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Can someone refer me on the Fourier transform of the fourier transform or clarify it for me?

It is known that the F.T. of the F.T. is some small modification of the original function itself, but I can't find anything about this online.

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    $\begingroup$ If i recall correctly, Fourier transform of a Fourier transform of $f(x)$ gives us the function $f(-x)$ $\endgroup$ – Jakobian May 31 '18 at 19:35
  • $\begingroup$ @Adam That's exactly what i'm looking for. Where can I find a more detailed source with explanation? ALso, this doesn't make sense to me intuitively $\endgroup$ – JobHunter69 May 31 '18 at 19:36
  • $\begingroup$ I'll check my notes. It's like i said. It's a direct conclusion from the formula for Fourier inverse transform $\endgroup$ – Jakobian May 31 '18 at 19:39
  • $\begingroup$ en.wikipedia.org/wiki/Fourier_inversion_theorem $\endgroup$ – Hans Lundmark May 31 '18 at 19:46
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    $\begingroup$ Also notice that it is not always possible to take the Fourier transform twice. For instance, if $f\in L^1$, then $\hat{f}\notin L^1$ in general. $\endgroup$ – Lorenzo Quarisa May 31 '18 at 19:53
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The Fourier transform of $f$ is: $$F(\omega)=\mathscr{F}(f(t))\{\omega\}=\int_{-\infty}^{\infty} \mathrm{d}t f(t) e^{-i \omega t}$$ Transforming it again: $$g(\tau)=\mathscr{F}(F(\omega))\{\tau\}=\int_{-\infty}^{\infty} \mathrm{d} \omega e^{-i \tau \omega}\int_{-\infty}^{\infty} \mathrm{d}t f(t) e^{-i \omega t}$$ Changing the order of integrations: $$=\int_{-\infty}^{\infty} \mathrm{d}t f(t) \int_{-\infty}^{\infty} \mathrm{d} \omega e^{-i \omega t} e^{-i \omega \tau}$$ And $\mathscr{F}(e^{i a t})\{\omega\}=2 \pi \delta(\omega -a)$: $$=\int_{-\infty}^{\infty} \mathrm{d}t f(t) 2 \pi \delta(t + \tau)$$ $$=2 \pi\int_{-\infty}^{\infty} \mathrm{d}t f(t) \delta(t + \tau)$$ $$=2 \pi f(-\tau)$$ But you might get a different result with a different definition of the FT.

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We have that $$ f(x) = \int\limits_{\mathbb{R}^n} \mathfrak{F}(f)(s)e^{2\pi\cdot isx} ds $$ And from this $$ f(x) = \mathfrak{ F}\circ\mathfrak{ F}(f)(-x) $$ Because all we do, is take the Fourier transform of $\mathfrak{F}(f)(s)$ with respect to $-x$

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It may be easier to see in the finite case. The Discrete Fourier Transform is given by multiplication by the $\, n \!\times\! n\,$ DFT (Vandermonde) matrix. Define $\, e_n(x) := \exp(2\pi\sqrt{-1}x/n),\,$ and matrix $\, T(n) := \{\frac1{\sqrt{n}}e_n(i j)\}_{i,j}.\,$ The matrix $\,A(n) := T(n)^2\,$ is $\, A(n)_{i,j} = \frac1n\!\sum_{k=0}^{n-1} e_n((i\!+\!j)k)\,$ but $\, d_n(j) := \sum_{k=0}^{n-1} e_n(jk) \,$ evaluates to $\, d_n(j) = n\,$ if $\, j \equiv 0 \pmod n\,$ and $\, d_n(j) = 0\,$ otherwise. This implies that $\, A(n)_{i,j} = 1\,$ if $\, i \equiv -j \pmod n\,$ and $\,A(n)_{j,j} = 0\,$ otherwise.

Thus, $\,A(n)\,$ takes any vector $\,(x_0,x_1,\dots,x_{n-1})\,$ to $\,(x_0,x_{n-1},\dots,x_1).\,$ This corresponds to $\,\mathscr{F}^2\!: f(x) \mapsto f(-x)\,$ in the continuous Fourier transform case.

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