Page 386 - Fiber Optic Communications Fund

Performance Analysis 367

Eq. (8.248) becomes
( 2 ) ∞ ( ′ 2 ′ 2 ) ( ′ ′ )
1 E 1 ′ r + E ∕4 r E ′
1
1
1 1
P(0|‘1’ sent)= exp − r exp − I 0 dr 1
1
2 2 16 2 ∫ 0 2 2 2 2
F F F
( 2 ) ∞
1 E 1
= exp − p (r |‘1’ sent)dr 1
1
2 16 2 ∫ 0 r 1
F ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟
=1
( het )
1
= exp − . (8.251)
2 4
Owing to the symmetry of the problem, P(1|‘0’ sent) is the same as P(0|‘1’ sent). Therefore,
( het )
1
P = P(0|‘1’ sent)= exp − . (8.252)
b
2 4
8.4.6 Comparison of Modulation Schemes with Heterodyne Receiver
Fig. 8.19 shows the error probability as a function of het for various modulation schemes with the heterodyne
receiver. First consider the synchronous detection. OOK requires 3 dB more het (equivalently 3 dB more
power or 3 dB less noise) than PSK to reach the same BER. FSK outperforms OOK by roughly 0.85 dB. Next
consider the asynchronous detection. OOK performs slightly better than FSK. However, from the practical

10 0

OOK ASYNC
10 *2
FSK ASYNC

10 *4 OOK SYNC
P b

10 *6 FSK SYNC

10 *8
PSK SYNC

10 *10
*10 *5 0 5 10 15 20
γ het (dB)

het
Figure 8.19 Error probability in heterodyne receiver as a function of . SYNC = synchronous detection.
ASYNC =asynchronous detection.