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

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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:
Data is the acquired signal, and dq.Rate is the sampling frequency.
I thank you in advance for your help.

Answers (2)

Hassaan on 20 Mar 2024
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;
Xamp = abs(X);
plot(freq, Xamp);
xlabel('Frequency (Hz)');
ylabel('Amplitude (g)');
title('Amplitude Spectrum');
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Chunru on 20 Mar 2024
Edited: Chunru on 20 Mar 2024
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
dq.Rate = 1;

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