How to remove DC component in FFT?
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Wakeel Mohammed
le 9 Jan 2021
Commenté : Image Analyst
le 29 Juil 2023
I succesfully plotted my FFT with MATLAB discussion help. Now I could not remove the DC component at 0Hz. Which shows me a very high amplitude. Can any one suggest me an idea?
data1 = xlsread('Reading 1.xlsx') ; %Loading Sensor data from Excel file
t = data1 (1:512,2); %Selecting Time vector
s = data1 (1:512,3); %Selecting Z axis vibrations
L = numel(t); %Signal length
Ts = mean(diff(t)); %Sampling interval
Fs = 1/Ts; %Sampling frequency
Fn = Fs/2; %Nyquist frequency
FTs = fft(s)/L; %Fast fourier transform (s- data)
Fv = linspace(0,1, fix(L/2)+1)*Fn; %Frequency vector
Iv = 1:numel(Fv); %Index vector
subplot(2, 1, 1); %plotting top pane
plot(t,s); %Acceleration vs time
set(gca,'xlim',[1 50]); %Scale to fit
grid; %Grids on
title ('Acceleration vs time');
xlabel('time(s)');
ylabel('Acceleration');
subplot(2, 1, 2); %Plotting bottom pane
plot(Fv, abs(FTs(Iv))*2,'red'); %FFT - Amplitude vs Frequency
grid
title ('Fast fourier transform');
xlabel('Frequency (Hz)');
ylabel ('Amplitude (m)');
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Image Analyst
le 9 Jan 2021
In the spatial domain, before fft, you can subtract the mean
Iv = Iv - mean(Iv);
In the frequency domain, you can zero out the DC component by setting it to zero
ft = fft(Iv);
ft(1) = 0;
11 commentaires
Image Analyst
le 29 Juil 2023
@AMOS, at the point where you run this line of code:
ft = fft(Iv);
Plus de réponses (1)
Sateesh Kandukuri
le 20 Déc 2022
Is it possible to modify this behaviour from asymmetrical to symmetrical? And then performing FFT may resolve my issue.
11 commentaires
Image Analyst
le 23 Déc 2022
I had the window width be several wavelents long. How many indexes are between each of your peaks? Try having the window width be like 3 or 4 times that long.
Sateesh Kandukuri
le 26 Déc 2022
Using your suggestion, I used movmean() in the calculation of fft as
A = readmatrix('table.txt');
ts=1e-12;
My = A(:,3);
[peakValues, indexesOfPeaks] = findpeaks(My);
windowWidth = 2 * mean(diff(indexesOfPeaks));
MySmooth = movmean(My, windowWidth);
My = My - MySmooth;
N = 2^(nextpow2(length(My)));
freq = fft(My,N);
freq2 = abs(fftshift(freq));
freq3 = freq2/max(freq2);
I got the following result for My component
Is this the right way to use movmean() function?
I've tried to understand the working of movmean() function using some arrays, but I still need clarification. Can you briefly explain with an example?
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