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Let's say we want to extract 1Hz signal from a noisy signal by using a LPF.

And assume we have sampled it with 100Hz sampling rate.

If we would have sampled it with 1kHz sampling rate and use the same LPF would the SNR be better? Why?

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    \$\begingroup\$ A higher sampling rate means less noise from higher frequencies is aliased into the baseband after digitization. \$\endgroup\$ – Andy aka Jul 15 '18 at 10:27
  • \$\begingroup\$ I see, so if we sample at 100Hz we get more aliased high freq. band than if we did sample at 1kHz. So the best way is to sample at very high freq. but it means too much storage. There must be a trade off. \$\endgroup\$ – floppy380 Jul 15 '18 at 13:45
  • \$\begingroup\$ Yes, there's a trade off and the price is how good you make your anti-alias filter. \$\endgroup\$ – Andy aka Jul 15 '18 at 14:40
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I see quite a few misstatements and possible misconceptions both in the answers and in the question, so let’s break down what is meant by “noise.”

  1. Analog noise within the same bandwidth as the signal.
  2. Analog noise outside of the bandwidth of the signal.
  3. Quantization noise introduced by the ADC.

I interpreted your question as implying option 1. Increasing the sampling rate will do absolutely nothing regarding that noise. No ifs or buts. That noise is there to stay.

Other answers only apply to options 2 & 3.

If your analog filter is not steep enough and you still have strong noise components above the bandwidth of the signal (option 2), particularly near the sampling rate and its multiples, then increasing the sampling rate and using an appropriate digital filter will allow you to (1) relax your analog filter requirements and (2) completely remove that out of band noise.

Regarding option 3, that’s the basic operating principle that makes sigma-delta converters possible. Increasing sampling rate, using an appropriate digital filter at a higher resolution, and decimating the output of the filter to the needed rate, allows you to increase the resolution of your samples.

You can gain 1/2 bit per each doubling of sampling rate, as long as your signal is large enough to remain uncorrelated to the sampling noise. Interestingly, this is one place where out of band noise works in your favor, as it breaks the sampling noise correlation to the signal. I.e., it serves the function of dithering (or, as physicists call it, stochastic resonance).

For slow enough signals this is a great advantage, using this technique I have obtained ~20 bits of resolution out of 12 bit converters by purposefully designing the analog anti-alias filter to introduce several bits of out-of-band random noise and oversampling by a factor of 4096.

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The SNR would be better, firstly because noise mostly being a higher frequency component would have a lesser chance of being aliased of 1Hz when sampled at 1kHz than at 100Hz

Secondly SNR is directly proportional to bandwidth or just look at Hartley shannon’s law

C = Blog_2 (1+SNR)

Where C is the digital bandwidth and B is the analog bandwidth

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  • \$\begingroup\$ Analog bandwidth is the signal's bandwidth? And what is meant by digital bandwidth? \$\endgroup\$ – floppy380 Jul 15 '18 at 22:00
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You have a 1Hz signal, with superimposed noise out to high frequencies.

Now impose a 4-pole filter, lowpass filter, at 10Hz. Thus energy at 100 Hz is down 80dB. Or about 12 bits.

At 30Hz, the noise energy is only down 40dB, or 7 bits. If the antialias filter is only 4-poles.

Thus the ability of more samples to be useful will depend on how you rolloff the noise.

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Oversampling can give you a chance to run an additional high-pass filter digitally. This can reduce out-of-band noise, and you can then reduce your sampling rate.

If you don't do that, it still gives an advantage if the noise is not correlated with the signal. Your signal adds coherently, giving you 6dB of gain for doubling the sample rate. Your noise adds incoherently, giving your 3db for the same doubling. You just got 3dB SNR improvement for 2X the samples. Of course you have to sample at 4X to get 6dB improvement, and 8X for 9dB, so there's practical limitations.

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