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I understand that the F.T. of a constant signal is the Dirac. However, I cannot find anywhere showing the derivation or proof for this. I'm trying to do it myself and am getting lost. Can anyone give a worked-out derivation that the Fourier Transform of a constant signal is the Dirac? Thank You for any help!

jstraugh
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5 Answers5

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The derivation is very simple, provided you know what a distribution is. Very briefly: A distribution is a continuous linear functional $$ L:C_0^\infty(\mathbb{R})\rightarrow\mathbb{C}. $$ I will not specify what it means for $L$ to be continuous (it's complicated). The Dirac delta distribution is the linear functional $$ \delta(\varphi):=\varphi(0). $$ The Fourier transform is defined on a subset of the distributions called tempered distritution. The Fourier transform $\mathcal{F}(L)$ of a (tempered) distribution $L$ is again a (tempered) distribution. It is defined as the linear functional $$ \mathcal{F}(L)(\varphi):=L(\mathcal{F}(\phi)). $$ If you want to Fourier transform the constant 1, you first have to identify the constant 1 with a distribution $L_1$. This is done canonically via $$ L_1(\varphi):=\int_{\mathbb{R}} 1\cdot \varphi(x)\,dx. $$ Now you can compute the Fourier transform $\mathcal{F}(L_1)$ of $L_1$: $$ \mathcal{F}(L_1)(\phi) = L_1(\mathcal{F}(\phi)) = \int_{\mathbb{R}} 1\cdot \hat\phi(x) dx = \int_{\mathbb{R}} e^{2\pi ix\cdot 0}\cdot \hat\phi(x) dx = \mathcal{F}^{-1}(\hat\phi)(0) = \phi(0) = \delta(\phi). $$ That is it! We see that the Fourier transform for $L_1$ coincides with the Dirac delta distribution $\delta$. So in the sense of distributions, the the Fourier transform of 1 is the Dirac delta distribution.

StarBug
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  • Is the constant function a tempered distribution? It seems to me that it doesn't satisfy that the gradients are vanishing when the zero-th gradient (the function itself) is included, which seems to be the case in the definitions that I've seen? – pterojacktyl Oct 10 '23 at 11:30
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    Yes, the constant function is a tempered distribution. The gradients do not need to vanish, only there are limits to how fast they can grow. – StarBug Oct 11 '23 at 12:08
  • Oh thanks! I was confusing tempered distributions with functions in the Schwartz space, but tempered distributions are linear functionals on the Schwartz space. – pterojacktyl Oct 12 '23 at 13:41
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First, it's trivial that the Fourier transform of the Dirac $\delta$ is a constant function: $$ \mathcal{F}\{\delta(x)\} = \int_{-\infty}^{\infty} \delta(x) \, e^{-i\xi x} dx = \left. e^{-i\xi x} \right|_{x=0} = 1. $$

Then we use the Fourier inversion theorem, saying that if $\mathcal{F}\{f(x)\} = F(\xi)$ then $\mathcal{F}\{F(x)\} = 2\pi \, f(\xi)$: $$ \mathcal{F}\{1\} = 2\pi\,\delta(\xi). $$

md2perpe
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  • Operationally, certainly, but I have a minor quibble that computing Fourier coefficients of a distribution $u$ by applying $u$ to exponentials is only truly guaranteed to work under some (mild) hypotheses on the distribution, since, after all, the exponentials are not test functions or distributions, but are smooth, so $u$ could be compactly supported, and are moderate growth, so $u$ could be "rapid decay". Yes, due to the remarkable robustness of such questions applied to physically meaningful situations, one does not have to explicitly mention such things to benefit from their validity. :() – paul garrett Aug 30 '20 at 22:29
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    This might be a good approach. However, the Fourier inversion theorem is valid only for a subset of functions, so it seems that more caution is required. – leonbloy Aug 30 '20 at 22:31
  • @paulgarrett. You are correct, but the same laws are valid for distributions. I decided to go with integral notation since I think that the OP understands that while using for example $\langle \cdot, \cdot \rangle$ would be too abstract. – md2perpe Aug 30 '20 at 22:35
  • @leonbloy. See my answer to paul garrett. – md2perpe Aug 30 '20 at 22:36
  • I guess what I would tell my own students is that, OF_COURSE use the obvious formula, and if you can make sense of it, the sense you find is very likely correct. And yet be alert for a small-but-non-zero of something blowing up. That is, when it's obvious what you have to do, do it, and/but be prepared for a certain (small!?) rate of surprise failures... :) – paul garrett Aug 30 '20 at 22:39
  • @paulgarrett. StarBug just added an answer where (s)he makes a more rigorous calculation. – md2perpe Aug 30 '20 at 22:42
  • A good second answer to have, I think. There're several layers of these things, especially for physicists or engineers, and also for mathematicians: first, have some sense about how to obtain the right answer (and interpret it!). "Justification" is a very murky thing, as I myself have learned later in life: e.g., if one can't imagine various things "going wrong", why prove that they don't? :) Best... – paul garrett Aug 30 '20 at 22:44
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Another approach is to consider the function $e^{-\epsilon x^2/2}.$ Obviously $e^{-\epsilon x^2/2} \to 1$ as $\epsilon \to 0.$ The Fourier transform of $e^{-\epsilon x^2/2}$ is another Gaussian, $C(\epsilon) e^{-\xi^2/(2\epsilon)},$ that tends to $2\pi \, \delta(\xi).$

(Sorry, need to go to bed, so I don't have time to show the calculations.)

md2perpe
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Yet another solution

Here I first use the formula $\mathcal{F}\{f'(x)\} = i\xi \, \mathcal{F}\{f(x)\}$: $$ 0 = \mathcal{F}\{0\} = \mathcal{F}\{\frac{d}{dx}1\} = i\xi \mathcal{F}\{1\}. $$

Then I use the fact from distribution theory that the solutions to $x \, u(x) = 0$ are $u(x) = C\,\delta(x),$ where $C$ is a constant: $$ \mathcal{F}\{1\} = C \, \delta(\xi). $$

Here we unfortunately don't directly get the value of $C$.

md2perpe
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Another one...

It's clear that $\chi_{[-R,R]} \to 1$ as $R \to \infty.$ The Fourier transform is $2 \frac{\sin R\xi}{\xi},$ which tends to $2\pi\,\delta(\xi)$ as $R \to \infty.$

md2perpe
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