Page 274 - Fiber Optic Communications Fund

Optical Amplifiers 255

The noise power is
∞
2
< |n | > = (f)df
EF ∫ n EF
−∞
= 2 B , (6.52)
ASE e
where
∞
1 2
̃
B = |H (f)| df. (6.53)
e 2 ∫ e
−∞
Substituting Eq. (6.52) in Eq. (6.50), we find
2 2
= 4R P B . (6.54)
s−sp out ASE e
Eq. (6.54) may be rewritten as
2 = 2I I , (6.55)
s−sp out ASE
where I and I are the signal current and noise current in the electrical bandwidth of B , respectively,
out ASE e
given by
I out = RP , (6.56)
out
I ASE = 2R ASE e (6.57)
B .
When the electrical filter is an ideal low-pass filter with cutoff frequency f ,
e
̃
H (f)= 1 for |f| < f e
e
= 0 otherwise, (6.58)
Eq. (6.53) becomes
∞
1
B = df = f . (6.59)
e
e
2 ∫ −∞
Next, let us consider the case in which the bandwidths of optical and electrical filters are comparable. When
the optical filter is an ideal band-pass filter with full bandwidth f and the electrical filter is an ideal low-pass
o
filter with cutoff frequency f , the variance is (see Example 6.8)
e
2
2 s−sp = 4R P B , (6.60)
out ASE eff
B eff = min{f ∕2, f }. (6.61)
o
e
Example 6.2

The ASE PSD of an amplifier ASE is 1.3 × 10 −16 W/Hz. The gain of the amplifier G = 20 dB and the input
power of the amplifier is 10 μW. The output of the amplifier is incident on a photodetector with responsivity
R = 0.8 A/W. Calculate the variance of the signal–ASE beat noise. Assume that the receiver can be modeled
as an ideal low-pass filter with cutoff frequency f = 7 GHz. Ignore the optical filter.
e