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Assertion: The noise due to quantization error for an ideal 12-bit ADC gives a signal to noise ratio of 74dB, assuming a signal with full dynamic range at the input.

Suppose I was using an ideal ADC to sample a noiseless analogue signal (i.e. analogue circuitry did not introduce any additional noise to the original input signal). And suppose I was using a low pass filter such that signal attenuation at 1kHz was -74dB. Is it true to say that sampling at greater than 2kHz is called oversampling, and that oversampling will not improve the SNR?

Edit1: 'Low pass filter' is an analogue filter in the above question

Edit2: Bode plot added. fs=2kHz, fa=50Hz.Filter Bode plot

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  • \$\begingroup\$ Your assertion is wrong for a dynamic signal. If you sample at 1000 Sps and your signal is 400 Hz you will have a significant quantization noise and your actual 400 Hz spectrum will be several dB lower than if the signal were (say) 40 Hz (or) 4 Hz. \$\endgroup\$ – Andy aka Sep 25 '18 at 16:30
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    \$\begingroup\$ @Andyaka - I don't understand your comment. Surely, assuming a brick wall reconstruction filter the 400Hz component will be of full amplitude with the quantization noise spread over 0-500Hz at a -74dBfs level. \$\endgroup\$ – Kevin White Sep 25 '18 at 16:46
  • \$\begingroup\$ @Andyaka. Can I add the following caveat then, please: My bandwidth of interest only contains signals at 0dB. For example, bandwidth of interest is 0-50Hz and these are all attenuated 0dB. 400Hz may be attenuated (say -20dB for the sake of argument) but I'm not interested in signals more than 50Hz. \$\endgroup\$ – DB17 Sep 25 '18 at 16:54
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    \$\begingroup\$ Actually I already provided an answer here: dsp.stackexchange.com/questions/40259/… \$\endgroup\$ – Dan Boschen Sep 26 '18 at 3:10
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    \$\begingroup\$ If you are concerned only with total noise power then yes it will reduce that power by filtering to the extent we can model that noise as a distributed noise process (spurs in practical implementations will ultimately limit the ability to keep adding bits.) However many processes are sensitive to noise density and as such do filter the signal that may not be as obvious as an actual filter implementation. So a broader more accurate statement; oversamping will reduce the noise density (power/Hz). \$\endgroup\$ – Dan Boschen Sep 26 '18 at 15:03
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No, it isn't true at all.

Even if quantization noise is the only source of error in your system, oversampling allows you to spread that noise over a wider bandwidth, and noise shaping allows you to move most of the noise energy to a frequency band that you don't care about, improving the SNR in the band of interest.

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  • \$\begingroup\$ Thanks for your answer. I may not have framed my question correctly. Let's assume that my bandwidth of interest is 0-50Hz. I'm not interested in frequencies above 50Hz. My filter is 0dB up to 50Hz and then starts to roll off, such that it is -74dB at 1kHz. Would sampling at >2kHz reduce the noise in the 0-50Hz band? (Assume I am not doing any digital signal manipulation.) \$\endgroup\$ – DB17 Sep 25 '18 at 17:00
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First of all if your signal of interest is 50 Hz maximum then providing your sampling rate is greater than twice 50 Hz you don't need an anti alias filter. However, sampling at a higher rate does improve the ability to restore that signal back to an analogue waveform without loss of amplitude. Consider these two scenarios: -

enter image description here

In the top picture a waveform is sampled 20 times per cycle and in the lower waveform it is sampled at only 10 times per cycle. Clearly, there is more noise in the signal that is sampled fewer times per cycle even though the sampling rate is still significantly greater that the nyquist rate.

That noise is "out of band" and, when reconstructed with a DAC, it can be filtered away but, using a flat filter will result in an observed attenuation of higher frequency base band signals.

It probably can't be easily seen but, the RMS values of both sampled waveforms are identical but one contains more sampling noise than the other hence its baseband spectrum must be reduced.

Consider a sinewave sampled about \$2\frac{2}{3}\$ times per cycle: -

enter image description here

Because the RMS of the sampled waveform AND the RMS of the unsampled waveform are identical, AND, there is clearly more noise in the sampled waveform, the spectral content of the baseband signal that is contained in the sampled signal MUST have reduced.

This means that sampling at a higher frequency can avoid an anti-sinc filter.

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  • \$\begingroup\$ Whoa! I never knew about that relationship regarding r.m.s. values. Thank you so much for putting that together to explain, I don't recall seeing that in any textbook. Taking your 10Hz and 20Hz examples, one can see that, as the sampling frequency is increased, the difference between sample amplitudes in successive readings decreases. At some high sampling frequency F, this amplitude difference will be equal to the ADC resolution. Assuming that no digital averaging is performed, is there any point sampling faster than F? \$\endgroup\$ – DB17 Sep 25 '18 at 18:24
  • \$\begingroup\$ You need to sample at more than twice the highest frequency or you will get aliasing. Aliasing is the sampling noise moving into the baseband spectral area. \$\endgroup\$ – Andy aka Sep 25 '18 at 19:34
  • \$\begingroup\$ Why do you say noise is greater in low sampling frequencies? That’s not true. Your analysis of the visual appearance of higher quantization error in time domain is not accurate. Quantization noise is due to the fact that the original signal is sampled with an error of +- 1/2 LSB, this has nothing to do with sampling frequency. \$\endgroup\$ – PDuarte Sep 25 '18 at 23:49
  • \$\begingroup\$ @PDuarte I think you need to study more carefully what I’ve said and drawn. The 0.5 LSb thing you mention is preposterous. I’ve assumed perfect infinite quantisation and if your attitude is to downvote first and ask questions later then that won’t make you friends. \$\endgroup\$ – Andy aka Sep 26 '18 at 6:21
  • \$\begingroup\$ @Andyaka I’m sorry if I offended you. That was not my intention. Back to the problem, if you didn’t consider the quantization noise in your explanation then it’s even worse because nyquist theorem says you can perfectly reconstruct the signal in both cases of your first two charts. Why did you state “Clearly there is more noise in the signal that is sampled fewer times per cycle”? That’s what I’m saying is not accurate \$\endgroup\$ – PDuarte Sep 26 '18 at 9:52

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