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ANALYSIS  AND  PROCESSING  OF  RANDOM  PROCESSES            [CHAP  6



                    Again taking the inverse Fourier transform of both sides and using the result of part (a), we have
                                                     d           d
                                            RX*(z) = - - RXx,(z) = - -
                                                                dz'  R~(z)
                                                     dz
                (c)  From Eq. (6.63),
                                        Sxr(o)  = ( H(o) (2Sx(o) = (jo (2SX(o) = o2SX(w)
                    Note that Eqs. (6.159) and (6.160) were proved in Prob. 6.8 by a different method.




          FOURIER  SERIES AND  KARHUNEN-LOEVE EXPANSIONS
          6.33.  Verify Eqs. (6.80) and (6.81).

                    From Eq. (6.78),



                Since X(t) is WSS, E[X(t)] = p,,  and we have







                   Again using Eq. (6.78), we have








                Now








                Letting t - s = z, and using Eq. (6.76), we obtain








                Thus







          6.34.  Let z(t) be the Fourier series representation of X(t) shown in Eq. (6.77). Verify Eq. (6.79).
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