Page 310 - Fiber Optic Communications Fund

Optical Amplifiers 291

Example 6.10
Find an analytical expression for the mean and variance of ASE–ASE beat noise. Assume that the optical filter
is an ideal band-pass filter with bandwidth B and the samples of ASE noise n(t) are identically distributed
o
complex Gaussian random variables. Ignore the electrical filter.

Solution:
From Eqs. (6.21) and (6.28), it follows that
2
B ,
< I sp−sp > = R < |n (t)| >= R ASE o
F
(6.232)
2 2 4
< I > = R < |n (t)| >. (6.233)
sp−sp F
Let
n = n + in . (6.234)
F Fr Fi
With the assumption that the samples of n(t) are identically distributed complex random variables, it follows
that
2
B
< |n | > ASE o
F
2
2
< n > = < n >= = ,
Fr Fi
2 2 (6.235)
< n n > = 0,
Fr Fi
(6.236)
4 2 2 2 4 4 2 2
< |n | > = < (n + n ) >=< n > + < n > +2 < n n >. (6.237)
F Fr Fi Fr Fi Fr Fi
For Gaussian random variables N , N , N , and N , from the moment theorem, we have
1 2 3 4
< N N N N >=< N N >< N N > + < N N >< N N > + < N N >< N N >. (6.238)
1 2 3 4 1 2 3 4 1 3 2 4 1 4 2 3
If we choose N = n , i = 1, 2, 3, 4, using Eqs. (6.235) and (6.236), we find
i Fr
2
2
< n 4 >= 3(< n 2 >) = 3 2 B ∕4. (6.239)
Fr Fr ASE o
Similarly,
4
2
2
2
< n >= 3(< n >) = 3 2 B ∕4. (6.240)
Fi Fi ASE o
If we choose N = N = n Fr and N = N = n , we find
4
Fi
2
3
1
2
2
2
< n n >= 2 B ∕4. (6.241)
Fr Fi ASE o
Substituting Eqs. (6.239)–(6.241) in Eq. (6.237), and using Eq. (6.233), we obtain
2 2
< I 2 sp−sp > = 2R ASE o 2
B ,
(6.242)
2
2
sp−sp = < I 2 sp−sp > − < I sp−sp > ,
(6.243)
2
2 2
sp−sp = R ASE o 2 (6.244)
B .