Page 494 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 475

Substituting Eq. (10.381) in Eqs. (10.379) and (10.380), we find
EmL eff
(mL −, T)= , (10.383)
a
T
b
√
u(mL −, T)= Ep(T) exp [i(mL −)]. (10.384)
a a
Next, let us consider the case when there is only one amplifier located at mL that introduces ASE noise.
a
The optical field envelope after the amplifier is
u(mL +, T)= u(mL −, T)+ n(T). (10.385)
a a
We assume that two degrees of freedom in the noise field are of importance. They are the in-phase component
n and the quadrature component n ; we ignore other noise components. As mentioned in Section 10.10.1,
0i
0r
the noise field is fully described by these two degrees of freedom for a linear system. Gordon and Mollenauer
[51] assumed that these two degrees of freedom are adequate to describe the noise field even for a nonlinear
system. Using Eqs. (10.384) and (10.353) in Eq. (10.385), we find
√
u(mL +, T)= Ep(T) exp [i(mL −)] + n p(T)
0
a
a
√
′
=( E + n )p(T) exp [i(mL −)], (10.386)
0 a
where
′
n = n exp [−i(mL −)]. (10.387)
0 0 a
′
n is the same as n , except for a deterministic phase shift which does not alter the statistical properties, i.e.,
0
0
′
< n >= 0, (10.388)
0
′ ′ ∗
< n n >= , (10.389)
0 0
′ ′
< n n >= 0. (10.390)
0 0
From Eq. (10.386), we see that the complex amplitude of the field envelope has changed because of the
amplifier noise. Using u(mL +, T) as the initial condition, the NLSE (10.372) is solved to obtain the field at
a
the end of the transmission line as
{ }
L tot
2
u(L , T)= u(mL +, T) exp i|u(mL +, T)| 2 ∫ a (Z)dZ
a
a
tot
mL a +
√ √
′
′ 2
=( E + n )p(T) exp [i(mL −) + i| E + n | (N − m)L ∕T ], (10.391)
0 a 0 a eff b
where L tot = N L is the total transmission distance. The phase at L tot is
a a
√
⎧ ′ ⎫ ′ 2
n 0i ⎪ | E + n | (N − m)L eff EmL eff
a
0
⎪
= tan −1 + +
⎨ √ ⎬
′ T T
⎪ E + n ⎪ b b
0r
⎩ ⎭
n ′ 0i √
′
≈ √ + (E + 2 En )(N − m)L ∕T + EmL ∕T . (10.392)
0r a eff b eff b
E