Create 2D radial symmetric matrix from radius vector

Hello, I have a following problem. I calculate intensity distribution Bessel-Gauss beam using Hankel transform. Input variables

 f = 10000;
 a = 10;
 ff = 3.2;
 w0 = ff * a;
 lambda = 10^(-6);
 k = 2 * pi / lambda;
 N = 100;
 rmax = 1.8 * lambda * f / a;
 # observation plane
 x1 = linspace(-rmax,rmax,N);
 y1 = linspace(-rmax,rmax,N);
 [U,V] = meshgrid(x1,y1);
 W = sqrt(U.^2 + V.^2);
 # aperture plane
 xx = linspace(-a,a,N);
 yy = linspace(-a,a,N);
 [X,Y] = meshgrid(xx,yy);
 R = sqrt(X.^2 + Y.^2);

Hankel transform:

 A2 = zeros(N);
 for i = 1:N
    for j = 1:N
        fun = besselj(0,k/f * W(i,j) * R) .* R .* exp(-(R/w0).^2);
        A2(i,j) = trapz(y1,trapz(x1,fun));
    end
 end
 A2 = abs(A2).^2;
 imagesc(A2);

Output:

This method takes about 100 sec using N = 100. So for much beter resolution calculating takes a much more time. So I think that beter solution will be using 1D intensity distribution radius vector and create 2D matrix rotate in relation to the center. I dont find solution for that, so i created intensity distribution diameter vector and then i can use A' * A what i find in Matlab Answers. Need to change range from -a:a to 0:a and this same in rmax case. Code:

 A = zeros(1,N);
 for i = 1:N
    fun = besselj(0,k/f * x1(i) * xx) .* xx .* exp(-(xx/w0).^2);
    A(i) = trapz(fun);
 end
 A = [fliplr(A(2:end)) A];
 A = abs(A).^2;

Output:

Using A = A' * A /2 I received:

My question: There is any option to improve radial symmetry in output image or maybe there is different method to create 2D matrix from vector?

Answer 1

[Matt]

Hey! Just a thought... Does resizing the figure window help?