there is an error in describing the loop, please help me to resolve this one.

clear
x = linspace (-1,1,256);
y = linspace (-1,1,256);
[X,Y] = meshgrid(x,y);
r = sqrt(X.^2 + Y.^2);
z= 0;
w = 100;
E0 = 1;
O = rand(256,256);
E = E0*exp(-1i*(O)); 
A = (abs(E)).^2 ;

x=linspace(-1,1,256);
[X,Y]=meshgrid(x,x);
r=sqrt(X.^2+Y.^2);
R= 0.06;
t = 0.5;
l= 0:36:360;
M = cell(1,length(l));
for  i = 1:length(l)
    h= t*cosd(l(i));k= t*sind(l(i));
    p=sqrt((X-h).^2+(Y-k).^2);
    M{i}=(p<R);
end
M = M{1}+M{2}+M{3}+M{4}+M{5}+M{6}+M{7}+M{8}+M{9}+M{10};


Io=1.0; 
Y=linspace(1,-1,256); Z=Y ; 
A(1:length(Y),1:length(Z))=0;
for i=1:length(Y)
    for j=1:length(Z)
        q=(Y(i).^2+Z(j).^2).^0.5;
        w = 0.3;
        A(i,j)=Io.*exp(-((q.^2)./w^2));
    end
end



G0 = 15;
I = 1;
E = E0*exp(-1i*(O)); 
for i = 1:100
    G(i) = G0/(1+ (abs(E(i)))^2/I);
    E(i+1) = M *(ifft(A*(fft(G* E(i)))));
end

Unable to perform assignment because the left and right sides have a different number of elements.

imagesc(E)

[Karim]

M*(ifft(A*(fft(G* E(i)))))

is a matrix of size 256x256

you are trying to store this into E( i+1 ) which is a scalar (i.e. a 1x1) location.

as a result you get an error which indicates that the left side (E(i+1)) doesn't have the same size as the right side (a 256x256 matrix)

[Rik]

So the question is: what do you want to do?

Answering that question is very hard because you chose non-descriptive variable names and didn't write any comments. How do you expect anyone to understand what is going on in your code? Remember that you in three months time are a different person from current you. How is future you suposed to understand what you did and extend functionality or fix a bug?

[SAHIL SAHOO]

@Karim please check this one.

clear
x = linspace (-1,1,256);
y = linspace (-1,1,256);
[X,Y] = meshgrid(x,y);
r = sqrt(X.^2 + Y.^2);
z= 0;
w = 100;
E0 = 1;
O = rand(256,256);
E = E0*exp(-1i*(O)); 
A = (abs(E)).^2 ;

x=linspace(-1,1,256);
[X,Y]=meshgrid(x,x);
r=sqrt(X.^2+Y.^2);
R= 0.06;
t = 0.5;
l= 0:36:360;
M = cell(1,length(l));
for  i = 1:length(l)
    h= t*cosd(l(i));k= t*sind(l(i));
    p=sqrt((X-h).^2+(Y-k).^2);
    M{i}=(p<R);
end
M = M{1}+M{2}+M{3}+M{4}+M{5}+M{6}+M{7}+M{8}+M{9}+M{10};


Io=1.0; 
Y=linspace(1,-1,256); Z=Y ; 
A(1:length(Y),1:length(Z))=0;
for i=1:length(Y)
    for j=1:length(Z)
        q=(Y(i).^2+Z(j).^2).^0.5;
        w = 0.3;
        A(i,j)=Io.*exp(-((q.^2)./w^2));
    end
end

% i want to apply loop here by using numerate or any for loop but i didn't how i can do
% that.
G0 = 15;
I = 1;
E1 = E0*exp(-1i*(O));
G1 = G0./(1+ (abs(E1))^2/I);
E2 = M *(ifft(A*(fft(G1* E1))));
G2 = G0./(1+ (abs(E2))^2/I);
E3 = M *(ifft(A*(fft(G2* E2))));
G3 = G0./(1+ (abs(E3))^2/I);
E4 = M *(ifft(A*(fft(G3* E3))));
G4 = G0./(1+ (abs(E4))^2/I);
E5 = M *(ifft(A*(fft(G4* E4))));
G5 = G0./(1+ (abs(E5))^2/I);
E6 = M *(ifft(A*(fft(G5* E5))));
G6 = G0./(1+ (abs(E6))^2/I);
E7 = M *(ifft(A*(fft(G6* E6))));
G7 = G0./(1+ (abs(E7))^2/I);
E8 = M *(ifft(A*(fft(G7* E7))));
G8 = G0./(1+ (abs(E8))^2/I);
E9 = M *(ifft(A*(fft(G8* E8))));
G9 = G0./(1+ (abs(E9))^2/I);
E10 = M *(ifft(A*(fft(G9* E9))));
G10 = G0./(1+ (abs(E10))^2/I);
E = M *(ifft(A*(fft(G10* E10))));

imagesc((abs(E)).^2);colorbar

Answer 1

[Rik]

If you're not planning to do anything with the intermediate values, why would you store them?

init % Runs the first part of your code.

G0 = 15;
I = 1;
E = E0*exp(-1i*(O));

for iteration=1:10
    G = G0./(1+ (abs(E))^2/I);
    E = M *(ifft(A*(fft(G* E))));
end
imagesc((abs(E)).^2);colorbar

If you want to store the intermediate numbers, that is also possible: use cell vectors. That will you give more intutive control. Just remember that Matlab uses 1-indexing, not 0-indexing.

G0 = 15;
I = 1;
E = { E0*exp(-1i*(O)) };
G = { G0./(1+ (abs(E{1}))^2/I) };
for ind=2:10
    E{ind} = M *(ifft(A*(fft(G{ind-1}* E{ind-1}))));
    G{ind} = G0./(1+ (abs(E{ind}))^2/I);
end
E_final = M *(ifft(A*(fft(G{10}* E{10}))));
imagesc((abs(E_final)).^2);colorbar