Page 479 - Fiber Optic Communications Fund

460 Fiber Optic Communications

where u (T, Z) is the jth order solution. The nonlinear term in Eq. (10.246) may be written as
j
∞ 2 ∞
| ∑ | ∑
2 | n | n
n
n
|u(T, Z)| u(T, Z)= | u | u . (10.250)
| |
|n=0 | n=0
n
Using Eqs. (10.249) and (10.250) in Eq. (10.246) and separating terms proportional to , n = 0, 1, 2, … ,we
obtain [20]
2
u 0 u 0
2
0
( )∶ i − = 0, (10.251)
Z 2 T 2
2
u 1 u 1 2 2
2
1
( )∶ i − =−a (Z)|u | u , (10.252)
0
0
Z 2 T 2
2
u 2 u 2
2
2
2
2 ∗
2
( )∶ i − =−a (Z)(2|u | u + u u ). (10.253)
1
0
0 1
Z 2 T 2
Eq. (10.251) is the linear Schrödinger equation. The linear solution u (T, Z) can be obtained using the lin-
0
ear fiber transfer function as discussed in Chapter 2. Eqs. (10.252) and (10.253) represent the first- and
second-order corrections due to nonlinear effects. The first (second)-order term u (u ) corresponds to the first
1
2
n
(second)-order echo pulses shown in Fig. 10.22. When the nonlinear effects are small, the terms of order ,
n > 1, can be ignored. As an example, let us consider a single-span lossless ( = 0) zero-dispersion ( = 0)
2
fiber. Eq. (10.251) becomes
du
0
= 0, (10.254)
dZ
u = k(const.). (10.255)
0
2
Note that k maybeafunctionof t. Since a (Z)= 1 in this example, Eq. (10.252) becomes
u 1 2
i =−|k| k, (10.256)
Z
2
u (T, Z)= u (T, 0)+ i|k| kZ. (10.257)
1 1
The initial condition is
2
u(T, 0)= u (T, 0)+ u (T, 0)+ u (T, 0)+··· . (10.258)
0 1 2
Since u (T, 0) are arbitrary functions, one convenient choice would be
j
u (T, 0)= u(T, 0)= k,
0
u (T, 0)= 0for j > 1. (10.259)
j
n
Using Eqs. (10.255)–(10.259) and ignoring terms of order , n > 1 in Eq. (10.249), we obtain
2
u(T, Z)=(1 + i|k| Z) k. (10.260)
In this simple example, Eq. (10.246) can easily be solved directly:
du 2
i =−|u| u,
dZ
2
u(T, Z)= exp [i|u(T, 0)| ]u(T, 0). (10.261)