Page 469 - Fiber Optic Communications Fund
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450                                                               Fiber Optic Communications


            Eq. (10.218) is a first-order ordinary differential equation. The integrating factor is e Z∕2 . So, multiplying Eq.
            (10.218) by e Z∕2 , we find
                                       d( e Z∕2 )
                                          n               −(−iΔ jkln )Z+iΔ jkl
                                                = iA A A e           .                    (10.220)
                                                     j k l
                                          dZ
            Integrating Eq. (10.220) from 0 to L with the condition  (0)= 0, we obtain
                                                          n
                                                              L
                                     (L)= iA A A e −L∕2+iΔ jkl  e −(−iΔ jkln )Z dZ
                                     n       j k l         ∫
                                                             0
                                              [1 − e − jkln L ]
                                         = K jkl        ,                                   (10.221)
                                                  jkln
            where
                                             K jkl  = iA A A e iΔ jkl −L∕2            (10.222)
                                                     j k l
            and
                                                  =  − iΔ  .                          (10.223)
                                                 jkln       jkln
            The power of the FWM component is [23, 24]

                                                         2
                                                     |K | |1 − e − jkln L 2
                                                                   |
                                                       jkl
                                                  2
                                      P      = | | =               e −L
                                        FWM,n   n              2
                                                          | jkln |
                                                      2
                                                             2
                                                   =  P P P L   e −L ,                (10.224)
                                                        j k l eff jkln
                                                         2
                                               2
                                               + 4e −L sin (Δ jkln L∕2)∕L 2 eff
                                          =                          ,                    (10.225)
                                         jkln         2        2
                                                      +(Δ jkln )
                                                          2
                                                    P = A ,                                 (10.226)
                                                     j    j
                                                    1 − exp (−L)
                                               L  =             .                           (10.227)
                                                eff
                                                         
            Here,  jkln  represents the FWM efficiency. Fig. 10.16 shows the dependence of the efficiency on the dispersion
            coefficient  when j = 1, k = 2, and l = 3. When  = 0, the efficiency is maximum and this is known as
                      2
                                                      2
            phase matching.As | | increases, the FWM efficiency decreases and it becomes significantly smaller when
                              2
                     2
            | | > 6ps /km. When the fiber is sufficiently long, the second term in Eq. (10.225) may be ignored and
              2
            Eq. (10.225) may be approximated as
                                                          2
                                                jkln  ≅       .                           (10.228)
                                                      2
                                                      +(Δ  ) 2
                                                            jkln
            Let the channel spacing be Δf and Ω = j2Δf, j =−N∕2, −N∕2 + 1, … , N∕2 − 1. Now, Eqs. (10.209) and
                                          j
            (10.216) become
                                                  j + k − l = n,                            (10.229)
                                                         2
                                            Δ  =(2Δf)  [nl − jk].                     (10.230)
                                               jkln        2
            With j = 1, k = 2, and l = 3, we find n = 0 and
                                                          2
                                              1230  ≅   2       .                         (10.231)
                                                     2
                                                     + 4 (2Δf) 4
                                                          2
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