# Are resolution increase and noise reduction from oversampling mutually exclusive?

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?

• (amplitude) resolution increase and the (SQNR) noise reduction are the same things ? – Fat32 Nov 28 '18 at 14:38
• @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? – dsp_novice Nov 28 '18 at 15:06
• 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 :-) ? – Fat32 Nov 28 '18 at 15:23
• 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? – dsp_novice Nov 28 '18 at 16:06