How to vectorize loop of fft ?

How to vectorize a for loop of fft ?

      clear all
      W = ones(1,100)*1.5;
      y(1:200,1:100) = 0.3;
      y(1,:) = ones(1,100)*1.2;

      for j=1:200;
      y(j+1,:) = ifft(fft(y(j,:)).*W);
      end

Answer 1

[Geoff Hayes]

Ole - is this something that you need to vectorize or just simplified? If the above is a valid example with all elements of W being identical, then the line

 y(j+1,:) = (ifft((((fft((y(j,:)))).*W))));

can be simplified to

 y(j+1,:) = ifft(fft((y(j,:))).*W);
          = ifft(1.5*fft((y(j,:))));
          = 1.5*y(j,:);

due to the linearity of the inverse Fourier transform. And so your for loop becomes

 for j=1:200;
      y(j+1,:) = 1.5*y(j,:);
 end

The above implies that the nth column of y (for n>1) is simply

 (1.5)^(n-1)*y(1,:)

in which case you can replace your code with just

y = 1.5*ones(201,100);
y(1,:) = 1.2;
y = cumprod(z);

Try the above and see what happens!

[Ole]

Thank you. W is actually a function in fourier space, I just tried to make it simple. This the structure of a fft propagation, with W being the transfer function of free space exp(i*kz*dz).

y(2:201,:) = ifft(fft(y(1:200,:)).*W); instead of the for loop gives errors

[Geoff Hayes]

Ole - what is the error? Is it

 Error using  .* 
 Matrix dimensions must agree.

because of the line

 ifft(fft(y(1:200,:)).*W);

where W is just a 1x100 array and fft(y(1:200,:)) is a 200x200 array. You can only perform an element-wise multiplication on arrays/matrices of the same dimension.

A question that I should have asked sooner - why are you trying to vectorize this problem? Performance reasons? Because the recurrence relation (k+1 depends on k) in the above for loop may provide the best solution.