Page 460 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 460

Sec. 13.7   Fourier Transforms                                 447


                                  which is the parallel to Eq. (13.7-10). Unlike the autocorrelation, the cross-correlation
                                  and the cross-spectral density functions are, in general, not even functions; hence, the
                                  limits  -00 to  -hoc  are retained.
                              Example  13.7-6
                                  Using the  relationship

                                                      S(f)  = 2Î  R ( t )  cos 27rfr dr
                                  and the  results of Example  13.5-1,
                                                          R ( t )  = A ^ { T -  t )
                                  find  Sif) for the rectangular pulse.
                              Solution:  Because  Rir)  =  0 for  r  outside  ± 7, we  have
                                                 CT
                                         S(f)  = 2Î  A^(T -  r)cos27rfrdr
                                                •'()
                                              =  2 A ^ T  c o s 2 ttf 7  d r  —  2 A ^  j  T C O s 2 T r f r d T


                                              =  2A^T-       -   2A‘  cos 27t/ t  ---r sin 27t/'t
                                                      2 v f                 277/
                                                                   (2'n-/)
                                                 2A^
                                                     r ( l  - 0 0 8 2 7 7 / 7 ) =
                                                {27TfY
                                  Thus,  the power spectral  density of a rectangular pulse  using Eq. (13.7-11) is




                                                                                   \
                                  Note  from Example  13.7-3  that this  is  also equal  to  Xif)X*(f)  = X(f )\^.
                              Example  13.7-7
                                  Show  that  the  frequency  response  function  Hioj)  is  the  Fourier  transform  of  the
                                  impulse response  function  hit).
                              Solution:  From the convolution integral, Eq. (4.2-1), the response equation in terms of the
                                  impulse  response  function  is

                                                      x ( 0 -  r f i O K t - O d ^
                                                            d — ro
                                  where the lower limit has been extended to  -  oo to account for all past excitations.  By
                                  letting  T  =  (i  -  ^),  the  last integral becomes

                                                       ^ ( 0  =  f f { t - r ) h { r ) d T
                                                             •'n
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