Page 182 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 182

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
                                                                                                     171
               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 ZðsÞ¼ 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
                           n         n 1                        m         m 1
                          d y       d  y          dy           d x       d   x          dx
                        a n  n  þ a n 1  n 1  þ     þ a 1  þ a 0 y ¼ b m  m  þ b m 1  m 1  þ     þ b 1  þ b 0 x
                          dt        dt            dt           dt         dt            dt
                             st
                                           st
               When xðtÞ¼ Xe and yðtÞ¼ Ye ,
                               n      n 1               st      m       m 1               st
                            ða n s þ a n 1 s  þ     þ a 1 s þ a 0 Þe ¼ðb m s þ b m 1 s  þ     þ b 1 s þ b 0 Þe
               Then,

                                                       n       n 1
                                              YðsÞ   a n s þ a n 1 s  þ     þ a 1 s þ a 0
                                       HðsÞ¼      ¼    m       m 1
                                              XðsÞ  b m s þ b m 1 s  þ     þ b 1 s þ b 0
                   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

                                                     ðs   z 1 Þðs   z 2 Þ   ðs   z   Þ
                                             HðsÞ¼ k
                                                     ðs   p Þðs   p Þ   ðs   p Þ
                                                          1     2
               where k is some real number.  The complex constants z ðm ¼ 1; 2; ... ; Þ, the zeros of HðsÞ, and the
                                                                m
               p ðn ¼ 1; 2; ... ; Þ the poles of HðsÞ, assume particular importance when HðsÞ is interpreted as the ratio
                n
               of the response (in one part of the s-domain network) to the excitation (in another part of the network).
               Thus, when s ¼ z m , the response will be zero, no matter how great the excitation; whereas, when s ¼ p ,
                                                                                                      n
               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.  Obtain the network function for the
               current IðsÞ due to an input voltage VðsÞ.
                                                          IðsÞ   1
                                                    HðsÞ¼     ¼
                                                          VðsÞ  ZðsÞ

                                                      5s  20
                                                                    2
                                                      3   s        s þ 8s þ 12
               Since                       ZðsÞ¼ 2:5 þ       ¼ð2:5Þ
                                                                      2
                                                      5s  20         s þ 12
                                                        þ
                                                      3   s
               we have
                                                              2
                                                              s þ 12
                                                  HðsÞ¼ ð0:4Þ
                                                            ðs þ 2Þðs þ 6Þ










                                                        Fig. 8-13
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