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The signal quantization consists in assigning to the different values of the sampled signal a value among the range of the $2^p$ values allowed by the $p$ bits used.

As the different values of the sampled signal can be represented by coordinates (sample value and a sampling time), I have a question:

How are the sampling times stored ? I mean how is the information that one sampling time is the sampling time of, say, the first point of the sampled signal stored ?

Also (less importantly), where is the increment in sample value (the sampling period) stored ?

Thanks in advance.

niobium
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    We bonehead engineers use the term "sampling time" or "sampling instance" instead of "ordinate". And we use the term "sample value" instead of "abscissa". Almost 100% of the time (maybe 99.99999% of the time), sample values are assumed to be equally-spaced in time and *no* sampling time information goes along with the sample value. The are transmitted and received in chronological order and equally-spaced. (Of course, the sampling instances are not perfectly equally spaced. There is "jitter". But it's *very* small and that is a design issue for A/D converters.) – robert bristow-johnson Feb 28 '24 at 17:34
  • I understand they are received in chronological order, but I am wondering how is implemented a way to store equally spaced sampling times ? Is that a stack storage (I know almost nothing about stacks, but the word came to my mind)? Plus, how is time handled in this storage ? – niobium Feb 28 '24 at 18:29
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    @niobium you're overthinking this. The samples are just bits. and they are simply stored one after the other, just like everything else a computer stores. "Stack" really comes from the wrong category of thinking. – Marcus Müller Feb 29 '24 at 01:36
  • Listen, I have never worked with such a system, myself, but somewhere on the planet are sampled-data systems with *non-uniform* sampling. Then for each sample there are two words instead of just one. There is the sample value (just like you would have for uniform sampling) and there is a timestamp for when the sample was drawn. The samples (value and timestamp) would still be put in a queue or buffer in chronological order, but they would not be considered equally-spaced in time. I imagine the math for dealing with non-uniformly sampled samples is messier than with uniform sampling. – robert bristow-johnson Feb 29 '24 at 02:17
  • You are mixing quantization and sampling, and method (how is it stored) with theory (quantization and sampling). I suggest that you edit your question, starting with a title something like "How do practical sampled-time systems quantize and store data". Then, instead of leading with a "factual" paragraph that is wrong, try saying something like "sampling happens, then quantization must happen, but how?". At that point, then your question will be come answerable, and someone will probably address it. – TimWescott Mar 02 '24 at 21:50
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    @robertbristow-johnson: I haven't worked with non-uniformly sampled systems, much, but it's certainly something that can be handled with the right theory (i.e., the Kalman filter could be easily extended to handle non-uniformly sampled data, at a potentially considerable increase in computational complexity). I wouldn't want to build a high-fidelity audio system with randomly-timed samples, though. – TimWescott Mar 02 '24 at 21:52
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    It's just that in such a case of non-uniform, sorta randomly sampled values, there would have to be two words for each sample. Both the sample value and the sample time-stamp. – robert bristow-johnson Mar 02 '24 at 22:11
  • BTW, in this thread I brought up an old paper by a friend who was able to show if there are regular-spaced missing samples, that you can *still* get perfect reconstruction if the *average* sample rate exceeds twice the highest frequency. That's a special case of non-uniform sampling. – robert bristow-johnson Mar 02 '24 at 22:15
  • Yup. Nyquist didn't say you had to have uniform samples, or even samples of the same type. He just stated that you need an average rate of at least $2 B$. If you're sampling bandlimited signals it soon becomes apparent that the Nyquist condition for other than uniform sampling is a necessary, but not sufficient condition (sample something centered around $f$ at $2 f / n$ and you'll see what I mean). So theoretically you could sample the signal and all of its derivatives up to the 20$^{th}$ at 1/20 the bandwidth and you'd be fine. There may be practical difficulties, though. – TimWescott Mar 04 '24 at 15:32

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