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340                                                               Fiber Optic Communications


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            curve |H()| and, therefore,  decreases. If the filter is too narrow (Fig. 8.4(b)), a significant fraction of the
            signal component is truncated by the filter and, therefore, the numerator of Eq. (8.29) becomes too small. The
                                                                                               ∗
            optimum filter transfer function can be obtained by setting the variation of  with respect to H() and H ()
            to zero:
                                                       
                                                = 0  and       = 0.                           (8.30)
                                             H          H ∗
            To find the variations given by Eq. (8.30), let us replace the integrals of Eq. (8.29) by summations:
                                        [                                    ] 2
                                         ∑ [             ]
                                            ̃ x ( )− ̃x ( ) H( ) exp (−i T )Δ n
                                                             n
                                                       n
                                                                      n b
                                                n
                                             1
                                                    0
                                          n
                                = lim                ∑                       .              (8.31)
                                                               2
                                   Δ n →0        N 0  |H( )| Δ n
                                                             n
                                                       n
                                                                               ∗
             can be optimized by setting its partial derivatives with respect to H( ) and H ( ) to zero. Note that
                                                                                  n
                                                                        n
                       ∗
            H( ) and H ( ) are independent variables. Alternatively, Re[H( )] and Im[H( )] can also be chosen as
                                                                   n
                                                                               n
                n
                          n
            independent variables:
                                     √
                                2 N[̃x ( )− ̃x ( )] exp (−i T )  ∗  N
                                             n
                                          1
                                                              n b
                                                    n
                                                  0
                                  =                               − N H ( )  = 0,         (8.32)
                                                                       0
                                                                            n
                           H( )               D                          D 2
                                n
            where N and D denote the numerator and denominator of Eq. (8.29), respectively. Simplifying Eq. (8.32), we
            obtain
                                                 ∗      ∗
                                       H( )= k[̃x ( )− ̃x ( )] exp (i T ),
                                           n     1  n   0  n        n b
                                                 ∗      ∗
                                        H()= k[̃x ()− ̃x ()] exp (iT ),               (8.33)
                                                 1      0          b
            where                                   (    )
                                                      2D    1
                                                k =  √                                        (8.34)
                                                       N   N 0
            is an arbitrary constant, which we set to unity from now on. The same result can be obtained by setting the
                                     ∗
            variation of  with respect to H () to zero. The filter with the transfer function given by Eq. (8.33) is called
            a matched filter. Using Eq. (8.33) in Eq. (8.29), we obtain
                                                     ∞
                                                1                  2
                                          max  =  N ∫  |̃x ()− ̃x ()| d
                                                               0
                                                        1
                                                 0  −∞
                                               2    T b         2
                                             =       [x (t)− x (t)] dt,                       (8.35)
                                                             0
                                                       1
                                               N ∫ 0
                                                0
            where we have made use of Parseval’s relations. Let
                                                 T b
                                          E =  ∫   x (t)x (t) dt,  j = 1, 0,                  (8.36)
                                                    j
                                           jk
                                                       k
                                                0
                                                    E ≡ E .                                   (8.37)
                                                     jj   j
            Using Eqs. (8.36) and (8.37) in Eqs. (8.35) and (8.27), we obtain
                                                   2
                                              =    [E + E − 2E ],                           (8.38)
                                             max       1   0    10
                                                  N
                                                    0
                                                        (       )
                                                          √
                                                    1         max
                                            P b,min  =  erfc      .                           (8.39)
                                                    2         8
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