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How to remove DC component in FFT?

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Wakeel Mohammed
Wakeel Mohammed am 9 Jan. 2021
Kommentiert: Image Analyst am 29 Jul. 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)');

Akzeptierte Antwort

Image Analyst
Image Analyst am 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 Kommentare
AMOS
AMOS am 29 Jul. 2023
At what point is ft implemented?
Image Analyst
Image Analyst am 29 Jul. 2023
@AMOS, at the point where you run this line of code:
ft = fft(Iv);

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Weitere Antworten (1)

Sateesh Kandukuri
Sateesh Kandukuri am 20 Dez. 2022
Dear @Image Analyst, the behaviour of my system is asymmetrical to the excitation fields.
Is it possible to modify this behaviour from asymmetrical to symmetrical? And then performing FFT may resolve my issue.
  11 Kommentare
Image Analyst
Image Analyst am 23 Dez. 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
Sateesh Kandukuri am 26 Dez. 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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