I'm digitizing a zero-mean complex Gaussian white noise signal with certain variance, through independent I/Q baseband sampling (two ADCs).

The noise variance (power) depends on the thermal emission/signal of a certain microwave body, so out of the ADC I want to be able to compute the signal power to gain information of the sensed body.

I will average ADC output samples during a certain time to reduce noise; this integration time being certainly much larger than the ADC sampling period, so I expect to gain something there.

How to design the number of bits of the ADC?

I know that the input will be Gaussian (or let's assume so, in fact it will be band-limited) and that increasing the integration time or number of averaged samples will help in decreasing the variance in the estimation of the power.

Integrating for an infinite number of samples will make my estimation effectively constant (I guess I'm getting rid of quantization noise).

However, the number of bits will have a direct impact on the bias of this power estimation. Depending on the input signal power and the number of bits the output retrieved signal power will have different values.

There is probably little that I can do with averaging (I will not adjust the gain before the ADC), so that signals that occupy less of the ADC full scale will inevitably suffer from stronger biases. Am I missing something?

  • $\begingroup$ OK, this is quite a wall of text and assumptions. Let's break it down. I'm removing the things in your question you are actively not assuming $\endgroup$ Dec 20, 2023 at 22:25
  • $\begingroup$ Whether or not your signal is bandlimited has nothing to do with whether or not it's Gaussian distributed. "Bandlimitation" describes the correlation properties, "Gaussian" the instantaneous distribution. $\endgroup$ Dec 20, 2023 at 22:27
  • $\begingroup$ Integrating for an infinite number… constant (I guess I'm getting rid of quantization noise). Barely so: if you really can assume Gaussian distribution (you didn't seem sure about that), then the probability that by averaging you complete suppress quantization noise is 1. However, if your noise amplitude distribution is not symmetrical and larger than a quantization bin in support, you will not achieve that. So, this really hinges on how you model your noise; we can't do that for you: Is your input distribution really Gaussian? What's the physical reason to assume that? $\endgroup$ Dec 20, 2023 at 22:31
  • $\begingroup$ Am I missing something? Dithering, probably. $\endgroup$ Dec 20, 2023 at 22:32
  • $\begingroup$ But also consider that unless you have $0$ error in the frequency you use for mixing down, your infinite long averaging will kill the signal, because an arbitrarily small frequency offset will lead to an average signal of 0. To make matters worse, I/Q mixers typically have LO leakage and DC offset, making the signal at 0 Hz unusable, anyways. Soooo, I'm really not convinced about the physical modeling here! $\endgroup$ Dec 20, 2023 at 22:35

1 Answer 1


In order to correctly assess the power of your signals, the only parameter that you need to be able to measure correctly is the width (standard dev.) of the gaussian distribution.

This width gets larger, the larger your sample rate (because you capture more of the noise bandwidth).

Ignoring noise of your ADC (let's assume it will stay in a single value with shorted inputs), you are perfectly good, if the standard dev. of the gaussian spans at least 2-3 values.

This can be achieved either by a high enough sample rate, or higher bit depth of the ADC. If the signal is band-limited, then increasing the sample rate beyond this, will no further increase the noise standard deviation. At that point, the only way forward to measure lower powers would be more bit depth (or more front-end analog gain, if applicable).

  • 1
    $\begingroup$ or more gain in front of the ADC to measure lower powers. Agree if it is just white noise (or filtered white noise), only a few bits are needed as long as they are toggling due to the noise. The chart in this post may be of interest as it shows, for the case of AWGN, how much gain to add depending on the number of bits: dsp.stackexchange.com/questions/86991/… $\endgroup$ Dec 23, 2023 at 3:06
  • 1
    $\begingroup$ @DanBoschen yes more analog gain is a good way, but I didn't know how much the signal power range was so it might not be always possible. I have added a note on this. $\endgroup$
    – tobalt
    Dec 23, 2023 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.