Page 268 - Fiber Optic Communications Fund

Optical Amplifiers 249

The noise power per unit frequency interval is the power spectral density (PSD), which is given by
P ASE,dp
ASE,dp = = 2n (G − 1)hf . (6.8)
0
sp
Δf
Note that the PSD given by Eq. (6.8) is single-sided, i.e., the frequency components are positive. The power
spectral density is constant over the bandwidth Δf ≪ f , and the ASE can be considered as a white noise
0
process. The single-sided PSD per polarization is
ASE,sp = n hf (G − 1). (6.9)
0
sp
Example 6.1

An optical amplifier operating at 1550 nm has a one-sided ASE power spectral density of 5.73 × 10 −17 W/Hz
in both polarizations. Calculate the gain G. Assume n = 1.5.
sp
Solution:
From Eq. (6.8), we have
ASE,dp = 2n (G − 1)hf ,
sp
0
c 3 × 10 8
f = = = 193.55 THz,
0
1550 × 10 −9
ASE,dp
G = + 1
2n hf 0
sp
5.73 × 10 −17
= + 1
2 × 1.5 × 6.626 × 10 −34 × 193.55 × 10 12
= 150.

6.4 Low-Pass Representation of ASE Noise
The complex ASE noise field in a single polarization may be written as

(t)= n(t) exp[−i(2f t)], (6.10)
n
0
where n(t) is the slowly time-varying field envelope of noise. Taking the Fourier transform of Eq. (6.10),
we find
̃
ñ(f)= (f − f ). (6.11)
n 0
Note that (t) is a band-pass noise process, and n(t) is its low-pass equivalent. Fig. 6.2(a) and 6.2(b) shows
n
̃
the absolute of the Fourier transform of (t) and n(t), respectively. As can be seen, (f) occupies a spectral
n
n
region f − B ∕2 ≤ f ≤ f + B ∕2 and ñ(f) is band-limited to B ∕2. Let us first consider the ASE noise as a
0
0
0
o
o
band-pass process with the single-sided PSD given by Eq. (6.9),
ASE,sp = n hf (G − 1). (6.12)
sp
0
The noise power in a bandwidth of B is
o
N = ASE,sp o (6.13)
B .