Page 93 - Schaum's Outline of Theory and Problems of Signals and Systems
P. 93

LINEAR TIME-INVARIANT SYSTEMS                         [CHAP. 2



                 (a)  Find the eigenvalue of  the system corresponding to the eigenfunction  es'.
                 (b)  Repeat part (a) by  using the impulse function  h(t) of the system.

                 (a)  Substituting x(r) = e"  in Eq. (2.861, we obtain








                      Thus, the eigenvalue of  the system corresponding to eS' is




                 (b)  From  Eq. (2.75) in  Prob.  2.1 1 we have
                               h(t) = A [u(t + :)   - u(l - :)I   = (:IT    -T/2<tsT/2
                                      T
                                                                            otherwise
                      Using  Eq. (2.241, the eigenvalue  H(s) corresponding to eS' is given by




                      which  is the same as Eq. (2.87).


           2.17.  Consider  a stable continuous-time  LTI system with  impulse response  h(t) that is real
                 and even. Show that cos wt and sin wt are eigenfunctions of  this system with the same
                 real eigenvalue.
                     By  setting  s = jw  in  Eqs.  (2.23) and  (2.241, we  see  that  eJ"'  is  an  eigenfunction  of  a
                 continuous-time LTI system and the corresponding eigenvalue is

                                             A  = H( jw)  = )  h(r) e-"'dr                    (2.88)
                                                          - Z
                 Since the system is stable, that is,







                 since le-j"'J  = 1. Thus, H( jw) converges for any w. Using Euler's  formula, we  have

                                          Ti
                                H( jw)  = /-?(r)  e'~"'di = jW h(r)(cos or - sin wr) dr
                                                                           j
                                                           - x
                                                              w
                                       = j::(r)   cos wr dr -  I-.  h(r) sin or dr
                                                           j
                 Since cos or is an even function of r and sin or is an odd function of 7, and if  h(t) is real and
                 even, then  h(r)cos or is  even  and  h(r) sin wr  is odd. Then by  Eqs. (1.75~) and (1.77), Eq.
                 (2.89  becomes
                                                         m
                                              H( jo) = 2/  h(r) cos or dr                     (2.90)
                                                        0
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