Butterworth bandpass filter on a signal using buttord and butter

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José Santos Pérez Leal 2021-6-14

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评论： Star Strider 2021-6-14

Hi everyone.

I've been working on a butterworth bandpass filter on a signal, but I can't use the function buttord and butter properly together. The response frecuency of the signal I'm working with has been obtained using fft (shown in the image below). I want to make a bandpass filter between 5 and 450 Hz. Originally, I obtained my bandpass filter only using the function butter, but I select an arbitrary filter order and I need to obtain the order with the function buttord. The code I've been working with for the filter using buttord and butter is shown below the response frecuency image, but it didn't work properly. If someone could tell me what I'm doing wrong in the design of my Butterworth bandpass filter or if I'm designing it in a wrong way, I would be very thankful.

%Arbitrary values for the ripple in the passband and stopband
delta1=0.01;
delta2=0.005;
Ripplep=-20*log10(1-delta1); %Ripple in the passband
Ripples=-20*log10(delta2); %Ripple in the stopband
f=1260; %Sampling frecuency
fnyq=f/2; %Nyquist frecuency
wp=[10 450]/fnyq; %10 is the omegap frecuency for the inferior cut off frecuency portion. 450 is the omegap 
                  %frecuency for the superior cut off frecuency portion
ws=[5 455]/fnyq; %5 is the omegas frecuency for the inferior cut off frecuency portion. 455 is the omegap
                 %frecuency for the superior cut off frecuency portion
[N,WN]=buttord(wp,ws,Ripplep,Ripples) %Obtain filter order and natural frecuency
[e,c]=butter(N,WN) %butterworth coeficients for the filter
bandpass_filt_data=filtfilt(e,c,window_data) %Apply filter to the original signal (window_data), which has
                                             %a dimension of 15725X1 

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Answer 1

Star Strider 2021-6-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/856165-butterworth-bandpass-filter-on-a-signal-using-buttord-and-butter#answer_724755

在 MATLAB Online 中打开

The transfer function implementation created an unstable filter.

Instead, use zero-pole-gain output from butter, then implement it as a second-order-section filter with zp2sos.

And always use freqz to assess the filter characteristics first. That will tell you if it met the design requirements.

Try this —

%Arbitrary values for the ripple in the passband and stopband

delta1=0.01;

delta2=0.005;

Ripplep=-20*log10(1-delta1); %Ripple in the passband

Ripples=-20*log10(delta2); %Ripple in the stopband

f=1260; %Sampling frecuency

fnyq=f/2; %Nyquist frecuency

wp=[10 450]/fnyq; %10 is the omegap frecuency for the inferior cut off frecuency portion. 450 is the omegap

%frecuency for the superior cut off frecuency portion

ws=[5 455]/fnyq; %5 is the omegas frecuency for the inferior cut off frecuency portion. 455 is the omegap

%frecuency for the superior cut off frecuency portion

[N,WN]=buttord(wp,ws,Ripplep,Ripples); %Obtain filter order and natural frecuency

[z,p,k]=butter(N,WN); %butterworth coeficients for the filter

[sos,g] = zp2sos(z,p,k);

figure

freqz(sos,2^14,f)

bandpass_filt_data=filtfilt(sos,g,window_data) %Apply filter to the original signal (window_data), which has

Unrecognized function or variable 'window_data'.

%a dimension of 15725X1

.

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José Santos Pérez Leal 2021-6-14

Thank you very much, Star Strider, I will try to implement the zero-pole-gain to the filter design and then apply the filter to the signal. Thank you!!!