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304                                                               Fiber Optic Communications


                                     10 0


                                                          P in = 0 dBm
                                    10 –5
                                                       P in = 2 dBm
                                                                        9
                                                    P in = 4 dBm      10
                                    10 –10
                                   BER
                                    10 –15

                                    10 –20


                                    10 –25
                                      120     140    160    180     200    220
                                                    Fiber length (km)

            Figure 7.4 BER as a function of fiber length. B = 10 Gb/s, B = 7.5 GHz, R = 1kΩ, R = 1 A/W, T = 290 K, fiber loss
                                                         e
                                                                    L
            coefficient  = 0.2 dB/km.





                     Figure 7.5 A fiber-optic system consisting of a transmitter, a fiber, an amplifier, and a receiver.


            we introduce a preamplifier of gain G, as shown in Fig. 7.5. Now, the received power is GP exp (−L) when
                                                                                    in
            ‘1’ is sent. The preamplifier adds ASE noise with the PSD per polarization given by Eq. (6.17). The mean
            current for bit ‘0’ is given by Eq. (6.84),
                                                 I = 2R ASE o                               (7.15)
                                                           B .
                                                  0
            We assume that the optical filter is an ideal band-pass filter with bandwidth f , the electrical filter is an ideal
                                                                          o
            low-pass filter with bandwidth f , and f < f . In this case B = f . The variance of bit ‘0’ is
                                           e
                                               o
                                                            o
                                      e
                                                                o
                                            2   2      2        2
                                            =   +      +    .                         (7.16)
                                            0   shot,0  thermal,0  sp−sp
            Using Eqs. (5.72), (5.76), and (6.87), we find
                                                 4k Tf e   2 2
                                                   B
                                       2
                                       = 2qI f +     + 2R   (2f − f ) f .                 (7.17)
                                       0     0 e   R         ASE  o   e  e
                                                    L
            Similarly, the mean and variance for bit ‘1’ are
                                              I = RGP + 2R   f ,                            (7.18)
                                               1      in    ASE o
                                         2
                                         =  2  +  2  +  2  +  2  .                  (7.19)
                                         1   shot,1  thermal,1  s−sp  sp−sp
            Using Eqs. (5.72), (5.76), (6.83), and (6.87), we find
                                           4k Tf
                                2            B  e    2
                                = 2qI f +      + 2R   [2P  f +   (2f − f )f ].          (7.20)
                                1     1 e   R          ASE  out e  ASE  o   e e
                                              L
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