What is the frequency representation of nonuniform sampling?

Question

Uniform sampling can be thought of as multiplication of a function $x(t)$ with a Dirac comb function: $$\text{III}_T(t) = \sum_{k=-\infty}^{\infty}\delta(t-kT) \tag 1$$

Multiplication of $x(t)$ with $\text{III}_T(t)$ is equivalent to convolution with $\text{III}_{1/T}(f) $ in the Fourier domain, giving rise to copies of $X(f)$ spaced by $\Delta f = 1/T$ (the well known sampling theorem):

$$x(t) \cdot \text{III}_T(t) \xrightarrow{\mathcal{F}} \frac{1}{T} \sum_{k=-\infty}^{\infty} X(f-\frac{1}{T}k) \tag 2$$

Question: What is the frequency representation of a non-uniformly sampled $x(t)$? Or relatedly, what is the Fourier transform of a "non-uniform Dirac comb," assuming we know the time of each impulse?

The easy answer is that it is a sum of complex exponentials, but is there a way to simplify this, or see that it converges to a frequency domain Dirac comb as the time domain spacing approaches that of a uniform Dirac comb?

Many thanks for any insights!

Edit

Perhaps it helps to focus on the issue of aliasing. What is the spacing and weighting of spectral copies of $X(f)$ for non-uniform sampling, if any? For uniform sampling, we know that spectral copies are uniformly weighted and spaced by $1/T$ as described above.

@Fat32 suggests in the comments on his answer below that the copies are spaced evenly at $k\Omega_s/N$ (where $\Omega_s$ = Nyquist sampling rate), but have unusual weights. Can anyone confirm this, and give insight into the weights?

Answer 1

You might be interested in this particular reference paper:

• Digital Spectra Of Non-Uniformly Sampled Signals : Theories and Applications is a somewhat obscure reference (from 1988! can't go wrong!) that gives a general representation of the spectrum of a special case of non-uniformly sampled digital signals: periodically non-uniform sampled signal. It's pretty convincing and a very interesting read, which I think answers part of your question.

Other references I found but didn't spend much time on:

• Nonuniform sampling and spectral aliasing, even though it doesn't really go into the math, it has some examples of aliasing frequency and weight which depend on the sampling scheme.

• Aliasing in Spectrum of Non-uniformly Sampled Signals has a lot more math, and seems to arrive at some formula but I admit I only grazed through that one.

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EDIT: Adding details for the 1st reference

The paper deals with a special case of non-uniform sampling, Periodically nonuniform sampling, expanding on @Fats32’s answer regarding

sampling instants periodically repeated by blocks of $N$

It starts with a signal $g(t)$, band-limited to ($-1/2T, 1/2T$), with Fourier transform $G^a(\omega)$:

The signal $g(t)$ is sampled in such a structured way that the sampling time instances [...] have an overall period $MT$.

It then defines another band-limited sequence, $\bar{g}(t)$:

The sampled data sequence is then treated as if it were obtained by sampling another function $\bar{g}(t)$, [...] at uniform rate $1/T$.

and states the aim of the paper (which answers your question, at least for these types of non-uniformly sampled signals):

We are interested in finding the representation of the digital spectrum of $\bar{g}(t)$ in terms of the Fourier transform $G^a(\omega)$ of $g(t)$.

Using the framework described above, it then goes through a really nice derivation of $G(\omega)$ to arrive at two representations, from which you can see the effect of the non-uniform sampling on the spectrum:

$$G(\omega) = \frac{1}{MT} \sum_{m=0}^{M-1}\left[\sum_{k=-\infty}^{\infty}G^a\left[\omega-k(\frac{2\pi}{MT})\right]e^{j[\omega-k(\frac{2\pi}{MT})]t_m}\right]e^{-jm\omega T} \tag{2}$$

And re-writing (2) with $t_m = mT - r_mT$, $r_m$ being the ratio of $mT - t_m$ to the average sampling period $T$ gives (4):

$$G(\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty}\left( \frac{1}{M}\sum_{m=0}^{M-1}e^{-j\left[\omega-k(\frac{2\pi}{MT})\right]r_mT}e^{-jkm(\frac{2\pi}{M})}\right)G^a\left[\omega-k(\frac{2\pi}{MT})\right] \tag{4}$$

Answer 2

For the simplest case of $N$ random samples of $x(t)$ taken over a duration of $T$ and satisfying the average Nyquist rate criterion, then the resulting Fourier transform of the non-uniform samples will be an $N$-fold aliased (as if $N$ times undersampled) and weighted sum of $X(\Omega)$.

This is an approximate statement which would be ideally true if the sampling instants periodically repeated by blocks of $N$, an ideal situation. If this is not so; i.e., random sampling instants never repeat or do not extend from minus to plus infinity, or if there's no analytic expression for the non-uniform sampling instants, then a closed form expression for the Fourier transform of the non-uniform samples may not exist. (I cannot say it won't, but I think it won't)

Answer 3

If the samples are finite (e.g. not a running process) then you'd have to take the greatest common divider and resample all the vector, resulting in a "toothless" train of impulses.

For example, suppose the vector is $x=[1, 2, 1, 3]$ and it's sampled at $t=[0, 3, 5, 10]$. The time derivative will be $t'=[3, 2, 5]$, showing the step sizes. This means that the maximum sample size that would satisfy equisampling would be $T=1$, in which case the resulting vector will look like this: $y=[1, 0, 0, 2, 0, 1, 0, 0, 0, 0, 3]$. This is as "analytic" as you can get.

If $x$ is a live signal then the only thing you can do is to use chunks of the signal and perform an FFT on them as above, but any subsequent difference will mean that the first FFT will not be the same as any of the following chunks.

Answer 4

Matlab has the nufft which uses:

For a vector $X$ of length $n$, sample points $t$, and frequencies $f$, the nonuniform discrete Fourier transform of $X$ is defined as

$$Y(k)= \sum_{j=1}^{n} X(j)e^{−2\pi i t(j) f(k)}$$

where $k = 1, 2, \ldots, m$. When $t = 0, 1, \ldots, n-1$ and $f = (0, > 1, \ldots, n-1)/n$ (defaults for nufft), the formula is equivalent to the uniform discrete Fourier transform used by the fft function.

Answer 5

There's some analog non-uniform sampling theory and before microchips took over signal processing there were analog systems around to obtain the spectrum of non-uniformly sampled signals.

Analog signal processing is still active in some applications, actually this IET recent event mentions a shy revival of analog processing.

But the point is, it all ends up being processed with a CPU or GPU and unless you are doing Quantum processing research or developing a new analog processing computer, you have to work with one discrete sequence of numbers.

It's far more practical and common practice to 1st 'resample' the signal into a regular time interval dt and then apply for instance FFT.

1.- Import the signal and time reference into MATLAB

2.- Try this

https://uk.mathworks.com/help/signal/ug/resampling-nonuniformly-sampled-signals.html?s_tid=srchtitle_non-uniform%20sampling_1

3.- Unless the signal has really dispare time intervals, like a cluster of many samples pretty close each other with really low dt, followed by eons of time to next cluster, then it may be the case that some interpolation sufices to turn it into a quite 'regular' time interval signal.

You can do different 'resamplings' and measure coherence among obtained spectrums.

4.- If you supply the signal or a significant chunk of it I may be able to further help with MATLAB, free of charge, I like solving problems with MATLAB.

Awaiting answer