Phoebe guidesA field team has a real problem to settle: Why Averaging Buys Effective Bits They must decide what happens before they change samples in moving average on the device. Predict the direction first.
See the relationship first
The figure reads from left to right. The blue card is samples in moving average. The middle card uses this page's rule. The green card is snr gain. Follow the arrows: set the input, use the rule, then read the result and its unit.
The audit later on checks more than one number. Here, the added model uses the baseline named below and holds every other chapter value fixed. That sentence bridges the fixtures, so the numbers do not change without a reason.
Derive the baseline in four moves
- 1
Name the input. The chapter baseline for samples in moving average is 8.
- 2
Name the rule. noise after averaging = noise/√N; ΔSNR = 10 log₁₀N; Δbits = 0.5 log₂N
- 3
Put in the chapter value. Set samples in moving average to 8. The page rule gives snr gain as 9.03 dB.
- 4
Read the result. Keep dB next to the value. Use it only within the limits on this page.
Predict, then change samples in moving average
Try Predict what happens to snr gain. Move one control, calculate, then check your idea.
Observe The readouts use noise/√N, 10 log₁₀N, and 0.5 log₂N, exactly the formulas derived above. Reset to 8 and compare snr gain.
Explain Only samples in moving average moves here. The other chapter values stay fixed.
Check yourself
What should you do before you trust the result?
What does this small model leave out?
1. Begin with repeated rounding
An analogue-to-digital converter rounds each input to the nearest code. If the input and its noise move a little between samples, one rounding error may be high and the next low. An average lets those independent errors partly cancel while a slowly changing pressure remains.
2. Separate power from amplitude
Variance, written σ², measures noise power. Root-mean-square noise, written σ, is its amplitude. Averaging N independent samples divides variance by N, so amplitude falls only by √N.
A decibel compares powers with 10 log₁₀. Each ideal ADC bit is worth about 6.02 dB, so a dB improvement can be translated into an equivalent bit gain.
3. Derive the bit gain
Average independent varianceN equal noise powers divided by N² gives σ_avg² = σ²/N.
Take the square rootσ_avg = σ/√N.
Form the power improvementOriginal variance divided by averaged variance is N, so ΔSNR = 10 log₁₀N.
Convert dB to bitsDivide by 6.02 dB/bit; the same result simplifies to 0.5 log₂N.
4. Reproduce the chapter's numbers
The 12-bit, 3.3 V converter has q = 3.3/4096 = 0.806 mV. The 2 mV noise is 2.48 code steps. To push it below one step, N must exceed (2/0.806)² = 6.16, so N = 7 leaves 2/√7 = 0.756 mV and gains 8.45 dB or 1.40 bits. N = 8 gains 9.03 dB = 1.50 bits. N = 16 gains 12.0 dB = 2.00 bits, but settles over sixteen sample periods.
5. Try the same formula
TryMove the average from 1 to 16 samples and find the first count that pushes 2 mV below the 0.806 mV code step.
ObserveNoise falls slowly with √N: four samples halve it, while sixteen divide it by four.
ExplainThe readouts use noise/√N, 10 log₁₀N, and 0.5 log₂N, exactly the formulas derived above.
The widget holds the chapter's 2 mV input noise and 0.806 mV step fixed.
- It assumes independent noise and a slowly changing signal
- Needs separate evidence
Use field evidence or a deeper model before release.
6. What the result buys you
Choose the smallest average that puts random noise below the useful resolution. More samples keep improving the number on paper, but also delay real changes. A step test must confirm that the chosen window still passes the fastest event the product needs.
7. Check yourself
1. Why does noise amplitude fall by √N rather than N?
Answer: Independent noise power divides by N; amplitude is the square root of power.
2. What does an eight-sample average buy?
Answer: 9.03 dB, 1.50 effective bits, and 2/√8 = 0.707 mV residual noise.
3. What assumption can make the gain claim fail?
Answer: Repeating or correlated error does not cancel like independent error.
These are the chapter inputs, worked results, and named teaching assumptions.
- 12-bit
- Digital resolution or converter setting
- 3.3 V
- Voltage or voltage-step value
- 0.806 mV
- Voltage or voltage-step value
- 2 mV
- Voltage or voltage-step value
- N = 7/8/16
- Named physical or model constant
- 0.756 mV
- Voltage or voltage-step value
- 8.45/9.03/12.0 dB
- Gain, loss, margin, or level ratio
- 1.40/1.50/2.00-bit
- Digital resolution or converter setting
The model does not promise extra effective resolution for a fixed code, correlated noise, drift, clipping, or a signal that changes inside the averaging window; the chapter's Under the Hood section covers those limits.