If I have some symmetric vector, representing radial distribution of data (the Line Spread Function (LSF) of the optical system, in my case, and a priori I know that PSF has circular symmetry), how, from this vector, can I create a circularly symmetric matrix (representing the PSF of the system)? And vise versa?
For example, I have a vector representing 1D Gaussian, and I want matrix of 2D Gaussian.
Thank you in advance!
For the 2d-gaussian from the 1d-version, it is "easy" since the gaussian function is "separable"
G(x,y) = G(x)*G(y)
So, if g1d is a vector column of the gaussian, then, g2d = g1d*(g1d.') will give the 2d version.
For a non-separable radial-symmetric function, it could be done with a simple 1d-interpolation. Let f1d the 1d function with respect the radial value rho.
Let x and y the two vector in column and row dimensions. Then the following lines in Matlab will construct the 2d version of f1d with radial-symmetry.
[Mx,My] = meshgrid(x,y);
Mro = sqrt(Mx.^2+My.^2) ;
f2d = interp1(rho, f1d, Mro,'lin',nan);
good luck
For the 2d-gaussian from the 1d-version, it is "easy" since the gaussian function is "separable"
G(x,y) = G(x)*G(y)
So, if g1d is a vector column of the gaussian, then, g2d = g1d*(g1d.') will give the 2d version.
For a non-separable radial-symmetric function, it could be done with a simple 1d-interpolation. Let f1d the 1d function with respect the radial value rho.
Let x and y the two vector in column and row dimensions. Then the following lines in Matlab will construct the 2d version of f1d with radial-symmetry.
[Mx,My] = meshgrid(x,y);
Mro = sqrt(Mx.^2+My.^2) ;
f2d = interp1(rho, f1d, Mro,'lin',nan);
good luck
I tried some other technique. I created loop in which used imrotate function to rotate 2D matrix by a little angle and after each rotation I replaced the central row by my vector. In the process I noticed that if we really want the circularly symmetric result we must have the symmetrical vector.
This technique work well for not high matrices.
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