼ 1= LC. With L ¼ H and
ffiffiffiffiffi
p
3
C ¼ 1 F, ! ¼ 12 rad/s.
20
The zeros and poles of a network function HðsÞ can be plotted in a complex s-plane. Figure 8-14
shows the poles and zeros of Example 8.8, with zeros marked 8 and poles marked . The zeros occur
p ffiffiffiffiffi
in complex conjugate pairs, s ¼ j 12, and the poles are s ¼ 2and s ¼ 6.

Fig. 8-14

8.8 THE FORCED RESPONSE
The network function can be expressed in polar form and the response obtained graphically. Be-
fore starting the development, it is helpful to recall that HðsÞ is merely a ratio such as V ðsÞ=V ðsÞ,
0
i
I 2 ðsÞ=V 1 ðsÞ,or I 2 ðsÞ=I 1 ðsÞ. With the polynomials factored,
ðs z 1 Þðs z 2 Þ ðs z Þ
HðsÞ¼ k
ðs p Þðs z 2 Þ ðs p Þ
1
Now setting ðs z m Þ¼ N m m ðm ¼ 1; 2; ... ; Þ and ðs p Þ¼ D n n ðn ¼ 1; 2; ... ; Þ, we have
n
ðN 1 1 ÞðN 2 2 Þ ðN Þ N 1 N 2 N
HðsÞ¼ k ¼ k ð 1 þ þ Þ ð 1 þ þ Þ
ðD 1 1 ÞðD 2 2 Þ ðD Þ D 1 D 2 D

It follows that the response of the network to an excitation for which s ¼ þ j! is determined by
measuring the lengths of the vectors from the zeros and poles to s as well as the angles these vectors make
with the positive axis in the pole-zero plot.

st
EXAMPLE 8.9 Test the response of the network of Example 8.8 to an exponential voltage excitation v ¼ 1e ,
where s ¼ 1 Np/s:
Locate the test point 1 þ j0 on the pole-zero plot. Draw the vectors from the poles and zeros to the test point
and compute the lengths and angles (see Fig. 8-15). Thus,
p ffiffiffiffiffi p ffiffiffiffiffi
N 1 ¼ N 2 ¼ 13; D 1 ¼ 3; D 2 ¼ 7; 1 ¼ 2 ¼ 0; and 1 ¼ 2 ¼ tan 1 12 ¼ 73:98
p ffiffiffiffiffi p ffiffiffiffiffi
ð 13Þð 13Þ
Hence, Hð1Þ¼ð0:4Þ 08 08 ¼ 0:248
ð3Þð7Þ