Page 315 - Fiber Optic Communications Fund

296 Fiber Optic Communications

Example 6.12
The electrical SNRs at the amplifier input and output are 30 dB and 25 dB, respectively. The signal power
at the input and output of the amplifier are −13 dBm and 2 dBm, respectively. Find the ASE power spectral
density per polarization, ASE . Assume that the carrier frequency f = 195 THz.
0
Solution:
SNR in
F =
n
SNR
out
F (dB)=(SNR) (dB)−(SNR) (dB)
n
in
out
=(30 − 25) dB
= 5dB,
F = 10 F n (dB)∕10 = 3.1623.
n
The amplifier gain is
P out
G = ,
P in
G(dB)= P (dBm)− P (dBm)= 2dBm −(−13) dBm = 15 dBm,
in
out
G = 10 G(dB)∕10 = 31.62.
The ASE PSD is given by Eq. (6.102),
hf (GF − 1)
0
n
=
ASE
2
12
= 6.626 × 10 −34 × 195 × 10 ×(31.62 × 3.162 − 1)∕2W/Hz
= 6.39 × 10 −18 W/Hz.

Example 6.13
A Fabry–Perot amplifier has a peak gain G of 20 dB and a single-pass gain G of 5 dB. Calculate the
max s
geometric mean of the facet reflectivity R. Assume R = R .
1 2
Solution:
From Eq. (6.127), we have
√
(1 − R) G
√ s
G = .
max
1 − RG
s