Re: image registration problem

Your way to convert x,y to rho and theta makes sense. Assume your
images are 256x256 big, rho will be in range 0~181 (=256*sqrt(2)/2) and
theta in -180~180 if we take the image center as the polar coordinate
origin. The final polar image is 362x360 big. Then the peak in the
cross power spectrum will give a rho between 0 and 181 and a theta in 0
and 359. Remap these values back to get x and y. That's my thought.
What do you think?

On Jan 24, 2:33 am, "ASN" <asnag...@xxxxxxxxx> wrote:
On Jan 23, 7:01 pm, "ASN" <asnag...@xxxxxxxxx> wrote:

Have you checked that one ?
http://archeoguide.intranet.gr/papers/publications/ARCHEOGUIDE-VAST01...This reference uses the phasecorrelation method described in the
references i cited earlier, for a tracking application.
My idea is to use the registration for an industrial inspection
problem.
I still dont understand how to get the rotation angle.

After getting the ifft2 of the cross power spectrum, the peak was at
the origin ALWAYS. So i zeroed out the first value [I got this idea
from the reference you pointed me to] and now i do get peaks at
different positions for images which are rotated at different angles.
But I dont understand how to interpret these positions.
Since we are in polar coordinates,will the position of the peak value
which is a 2-tuple will give me the Rho and Theta values which I am
looking for?
The Rho and Theta values will be a number between 256 (if my image is
of size 256X256). How do i correspond this to the required rotation
angle?

Greateful for any guidance i can get.
Thanks,
ASN

.

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