Posted by
Gluender-2 on
Dec 24, 2008; 4:28pm
URL: http://imagej.273.s1.nabble.com/FFT-filter-conversion-to-real-space-tp3694163p3694164.html
>Hello all,
>
>I have a slit-scanning confocal in my facility that sometimes introduces
>vertical line artifacts into images. I use ImageJ to correct these defects
>in frequency space; I make a thin horz. rectangle through the middle of the
>FFT map, except through the very center. How can I do the same filtering in
>real space? I'm looking for a kernel that I can use in other programs that
>don't have Fourier capabilities. (sample images and mask at
>
http://www.missouri.edu/~fernandezg/FFT)
>
>On a related note, can people suggest a resource/reference for converting
>Fourier filters to real space in general?
>
>Thanks and Happy Holidays!
>
>-Esteban
>
>--
>G. Esteban Fernandez, Ph.D.
>Associate Director
>Molecular Cytology Core Facility
>University of Missouri
>120 Bond Life Sciences Center
>Columbia, MO 65211
>
>
http://www.biotech.missouri.edu/mcc/>
>573-882-4895
>573-884-9395 fax
By "real space" you mean the image- or 2-dimensional space-domain, do you?
Filtering in the frequency- or Fourier-domain means multiplication of
the complex-valued Fourier-spectrum of your image with the filter
function (mask), in your case the horizontal bar with a gap around
the DC-term.
Filtering in the space-domain means convolution with the filter's
point-response (convolution kernel) which is the Fourier-retransform
of the filter function.
Consequently, you could try to Fourier-retransform the horizontal bar
with a gap around the DC-term. However the resulting convolution
kernel may turn out impractical for several reasons:
1) It may be complex-valued (in any case it will be bipolar)
2) It may be too big (extended)
But the result may be a starting point for what you want.
If your bar were line-like and without the gap, the
Fourier-retransform would be confined to the y-axis in space domain.
It would be a negative constant on this line with a positive peak at
the origin. To mimic the gap, you could add a constant to the result
of the convolution until your image is strictly positive.
Of course, there are better solutions to the problem but they are
more involved.
HTH
--
Herbie
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