Remove DC component from EEG signals
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altaf adil
am 23 Feb. 2011
Kommentiert: Jinrong Tang
am 11 Sep. 2019
I have tried two methods to remove the DC component from the EEG signal.
Method 1: Data_without_dc = Data_with_dc - mean (Data_with_dc)
Method 2: Data_fft = fft(Data_with_dc); Data_fft(1) = 0; Data_without_dc = ifft (Data_fft);
I think the first component of signal's FFT is actually the mean of whole signal and it represents signal amplitude at zero frequency, so by making it to zero, we can remove DC component.
But from both the methods, I am getting different results. Please tell me which method is preferable if I have to remove the DC component whose amplitude is unknown in case of EEG data.
1 Kommentar
Edith
am 28 Mär. 2015
Please why do you want to remove the DC component from the EEG signal? I also trying to show in a plot my signals from 16 electrodes.
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Jan
am 23 Feb. 2011
You can remove the DC component by removing the MEAN. This is fast, accurate and cheap.
I cannot see a difference between the methods, if you remove the imaginary roundoff errors:
x = rand(100, 1);
f = fft(x);
f(1) = 0;
x_ac = real(ifft(f));
[x_ac, x - mean(x)]
>> -0.4723 -0.4723
0.4304 0.4304
-0.4166 -0.4166
etc.
The difference is as expected in the magnitude of EPS(1).
BTW: I prefer "SUM(x)./SIZE(x,1)", because it is faster than MEAN.
1 Kommentar
Jinrong Tang
am 11 Sep. 2019
Hi, I don't think remove the Mean is suitable for the ECG signal since it's asymmetrical and not like the sine wave. So removing the Mean will make the signal's baseline a little below the x-axis.
Weitere Antworten (2)
Narayan Puthanmadam Subramaniyam
am 29 Mär. 2012
Hey altaf, if you data is in form of a matrix, then make sure to remove the entire first row and not just the first element. For example, make Data_fft(1,:) = 0. Then your answers should be identical!
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VENKATA PHANIKRISHNA B
am 7 Jun. 2017
Removal of DC component means Mean removal from the signal. By doing this we can eliminate the dc, i.e., 0Hz noise. Here DC component means, the signal positive half cycles average and the negative half cycle is not zero. if x is the signal then xm=x-mean(x).
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