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Sec. 7–11 Study-Aid Examples 557
by V 2 RMS = v2 (t) = P. Because the average input signal power is P s = A 2 /2, the RMS
signal voltage at the receiver input is
V s = 2A 2 >2 = A> 12 = 5> 12 = 3.54 V RMS signal
The RMS noise voltage (measured in a bandwidth of 2R) at the receiver input is
V n = C
N 0
2 (2B) = 12N 0R = 32(6 * 10 -5 ) (9600) = 1.07 V RMS noise
Thus, the input SNR is
a S N b dB
= 20 log a V s
b = 20 log a 3.54 b = 10.4 dB
V n 1.04
Also, the SNR can be evaluated by using
a S N b dB
A2
= 10 log a P s
b = 10 log a
P n 2N 0 B b
(7–150)
which, for B = 2/T = 2R, becomes
a S N b dB
= 10 log a A2
b = 10.4 dB
4N 0 R
The BER is obtained by using Eq. (7–24a) for the case of a RC LPF, where s 01 (t 0 ) ≈ A
and B = 2/T = 2R. So, we get
A 2
P e = Q¢ B 8N 0 R ≤ = Q¢ (5) 2
≤ = 9.9 *
B 8(6 * 10 -5 10-3
)(9600)
(b) For a receiver that uses a RC LPF with B = 1/T = R, the RMS noise voltage (measured in
a bandwidth of R) at the receiver input is
V n = B
N 0
2 (2B) = 1N 0R = 2(6 * 10 -5 )(9600) = 0.76 V RMS noise
Thus, the input SNR is
a S N b dB
= 20 log a V s
b = 20 log a 3.54 b = 13.4 dB
V n 0.76
Also, using Eq. (7–150) with B = R, we get
a S N b dB
= 10 log a A2
b = 13.4 dB
2N 0 R