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I have posted a similar question before: ADC resolution, considering DSP. And finally I think I know the answer. Could you please have a check on it?

Here is the set up:

  • The input signal of the ADC has a noise floor with noise density of \$ 5\times10^{-4} V/ \sqrt{Hz}\$ already.
  • The input signal has been amplified to full scale of ADC.
  • The ADC has a range of 3V with 12 bits.
  • The sampling rate will be 10,000Hz.

Is it correct to say that:

  • The quantization noise RMS of the ADC will be \$ {q\over\sqrt{12} }={3V \over {2^{12} \times \sqrt{12}} } ={2.11 \times 10^{-4} V}\$
  • The noise density created by ADC quantization will be \${{2.11\times 10^{-4}}V\over \sqrt{5,000Hz}} = {{2.984\times 10^{-6}}V/\sqrt{Hz}} \$
  • The noise density is dominated by ADC input signal noise, instead of ADC quantization noise. Therefore, 12 bit ADC is sufficient in this application.

Edit: What I want to know

Sorry for the confusion. I will try to put "what I really want to know" here:

  • Lets assume that ADC is perfect and the only enemy will be analog noises.
  • Under this condition, we are employing very high Q band pass filters in DSP, with bandwidth around 0.01Hz. Therefore the \$ 0.5mV/ \sqrt{Hz} \$ noise becomes 0.05mV rms here and is acceptable for us.
  • The measurement precision will corresponds to \$ 3V/0.05mV = 2^{16} \$, which has 16 bit resolution.
  • Now let us come back to the real case. When a 12-bit ADC is to be employed, could the 12-bit resolution be simply treated as quantization noise? If this is the case, 12 bit ADC can also lead to a 16 bit resolution result.
  • What bothers me is "Can I really get a more precised result than ADC resolution WITHOUT oversampling?"

Thank you very much.

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  • \$\begingroup\$ Typically if you are talking about ADC resolution and accounting for noise sources (quantisation, system etc), one would calculate the ENOB (effective number of bits). Take a look at the wiki pages for ENOB. It will derate your 12bit ADC to something less, and it allows you to say "Accounting for noise sources x and y, we can assume that we will be able to reliably measure the ADC with z bits of accuracy". That is to say a noisy 12bit ADC performs as well as a perfect 10bit ADC. \$\endgroup\$ – Oliver Sep 23 '14 at 12:05
  • \$\begingroup\$ Wow, 0.5mV/rtHz of input noise, that's a lot! If the sample rate is 10kHz then the ADC bandwidth is 5kHz. But that's only a factor of two error. Is there any low pass filter on the input? If the input was filtered to (say) 1kHz then that would change things. \$\endgroup\$ – George Herold Sep 23 '14 at 12:48
  • \$\begingroup\$ What exactly do you mean by the DSP "handling" the noise? \$\endgroup\$ – Dave Tweed Sep 23 '14 at 15:49
  • \$\begingroup\$ @DaveTweed I have edited the post. Could you have a look? \$\endgroup\$ – richieqianle Sep 24 '14 at 10:28
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The statement:

  • The noise density created by ADC quantization will be \${{2.11\times 10^{-4}}V\over \sqrt{10,000Hz}} = {{2.11\times 10^{-6}}V} \$

is incorrect. The analog bandwidth is going to be no more than half the sampling rate. This calculation is not necessary anyway, since you already have the RMS value for this noise.

What you need to do is compute the corresponding RMS value for the analog noise at the ADC input, which is \$5\times10^{-4}\frac{V}{\sqrt{Hz}}\times\sqrt{5000 Hz} = 3.5\times10^{-2}V\$. It will be less if you can band-limit the input signal to something less than the Nyquist bandwidth.

But this gives you a worst-case scenario. It basically says that you have roughly a 100:1 (40 dB) SNR (relative to a full-scale signal) at the ADC input, which would suggest that anything over about 7 bits will be enough.

To address the broader issues you raise: The real question is what is the probability distribution that each source of noise introduces into the stream of samples. The quantizaiton noise is uniformly distributed, and has a peak-to-peak amplitude that's exactly equal to the step size of the ADC: 3V/4096 = 0.732 mV.

