Page 404 - Fiber Optic Communications Fund

Performance Analysis 385

Substituting Eqs. (8.354) and (8.355) in Eq. (8.353), we find
1 √
P = erfc( N )
s
b
2
−9
= 10 . (8.356)
When = 1,
√ −1 −9
N = erfc (2 × 10 )
s
= 4.2411 (8.357)
N ≅ 18. (8.358)
s
(ii) From Eq. (8.144), we have
(√ )
1 E av
P = erfc . (8.359)
b het
2 4N
0
From Eq. (8.123), we find
hf
het
N = . (8.360)
0 2
Substituting Eqs. (8.360) and (8.355) in Eq. (8.359), we obtain
1 √ −9
P = erfc( N ∕2)= 10 (8.361)
b s
2
√
√ −1 −9
N = erfc (2 × 10 ) 2 (8.362)
s
N ≅ 36. (8.363)
s

Example 8.7
Show that the error probability in a fiber-optic system based on PSK that uses a heterodyne receiver with
synchronous demodulator and a filter matched to the transmitted signal s(t) (see Fig. 8.14) is given by
(√ )
1 E av
P = erfc . (8.364)
b
2 2N het
0
Solution:
Let the signal output of the synchronous demodulator be
2
x(t)= s(t)cos ( t +Δ)
IF
s(t)
= [1 + cos (2 t + 2Δ)]. (8.365)
IF
2