Perhaps I'm sleepy but I can't quite get my head around how a sigma delta converter takes the noise that is in a bandwidth of (X) Hz and by over sampling spread, that same amount of noise over 2(X)Hz.

My understanding is that basically we're doing two filtering stages.

First stage is taking the input and passing it through a relaxed analog filter ( something like a 2-pole butterworth??) with a low cut-off frequency and this allows us to use a uber-high sample rate.

Next we do some funky noise shaping to boost the high frequencies which are already outside the bandwidth we're interested in. Then we pass that through a second digital filter that has a very sharp cut-off at our corner frequency.

Lastly we use a much lower sample rate (2-3 times the input frequency) when we rebuild the signal to save on the bit count.

Do I have that correct?

So why would the noise density that's present at 0-3 kHz using a Nyquist/Shannon sample rate, spread out over the whole spectrum and effectively become less if we are using a more relaxed filter and over-sampling?

Doesn't make sense to me.

enter image description here

  • \$\begingroup\$ It would help if you have some illustrations to go with your words. \$\endgroup\$
    – jippie
    Mar 9, 2013 at 9:27

1 Answer 1


A delta-sigma modulator, used in both ADCs and DACs, comprises a difference (delta) circuit that measures the error between the input signal and the feedback signal, followed by an integrator (sigma), a quantizer (often just a comparator that yields one bit of information) and a time-domain sampler. The output of the sampler is fed back to the difference circuit through a suitable inverse quantizer (i.e., 1-bit DAC).

The trick to understanding noise shaping is to consider the quantizer as linear summing circuit that adds a "quantization noise" signal to the output of the integrator. You can then use superposition to separately evaluate the effect of the circuit on both the original signal and the quantization noise "signal".

For the original signal, the integrator is in the feed-forward path, and as you might expect, it acts as a low-pass filter for that signal — high frequencies have lower gain than low frequencies.

However, for the quantization noise signal, the integrator is in the feedback path, which means that overall, the circuit functions as a high-pass filter for the noise, reducing its gain in the low frequencies (where the desired signal is) and increasing it at the higher frequencies, where it will be subsequently removed by another filter.

It is this noise shaping that accounts for the reduction in noise density in the final passband.

  • \$\begingroup\$ So if I understand correctly, graph#2 is showing the net effect after the whole process has occured not just at an early stage of the process. The noise density would be similar to graph#1 even if you increase the sampling rate but due to the additional feedback/filtering in conjunction with oversampling you end up with a noise density like graph#2? The instructor sorta made things out as if you could get noise reductions using individual elements of the SigmaDelta converter but it appears that I misunderstood and you need all elements together to get your desired result. \$\endgroup\$ Mar 9, 2013 at 21:04
  • \$\begingroup\$ Yes. The amplitude of the quantization noise signal is constant, regardless of the sample rate. This means that the "noise density" over any given bandwidth does not vary. \$\endgroup\$
    – Dave Tweed
    Mar 10, 2013 at 2:36

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