Page 336 - Fiber Optic Communications Fund

Transmission System Design 317

In this section, we assume ideal in-line amplifiers. If EDFAs are used as in-line amplifiers, the saturation
power of each successive amplifier has to be increased slightly to compensate for the gain saturation caused
by the build-up of the ASE [5]. The optical signal-to-noise ratio is defined as (Eq. (6.107))
mean signal power
OSNR = , (7.109)
mean noise power in a bandwidth of B opt

where B opt is the reference bandwidth, typically chosen to be 12.5 GHz. The noise power in both polarizations
is twice that given by Eq. (7.107). Using Eqs. (7.103) and (7.107), we find the OSNR at the receiver to be

P in
OSNR = . (7.110)
2Nn hf(G − 1)B opt
sp
Sometimes, it is convenient to express OSNR in dB units. Assuming G ≫ 1 and noise figure F ≅ 2n , and
sp
n
dividing the numerator and denominator of Eq. (7.110) by 1 mW, it can be written in dB units as
( )
hfB opt
OSNR(dB)= P [dBm]− N[dB]− G[dB]− F [dB]− 10 log 10
in
n
1mW
= P [dBm]− N[dB]− G[dB]− F [dB]+ 58, (7.111)
in n
wherewehaveused f = 194 THz.

7.4.1 Equivalent Noise Figure
Fig. 7.14(a) shows a two-stage amplifier with loss element, such as a dispersion compensation module
between the two stages. Fig. 7.14(b) shows the equivalent amplifier with gain G eq and noise figure F n,eq .
From Fig. 7.14(a), it is easy to see that
P out = G HG P . (7.112)
1
2 in
Therefore,
G eq = G HG . (7.113)
2
1
The ASE PSD of the amplifier j is given by Eq. (6.102),
=(G F − 1) hf∕2, j = 1, 2. (7.114)
ASE,j j n, j
The noise power per polarization due to the amplifier 1 in a bandwidth of Δf is

N = Δf. (7.115)
1 ASE,1

(a) Two-stage amplifier (b) Equivalent amplifier

Figure 7.14 Two-stage amplifier with a loss element in between. (a) Two-stage amplifier and (b) Equivalent amplifier.