FFT of a function

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ong
ong 2013-4-11
I would like to do a FFT of x, x=cos(2*pi*f*t)
How can i replace X(W) = FFT(X) --> X(W/a) = ?
Thanks.

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回答（1 个）

themaze
themaze 2013-4-11
Here is a nice way to accomplish that
I am assuming that w = 2*pi*f. Also you need to define your t
Xw = @(w, t) fft(cos(w*t));
to call the function, just use
Xw(w, t) or Xw(W/a, t)
  1 个评论
ong
ong 2013-4-11
Hi, thanks for the respond. What does the @ component do?

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