core24 min studyLesson 958 in Physics

Analog-to-digital resolution and quantization

Separate converter step size from uncertainty, dynamic performance, and the assumptions that make quantization resemble white noise.

Before you begin

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The question for today

What you’ll understand

Calculate an ideal converter step and an input-referred quantization contribution.
State the conditions behind the uniform-error and ideal-sine SNR formulas.
Combine independent acquisition noise without averaging away a fixed offset.

A converter's bit count describes how many output codes are available. It does not by itself specify the accuracy of the corresponding voltages, the noise of the front end, or the stability of the voltage reference. An ideal quantizer provides a useful starting model, but its error is a deterministic function of the input. Treating that error as random white noise needs additional signal or dither assumptions. Keeping these statements separate helps determine whether extra bits improve the experiment or merely produce a more finely formatted record of the same analog uncertainty.

A code step is not an uncertainty certificate

For an ideal uniform N-bit converter spanning a total input range Vspan, define one code step q as Vspan divided by two to the power N. Under nearest-level rounding and away from overloaded endpoints, the quantization error lies within half a step. Code-bin and endpoint conventions must be stated when translating an actual code. A converter with a different reference range has a different voltage step even if its number of bits is unchanged.

If the error is adequately modeled as uniformly distributed over that interval, its mean is zero and its variance is q squared over twelve. This follows by integrating the squared error against a uniform density. A constant input well inside one code bin does not generate a fresh independent uniform error every time it is sampled. Repeated copies of the same code therefore do not automatically average the quantization error toward zero.

q=\frac{V_{\mathrm{span}}}{2^N},\qquad -\frac q2\leq e_q\leq\frac q2,\qquad \sigma_q^2=\frac1q\int_{-q/2}^{q/2}e^2\,de=\frac{q^2}{12}

q and eq have voltage units. The bound assumes ideal nearest-level rounding within the unsaturated interior. The variance additionally assumes the stated uniform-error statistical approximation; it is not an automatic distribution for every input waveform.↔ Scroll wide equations horizontally.

Ideal signal-to-noise formulas have a specific stimulus and band

For a full-scale sinusoid using the available span without clipping, the AC RMS amplitude is Vspan divided by twice square root two. Combining it with the uniform quantization variance yields the familiar approximately six decibels per bit result, plus a constant. The result integrates quantization noise over the full Nyquist band and excludes thermal noise, distortion, reference fluctuations, and timing errors. It is an ideal-model benchmark rather than a guarantee for a converter bearing that bit label.

A smaller sinusoid has a lower signal-to-noise ratio even if the quantization floor is unchanged. Oversampling and digital filtering can reduce in-band uncorrelated noise, while noise shaping changes its spectral distribution. Harmonic coherence between a simple input and sampling clock can instead concentrate quantization error into tones. Effective-number-of-bits values derived from a sine-wave SINAD measurement include noise and distortion under specified test conditions; they do not certify DC accuracy.

\mathrm{SNR}_{q}\simeq6.02N+1.76\ \mathrm{dB},\qquad \sigma_{q,\mathrm{input}}=\frac{q}{\sqrt{12}\,|G|},\qquad \sigma_{\mathrm{mean}}=\frac{\sigma}{\sqrt{M}}\quad\text{for independent equal-variance errors}

The SNR benchmark uses a full-scale sinusoid and the entire 0-to-fs/2 noise band under the uniform-error approximation. G is the preceding linear voltage gain. M is a number of independent samples, not the converter bit count N; the averaging relation requires uncorrelated errors.↔ Scroll wide equations horizontally.

Budget the whole digitizing chain

Analog gain can reduce the converter's quantization contribution when referred to the sensor input, but it also reduces the allowable sensor excursion before clipping. Input noise and offsets already present before gain are not improved by that operation. Reference drift, integral nonlinearity, missing codes, and input acquisition settling are additional mechanisms. Some can be calibrated; their remaining uncertainties still belong in the result.

