Page 98 - Schaum's Outline of Theory and Problems of Signals and Systems
P. 98

CHAP. 21                  LINEAR TIME-INVARIANT SYSTEMS                               87



           2.22.  Consider  the  system in  Prob. 2.20.  Show that  the  initial  rest  condition  y(0) = 0 also
                 implies that the system is time-invariant.
                    Let  y,(t) be  the response  to an input  x,(t) and
                                                 xI(t) = 0     t10                          (2.114)


                 Then

                 and                                  Y do) = 0                             (2.116)
                 Now, let  y2(t) be the response  to the shifted input  x2( t ) = x ,(t  - T). From Eq. (2.114) we have
                                                 x2(t) = 0     tsr                          (2.117)

                 Then  y ,(  t  must satisfy




                 and                                  ~~(71
                                                             0
                                                           =
                 Now, from Eq. (2.115) we  have



                 If we  let  y2(t) = yl(t - T), then by  Eq. (2.116) we have

                                             yz(7) = Y,(T - 7) = ~1(0) = 0
                 Thus, Eqs. (2.118) and (2.119) are satisfied and we conclude that the system is time-invariant.


           2.23.  Consider the system in Prob. 2.20.  Find the impulse response h(r) of  the system.
                    The impulse response  h(t) should satisfy the differential equation





                 The homogeneous solution  hh(t) to Eq. (2.120) satisfies




                 To obtain  hh(t) we  assume
                                                    hh(t) = ceS'
                 Substituting this into Eq. (2.121) gives

                                             sces' + aces' = (s + a)ces' = 0
                 from which we have  s = -a  and
                                                  hh(t) = ce-"'u(t)

                 We predict that the particular solution hp(t) is zero since hp(t) cannot contain  Nt). Otherwise,
                h(t) would have a derivative of  S(t) that is not part of the right-hand side of  Eq. (2.120). Thus,

                                                  h(t) =ce-"'u(t)                           (2.123)
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