Math Bridge: Averaging and ADC Resolution

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Math BridgeSensorsStruggle-friendly runway

How averaging recovers useful ADC resolution

One thread from converter bins to the chapter's ten-reading beehive average, with every square-root step shown.

Phoebe, the physics guidePhoebe guides
The one targetExplain why independent noise falls as 1/√N.
The chapter case50,000 g across a 12-bit ADC and N = 10.
What it buys youSeparate honest averaging gain from impossible anti-alias repair.

See the relationship before changing it

The figure reads from left to right. The blue card is independent samples averaged. The middle card applies this page's rule. The green card is rms input noise. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only independent samples averaged, so the numeric fixture does not switch without explanation.

Independent samples averaged changes rms input noise An input card leads through the rule averaged noise = 4.41 mV / sqrt(sample count) to the rms input noise result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. Independent random noise falls with the square root of sample count.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 10 samples.

  2. 2

    Name the relationship. averaged noise = 4.41 mV / sqrt(sample count)

  3. 3

    Substitute with units. 4.41 / sqrt(10) = 1.394 mV

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change independent samples averaged

Try Predict the direction of averaged noise = 4.41 mV / sqrt(sample count). Test another independent samples averaged, then compare rms input noise.

10 samples
Chapter baseline
RMS input noise

Observe Independent random noise falls with the square root of sample count. Reset independent samples averaged to 10 and compare rms input noise.

Explain Independent random noise falls with the square root of sample count.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only independent samples averaged moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only independent samples averaged. It is not a field prediction.

1. Start with converter bins

An analog-to-digital converter (ADC) cannot return every possible voltage. A 12-bit result has 2¹² = 4,096 codes. Splitting 50,000 g evenly gives the ideal least-significant-bit step.

q = 50,000 g / 4,096 = 12.2 g/count

The lab's measured span uses 3,275 counts, so its actual step is 50,000/3,275 = 15.27 g/count.

2. Turn a step into RMS noise

When quantisation error wanders evenly between half a step below and half a step above, its root-mean-square (RMS) size is the step divided by √12.

e_rms(1) = q/√12 = 15.27/√12 = 4.41 g
Phoebe: RMS is one number for the typical size of a changing error. It does not say every reading is wrong by exactly 4.41 g.

3. Derive the averaging rule

1

Add independent errorsFor N uncorrelated readings, error variances add: variance(sum) = Nσ².

2

Divide to form the meanThe average divides the sum by N, so its variance divides by N²: Nσ²/N² = σ²/N.

3

Take the square rootRMS is the square root of variance, so e_rms(N) = e_rms(1)/√N.

4. Try the chapter's averaging window

e_rms(N)=e_rms(1)/√N; SNR gain=10 log₁₀N; bit gain=½ log₂N

TryMove the window from 1 to 64 readings and stop at the chapter's N = 10.

RMS error
SNR gain
Effective-bit gain
Below one raw LSB?

ObserveAt N = 10 the RMS error is 1.39 g, a √10 = 3.16-fold tightening.

ExplainEvery readout uses the same 1/√N, logarithm, and bit formulas derived above.

Technical boundaries.

The calculation assumes independent dither.

Correlated drift and aliased motion do not average away by this rule
Needs separate evidence

Use field evidence or a deeper model before release.

5. Work both chapter cases

Ideal: 12.2/√12 = 3.52 g RMS, then 3.52/√10 = 1.11 g. Measured: 15.27/√12 = 4.41 g RMS, then 4.41/√10 = 1.39 g. Both improve by √10 = 3.16 because the window is the same.

6. Name what the result buys

The average makes a slowly changing beehive mass more repeatable without changing hardware. It does not create a smaller physical ADC bin, and it cannot recover a fast signal that folded during sampling.

7. Check yourself

Why are there 4,096 codes?
Answer: A 12-bit code has 2¹² = 4,096 possible values.
What does N = 10 do to 4.41 g RMS?
Answer: 4.41/√10 = 1.39 g.
Would this repair an aliased vibration?
Answer: No. Averaging reduces independent noise after conversion; an anti-alias filter must act before sampling.
Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

50,000 g span
Sensor scale, pressure, or digital result
4,096 levels
Sensor scale, pressure, or digital result
3,275-count measured span
Sensor scale, pressure, or digital result
N = 10
Named physical or model constant
stated
Chapter input or worked result

The independent-noise model is appropriate for dither, not drift, interference, or aliased motion.

Return to the chapter with the maths unpacked →