Page 363 - Fiber Optic Communications Fund

344 Fiber Optic Communications

From Eq. (8.33), the matched filter is

∗
∗
H()=[̃x ()− ̃x ()] exp (−iT )
1 0 B
4A sin (T ∕2)
b
= exp (−iT ∕2). (8.58)
b

Using Eq. (8.6), the signal sample at t = T is
B
2
8A 2 ∞ sin (T ∕2)
B
u(T )=± d
B 2 ∫ −∞ 2
2
=±2A T . (8.59)
B
2
Note that A T is the pulse energy, E . Since x (t)=−x (t), the matched filter can also be realized using a
b 1 0 1
correlation receiver as shown in Fig. 8.8:
T B
r(T )= y(t)x (t) dt, (8.60)
B ∫ 1
0
T B
2
u(T )= x(t)x (t) dt =±A T , (8.61)
B ∫ 1 b
0
r(T )= u(T )+ n (T ). (8.62)
b b F b
If r(T ) is positive (negative), the threshold device selects bit ‘1’ (‘0’). Note that the signal value given by
B
Eq. (8.61) is half of that given by Eq. (8.59). As far as the BER is concerned, this makes no difference since
the noise sample corresponding to Fig. 8.8 is half of that corresponding to Fig. 8.5.
(b) Since T << T , the signal power outside the bit interval is negligible. So, we approximate Eq. (8.55) as
0 B
x(t)= s(t − T ∕2), (8.63)
B
( )
t 2
s(t)=±A exp − . (8.64)
2T 2
0
From Eq. (2.152), we have
[ ]
A 2
̃ s()=± exp − , (8.65)
a 4a 2
1
a = √ . (8.66)
2T
0
Using the time-shifting property, we find
̃ x()= ̃s()e iT b ∕2 , (8.67)
[ 2 ]
A
̃ x ()=−̃x ()= exp − + iT ∕2 , (8.68)
1 0 2 b
a 4a
[ ]
2A 2 iT b
H()= exp − + . (8.69)
a 4a 2 2