FFT of a function

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.

답변 (1개)

themaze
themaze 2013년 4월 11일

0 개 추천

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)

댓글 수: 1

ong
ong 2013년 4월 11일
Hi, thanks for the respond. What does the @ component do?

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도움말 센터File Exchange에서 Fourier Analysis and Filtering에 대해 자세히 알아보기

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ong
2013년 4월 11일

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