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Performance of Communication Systems Corrupted by Noise Chap. 7
SA7–2 BER for an RC LPF Receiver
filter with
Rework SA7–1 for the case when the LPF is a RC
H(f) =
1
1 + ja f
B 3dB
b
(7–148)
where B 3dB = 1/(2p RC). Note that, using Eq. (6–104), we can see that the equivalent bandwidth
of the RC LPF is B = (p/2) B 3dB .
Solution The MATLAB solution is shown by the plots in Fig. 7–29. Fig. 7–29a shows the
unfiltered rectangular pulse of amplitude A = 1 and width T = 1. Figs. 7–29b, 7–29c, and 7–29d
show the filtered pulse when the equivalent bandwidth of the RC LPF is B = 1/T, B = 2/T, or
B = 3/T, respectively. For B = 1/T, s 01 (t 0 ) = 1 = A only if the sampling time, t 0 , can be set near
the end of the pulse precisely at the point where the filtered pulse is a maximum. This requires
the use of an accurate bit synchronizer. For B Ú 2/T, it is seen that s 01 (t 0 ) = 1 = A and that the
sampling time, t 0 , is not too critical because the filtered pulse is flat (with a value of A) over a
significant portion of the bit (pulse) interval. Also, the ISI is negligible when B Ú 2/T; that is,
the solution to homework Problem 7–9d demonstrates that the worst case signal to ISI ratio is
about 70 dB. Consequently, Eq. (7–24a), which assumes s 01 (t 0 ) L A is approximately correct
for B Ú 2/T.
SA7–3 Comparison of BER for RC and MF Receivers Compare the performance of a digital
receiver that uses a RC LPF with the performance of a receiver that uses a matched filter (MF).
Referring to Fig. 7–4a, let the input to the receiver be a unipolar signal plus white Gaussian noise.
Assume that the unipolar signal has a data rate of R = 9600 bits/s and a peak value of A = 5.
The noise has a PSD of n ( f ) = 3 × 10 - 5 .
(a) If the receiver filter is a RC LPF with an equivalent bandwidth of B = 2/T = 2R,
evaluate the SNR at the RC filter input where the bandwidth of the noise is taken to be
the equivalent bandwidth of the RC LPF. Also, evaluate the BER for the data at the
receiver output.
(b) Repeat (a) for the case of B = 1/T = R.
(c) If the receiver filter is a MF, evaluate the SNR at the MF input where the bandwidth of the
noise is the equivalent bandwidth of the MF. Also, evaluate the BER for the data at the
receiver output.
(d) Compare the BER performance of the receiver that uses the RC LPF with the performance
of the MF receiver.
Solution The average energy per bit for the signal at the receiver input is E b = (A 2 /2)T =
A 2 /(2R). n ( f ) = N 0 /2 = 3 × 10 -5 so N 0 = 6 × 10 -5 . Thus,
a E b
b = 10 log a A2
N 0 dB
2N 0 R b = 10 log a (5) 2
2(6 * 10 -5 b = 13.4 dB
) (9600)
(7–149)
(a) For a receiver that uses a RC LPF with B = 2/T = 2R, the noise power in the bandwidth
B is P n = (N 0 /2) (2B). The RMS signal or noise voltage is related to its average power