Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We know that the Fourier transform of a Gaussian function is Gaussian function itself. Can anyone give one or more functions which have themselves as Fourier transform?

share|improve this question
    
Yes we can. But why do you need? –  user17762 Mar 9 '12 at 2:42
5  
The key here was the phrase "fixed point" or "eigenfunction", then Google performed the rest of the work. mathoverflow.net/questions/12045/… and en.wikipedia.org/wiki/Fourier_transform#Eigenfunctions –  dls Mar 9 '12 at 2:49
    
Also, have a look at the answers here. –  kuch nahi Mar 9 '12 at 2:59
    
In the old days the "fixed-point functions" were called self-reciprocal functions and investigated by the likes of Hardy and Titchmarsh. See "On self-reciprocal functions under a class of integral transforms" by Kurt Wolf, fis.unam.mx/~bwolf/Articles/28.pdf . You might follow the references in this paper, particularly those of Titchmarsh– Tom Copeland 1 min ago –  Tom Copeland Apr 7 '12 at 22:30

2 Answers 2

Well, the Dirac delta function, strictly speaking, is not a 'function' but it serves your purpose.

The Fourier transform of a Dirac comb (or an impulse train, as it is called in Electrical Engineering) is another Dirac comb.

$$ \sum_{n=-\infty}^{\infty} \delta(t-nT_o) = \frac {2\pi} {T_o} \sum_{m=-\infty}^{\infty} \delta(\omega - m\omega_o) $$

For the proof, refer http://nptel.ac.in/courses/IIT-MADRAS/Principles_Of_Communication/pdf/Lecture07_FTPeriodic.pdf

share|improve this answer

Pick a function $f$ that is reasonable enough for the inversion formula to hold (e.g. take $f$ in Schwartz space, which contains the Gaussian among other functions). If $\mathcal{F}$ denotes the linear transformation which takes $f$ to its Fourier transform, then it's easy to check that $\mathcal{F}^{4}$ is the identity map. In particular, by playing some games you find that $$g \ = \ f + \mathcal{F}(f) + \mathcal{F}^{2}(f) + \mathcal{F}^{3}(f) $$ is fixed by $\mathcal{F}$. So $g$ is its own Fourier transform.

This argument doesn't produce a concrete function, but it at least shows you that the Gaussian is far from the only function that is equal to its own Fourier transform. If you want a more specific example, you can show that $(\cosh \pi x)^{-1}$ is its own Fourier transform (use contour integration and the residue theorem).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.