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CHAP. 61  FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS AND SYSTEMS



                  (a)  From Eq. (6.28)
                                            1                                    sin Wn
                                                                    IW
                                    x[n] = - IT X(R)       dfi = -     elnn do = -
                                           27  -,               27  -w            rn
                      Thus, we  obtain
                                           sin Wn                     IRl  s w
                                                 -x(n)  =
                                            7n                        WCI~IST

                  (b)  The sequence x[n] is plotted in Fig. 6-13(b) for  W = 7/4.

            6.16.  Verify the frequency-shifting property (6.44, that is,
                                               eJnonx[n] tt X(n - 0,)
                      By Eq. (6.27)
                                                     m
                                    ~  (  ~  j  f  l  t  ~ =  ~  [C   ]e~fl~"X[n] e-~fl"
                                                          )
                                                      ~
                                                   ,,=  -m
                                                     m
                                                 =  C  x[n] e-~(fl-flt~)n =  X(fl -a,)
                                                   ,,=  -m
                  Hence,
                                                e~~o"x[n] -X(R  - a,)


            6.17.  Find the inverse Fourier transform x[n] of
                                                       -
                                       x(n) = 2~qn no)             WI, lfiol 5
                     From  Eqs. (6.28) and (1.22) we have




                  Thus, we  have
                                        ejnon - 2rS(R - R,)

           6.18.  Find the Fourier transform of
                                                 x[n] = 1       all n
                     Setting R, = 0 in  Eq. (6.1351, we get
                                           x[n] = 1 o 2nS(R)        In1  I
                  Equation (6.136) is depicted in Fig. 6-14.
















                               Fig. 6-14  A constant sequence and its Fourier transform.
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