Page 538 - Fiber Optic Communications Fund

Digital Signal Processing 519

scheme. The received field q(t, L)= q (t, 0).Wewishtofind q (t, Δz), which corresponds to q(t, L −Δz).The
b
b
operator M −1 in this propagation step can be approximated as
[ ] [ ]
Δz Δz
M −1 = exp [N (t, z)+ D (t)]dz ≅ exp [N (t, z)]dz exp [D (t)Δz], (11.128)
∫ b b ∫ b b
0 0
[ Δz ]
−1
l
q (t, Δz)= M q (t, 0)= exp ∫ N (t, z)dz q (t, Δz), (11.129)
b
b
b
b
0
where
l
q (t, Δz)= exp [D (t)Δz]q (t, 0). (11.130)
b b b
Eq. (11.130) is equivalent to solving the following equation:
2 l
q l i q
b l 2 b
= D q = , (11.131)
b b
z 2 t 2
l
with q (t, 0)= q (t, 0). To solve Eq. (11.131), we take the Fourier transform on both sides:
b b
l
d̃q (, z) −i 2
b 2 l
= ̃ q (, z), (11.132)
dz 2 b
2
−i Δz
2
l
l
̃ q (, Δz)= ̃ q (, 0). (11.133)
b b
2
l
The signal q (t, Δz) can be obtained by the inverse Fourier transformation
b
l
l
q (t, Δz)=  −1 [̃q (, Δz)]. (11.134)
b b
In other words, the initial spectrum ̃q (, 0) is multiplied by the inverse fiber linear transfer function to obtain
b
q (t, Δz) and, therefore, it represents the inverse linear response of the fiber. Computation of the Fourier
b
2
transform/inverse Fourier transform takes N complex additions/multiplications, where N is the number of
samples. To facilitate fast computation of the Fourier transform, fast Fourier transform (FFT) is used which
takes ∼ Nlog N complex additions/multiplications. Eq. (11.131) may also be solved using the FIR filter
2
approach [22], as discussed in Section 11.6. Next, let us consider the nonlinear operator in Eq. (11.129).
Eq. (11.129) is formally equivalent to the following equation:
q b ( 2 )
= N q = −i|q | + q , (11.135)
b
b
b b
z 2
l
with q (t, 0)= q (t, Δz).Let
b b
q = A exp (i). (11.136)
b
Substituting Eq. (11.136) into Eq. (11.135) and separating the real and imaginary parts, we find
dA
= A, (11.137)
dz 2
d 2
=−|A| . (11.138)
dz
Integrating Eq. (11.137), we obtain
( z )
A(t, z)= exp A(t, 0), (11.139)
2