Page 504 - Fiber Optic Communications Fund
P. 504

Nonlinear Effects in Fibers                                                        485





                                       Δ −2,1,−1  = 0.5 + 0.8 −(−0.7) rad
                                               = 2 rad,                                   (10.454)

                                              P(dBm)= 3 dBm,                              (10.455)

                                           P(mW)= 10 P(dBm)∕10  mW
                                                  = 2 mW.                                 (10.456)

           Substituting Eqs. (10.453), (10.454), and (10.456) in Eq. (10.452), we find
                                                                   √
                                     (−2,1,−1,0)                 −4
                                          (L)=(−0.8 + 1.6i)× 10    W.                   (10.457)
                                     0
           Similarly, the FWM fields due to other triplets are
                                                                       √
                                  (−2,2,0,0)         −7          −5
                                      (L)=(3.33 × 10  − 3.92 × 10  i)  W,               (10.458)
                                 0
                                                                   √
                                     (−1,−1,−2,0)                −4
                                           (L)=(2.03 + 1.9i)× 10   W,                   (10.459)
                                     0
                                                                  √
                                      (−1,1,0,0)                −4
                                          (L)=(2.28 − 1.6i)× 10   W,                    (10.460)
                                      0
                                                                   √
                                     (−1,2,1,0)                  −4
                                         (L)=(−1.4 − 1.12 i)× 10   W,                   (10.461)
                                     0
                                                                   √
                                     (1,1,2,0)                   −4
                                        (L)=(1.435 − 2.394 i)× 10  W.                   (10.462)
                                     0
           The total FWM field is
                        = 2 (−2,1,−1,0)  + 2 (−2,2,0,0)  + 2 (−1,1,0,0)  + 2 (−1,2,1,0)  +  (1,1,2,0)  +  (−1,−1,−2)
                        0
                                              √
                         =(3.64 − 3.509 i)× 10 −4  W.                                     (10.463)
           The FWM power at the fiber output is
                                                  2
                                        P     = | | = 2.56 × 10 −4  mW.                 (10.464)
                                         FWM     0
                                Table 10.2 FWM tones on the central channel.
                                  j          k             l            Type
                                −2             1           −1            ND
                                −2             2            0            ND
                                −1           −1            −2             D
                                −1             1            0            ND
                                −1             2            1            ND
                                 1           −2            −1            ND
                                 1           −1             0            ND
                                 1             1            2             D
                                 2           −2             0            ND
                                 2           −1             1            ND
   499   500   501   502   503   504   505   506   507   508   509