Network Synthesis                                     349
?+ 2s s 48 +3 |1 Z,
                              s² + 2s
                                      2s + 3 s + 2s                   Y
                                                      1
                                                             2s +3 (4> 23
                                                             28
                                                                  1       1
                                                                   X
   Therefore, the synethesized network is shown in figure 8.24.
                                                     12           42
                                        Z(9)-’
                                                          Fig. 8.24.
EXAMPLE 8.34 Synthesize
                            (s +5)
             Zs) =                          in Foster's II form.                                (U.P.T.U, 2001)
                      (s +1)(s +6)
Solution :
                                        8+5
                     Z(s) =
                                (s+ 1)(s + 6)
                                (s+1) (s+6)
                     Ys) =
                                   (s+5)
                 8+5)8 + 7s + 6 |s + 2
                              g2 + 5s
                                 2s + 6
                                 2s + 10
                                            -4
                                              -4
                     Y(8) = 8+2+
                                             s+5
                     Y(s)        g +7s +6                                                               5
                                     s+58                                                               4
                                                                              Zs)’   1F
                                                                                            6
                g + 5sy s + 7s + 6 |1                                                                       4
                                                                                                         25
                                     + 5s
                                      28 + 6                                         Fig. 8.25.
                                                                 Synthesis
                                       Network Analysia and
 348
                                                2r-6
                      A,(x) 3
                                            A,         Ay           No. of Sign Changes
                                 Ao
                      0
                            8)   2
    Now
   Therefore, A(o) 0 for
                           all o
                            not a p.r.f.
   Hence, given function is
                                  g+6s +5          (8+1)(8+5)
                                      +9s +14
                                                   (s+2)(8 +7)
   (ü)
                                   N(s) = 9s
   Condition   (1): M(s) = g + 14;
                                       9s
                             14) 9s 14
                                 9s
   Therefore, Dls) is Hurmitz polynomial.
   Condtion (2): doesnot exist.
   Condition (3): M, (s) =g²+ 5,Mals) =² + 14
                   N,(s) = 6s, Ngls) = 9s
                           M,M, -N, N, 20
      (s2 + 5) (s² + 14) -(6s) (9s) 20
         s + 19s2+ 70 - 54 s² 0
         s-35s+ 70 20
               A(o) = a+35 w + 70 >0
   Since allcoefficients of A(o?) are positive therefore, A(@²) >0for all o.
   Hence given function is a p.r.f.
EXAMPLE 8.33. Synthesize
                          (s +1) (s +3)
              Zs) =                     in Cauer'sI form.                    (U.P.T.U,, 2001, 2003/C.0.)
                             s (s+2)
Solution:             Z(s) =
                                 s+4s +3
                                     s+2s