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Sec. 7–7 Output Signal-to-Noise Ratio for PCM Systems 529
Thus, Eq. (7–80) reduces to
e b 2 = 4V 2 P e a
n
j=1
A 1 4 B j
which, from Appendix A, becomes
or
e 2 b = 4 3 V2 P M2 - 1
e (7–81)
M 2
Substituting Eqs. (7–76) and (7–81) into Eq. (7–72) with the help of Eq. (7–75), we have
which reduces to Eq. (7–70).
A 1 4B n - 1
e 2 b = V 2 P e 1
4 - 1 = 4 3 V2 P (2n ) 2 - 1
e
(2 n ) 2
V 2
a S N b =
pk out (V 2 /3M 2 ) + (4V 2 /3M 2 )P e (M 2 - 1)
Example 7–8 (S/N) FOR THE RECOVERED ANALOG SIGNAL
AT THE OUTPUT OF A PCM SYSTEM
Using Eq. (7–70), evaluate and plot the peak S/N of the recovered analog signal at the output of a
PCM digital-to-analog converter as a function of the BER for the cases of M = 256 and
M = 8. See Example7_08.m for the solution. Compare these results with those shown in Fig. 7–17.
Equation (7–70) is used to obtain the curves shown in Fig. 7–17. For a PCM system
with M quantizing steps, (SN) out is given as a function of the BER of the digital receiver. For
P e 6 1(4M 2 ), the analog signal at the output is corrupted primarily by quantizing noise. In
fact, for P e = 0, (SN) out = 3M 2 , and all the noise is quantizing noise. Conversely, for
P e 7 1(4M 2 ), the output is corrupted primarily by channel noise.
It is also stressed that Eq. (7–70) is the peak signal to average noise ratio. The averagesignal-to-average-noise
ratio is obtained easily from the results just presented. The averagesignal-to-average-noise
ratio is
a S N b out
= (x k) 2
n k
2
= V2
3n k
2
where x 2 k = V 2 >3 because x k is uniformly distributed from -V to +V. Thus, using Eqs. (7–72)
and (7–70), we calculate
a S (7–82)
N b =
out 1 + 4(M 2 - 1)P e
when (SN) out is the average-signal-to-average-noise power ratio at the output of the system.
M 2
= 1 3 a S N b pk out