Page 368 - Matrix Analysis & Applied Linear Algebra
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364              Chapter 5                    Norms, Inner Products, and Orthogonality

                                    symmetric about the point n/2in the frequency domain, and the information in
                                    just the first (or second) half of the frequency domain completely characterizes
                                    the original waveform—this is why only 512/2=256 points are plotted in the
                                    graphs shown in Figure 5.8.4. In other words, if
                                             2
                                         y =   F n x =  α k (e f k  + e n−f k )+i  β k (−e f k  + e n−f k  ) ,  (5.8.8)
                                             n
                                                      k                    k
                                    then the information in

                                               y n/2 =  α k e f k  − i  β k e f k  (the first half of y )
                                                      k           k
                                    is enough to reconstruct the original waveform. For example, the equation of the
                                    waveform shown in Figure 5.8.7 is
                                                          x(τ)=3 cos 2πτ +5 sin 2πτ,               (5.8.9)

                                                   6
                                                   5
                                                   4
                                                   3
                                                   2 1
                                                 Amplitude  -1 0  .25  Time      .75        1
                                                                        .5
                                                  -2
                                                  -3
                                                  -4
                                                  -5
                                                  -6
                                                                  Figure 5.8.7
                                    and it is completely determined by the four values in
                                                                x(0)          3
                                                                             
                                                                                 .
                                                               x(1/4)     5 
                                                                x(1/2)   =  −3
                                                          x = 
                                                                x(3/4)       −5
                                    To capture equation (5.8.9) from these four values, compute the vector y defined
                                    by (5.8.8) to be
                                                           1    1    1   1      3         0
                                                                                        
                                                  2       1   −i −1
                                              y =                         i   5     3 − 5i 
                                                  4  F 4 x =  1  −1  1  −1     −3   =   0  
                                                           1    i −1    −i     −5       3+5i
                                                    0         0
                                                            
                                                                =3(e 1 + e 3 )+ 5i(−e 1 + e 3 ).
                                                   3     −5 
                                                    0         0
                                                =   +i 
                                                    3         5
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