enter image description here [Marcel Pelgrom. Analog-to-Digital Conversion]

I'm having a hard time understanding how quantization error power changes with a different signal bandwidth? How come when we have a larger signal bandwidth, we have more quantization error power? Can someone explain it intuivitely?


1 Answer 1


The modeling of quantization noise as 'white noise' leads to the result that the quantization noise power has a flat spectrum. If you have a finite measurement bandwidth due to sampling, then the total quantization noise power is simply spread out evenly over this measurement bandwidth, leading to equation 4.7 in the text you posted.

If your signal has a lower bandwidth than your measurement bandwidth (which it should if you want to prevent aliasing), then you could simply filter out the quantization noise power that is outside your signal bandwidth, and it would not contribute to your measurement error. You can't filter out the quantization noise power that is within your signal bandwidth without also filtering out your signal. So you are 'stuck' with that part, and so that is the quantization noise power that should properly be associated with your signal measurement -- given simply by the product of the quantization noise power density and your signal bandwidth.

  • \$\begingroup\$ Ah. I see. So lower bandwidth of signal means lower quantization noise power in that? Does that mean that if I input a pure single-tone sinusoid (hence 0 bandwidth) to my ADC at a certain frequency, then I should have no quantization error? \$\endgroup\$ May 9, 2020 at 21:08
  • \$\begingroup\$ For a single-tone sinusoid, it means that you could average the quantization error to arbitrarily-low levels, such as with a lock-in amplifier. Just a bit of caution is warranted here, though, in that we are assuming uncorrelated sampling and signal. So if your sampling rate in some way to your sinusoidal frequency (maybe a rational multiple?) then the fundamental assumption of uncorrelated quantization noise breaks down. \$\endgroup\$
    – rpm2718
    May 9, 2020 at 21:15
  • \$\begingroup\$ Hmm, still a bit confused. So if I test my ADC with a single-tone sinusoid, my quantization noise will be a lot lower than that when testing a sinusoid with a larger bandwidth. Have a look at the plot in this post: electronics.stackexchange.com/questions/473330/… It is with a 900kHz input sinusoid. \$\endgroup\$ May 9, 2020 at 21:21
  • \$\begingroup\$ I don't think I can respond adequately in comments, and it is not really what they are for. A brief attempt, though -- I think the answer depends on how you frame the question....do you know it is a sine wave of a certain frequency, and want to measure the amplitude, or do you know nothing about the signal that just happens to be a sine wave, so you can't apply filtering? I consider the answer different for those two cases. \$\endgroup\$
    – rpm2718
    May 9, 2020 at 21:41
  • \$\begingroup\$ I think I understand it now. It looks to be like it's more dependent on the measurement range. So if I input a single-tone frequency and I know that my ADC needs to work with a banwdith from 0 to fs/2, then I view the spectrum from 0 to fs/2 and will see the total quantization noise. However, if I know my ADC needs to work with only fs/4 bandwidth, then I can still input a single-tone sinusoid, but view/filter my output FFT from 0 to fs/4 and hence have lower quantization error power. \$\endgroup\$ May 9, 2020 at 21:46

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