Page 403 - Fiber Optic Communications Fund

384 Fiber Optic Communications

From Eq. (8.341),
n ASE = n + n , (8.348)
1
2
ASE
= for |f − f | < B ∕2
IF
o
4
n ASE
ASE
= for |f + f | < B ∕2
IF
o
4
= 0 otherwise, (8.349)
where we have used the fact that n and n are statistically independent random processes. Since the signal
2
1
spectrum is centered around f IF extending from f − B ∕2to f + B ∕2, n ASE may be approximated as a
o
o
IF
IF
white noise process over the band of interest,
ASE
= . (8.350)
4
n ASE
The PSD of n (t)∕2RA LO is
d

n d shot,eff
= . (8.351)
2 2
2
4R A 4R A
LO LO
Combining Eqs. (8.350) and (8.351), we find the PSD of n (t) is
het
N het ASE shot,eff
0
= = + . (8.352)
n het 2 2
2 4 4R A
LO
Example 8.6
−9
To reach an error probability of 10 , find the mean number of signal photons required in a shot noise-limited
coherent communication system based on OOK for the following cases: (i) balanced homodyne receiver; (ii)
balanced heterodyne receiver. Assume quantum efficiency, = 1.
Solution:
(i) Let us first consider the case of the homodyne receiver with OOK. From Eq. (8.113), we have
(√ )
1 E av
P = erfc . (8.353)
b
2 2N homo
0
For a shot noise-limited system, the PSD of ASE can be ignored. From Eq. (8.86), we have
q hf
homo
N = = . (8.354)
0 2R 2
The mean number of signal photons is
E av
N = . (8.355)
s
hf