MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have a 2D order-2 polynomial, a Gaussian and a 'box' indicator function. Let:

$\begin{eqnarray} p(x,y) &=& c_0+c_1x+c_2y+c_3xy+c_4x^2+c_5y^2+c_6xy^2 \\ G(x,y) &=& c_k\cdot\exp\left(\frac{-(x^2+y^2)}{2\sigma^2}\right) \\ \square_a(x,y) &=& \mathbf{1}_{[-a,a]\times[-a,a]}(x,y) \end{eqnarray}$

all functions of reals, all real constants, and $a,\sigma>0$.

Let $\otimes$ denote convolution, and $\mathcal{F\{\dots\}}$ denote the Fourier transform.

Is there a "closed form" for the deconvolution of $\square_a\otimes G$ from $p$ (when $\mathcal{F}\{\square_a\otimes G\}$ is away from 0)? In other words, can the following expression be significantly reduced:

$$\mathcal{F}^{-1}\left\lbrace\frac{\mathcal{F}\{p(x,y)\}}{\mathcal{F}\{\square_a\}\cdot \mathcal{F}\{G\}}\right\rbrace $$

I am trying to start computing this in Maple.

This looks like the inverse Fourier transform of a bunch of derivatives delta functions divided by a sinc scaled by a Gaussian. Is there a better way of performing the deconvolution?

Thank you very much

EDIT: Maple says its nothing but a polynomial

$$-\frac{1}{8}\frac{(c_4+c_5)\cdot \sigma^2-c_0}{\pi\cdot c_g\cdot \sigma^2\cdot a^2}$$

share|cite|improve this question
Oops. Maple says the result is a polynomial of $x$ and $y$, and what I posted is that polynomial's evaluation at $0$. (i.e. the result of integrating the quotient of Fourier transforms over the entire plane) – enthdegree Mar 13 '14 at 18:49
With reference to the flag, of course it is a polynomial; there are deltas. Had I been exposed to the theory of distributions before attempting this it would have been textbook. – enthdegree Aug 23 '14 at 20:20
Instead of closing, could you just write a short answer? – S. Carnahan Aug 23 '14 at 23:15

Your Answer


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

Browse other questions tagged or ask your own question.