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Analysis of A/D Quantization Noise 543
Distribution of e 1
Distribution of e 2
4128
3128
2128
1128
4128
3128
2128
1128
SAMPLE SIZE N = 500000
ROUNDED T0 B = 6 BITS
MEAN = −4.1044e–006
MIN PROB BAR HEIGHT = 0.007478
MAX PROB BAR HEIGHT = 0.00814
SIGMA = 0.0045076
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 = 6 BITS
MEAN = −4.1044e–006
MIN PROB BAR HEIGHT = 0.000112
MAX PROB BAR HEIGHT = 0.01572
SIGMA = 0.0031818
0
−0.5 −0.375 −0.25 −0.125 0 0.125 0.25 0.375 0.5
Normalized error e 2
FIGURE 10.4 Quantization error distribution for the sinusoidal signal in
Example 10.1, B =6bits
Figure 10.4, from which we observe that the quantization error sequence appears
to satisfy the model assumptions for B ≥ 6 bits.
□
□ EXAMPLE 10.2 Let x(n) beanindependent and identically distributed random sequence whose
samples are uniformly distributed over the [−1, 1] interval. Determine the error
distributions for B = 2 and 6 bits.
Solution
We again choose 500,000 samples to minimize any statistical variations. The
following MATLAB fragment computes the distributions for B =2bits.
clear; close all;
% Example parameters
B = 2; N = 500000; xn = (2*rand(1,N)-1);
% 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 10.5. The corresponding
plots for B =6bits are shown in Figure 10.6. From these plots we observe
that even for B =2bits the quantization error samples are independent and
uniformly distributed.
□
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