I would like to Fourier-Transform the following function:

$\tau\left(x\right) = e^{i\left(H\left(x\right) + G\left(x\right)\right)}$.

$H\left(x\right)$ and $G\left(x\right)$ are Fourier-Series in exponential form, given by

$H\left(x\right) = \sum_{n=-\infty}^{+\infty} a_n e^{inx/g_1}$ and

$G\left(x\right) = \sum_{n=-\infty}^{+\infty} b_n e^{inx/g_2}$.

$G\left(x\right)$ is a known function, so are the corresponding coefficients $a_n$. However, nothing is known about $H\left(x\right)$ and $b_n$ expect that the coefficients are real numbers. $g_1$ and $g_2$ are positive real numbers, $x$ is also real-valued.

I would like to Fourier-Transform $\tau\left(x\right)$ into $T\left(k\right)$. Here I would like to tell which coefficients of $H$ and $G$ contribute to two certain orders of $T\left(k\right)$.

I tried rewriting $\tau\left( x\right)$ into a product-series, in order to get rid of that "double-exponential". But from this point on, I have no idea how to proceed further.

I realize that things are very easy in case that $\tau\left( x\right) = H\left( x\right) + G\left( x\right)$, without that exponential function.

Online-research did not bring me any further. Is this a well-known problem? Is there a solution?


1 Answer 1


The characteristic function, https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory), for an $X_\lambda \in Po(\lambda)$ variable has the form, $$ e^{\lambda (e^{it} - 1)} $$ Which looks close to one of your factors. the characteristic function. The characteristic function for $aX_\lambda$ is $$ e^{\lambda (e^{i a t} - 1)} $$ So let's rewrite your formula as $$ \tau = \exp \left ( \sum a_n(e^{i(nt/g_1 + \pi/2)}-1) + \sum b_n(e^{i(nt/g_2+\pi/2)}-1)\right ) \exp \left ( \sum (a_n+b_n) \right ) $$ $$ \exp(a_n(e^{i(nt/g_1 + \pi/2)}-1)) = E(\exp(i \frac{n(t + \frac{\pi g_1}{2 n})}{g_1}X_{a_n})) = \phi_{a_n}(t+\frac{\pi g_1}{2 n}) $$ with $\phi_{a_n}$ the characteristic function of $\frac{n}{g_1}X_{a_n}$ Fourier transform of this lead to a complex distribution $Y_n$ of atoms so that the weight is essentially, also note that the fourier transform and the inverse fourier transform is differnt in sign $$ f(Y = -\frac{n k}{g_1}) = C e^{-a_n}\frac{(i a_n)^k}{k!} $$ Now the final distribution of this is the convolution of all those distributions.

  • $\begingroup$ Thanks for the hint. I lost track at the beginning of the second line after "let's rewrite your formula". Could you please advise what's happening here? $\endgroup$ Commented Jun 21, 2018 at 19:00
  • $\begingroup$ I use $i=e^{i\pi}$ and I add and subtract $a_n, b_n$ I also break out those terms as a separate factor. Also the inverse fourier transform as a little different from the fourier transform and a correction for that is needed. $\endgroup$
    – Stefan
    Commented Jun 21, 2018 at 19:11
  • $\begingroup$ ouch $i = e^{i \pi / 2}$ I'll correct the analysis $\endgroup$
    – Stefan
    Commented Jun 21, 2018 at 19:12
  • $\begingroup$ Sorry to ask another one: where do you get that $exp(a_n)$ in the exponent of the second expression from? $\endgroup$ Commented Jun 23, 2018 at 5:58
  • $\begingroup$ good catch, it's a typo, I'll fix it $\endgroup$
    – Stefan
    Commented Jun 23, 2018 at 10:39

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