-2
$\begingroup$

This question has bugged me for a while as it's not addressed in any books I've read. I was wondering how folks from this community interpreted it. Below are my personal notes on the matter (please excuse the formal tone).

My Notes

In the context of signal sampling, one cycle of an event represents a single, complete, sampling (measurement) of a value of a quantity. The interpretation of (positive) integer sampling rates is intuitive: fs = 5 Hz means 5 samples are taken every second. But how to do we interpret fractional sampling rates (e.g., fs = 1/3 = .333... Hz)?

By definition, a sampling frequency of 1/3 means one-third of a sample is taken every second -- but what does one-third of a sample mean? The idea of a third of a sample of a measured value is indeed absurd (is the measured value "a third-valid" and "two-thirds invalid"?) but one way to make sense of it is to consider the imperfect nature of a sampling operation.

Though we can conceive of sampling as an operation that occurs instantaneously in the abstract, we must realize that all system takes some non-zero amount of time to acquire a value of a quantity (a delay that manifests as a physical limitation of the measurement procedure or is introduced by hardware that carries out the analog sampling).

So it becomes conceivable that a unit of time (one second) might not be enough time for a single sample of a quantity to be available (or perhaps only one-third of the sample is available every second). We can interpret fractional frequency as an incomplete sample per unit time or the fraction of a sample that is available per unit time.

Another equivalent (perhaps more sensible) way to interpret a fractional sampling frequency is to interpret its mathematical inverse, the sampling period, which is the number of seconds required for a single, complete measurement of a value of a quantity.

Edit: Not sure why I was downvoted. I think this is a legitimate question. Please specify how I can improve the question.

$\endgroup$

closed as unclear what you're asking by Marcus Müller, Stanley Pawlukiewicz, MBaz, lennon310, Peter K. May 30 at 15:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ 1/3 Hz= it takes 3 seconds for one cycle. $\endgroup$ – Stanley Pawlukiewicz May 21 at 20:40
  • 1
    $\begingroup$ By definition, a sampling frequency of 1/3 means one-third of a sample is taken every second: No. That understanding is wrong: it just means that there's one sample every three seconds; since the rest of your question builds upon this false premise, I don't know what your question is. By the way, the notion of having one third of the info of one sample is contrary to your opinion not absurd – information theory and noisy communications deal with that problem very thoroughly. $\endgroup$ – Marcus Müller May 21 at 21:15
  • 1
    $\begingroup$ a better question would be what does 1/\pi Hz mean? $\endgroup$ – Stanley Pawlukiewicz May 22 at 0:49
  • $\begingroup$ I believe I was consistent. If 10 Hz sampling frequency meant 10 samples per sec, then one would think 1/3 Hz meant one-third of a sample per second -- or does Hz only apply to integer frequency values? $\endgroup$ – Minh Tran May 22 at 2:42
  • 2
    $\begingroup$ @MinhTran exactly this comparison is what's wrong. Samples are discrete, unlike water. You might calculate an average sample per time, but that doesn't mean anything. $\endgroup$ – Marcus Müller May 22 at 6:35
1
$\begingroup$

You better look at the sampling operation, from the sampling period point of view.

So a sampling frequency of $5$ Hz (which you consider as 5 integral samples per second) can also be understood as a uniform sampling with a sampling period of $T = 1/5 = 0.2$ seconds. Which means that every other sample is $0.2$ second apart.

So from this pint of view, if you consider a sampling frequency of $F = 5.1$ Hz, then you would be talking about a sampling period of $T = 1/5.1 = 0.1961$ seconds, which means that every next sample will be $0.1961$ seconds away.

Since time is a continuous quantity, then you won't fall into the trap of asking whether there are integral samples or not.

Of course you can equally interpret is as you did before: you will be taking $5.1$ samples per one second. But that's counter untuitive to your integral sample concept, then be aware that it doesn't mean that there is a fractional sample rather it actually means that in one second you take $5$ net samples and spend $0.1$ sampling time empty (and wait a little more to get the next (6th) sample). So actually $5$ samples would take $5 \times 0.1961 = 0.9804$ seconds (slightly less than a second). But if you perform a sampling of 10 seconds, then you would get $51$ integral samples...

And from this last thing you would deduce that $51$ samples in $10$ seconds makes a sampling rate of $5.1$ samples per second. But no, there is no fractional sample.

Here I add a simple plot of continuous signal and its samples taken at a period of $0.35$ seconds or the sampling frequency of $1/0.35 = 2.8571$ samples per second.

enter image description here

In this figure, the continuous curve represents the continuous-time signal $x(t)$ and the red squares represents samples of it at a rate of $F_s = 2.8571$ samples per second, or equivalently a period of $0.35$ seconds between two samples. So when asked to compute the rate of samples you would use the conventional formula for (time) rate of anything:

$$ \text{rate of samples} = \frac{ \text{number of samples} -1 } {\text{duration}} $$

As an example, compute this rate for the first $3$ samples, the duration is (from figure) $T_d = 0.7 $ seconds. And we drop the last sample from the count which yields actually $2$ samples contained in that duration. (more specifically the duration interval includes the left endpoint but discludes the right endpoint; $T_d = [0,0.7)$, hence the sample at the right endpoint is discluded from the sample count but the duration will be same for either $T_d = [0,7]$ or $T_d = [0,0.7)$)

So we have:

$$ \text{rate of samples} = \frac{ 3 -1 } {0.7} = 2.8571 $$

samples per one second. Note again that according to this rate of $2.8571$ samples per second, during a 1 second interval expect about $ \lfloor 2.8571 \rfloor + 1 = 3 $ complete samples to be included. And there's some empty time left after the third sample, and that's where the illusion of a $0.8571$ fractional sample appears... of course there's no fractional sample and the next sample will be at time $t= 1.05$ seconds (outisde of the first second interval).

$\endgroup$
  • 1
    $\begingroup$ I think I understand your point (and perhaps Marcus') -- it's not helpful to think about frequency in terms of a "fraction of sample" because that doesn't tell us what the value for that sample is for computation. $\endgroup$ – Minh Tran May 22 at 2:58
  • $\begingroup$ yes quite similar... $\endgroup$ – Fat32 May 22 at 16:54
  • $\begingroup$ I wouldn't quite agree, sorry! It doesn't "not tell us the value", there is no value; for all purposes here, the sampling is instant; what happens directly after one sample is taken and for the next 2.99999 s doesn't matter; what matters is the value the analog signal has exactly three seconds later. That's an important difference! The physical and mathematical model is that you multiply your analog signal with a Dirac delta comb, and not with something that integrates your signal between sampling instants, because otherwise, you couldn't describe aliasing. $\endgroup$ – Marcus Müller May 22 at 17:18
  • $\begingroup$ Hi @MarcusMüller ! You don't agree with whom? "me" or "him" ? $\endgroup$ – Fat32 May 22 at 18:53
  • 1
    $\begingroup$ @MarcusMüller yeah ok, but as you know sometimes it's best to leave things as they are... It seems his confusion can best be addressed this (or a similar) way. Perhaps I could make a few sketches and plot the samples to get the ultimate concrete picture, but frankly I'm a bit lazy for that now ;-) quite similar is the key ;-) $\endgroup$ – Fat32 May 22 at 19:53

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