Page 480 - Fiber Optic Communications Fund

Nonlinear Effects in Fibers 461

Taylor series expansion of the exponential function in Eq. (10.261) yields
( 2 2 )
Z 4
2
u(T, Z)= 1 + i|k| Z − |k| +··· k. (10.262)
2!
Comparing Eqs. (10.260) and (10.262), we see that they match up to the first-order term in .Ifwesolve
Eq. (10.253) and add the second-order correction in Eq. (10.260), we find that Eqs. (10.260) and (10.262)
would match up to second order in (see Example 10.11).
Now, let us consider a more general problem in which neither nor is zero. Let the optical field envelope
2
at the fiber input be
∞
∑
q(T, 0)= a(0)u(T, 0)= b f(T − nT ), (10.263)
n s
n=−∞
where T is the symbol interval and f(T) is the pulse shape. For systems based on OOK,
s
{
1for ‘1’,
b = (10.264)
n
0for ‘0’.
For systems based on PSK or DPSK,
{
1 for ‘1’,
b = (10.265)
n
−1 for ‘0’.
Eq. (10.251) can be solved using the Fourier transform technique (see Chapter 2). The solution is

2
u (T, Z)=  −1 {̃u (, 0) exp [i S(Z)∕2]}, (10.266)
0
0
where ̃u (, 0)= [u (T, 0)], u (T, 0)= u(T, 0), and S(Z) is the accumulated dispersion
0
0
0
Z
S(Z)= (x)dx. (10.267)
∫ 2
0
Taking the Fourier transform of Eq. (10.252), we obtain
d̃u 1 (Z) 2 2
2
̃
i + ̃u (, Z)=−a (Z)F(, Z), (10.268)
1
dZ 2
̃
2
where ̃u (, Z)= [u (T, Z)] and F(, Z)= [|u (T, Z)| u (T, Z)]. Eq. (10.268) is a first-order ordinary dif-
1
0
0
1
ferential equation which can be solved to yield
L tot
̃
2
̃ u (, L )IF(L )= i ∫ a (x)F(, x)IF(x)dx, (10.269)
1
tot
tot
0
where the integrating factor is
[ 2 ]
IF(Z)= exp iS(Z) ∕2 , (10.270)
and L is the total transmission distance. We assume that the dispersion is fully compensated either in the
tot
optical or the electrical domain before the decision device. So, S(L )= 0 and Eq. (10.269) becomes
tot
L tot
2
̃
̃ u (, L )= i ∫ a (x)F(, x)IF(x)dx. (10.271)
1
tot
0