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 $g(x,y)$ using some mathematical expression?


In the following the convolution operation is performed in Frequency domain: I first take the FFT of the ideal image (top left) which we call $F(\nu_x, \nu_y)$, similarly take the FFT of the gaussian blur Kernel which we call $G(\nu_x, \nu_y)$. Then take the product $F(\nu_x, \nu_y) G(\nu_x, \nu_y)$ and perform the iFFT which gives me the blurred image (bottom left).

The crosssection of the blurred and ideal image is plotted in the bottom right corner.

We can see that with increasing width of the blur kernel, the amplitude of the convolved sin function decreases.

The questions which I have in mind are

  • Does convolving a sinusoidal function with a circular (or gaussian) blur kernel change the phase of the convolved sinusoidal function?

  • Can I determine the parameters of the convolved sine function such as its amplitude, frequency, phase based on the width of the blur kernel?

enter image description here

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    $\begingroup$ Your question is written with clarity and details. What might be missing for this forum is a beginning of your computations, and where you block. Hence, you will find in an answer some starting points, which you may elaborate on $\endgroup$ Commented Sep 29, 2021 at 15:27
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    $\begingroup$ Thanks for the tips Laurent : ) $\endgroup$ Commented Sep 30, 2021 at 7:04
  • $\begingroup$ Did you get further? $\endgroup$ Commented Oct 29, 2021 at 19:53
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    $\begingroup$ Unfortunately not : ( $\endgroup$ Commented Nov 1, 2021 at 15:38
  • $\begingroup$ I have it partially. I used the library for symbolic computing in MATLAB to perform the convolution in Fourier Domain. By that I get a new expression for the convolved image whose amplitude depends on the amplitude and frequency of the signal and the radius of the circular blur kernel. $\endgroup$ Commented Nov 3, 2021 at 19:16

1 Answer 1


Here is a starting point. The $1/2$ part is easily computed. For the second part, it looks like a Fourier sine transform, which can be computed from a standard Fourier transform, and taking properly the real or imaginary part.

What you might need is thus the Fourier transform of a unit disk. The blog post 2D Fourier transform (of) a disk provides this information, related to Bessel functions.


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