How much gain in dynamic range and SNR can be expected if we are to oversample a signal with fixed analog input bandwidth. For Example if I have a analog filter at the input which limits the bandwidth to 100 MHz. The noise power will also be limited to the 100 MHz bandwidth ($kTB$ formula for noise).

Now if I were to oversample the data by a factor of 2 how much gain should I expect and why? I know we get 3 dB gain due to spreading of quantization noise, will a similar gain be achieved also due to limiting the noise bandwidth at the input?

  • $\begingroup$ So is this solely about quantization noise? Is the "N" in SNR quantization noise? $\endgroup$ – robert bristow-johnson Oct 13 '20 at 5:00
  • $\begingroup$ No I mean SNR in the general sense including all noise sources $\endgroup$ – malik12 Oct 13 '20 at 5:04
  • $\begingroup$ well oversampling will not help regarding noise sources other than quantization noise. $\endgroup$ – robert bristow-johnson Oct 13 '20 at 5:10

OP clarified that the question in the comments as follows:

If we ignore any modulation for now and assume that we are receiving pure tones plus the band limited noise and we try to improve the SNR in post processing how much improvement can we expect by oversampling and is there a limit to it? My original question was about this aspect.

First consider the case of sampled white noise (no band-limiting of the sampled signal other than anti-alias filtering prior to the A/D converter), the improvement in SNR due to oversampling a pure tone would be the processing gain in a matched filter such as a correlator to $e^{j\omega n}$ given as $\sum x[n]e^{-j\omega n}$:

$$PG = 10\log_{10}(N) \text{ dB}$$

This is because as we sum the samples in the correlation process, the signal component of each sample which is correlated from sample to sample will increase at $20\log(N)$ dB while the noise components will increase at $10\log(N)$ dB: The total signal magnitude would be $x_1+x_2+x_3 +\ldots$ while the total noise rms magnitude would be $\sqrt{x_1^2+x_2^2 + x_3^2 + \ldots}$

But with band-limiting, as we increase the sampling rate beyond the Nyquist frequency for the bandwidth of the signal, there will be no further increase in SNR: since the signal is bandlimited (no longer white), adjacent samples will be correlated both in signal and in noise, and as we sum the samples in an attempt to increase SNR, both the signal and noise components of the signal will increase at the same rate ($20\log_{10}(N)$).

Assuming a constant noise out to $f_b$ and then rolling off after that, we would increase SNR after correlation up to the point where $f_s/2 = f_b$ (for a real signal) Increasing the sampling rate beyond that will not increase SNR. Further we have a practical limit for long averaging situations from ADC spurious free dynamic range (SFDR), phase noise from clocks and local oscillators, and similar $1/f$ noise sources where the signal is no longer stationary and increasing the averaging duration further will begin to degrade SNR. The Allan Deviation (ADEV) is an excellent statistical tool for determining the optimum duration to average given we have no other constraints on the time interval. (For more on ADEV for this application see What determines the accuracy of the phase result in a DFT bin? )

If the input signal has a fixed analog bandwidth, and you are properly not limited by the quantization noise, and front-end filtering has been sufficient to not have aliasing to occur, then oversampling will not have any effect on the SNR in band.

The idea of oversampling is to spread the quantization noise out so that it is no longer swamping out the noise floor of your signal which should be the limiting factor to SNR, and to allow for appropriate filter design to eliminate noise out of band from folding in at a lower sampling rate. These are the two most prominent reasons to oversample: increasing the effective resolution by spreading out the same total quantization noise over the full sampling bandwidth, and increasing the total frequency space in which we can do digital radio processing of signals outside our primary band of interest, for filter rejection and multi-carrier operations. For more details on the first and it's limitations please see: What are advantages of having higher sampling rate of a signal?

It is assumed that the proper receiver design will filter out all noise that is outside of your signal bandwidth (the matched filter), so given you are sampling sufficiently high according to Nyquist (2x the bandwidth, plus some margin for realistic filter realization) then there is no reason to sample any higher once the signal is properly filtered (and this only increases power dissipation, increased resources, etc).

The strategy should be to sample sufficiently high at the front-end of the digital receiver for the filter designs involved and in consideration with the analog filtering that may be done for anti-aliasing, then once out of band noise, interference and other channels are filtered out, reduce the sampling rate to be as small as possible (typically 2 samples/symbol).

This may add a further intuitive explanation: Consider a band-limited random noise process with the sampled time domain signal shown below along with a histogram of those samples along the right axis, and then immediately below that is the spectrum. As long as we sample that spectrum more than twice the bandwidth shown, no matter how much faster we sample we will still get the same histogram, from which we get the standard deviation of that noise.

