Page 481 - Fiber Optic Communications Fund
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462                                                               Fiber Optic Communications


            The first-order correction u (T, Z) is obtained by performing the inverse Fourier transformation of ̃u (, Z).
                                  1
                                                                                            1
            Typically, in quasi-linear systems, the nonlinear effects are smaller than the dispersive effects and the
            first-order correction u (T, Z) is often adequate to describe the nonlinear propagation. However, when
                               1
            the transmission distance is long and/or the launch power is large, a second-order perturbation theory is
            needed [20].
              A closed-form expression for u (T, Z) and u (T, Z) can be obtained if we assume that the pulse shape f(T)
                                       0
                                                 1
            is Gaussian, i.e.,
                                                          (     )
                                                  √          T 2
                                            f(T)=   P exp −    2  ,                         (10.272)
                                                     0
                                                             2T
                                                               0
            where P is the peak power. The linear propagation of this pulse is described by (see Eq. (2.158))
                   0
                                               √       [        ]
                                             T 0  P 0       T 2
                                                    exp −        ,                          (10.273)
                                              T (Z)       2T (Z)
                                                            2
                                               1
                                                            1
                   2
                        2
            where T = T − iS(Z). When a long bit sequence is launched to the fiber, u (T, 0) is given by Eq. (10.263).
                   1   0                                                 0
            In this case, the linear solution is
                                               √     ∞       [           ]
                                             T 0  P 0  ∑        (T − nT ) 2
                                                                      s
                                    u (T, Z)=           b exp −           ,                 (10.274)
                                                         n
                                     0
                                                                    2
                                               T 1  n=−∞          2T (Z)
                                                                    1
                                                                ∞
                                                                    ∞
                                                           ∞
                                                     T 0 3  ∑ ∑ ∑
                                                 3∕2
                                      2
                      F(T, Z)= |u (T, Z)| u (T, Z)= P 0  2             b b b
                                        0
                                0
                                                                        l m n
                                                   |T | T
                                                     1   1 l=−∞ m=−∞ n=−∞
                                                     [                                 ]
                                                        (T − lT ) 2  (T − mT ) 2  (T − nT ) 2
                                                                         s
                                                                                    s
                                                              s
                                                × exp −         −           −           .   (10.275)
                                                                                   2 ∗
                                                          2T 2       2T 2       2(T )
                                                            1          1          1
            The Fourier transform of F(T, Z) is (Example 10.13)
                                   P 3∕2 3 √    ∑
                                       T
                                        0
                                    0
                          ̃
                                                                             2
                          F(, Z)=                b b b exp [−g(Z)+[i − d(Z)] ∕4C(Z)],   (10.276)
                                                    l m n
                                      2
                                   |T | T   C(Z)
                                     1  1       lmn
            where
                                                         2
                                                       3T + iS
                                                         0
                                                C(Z)=    4     ,                            (10.277)
                                                             2
                                                      2(T + S )
                                                         0
                                                         2
                                             T [(l + m + n)T + i(l + m − n)S]
                                              s
                                                         0
                                       d(Z)=                            ,                   (10.278)
                                                        4
                                                       T + S 2
                                                        0
                                                                       2
                                                   2
                                              2
                                            2
                                                              2
                                                                   2
                                                       2
                                                          2
                                          T [(l + m + n )T + i(l + m − n )S]
                                           s
                                                         0
                                    g(Z)=                4                .                 (10.279)
                                                             2
                                                      2(T + S )
                                                         0
            Substituting Eq. (10.276) in Eq. (10.269), and after performing the inverse Fourier transformation, we find
                                                   ∑
                                         u (T, L )=   u lmn (T, L )b b b ,                (10.280)
                                              tot
                                                                 l m n
                                                              tot
                                          1
                                                   lmn
                                              2
                                                               2
                                          L tot  a (x) exp [−(2CT − d) ∕(4C(1 + i2SC)− C)]
                    u  (T, L )= iP 3∕2 3                                       dx.        (10.281)
                                     T
                      lmn    tot   0   0  ∫       √
                                         0                     2      4   2
                                                    (1 + i2SC)(T − iS)(T + S )
                                                               0      0
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