Page 349 - Fiber Optic Communications Fund

330 Fiber Optic Communications

5.01
F n,eq = 3.54 + = 3.56,
39.81 × 5.01
F (dB)= 5.52 dB.
n,eq
Thus, the F of the equivalent amplifier is roughly the same as that of Amp2. The net gain is
n
G eq = G HG ,
1
2
G (dB)= G (dB)+ H(dB)+ G (dB)
eq
1
2
= 16 − 7 + 8dB
= 17 dB.

If we choose Amp1 as the first amplifier, from Eq. (7.122) we find
F n,2
F n,eq = F n,1 +
G H
1
3.54
= 5.01 +
6.309 × 5.52
= 5.11,
F (dB)= 7.08 dB.
n,eq
In this case, the equivalent noise figure is roughly the same as that of Amp1. Therefore, the optimum config-
uration is the one in which the first amplifier is Amp2.

Example 7.8
In the presence of ASE noise and shot noise, show that the Q-factor of a fiber-optic system based on PSK
with balanced detection is given by
2N s
2
Q = , (7.169)
N + 1∕2
n
where N and N are the mean number of signal photons and noise photons, respectively. Assume = 1.
s n
Solution:
For PSK, I =−I . Ignoring the thermal noise, Eq. (7.158) is modified as
1
0
√
2R P LO 1r
P
Q PSK = √ . (7.170)
2
2R P LO ASE + 2qB RP LO
P
e
Since R = q∕hf and = 1, Eq. (7.170) becomes
√
2P 1r
Q PSK = . (7.171)
P + hfB
ASE e