Page 388 - Fiber Optic Communications Fund

Performance Analysis 369

The output of the matched filter passes through the photodetector and the signal component of the photocur-
rent is
2
s F b
I (t)= R|s (T )|
= RE 2 when ‘1’ is sent
1
= 0 otherwise. (8.258)
The total current at the decision instant T is
b
2
b
b
b
F
F
I(T )= R|s (T )+ n (T )|
2
2
= R[(s (T )+ n (T )) +(s FQ (T )+ n FQ (T )) ], (8.259)
FI
FI
b
b
b
b
where y = Re[y] and y = Im[y], y = s , n . Without loss of generality, we can assume s (T ) to be real so
I Q F F F b
√ √
that s FQ (T )= 0 and s (T )= E . Rn (T ) and Rn FQ (T ) are independent Gaussian random variables
b
FI
b
FI
b
1
b
with zero mean and variance,
2
2
= R < n 2 >= R < n 2 >= R < |n | >
FI FQ 2 F
∞
R 2 ASE
= |H()| d = E R. (8.260)
4 ∫ ASE 2 1
−∞
When a bit ‘0’ is transmitted, s (T )= s (T )= 0 and in this case the pdf of I is given by the central
FI b FQ b
chi-square distribution
1 ( I )
p(I|‘0’ sent) ≡ p (I)= exp − . (8.261)
0
2 2 2 2
2
When a bit ‘1’ is transmitted, the current in the absence of noise is RE and in this case the pdf is given by
1
the non-central chi-square distribution
√
( 2 ) ⎛ 2 ⎞
1 RE + I ⎜ IRE 1 ⎟
1
p(I|‘1’ sent) ≡ p (I)= exp − I . (8.262)
1
2 2 2 2 0 ⎜ 2 ⎟
⎜ ⎟
⎝ ⎠
The threshold current is determined by the interSection of two curves p (I) and p (I):
1
0
p (I )= p (I ) (8.263)
1 T 0 T
or
√
( 2 ) ⎛ 2 ⎞
RE 1 ⎜ I RE 1 ⎟
T
exp − I ⎟ = 1. (8.264)
2 2 0 ⎜ 2
⎜ ⎟
⎝ ⎠
This equation is satisfied to a close approximation [1]
( )
2
RE 1 8 2
I = 1 +
T
4 RE 2
1
RE 1 2 ( 4 ASE )
= 1 + . (8.265)
4 E 1