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Performance Analysis                                                               337


                              p







                                   p (r)                           p (r)
                                    0                               1





                                   P(0∣1)                         P(1∣0)







                                                    r                        r
                                             u (T )  T  u (T )
                                              0
                                                b
                                                         1
                                                           b
           Figure 8.2 Conditional pdfs. P(0|1) is the chance of mistaking bit ‘1’ as bit ‘0’. P(1|0) is the chance of mistaking bit
           ‘0’ as bit ‘1’.
           Similarly, when a bit ‘1’ is sent, the conditional pdf is

                                                                         2
                                                           { [          ] }
                                                    1           r − u (T )
                                                                    0
                                                                      b
                                            0
                               p(r|‘0’ sent) ≡ p (r)= √  exp  −      2      .               (8.12)
                                                    2          2
           The probability of mistaking a bit ‘0’ as a bit ‘1’ is the area under the curve p (r) from r to ∞ (shown as
                                                                          0       T
           slanted lines in Fig. 8.2):
                                                                    2
                                                      { [          ] }
                                                  ∞        r − u (T )
                                            1                  0  b
                                  P(1|0)= √    ∫   exp   −      2      dr.                  (8.13)
                                           2  r T         2
           The total BER is given by
                                         P = P(0|1)P(1)+ P(1|0)P(0),                        (8.14)
                                           b
           where P(j) is the probability of sending bit ‘j’, j = 0, 1. Assuming that the bits ‘1’ and ‘0’ are equally probable,
           we obtain
                                                1
                                           P =   [P(0|1)+ P(1|0)].                          (8.15)
                                             b
                                                2
           Substituting Eqs. (8.13) and (8.11) in Eq. (8.15), we find
                          [              ∞       ]
                        1    r T
                   P =  2 ∫ −∞  p (r)dr +  ∫  p (r)dr                                       (8.16)
                     b
                                1
                                            0
                                                                                  2
                               {        r T { [      ] }       ∞    { [          ] }
                                                      2
                           1       r T       r − u (T )                  r − u (T )
                                                                             0
                                                 1
                                                                                b
                                                   b
                      = √        ∫   exp  −       2      dr +  ∫  exp  −      2      dr.    (8.17)
                        2 2   −∞            2           r T           2
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