If we had additional noise energy outside the signal's bandwidth, then oversampling can of course help us to eliminate that interference through filtering where it could otherwise fold into band with a lower sampling rate-- but this isn't improving SNR by oversampling, this would be implementing proper receiver design by filtering out interference (which that task alone may require a higher sampling rate).

band limited noise

Frequency Spectrum

  • $\begingroup$ Thanks for your quick response Dan. You mentioned at the start that if the analog bandwidth is fixed...oversampling will not effect the SNR. This is just by oversampling alone however if we were to decimate the oversampled signal after digital filter at required nyquist rate would that not improve the SNR? $\endgroup$ – malik12 Oct 13 '20 at 4:52
  • $\begingroup$ @malik12 It will certainly improve SNR if we have excess bandwidth (such as processing gain in a spread spectrum system). But if we are to define SNR as the Signal to Noise Ratio within the bandwidth of the waveform (as we should), and if we used sufficient quantization to be limited by the SNR of the signal itself and not noise that we are adding, then increasing that sampling rate further will not help. Make sense? $\endgroup$ – Dan Boschen Oct 13 '20 at 4:56
  • $\begingroup$ @malik12 I think the confusion may be when we consider the signal as a constant value, with white noise, in which case we can average for all time and therefore always get an increase in the SNR-- but with a modulated waveform at a symbol period T, the longest we can average for is T -- We filter down to that duration so that all samples over T are highly correlated - so you get no improvement if you average ten samples or two-hundred samples over that duration due to that correlation of the samples within time T. If you see an improvement it means you have out of band noise. $\endgroup$ – Dan Boschen Oct 13 '20 at 5:18
  • $\begingroup$ Yes this is also what I had understood. If we ignore any modulation for now and assume that we are receiving pure tones plus the band limited noise and we try to improve the SNR in post processing how much improvement can we expect by oversampling and is there a limit to it? My original question was about this aspect. $\endgroup$ – malik12 Oct 13 '20 at 5:22
  • $\begingroup$ @malik12 I updated the answer; it really doesn't change it-- the key characteristic is if the noise is white or not. $\endgroup$ – Dan Boschen Oct 13 '20 at 10:59

When referring to the SNR achieved by oversampling, we have to be careful in using the term "SNR". There are essentially two SNR's to consider:

  1. The SNR that is the signal-to-quantization-noise ratio.
  2. The SNR that is your signal-to-noise ratio. Here the noise is produced by your system. For simplicity this can be modeled by the kTB relation that you already identified.

Signal-to-Quantization-Noise Ratio

When doing research on SNR gains due to oversampling for an $N$-bit ADC, you will see this ubiquitous equation:

$$SNR = 6.02N + 1.76 + 10\log10 \big (\frac{f_s}{2B} \space \big) \space dB \space\text{over the bandwidth B}$$

Setting $f_s = 2B$, which is the theoretical Nyquist rate, gives you the simplified expression for the SNR of the ADC, which is a function of $N$ only. Increasing $f_s$ further, to let's say double to give us $f_s = 4B$, we get an additional 3 dB gain in "SNR".

Even though we've used the term "SNR", this is the ratio of the signal against the quantization noise of the ADC and not against any other sources of noise.

Signal-to-Noise Ratio

This is the "SNR" where the signal is competing against the noise produced by the system. In your case we can model the system noise captured over a bandwidth $B$:

$$P_n = kTB$$

And the SNR (linear) given a signal with power $P_s$ is simply

$$SNR = \frac{P_s}{P_n}$$

Here, the "SNR" is different than the SNR described above concerning the quantization noise. The SNR is determined by the power of the signal and the noise produced by the system only. Once you've chosen to receive a signal with a bandwidth $B$, your fate is sealed on the noise performance of your system.

Ok, then why do we oversample?

Oversampling is done when within a certain bandwidth, the quantization noise is higher than the system noise. You don't want the quantization noise to be the limiting factor on your final SNR because you can usually do something about it (oversample). The idea is to oversample so that the quantization noise is below the system noise, thus making the system noise the limiting factor as it should be.

So, as others have mentioned, if your quantization noise is already well below the system noise at a nominal sample rate, then oversampling will not increase your final SNR. Doing so is a waste of resources.

  • $\begingroup$ Thanks for your response, I have marked Dan's answer as correct because it also covers the part about processing gain. $\endgroup$ – malik12 Oct 14 '20 at 6:16
  • $\begingroup$ @malik12 Of course! I tried simplifying the general theory without getting into the processing gain since it differs between what the ADC is doing and the processing gain you can achieve after the ADC. $\endgroup$ – Envidia Oct 14 '20 at 7:01

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