In a sigma-delta ADC, the sigma-delta modulator outputs a digital stream of 0's and 1's which are then low pass filtered and decimated. Where does the bit resolution of ADC come into the picture? What is the input to the low pass filtering operation? Remember the output of sigma-delta modulator is a stream of 0's and 1's at the oversampled rate. How is this bit stream converted to numbers which is then passed to the low pass filter?



1 Answer 1


A key concept with Sigma Delta is "noise shaping". With a typical data converter the quantization noise is uniformly distributed across the primary Nyquist frequency range of $-f_s/2$ to $+f_s/2$. Oversampling results in the same noise (which is dependent on the number of bits used) spread over a wider frequency range, such that the power spectral density (watts/Hz) reduces. Filtering thus reduces the overall noise, which is the equivalent of increasing the effective number of bits. That said we can gain 1/2 a bit for every doubling of frequency: If we double the frequency, the noise power is spread by a factor of two, so the noise density drops by a factor of 2 in power which is 3 dB, assuming we properly filter out the higher frequency noise.

With the Sigma Delta, the same thing occurs, but a feedback processes shapes the noise such that it is much lower as the spectrum approaches the lower frequencies (DC), and higher at the higher frequencies (same total noise power, but shaped to favor the lower frequencies). Thus when we low pass filter (with a digital filter for a sigma delta ADC), the output of the filter can be a much higher number of bits, given the reduced quantization noise in the spectrum of the filters output.

Every order adds another 6 dB/octave of SNR improvement. A direct ADC has 3 dB/octave as described above. A first order sigma delta has 9 dB/octave (but suffers from pattern noise, not advised to use), a second order sigma delta has 15 dB/octave, a third order has 21 dB/octave etc. 5th order sigma deltas are not uncommon as commercial solutions.

  • $\begingroup$ Thanks. My question is how does low pass filtering of 0's and 1's give a higher precision number which is a good quantisation of original input? $\endgroup$ Commented Jun 22, 2023 at 14:30
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    $\begingroup$ @user3005720. The filter coefficients are not just 0's and 1's $\endgroup$
    – Hilmar
    Commented Jun 22, 2023 at 15:25
  • $\begingroup$ The output of the modulator really is $x[n]=(-1)^{a[n]}$ which is -1 and +1 ($a[n]$ are the 0,1 bits). That binary signal is simply low-pass filtered like any other signal is. But the sample rate is high. Like 3 MHz for audio sigma-delta. $\endgroup$ Commented Jun 22, 2023 at 16:18
  • $\begingroup$ Yes, the output of the SD converter is 1's and 0's only. (or whatever your 1 bit binary weight is). Averaging is a form of low pass filtering, so that may be the simplest explanation for you: If we get an output of1 0 0 0, the average of that is 0.25 which is higher precision than 1 bit. We can get better filtering than a simple average which gives us even better performance. I'm hoping that clears it up for you? $\endgroup$ Commented Jun 22, 2023 at 17:54
  • $\begingroup$ See this post where I detail that further: dsp.stackexchange.com/questions/18267/… $\endgroup$ Commented Jun 22, 2023 at 17:57

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