Math Bridge: Averaging and Aliasing

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Math BridgeSensorsStruggle-friendly runway

Why do four samples buy only one extra bit?

One thread from random rounding error to square-root averaging—and the coherent alias it cannot remove.

Phoebe, the physics guidePhoebe guides
The one targetKnow exactly what averaging can reduce.
The chapter caseM = 4, 6.02 dB, one extra bit.
What it buys youNever mistake an alias for random noise.

A field team faces an unresolved physical question: Why do four samples buy only one extra bit? They must answer it before changing number of averaged samples on the real device. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is number of averaged samples. The middle card applies this page's relationship. The green card is snr gain. 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 added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Number of averaged samples changes snr gain An input card leads through the page relationship to the snr gain result. SET INPUT ONE CONTROL APPLY RULE predict calculate check units READ RESULT
Walk the arrows. The square-root law needs independent error. The aliased tone is coherent: each record repeats the same false 10 Hz pattern, so averaging preserves it.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for number of averaged samples is 4.

  2. 2

    Name the relationship. reduction=√M; gain=10log10(M); bits=gain/6.02; falias=|f-kfs|

  3. 3

    Substitute the chapter fixture. Set number of averaged samples to 4. The page ledger gives snr gain as 6.02 dB.

  4. 4

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

Predict, then change number of averaged samples

Try Predict the direction of snr gain. Move one control, calculate, then check your prediction.

4
Chapter baseline
SNR gain

Observe The square-root law needs independent error. The aliased tone is coherent: each record repeats the same false 10 Hz pattern, so averaging preserves it. Reset the control to 4 and compare snr gain.

Explain Only number of averaged samples moves here. The other chapter fixtures remain fixed.

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 number of averaged samples moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only number of averaged samples. It replays the chapter fixture; it is not a field prediction.

1. One code is a rounding interval

An N-bit ADC divides its input span into 2ᴺ code regions. The reported code rounds the true input into one region, leaving an error somewhere within one step q.

Phoebe: Random-looking errors can partly cancel. A repeated wrong waveform cannot.

2. Turn a step into RMS noise

1

Find the stepq = Vref/2ᴺ.

2

Use uniform-error RMSIf error is spread evenly from −q/2 to +q/2, its RMS is q/√12.

erms(1) = q/√12

3. Average independent errors

3

Add variancesM independent errors add in variance, giving Mσ².

4

Divide the mean by MThe average has variance σ²/M, so RMS becomes σ/√M.

erms(M)=erms(1)/√M; gain=10 log10(M) dB

4. Try the sample count

reduction=√M; gain=10log10(M); bits=gain/6.02; falias=|f−kfs|

TryRaise M while the chapter's 100 samples/s and 90 Hz interference stay fixed.

RMS noise reduction
SNR gain
Equivalent bits
90 Hz still aliases to

ObserveAt M = 4, random RMS noise halves, SNR rises 6.02 dB, and the equivalent gain is one bit. The 90 Hz tone remains a 10 Hz alias for every M.

ExplainThe square-root law needs independent error. The aliased tone is coherent: each record repeats the same false 10 Hz pattern, so averaging preserves it.

Technical boundaries.

The improvement assumes uncorrelated, zero-mean noise and enough analogue dither.

Correlated noise, drift, clipping, deterministic interference, reference error, and aliases do not follow 1/√M
Needs separate evidence
Oversampling also trades bandwidth, memory, power, and delay
Needs separate evidence

Use field evidence or a deeper model before release.

5. Reproduce the chapter milestones

M=4 ⇒ √M=2, gain=6.02 dB, extra bits=1.00
M=16 ⇒ √M=4, gain=12.0 dB, extra bits=2.00

Each additional bit needs four times as many independent samples, not twice as many.

6. Protect the first sample

Use analogue bandwidth limiting before the ADC so a 90 Hz component cannot enter a 100 samples/s record as 10 Hz. Then average broadband noise only after the signal is honestly represented.

7. Check yourself

Why does M = 4 give a factor of two?
Answer: RMS noise falls by √M, and √4 = 2.
How many samples buy two ideal effective bits?
Answer: Sixteen, because 10 log10(16) ≈ 12.0 dB and 12.0/6.02 ≈ 2.
Why does the 10 Hz alias survive averaging?
Answer: It is a coherent repeated waveform, not independent random error.
Honesty boundary.

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

M = 4
Distance, wavelength, or size
M = 16
Distance, wavelength, or size
6.02 dB-per-bit relationship
Gain, loss, margin, or level ratio
the chapter's 90 Hz to 10 Hz example are reproduced exactly
Named teaching assumption

The derivation is an ideal independent-noise model, not a promise that real oversampling always adds bits; Under the Hood keeps the analogue and system constraints.

Return to the chapter with the maths unpacked →