How to remove DC component in FFT?

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Wakeel Mohammed 2021-1-9

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评论： Image Analyst 2023-7-29

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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采纳的回答

Image Analyst 2021-1-9

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https://ww2.mathworks.cn/matlabcentral/answers/712808-how-to-remove-dc-component-in-fft#answer_594373

在 MATLAB Online 中打开

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;

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AMOS 2023-7-29

At what point is ft implemented?

Image Analyst 2023-7-29

在 MATLAB Online 中打开

@AMOS, at the point where you run this line of code:

ft = fft(Iv);

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Sateesh Kandukuri 2022-12-20

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https://ww2.mathworks.cn/matlabcentral/answers/712808-how-to-remove-dc-component-in-fft#answer_1132307

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.

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Image Analyst 2022-12-23

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 2022-12-26

Dear @Image Analyst,

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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