How can I deconvolute image data?

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Joo
Joo on 13 Apr 2013
Commented: Walter Roberson on 6 Jul 2018
Hi experts,
i have a question concerning the topic deconvolution in MATLAB with the Image Processing Toolbox.
As a result of a test bench I have the two-dimensional distribution of the force of a fluid flux. This distribution is captured with a measuring pin which is guided through the fluid perpendicular to the flux. In this way I gain a pseudocolor image of the force 2D-distribution.
The problem is now that by scanning the fluid (cross section about 5 mm) with the measuring pin (which has a diameter of 1 mm) the gained distribution is broadened respectively blurred in relation to the real distribution. For me it is a problem concerning deconvolution to gain the real profile out of the measured profile.
Enclosed a image which represents my ideas to the problem in 1D: Link: http://i.imgur.com/tehUhSQ.jpg You can see that the measured profile is broader than the real profile because of scanning with the measuring pin of finite diameter. In my opinion the pin can be described as a rectangular function in this case.
In the Image Processing Toolbox I´ve found the features deconv and deconvreg for deconvolution of data. At the moment I have managed to convolute/deconvolute two functions with each other by conv/deconv, but only in 1D. I think the key to my problem for 2D-dimensional problems is the deconvreg-function with the PSF, but I don´t know how to transfer my problem to this function.
It would be very helpful if you, experts, have some hint for me to better understand how to use the features mentioned above for my problem. Or if this problem could be solved in an easier way.
Thanks a lot, Joo

Accepted Answer

Image Analyst
Image Analyst on 13 Apr 2013
Edited: Image Analyst on 13 Apr 2013
Try deconvlucy() and try some different PSFs until you get one that produces an image that is sharp enough for you to continue your analysis.
  11 Comments
Rahim Gholami
Rahim Gholami on 6 Jul 2018
yeah i check it. H is defining as this : d=pi*(u+v); H=(sin(d).*exp(-1i*d))./d; u & v are variables in frequency domain. And as you said H matrix arrays are not all oure zero.
Walter Roberson
Walter Roberson on 6 Jul 2018
Completely black images typically indicate that you either calculated in an integer data type when you should have calculated in floating point, or else that you calculated in floating point and did not correctly convert back to integer. But when fft*() is involved it can also hint that you might have ended up with imaginary components due to not properly handling the filtering on the conjugate symmetric coefficients you get from fft*() of real-valued systems.

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