Sampling rate vs ADC noise tradeoff 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 &lt;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:


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?)
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).
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.