energy of a signal in t and f domain

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?

Answer 1

[Star Strider]

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.

[Ray Lee]

Before, I got ey/ex=16

after normalization, I got ey/ex=1.6000e-07

[Ray Lee]

it seems y = fft(x) / fs

Answer 2

[Ray Lee]

I found the solution myself.

Normalizing spectral amplitude by fs will work.

But I don't know why to do this.

Answer 3

[Paul]

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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