How to implement this without using for?

8 次查看(过去 30 天)
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?

采纳的回答

Sean de Wolski
Sean de Wolski 2011-5-6
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);
  3 个评论
Sean de Wolski
Sean de Wolski 2011-5-6
repmat(1:N,[ntimes,1]); %But this can often be avoided with bsxfun!

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