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Optical Amplifiers 289
Using Parseval’s theorem, the mean noise power is
T∕2
1 2
N = lim < I (t) > dt
s−sp T→∞ T ∫ F
−T∕2
∞
1 2
̃
= lim < |I (f)| > df. (6.216)
T→∞ T ∫ F
−∞
From Eq. (6.215), we have
2
̃
2
2
2 ̃
2
2
2
< |I (f)| > = R |H (f)| [| | < |̃n (f)| > +| | < |̃n (−f)| >
F
F
out
e
out
F
∗
∗
2
∗
+( ) <̃n (f)̃n (−f) > + 2 <̃n (−f)̃n (f) >]. (6.217)
out F F out F F
The contribution of the third and fourth terms on the right-hand side of Eq. (6.217) is zero. This can be shown
by writing
n (f)= |n (f)| exp[i (f)], (6.218)
F
F
F
out = | | exp[i ], (6.219)
out
out
∗ 2
( ) <̃n (f)̃n (−f) > = < |̃n (f)||̃n (−f)| exp(i[ (f)+ (−f)− 2 ] >)
out F F F F F F out
2 ∗ ∗
<̃n (−f)̃n (f) > = < |̃n (f)||̃n (−f)| exp(i[− (f)− (−f)+ 2 ]) >,
out F F F F F F out
(6.220)
∗ 2 2 ∗ ∗ 2
( ) <̃n (f)̃n (−f) > +
out F F out <̃n (−f)̃n (f) > = 2| | |̃n (f)||̃n (−f)|
out
F
F
F
F
< cos [ (f)+ (−f)− 2 ] >
F F out
= 0, (6.221)
since (f) is a random variable with uniform distribution in the interval [0, 2]. By definition, the power
F
spectral densities are
2
< |̃n(f)| >
= lim , (6.222)
ASE
T→∞ T
2
< |̃n (f)| > 2
F
̃
= lim = |H (f)| , (6.223)
n F ASE opt
T→∞ T
̃
2
< |I (f)| >
F
= lim . (6.224)
I F
T→∞ T
Using Eqs. (6.217), (6.223), and (6.224), we find
2 ̃ 2 2
= 2R |H (f)| | |
I F e out n F
2 2
̃
= 2R P |H (f)| , (6.225)
out ASE eff
where
̃
̃
̃
H (f)= H (f)H (f) (6.226)
eff opt e