Page 392 - Fiber Optic Communications Fund

Performance Analysis 373

I (T ) is zero if
0
b
ΔfT = n, n = 1, 2, … (8.285)
b
or
n
Δf = . (8.286)
T b
Comparing Eqs. (8.286) and (8.239), we find that the orthogonality conditions for the asynchronous receiver
and direct detection receiver are the same. In this section, we assume that the orthogonality condition is
satisfied so that the output of PD0 (PD1) is zero when bit ‘1’ (bit ‘0’) is transmitted in the absence of noise.
Since the photo-detector output is not sensitive to the phase factor , we ignore it from now on.

Case (i): bit ‘1’ transmitted. Expanding Eq. (8.278), we obtain
2
I (T )= R[u (T )+ 2u (T )n (T )+ n 2 (T )+ n 2 (T )]
1 b 1 b 1 b F 1r b F 1r b F 1i b
2
= R{[E + n (T )] + n 2 (T )}, (8.287)
1 F 1r b F 1i b
I (T )= R[n 2 (T )+ n 2 (T )], (8.288)
0 b F 0r b F 0i b
wherewehaveused u (T )= E and subscripts r and i denote the real and imaginary parts, respectively. Let
1
b
√ √
1 √
us first consider I (T ). R[E + n (T )] and Rn are Gaussian random variables with means RE and
1 b 1 F 1r b F 1i 1
zero, respectively. The variances of these two random variables are equal and given by
∞
E
R ASE 2 R ASE 1
2
= |H ()| d = . (8.289)
2 ∫ −∞ 2 2
1
The pdf of I (T ) is given by the non-central chi-square distribution
1
b
√
[ ( 2 )] ⎛ I RE 2 ⎞
1 RE + I 1 ⎜ 1 1 ⎟
1
p (I )= exp − I . (8.290)
1 1 2 2 0 ⎜ 2 ⎟
2 2
⎜ ⎟
⎝ ⎠
The pdf of I (T ) is given by the central chi-square distribution
0
b
( )
1 I 0
p (I )= exp − . (8.291)
1 0
2 2 2 2
If I (T ) > I (T ), it will be decided that the bit ‘1’ is transmitted. Therefore, an error is made if I (T ) < I (T )
1 b 0 b 1 b 0 b
when s (t) is transmitted:
1
P(0|‘1’ sent)= P(I (T ) > I (T )|‘1’ sent). (8.292)
0 b 1 b
The chance that I (T ) < I (T ) can be found as follows. Since I (T ) and I (T ) are independent random
1 b 0 b 1 b 0 b
variables, the joint pdf of I and I can be written as
1 0
p (I , I )= p (I )p (I ). (8.293)
1 1 0 1 1 1 0
The chance that I (T ) < I (T ) is the same as that I (T ) has a value i and I (T ) has a value i greater than
1
0
b
0
1
b
0
b
b
1
i . Since I (T ) can take any value in the range (0, ∞),wehave
1 1 b
∞ { ∞ }
( )
P(I (T ) > I (T )|‘1’ sent)= ∫ ∫ p 1 i , i 0 di 1 di . (8.294)
b
1
0
0
1
b
i 1 0