Page 327 - Fiber Optic Communications Fund

308 Fiber Optic Communications

2 = 2qB RP , (7.45)
0,shot e LO
2
= 2qB RP + 4k TB ∕R . (7.46)
0 e LO B e L
The Q-factor is calculated as
I − I 0
1
Q = . (7.47)
OOK
+
1 0
Some approximations can be made to Eq. (7.47) to gain some insight. When P ≫ P , from Eqs. (7.42)
LO 1r
and (7.45) we have
2 2
= = 2qB RP . (7.48)
1,shot 0,shot e LO
Let the photocurrent due to LO be I . Eq. (7.48) may be rewritten as
LO
2 2
≡ = 2qI B . (7.49)
shot, eff 1,shot LO e
Comparing Eq. (7.49) with Eq. (5.72), the effective PSD of shot noise in balanced detection is
= qI (7.50)
shot, eff LO
If P LO is sufficiently large, the shot noise will dominate the thermal noise and it may be ignored in Eqs. (7.43)
and (7.46),
2
2
= = 2qB RP . (7.51)
1 0 e LO
Now Eq. (7.47) reduces to
√
2R P P
1r LO
Q OOK = √ ,
2 2qB RP
e LO
√
RP 1r
= . (7.52)
2qB e
Note that the Q-factor is independent of P under these conditions. Using Eq. (5.17), the Q-factor may be
LO
rewritten as
√
P 1r
Q = . (7.53)
OOK
2hfB
e
where f is the mean frequency. The energy of a bit ‘1’ at the receiver is E = P T . If the receiver filter is
1r b
1r
an ideal Nyquist filter (B = 1∕(2T )), Eq. (7.53) becomes
b
e
√
Q = N , (7.54)
OOK 1r
where N = E ∕hf is the number of signal photons of bit ‘1’. For an OOK signal, the mean number of
1r
1r
received photons per bit, N = N ∕2. So, Eq. (7.54) becomes [1]
rec 1r
√
Q OOK = 2N . (7.55)
rec
−9
For an ideal photondetector, = 1. To have a BER of 10 , Q = 6 and from Eq. (7.55), we see that the average
number of signal photons per bit, N , should be 18. In other words, if the mean number of signal photons is
rec