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

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.

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

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• 1/3 Hz= it takes 3 seconds for one cycle. – Stanley Pawlukiewicz May 21 at 20:40
• 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. – Marcus Müller May 21 at 21:15
• a better question would be what does 1/\pi Hz mean? – Stanley Pawlukiewicz May 22 at 0:49
• 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? – Minh Tran May 22 at 2:42
• @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. – Marcus Müller May 22 at 6:35

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).

• 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. – Minh Tran May 22 at 2:58
• yes quite similar... – Fat32 May 22 at 16:54
• 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. – Marcus Müller May 22 at 17:18
• Hi @MarcusMüller ! You don't agree with whom? "me" or "him" ? – Fat32 May 22 at 18:53
• @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 ;-) – Fat32 May 22 at 19:53