A general rule of thumb is that is you want something to not contribute to your noise budget, that it must be at least a factor of 10 higher SNR than the dominant noise source in your signal chain. As an example, if you have a signal source that is at 300 :1 SNR, run your ADC at 3000:1 and for all intents and purposes you can ignore the ADC.
The only way to do this properly is to do a noise analysis.
Post processing (via in DSP for example) has the potential to extract out salient features from above the noise but you have to be careful. You have to have sufficient bit depth so you don't introduce rounding/truncation errors. You have to ensure that you are conserving the nature of the noise (gaussian/poisson pdf) or else the noise floor may rise in an unpredictable way and may not be amenable to DSP techniques. These sorts of steps (matched filters etc.) typically at best can improve the SNR by factors of \$ \sqrt{N} \$ and often the processing cost (# of operations) often follows \$ N^2 \$ so these sorts of steps often become rapidly very expensive. But agains a proper analysis will show this.
I would caution you against assuming that a DSP technique will automatically reduce your noise. It is very important that you lot at your noise sources via histogram analysis to ensure that the PDF (Probability Density Function) is amenable to processing. I.e. it appears well behaved, Gaussian or Poisson, is not multivariate and is stationary
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.
Best Answer
You shouldn't believe every answer you see to questions, though they are often worth thinking about.
Question - Should an ADC have better resolution than the noise of the input signal?
Answer - Yes, but the answer is complicated by the type of noise, and the bandwidth of the signal and the noise. It also leaves open the question of how much better.
Let's assume a noise like the well defined noise you've drawn rather than random noise, it won't make too much difference when handwaving, but it is important when you get down to detailed calculations.
It's probably better to ask the question in a slightly different way, 'when should I stop striving to increase the ADC resolution?'
Obviously if the ADC resolution is 10mV, with 1mV of noise on the signal, then adding an extra bit to the ADC will always yield essentially an extra bit of answer accuracy. The ADC is the limiting factor, not the input noise. We can take a single reading, or take many and average them, that conclusion is unchanged.
If the ADC resolution is 1mV and the noise is 10mV, then adding an extra bit to the ADC will never improve the answer accuracy significantly. The noise is the limiting factor. If we take a single reading, then the uncertainty is limited by the noise. If we take multiple readings and average, then the ADC already has sufficient resolution to reduce the noise.
If the ADC resolution and the noise are comparable, then when taking a single reading, we can't improve the reliability of the answer (the likelihood that it will fall within the limits of our published specification for the equipment) by increasing the ADC resolution. In this circumstance, an extra bit of ADC resolution may not be worth it.
However, if we take multiple readings and average, then we really need more samples of the noise to make that process work well (reliably). We really need to resolve the noise into 4 or 5 levels, with a constant input signal, to get the most benefit from averaging it. Ten levels would be a little better, but not a factor of two better.