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I have an image with few pixels in length and height. For this image I calculated the two dimensional Fourier transformation. What I got for the frequency spectrum in one direction was a very frequency spectrum with very few points (no continuous first derivation). To increase the number of points of the frequency spectrum, I used a window function, in my case a Hanning-filter/rectangular filter with zeros outside of the picture so that the filtered picture has much more pixels in length (horizontal) and height (vertical). This resulted in a frequency spectrum with more points so that the frequnecy spectrum in e. g. the horizontal direction looked more continuous but there were in contrast to the unfiltered image higher peaks for low frequencies. What I want to do now is to find a possibility to reduce that. My approach was the convolution theorem (but I have no idea how to use it in two dimensions) which says, that $F(f) * F(g)$ = const $F(gf)$. I have $f$ (original image) and $g$ (multiplication by rectangular window: some $1$ in the centre, $0$ outside it) and in that way the fourier spectrum that I get is $F(gf)$. To get $F(f)$ I have to deconvolve $F(f)$ and $F(g)$. But the problems are at first:

  1. Holds the convolution theorem for this?

  2. Will the matrix $F(f)$ that I will finally get, the same that I got by Fourier transformation without the window function of the rectangular but with more points in the frequency spectrum? In other words, will it reduce effectively the influence of the filter in the frequency spectrum?

  3. Is there an effective way to calculate the deconvolution with e.g. matlab?

If there are problems, is there an alternative way to increase the number of points in the frequency spectrum without changing it at all? If not, how can I fit the points in the frequency spectrum in the best way?

I would be very grateful for any helpful hints.

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    $\begingroup$ What are you trying to do with the spectrum? $\endgroup$ – Conrad Turner Mar 20 '15 at 5:07
  • $\begingroup$ Thank you for answers. I have muscle cells of C. elegans. I used GFP to mark a specific protein and this protein binds in different states on different positions of the cell. I also used mutants to suggest the meaning of domains. The problem is that one cell usually has only a limited number of periods and the resolution of the microscope is finite. That implies a discrete frequency spectrum. In fact, the frequency spectrum isn't discrete of course. To publish this I wanted to increase the resolution of the frequency spectrum with a fit or the technique of window functions that I have read. $\endgroup$ – Varou Mar 20 '15 at 9:40
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You can also interpolate as many points as you need using a Sinc or windowed Sinc interpolation kernel.

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  • $\begingroup$ I have tried to interpolate the frequency spectrum with a sinc of funktion but why does it make sence? In my frequency spectrum without a a window function the highest maximum is not in for the frequency 0. But the main maximum of the sinc-function is there. So the interpolated curve for the data does not fit with the data.I used as the sinc-function a * sin(b * x - c) / x and fitted with origin with a, b, c as independent parameters. $\endgroup$ – Varou Mar 20 '15 at 17:22
  • $\begingroup$ can you please explain how to do an interpolation with sinc of a frequency spectrum and why should it make sence? To specify this question, I've sent you a link to such a frequncy spectrum without window function:dl.dropboxusercontent.com/u/75653143/… and with vonHanning filter:dl.dropboxusercontent.com/u/75653143/window%20function.pdf I hope the problem will become apparent. $\endgroup$ – Varou Mar 23 '15 at 17:44
  • $\begingroup$ I can explain why it makes sense, the way to put more points to an FFT is to feed a longer time domain signal into it. When the added data is all zeroes, that is equivalent to a sinc interpolation/convolution. Why? Because it's a rectangular window and FFT of rectangular is a sinc. Or you can just double the signal length (by zero padding) and calculate the FFT of that... same difference. $\endgroup$ – Dole Feb 13 '16 at 21:58
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Zero pad the time domain signal; this will interpolate more samples in the FFT.

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