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I s 1 V s Z s    5s 20 s2 8s 12 3 s Z s 2:5 2:5 5s 20 s2 12 3 s
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we have H s 0:4 s2 12 s 2 s 6
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p The numerator of H s in Example 8.8 is zero when s j 12. Consequently, a voltage function at this frequency results in a current of zero. In 12 where series and parallel resonance are p discussed, it will found that the parallel LC circuit is resonant at ! 1= LC . With L 5 H and 3 p be 1 C 20 F, ! 12 rad/s. The zeros and poles of a network function H s can be plotted in a complex s-plane. Figure 8-14 shows the poles and zeros of Example with zeros marked 8 and poles marked . The zeros occur p 8.8, in complex conjugate pairs, s j 12, and the poles are s 2 and s 6.
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The network function can be expressed in polar form and the response obtained graphically. Before starting the development, it is helpful to recall that H s is merely a ratio such as V0 s =Vi s , I2 s =V1 s , or I2 s =I1 s . With the polynomials factored, H s k s z1 s z2 s z s p1 s z2 s p
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Now setting s zm Nm m m 1; 2; . . . ;  and s pn Dn n n 1; 2; . . . ;  , we have H s k N1 1 N2 2 N  D1 1 D2 2 D  k N1 N2 N 1  1  D1 D2 D
It follows that the response of the network to an excitation for which s  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 1est , where s 1 Np/s: Locate the test point 1 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 p N1 N2 13; D1 3; D2 7; 1 2 0; and 1 2 tan 1 12 73:98 Hence, H 1 0:4 p p 13 13 08 08 0:248 3 7
CHAP. 8]
HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
Fig. 8-15
The result implies that, in the time domain, i t 0:248v t , so that both voltage and current become in nite according to the function e1t . For most practical cases,  must be either negative or zero. The above geometrical method does not seem to require knowledge of the analytic expression for H s as a rational function. It is clear, however, that the expression can be written, to within the constant factor k, from the known poles and zeros of H s in the pole-zero plot. See Problem 8.37.
THE NATURAL RESPONSE
This chapter has focused on the forced or steady-state response, and it is in obtaining that response that the complex-frequency method is most helpful. However, the natural frequencies, which characterize the transient response, are easily obtained. They are the poles of the network function.
EXAMPLE 8.10 The same network as in Example 8.8 is shown in Fig. 8-16. source V s is inserted at xx 0 . Obtain the natural response when a
Fig. 8-16 The network function is the same as in Example 8.8: H s 0:4 The natural frequencies are then 2 Np/s and 6 Np/s. is of the form s2 12 s 2 s 6 Hence, in the time domain, the natural or transient current
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