Page 509 - Fiber Optic Communications Fund

490 Fiber Optic Communications

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2
(T − lT ) −(T − mT ) 2 (T − nT ) 2
s
s
s
∗
r r r = exp − 2 −
l m n
2 ∗
2T 2(T )
1 1
2
= exp [−(CT − dT)− g], (10.493)
where
2
3T + iS
0
C = 4 , (10.494)
2
2(T + S )
0
2
[(l + m + n)T + i(l + m − n)S]T s
0
d = , (10.495)
4
T + S 2
0
2
2
2
2
2
2
2
[(l + m + n )T +(l + m − n )iS]T 2 s
0
g = . (10.496)
4
2
2(T + S )
∞ 0
2
∗
{r r r }= exp (−g) ∫ exp [−(CT − dT)− iT]dT
l m n
−∞
∞
2
2
2
= exp (−g) exp [−C(T + 2xT + x )+ Cx ]dt, (10.497)
∫
−∞
where
i − d
x = . (10.498)
2C
∞
2
2
∗
{r r r }= exp (−g + Cx ) ∫ exp [−C(T + x) ]dT. (10.499)
l m n
−∞
Let
√ √
u = C(T + x) and du = CdT, (10.500)
∞
∗ 2 1 2
l m n ∫
{r r r }= exp (−g + Cx ) √ exp (−u )du
C −∞
√
2
= exp [−g +(i − d) ∕4C], (10.501)
C
T
P 3∕2 3 √ ∑
0
0
2
2
{|u | u }= b b b exp [g +(i − d) ∕4C]. (10.502)
0
0
l m n
2
|T | T C
1 1 lmn
Example 10.14
Find the variance of ‘1’ in a direct detection OOK system due to linear and nonlinear distortion.
Solution:
From Eq. (10.308), we obtain
∞ ∞
∑ ∑
2
<I >= 4P < b b >u u . (10.503)
lin 0 n m lin,n lin,m
n=−∞ m=−∞
n≠0 m≠0