172
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
[CHAP. 8
pffiffiffiffiffi 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 are pffiffiffiffiffiffiffiparallel resonance 5 H and discussed, it will be found that the parallel LC circuit is resonant at ! ¼ 1= LC . With L ¼ 3 pffiffiffiffiffi 1 C ¼ 20 F, ! ¼ 12 rad/s. 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 pffiffiffiffiffi in complex conjugate pairs, s ¼ j 12, and the poles are s ¼ 2 and 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. Before starting the development, it is helpful to recall that HðsÞ is merely a ratio such as V0 ðsÞ=Vi ðsÞ, I2 ðsÞ=V1 ðsÞ, or I2 ðsÞ=I1 ðsÞ. With the polynomials factored, HðsÞ ¼ k
ðs z1 Þðs z2 Þ ðs z Þ ðs p1 Þðs z2 Þ ðs p Þ
Now setting ðs zm Þ ¼ Nm m ðm ¼ 1; 2; . . . ; Þ and ðs pn Þ ¼ Dn n ðn ¼ 1; 2; . . . ; Þ, we have HðsÞ ¼ k
ðN1 1 ÞðN2 2 Þ ðN Þ ðD1 1 ÞðD2 2 Þ ðD Þ
¼k
N1 N2 N ð 1 þ þ Þ ð 1 þ þ Þ D1 D2 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. EXAMPLE 8.9 Test the response of the network of Example 8.8 to an exponential voltage excitation v ¼ 1est , 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, pffiffiffiffiffi pffiffiffiffiffi N1 ¼ N2 ¼ 13; D1 ¼ 3; D2 ¼ 7; 1 ¼ 2 ¼ 0; and 1 ¼ 2 ¼ tan 1 12 ¼ 73:98 Hence,
Hð1Þ ¼ ð0:4Þ
pffiffiffiffiffi pffiffiffiffiffi ð 13Þð 13Þ 08 08 ¼ 0:248 ð3Þð7Þ