Page 375 - Fiber Optic Communications Fund

356 Fiber Optic Communications

(b) For a balanced heterodyne receiver with PSK signal, the error probability is given by Eq. (8.130),
(√ )
1 E av
PSK
P = erfc
b 2 het
2N
0
(√ )
1 3.16 × 10 −15
= erfc
2 2 × 9.617 × 10 −17
−9
= 4.901 × 10 . (8.164)
If the signal is OOK, from Eq. (8.144) we have
(√ )
1 E av
OOK
P = erfc
b 2 4N het
0
−3
= 2.07 × 10 . (8.165)

8.4.3 FSK: Synchronous Detection
To transmit bit ‘1’ (‘0’), the frequency of the optical carrier is shifted by Δ∕2 (−Δ2). The complex field
envelopes corresponding to bits ‘1’ and ‘0’ are
( −iΔt )
s (t)= A exp ,
1
2
( )
iΔt
s (t)= A exp , (8.166)
0
2
for a duration of T . The photocurrents are
b
I = I (t) for bit ‘1’ (8.167)
1
= I (t) for bit ‘0’ (8.168)
0
where
[( Δ ) ]
I (t)= Re[s (t) exp (−i t)] = A cos + t ,
1
1
IF
IF
2
[( Δ ) ]
I (t)= A cos − t . (8.169)
IF
0
2
Here, we have ignored the constant factor 2RA LO and Δ is set to zero. The filter matched to I(t) can be
realized as a correlator, as shown in Fig. 8.15. Replacing x (t) and x (t) of Eq. (8.36) by I (t) and I (t),
1
0
1
0
respectively, we obtain
T b
e
E = I (t)I (t) dt
10 ∫ 1 0
0
A 2 T b
= [cos (Δt)+ cos ( t)] dt
IF
2 ∫ 0
2
A T b
= [sinc(ΔT ∕)+ sinc( T ∕)], (8.170)
b
IF b
2