Page 192 - Schaum's Outline of Theory and Problems of Electric Circuits
P. 192

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
               CHAP. 8]
                     At the input terminals, KVL gives                                               181

                                                              0        15
                                                VðsÞ¼ 2sIðsÞþ V ðsÞ¼ 2s þ  IðsÞ
                                                                        s
                                                                   2
                                                            VðsÞ  2s þ 15
                     Then                             HðsÞ¼    ¼
                                                            IðsÞ    s


               8.15  For the two-port network shown in Fig. 8-24 find the values of R 1 , R 2 , and C, given that the
                     voltage transfer function is
                                                       V o ðsÞ   0:2
                                                H v ðsÞ     ¼  2
                                                       V i ðsÞ  s þ 3s þ 2














                                                        Fig. 8-24
                                                  0
                         The impedance looking into xx is
                                                   ð1=sCÞðR 1 þ R 2 Þ
                                                0                    R 1 þ R 2
                                              Z ¼               ¼
                                                   ð1=sCÞþ R 1 þ R 2  1 þðR 1 þ R 2 ÞCs
                     Then, by repeated voltage division,
                                                                   0
                                   V o    V o  V xx 0    R 2     Z         R 2 =ðR 1 þ R 2 ÞC
                                      ¼             ¼            0     ¼
                                                                         2
                                    V i  V xx  0  V i  R 1 þ R 2  Z þ s1  s þ   1    s þ  1
                                                                            ðR 1 þ R 2 ÞC  C
                         Equating the coefficients in this expression to those in the given expression for H v ðsÞ, we find:

                                                  1           3             1
                                              C ¼  F      R 1 ¼  
     R 2 ¼
                                                  2           5            15



               8.16  Construct the pole-zero plot for the transfer admittance function

                                                              2
                                                       I o ðsÞ  s þ 2s þ 17
                                                HðsÞ¼      ¼  2
                                                      V i ðsÞ  s þ 3s þ 2
                         In factored form,

                                                         ðs þ 1 þ j4Þðs þ 1   j4Þ
                                                   HðsÞ¼
                                                            ðs þ 1Þðs þ 2Þ
                     Poles exist at  1 and  2; zeros at  1   j4.  See Fig. 8-25.
   187   188   189   190   191   192   193   194   195   196   197