How to implement this without using for?

This is my code:

function Y=FCOS(win)

   N=size(win,2)-1;
   teta=2*pi/N;
   d=2*pi/N;
   win1=win(:,1:N);
   win2=win(:,2:N+1);
   C1=0; C2=0;

   for j=1:N
      C1=C1+(2/N)*win1(:,j)*cos(j*teta);
      C2=C2+(2/N)*win2(:,j)*cos(j*teta);
   end

   Yc=C1;
   Ys=(C1*cos(d)-C2)/sin(d);

Y=complex(Yc,Ys);

My idea was to do the following:

C1=2/N*sum(win1.*cos((1:N)*teta),2);
C2=2/N*sum(win2.*cos((1:N)*teta),2);

But it will only work if my variable 'win' has only 1 line. For this to work, I would have to use, instead of (1:N), [1:N; 1:N; 1:N ... ;1:N], where this matrix should have the same number of lines as 'win'. But I can only think of a way of doing this with a for.

Anyone have a better solution?

Answer 1

[Sean de Wolski]

Close!

Use bsxfun:

C1 = sum(2/N*bsxfun(@times,win1,cos((1:N)*teta)),2);
C2 = sum(2/N*bsxfun(@times,win2,cos((1:N)*teta)),2);

And to make it faster do the cos(1:N)*teta only once:

cosnxteta = cos((1:N)*teta));
C1 = sum(2/N*bsxfun(@times,win1,cosnxteta,2);
C2 = sum(2/N*bsxfun(@times,win2,cosnxteta,2);

[Bernard Küsel]

Thanks! It really sped up the code. I will surely use more of bsxfun for future coding!

Would you know how to create that [1:N; 1:N; 1:N ... ;1:N] matrix without using for?

[Sean de Wolski]

repmat(1:N,[ntimes,1]); %But this can often be avoided with bsxfun!

[Bernard Küsel]

Thanks for the tips!