Page 367 - Fiber Optic Communications Fund
P. 367

348                                                               Fiber Optic Communications


            Using Eq. (8.93) in Eq. (8.39), we obtain
                                                    1    √
                                               P =   erfc(  homo ),                         (8.96)
                                                b
                                                    2
                                                         E av
                                                  homo  =  homo  .                          (8.97)
                                                        N
                                                          0
            The parameter  homo  represents the normalized energy per bit, which serves as a figure of merit in digital
            communication.
              When  homo  is much larger than unity, Eq. (8.96) can be approximated as
                                                    exp (− homo )
                                                P ≅            .                              (8.98)
                                                 b    √
                                                     2  homo


            8.3.1.1  Relation between Q-factor and BER

            The BER and Q-factor are related as follows. From Eq. (7.8), we have
                                                       I − I 0
                                                       1
                                                  Q =       .                                 (8.99)
                                                       +  0
                                                       1
            For a PSK signal, I =−I and  =  . So,
                                       1
                                 0
                                           0
                            1
                                                        I 1
                                                    Q =   .                                  (8.100)
                                                        
                                                         1
            Suppose that the correlator shown in Fig. 8.8 is used as the matched filter. The mean of bit ‘1’ after the
            correlator is (after setting the scaling factor 2RA  to unity in Eq. (8.83))
                                                   LO
                                                       T b
                                                         2
                                                 I =     s (t) dt
                                                 1  ∫    1
                                                      0
                                                   = E .                                     (8.101)
                                                      av
            In this case, we have
                                                 h(t)= s (T − t),                            (8.102)
                                                       1  b
                                                    ∗
                                             H()= s () exp (−iT ).                     (8.103)
                                                    1           b
            The variance of bit ‘1’ (or bit ‘0’) after the correlator is (see Eq. (8.7))
                                 N homo  1  ∞           N 0 homo  T b     N homo
                                  0
                                                                           0
                                                                   2
                                                  2
                             2
                             =             |H()| d =        s (t)dt =    E .           (8.104)
                             1        2 ∫                   ∫    1            av
                                   2      −∞              2    0            2
            Here, we have used Parseval’s relations. Substituting Eqs. (8.101) and (8.104) in Eq. (8.100), we find
                                                     √
                                                        2E av
                                                  Q =        .                               (8.105)
                                                        N homo
                                                         0
            From Eqs. (8.97) and (8.96), we have
                                                          Q 2
                                                   homo
                                                     =    ,                                (8.106)
                                                          2
                                                         (   )
                                                    1      Q
                                               P =   erfc  √   .                             (8.107)
                                                b
                                                    2       2
            Eq. (8.107) holds true even when the matched filter is not used (see Section 8.2.2).
   362   363   364   365   366   367   368   369   370   371   372