# How to remove DC component in FFT?

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Wakeel Mohammed am 9 Jan. 2021
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?
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)'); ##### 0 Kommentare-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden

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### Akzeptierte Antwort

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;
##### 8 Kommentare7 ältere Kommentare anzeigen7 ältere Kommentare ausblenden
Image Analyst am 12 Dez. 2022
Glad it is working now. The DC component is now removed, but as you can see there is power in frequencies near pure zero. You may want to erase a longer section at the beginning and end of your unshifted spectrum array to get rid of those. Like
freq(1:30) = 0;
freq(end-30 : end) = 0;
Or you can do it in the middle of the array if you're dealing with the shifted array.

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### Weitere Antworten (1)

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.
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Sateesh Kandukuri am 26 Dez. 2022
Using your suggestion, I used movmean() in the calculation of fft as
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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