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I have a digital sensor that can output at two frequencies (250Hz, 1000Hz), but with different RMS AWGN noise (.35 units RMS, .5 units RMS respectively). The signal of interest has a single frequency <125Hz (i.e. Nyquist is satisfied at either sampling mode) of which I am trying to determine the amplitude, and the window function has fixed time duration (i.e. 4x more samples for the 1000Hz mode). Which sample rate/RMS noise mode yields better SNR?

There are 2 relevant phenomena I am aware of:

  1. Oversampling can reduce quantization noise if much of the resulting spectrum increase is bandpassed (See What are advantages of having higher sampling rate of a signal?)

  2. For a fixed RMS AWGN noise, a higher frequency would mean a lower PSD at all frequencies, and thus lower noise power at the frequency of interest as well.

Are there other considerations I'm missing?

I would guess that the 4x increase in sampling rate for only a 43% increase in RMS noise would give better SNR, but I am not sure how to quantify this.

Edit: To add further clarification, assume that the samples just 'show up' in my DSP, and I have no means to provide feedback to the ADC/quantizer of the sensor. (This is because the sensor measures a wireless signal, which I'm not sure how to properly simulate locally.) Also, even though quantization noise may be 'non-white' since the signal of interest is periodic and low in amplitude with respect to the quantization intervals, let's assume it is white for simplicity. I am then simply asking for the equation that relates sampling rate and RMS noise to SNR.

I believe AWGN noise difference can be accounted for as: $SNR_{1000Hz} = (\frac{.35}{.5})^2(\frac{1000}{250})SNR_{250Hz}$, which implies that the SNR of the 1000Hz mode is 1.96x or $10 log_{10}(1.96) \approx 2.92$dB better. This is because the variance of a rectangular-windowed AWGN signal is $\frac{noise^2_{RMS}}{n}$ (not proven here).

It still remains to account for quantization noise difference (if any).

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    $\begingroup$ Are you absolutely certain the noise in both cases is white? If not that can have a significant impact on the answer and the approach you would take. Also you said the signal of interest is <125 Hz, but do you know what the actual bandwidth is? What do you mean by "Window Function"? Is this a window you are taking on a block of data prior to taking an FFT? $\endgroup$
    – Dan Boschen
    Commented Apr 15, 2017 at 22:31
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    $\begingroup$ There must be some bandwidth of the information you are interested in.... if it wasn't changing there would be no reason to sample it (?). Are you trying to determine if the tone is present or not? In which case if you were not limited by the duration of the tone, 1/f noise (phase noise etc on your sampling clock or sensor) would ultimately limit your SNR which would otherwise go up proportional to your observation time. As long as you are not running into those issues then I believe FAT32's answer is correct in that you stand to gain a 3dB SNR advantage by going with the higher rate. $\endgroup$
    – Dan Boschen
    Commented Apr 17, 2017 at 16:19
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    $\begingroup$ Well that on its own sounds like another good SE question! You should post it as that (a little different from this one which was also good as is) $\endgroup$
    – Dan Boschen
    Commented Apr 17, 2017 at 16:24
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    $\begingroup$ Also in case you did not mention anywhere do clarify if the signal from the sensor is real or complex (I assume real but good to clarify that) $\endgroup$
    – Dan Boschen
    Commented Apr 17, 2017 at 16:37
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    $\begingroup$ 10Log(1.96)= 2.92 dB. Looks like you used natural log, but decibels by definition is $10Log_{10}(x)$ $\endgroup$
    – Dan Boschen
    Commented Apr 27, 2017 at 3:54

1 Answer 1

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In theory, a 4x oversampled quantizer with n-th order noise shaping feedback, is capable of providing you equivalent bit savings (wrt a direct quantization of the critically sampled signal) of 1 bits for n=0, 2.2 bits for n=1, 2.9 bits for n=2 and 3.5 bits for n=3 (this list goes further but in practice high order noise shaping feedback is difficult to implement due to potential instability) Note that spectral shape of quantization noise (or analog noise present before sampling) is an important factor in achieving these gains. (Table excerpted from ch.4,section 4.9.2, Discrete-Time Signal Processing 2e, A.Oppenheim)

The mathematical exposition of this fact is a little long but you can find them in many DSP books which include an indepth discussion of sampling and ADC strategies...

In rough terms doubling the awgn noise power would yield 3dB SNR loss which is equivalent to a loss of less than 1 bit of linear uniform ADC. So you could expect SNR gains for feedback orders of n=1 or higher.

However the question is whether you can embed that sensor into the noise shaping quantization loop in a practically convenient way or not? If you can, then certainly go for 4X oversampled case. If your project is restricted in resources such that you cannot afford for such added DSP cost, then stay at the 1X rate.

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    $\begingroup$ So with n=0, he would have a +2.9 dB SNR gain, correct? 4x sampling rate = +6 dB gain, 0.5/.35 = -3.1 dB loss... so even without the complexity of noise shaping (which were good comments!) that is a reasonable gain to go forward with the 4x sampling case, don't you think? $\endgroup$
    – Dan Boschen
    Commented Apr 15, 2017 at 22:35
  • $\begingroup$ If I'm not missing a secret point then assuming white noise and uniform quantizers, 4x oversampling with n=0 (i.e., no feedback noise shaping) would still yield about 3 dB SNR gain, as you pointed out. $\endgroup$
    – Fat32
    Commented Apr 15, 2017 at 22:40
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    $\begingroup$ Good - wanted to make sure I didn't miss a secret point ;) $\endgroup$
    – Dan Boschen
    Commented Apr 15, 2017 at 22:57
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    $\begingroup$ @abc- Filter with whatever reasonable analog filter you can do prior to ADC sampling and then sample at 2x the bandwidth of your analog filter; there will be no advantage to sampling faster than that (and at some point ADC performance will degrade vs sampling rate). Observe the signal as long as possible; the rms noise will go down at the square root of N where N is the number of samples, assuming the noise is indeed white. (And don't window your data, assuming the noise is indeed white--other than the inherent rectangular window). $\endgroup$
    – Dan Boschen
    Commented Apr 17, 2017 at 17:02
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    $\begingroup$ as mostly expected...(unless your designing the chip itself or so) So you will be using 4x oversampling mode of the sensor & ADC combination and expecting about 3dB SNR gain... Btw It's best to measure the resulting SNR gain in an actual test bench... $\endgroup$
    – Fat32
    Commented Apr 17, 2017 at 22:38

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