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510
Performance of Communication Systems Corrupted by Noise Chap. 7
Substituting for s 01 , s 02 , and s 0 into Eq. (7–17), we have
A
P e = Q¢ 2
≤ (bandpass filters)
(7–45)
C 4N 0 B
Comparing the performance of FSK with that of BPSK and OOK on a PEP basis, we see that
FSK requires 3 dB more power than BPSK for the same P e , but 3 dB less power than OOK.
Comparing the performance on an average power basis, we observe that FSK is 3 dB worse
than BPSK, but is equivalent to OOK (since the average power of OOK is 3 dB below
its PEP).
The performance of FSK signaling with matched-filter reception is obtained from
Eq. (7–20). The energy in the difference signal is
T
E d = [A cos (v 1 t + u c ) - A cos (v 2 t + u c )] 2 dt
3
0
T
= [A 2 cos 2 (v 1 t + u c ) - 2A 2 cos (v 1 t + u c )
3
0
* cos (v 2 t + u c ) + A 2 cos 2 (v 2 t + u c )] dt
or † T
E d = 1 2 A2 T - A 2 [cos(v 1 - v 2 )t] dt + 1 2
(7–46)
3
A2 T
0
Consider the case when 2∆F = f 1 - f 2 = n/(2T) = nR/2. Under this condition the integral
(i.e., the cross-product term) goes to zero. This condition is required for s 1 (t) and s 2 (t) to
be orthogonal. Consequently, s 1 (t) will not contribute an output to the lower channel (see
Fig. 7–8), and vice versa. Furthermore, if (f 1 - f 2 ) R, s 1 (t) and s 2 (t) will be approximately
orthogonal, because the value of this integral will be negligible compared with A 2 T.
Assuming that one or both of these conditions is satisfied, then E d = A 2 T, and the BER for
FSK signaling is
A 2 T E b
P e = Q¢ ≤ = Q¢ ≤ (matched filter)
(7–47)
C 2N 0 C N 0
where the average energy per bit is E b = A 2 T/2. The performance of FSK signaling is equivalent
to that of OOK signaling (matched-filter reception) and is 3 dB inferior to BPSK signaling.
(See Fig. 7–5.)
As we demonstrate in the next section, coherent detection is superior to noncoherent
detection. However, for coherent detection, the coherent reference must be available. This
† For integrals of the type 1 T the second
0 A2 cos 2 (v 1 t + u c ) dt = 1 2 A2 C1 T 0 dt + 1 T
0 cos(2v 1t + 2u c ) dtD,
integral on the right is negligible compared with the first integral on the right because of the oscillation in the second
integral (Riemann–Lebesque lemma [Olmsted, 1961]).