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To increase the ADC resolution, oversampling technique can be used which must be followed by decimation (average) method.

Oversampling is just take several samples within time period. Unlike the normal averaging in which you might lose all the extra bits from oversampling, Decimation can remove only the lowest of low bits leaving some of the higher low bit to contribute to final result.

  1. What is the lowest and higher low bit? And why the lowest low bit usually most noisy than the rest of the bits?

  2. From my understanding is that the higher the extra bit can we get, a better ADC resolution can we have. In this case why can't we just use oversampling without decimation?

    For example, a summation of 16 10-bit ADC result in 14-bit ADC and when we use decimation method (divide by 4) result in 12-bit resolution.

So why can't we just use oversampling and gain 14-bit resolution instead of 12-bit resolution with decimation?

Is that to reduce the noise after the oversampling stage?

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marked as duplicate by Scott Seidman, The Photon, Phil G, Finbarr, Dave Tweed May 17 at 21:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ If the analog noise level is properly arranged relative to the ADC resolution, then you can integrate (add up) multiple samples to improve resolution (assuming you have the bits in the integration variable to support it.) This fact alone means that you have fewer integral samples, since you had to consume more than 1 ADC sample to create each sum. If the noise is Poisson (usually is), the noise bits increase too but increase at the sqrt(N) whereas the signal increases by N. So in this case you get 4 more bits of signal but then also 2 more bits of noise. A net of 2 bits. \$\endgroup\$ – jonk May 16 at 18:07
  • \$\begingroup\$ This only works where you have a 'white noise' spectra on the input: silabs.com/documents/public/application-notes/an118.pdf \$\endgroup\$ – Jack Creasey May 16 at 18:10
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    \$\begingroup\$ Thus ADC must be linear. The noise is reduced by averaging with a tradeoff of bandwidth. If the resulting average improves SNR in dB then resolution can also improved by decimation by 1bit per 6dB improvement in SNR. The noise does not have to be random , it just needs to be reduced by averaging. This would not work say for sine noise like 50Hz if your signal frequency was great than orequal to 50 Hz. The higher frequency noise only needs to be reduced by averaging and BW reduction, so it can be sinusoidal to dither the lower frequency signal. \$\endgroup\$ – Sunnyskyguy EE75 May 16 at 18:45
  • \$\begingroup\$ There is a relationship between BW reduction and samples averaged and SNR increase. Do you know? It depends on the type of noise. \$\endgroup\$ – Sunnyskyguy EE75 May 16 at 18:49
  • \$\begingroup\$ Consider a binary 1bit ADC. measuring DC with some AC sine wave that crosses the threshold. How many samples would it take to get 10 bit accuracy of DC assuming it was this stable (<0.1%}, and the sine is not harmonically related to sampling rate but less than the Nyquist bandwidth. just using a comparator and averaging the binary outputs relative to some Vref. \$\endgroup\$ – Sunnyskyguy EE75 May 16 at 18:54
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What is the lowest and higher low bit

Here is your typical binary number:

1001101001
^... ^ ^ ^
|    | | |______ Lowest bit
|    | |________ A "higher" low bit than the lowest bit
|    |__________ A higher bit than the two above
|________________ The highest bit

Why the lowest low bit usually most noisy than the rest of the bits

Let's take a look at say.. 3 bit number:

Decimal   Binary
      0   000
      1   001
      2   010
      3   011
      4   100
      5   101
      6   110
      7   111

Note that if we have a continuous transition between numbers A and B the lowest bit will have the most changes from 0 to 1 and vise versa. And the noise is nothing but a bunch of random (or not) continuous transitions from As to Bs.

So why can't we just use oversampling and gain 14-bit resolution instead of 12-bit resolution with decimation?

Is that to reduce the noise after the oversampling stage?

Yes. The "noisy" bits don't provide any useful information (given they are just uniformly random).

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  • \$\begingroup\$ Or you mean Gaussian-ly random. most electronic noise is gaussian \$\endgroup\$ – Voltage Spike May 16 at 20:31
  • \$\begingroup\$ @laptop2d What would be the meaning of Gaussian distribution of the 0 and 1 values? I think it is still giving equal probability of having 0 or 1 - same as uniform. \$\endgroup\$ – Eugene Sh. May 16 at 20:59
  • \$\begingroup\$ Gaussian distributions can have only 0 and 1 values, but they end up being more like binomial. If you have a sampled Gaussian with only two values, you can actually do statistics on that they just won't be very good. The original distribution is mostly gaussian (with some 1/f noise, which is not gaussian). We are trying to find the mean value of the original signal that has been sampled and binned. Its like taking a histogram and then trying to find the true value of the original distribution. A binary system can be modeled as uniform, in this case it wouldn't tell you much. \$\endgroup\$ – Voltage Spike May 16 at 21:12
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So why can't we just use oversampling and gain 14-bit resolution instead of 12-bit resolution with decimation?

If the noise is gaussian and there are no other noise sources, you can find a better average of the signal. However, there are other errors that prevent you from finding a better average. ADC's have DNL and INL, in short, this means all of the bits are not spaced evenly apart. This leads to error, when oversampling this error is integrated.

If you had an ideal ADC and an ideal gaussian noise source, you would be able to oversample infinitely to find a better average. With non ideal ADC's there is intrinsic error in the measuring of the signal, which prevents us from reaching a true mean by oversampling.

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