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Nonlinear Effects in Fibers                                                        429


                               0.1

                              0.08
                              0.06    Blue shift                     Red shift
                            Electrical field (arb. units)  –0.02 0
                              0.04
                              0.02




                             –0.04
                             –0.06

                             –0.08
                              –0.1
                               –1500  –1000    –500     0      500    1000    1500
                                                    Time, T (ps)
           Figure 10.3  The electric field distribution at the fiber output. The broken line shows the field envelope and the rapid
           oscillation shows the actual field.  < 0.
                                    2



           Eq. (10.77) becomes
                                                  2
                                         q    q    2      q
                                               2
                                         i  −       + |q| q =−i  .                       (10.81)
                                         Z   2 T 2          2
           Eq. (10.81) is known as the nonlinear Schrödinger equation (NLSE); it is of significant importance in mod-
           eling fiber-optic transmission systems and can not be solved analytically for arbitrary inputs. Numerical
           techniques such as the split-step Fourier scheme (SSFS) are used to solve the NLSE (see Chapter 11).




           Example 10.1
                                                                                    2
                                                           2
           The Kerr coefficient of a single-mode fiber is 2.5 × 10 −20  m ∕W. Itseffectiveareais80 μm . Find the non-
           linear coefficient  at the wavelength 1550 nm.
           Solution:
           From Eq. (10.78), we have

                                             n  0  n 2f 0  2n 2
                                              2
                                                     2
                                          =     =       =      ,                         (10.82)
                                             cA eff  cA eff   A
                                                             0 eff
                                             2 × 2.5 × 10 −20  −1  −1
                                       =                    W   m
                                         1550 × 10 −9  × 80 × 10 −12
                                                        −1
                                       = 1.26 × 10 −3  W −1  m .                           (10.83)
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