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Performance of Communication Systems Corrupted by Noise Chap. 7
envelope of the OOK signal is (approximately) preserved at the filter output. The baseband
analog output will be
A, 0 6 t … T, binary 1
r 0 (t) = e (7–30)
0, 0 6 t … T, binary 0 f + x(t)
where x(t) is the baseband noise. With the help of Eq. (6–133g), we calculate noise power as
x 2 (t) = s 2 0 = n 2 (t) = 2(N 0 >2) (2B) = 2N 0 B. Because s 01 = A and s 02 = 0, the optimum
threshold setting is V T = A/2. When we use equation (7–17), the BER is
P e = Q¢ C
A 2
8N 0 B ≤
(narrowband filter)
(7–31)
B is the equivalent bandwidth of the LPF. The equivalent bandpass bandwidth of this receiver
is B p = 2B.
The performance of a matched-filter receiver is obtained by using Eq. (7–20). The
energy in the difference signal at the receiver input is †
T
E d = [A cos (v c t + u c ) - 0] 2 dt = A2 T
(7–32)
L 2
Consequently, the BER is
0
P e = Q¢ C
A 2 T
4N 0
≤ = Q¢ C
E b
N 0
≤ (matched filter)
(7–33)
where the average energy per bit is E b = A 2 T/4. For this case, where s 1 (t) has a rectangular
(real) envelope, the matched filter is an integrator. Consequently, the optimum threshold value is
V T = s 01 + s 02
2
= 1 2 s 01 = 1 2 c L
0
T
2A cos 2 (v c t + u c ) dt d
which reduces to V T = AT/2 when f c R. Note that the performance of OOK is exactly the
same as that for baseband unipolar signaling, as illustrated in Fig. 7–5.
Binary-Phase-Shift Keying
Referring to Fig. 7–7, we see that the BPSK signal is
s 1 (t) = A cos (v c t + u c ), 0 6 t … T (binary 1)
and
s 2 (t) = -A cos (v c t + u c ), 0 6 t … T (binary 0)
(7–34a)
(7–34b)
† Strictly speaking, f c needs to be an integer multiple of half the bit rate, R = 1/T, to obtain exactly A 2 T/2
for E d . However, because f for all practical purposes E d = A 2 c R,
T/2, regardless of whether or not f c = nR/2.