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.