In comparison, the AWGN over a 5000 Hz bandwidth has an RMS value of 35 mV, which means that the peak-to-peak value is going to be less than 140 mV 95% of the time and less than about 210 mV 99.7% of the time. In other words, your digital sample words will have a distribution of ±70 mV/0.732 mV = ±95 counts around the correct value, 95% of the time.

EDIT:

  • The measurement precision will corresponds to \$ 3V/0.05mV = 2^{16} \$, which has 16 bit resolution.

Be careful — you're comparing a peak-to-peak signal value to an RMS noise value. Your actual peak-to-peak noise value is going to be about 4× the RMS value (95% of the time), so you're really getting about 14 bits of SNR.

  • Now let us come back to the real case. When a 12-bit ADC is to be employed, could the 12-bit resolution be simply treated as quantization noise? If this is the case, 12 bit ADC can also lead to a 16 bit resolution result.

The 12-bit resolution is quantization noise. And yes, its effects are reduced by subsequent narrow-bandwidth filtering.

  • What bothers me is "Can I really get a more precised result than ADC resolution WITHOUT oversampling?"

Yes. Narrow-bandwidth filtering is a kind of long-term averaging. And the wide-bandwidth sampling is oversampled with respect to the filter output. Since the signal contains a signficant amount of noise prior to quantization, this noise serves to "dither" (randomize) the signal, which, when combined with narrowband filtering in the digital domain, effectively "hides" the effects of quantization.

It might be a little more obvious if you think about it in terms of a DC signal and a 0.01-Hz lowpass (averaging) filter in the digital domain. The mean output of the filter will be the signal value plus the mean value of the noise. Since the latter is zero, the result will be the signal value. The quantization noise is "swamped out" by the analog noise. In the general case, this applies to any narrowband filter, not just a low-pass filter.

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  • \$\begingroup\$ "The statement: ... is incorrect. Quantization noise has a constant amplitude at all frequencies." For that calculation to be valid the assumption must be that the noise is constant. So what is the issue? As far as the units, The OP was simply trying to put things into the same units for direct comparison, you are right that they need not do that but it's not a bad start. \$\endgroup\$ – placeholder Sep 23 '14 at 14:45
  • \$\begingroup\$ @placeholder: You're right. I misspoke, because I frequently confuse issues related to probability distribution and those of frequency distribution. I've edited my answer. \$\endgroup\$ – Dave Tweed Sep 23 '14 at 15:26
  • \$\begingroup\$ It is really helpful. Really appreciate. \$\endgroup\$ – richieqianle Sep 24 '14 at 12:17
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If your sampling rate is f then your antialiasing filter should be set to something like (1/2.2)f, and the BW used for your calculation (assuming a 1st order filter) will be the noise bandwidth which is \$ \dfrac{\pi}{2}\$

So the BW you want to use is \$ \dfrac{10}{2.2} \dfrac{\pi}{2} = 7.14 \$ [KHz]

Given the input noise of \$ 5 x 10^{-4} * (7140)^{-0.5} = 42.25 mV\$ it looks like you can get away with a 12 bit convertor.

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  • \$\begingroup\$ You need to start by setting the bandwidth to f/2.2 (not f*2.2), and then rework your numbers from there. \$\endgroup\$ – Dave Tweed Sep 23 '14 at 13:16
  • \$\begingroup\$ You also seem to have confused the analog noise density and the quantization noise RMS value. \$\endgroup\$ – Dave Tweed Sep 23 '14 at 15:48
  • \$\begingroup\$ @DaveTweedYOu're right, I grabbed the wrong #, I meant to grab the 5 e-4 but grabbed 2.11e-4 instead. I gotta stop editing early in the Morning. Thanks for the catch. \$\endgroup\$ – placeholder Sep 23 '14 at 16:39
  • \$\begingroup\$ Sorry, my English is not native. What do you mean by get away? \$\endgroup\$ – richieqianle Sep 24 '14 at 10:11
  • \$\begingroup\$ I have edited the post. Could you have a look? \$\endgroup\$ – richieqianle Sep 24 '14 at 10:29

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