0
$\begingroup$

I have an image of a 2D-sinusoidal pattern $f(x, y)$ with wavelength $\lambda$ which I would like to convolve with a 2D circular pill-box function $h(x,y)$ of radius $r$.

The image is given by $$ f(x,y) = \dfrac{1}{2}\bigg[1 + \sin\bigg(\dfrac{2\pi x}{\lambda}\bigg)\bigg]. $$ Similarly the circular pill-box function is given by $$ h(x,y)= \dfrac{1}{\pi r^2} \begin{cases} 1& \text{if } x^2 + y^2 \leq r^2\\ 0 & \text{otherwise} \end{cases} $$ $\qquad$ where the scaling constant $\dfrac{1}{\pi r^2}$ ensures that the area of the filter is one.

The 2D convolution integral may be expressed as $$ g(x,y) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\tau_u, \tau_v)\, h(x - \tau_u, y-\tau_v) \, \mathrm{d}\tau_u \, \mathrm{d}\tau_v $$

$\qquad$ where $\tau$ is a dummy variable to represent the shift of one function with respect to the other.

How do I evaluate the convolution integral so that I can express the convolved image $g(x,y)$ in analytic form.

Thanks in advance : )

$\endgroup$
7
  • $\begingroup$ You generically can’t simplify convolution integrals to a special form $\endgroup$ Sep 27, 2021 at 11:00
  • $\begingroup$ What do you mean by simplifying in to a special form; could you please be more explict : ) If i plug in the equations of $f(x,y)$ and $h(x,y)$ in the convolution integral, shouldn't I get an expression for the convolved image? $\endgroup$ Sep 27, 2021 at 11:04
  • $\begingroup$ For eg a new sinusoidal pattern with different amplitude.. $\endgroup$ Sep 27, 2021 at 11:04
  • $\begingroup$ Well what do you mean first by ‘into analytic form’, if not some sort of special form? You get an expression for the convoluted image, it is what you already wrote down $\endgroup$ Sep 27, 2021 at 11:08
  • $\begingroup$ If I plug in the expressions for $f(x,y)$ and $g(x,y)$ and evaluate the integral, shouldn't I obtain an expression for $g(x,y)$. For eg $g(x,y) = \mathrm{sinc\bigg(\dfrac{A}{B}\bigg)} \dfrac{1}{2}\bigg[1 + \sin\bigg(\dfrac{2\pi x}{\lambda}\bigg)\bigg]$, where $A$ and $B$ are some constants. $\endgroup$ Sep 27, 2021 at 11:14

0

Your Answer

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

Browse other questions tagged or ask your own question.