Page 488 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 469

Example 10.6
A rectangular pulse of peak power 6 mW is transmitted over a dispersion-free fiber of length 40 km. Find the
nonlinear phase shift at the center of the pulse. Compare the exact results with those obtained using first- and
−1
second-order perturbation theory. Assume fiber loss = 0.2 dB/km and = 1.1W −1 km .
Solution:
In the absence of dispersion, the field envelope evolution is given by the NLSE in lossless form (Eq. 10.246),
u −Z 2
i =−e |u| u. (10.317)
Z
Let
i
u = Ae . (10.318)
Substituting Eq. (10.318) in Eq. (10.317), we find

dA i d i −Z 2 i
i e − Ae =−e |A| Ae . (10.319)
dZ dZ
Comparing the real and imaginary parts of Eq. (10.319), we find
dA
= 0, A = constant, (10.320)
dZ
d −Z 2
= e |A| , (10.321)
dZ
Z
2 −Z
(T, Z)− (T, 0)= |A| e dZ
∫
0
2
= |A| Z (10.322)
eff,
1 − e −Z
Z = . (10.323)
eff

Note that A does not change as a function of Z, but it may depend on T. Since |u(T, Z)| = A does not change
with Z,wehave
|u(T, 0)| = |u(T, Z)| = A. (10.324)
Using Eq. (10.324) in Eqs. (10.322) and (10.318), we find
2
(T, Z)= (T, 0)+ |u(T, 0)| Z , (10.325)
eff

2
u(T, Z)= |u(T, 0)|e i[(T,0)+|u(T,0)| Z eff ] (10.326)
2
= u(T, 0)e i|u(T,0)| Z eff . (10.327)
From Eqs. (2.177) and (10.323), we have
0.2 −1 −2 −1
= km = 4.605 × 10 km , (10.328)
4.343
1 − exp (−4.605 × 10 −2 × 40)
Z = = 18.27 km. (10.329)
eff −2
4.605 × 10