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Section 3.7  The Time Response and the State Transition  Matrix     191

                                     •V|<(»    Initial    *»((>)



                        U(s) O                                O  VJs)

       FIGURE 3.23
       Flow graph of the
       RLC network.

                               Al(0)                A ,(0)




       FIGURE 3.24                                        X,ls)
       Flow graph of the
       RLC network with
       U(s)  = 0.


                        signal-flow  gain formula, we obtain  X\(s)  in terms  of x^O)  as

                                                       1-^(5)-[^(0)/5]
                                               X^s)  =                                    (3.87)
                                                             A(5)
                        where  A(5)  is  the  graph  determinant,  and  A T(5)  is  the  path  cofactor.  The  graph
                        determinant  is
                                                                     2
                                                 A (5)  =  1  +  3s"  1  +  2s" .
                                                     _1
                        The path  cofactor  is  A 2  =  1  +  35  because  the path  between  *i(0)  and  Xi(s)  does
                                                        - 1
                                                     -
                        not touch the loop with the factor 35 . Therefore, the first element  of the  transition
                        matrix is
                                                           l
                                                  (1  +  3s' )(l/s)   5  +  3
                                         <f>u(s)         1     -2   2                     (3.88)
                                                  1  +  35"  +  25  "  5  +  3s  +  2
                       The  element  4>n(s) is  obtained  by  evaluating  the  relationship  between  X](s)  and
                        x 2(0)  as
                                                            1
                                                        (-25' )(x 2 (0)/5)
                                                X l(s)  =     _1      2
                                                        1  +  35  +  25" '
                       Therefore, we obtain
                                                              - 2
                                                                                          (3.89)
                                                          s*  +  3s  +  2
                       Similarly, for <f>2i(s) we have
                                                    (5^)(1/5)           1
                                         02i (s)  =      1     2    2                     (3.90)
                                                  1  +  35"  +  25"  5  +  35  +  2

                       Finally, for  cf> 22(s), we obtain
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