If I have a signal that is lower than the noise power, I may want to use signal averaging over time to increase SNR.

However, when the signal has been obtained after digitization, there is quantization noise and signal averaging is not possible anymore.


Are there any techniques to overcome this problem?

  • $\begingroup$ See dither: en.wikipedia.org/wiki/Dither. $\endgroup$
    – Ed V
    Aug 9, 2023 at 19:14
  • $\begingroup$ This makes little sense. Can you give rough numbers of noise level, signal level and quantization noise level. If your noise level is fairly high (which it sounds like), than the quantization noise shouldn't matter. Please add more detail and specifics $\endgroup$
    – Hilmar
    Aug 9, 2023 at 20:28

1 Answer 1


I believe the statement "However, when the signal has been obtained after digitization, there is quantization noise and signal averaging is not possible anymore." is a misunderstanding. In fact, such averaging is used to increase the effective number of bits for lower level converters through oversampling and averaging. Increasing the number of bits, and decreasing the effective quantization noise are synonymous, so in that construct the noise due to quantization is being reduced through averaging (or similarly low pass filtering since averaging is a low pass filter).

Quantization noise is an additive and independent noise source, so we can use the same techniques that are used for signals that are received below thermal noise (such as spread spectrum waveforms, including GPS) to achieve a higher SNR result.

The various techniques are typically a correlation process, which is "multiply and accumulate" such that $N$ samples are summed together such that their correlated component (the signal) grows by $20\log_{10}(N)$ while the uncorrelated component (noise that is independent from sample to sample) grows by $10\log_{10}(N)$. To the extent the noise (from any source) is independent from sample to sample, and stationary, we would get a processing gain (increase in SNR) of $10\log_{10}(N)$. Quantization noise will have this property when the sampling rate is independent of the waveform and the waveform being sampled (which itself can be signal + noise) is large enough to cross a few quantization levels.

Here's a very simple example to see this intuitively: Imagine I have a constant waveform where the true value is $0.625$. We bury that DC signal in external noise (amplified thermal noise) sufficiently to exceed a quantization threshold of a simple 1 bit A/D converter (a comparator with a threshold of 0.5). We get $N$ samples that will vary with values of $0$ or $1$. This result is the signal that has a value of 0.625 together with additive external noise that caused the signal to sufficiently exceed past the 0.5 threshold, as well as the additive quantization noise that results in values of only 0 or 1. If we take enough samples, the average of the result will approach $0.625$. There will be a small deviation from truth every time we do this experiment, and if we did that enough times we could compute the variance of our error, which is a measure of the output noise level. As we increase $N$, we will decrease this noise to the extent all noise is independent from sample to sample (and stationary, this little chink in the armor will eventually stop our windfall of noise reduction!)

Getting deeper into the math and statistics involved, when the sampling rate is independent of the waveform being sampled (which can be signal + noise itself), and the waveform crosses a few levels from sample to sample, the quantization noise will be well approximated as a noise that is independent from sample to sample (and thus "white") with a uniform distribution that has a variance of $q^2/12$ where $q$ is the quantization step size. (I detail this in this post). When we add independent noise samples, and then divide by the sum (which is specifically an average) the resulting mean will have a variance that goes down by $N$. Thus if we extend to the above example to be a multi-level converter (to avoid me having to introduce clipping noise), with quantization levels $q=1$ (such that the output may actually be -1, 0, 1, 2, -2, 1, -1 ... ) the signal as $0.625$ relative to the noise where $q=1$ will have a noise that is approximately $1/(12N)$. The $q^2/12$ approximation holds up really well for multi-level conversion when we are assured there is insignificant clipping.


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