Plot the phase structure function of a phase screen by definition - matlab

Question

I have already had a phase screen (a 2-D NxN matrix and LxL in size scale, ex: N = 256, L = 2 meters).
I would like to find phase structure function - D(r) defined by D(delta(r)) = <[x(r)-x(r+delta(r))]^2> (<.> is ensemble averaging, r is position in phase screen in meter, x is phase value at a point in phase screen, delta(r) is variable and not fix) in Matlab program. Do you have any suggestion for my purpose?
P/S: I tried to calculate D(r) via the autocorrelation (is defined as B(r)), but this calculation still remaining some approximations. Therefore, I want to calculate precisely the result of D(r). May you please see this image to better understand the definition of D(r) and B(r). Below is my function code to calculate B(r).
% Code copied from "Numerical Simulation of Optical Wave Propagation with Examples in Matlab",
% by Jason D. Schmidt, SPIE Press, SPIE Vol. No.: PM199
% listing 3.7, page 48.
% (Schmidt defines the ft2 and ift2 functions used in this code elswhere.)
function D = str_fcn2_ft(ph, mask, delta)
% function D = str_fcn2_ft(ph, mask, delta)
N = size(ph, 1);
ph = ph .* mask;
P = ft2(ph, delta);
S = ft2(ph.^2, delta);
W = ft2(mask, delta);
delta_f = 1/(N*delta);
w2 = ift2(W.*conj(W), delta_f);
D = 2 * ft2(real(S.*conj(W)) - abs(P).^2, delta) ./ w2 .*mask;`
%fire run
N = 256; %number of samples
L = 16; %grid size [m]
delta = L/N; %sample spacing [m]
F = 1/L; %frequency-domain grid spacing[1/m]
x = [-N/2 : N/2-1]*delta;
[x y] = meshgrid(x);
w = 2; %width of rectangle
%A = rect(x/2).*rect(y/w);
A = lambdaWrapped;
%A = phz;
mask = ones(N);
%perform digital structure function
C = str_fcn2_ft(A, mask, delta);
C = real(C);

Answer 1

One way of directly computing this function D(r) is through random sampling: you pick two random points on your screen, determine their distance and phase difference squared, and update an accumulator:
phi = rand(256,256)*(2*pi); % the data, phase
N = size(phi,1); % number of samples
L = 16; % grid size [m]
delta = L/N; % sample spacing [m]
D = zeros(1,sqrt(2)*N); % output function
count = D; % for computing mean
for n = 1:1e6 % find a good amount of points here, the more points the better the estimate
coords = randi(N,2,2);
r = round(norm(coords(1,:) - coords(2,:)));
if r<1
continue % skip if the two coordinates are the same
end
d = phi(coords(1,1),coords(1,2)) - phi(coords(2,1),coords(2,2));
d = mod(abs(d),pi); % you might not need this, depending on how A is constructed
D(r) = D(r) + d.^2;
count(r) = count(r) + 1;
end
I = count > 0;
D(I) = D(I) ./ count(I); % do not divide by 0, some bins might not have any samples
I = count < 100;
D(I) = 0; % ignore poor estimates
r = (1:length(D)) * delta;
plot(r,D)
If you need even more precision, consider interpolating. Compute random coordinates as floating-point values, and interpolate the phase to get the values in between samples. D then needs to be longer, indexed as round(r*10) or something like that. You will need many more random samples to fill up that much larger accumulator.