Page 364 - Fiber Optic Communications Fund
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Performance Analysis                                                               345



           Substituting Eqs. (8.67) and (8.69) in Eq. (8.6), we find

                                                  A 2   ∞  −   2
                                          u(T )=± a ∫ −∞  e  2a 2  d
                                             b
                                                    2
                                                     √
                                                    2
                                               =±2A    T ,                                (8.70)
                                                         0
                  √
           where A 2  T is the pulse energy E .
                      0                 1
           8.2.2  Error Probability with an Arbitrary Receiver Filter
           From Eqs. (8.27) and (8.28), we have
                                                       (√ )
                                                  1        
                                             P =   erfc       ,                             (8.71)
                                              b
                                                  2        8
                                                [u (T )− u (T )] 2
                                                           b
                                                    b
                                                  1
                                                         0
                                             =               ,                            (8.72)
                                                       2
                         2
           where u (T ) and  are given by Eqs. (8.6) and (8.7), respectively. Eqs. (8.71) and (8.72) are valid for arbitrary
                 j
                   b
           filter shapes. From Eq. (7.8), we have
                                                     I − I 0
                                                     1
                                                Q =        .                                (8.73)
                                                     +  0
                                                     1
           Since the mean of bit ‘1’ (‘0’) is u (T )(u (T )), Eq. (8.72) becomes
                                      1  b  0  b
                                                  (       ) 2
                                                    I − I 0
                                                     1
                                                =         .                               (8.74)
                                                      
           When  =  ≡ , from Eqs. (8.74) and (8.73), we find
                 1
                     0
                                                        2
                                                   = 4Q ,                                 (8.75)
                                                       (    )
                                                  1      Q
                                             P =   erfc       .                             (8.76)
                                               b         √
                                                  2
                                                          2
           8.3  Homodyne Receivers
           Consider a fiber-optic transmission system with a homodyne balanced receiver, as shown in Fig. 8.9. Its
           mathematical representation is shown in Fig. 8.10. Let the transmitted optical field distribution be
                                             q = s(t) exp (−i t),                         (8.77)
                                              s
                                                            c
           where s(t) is the complex field envelope,  is the angular frequency of the optical carrier. We make the
                                              c
           following assumptions to find the best achievable performance.
           (1) Fiber dispersion, PMD, and nonlinearity are absent so that the fiber channel can be modeled as an AWGN
           channel. In fact, fiber dispersion and PMD cause distortion, and nonlinearity causes both distortion and noise
           enhancement, which will be discussed in Chapters 10 and 11.
           (2) The frequency, phase, and polarization of the local oscillator are exactly aligned with the received signal.
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