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How to interpret FFT output signal?

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Melvin Corvers
Melvin Corvers am 6 Aug. 2020
Kommentiert: Rena Berman am 9 Okt. 2020
Hi,
I am analyzing a signal using FFT to obtain its frequency spectrum. Raw data is shown below.
The machine should measure at a frequency 0.316 Hz.
Code spectral analysis:
x = t2(:,3);
%x = x-mean(x);
n = length(x);
dt = 3;
fs = 1/dt;
t = (0:n-1)*dt;
y = fft(x);
f = (0:n-1)*(fs/n);
power = abs(y).^2/n;
figure()
plot(f,power)
xlabel('Frequency [Hz]')
ylabel('Power')
grid on
After spectral analysis I end up with two peaks: one at 0 Hz and one at 0.3316 Hz.
If you uncomment the second line you end up with this: a peak at 0.0017 and 0.3316 Hz.
My question is how to interpret the first peak? Has it a physical meaning? I've read somewhere that a large peak at zero frequency means that you have a massive DC offset. What does that mean? And could it be that through discretization the actual value (0.316 Hz) and found value (0.3316 Hz) are slightly different?
  3 Kommentare
Rik
Rik am 7 Sep. 2020
Question posted by Melvin Corvers recovered from Google cache (permalink, should be available after a few hours):
Hi,
I am analyzing a signal using FFT to obtain its frequency spectrum. Raw data is shown below.
The machine should measure at a frequency 0.316 Hz.
Code spectral analysis:
x = t2(:,3);
%x = x-mean(x);
n = length(x);
dt = 3;
fs = 1/dt;
t = (0:n-1)*dt;
y = fft(x);
f = (0:n-1)*(fs/n);
power = abs(y).^2/n;
figure()
plot(f,power)
xlabel('Frequency [Hz]')
ylabel('Power')
grid on
After spectral analysis I end up with two peaks: one at 0 Hz and one at 0.3316 Hz.
If you uncomment the second line you end up with this: a peak at 0.0017 and 0.3316 Hz.
My question is how to interpret the first peak? Has it a physical meaning? I've read somewhere that a large peak at zero frequency means that you have a massive DC offset. What does that mean? And could it be that through discretization the actual value (0.316 Hz) and found value (0.3316 Hz) are slightly different?
Rena Berman
Rena Berman am 9 Okt. 2020
(Answers Dev) Restored edit

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Antworten (1)

Peng Li
Peng Li am 6 Aug. 2020
Based on your code, your sampling frequency is 1/3 Hz. There is no way you can detect a component of 0.316 Hz using 1/3 Hz sampling frequency.
The spectrum is symmetrical to 0 Hz. By default MATLAB gives a shifted spectrum that is symmetrical Fs/2. You can fftshift it or you can simply plot the first half. The second peak on the righthand side is not what you want to detect; it is simply a mirror of the peak on the left hand side.

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