filtering problem, need help

Question

Ali Asghar 2020-2-1

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https://ww2.mathworks.cn/matlabcentral/answers/503097-filtering-problem-need-help

评论： Star Strider 2020-2-8

采纳的回答： Star Strider

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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

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Walter Roberson 2020-2-1

编辑：Walter Roberson 2020-2-1

https://www.mathworks.com/help/signal/ref/bandpass.html

bandpass uses a fir filter if it can achieve the desgign goals with one, and otherwise uses a iir filter.

For Butterworth filters see https://www.mathworks.com/help/signal/ref/butter.html

Ali Asghar 2020-2-1

Thank you

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Answer 1

Star Strider 2020-2-1

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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

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Star Strider 2020-2-1

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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