Page 281 - Fiber Optic Communications Fund

262 Fiber Optic Communications

Example 6.4
An optical amplifier at 1550 nm has a noise figure of 4.5 dB. The signal output of the amplifier is 0 dBm,
which is incident on a photodetector. Calculate the amplifier gain if the standard deviation of the signal–ASE
beat noise current is 0.066 mA. Assume R = 0.9A/W, B = 7.5 GHz, and the optical filter is absent.
e
Solution:
P (dBm)= 0dBm,
out
P out = 10 P out (dBm)∕10 mW = 1mW.

From Eq. (6.54), we have
2
s−sp (0.066 × 10 )
−3 2
ASE = = W/Hz
9
2
2
4R B P 4 × 0.9 × 7.5 × 10 × 1 × 10 −3
e out
= 1.79 × 10 −16 W/Hz,
F = 10 F n (dB)∕10 = 2.818,
n
c
f = = 193.55 THz.
0

Using Eq. (6.102), we find
( ) ( )
1 2 ASE 1 2 × 1.79 × 10 −16
G = + 1 = + 1 = 992.
F hf 2.818 6.626 × 10 −34 × 193.55 × 10 12
n 0

6.5.6 Optical Signal-to Noise Ratio
The noise added by an amplifier is characterized by the noise figure, which is the ratio of electrical SNRs
at the input and output of the amplifier. The noise added by the amplifier may also be characterized by the
optical signal-to-noise ratio (OSNR), defined as
mean signal power
OSNR = . (6.107)
mean noise power in a bandwidth of 0.1 nm
At 1550 nm, 0.1 nm corresponds to B = 12.49 GHz and the mean noise power in the bandwidth of B is
opt opt
P = 2 B , (6.108)
ASE ASE opt
P out
OSNR = . (6.109)
P ASE
Or in decibels,
OSNR (dB) = 10 log OSNR, (6.110)
10
The factor 2 is introduced in Eq. (6.108) to account for two polarizations. Note that B is not the same as
opt
the effective bandwidth of the optical filter B defined in Eq. (6.22). B is a reference bandwidth used in the
o opt
definition of OSNR, which may or may not be equal to B .
o