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I was reading about delta sigma modulators. I read that to make the data more accurate it performs digital low pass filtering on the quantized data. So like, if we have a one bit quantizer which is pushing data out at some rate, let's assume that its a DC signal,and we take average after every four samples we increase its accuracy to 2 bits. Similarly, if we take the average every 16 samples we increase it to 4 bits.
For a single bit quantizer, it makes sense since adding 4 one bit values would result in 4 different results which could be expressed in 2 bits. Firstly, is this reasoning correct? But suppose instead, I have a 2 bit quantizer, and I average every 4 samples, what should be the resolution of the resulting output data?
Here is the link from which I was reading: http://ewh.ieee.org/r5/denver/sscs/References/2002_07_Analog_AN-283.pdf

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2 Answers 2

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If you averaged 4 samples from a standard ADC connected to a source that has band-limited (but spectrally flat) white noise then you would increase the resolution by 1 bit. This is the standard approach.

But, you are talking about a sigma delta ADC and the returns are much better than one-bit per four samples because this type of ADC produces quantization noise that is much greater at higher frequencies thus, averaging is much more effective.

enter image description here
(source: eet.com)

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  • \$\begingroup\$ Thanks for the answer....This is true but how much the resolution will increase by averaging four 2 bit values rather than 1 bit. Will it double too? \$\endgroup\$
    – sarthak
    Commented Jun 28, 2015 at 10:30
  • \$\begingroup\$ For a standard ADC averaging four n bit numbers will produce n+1 bits of resolution. \$\endgroup\$
    – Andy aka
    Commented Jun 28, 2015 at 10:33
  • \$\begingroup\$ Could you please explain why this is so?? \$\endgroup\$
    – sarthak
    Commented Jun 28, 2015 at 10:36
  • \$\begingroup\$ Atmel can: atmel.com/images/doc8003.pdf \$\endgroup\$
    – Andy aka
    Commented Jun 28, 2015 at 10:38
  • \$\begingroup\$ Thanks... a good read. And I read that it is a statistical result that if you take N measurements then the average value is sqrt(N) times more accurate. Thus, if we average 4 values the result would be twice as accurate which translates to addition of one accuracy bit.... \$\endgroup\$
    – sarthak
    Commented Jun 28, 2015 at 13:23
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You don't mention what happens to the sample rate in your question but if I assume that the 2 bit quantizer works at half the speed (compared to the one bit quantizer) then I think your reasoning is correct.

Using a 2 bit quantizer (which outputs 00, 01, 10 or 11) instead of a 1 bit quantizer (which outputs 0 or 1) at double the sample rate (!!) is equivalent I think.

The resolution only does not tell you everything, the sample rate is also important. If I half the resolution but double the sample rate my output datarate is the same !

You can just exchange resolution with sample rate and keep the same datarate.

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  • \$\begingroup\$ Could you please elucidate what you mean by datarate? My question was about the resolution of ADC with the new quantizer (2 bits). I think it should now be 4 bits. \$\endgroup\$
    – sarthak
    Commented Jun 28, 2015 at 10:14
  • \$\begingroup\$ The datarate is simply the amounts of bits of information you get out of the quantiser per second. You keep going on about accuracy of the quantiser but that by itself does not say much if you ignore the sample rate. Why can a 1 bit AD converter still have a 16 bit accuracy ?? Because it oversamples ! Taking more samples and averaging these also increases the resolution. \$\endgroup\$ Commented Jun 28, 2015 at 19:06

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