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Transmission System Design                                                         305


                                   10  0



                                                 w/o preamp.
                                   10  –5
                                                             with preamp.
                                 BER  10  –10     10 –9




                                  10  –15



                                  10  –20
                                     120    140    160     180    200    220
                                                   Fiber length (km)
           Figure 7.6  BER as a function of fiber length with and without preamplifier. P = 2dBm, n = 1.5, and G = 20 dB.
                                                                     in        sp
           Other parameters are the same as in Fig. 7.4.

           Using Eqs. (7.15), (7.17), (7.18), and (7.20) in Eqs. (7.8) and (7.14), the BER is calculated and shown in
           Fig. 7.6. Using a preamplifier of 20-dB gain, the maximum transmission distance at a BER of 10 −9  is about
           200 km when the fiber launch power P = 2 dBm. Note that for the same launch power, the maximum trans-
                                         in
           mission distance is limited to about 130 km when no preamplifier is used. It can be increased to about 140 km
           if the launch power is 4 dBm (see Fig. 7.4). Here, we have ignored the fiber nonlinear effects. In the pres-
           ence of fiber nonlinearity, as the launch power increases, nonlinear distortions limit the maximum achievable
           transmission distance (see Chapter 10).



           Example 7.1

           In the fiber-optic system of Fig. 7.1, it is desired that Q ≥ 6 at the receiver. Fiber loss coefficient
                                                                                              ∘
            = 0.046 km −1  and length = 130 km. Find the lower limit on the transmitter power. Assume T = 25 C,
           R = 50 Ω, R = 1 A/W, and B = 7 GHz.
            L                     e
           Solution:
           From Eq. (7.9), we have
                                                     RP 1r
                                            Q =            √ ,                              (7.21)
                                                √
                                                  aP + b +   b
                                                     1r
                                             P = P exp (−L),                              (7.22)
                                                    in
                                               1r
                                     a = 2qRB ,                                             (7.23)
                                            e
                                                                   2
                                                                 9
                                      = 2 × 1.602 × 10 −19  × 1 × 7 × 10 A ∕W
                                                   2
                                      = 2.24 × 10 −9  A ∕W,
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