Page 406 - Fiber Optic Communications Fund

Performance Analysis 387

or
()
n het,I
()= . (8.380)
n
4
Since
()= 2 = N het (8.381)
n het,I n het 0
then
N 0 het
()= . (8.382)
n
4
Using Eq. (8.7 ), the variance of noise after the filter is

N het 1 ∞
0
2
2
= |H ()| d. (8.383)
4 2 ∫ −∞ s
Using Eq. (8.367) in Eq. (8.383), we obtain
2
het
= N E . (8.384)
1
0
Using Eqs. (8.369) and (8.384) in Eqs. (8.28) and (8.27), we find
4E 2 1 4E 1
= = , (8.385)
het
N E 1 N het
0 0
(√ )
1 E av
P = erfc . (8.386)
b
2 2N het
0

Exercises

8.1 The transmitted signal is
( )
t − T ∕2
B
x(t)= A rect , (8.387)
T B
where
rect(x)= 1if |x| < 1∕2
= 0 Otherwise (8.388)
Show that the filter matched to the transmitted signal is an integrator with the limits of integration
from 0 to T (integrate-and-dump filter).
B
8.2 Explain the meaning of a matched filter.
8.3 In a 25-Gb/s homodyne fiber-optic system operating at 1530 nm, the PSD of the ASE at the receiver
ASE is 7.78 × 10 −16 W/Hz. Find the average signal power required at the receiver to reach the BER