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Analysis of A/D Quantization Noise 547<br />

Similarly, for B =6computed statistical values are shown in Figure 10.4.<br />

The computed values of the two means are −4.1044 × 10 −6 , which agree very<br />

well with the model. The standard deviation of e(n), from (10.9), is 0.0045105,<br />

while that from the top plot in Figure 10.4 is 0.0045076, which again agrees<br />

closely with the model. The standard deviation of the average of the two consecutive<br />

samples, from (10.10), is 0.0031894, while from the bottom plot in<br />

Figure 10.4 it is 0.00318181, which clearly agrees with the model. Hence the<br />

samples of e(n) for B =6are independent. This was also confirmed by the<br />

bottom plot in Figure 10.4.<br />

□<br />

Similar calculations can be carried out for the signal in Example 10.2.<br />

The details are left to the reader.<br />

10.1.6 A/D QUANTIZATION NOISE THROUGH DIGITAL FILTERS<br />

Let a digital filter be described by the impulse response, h(n), or the frequency<br />

response, H(e jω ). When a quantized input, Q[x(n)] = x(n)+e(n),<br />

is applied to this system, we can determine the effects of the error sequence<br />

e(n) onthe filter output as it propagates through the filter, assuming<br />

infinite-precision arithmetic implementation in the filter. We are generally<br />

interested in the mean and variance of this output-noise sequence, which<br />

we can obtain using linear system theory concepts. Details of these results<br />

can be found in many introductory texts on random processes,<br />

including [27].<br />

Referring to Figure 10.8, let the output of the filter be ŷ(n). Using<br />

LTI properties and the statistical independence between x(n) and e(n),<br />

the output ŷ(n) can be expressed as the sum of two components. Let y(n)<br />

be the (true) output due to x(n) and q(n) the response due to e(n). Then<br />

we can show that q(n) isalso a random sequence with mean<br />

m q<br />

△<br />

=E[q(n)] = me<br />

∞<br />

∑<br />

−∞<br />

h(n) =m e H(e j0 ) (10.12)<br />

where the term H(e j0 )istermed the DC gain of the filter. For truncation,<br />

m eT = −∆/2, which gives<br />

m qT = − ∆ 2 H(ej0 ) (10.13)<br />

For rounding, m eR =0or<br />

m qR =0 (10.14)<br />

x(n) ˆ = x(n) + e(n)<br />

h(n), H(e jω )<br />

y(n) ˆ = y(n) + q(n)<br />

FIGURE 10.8<br />

Noise through digital filter<br />

Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).<br />

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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