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If I want to Fourier transform a function $$t \in [-1,1] \to f $$ but this function and it's first $n$ differentials are not equal at the edges : $$f^{(k)}(-1) \neq f^{(k)}(1) , k \in \{0,1,\cdots,n\}$$, which function $g$ can I additively adjust $f$ with so that the resulting function $h(t) = f(t)+g(t)$ has as sparse a Fourier transform as possible? Let us for simplicitys sake assume that f is awfully nicely behaving on the inside of the interval $]-1,1[$ having bounded $0$:th to $k$:th derivatives there.

Now the trivial solution is to select $g(t)=-f(t)$. This one I am not interested in as it does not help me in any sense. See I still need to store the information about $g$ as this is for data compression purposes.

Own work One simple approach I have thought about is to use a linear compensatory function $$2g(t) = -f(-1)-f(1) + t\cdot(f(-1)-f(1))$$ This will remove the step discontinuity as $h(t)$ should be 0 at the edges, but probably leave us having discontinuities in all the derivatives.

Having started with a polynomial... for the purpose of expanding into being able to attack and remove higher order discontinuities it could be tempting to continue into the realm of polynomial splines or Bezier curves or something like that?

Joako
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mathreadler
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  • A linear $g(t)$ was exactly my first thought. Why do you think it would introduce discontinuities in the derivatives? – aschepler Jun 29 '23 at 23:02
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    Not introduce new ones. I mean it may not remove any discontinuities already present in the derivatives. Derivative from the right $f'(-1)$ may differ from derivative from the left $f'(1)$ and this may still be the case for $h$ in those points if choosing such a linear $g$. But allowing $g$ to be a higher order polynomial maybe we can smooth those out as well. – mathreadler Jun 29 '23 at 23:42
  • Your idea of a compensatory function seems like a good one. Your linear function has the right impact on the function values. You want a function that also compensates the derivatives. You just need a polynomial with suitable derivatives at the two ends of the interval. Hermite interpolation will give you this. – bubba Jun 30 '23 at 08:10

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Lets imaging the following: you have a function $f(t)$ which is defined for any time $t$, such as you aim to look on their Fourier Transform of its derivative, but just on a specific time interval $[t_0;\ t_f]$ ($t_0<t_f$).

As you noted, while $f(t_0)=f(t_f)=0$ is not fulfilled, when taking the standard Fourier Transform $F(iw)$ with integration limits $t_0$ and $t_f$ will take "inside" the discontinuities due the jumps made by the edges of the domain $f(t_0)\neq 0$ and $f(t_f)\neq 0$.

This situation could be avoided by making the function $x'(t)=f'(t)\cdot \left(\theta(t-t_0)-\theta(t-t_f)\right)$ where $\theta(t)$ is the Heaviside step function, finding $X(iw)$, matching it with $F(iw)$ and matching the corresponding function you are interested in (it is a bit long but the details are explained in this other answer I made to my own question, but caution since its old and have many conceptual misconceptions - but the math its well done), this would lead to the following correction terms: $$\int\limits_{t_0}^{t_f} f'(t)e^{-iwt}\ dt = iw\left[\int\limits_{t_0}^{t_f} f(t)e^{-iwt}\ dt\right]+f(t_f)e^{-iwt_f}-f(t_0)e^{-iwt_0}$$

Later I found this formula in a math book almost like a footnote and with a sign differences, so maybe there is something defined different among what I did and the author (be cautious since the Fourier Transform have several definitions among authors). I don't know why is so underrated but I am confident it works because of the examples I solve within the mentioned answer. Hope it helps.

book: "The Fourier Transform and its applications (3rd Edition)" by Ronald N. Bracewell, Chapter 10 section "Finite Fourier Transform" (page numbered 243).

Joako
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  • I am not completely sure about this, but maybe you could use the following: let $f(t)$ the function you are interested in, and $F(t)$ its antiderivative ($F(t)=\int_{-\infty}^t f(u)\ du$), then, instead of taking the Fourier Transform of $f(t)$ alone within the interval $[t_0,\ t_f],\ t_0<t_f$, take the transform of: $$\int\limits_{t_0}^{t_f} e^{-iwt}\left[ f(t)\color{red}{+}\left(F(t_0)\delta(t-t_0)-F(t_f)\delta(t-t_f)\right)\right]\ dt$$

    and let me know if it works, I got messed up looking for this so maybe there is a sign mistake where is red.

     – Joako Nov 06 '23 at 20:21
  • I have realized later that it is subtracting the DC component, so it is equivalent under the Finite Duration Fourier Transform to: $$\int\limits_{t_0}^{t_f} e^{-iwt}\left[f(t)-\int\limits_{t_0}^{t_f} f(u)\ du \right]\ dt = \int\limits_{t_0}^{t_f} e^{-iwt}\left[f(t)-\hat{f}(iw)\Biggr|_{w=0} \right]\ dt$$ – Joako Nov 06 '23 at 23:58
  • Caution!: I have been testing the formulas from my last the comments and I don't know if they are right, since I found results which don't match. Maybe I messed something up when doing factorizations by $iw$ since some Cauchy P.V. could rise and I missed them (my knowledge on working with distributions is very limited). – Joako Nov 07 '23 at 03:34
  • I found the problem: actually the effective subtraction is $$\int\limits_{t_0}^{t_f} e^{-iwt}\left[f(t) -\delta(t-t_f)\int\limits_{t_0}^{t_f}f(u)\ du +\left(\delta(t-t_0)-\delta(t-t_f)\right)\int\limits_{-\infty}^{t_0}f(u)\ du\right]\ dt$$ but I am not sure what its is the last term, thinking in making a spectrogram, as example. – Joako Nov 08 '23 at 21:35