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Performance Analysis                                                               347


           where
                                                          n (t)
                                                           d
                                             n(t)= n (t)+                                   (8.84)
                                                   cr
                                                         2RA
                                                            LO
           is a white noise process with Gaussian distribution. Its PSD is
                                              N homo   ASE   shot,eff
                                               0
                                          =      =      +                                 (8.85)
                                          n
                                                2      2    4A 2  R 2
                                                              LO
                                                      ASE  q
                                                   =      +   .                             (8.86)
                                                       2    4R
           The factor 1∕2 is introduced in the first term of Eq. (8.85) since the PSD of the real part of n (t) is half of that
                                                                                   c
           of n (t). The scaling factor 2RA LO  appearing in Eq. (8.83) multiplies both signal and noise and, hence, it is of
              c
           no consequence in evaluating the performance. Dropping this term, we write the normalized signal used for
           decision as
                                                I = s(t)+ n(t).                             (8.87)
                                                d
           8.3.1  PSK: Homodyne Detection

           The optical field envelope may be written as
                                             {
                                                  s (t) for bit ‘1’
                                        s(t)=      1                                        (8.88)
                                               s (t)=−s (t) for bit ‘0’.
                                               0       1
           We assume that s(t) is real and the filter shown in Fig. 8.6 is matched to s(t). Replacing x(t) by s(t) in Eq. (8.36),
           we obtain
                                             T b
                                                2
                                       E =     s (t)dt = E ,                                (8.89)
                                        1  ∫    1       0
                                            0
                                               T b
                                                  2
                                      E =−   ∫   s (t)dt =−E ,                              (8.90)
                                                           1
                                       10
                                                  1
                                              0
                                             2                    8E 1
                                      max  =  (E + E − 2E )=        .                     (8.91)
                                                            10
                                                  1
                                                      0
                                           N homo                N homo
                                             0                    0
           Since the bits ‘1’ and ‘0’ are equally probable, the average energy transmitted is
                                                   E + E 0
                                                    1
                                             E =          = E .                             (8.92)
                                                             1
                                              av
                                                     2
           The average energy forms a basis for comparison of various modulation formats and Eq. (8.91) can be written
           as
                                                       8E av
                                                 max  =  homo  .                          (8.93)
                                                      N
                                                       0
           The matched filter is given by (Eq. (8.40))
                                      ∗      ∗                 ∗
                              H()=[̃s ()− ̃s ()] exp (−iT )= 2̃s () exp (−iT ),  (8.94)
                                      1      0            b    1           b
           and the threshold r is (Eq. (8.47))
                          T
                                                   E − E 0
                                                    1
                                              r =         = 0.                              (8.95)
                                               T
                                                     2
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