filtering problem, need help

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

[Walter Roberson]

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]

Thank you

Answer 1

[Star Strider]

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

[Ali Asghar]

dear

wrote code for following work on data 30000x4

1- apply notch filter of 50Hz

2- apply band pass of 2nd order, zero-phase shift of 20-500hz for column 1-2.

3- apply band pass of 2nd order, 60-2000hz for column 3-4.

[b,a] = butter(2,[49.9 50.1]/(10000/2),'stop');
filter1 = filter(b,a,dath001);
[b,a] = butter(2,[20 500]/(10000/2),'bandpass');
filter2 = filter(b,a,filter1(:,1:2));
[b,a] = butter(2,[60 2000]/(10000/2),'bandpass');
filter3 = filter(b,a,filter1(:,3:4));
filter_signal = [filter2 filter3];  

I got the values without error but am i did it correctly?

Thank you

[Star Strider]

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]

Dear

Thank you for your advice....

Actually, in column 1-2 i have sEMG data and in column 3-4 i have iEMG data.

As per my literature review butterworth give good result.

So kindly check that the above coding is correct or not.....

[Star Strider]

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]

Thanks dear

[Star Strider]

As always, my pleasure!

[Ali Asghar]

Hey star

signal is 38000x1

I need to find the rms values of signal range 1-2000.

then rms of 2001-4000

till 38000 values of signal

means i get 19 values of rms.

How can i do this?

[Star Strider]

Try this:

sigseg = buffer(signal, 2000);
RMS_sigseg = sqrt(mean(sigseg.^2));

See the documentation for buffer for informaiton about it.

[Ali Asghar]

Dear

It works but when i use it in function then it give only one value

function MAV=jMAV(X)

y=buffer(X, 2000);

MAV=mean(abs(y));

end

MAV=jMAV(x);

give one value, instead of giving multiple values.

[Star Strider]

It gives one value for each column of the matrix. That is what it is defined to do.

[Ali Asghar]

Dear, Can I find mutiple values of mav with this function?

[Star Strider]

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.