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444                                                               Fiber Optic Communications


                                   1



                                  0.8

                                XPM efficiency  0.6



                                  0.4


                                  0.2


                                   0
                                    0          2          4         6          8
                                               Walk-off parameter, |d p | (ps/km)

                                                                                      −1
            Figure 10.13  XPM efficiency versus absolute of the walk-off parameter. Parameters:  = 0.046 km , fiber length =
            80 km, modulating frequency Ω∕2 = 5 GHz.

            pump and the probe decreases and the XPM efficiency decreases. Fig. 10.13 shows the FWM efficiency as a
            function of the absolute walk-off parameter. From Eq. (10.147), we see that as the channel spacing increases,
            the walk-off increases and the XPM efficiency decreases. In other words, in a WDM system, the impact of the
            XPM due to the nearest-neighbor channels is the greatest. An arbitrary pump may be written as a superposition
            of sinusoids of the form given by Eq. (10.168), and the total XPM-induced phase shift can be calculated by
            adding terms of the form given by Eq. (10.172) due to each frequency component.



            Example 10.4
            A pump is sinusoidally modulated with modulating frequency 10 GHz. The fiber-optic system has the follow-
                                                 −1
            ing parameters: loss coefficient  = 0.046 km , length L = 50 km, dispersion coefficient D = 17 ps/nm⋅km,
                                          2
            and dispersion slope S = 0.06 ps/nm /km. The signal wavelength is 1550 nm and the pump wavelength is
            1549.6 nm. Calculate the XPM efficiency.
            Solution:
            From Eqs. (2.216) and (2.202), we have
                                                  (  2  ) 2     3
                                                            
                                             = S        + D     ,                         (10.175)
                                             3
                                                               2 2
                                                   2c      2 c
                                                           2
                                                   =−D     ,                              (10.176)
                                                   2
                                                         2c
                                                               2
                                               D = 17 × 10 −6  s∕m ,                        (10.177)
                                                               3
                                                          3
                                               S = 0.06 × 10 s∕m ,                          (10.178)
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