Page 342 - Fiber Optic Communications Fund

Transmission System Design 323

Proceeding as in Section 6.5.1, Eq. (7.145) is simplified as
2
2
<I >= 2R P P , (7.146)
LO ASE
where P ASE is the mean noise power within the receiver bandwidth. The variance given by Eq. (7.146)
represents the signal–spontaneous beat noise due to the interaction of the LO signal and ASE. Note that
spontaneous–spontaneous beat noise is absent when the balanced coherent receiver is used. We assume that
P LO ≫ P . So the variances of ‘1’ due to ASE, shot noise, and thermal noise are
1r
2 2
= 2R P P , (7.147)
1,ASE LO ASE
2 = 2qB RP , (7.148)
1,shot e LO
2
= 4k TB ∕R , (7.149)
1,thermal B e L
2
= 2 + 2 + 2 . (7.150)
1 1,ASE 1,shot 1,thermal
Similarly, for bit ‘0’, we have
2 = 2 , (7.151)
0,ASE 1,ASE
2 2
= , (7.152)
0,shot 1,shot
2 = 2 , (7.153)
0,thermal 1,thermal
2 2
= . (7.154)
0 1
The mean currents of bit ‘1’ and ‘0’ are
√
I = 2R P P , (7.155)
1 LO 1r
I = 0. (7.156)
0
The Q-factor is
I 1
Q = (7.157)
OOK
2
1
√
R P P
LO 1r
= √ . (7.158)
2
2R P P + 2qB RP + 4k TB ∕R
LO ASE e LO B e L
For a long-haul fiber-optic system, the ASE noise due to amplifier chains is dominant. Hence, the shot noise
and thermal noise can be ignored in Eq. (7.157) to obtain
√
P 1r
Q OOK = . (7.159)
2P
ASE
For a fiber-optic system with the loss fully compensated by amplifier gain, we have
P = P = 2P . (7.160)
1r
in
in
We assume that the electrical filter bandwidth B is smaller than the optical filter bandwidth, so that (see
e
Eq. (6.52))
eq
B .
P ASE = 2 ASE e (7.161)