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Why does the fft function cut in half the amplitude of my signal?

조회 수: 38 (최근 30일)
Claudio Gerosa
Claudio Gerosa 2024년 3월 20일
댓글: Paul 2024년 3월 21일
Hi, I have a question I would like to submit. I'm working on a project for uni in which i have to measure vibrations using an accelerometer. Now what I do is that I process the acquired datas using the fft function, in order to gain information about the frequency and the amplitude of the signal. The problem that I'm facing is that, by looking at the signal you can see that it has an amplitude of approximatively 40 g, but when I plot the Fourier Transform, it shows a maximum amplitude of about 20 g, which is half the amplitude of the original signal. Is there something that I'm missing?
This is the code that I'm using to plot the Fourier Transform:
N=numel(data);
freq=[0:N-1]/N*dq.Rate;
X=2*fft(data)/N;
X(1)=X(1)/2;
Xamp=abs(X);
figure()
plot(freq,Xamp);
Data is the acquired signal, and dq.Rate is the sampling frequency.
I thank you in advance for your help.
  댓글 수: 1
Paul
Paul 2024년 3월 21일
Can you upload your data in a file using the Paperclip icon on the Insert menu?

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답변 (2개)

Hassaan
Hassaan 2024년 3월 20일
N = numel(data);
freq = [0:N-1] / N * dq.Rate;
X = fft(data);
% Apply the window correction if you used a window function
% e.g., for a Hann window: window_correction = sum(hann(N))^2;
% X = X / window_correction;
% Doubling the FFT values, except for the DC and Nyquist (if N is even)
X = X / N;
X(2:end-1) = 2 * X(2:end-1);
% If N is even, the last element is the Nyquist frequency and should not be doubled
if mod(N, 2) == 0
X(end) = X(end) / 2;
end
Xamp = abs(X);
figure();
plot(freq, Xamp);
xlabel('Frequency (Hz)');
ylabel('Amplitude (g)');
title('Amplitude Spectrum');
-----------------------------------------------------------------------------------------------------------------------------------------------------
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Chunru
Chunru 2024년 3월 20일
편집: Chunru 2024년 3월 20일
Note that a sinosoidal signal can be expressed as:
FFT of (normalized by N) will have a peak at f0 with amplitude of 1 (similar for negative freq as well). The half is evident from the above equation.
A = 1; % amplitude
f0 = 1/16; % normalized freq
data = cos(2*pi*f0*(0:127))
data = 1×128
1.0000 0.9239 0.7071 0.3827 0.0000 -0.3827 -0.7071 -0.9239 -1.0000 -0.9239 -0.7071 -0.3827 -0.0000 0.3827 0.7071 0.9239 1.0000 0.9239 0.7071 0.3827 0.0000 -0.3827 -0.7071 -0.9239 -1.0000 -0.9239 -0.7071 -0.3827 -0.0000 0.3827
N=numel(data);
dq.Rate = 1;
freq=[0:N-1]/N*dq.Rate;
X=fft(data)/N;
X(1)=X(1)/2;
Xamp=abs(X);
figure()
plot(freq,Xamp);

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