energy of a signal in t and f domain

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Ray Lee
Ray Lee 2014년 12월 3일
댓글: Ray Lee 2014년 12월 5일
The energy of a signal is expected to be the same in t and f domain.
n = 1e4;
dx = 0.25;
x = rand(n,1) -0.5;
ex = sum(x.^2) *dx; % energy in t domain
y = fft(x);
fs = 1/dx;
df = fs/n;
ya = abs(y);
ey = sum(ya.^2) *df; % energy in f domain
but from the code, ey/ex=16, exactly the squared fs.
what's the problem?

채택된 답변

Star Strider
Star Strider 2014년 12월 3일
You need to normalise the fft by dividing it by the length of the signal:
y = fft(x)/length(x);
See the documentation for fft for details.
  댓글 수: 2
Ray Lee
Ray Lee 2014년 12월 4일
Before, I got ey/ex=16
after normalization, I got ey/ex=1.6000e-07
Ray Lee
Ray Lee 2014년 12월 5일
it seems y = fft(x) / fs

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추가 답변 (1개)

Ray Lee
Ray Lee 2014년 12월 5일
I found the solution myself.
Normalizing spectral amplitude by fs will work.
But I don't know why to do this.

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