2
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ 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) $\endgroup$ Mar 13 '14 at 18:49
  • $\begingroup$ 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. $\endgroup$ Aug 23 '14 at 20:20
  • $\begingroup$ Instead of closing, could you just write a short answer? $\endgroup$
    – S. Carnahan
    Aug 23 '14 at 23:15
0
$\begingroup$

In a lax enough environment, the Fourier transform of $x^n$ is the distributional derivative of the $\delta$ distribution. The relevant behavior of the resulting distribution is: $$\delta^{(n)}[\phi]=(-1)^{n}\cdot \phi^{(n)}(0).$$

From this the result quickly follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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