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Why is it, that if I have a convolution and I take the Fourier Transform, it becomes a multiplication?

Ali Caglayan
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henry
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  • @Rohan Thanks for the edit ! – henry Dec 03 '16 at 13:36
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    You could make your own research, it is in every course on the Fourier transform – reuns Dec 03 '16 at 13:38
  • @user1952009: this is a quite legitimate question. I have learnt the analytical arguments, but I realize that I still have never been truly explained this behavior. –  Dec 03 '16 at 13:47
  • @YvesDaoust ?? The answer I wrote is in every course on the Fourier transform. See for example there https://en.wikipedia.org/wiki/Convolution_theorem – reuns Dec 03 '16 at 13:49
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    @user1952009: right, this is precisely what I learnt as well and which sheds no light. –  Dec 03 '16 at 13:50
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    @YvesDaoust What do you mean with "sheds no light" ? On what ? What do you want more ? – reuns Dec 03 '16 at 13:51
  • @user1952009 Thanks for the link, but I agree with YvesDaoust, it's not so trivial. – james Dec 03 '16 at 13:52
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    @totyped I didn't say it is trivial, I said "you could make your own research, it is in every course"... And stop trolling, thank you. – reuns Dec 03 '16 at 13:53
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    Guys... calm down. Haha... Thanks a lot to user1952009 for taking you time . – henry Dec 03 '16 at 13:58

3 Answers3

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Basically, this is because Fourier modes are eigenfunctions of translation operators: $$\underbrace{e^{ik(x-p)}}_{\text{translated function}} = \underbrace{e^{-ipk}}_{\text{scalar}} ~ \underbrace{e^{ikx}}_{\text{original function}}.$$A convolution operator is build up from sums or integrals of translation operators, and hence Fourier modes also diagonalize convolution operators. The function-space analog of a diagonal matrix is a multiplication operator.

Making this precise requires some functional analysis, but this is the basic idea and the correct intuition.

Nick Alger
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    In that case it is a good idea to start by looking at how it works for the discrete Fourier transform, where you can write that the (circular) convolution operator $g[n] \mapsto g \ast h[n]$ is a linear operator i.e. a matrix $g \ast h = M g$ and it can be diagonalized $M = W H W^*$ where $W$ is the discrete FT operator (again a matrix) and $H$ is a diagonal matrix with diagonal the discrete FT of $h[n]$ – reuns Dec 03 '16 at 14:10
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Assuming everything converges absolutely, with the change of variable $y = t-x$ : $$\mathcal{F}[g(x)](\xi)\ \mathcal{F}[h(x)](\xi) = \left(\int_{-\infty}^\infty g(x) e^{-2i \pi\xi x}dx\right)\left( \int_{-\infty}^\infty h(x) e^{-2i \pi\xi x}dx\right)$$ $$= \int_{-\infty}^\infty \int_{-\infty}^\infty g(x) h(y) e^{-2i \pi\xi x}e^{-2i \pi\xi y}dxdy= \int_{-\infty}^\infty \int_{-\infty}^\infty g(x) h(t-x) e^{-2i \pi\xi x}e^{-2i \pi\xi (t-x)}dxdt$$ $$ = \int_{-\infty}^\infty \left(\int_{-\infty}^\infty g(x) h(t-x) dx\right)e^{-2i \pi\xi t} dt=\mathcal{F}[g\ast h(x)](\xi)$$

The opposite direction needs the Fourier inversion theorem, with $G(\xi) = \mathcal{F}[g(x)](\xi), H(\xi ) =\mathcal{F}[h(x)](\xi)$, and since the inverse Fourier transform is essentially the same operator as the Fourier transform, it means that $$g \ast h(x) = \mathcal{F}^{-1}[G(\xi)H(\xi)](x) \quad \implies \quad G \ast H(\xi) = \mathcal{F}[g(x) h(x)](\xi)$$

reuns
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    looks good, but needs a bit more work on the integral. dx missing in first tow integrals and suppsitiution from x to y not explained. – james Dec 03 '16 at 13:48
  • @totyped Come on.. – reuns Dec 03 '16 at 13:50
  • @totyped So if you want more details that are a lot to say : $g,h \in L^1$, look at the Fubini theorem, $L^1 \cap L^2$ is dense in $L^2$, the FT is a bounded (continuous) operator on $L^2$ and also on the space of tempered distributions so the convolution theorems stay true also for $g,h \in L^2$ or $g,h \in S'(\mathbb{R})$ the space of tempered distributions – reuns Dec 03 '16 at 14:05
  • @user1952009 Well, you start of with h(x), then you take it's Fourier Transform, within the Fourier Transform, you change x to y, but only for one x, since you keep the x from g(x). – james Dec 03 '16 at 14:13
  • @totyped Useless detail. – reuns Dec 03 '16 at 14:25
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The Fourier transform is the decompostion of a signal as a sum of sinuoids.

When you convolve two such sums, by the superposition principle you get a double sum of convolved sinusoids. By orthogonality, the individual convolutions amount to Dirac deltas (modulated by the phase shift). Hence, the Fourier decomposition of a convolution is given by the product of the coefficients where the frequencies match.

Informally, for every component of the first signal $$a_ie^{i\omega t}*\sum_j b_je^{j\omega t}=\sum_j a_i b_j(e^{i\omega t}*e^{j\omega t})=\sum_j a_ib_j\delta_{ij}=a_ib_i.$$

(Summations are understood in a broad sense.)

  • Let $f(t) = e^{i \omega_1 t},g(t) = e^{i \omega_2 t}$, how do you define $f \ast g(t)$ ? – reuns Dec 03 '16 at 14:28
  • @user1952009: don't try trolling. –  Dec 03 '16 at 14:33
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    Not at all, this is a deep question – reuns Dec 03 '16 at 14:34
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    I don't think you know the answer (which is complicated). So how do you define $f \ast g(t)$ ? – reuns Dec 03 '16 at 14:35
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    Also see what Nick Alger wrote, which means that with $f(t)=e^{i \omega t}$ then $f \ast \delta(t-a)= e^{i \omega a} f(t)$ and which is correct (in the context of the theory of distributions). But what you wrote isn't. – reuns Dec 03 '16 at 14:43
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    1st of all what you wrote is not correct, did you mean $\sum_k b_k e^{i k t}$ ? In the context of Fourier series, the convolution is circular, that is with $g,h$ both $1$-periodic : $g \ast h (t) = \int_a^{a+1} g(x)h(t-x)dx$ which is $1$-periodic too. And $e^{2i \pi n t} \ast e^{2i \pi k t} = \int_a^{a+1} e^{2i \pi n x} e^{2i \pi k (t-x)}dx = e^{2i \pi k t}\int_a^{a+1} e^{2i \pi (n-k) x}dx = e^{2i \pi k t} 1_{n = k}$. But in the context of the Fourier transform you can't write anything like this, because the Fourier transform isn't directly the limiting case of the Fourier series. – reuns Dec 03 '16 at 16:07
  • Even in with of the theory of distributions, trying to generalize this to the Fourier transform is really not easy. – reuns Dec 03 '16 at 16:09