3
\$\begingroup\$

Quantization noise refers to the modeling of the difference between an analog signal and its quantized version as an additive noise process. The power of the quantization noise can be calculated by the assumption that the difference between the analog signal and the quantized signal is uniformly distributed inside the quantization level (the power is proportional to the variance of this distribution). Under this assumption, if you quantize to N bits, with uniformly separated levels, you get that the signal to quantization noise power ratio is about 6N dB.

Now, so far I haven't spoken at all about frequency or spectrum, since the quantization noise is added to each sample of the signal without regard to the sampling rate, or the frequency of the analog signal itself.

A Sigma-Delta ADC is known to reduce quantization noise. The explanation usually involves talking about high-pass filtering of the noise, and stuff like that. But I don't understand how anything can change the power of the noise, as it is still true that the quantized signal has an error somewhere within the quantization level, and by definition of quantization you can't get a smaller error than that (maybe the answer is that Sigma-Delta ADC makes the error not uniformly distributed?). And how can the frequency domain help to analyze this?

I would appreciate to see a concrete worked-out example with numbers/parameters that show how the Sigma-Delta ADC operates on a given signal, giving low quantization noise.

\$\endgroup\$
4
\$\begingroup\$

Any quantized signal has a certain minimum quantization noise level, which is related to the number of bits per sample as you have noted. Even a delta-sigma converter cannot reduce the noise below this theoretical floor.

The noise reduction (noise shaping) that is talked about in these converters relates to the fact that they use a very coarse quantizer (very often a single comparator yielding just one bit a time), which has a very high quantization noise level at its output. This noise is uniformly distributed throughout the spectrum defined by its sample rate (which is very high).

It is the function of the noise-shaping circuitry to shift as much of this noise energy as possible to a frequency band outside the band occupied by the signal of interest, in order to get as close to the minimum noise level in the passband as possible. The noise is then removed by digital filtering. At the same time, the sample rate is reduced to the desired output sample rate (decimation).

Ultimately, a very noisy string of high-speed 1-bit samples is converted to a nearly-optimal string of N-bit samples at a lower speed. The key advantages of this approach are that it's possible to get both high resolution and high linearity with relatively low cost — properties that are particularly valuable in applications such as digital audio.

\$\endgroup\$
  • \$\begingroup\$ you might emphasise that the cleaned passband is very much narrower than the input bandwidth \$\endgroup\$ – Neil_UK May 11 '16 at 20:51
  • \$\begingroup\$ @Neil_UK: If you think that's true, then you should write up your own answer explaining why. But high-order delta-sigma modulators can do a good job of partitioning the noise from the signal, so it isn't a universal truth. \$\endgroup\$ – Dave Tweed May 11 '16 at 20:56
  • \$\begingroup\$ With single bit low order, it's orders of magnitude ratio, I agree that with higher order and multibit quantisers it can drop to one to two orders of maghnitude depending on the amount of improvement that has to be gained. I upvoted your answer rather than write my own, just suggesting that if you want to complete yours, that's the bit it's missing. I can point you to my sigma delta patents to get some initial references if you like. \$\endgroup\$ – Neil_UK May 12 '16 at 5:46
  • \$\begingroup\$ So, if I have, for example, a signal sampled at 8kHz with a 16 bit ADC, should I expect theoretically that the output for a simple rounding ADC and for a Sigma-Delta ADC to be similar (same quantization noise distribution), the differnce only being that the latter has lower cost? \$\endgroup\$ – Lior May 12 '16 at 7:17
  • \$\begingroup\$ @Lior compare the specifications of the two actual converters. You can get good and bad, high and low resolution, in all types. The choice really depends on what performance you want. If it's 16 bit over an audio bandwidth, you can find either. If you need 24 bit, you'll only find sigma delta, with sample rates getting into the 100s. If you need 100Ms/s, you'll only find pipeline SAR, 12 to 14 bits pushing 16 these days. \$\endgroup\$ – Neil_UK May 12 '16 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.