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366                                                               Fiber Optic Communications


            where n jFI  and n jFQ  are the in-phase and quadrature components of the noise output of the matched filter,
            H (), respectively. The variance of n jFI  and n jFQ , j = 1, 0 is given by Eq. (8.207)
              Ij
                                                          het
                                                        N E
                                         2    2    2      0  1
                                         ≡   =   =      ,  j = 0, 1.                   (8.242)
                                         F    jFI  jFQ     4
            The pdf of the envelope r (T ) when s (t) is transmitted is given by the Rayleigh distribution
                                 0  b      1
                                                             (    2  )
                                                        r 0      r 0
                                         p (r |‘1’ sent)=  exp −      .                      (8.243)
                                          r0  0          2         2
                                                              2
                                                        F          F
            The pdf of the envelope r (T ) is given by the Rician distribution
                                   b
                                 1
                                                      (   2    2  )   (    )
                                                 r 1      r + E ∕4      rE 1
                                                               1
                                  p (r |‘1’ sent)=  2  exp −  2     I 0   2  .               (8.244)
                                      1
                                   r1
                                                          2        2
                                                  F           F          F
            If r (T ) > r (T ), it will be decided that ‘1’ is transmitted. Therefore, an error is made if r (T ) < r (T )
               1
                                                                                          b
                                                                                       1
                                                                                                 b
                                                                                              0
                       0
                         b
                  b
            when s (t) is transmitted. So, the probability of mistaking ‘1’ as ‘0’ is
                  1
                                      P(0|‘1’ sent)= P(r (T ) < r (T )|‘1’ sent).            (8.245)
                                                             0
                                                        b
                                                               b
                                                      1
            The probability that r (T ) < r (T ) can be found as follows. Since r (T ) and r (T ) are independent random
                                                                           0
                                                                             b
                                b
                             1
                                     0
                                                                  1
                                                                     b
                                       b
            variables, the joint pdf of r (T ) and r (T ) can be written as
                                             b
                                  1
                                    b
                                          0
                                  p   (r , r |‘1’ sent)= p (r |‘1’ sent)p (r |‘1’ sent).     (8.246)
                                   r 1 r 0  1  0     r 0  0      r 1  1
            The chance that r (T ) < r (T ) is the same as that r (T ) has a value r in the range 0 < r < ∞ and r (T )
                          1  b    0  b                1  b           1               1        0  b
            has a value greater than r ,
                                1
                                                    ∞  {  ∞                     }
                                                               (            )
                          P(r (T ) < r (T )|‘1’ sent)=  ∫  ∫  p r 1 r 0  r , r |‘1’ sent dr 1  dr .  (8.247)
                                      b
                                                                                    0
                                    0
                            1
                               b
                                                                    0
                                                                 1
                                                   r 1   0
            Using Eqs. (8.243), (8.244), and (8.246), Eq. (8.247) can be simplified as
                                             ∞      {   ∞         }
                              P(0|‘1’ sent)=   p (r )     p (r )dr 0  dr 1
                                                  1
                                           ∫    r 1   ∫    r 0
                                                              0
                                            0          r 1
                                                       (      )
                                             ∞             r 2
                                         =     p (r ) exp −  1  dr 1
                                                  1
                                           ∫    r 1          2
                                            0              2
                                                             F
                                                       (          )   (     )
                                                ∞          2   2
                                           1              r + E ∕8      r E
                                                                         1 1
                                                           1
                                                               1
                                         =        r exp −           I 0       dr .           (8.248)
                                                  1
                                                                               1
                                            2 ∫ 0           2       2 2
                                            F                 F           F
            Let
                                                        √
                                                   ′
                                                   r = r  2,                                 (8.249)
                                                   1   1
                                                          √
                                                   ′
                                                  E = E ∕ 2.                                 (8.250)
                                                        1
                                                   1
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