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I have a Fourier transformable complex function that is a function of independent real variable a. Now I take the Fourier transform of it, giving me a complex function of real variable b. Now I treat the resulting function as if it is in the original domain of a and again take Fourier transform of it - in stead of inverse Fourier transform as is usually done to get the original fucntion. May be I will do this repeatedly. What are relations between the original function and the transform function at various stages?

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if you define the continuous Fourier Transform in a unitary manner, my preferred unitary definition is

$$ X(f) \triangleq \mathscr{F}\{ x(t) \} = \int\limits_{-\infty}^{+\infty} x(t) \, e^{-j 2 \pi f t} \ dt $$

$$ x(t) \triangleq \mathscr{F}^{-1}\{ X(f) \} = \int\limits_{-\infty}^{+\infty} X(f) \, e^{+j 2 \pi f t} \ df $$

then you can see a lotta symmetry and isomorphy in the forward and inverse transformation. in fact they are exchangeable since $-j$ and $+j$ both have equal claim to squaring to be $-1$. (i.e. if, in all of our textbooks and technical and math lit, every $j$ was replaced by $-j$ and vise versa, all of our theorems would be just as valid. the choice of which imaginary unit to go with which direction of Fourier transformation is a convention.)

now, it's not hard to see where the duality theorem comes from. Given the above, then

$$ x(-f) = \mathscr{F}\{ X(t) \} = \int\limits_{-\infty}^{+\infty} X(t) \, e^{-j 2 \pi f t} \ dt $$

$$ X(-t) = \mathscr{F}^{-1}\{ x(f) \} = \int\limits_{-\infty}^{+\infty} x(f) \, e^{+j 2 \pi f t} \ df $$

from that, it's not hard to see that

$$ x(-t) = \mathscr{F} \Big\{ \mathscr{F}\{ x(t) \} \Big\} $$

and that

$$ x(t) = \mathscr{F} \Big\{ \mathscr{F}\{ x(-t) \} \Big\} $$

so it's not hard to see that if $x(t)$ is even symmetry ($x(-t) = x(t)$) then transforming it twice gets you back to the original.

and this

$$ x(t) = \mathscr{F} \bigg\{\mathscr{F} \Big\{\mathscr{F} \big\{ \mathscr{F}\{ x(t) \} \big\}\Big\}\bigg\} $$

so it's sorta like multiplying by $j$. do it four times and you wind up with the thing that you started with.

you can use this fact to create an infinite number of Fourier transform pairs that are exactly equal to each other. all you have to do is create the Fourier transform three levels deep and add each to the original. each time you FT that, you get the same thing.

there are two simple functions that i can think of that have themselves as their own FT. one is the Gaussian function and the other is the Dirac comb (both properly scaled).

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  • $\begingroup$ For a discussion of Fourier transform pairs that are equal to each other, see this answer of mine over on stats.SE. $\endgroup$ – Dilip Sarwate Jun 4 '16 at 17:17
  • $\begingroup$ it's an interesting answer @DilipSarwate. i would say that, as long as you make sure that the distributions are non-negative, even, and scale them so that their integral is 1, you can make an infinite set of distributions that equal their own Fourier transform. it need not be based on the sums of Gaussian $x_1$ or sum of triangular and sinc^2 $x_2$. can be more general than that. $\endgroup$ – robert bristow-johnson Jun 4 '16 at 20:31

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