Page 452 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
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Sec. 13.6   Power Spectrum and Power Spectral Density.        439




                         xit)
                                                             ^ S { f ) A f



                                                                     Figure  13.6-7.  Power spectral  den­
                                                                     sity analyzer.
                              The  band-pass  filter of passband  B  =   passes  x(t)  in  the  frequency  interval  /
                              to  /  +  A/,  and  the output  is squared,  averaged,  and divided by A/.
                                  For high resolution.  A / should be made  as  narrow as possible;  however,  the
                              passband  of the  filter cannot  be  reduced  indefinitely without  losing  the  reliability
                              of the  measurement.  Also,  a  long  record  is  required  for  the  true  estimate  of the
                              mean square value, but actual records are always of finite length.  It is evident now
                              that  a  parameter  of  importance  is  the  product  of  the  record  length  and  the
                              bandwidth,  2BT, which must be sufficiently large.^
                              Example  13.6-1
                                  A random  signal has a  spectral  density that  is  a constant

                                                        S(f)  = 0.004 cm^/cps
                                  between 20 and  1200 cps and that is zero outside this frequency range. Its mean value
                                  is 2.0 cm.  Determine  its  rms value  and  its standard deviation.
                              Solution:  The  mean  square value  is found  from
                                                  —    /-oo      /"1200
                                                  X^=  f  S{f )df =  f    0.004 rf/= 4.72
                                                      •'o       ■'20
                                  and  the  rms value  is
                                                     rms  =   =  \/4.72  =  2.17 cm

                                  The variance  cr^  is defined by Eq.  (13.2-6):



                                                           =  4.72  -  2^ =  0.72
                                  and the  standard  deviation becomes

                                                         or =  \/0.72  =  0.85 cm
                                  The problem is graphically displayed by Fig.  13.6-8, which shows the time variation of
                                  the  signal and  its probability distribution.

                                  ^See  J.  S.  Bendat,  and  A.  G.  Piersol,  Random  Data  (New  York:  John  Wiley  &  Sons,  1971),
                              p.  96.
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