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Performance of Communication Systems Corrupted by Noise Chap. 7
The BER is obtained by using Eq. (7–24a) with B = 1/T = R. So, we get
A 2
P e = Q¢ B 4N 0 R ≤ = Q¢ (5) 2
≤
B 4(6 * 10 -5 = 4.9 * 10-4
)(9600)
(c) To obtain the SNR for the MF receiver, we need to first evaluate the equivalent bandwidth
of the MF. Using Eq. (6-155), we see that the transfer function for the MF that is matched
to the rectangular pulse s(t) = 5P(t/T) is
H(f) = K S*(f)
n (f) e-jvt 0
= CTa sinpTf
pTf be-jvt 0
where C = 5K/(N 0 /2). From Eq. (2–192), the equivalent bandwidth of the MF is
B =
1
ƒ H(0) ƒ 2 L
q
-q
q
ƒ H(f) ƒ 2 df = a
L
0
sin pTf
pTf b 2
df
To evaluate this integral, make a change in variable. Let x = pTf, so that with the aid of
Appendix A, we get
B = 1
pT L0
q
a sin x
x b 2
dx = a 1
pT bap 2 b = 1
2T = 1 2 R
(7–151)
Thus, the RMS noise voltage (measured in a bandwidth of B = R/2) at the receiver
input is
V n = C
N 0
2 (2B) = 3N 0R>2 = 36 * 10 -5 ) (9600)/2 = 0.54 V RMS noise
As previously obtained, the RMS signal voltage is V s = 3.54 V. So, the input SNR is
a S N b dB
= 20 log a V s
b = 20 log a 3.54 b = 16.4 dB
V n 0.54
Also, using Eq. (7–150) with B = R/2, we get
a S N b dB
= 10 log a A 2
b = 16.4 dB
N 0 R
The BER is obtained by using Eq. (7–24b):
P e = Q¢ B
E b
N 0
≤ = Q¢ B
A 2 T
2N 0
≤ = Q¢ B
A 2
2N 0 R ≤
(5) 2
= Q¢ ≤ B 2(6 * 10 -5 = 1.6 * 10-6
)(9600)
This result may be verified by using Fig. 7–14 for the case of unipolar signaling with
(E b N 0 ) dB = 13.4 dB.