energy of a signal in t and f domain

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Ray Lee
Ray Lee 2014-12-3
回答： Paul 2024-11-10
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?

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采纳的回答

Star Strider
Star Strider 2014-12-3
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.
  2 个评论
Ray Lee
Ray Lee 2014-12-4
Before, I got ey/ex=16
after normalization, I got ey/ex=1.6000e-07

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更多回答（2 个）

Ray Lee
Ray Lee 2014-12-5
I found the solution myself.
Normalizing spectral amplitude by fs will work.
But I don't know why to do this.
Paul
Paul 2024-11-10
Ran in:
For a finite duration signal x[n] of length N, and its Discrete Fourier Transform (DFT) X[k] (as computed by fft), the energy relationship is given by Parseval's Theorem: sum(abs(x[n]^2)) = sum(abs(X[k])^2))/N
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
Parseval's Theorem:
[sum(abs(x).^2) sum(abs(y).^2)/n]
ans = 1×2
825.5910 825.5910
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If the first term is multiplied by dx, then the second must also be multiplied by dx = 1/fs
[sum(abs(x).^2)*dx sum(abs(y).^2)/n/fs]
ans = 1×2
206.3977 206.3977
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