Page 542 - Fiber Optic Communications Fund

Digital Signal Processing 523

Example 11.3
Show that the impulse response of a dispersion-compensating filter is
√ [ 2 ]
1 it
W(t)= exp . (11.155)
2i L 2 L
2
2
Solution:
For a Gaussian pulse, we have the following relations:

2
2
[exp (−t )] = exp [−f ], (11.156)
1
̃
[W(at)] = W(f∕a), Re(a) > 0. (11.157)
a
The transfer function of a dispersion-compensating filter is
̃
2 2
W(f)= exp (−i2 f L). (11.158)
2
2
Choosing a = 1∕(2 Li), Eq. (11.158) can be written as
2
2
̃
2
W(f)= exp (−f ∕a ). (11.159)
Using Eqs. (11.156), (11.157), and (11.158), we find
[ 2 ]
2 1 it
W(t)= a exp [−(at) ]= √ exp . (11.160)
2 Li 2 L
2
2

Example 11.4

Find the LMS for the updated weights of the PMD equalizer.
Solution:
Following the notation of Section 11.7, let
e [n]= [n]− ̂ [n], r = x, y, (11.161)
r r,in r,in
∗
∗
2
∗
2
J =< |e [n]| >=< | r,in [n]| − r,in [n] ̂ r,in [n]− r,in [n] ̂ r,in [n]+ ̂ r,in [n] ̂ r,in [n] >. (11.162)
r
r
Using Eq. (11.76), the gradient vectors are
J x ∗ ∗ ∗
2 ∗ = 2 < − x,in [n] x,out [n − k]+ x,out [n − k] ̂ x,in [n] >=−2 < x,out [n − k]e [n] >, (11.163)
x
W [n]
xx
J x ∗
2 =−2 < y,out [n − k]e [n] >. (11.164)
x
∗
W [n]
xy