filtering problem, need help
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Dear
I have a EMG signal of 30000x4. with sampling frequency of 10KHz.
I filter noise by below code
NotchFilter = bandstop(Data01Raw,[49.9 50.1],Fs);
sEmgFilter = bandpass(NotchFilter(:,1:2),[20 500],Fs);
iEmgFilter = bandpass(NotchFilter(:,3:4),[600 2000],Fs);
Data02Filtered = [sEmgFilter iEmgFilter];
Questions
in above code bandpass or bandstop occur by which filer like is it butterwork or cheby or ellip?
I need to fiter the signal using butterworth 2nd order. How can i do this?
Thank you
2 个评论
Walter Roberson
2020-2-1
编辑：Walter Roberson
2020-2-1
bandpass uses a fir filter if it can achieve the desgign goals with one, and otherwise uses a iir filter.
Ali Asghar
2020-2-1
采纳的回答
Star Strider
2020-2-1
Request two outputs from the bandpass and bandstop functions. The second output is the digital filter object. Displaying the digital filter object in your Command Window will tell you everything you need to know about it.
For example —
Fs = 10E+3;
[NotchFilter, df] = bandstop(Data01Raw,[49.9 50.1],Fs);
then:
df
displays:
df =
digitalFilter with properties:
Coefficients: [8×6 double]
Specifications:
FrequencyResponse: 'bandstop'
ImpulseResponse: 'iir'
SampleRate: 10.0000e+003
StopbandAttenuation: 60.0000e+000
PassbandRipple2: 100.0000e-003
StopbandFrequency2: 50.0843e+000
PassbandRipple1: 100.0000e-003
PassbandFrequency2: 50.1000e+000
PassbandFrequency1: 49.9000e+000
StopbandFrequency1: 49.9157e+000
DesignMethod: 'ellip'
Use fvtool to visualize filter
Use designfilt to edit filter
Use filter to filter data
12 个评论
Ali Asghar
2020-2-1
Star Strider
2020-2-1
I do not understand the overlapping bandpass filters. (The notch filter for the 50 Hz mains interference is obviously required.)
I would instead use elliptic filters, because they are shorter and therfore more computationally efficient, and among other things, they will produce the result you want.
Example —
[n,Wp] = ellipord([20 500]/Fn, [5 515]/Fn, 1, 50);
[z,p,k] = ellip(n, 1, 50, Wp,'bandpass');
[sos2,g2] = zp2sos(z,p,k);
figure('Name','Filter #2')
freqz(sos2, 2^24, Fs)
set(subplot(2,1,1), 'XLim',[0 1000]) % Optional
set(subplot(2,1,2), 'XLim',[0 1000]) % Optional
Then use the filtfilt function to do the actual filtering:
filter1 = filtfilt(sos2,g2,filter1(:,1:2));
and so for the others. The filtfilt function produces a phase-neutral result. so the signal has no phase distortion.
My apologies for the delay. I am having problems with MATLAB being very slow this morning for some reason.
Ali Asghar
2020-2-1
Star Strider
2020-2-1
It is, however it could be improved. Specifically, the filter order needs to be increased, since order 2 filters will not do what you want them to do. Use the buttord function to determine the optimal filter order, and specify a stopband as well as a passband in the filter design, such as I did with the second filter here. The second-order-section representation will be much more stable than the transfer-function representation. Also, it is best to use the freqz function to be certain the filters are doing what you want them to do.
Consider this version:
Fs = 10000;
Fn = Fs/2;
[z,p,k] = butter(2,[49.9 50.1]/(10000/2),'stop');
[sos1,g1] = zp2sos(z,p,k);
filter1 = filtfilt(sos1,g1,dath001);
[n,Wn] = buttord([20 500]/Fn, [5 515]/Fn, 1, 50);
[z,p,k] = butter(n, Wn,'bandpass');
[sos2,g2] = zp2sos(z,p,k);
filter2 = filtfilt(sos2,g2,filter1(:,1:2));
[z,p,k] = butter(2,[60 2000]/(10000/2),'bandpass');
[sos3,g3] = zp2sos(z,p,k);
filter3 = filtfilt(sos3,g3,filter1(:,3:4));
filter_signal = [filter2 filter3];
figure('Name','Filter #1')
freqz(sos1, 2^24, Fs)
set(subplot(2,1,1), 'XLim',[0 100]) % Optional
set(subplot(2,1,2), 'XLim',[0 100]) % Optional
figure('Name','Filter #2')
freqz(sos2, 2^24, Fs)
set(subplot(2,1,1), 'XLim',[0 700]) % Optional
set(subplot(2,1,2), 'XLim',[0 700]) % Optional
figure('Name','Filter #3')
freqz(sos3, 2^24, Fs)
set(subplot(2,1,1), 'XLim',[0 2500]) % Optional
set(subplot(2,1,2), 'XLim',[0 2500]) % Optional
Ali Asghar
2020-2-6
Star Strider
2020-2-6
As always, my pleasure!
Ali Asghar
2020-2-7
Star Strider
2020-2-7
Try this:
sigseg = buffer(signal, 2000);
RMS_sigseg = sqrt(mean(sigseg.^2));
Ali Asghar
2020-2-8
编辑：Ali Asghar
2020-2-8
Star Strider
2020-2-8
It gives one value for each column of the matrix. That is what it is defined to do.
Ali Asghar
2020-2-8
Star Strider
2020-2-8
I am not certain what you are asking. You can have buffer break the signal up into shorter segments if you want.
With respect to calculating the RMS value, if you have R2016a or later, you can use the movmean function instead of mean.
This computes the RMS value over a sliding window of 200 samples:
RMS_sigseg = sqrt(movmean(sigseg.^2, 200));
It results in a matrix the same size as the original matrix.
Note that the code you posted gives the mean of the absolute value. This is not the same as RMS value.
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