I am given a noisy signal $y(t) = x(t) + w(t)$, where $x(t)$ is my desired signal and $w(t)$ is the noise. In my scenario, the noise is very strong, much stronger than the desired signal $x(t)$. However, I know it is zero mean. Hence, if I can oberve $y(t)$ directly, I can recover $x(t)$ by averaging since ${\mathbb E}\{y(t)\} = x(t)$.

However, in practice I need to replace the expected value by averaging over realizations and this can only be done in digital domain. Hence, I only have access to a quantized version of $y(t)$, quantized to some finite number of quantization levels $L = 2^b$. Now my question is, can I hope to still recover $x(t)$ from averaging a quantized version of $y(t)$, i.e., is ${\mathbb E}\{Q\{y(t)\}\} = x(t)$ under some conditions on the quantizer (and, maybe the noise)?

Numerical evidence seems to suggest it works, even under very coarse quantization, i.e., when one level of the quantizer is much bigger than my desired signal $x(t)$. My intuition here is that it kind of works like dithering: the small addditive $x(t)$ alters the statistics of $w(t)$ slightly such that the quantization levels shift slightly, which allows to recover $x(t)$.

My other thought in this direction is that under some conditions, quantization can be treated like additive noise and the randomness of $w(t)$ may lead to noise realizations being uncorrelated over realizations in the ensemble.

Can we back this intuition by a more rigorous analysis and what would be the right tool for this? Are there limits to when and how this can work? Are there other reasons that could prevent this from working in practice?

In case the background matters: I need to detect very weak signals in measurements but I can repeat the measurements many times and each time, the signal will appear in the same spot (in time) so in theory even if in each measurement my desired signal is much below the receiver's sensitivity, I should be able to recover it by averaging enough trials. I think people do things like this in areas where we need to go to the extremes regarding sensitivity, e.g., astronomy/physics and the like.

  • $\begingroup$ okay, if you repeat the measurement many times, the signal will appear in the same spot in time because it is synchronized to a signal that your equipment is generating, right? $\endgroup$ Commented Feb 17, 2021 at 23:44
  • $\begingroup$ see, with astronomy, the measurements are not the consequence of a human-generated driving signal. then to synchronize many measurements (so you can average the results) you have to derive the epoch or event clock from the data itself. $\endgroup$ Commented Feb 17, 2021 at 23:47
  • $\begingroup$ Yes, maybe astronomy wasn't the same example. I am triggering the transmission of a signal and basically listening for echos. They appear after a fixed duration (which has to do with the signal's travel time) but they may be very weak. In fact I will have a strong signal overlaying a weak one and I hope that I'll be able to cancel the strong one. $\endgroup$
    – Florian
    Commented Feb 18, 2021 at 7:39
  • $\begingroup$ So Florian, what you are essentially doing is measuring the system that exists between the transmitted signal and the echoes that you are hearing come back. Is that not correct? $\endgroup$ Commented Feb 18, 2021 at 19:02
  • $\begingroup$ Yeah kinda. At the end the parameter of interest is the exact time of flight between transmitting a known signal and receiving a weak echo. $\endgroup$
    – Florian
    Commented Feb 18, 2021 at 20:10

1 Answer 1


This will should work fine.

For each realization $k$, we can write

$$y_k[n] = x[n] + w_k[n] + q_k[n] = x[n] + v_k[n]$$

So basically define the "effective" noise $v[n]$ as the sum of the analog noise and the quantization noise $q[n]$. Averaging will converge towards the desired signal $x[n]$ as long as two conditions are met:

  1. $v[n]$ is uncorrelated with $x[n]$
  2. $v_k[n]$ is uncorrelated with $v_m[n]$ for $k \neq m$

So as long as the quantization doesn't create any correlation with the desired signal or across realizations, your good to go.

In general this will always be the case if the analog noise is significantly larger than the smallest quantization step and also uncorrelated across realizations. This will even work with a 1 bit quantizer, it just going take a really long time to get any decent signal to noise ratio.


The first condition will NOT be met if the analog noise is smaller than the quantization noise, i.e. $ \sum w_k^2[n] < \sum q_k^2[n] $

In this case the quantization noise dominates and is indeed correlated to the signal. You can just dither it: add more noise to decorrelate. The choice of the dither is primarily determined by the quantizer and not by the signal itself. If you know roughly the spectrum of your signal you can throw in a noise shaper as well to help things go faster.

It's a bit counterintuitive: by adding noise, the final effective quantization noise gets smaller.

  • 1
    $\begingroup$ Thanks that maps my intuition. The first condition is the one that's potentially troublesome as $v[n]$ contains $q[n]$ which is the result of applying the quantizer so it may not be entirely independent of $x[n]$. Would this still work of $x[n]$ were a composition of a strong and a weak signal (such that, say, the noise power is between the two)? I am really trying to find all conditions on $x$ and $w$ that must be satisfied for this to work. $\endgroup$
    – Florian
    Commented Feb 18, 2021 at 7:43
  • 1
    $\begingroup$ Good point. See edit $\endgroup$
    – Hilmar
    Commented Feb 19, 2021 at 12:40
  • $\begingroup$ Thanks for the helpful edit! "by adding noise, the final effective quantization noise gets smaller" - you're right, it sounds counter-intuitive. Then again, I think this is very similar to the idea of dithering. $\endgroup$
    – Florian
    Commented Feb 19, 2021 at 21:06
  • 1
    $\begingroup$ It is EXACTLY the idea of dithering $\endgroup$
    – Hilmar
    Commented Feb 20, 2021 at 14:03

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