Dither can help decorrelate quantization error and reveal sub-step averages, at the cost of a controlled noise contribution and an appropriate reconstruction model. Averaging then addresses random components while a fixed offset persists unless corrected. A useful design compares input-referred analog noise, quantization, calibration uncertainty, and available range. That comparison can reveal either a genuinely limiting code step or a system in which buying more nominal resolution would leave the main physical error unchanged.

Work it through

Refer a declared ideal quantizer back to the sensor

A hypothetical ideal 12-bit ADC spans 4.096 V and is preceded by a noiseless linear voltage gain of 100. Assume interior nearest-level rounding and the uniform-error model for this calculation. Find the ADC step, its RMS quantization contribution, and both the input-referred RMS value and half-step error bound.

There are 4096 codes, so $q=4.096/4096=0.001$ V, or 1 mV per step at the ADC input.
The uniform-model RMS contribution is $q/\sqrt{12}\simeq0.288675$ mV at the ADC. The interior rounding bound is ±0.500 mV there.
Dividing by gain 100 gives 2.88675 µV RMS and an input-referred rounding bound of ±5.00 µV. These quantities refer to different statistical statements and should not share an unexplained error label.

The result

The step is 1 mV at the converter; the sensor-referred contribution is about 2.887 µV RMS with an ideal interior rounding bound of ±5 µV.

The living notebook

Explore the model

Interactive lab

Interactive lab · sampling

Averaging improves only under an independence model

For each fresh comparison, Reset and set BOTH Success probability p=0.50 and Independent samples n. Compare n=4 and n=16, predicting sample-mean variances 0.0625 and 0.015625, hence standard deviations 0.25 and 0.125. Playback highlights exact binomial outcomes rather than drawing new random data. This Bernoulli model illustrates independent averaging, not ADC quantization, voltage-code linearity, reference drift, or a dither implementation.

Exact binomial massNormal interval massHighlighted outcome / mean

Mean sample proportion: 0.5
Sample-mean variance: 0.0125
Uniform CDF error bound: ≤ 0.1062

Success probability p0.5Independent samples n20

t = 0 s

Exact binomial masses for the mean of n independent Bernoulli trials, compared with normal mass between adjacent half-count boundaries; normal tails outside the finite support remain outside the bars. Playback highlights outcomes, not random samples. The Berry–Esseen bound applies to standardized CDFs, not individual bars or continuity-corrected errors. At p = 0 or 1, normal comparison and that bound are disabled. Normal-CDF evaluation uses a numerical approximation with absolute error below 7.5×10⁻⁸ (at most 1.5×10⁻⁷ per interval, before rounding). Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFour times as many independent observations halves this model’s standard deviation; fixed or correlated errors need a different analysis.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The RMS formula is the variance of a uniform error model. A constant or coherently sampled input need not produce independent white quantization errors.

Your turn

Try it yourself

A separate budget has independent sensor-referred random components of 3.00 µV RMS from the analog front end and 2.88675 µV RMS from a valid decorrelated quantization model. Compute their combined RMS and the RMS of an average of sixteen independent samples. A fixed uncorrected offset of +10 µV is also present: does that offset shrink with the average?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add independent variances before taking the square root. Average only the components explicitly declared independent between samples.

Compare with the solution

The combined random RMS is $\sqrt{3^2+2.88675^2}\simeq4.16333$ µV. Sixteen independent samples reduce it by four, to approximately 1.04083 µV RMS. The fixed +10 µV offset remains +10 µV and can dominate the averaged result. It needs a justified correction and an uncertainty for that correction, not an additional division by four. If the samples are correlated or the uniform quantization model fails, the quoted averaging improvement also need not hold.

Keep the idea

Take it with you

Code resolution, rounding bounds, RMS models, and accuracy are different statements.
Ideal converter SNR assumes a specified sinusoid, bandwidth, and error distribution.
Gain and averaging help selected contributions but do not eliminate clipping or fixed offsets.

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