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Transmission System Design                                                         311






                                   10 –5
                                          –9
                                         10
                                                           3 dB
                                  BER  10 –10
                                                                 OOK
                                                     PSK
                                   10  –15



                                   10  –20
                                      52   51   50   49   48   47   46   45
                                                Av. received power (dBm)

                      Figure 7.10 BER as a function of the mean received power. Thermal noise is ignored.



           Example 7.2
           In a 1.55-μm coherent fiber-optic system as shown in Fig. 7.7, the parameters are as follows. Mean fiber
           launch power = 1 dBm, fiber loss = 0.2 dB/km, fiber length = 240 km, quantum efficiency  = 0.7, T = 290 K,
           R = 100 Ω, P LO  = 10 dBm, and B = 7.5 GHz. Calculate exact and approximate Q-factor if the signal is (a)
                                      e
            L
           OOK, (b) PSK. Assume that the NRZ format with rectangular pulses is used.
           Solution:
           The mean frequency
                                           c    3 × 10 8
                                       f =  =            = 193.54 THz.                      (7.64)
                                            1.55 × 10 −6
           The responsivity is given by Eq. (5.17),

                                      q       0.7 × 1.602 × 10 −9
                                   R =   =          −34           12  A/W.
                                      hf   6.626 × 10  × 193.54 × 10
                                    = 0.874 A/W.                                            (7.65)
           (a) For OOK, the peak power is twice the average power P when the NRZ with rectangular pulses is used:
                                                         in
                                                 P = 2P .                                   (7.66)
                                                  in
                                                        in
           Eq. (7.66) may be rewritten in dB units as
                                        P (dBm)= 10 log 2 + P (dBm)                         (7.67)
                                                       10
                                         in
                                                             in
                                                = 4dBm.
           The received peak power is
                                              P = P exp(−L),                              (7.68)
                                               1r   in
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