It is clear that oversampling and noise shaping in A/D conversion can help in shaping quantization noise. But for D/A conversion, normally there is no quantization and hence I did not understand its use. What is the gain (advantage) of oversampling in D/A conversion?
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2$\begingroup$ "But for D/A conversion, normally there is no quantization" I think there still will be quantization noise because you are representing each sample before D/A using finite number of bits. If D/A output is $x(t) = \sum \tilde{x}[n] sinc(t-nT)$, $\tilde{x}[n] = x[n] + q[n]$ where $q$ is the quantization noise due to finite number of bits used to represent $x$. $\endgroup$– jithinDec 22, 2020 at 12:39
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2$\begingroup$ You appear to be talking about two different things at once. Are you talking about plain old DACs with some oversampling (as @justme is doing in his answer), or -- given you're talking about noise shaping -- are you talking about sigma-delta DACs? $\endgroup$– TimWescottDec 22, 2020 at 16:02
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$\begingroup$ @TimWescott , I was talking for the second. $\endgroup$– DSPinfinityDec 22, 2020 at 22:27
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1$\begingroup$ Could you edit your question to make that clear, please? $\endgroup$– TimWescottDec 22, 2020 at 23:11
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3 Answers
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If you play 16-bit audio at 48kHz, you need the DAC analog reconstruction filter to pass 20kHz and attenuate 96dB at 24kHz, which is quite steep and requires complex multistage analog filter.
The advantage of using oversampling is moving the sampling rate much higher, for example oversampling by 4x means the DAC runs at 192kHz, and the analog filter only needs to pass 20kHz and block 96dB at 96kHz, which allows for much simpler filter.
The noise shaping is much like dithering after signal processing, but instead of adding white noise uniformly to the whole band, it weights the noise to high frequencies above the audio band, which can then be filtered away by the DAC reconstruction filter, so less noise remains on the audio band, thus increasing the apparent signal-to-noise ratio of the audio band.
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But for D/A conversion, normally there is no quantization
Well, yes there is. In general, because DACs have a certain resolution (8-bit, 10-bit, etc.). But specifically to your question, in a sigma-delta modulator, there's a lot of quantization -- a sigma-delta modulator is based on the notion of a "one-bit" DAC whose output is either $v_{max}$ or $v_{min}$, and nothing in between*.
The science and art of implementing a sigma-delta modulated DAC is switching the thing between $v_{max}$ and $v_{min}$ in such a way that the desired signal is present in the output spectrum of the sigma-delta modulator, while the quantization noise is all pushed up into the higher frequencies where it's easy to filter out.
* At least, that's what we'd like.
answered Dec 23, 2020 at 0:05
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i would suggest that there is a definite advantage to 1-bit DACs that have a good balanced output (that is if one terminal is high the other must be low) going into an analog difference amp, that the 1-bit DAC suffers no non-linearity distortion. with a 1-bit, two-level output DAC, there are no intermediate steps in the DAC input/output relationship. there are no "missing codes" or non-monotonicity like you can get with a conventional DAC. the analog value is determined solely by the local duty cycle of the output that is switching between -1 and +1.
answered Dec 23, 2020 at 0:30