# 0.1 Amplitude quantization

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Analog-to-digital conversion.

The Sampling Theorem says that if we sample a bandlimitedsignal $s(t)$ fast enough, it can be recovered without error from its samples $s(n{T}_{s})$ , $n\in \{\dots , -1, 0, 1, \dots \}$ . Sampling is only the first phase of acquiring data into acomputer: Computational processing further requires that the samples be quantized : analog values are converted into digital form. In short, we will have performed analog-to-digital (A/D) conversion . A three-bit A/D converter assigns voltage in the range -1 1 to one of eight integers between 0 and 7. For example, allinputs having values lying between 0.5 and 0.75 are assigned the integer value six and, upon conversion back to an analogvalue, they all become 0.625. The width of a single quantization interval Δ equals 2 2 B . The bottom panel shows a signal going through theanalog-to-digital converter, where B is the number of bits used in the A/D conversion process (3 inthe case depicted here). First it is sampled, then amplitude-quantized to three bits. Note how the sampledsignal waveform becomes distorted after amplitude quantization. For example the two signal values between 0.5and 0.75 become 0.625. This distortion is irreversible; it can be reduced (but not eliminated) by using more bits inthe A/D converter.

A phenomenon reminiscent of the errors incurred in representing numbers on a computer prevents signal amplitudesfrom being converted with no error into a binary number representation. In analog-to-digital conversion, the signal isassumed to lie within a predefined range. Assuming we can scale the signal without affecting the information itexpresses, we'll define this range to be $\left[-1 , 1\right]$ . Furthermore, the A/D converter assigns amplitude values inthis range to a set of integers. A $B$ -bit converter produces one of the integers $\{0, 1, \dots , 2^{B}-1\}$ for each sampled input. [link] shows how a three-bit A/D converter assigns input values tothe integers.We define a quantization interval to be the range of values assigned to the same integer. Thus, for our examplethree-bit A/D converter, the quantization interval $\Delta$ is $0.25$ ; in general, it is $\frac{2}{2^{B}}$ .

Recalling the plot of average daily highs in this frequency domain problem , why is this plot so jagged? Interpret this effect interms of analog-to-digital conversion.

The plotted temperatures were quantized to the nearest degree. Thus, the high temperature's amplitude wasquantized as a form of A/D conversion.

Because values lying anywhere within a quantization interval are assigned the same value for computer processing, the original amplitude value cannot be recovered without error . Typically, the D/A converter, the device that converts integers to amplitudes, assigns anamplitude equal to the value lying halfway in the quantization interval. The integer 6 would be assigned to the amplitude0.625 in this scheme. The error introduced by converting a signal fromanalog to digital form by sampling and amplitude quantization then back again would be half the quantizationinterval for each amplitude value. Thus, the so-called A/D error equals half the width of a quantization interval: $\frac{1}{2^{B}}$ . As we have fixed the input-amplitude range, the more bitsavailable in the A/D converter, the smaller the quantization error.

To analyze the amplitude quantization error more deeply, we need to compute the signal-to-noise ratio, which equals the ratio of the signal power and the quantizationerror power. Assuming the signal is a sinusoid, the signal power is the square of the rms amplitude: $\mathrm{power}(s)=\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{2}$ . The illustration details a single quantization interval. A single quantization interval is shown, along with atypical signal's value before amplitude quantization s n T s and after Q s n T s . ε denotes the error thus incurred. Its width is $\Delta$ and the quantization error is denoted by $\epsilon$ . To find the power in the quantization error, we note that no matter into whichquantization interval the signal's value falls, the error will have the same characteristics. To calculate the rms value, wemust square the error and average it over the interval.
$\mathrm{rms}(\epsilon )=\sqrt{\frac{1}{\Delta }\int_{-\left(\frac{\Delta }{2}\right)}^{\frac{\Delta }{2}} \epsilon ^{2}\,d \epsilon }=\left(\frac{\Delta ^{2}}{12}\right)^{\left(\frac{1}{2}\right)}$
Since the quantization interval width for a $B$ -bit converter equals $\frac{2}{2^{B}}=2^{-(B-1)}$ , we find that the signal-to-noise ratio for theanalog-to-digital conversion process equals
$\mathrm{SNR}=\frac{\frac{1}{2}}{\frac{2^{-(2(B-1))}}{12}}=\frac{3}{2}2^{(2B)}=6B+10\lg 1.5\mathrm{dB}$
Thus, every bit increase in the A/D converter yields a 6 dB increase in the signal-to-noise ratio.The constant term $10\lg 1.5$ equals 1.76.

This derivation assumed the signal's amplitude lay in the range $\left[-1 , 1\right]$ . What would the amplitude quantization signal-to-noiseratio be if it lay in the range $\left[-A , A\right]$ ?

The signal-to-noise ratio does not depend on the signal amplitude. With an A/D range of $\left[-A , A\right]$ , the quantization interval $\Delta =\frac{2A}{2^{B}}$ and the signal's rms value (again assuming it is a sinusoid) is $\frac{A}{\sqrt{2}}$ .

How many bits would be required in the A/D converter to ensure that the maximum amplitude quantization error wasless than 60 db smaller than the signal's peak value?

Solving $2^{-B}=.001$ results in $B=10$ bits.

Music on a CD is stored to 16-bit accuracy. To what signal-to-noise ratio does this correspond?

A 16-bit A/D converter yields a SNR of $6\times 16+10\lg 1.5=97.8$ dB.

Once we have acquired signals with an A/D converter, we canprocess them using digital hardware or software. It can be shown that if the computer processing is linear, the result ofsampling, computer processing, and unsampling is equivalent to some analog linear system. Why go to all the bother if thesame function can be accomplished using analog techniques? Knowing when digital processing excels and when it does not isan important issue.

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