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Oversampling a signal means sampling it with a significantly higher sampling frequency than the Nyquist rate. As far as I know, there are three advantages:

  1. Easier design of anti alias filter
  2. Increase in resolution in the presence of a white noise signal (I was told to assume the quantization error can under the right conditions be taken as this noise). While the uncorrelated noise adds up destructively (amplitude raises by $\sqrt N$), the signals amplitude increases by a factor of $N$. This means the SNR increases by $\sqrt N$, acting as an increase in resolution.
  3. Decrease in noise. The quantization noise power is independent of the sample rate and only depends on the amount of bits. So the same noise power is distributed over a larger frequency band, resulting in less noise power per frequency. The a low pass filter can be used to remove the high frequency part of the signal.

If I average over multiple values I am back to the lower sample rate I started with, so I don't think the noise would decrease as described in 3. This leads me to think 2. and 3. are exclusive. My question is, are 2. and 3. exclusive or can you benefit from both when oversampling?

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  • $\begingroup$ (amplitude) resolution increase and the (SQNR) noise reduction are the same things ? $\endgroup$
    – Fat32
    Nov 28, 2018 at 14:38
  • $\begingroup$ @Fat32 When I hear resolution increase I would think more bits at the output. Of course the effective resolution increases with lower noise, but the actual one doesn't, right? $\endgroup$
    – dsp_novice
    Nov 28, 2018 at 15:06
  • $\begingroup$ OK let me put it this way. Assume you have an analog signal $x_c(t)$ and you sample it with an ADC with $M = 256$ times oversampling at say $N_{adc} = 8$ bits. Then after LPF filtering at $w_c = \pi/M$ and downsampling by $M=256$, inside the digital system you can store those data samples $x_d[n]$ into $N_d = 8 + 4 = 12$ bit variables, as if you had initially sampled them at $N_d=12$ bits at critical sampling rate. Now I don't know how you consider this as a noise reduction or a resolution increase, or both or exclusive :-) ? $\endgroup$
    – Fat32
    Nov 28, 2018 at 15:23
  • $\begingroup$ Where do you get the extra 4 bits in the process? I only know downsampling by $M=256$ as keeping every 256th bit, which doesn't make sense in this application. Or do you average the values? $\endgroup$
    – dsp_novice
    Nov 28, 2018 at 16:06

2 Answers 2

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If you just average multiple samples, you are using a very low quality anti-aliasing low-pass filter which can add a ton of noise. If you then re-quantize after averaging, you add still more noise.

If you don't (re)quantize after a high quality anti-aliasing low-pass filter (but keep a ton more bit of resolution), then noise could still decrease as per (3). And if you FFT before your resampling to a lower sample rate, you still get the higher bin resolution of a longer FFT. So (2) and (3) need not be mutually exclusive.

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    $\begingroup$ If you want to reduce white noise, then averaging is probably the best quality filter you can use. . $\endgroup$
    – Ben
    May 28, 2019 at 2:34
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Your first point is true: if you oversample, your anti-aliasing low pass filter doesn't have to be as much of a "brick wall."

The second point isn't exactly true. First of all, SNR and resolution are distinct. True resolution depends on the signal bandwidth, not SNR or sampling rate. Increasing the sampling rate beyond Nyquist will only interpolate your signal, perhaps making it look smoother, but not technically increasing the resolution and/or information content of the signal.

However, oversampling by a factor of N can reduce quantization noise by a factor of N (see this excellent answer). If quantization noise is your dominant noise source, this improves your SNR. I'm not a hardware guy, but I think thermal noise (probably mainly from the antenna) typically dominates (I didn't follow this detailed answer entirely, but I think it makes this point as well). In that case, increasing the bandwidth of the receiver also increases the total noise. Therefore, if your receiver has a higher bandwidth than your signal, you are actually decreasing your total energy SNR, because all those extra samples are just more noise (if you think about it in the frequency domain). Depending on your application and how you process your signal, that noise can be folded into your signal and decrease your SNR if you aren't careful. Typically we would bandpass filter the received signal down to near Nyquist to avoid this.

As you've probably gathered at this point, your third conjecture is true with respect to quantization noise, but not thermal noise. So it depends on whether thermal or quantization noise dominates the receiver.

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