Page 309 - Fiber Optic Communications Fund

290 Fiber Optic Communications

is the transfer function of an effective filter, which is the cascade of optical and electrical filters. From
Eqs. (6.216), (6.224), and (5.79), we have
∞
N = 2 R = R (f)df. (6.227)
s−sp s−sp L L ∫ I F
−∞
Therefore,
∞
2
2
̃
2
s−sp = 2R P ∫ |H (f)| df
out ASE
eff
−∞
2
= 4R P B , (6.228)
out ASE eff
where B eff is the effective bandwidth of the filter obtained by cascading the optical and electrical filters.
Let us consider two limiting cases of Eq. (6.228). When the electrical filter bandwidth is much larger than the
̃
̃
̃
optical filter bandwidth, i.e., H (f)≅ 1, from Eq. (6.226), we have H (f)= H (f) and Eq. (6.228) reduces to
e
opt
eff
̃
Eq. (6.42). When the optical filter bandwidth is much larger than the electrical filter bandwidth, i.e., H (f)≅
opt
̃
̃
1, H (f)= H (f), and Eq. (6.228) becomes
e
eff
2 2
= 4R P B , (6.229)
s−sp out ASE e
where ∞
1 2
̃
B = 2 ∫ −∞ |H (f)| df. (6.230)
e
e
When the optical filter is an ideal band-pass filter with full bandwidth f and the electrical filter is an ideal
o
low-pass filter with bandwidth f , it is easy to see that
e
B eff = min(f , f ∕2). (6.231)
o
e
Example 6.9
An amplifier has an output signal power of 0 dBm and a noise power per polarization per unit frequency
interval (single-sided) of −126 dBm/Hz. Calculate the variance of the signal–ASE beat noise current. Assume
B = 20 GHz, B ≫ B , and R = 0.9A/W.
o
o
e
Solution:
P out = 10 P out (dBm)∕10 = 1mW.
The noise power per polarization per unit frequency interval is the single-sided power spectral density,
ASE,sp = 10 ASE (dBm/Hz)∕10 mW/Hz
= 1.58 × 10 −13 mW/Hz.
From Eq. (6.17), we have = 1.58 × 10 −13 mW/Hz. Noise power P = B = 3.17 mW.
ASE ASE ASE o
The variance of the signal–ASE beat noise current is
2 2 −9 2
= 2R P P = 5.13 × 10 A .
s−sp out ASE