What does a fractional frequency (for discrete-time signals) mean/how to interpret it? [closed]

Question

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.

Answer 1

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.

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).