Page 391 - Fiber Optic Communications Fund

372 Fiber Optic Communications

The transfer function of the matched filter is

∗
H ()= x ()e i+iT b , j = 0, 1
j j
∗
= ̃s ( − )e i+iT b , (8.277)
j c
where is an arbitrary phase factor. The output of the photodetector j at t = T is
b
2
I (T )= R[|u (T )+ n (T )| ], j = 0, 1. (8.278)
Fj
b
b
j
b
j
First let us consider the photodetector outputs in the absence of noise. Suppose a bit ‘1’ is transmitted so that
x(t)= x (t). The output of PD1 is
1
2
I (T )= R[|u (T )| ]. (8.279)
1 b 1 b
From Eqs. (8.275) and (8.277), we have
e i ∞ 2 i
u (T )= 2 ∫ −∞ |x ()| d = E e , (8.280)
1
1
b
1
2
I (T )= RE . (8.281)
b
1
1
For this case, the output of PD0 is
2
b
0
0
I (T )= R|u (T )|
b
∞ 2
| 1 ∗ |
= R | ̃ x ()̃x () d . (8.282)
|
| 2 ∫ −∞ | |
1
0
|
Using Parseval’s relations,
∞ ∞
1 ∗ ∗
1
2 ∫ −∞ ̃ x ()̃x ()d = ∫ −∞ x (t)x (t) dt, (8.283)
1
0
0
Eq. (8.282) may be rewritten as
2
| T b |
| ∗ |
b
1
0
0
I (T )= R| |∫ x (t)x (t) dt|
| 0 | |
2
| T b |
| ∗ |
= R| s (t)s (t) dt|
|∫ |
1
0
| 0 |
2
| T b |
| |
= R| exp (−i2Δft) dt|
|∫ 0 |
| |
2
|exp (−iΔfT ) sin (ΔfT )|
b
b
= R | |
| Δf |
| |
2
Rsin (ΔfT )
b
= . (8.284)
2
Δf 2