Page 378 - Fiber Optic Communications Fund

Performance Analysis 359

The FSK scheme using the frequency separation given by Eq. (8.180) is called minimum shift keying (MSK).
Since E e is zero for orthogonal FSK,
10
2 e e 2E av
= (E + E )= (8.181)
max het 1 0 het
N N
0 0
and (√ )
1 het
P = erfc . (8.182)
b
2 4
Comparing Eqs. (8.182) and (8.144), we see that the expression for P for systems based on OOK is the same
b
as that for systems based on orthogonal FSK when synchronous detection is used in both systems.

8.4.4 OOK: Asynchronous Receiver

In Section 8.4.2, we discussed the error probability for the OOK signal with the receiver filter matched to I(t)
exactly. Here, we consider the case in which the receiver filter is matched to I(t) except for the phase. The
transmitted optical field envelope can be written as
{
s (t) for bit ‘1’
1
s(t)= (8.183)
0 for bit ‘0’
Without loss of generality, we assume that s(t) is real. The received signal passes through a filter matched to
s (t) cos ( t). From Eq. (8.120), the photocurrent is
1 IF
I (t)= s (t) cos ( t +Δ),
1
1
IF
[̃s ( − ) exp (−iΔ)+ ̃s ( + ) exp (iΔ)]
1
IF
IF
1
̃
I ()= . (8.184)
1
2
As before, we have ignored the scaling factor 2RA . The matched filter need not match the phase Δ of I (t).
LO 1
One possible way of realizing this type of matched filter is to use the envelope detector which detects only
the envelope and ignores the phase, i.e., it is matched to s (t) cos ( t). The transfer function of the matched
IF
1
filter is
∗
H ()={[s (t) cos ( t)]} exp (iT )
I 1 IF b
∗
∗
[̃s ( − )+ ̃s ( + )] exp (iT )
b
IF
IF
= 1 1 . (8.185)
2
The impulse response of the matched filter can be found from Eq. (8.42) by replacing x (t) with s (t) cos ( t)
1
1
IF
and x (t) with zero:
0
h (t)= s (T − t) cos [ (T − t)]. (8.186)
IF
b
b
I
1
The matched filter can be realized as a correlator (see Fig. 8.6), and the signal component of its output is
(Eq. (8.48))
T b
I (t)= ∫ s() cos ( +Δ)s (T + − t) cos [ (T + − t)]d (8.187)
IF
b
b
1
IF
F
0
1 T b
= s()s (T + − t){cos [ (t − T )+Δ]
IF
1
b
b
2 ∫ 0
+ cos ( (2 + T − t)+Δ)}d. (8.188)
IF
b