Page 527 - Fiber Optic Communications Fund

508 Fiber Optic Communications

The impulse response function of the dispersion-compensating filter is given by [16] (see Example 11.3)
√
1
W(t)= exp [i(t)], (11.56)
2i L
2
2
(t)= t ∕2 L. (11.57)
2
A dispersion-compensating filter is an all-pass filter and its impulse response W(t) is infinite in duration. The
summation in Eq. (11.50) may be truncated to a finite number of terms, known as a finite impulse response
(FIR) filter. Now, Eq. (11.50) becomes
K
∑
x[n]= W[k]y[n − k], (11.58)
k=−K
[ ]
√ ik T
2 2
1 samp
W[k]= T exp . (11.59)
samp
2i L 2 L
2 2
Fig. 11.13 shows a schematic of the FIR filter. The number of taps, 2K + 1, has to be decided based on the
Nyquist sampling theorem, which states that if the signal is band-limited to B, the sampling rate, R , has to
samp
be greater than or equal to 2B. Otherwise, aliasing could occur. From Eq. (11.57), the instantaneous frequency
of W(t) is
−1 d −t
f = = . (11.60)
i
2 dt 2 L
2
From Eq. (11.60), we see that the magnitude of instantaneous frequency increases with t. When the summation
in Eq. (11.50) is truncated to 2K + 1 terms (see Eq. (11.58)), the highest-frequency component occurs at
t = KT samp :
KT samp
B = |f | = . (11.61)
i,max
2| |L
2
y[n + K ] y[n + K * 1] y[n * K ]
Delay Delay
T samp
T samp

W [*K ] W [*K + 1] W [K ]
× × X

∑
xˆ(n)

Figure 11.13 Schematic of the FIR dispersion-compensating filter.