Page 389 - Fiber Optic Communications Fund
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370                                                               Fiber Optic Communications


            When a bit ‘0’ is transmitted, if I > I , the bit ‘0’ is mistaken as the bit ‘1’ and this probability is
                                          T
                                             ∞          1    ∞    (   I  )
                                    P(1|0)=    p (I)dI =       exp −      dI
                                                0
                                           ∫           2 ∫         2 2
                                                          2
                                            I T             I T
                                              (     )
                                                  I T
                                         = exp −      .                                      (8.266)
                                                 2 2
            Using Eqs. (8.260) and (8.265), we obtain
                                                    [  DD  (       )]
                                                              2
                                         P(1|0)= exp −      1 +    ,                         (8.267)
                                                        2       DD
            where  DD  is given by
                                                         E av
                                                   DD
                                                    =     .                                (8.268)
                                                         ASE
            For OOK, E = E ∕2. So
                      av
                           1
                                                         E 1
                                                   DD
                                                    =      ,                               (8.269)
                                                        2
                                                          ASE
            When  DD  ≫ 1, P(1|0)≅ exp (− DD ∕2).
              Similarly, the probability of mistaking bit ‘1’ as bit ‘0’ is
                                            I T
                                  P(0|1)=    p (I)dI
                                              1
                                         ∫
                                          0
                                                                   √
                                                    (     2   )   ⎛  IRE 2 ⎞
                                          1    I T     RE + I     ⎜     1 ⎟
                                                          1
                                       =         exp −          I          dI.               (8.270)
                                         2 ∫ 0         2 2   0 ⎜ ⎜   2  ⎟
                                            2
                                                                         ⎟
                                                                  ⎝      ⎠
            Changing the variable of integration from I to x, where
                                                         I
                                                     2
                                                    x =    ,                                 (8.271)
                                                         2
                            2
                      2
                               2
            and letting a = RE ∕ , Eq. (8.270) becomes
                            1
                                               √
                                                I T ∕  (  a + x 2  )
                                                            2
                                     P(0|1)=        x exp −        I (ax) dx
                                             ∫               2      0
                                              0
                                                  (   √ )
                                                        I T
                                           = 1 − Q 1  a,                                     (8.272)
                                                       
                       √
            where Q (a,  I ∕) is the generalized Marcum’s Q-function given by Eq. (8.228). Using Eqs. (8.260),
                    1    T
            (8.265), and (8.268), Eq. (8.272) can be rewritten as
                                                          √
                                                                (       )
                                                  ⎛                      ⎞
                                                    √                 2
                                     P(0|1)= 1 − Q 1  ⎜ 2  DD ,   DD  1 +  ⎟             (8.273)
                                                  ⎜                    DD ⎟
                                                  ⎝                      ⎠
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