Page 337 - Fiber Optic Communications Fund

318 Fiber Optic Communications

The ASE due to amplifier 1 is attenuated by the loss element and, then, it is amplified by the second amplifier.
Therefore, the noise power due to amplifier 1 at the output end is
N out,1 = ASE,1 ΔfHG . (7.116)
2
Similarly, the noise power due to amplifier 2 at the output end is ASE,2 Δf, and the total noise power at the
output is
N =( HG + )Δf. (7.117)
out ASE,1 2 ASE,2
Therefore, the PSD at the output is
N out
= = HG + . (7.118)
ASE,eq ASE,1 2 ASE,2
Δf
Using Eq. (7.114), we obtain
=[G F G H + G F − HG − 1] hf∕2. (7.119)
ASE,eq 1 n,1 2 2 n,2 2
For an equivalent amplifier shown in Fig. 7.14(b), we have

=(G F − 1)hf∕2. (7.120)
ASE,eq eq n,eq
Comparing Eqs. (7.119) and (7.120), we obtain
F n,2 1
F = F + − . (7.121)
n,eq n,1
G H G
1 1
Note that in Eq. (7.121), the noise figure of the second amplifier is divided by the gain of the first ampli-
fier and the loss, H. Therefore, in practice, an amplifier with higher noise figure is used as the second
stage and/or an amplifier with higher gain is used as the first stage. Typically, G ≫ 1 and Eq. (7.121)
1
reduces to
F n,2
F n,eq ≅ F n,1 + . (7.122)
G H
1
If two amplifiers are cascaded without a loss element in between, H = 1 and Eq. (7.122) becomes [4, 6]
F n,2
F n,eq ≅ F n,1 + . (7.123)
G 1
The effective noise figure of a cascaded chain of k amplifiers is given by (Exercise 7.9)
F n,2 − 1 F n,3 − 1 F n,k − 1
F = F + + +···+ . (7.124)
n,eq n,1
G G G G G ··· G
1 1 2 1 2 k−1
7.4.2 Impact of Amplifier Spacing
Consider the long-haul fiber-optic system with identical amplifier gains and noise figures. The power spectral
density of ASE at the output is given by (Eq. (7.108))
eq
= n hf[exp (L )− 1]N, (7.125)
sp
a
ASE,1
where L is the amplifier spacing and G = exp (L ). Let the total transmission distance, L ,be NL.Eq.
a
tot
(7.125) can be rewritten as
L
eq tot
= n hf[exp (L )− 1] . (7.126)
ASE,1 sp a L
a