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)?