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The Fourier transform of this $e^{-2\pi i f_0 t}$ is:

$$\delta(f-f_0)$$

One can just imagine this as a vertical line in Fourier Space picking out the specific frequency of that exponential.

The Fourier Transform of that complex exponential times $t$ is:

$$\frac{i}{2\pi}\delta'(f-f_0) =\frac{i}{2\pi}\delta(f-f_0)\frac{d}{df} $$

Clearly this is an operator, but I would like to imagine how this would look when plotted. Any ideas?

Φίλ λιπ
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1 Answers1

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Distributions are limits of sequences of functions, the limit being taken in the sense of distributions

$$\delta(x)=\lim_{n\to \infty} n e^{-\pi n^2 x^2},\qquad\delta'(x)=\lim_{n\to \infty} (n e^{-\pi n^2 x^2})'=\lim_{n\to \infty} - \pi n^3 2x e^{-\pi n^2 x^2}$$ Alternatively you can visualize it as two very close $\delta$ $$\delta'(x) = \lim_{h\to 0} \frac{\delta(x+h)-\delta(x-h)}{2h}$$

reuns
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  • Idk if it's possible, but I'm looking for something more exact. The same way how when you calculate the derivative of $x^2$ you get a line of slope 2x, I am looking for an explicit evaluation of one of these two limits. – Φίλ λιπ Feb 26 '20 at 20:37
  • $<\delta',f>=-f'$ there is nothing more exact. My answer is about representing it as the limit of some sequence of graphs. Also finding if such limits converge is at the core of the problems in harmonic analysis. – reuns Feb 26 '20 at 20:42
  • I can also say $<\delta, f> = f$ and yet there is a simple way of representing $\delta$ in a plot. – Φίλ λιπ Feb 27 '20 at 00:52
  • There is no way to represent $\delta$ in a plot. You need a special symbol for it exactly as for $\delta'$, and give a way to translate this symbol into the limit of a sequence of functions. – reuns Feb 27 '20 at 20:32
  • You're just wrong on that one. You can represent the delta function $\delta(x-x_0)$ as a vertical line at $x_0$ despite not being a function. It provides a decent illustration of what it is and is a good heuristic for Fourier Transforms, for instance. – Φίλ λιπ Feb 28 '20 at 08:53