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Below are discrete samples { t1, f(t1) }, { t2, f(t2) }, ... ,{ tn, f(tn) } using Mathematica syntax.

{{7.0,0.354887404925574},{7.3,0.4003399403324751}, {7.6,0.5849632195845474},{7.9,0.8785638270289906}, {8.2,1.235637591701261},{8.5,1.6025351028031707}, {8.8,1.925797159944803},{9.1,2.160385431197419}, {9.4,2.276553177467603},{9.7,2.2643104784151262}, {10.0,2.1348061618705962},{10.3,1.9184189440030162}, {10.6,1.6598520161702173},{10.9,1.4109826017641867}, {11.2,1.2225616685298717},{11.5,1.1360369427321668}}

Now I plot the samples.

enter image description here

The samples are from a bandlimited function f(t) that I defined. However, suppose they are discrete samples of an unkown function, but we know the Fourier spectrum of f(t) is band-limited to frequencies 0.25 Hz and below. What would be a good method to use for interpolating between the samples and extrapolate beyond the samples? Here I mentioned that polynomial interpolation has less error than cubic spline interpolation in a case like this? What method would have even less approximation error in a case like this?

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  • $\begingroup$ for band-limited functions, sinc-interpolation yields the exact function. However, it does not work for extrapolation. Note, that sinc-interpolation also does not work nicely at the edges of the sampled interval, due to its non-causality and slow decay. Further note, that you might need much more samples (a larger sampled interval), since a max. frequency of 0.25Hz is already 4seconds period (your interval is just 5 seconds), so smaller frequencies will not even have a full period of sampled signal. This might make it more difficult. $\endgroup$
    – Maximilian Matthé
    Apr 10, 2017 at 6:53

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@Maximilian Matthé, The discrete samples above were made using the function

$$ \mathrm{Truth}(t) = 1.4 + \frac{\sin(13-t)}{13-t} - \sin\big((1.3)(13-t)\big) $$ In this case, the sample rate is several times the Nyquist frequncy of the signal. When that is the case, we can get a very good interpolation over the samples by finding a linear combination of eight sin(f t) terms and eight cos(f t) terms that interpolate the sixteen samples. Results using eight frequencies equally spaced from ($\tfrac{\pi}{16}$ = 0.19635 radians/second) to ($\tfrac{\pi}{2}$ = 1.5708 radians/second) are shown below.

enter image description here

This method is far better for extrapolation (i.e. beyond the samples) than any other method I know of, and the interpolation error (i.e. between samples) is less than $4 \times 10^{-12}$.

enter image description here

This example was not selected to make sure the results look very good, but it is possible to find examples where the approach doesn't excel. If the same frequencies are used and the function is nearly linear over the same sample times, then cubic spline interpolation is better than an interpolating trigonometric series. However, that issue can be overcome by selecting the range of frequencies based on how much oscillation there is between the samples. If the samples are nearly linear, then a set of frequencies closer to zero could be used. Then an interpolating trigonometric series would still work very well if the range of frequencies is appropriate. Also, the samples used in this example are equally spaced, but that is not necessary for an interpolating trigonometric series to work well.

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