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542 Chapter 10 ROUND-OFF EFFECTS IN DIGITAL FILTERS
Distribution of e 1
Distribution of e 2
4128
3128
2128
1128
4128
3128
2128
1128
SAMPLE SIZE N = 500000
ROUNDED T0 B = 2 BITS
MEAN = 3.4239e–005
MIN PROB BAR HEIGHT = 0.007446
MAX PROB BAR HEIGHT = 0.00828
SIGMA = 0.072073
0
−0.5 −0.375 −0.25 −0.125 0 0.125 0.25 0.375 0.5
Normalized error e 1
SAMPLE SIZE N = 500000
ROUNDED T0 B = 2 BITS
MEAN = 3.4396e–005
MIN PROB BAR HEIGHT = 0.000334
MAX PROB BAR HEIGHT = 0.015212
SIGMA = 0.063851
0
−0.5 −0.375 −0.25 −0.125 0 0.125 0.25 0.375 0.5
Normalized error e 2
FIGURE 10.3 A/D quantization error distribution for the sinusoidal signal in
Example 10.1, B =2bits
Through these examples we hope to learn how small error e must be (or
equivalently, how large B must be) for the above assumptions to be valid.
□ EXAMPLE 10.1 Let x(n) = 1 {sin(n/11) + sin(n/31) + cos(n/67)}. This sequence is not periodic,
and hence its samples never repeat using infinite-precision representation.
3
However, since the sequence is of sinusoidal nature, its continuous envelope is
periodic and the samples are continuously distributed over the fundamental
period of this envelope. Determine the error distributions for B = 2 and 6 bits.
Solution
To minimize statistical variations, the sample size must be large. We choose
500,000 samples. The following MATLAB script computes the distributions for
B =2bits.
clear; close all;
% Example parameters
B = 2; N = 500000; n = [1:N];
xn = (1/3)*(sin(n/11)+sin(n/31)+cos(n/67)); clear n;
% Quantization error analysis
[H1,H2,Q, estat]] = StatModelR(xn,B,N); % Compute histograms
H1max = max(H1); H1min = min(H1); % Max and Min of H1
H2max = max(H2); H2min = min(H2); % Max and Min of H2
The plots of the resulting histogram are shown in Figure 10.3. Clearly, even
though the error samples appear to be uniformly distributed, the samples
are not independent. The corresponding plots for B = 6 bits are shown in
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