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CHAP. 8]



s-domain where only magnitudes and phase angles are shown. In the s-domain, inductances are expressed by sL and capacitances by 1=ðsCÞ. The impedance in the s-domain is ZðsÞ ¼ VðsÞ=IðsÞ. A network function HðsÞ is defined as the ratio of the complex amplitude of an exponential output YðsÞ to the complex amplitude of an exponential input XðsÞ If, for example, XðsÞ is a driving voltage and YðsÞ is the output voltage across a pair of terminals, then the ratio YðsÞ=XðsÞ is nondimensional. The network function HðsÞ can be derived from the input-output differential equation an

dny d n1 y dy d mx d m1 x dx þ a0 y ¼ bm m þ bm1 m1 þ    þ b1 þ b0 x þ an1 n1 þ    þ a1 n dt dt dt dt dt dt

When xðtÞ ¼ Xest and yðtÞ ¼ Yest , ðan sn þ an1 sn1 þ    þ a1 s þ a0 Þest ¼ ðbm sm þ bm1 sm1 þ    þ b1 s þ b0 Þest Then, HðsÞ ¼

YðsÞ a sn þ an1 sn1 þ    þ a1 s þ a0 ¼ nm XðsÞ bm s þ bm1 sm1 þ    þ b1 s þ b0

In linear circuits made up of lumped elements, the network function HðsÞ is a rational function of s and can be written in the following general form HðsÞ ¼ k

ðs  z1 Þðs  z2 Þ    ðs  z Þ ðs  p1 Þðs  p2 Þ    ðs  p Þ

where k is some real number. The complex constants zm ðm ¼ 1; 2; . . . ; Þ, the zeros of HðsÞ, and the pn ðn ¼ 1; 2; . . . ; Þ the poles of HðsÞ, assume particular importance when HðsÞ is interpreted as the ratio of the response (in one part of the s-domain network) to the excitation (in another part of the network). Thus, when s ¼ zm , the response will be zero, no matter how great the excitation; whereas, when s ¼ pn , the response will be infinite, no matter how small the excitation. EXAMPLE 8.8 A passive network in the s-domain is shown in Fig. 8-13. current IðsÞ due to an input voltage VðsÞ.

IðsÞ 1 ¼ VðsÞ ZðsÞ    5s 20 s2 þ 8s þ 12 3 s ZðsÞ ¼ 2:5 þ ¼ ð2:5Þ 5s 20 s2 þ 12 þ 3 s HðsÞ ¼


Obtain the network function for the

we have HðsÞ ¼ ð0:4Þ

s2 þ 12 ðs þ 2Þðs þ 6Þ

Fig. 8-13