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I'm trying to understand how oversampling can help to reduce the signal to noise ratio, so I looked this document, and at page 6 it is said that:

Along with the input signal, there will be a noise signal (present in all frequencies as white noise) that will fold or alias into the measured frequency band of interest (frequencies less than one-half of fs)

After that, it is shown the equation for the Energy Spectral Density of InBand Noise:

$$E(f) = e_{RMS} \sqrt{\frac{2}{fs}}$$

So this would mean that the higher it is $f_s$, the lower will be the spectral density of the noise spectrum. But still I can't interpret properly this relation because I don't see any: The white noise should be an intrinsic property of the signal given by the physical medium where is, so it should be independent of $f_s$.

Assuming $E(f)$ constant (uniformely distributed over all frequencies), the noise should remain the same if I sample at the Nyquist frequency or at a higher one because are the anti aliasing filters what eliminate most of the noise of the spectrum by attenuating frequencies higher than the Nyquist frequencies, while the noise in the lower frequencies remains impossible to mitigate.

Why this would be true (or not)?

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    $\begingroup$ What is the source of the noise? Is it due to quantization error? If that's the case, then the total quantization noise power is $\frac{\Delta^2}{12}$ and that is the area under the curve from -Nyquist to +Nyquist. Increasing the sample rate will cause the height of the noise spectrum to lower. $\endgroup$
    – robert bristow-johnson
    Jun 20 at 18:46

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Oversampling does not reduce the Signal to Noise ratio (SNR), but specifically can increase the Effective Number of Bits (ENOB) for the sampling process- we can for example get 16 bits of precision from a 12 bit converter through the use of oversampling. If our SNR was limited by quantization noise, then oversampling would increase the SNR.

Quantization noise from the analog to digital conversion process is well approximated as a uniform white noise process. This means it's distribution in magnitude is uniform (in contrast to "thermal noise" which is Gaussian) and it's distribution in frequency as "white" is also uniform. This approximation holds well when the sampling clock is incommensurate or independent with the signal being sampled, and the waveform transitions multiple quantization boundaries from sample to sample.

As detailed in great length in this other post, the signal to noise ratio as limited by quantization noise alone of a full scale sine wave is given as:

$$SNR = 6.02 \text{ dB/bit} + 1.76 \text{ dB}$$

This is the total noise power, distributed evenly over the spectrum extending from DC to the sampling rate $f_s$ (or equivalently $-f_s/2$ to $+f_s/2$. If we increase the sampling rate, the total noise power is the same, but it will be distributed across a wider spectrum and thus the power spectral density (in power/Hz) will be reduced.

As the OP suspects, if we were to sample a white noise source itself (rather than the "self-noise" the A/D creates), such that the power spectral density of the signal we were sampling was well above the power spectral density of the quantization noise, and if we properly filtered the signal prior to sampling to prevent aliasing, then increasing the sampling rate would have no effect.

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  • $\begingroup$ Remember Dan that you get 1 additional meaningful bit every time you quadruple the sample rate. So, to go from 12 bits to 16 bits (and get the 24.08 dB additional dynamic range) you would have to quadruple the sample rate 4 times. That's $4^4$ or 256x oversampling. That's a lotta oversampling. $\endgroup$
    – robert bristow-johnson
    Jun 21 at 2:37
  • $\begingroup$ @robertbristow-johnson Yup-- that's consistent with the "10Log(N)" in the linked post. I think 1/2 a bit for every doubling of frequency (same thing). Unless we go Sigma Delta, but then we're clearly not talking about uniform white quantization noise-- but we ARE talking about very significant increases in ENOB for a modest increase in sampling rate (4th order Sigma Delta approximates 27 dB/octave or 4.5 bits every time we double!--- I believe this is closely related to your "fraction-saving" technique, as a first order SD. $\endgroup$
    – Dan Boschen
    Jun 21 at 2:48
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    $\begingroup$ The "fraction saving" is first-order noise shaping with a zero (in the quantization-to-output transfer function) right on DC (like $z=1$). ΣΔ doesn't necessarily mean 1-bit. But for audio it usually is a 1-bit quantizer. But any oversampling that isn't white quantization noise does have to have noise shaping, and if it does, then it's ΣΔ. $\endgroup$
    – robert bristow-johnson
    Jun 21 at 3:34

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