Page 382 - Fiber Optic Communications Fund

Performance Analysis 363

This equation is satisfied to a close approximation [1]
√
√ 2
E √ 8 F
r = 1 √ 1 + (8.217)
T
2
4 E ∕4
1
√
E 1 8N het
0
= 1 + , (8.218)
4 E
1
2
where we have used Eq. (8.207) for . When a bit ‘0’ is transmitted, if r > r , ‘0’ is mistaken as ‘1’ and this
F T
probability is
( )
∞ 1 ∞ r 2
P(1|0)= p(r|‘0’ sent)dr = r exp − dr (8.219)
∫ 2 ∫ 2
r T r T 2
F F
( 2 ) [ ( )]
r T het 4
= exp − 2 ≅ exp − 1 + het , (8.220)
2 4
F
where we have used Eq. (8.218) for r and
T
E av
het = het . (8.221)
N
0
For OOK, E = E ∕2, so
1
av
E
het 1
= . (8.222)
2N het
0
When het ≫ 1,
( het )

P(1|0)≅ exp − . (8.223)
4
Similarly, the probability of mistaking ‘1’ as ‘0’ is
( ) ( )
2
2
r T r T r + E ∕4 rE
1 1 1
P(0|1)= p(r|‘1’ sent)dr = r exp − I dr. (8.224)
∫ 2 ∫ 2 0 2
0 0 2 2
F F F
Let
r
x = , (8.225)
F
E 1 √
a = = 2 het . (8.226)
2 F
Now, Eq. (8.224) becomes
( )
r T ∕ F 2 2
a + x
P(0|1)= x exp − I (ax) dx
∫ 2 0
0
[ ( )]
r T
= 1 − Q 1 a, , (8.227)
F