How does oversampling improve the measurement precision?

Question

Regarding improving the precision of a measurement, I have heard that a measurement precision can be improved by filtering or oversampling.

When I think of filtering I can make sense because filtering reduces outlier measurement points and noise amplitude so it is easy to see that it reduces the spread around the mean value of our measurement. Less fluctuation results as narrower standard deviation hence better precision.

But I couldn't come to a similar qualitative understanding regarding oversampling resulting better precision. How could this be explained?

Answer 1

If you had two identical recordings of a piece of music both with their own background noise (assumed to be non correlated) and you added numerically those two recordings together, the music amplitude would increase by 6 dB (assuming you got things exactly aligned) but, because the noises are uncorrelated the resultant noise amplitude would be only 3 dB larger. This is because uncorrelated noises add like this: -

$$N_T = \sqrt{N_1^2 + N_2^2}$$

That's a signal-to-noise improvement of 3 dB.

In effect, the scenario above is the same as sampling a noisy signal twice. You get a 3 dB improvement in SNR. If you sampled 4 times you'd get a 6 dB improvement in SNR. If you sampled 16 times you'd get a 12 dB improvement in SNR.

Another name for this is dithering.

Answer 2

Oversampling provides more measuring points allowing averging over a higher number of samples to improve precision.

Elliot's answer is correct that oversampling helps in the presence of suitable noise but I disagree with the word "only".

As a counter-example, assume a 1 LSB peak-peak sawtooth waveform, period = final sample rate, is added to the input signal. Then a signal exceeding quantisation interval N by 0.25 LSBs will generate value N until the sawtooth reaches 0.75 LSB, and then generate N+1 for the final 25% of samples. Average enough samples (N * 0.75) + (N+1 * 0.25) and you will see N + 0.25 at he final sample rate. (I have shipped product that does this, it works).

It should be clear that other distributions including noise can produce similar increase in resolution.

A further approach is to subtract the residual error after quantization, in a feedback loop, from the input signal. This is often done with low-bit esp. 1 bit quantizers and high oversampling factors. With the right design of feedback loop it can arbitrarily reduce noise level within a specific bandwidth of interest, at the expense of noise outside that band (which is later filtered out during the decimation process). This "noise shaping" is common in higher order sigma-delta converters.

Answer 3

Oversampling only helps improve precision if your measurements are subject to randomly distributed, zero-mean noise. If the noise is limiting the effective precision of your measurements then averaging multiple samples will improve the precision of your results. I have heard (but can not provide a citation) that random noise may be intentionally introduced into a system so that oversampling can be done.

It's important to note that precision is not the same as accuracy. You won't necessarily increase the accuracy of a measurement by oversampling. Any systematic errors and uncertainty will remain.

Answer 4

This can be explained with the following analogy. Suppose you are doing an experiment where you measure the time period of a pendulum using a stop watch. Assume the watch has a resolution of 1s, implying that your individual measurements cannot be more accurate than 1s. So, for instance, if you measure a time interval of 9s then, you will be sure that the real time period lies between 8s and 10s i.e., \$9 \pm 1s\$. To ensure that your measurement are correct, you repeat the experiment multiple times.
Suppose you make 10 measurements as follows: 9s, 7s, 11s, 8s, 10s, 9s, 11s, 12s, 9s, 11s. You would expect that the true value of the time period will be given by the mean of these measurements, which is 9.7s. But should we round this number to 10s since the accuracy of our instrument is only 1s and fraction of a second cannot be measured? It turns out actually that the uncertainty in the mean goes down as \$\frac{1}{\sqrt{N}}\$
with the number of measurements. So, after averaging 10 measurements your uncertainty goes down to \$\frac{1}{\sqrt{10}}.1s = 0.3s\$. Thus, your measurement result should indeed be: \$9.7 \pm 0.3s\$, implying better accuracy (or resolution) compared to the individual measurements.
Same principle holds for oversampling. When you oversample a signal, you sample it at a very fast rate, much higher than the rate at which the signal is changing. Since the signal is not changing much in the consecutive samples, oversampling is identical to taking multiple measurements of the signal at a particular time. Thus an oversampling ratio (OSR) of 4 implies making 4 measurements for the signal value. The following low pass filtering stage is nothing but an averaging stage, which takes the average of all the extra samples from the previous stage, thus average of 4 samples in case OSR is 4. As explained above, for OSR of 4 the uncertainty (given by the input noise standard deviation or square root of the noise power input to the ADC) will go down by \$\frac{1}{\sqrt{4}} = 0.5\$. Thus, your SNR increases 4 times and you gain an additional bit of accuracy.