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486                                                               Fiber Optic Communications



                            Table 10.3  FWM tones on the central channel with the degeneracy factor.
                            Tone number   Number of tones  j      k      l     Type

                            1                   2          −2      1     −1     ND
                            2                   2          −2      2      0     ND
                            3                   2          −1      1      0     ND
                            4                   2          −1      2      1     ND
                            5                   1           1      1      2      D
                            6                   1          −1     −1     −2      D




            Example 10.11

            For a single-span dispersion-free fiber, find the nonlinear distortion up to second order using the perturbation
            theory.

            Solution:
            When  = 0, from Eq. (10.252), we have
                   2
                                               du 1    2      2
                                              i    =−a (Z)|u | u .                          (10.465)
                                                            0
                                                               0
                                               dZ
                                  2
            For a single-span system, a (Z)= exp (−Z). From Eq. (10.255), we have u = k. Integrating Eq. (10.465)
                                                                          0
            and using u (T, 0)= 0, we obtain
                      1
                                                              2
                                               u (T, Z)= iZ |k| k,                          (10.466)
                                                1
                                                          eff
            where
                                                    1 − exp (−Z)
                                               Z eff  =         .                           (10.467)
                                                         
            From Eq. (10.253), we have
                                       du
                                         2                  4         4
                                          = i exp (−Z)(2iZ |k| k − iZ |k| k)
                                                         eff
                                                                  eff
                                       dZ
                                            exp (−Z)− exp (−2Z)  4
                                          =                     (−|k| k).                   (10.468)
                                                      
            Integrating Eq. (10.468), we obtain
                                            [                           ]
                                             1 − exp (−Z)  1 − exp (−2Z)  4
                                  u (T, Z)=−             −               |k| k
                                   2               2              2
                                                              2
                                              4
                                            |k| k  2
                                        =−      Z .                                         (10.469)
                                             2   eff
            The total solution up to second order is
                                                      2
                                        u = u + u +  u 2
                                                  1
                                             0
                                             (               2  4  2  )
                                                             |k| Z eff
                                                     2
                                          = k 1 + i|k| Z  −         .                     (10.470)
                                                       eff
                                                               2
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