Band-pass Butterworth filter

Hello, I'm trying to make a band-pass Butterworth filter in order to filter a signal. With the help of Star Strider I already made a high-pass filter:

Fcp=1;  %cutoff frequency
  [z,p,k]=butter(8,Fcp/(Fsp/2),'high');
  [sos,g]=zp2sos(z,p,k);
  %fvtool(sos,'Analysis','freq')

    deltap1filtfilt=filtfilt(sos,g,deltap_1);

I tried to make a band-pass Butterworth filter with this indications (https://it.mathworks.com/matlabcentral/answers/272316-how-to-butterworth-filter-with-bandpass-10-500-with-sampling-rate-1000) but it doesn't work. My sampling frequency is 600 Hz and I'd like to make visible only frequency contributes from 1 Hz to 50 hz. Thank u very much!

[Jan]

Please post, what you have tried and explain "doesn't work" with details. Then suggestion an improvement is much easier.

[Guglielmo Giambartolomei]

Hello Jan Simon, I tried with the indications of the link i posted; I wrote:

Wp = [1 49]/Fsp/2;
Ws = [0.5 49.5]/Fsp/2;

Rp=1;
Rs=25; 
[n,Wn] = buttord(Wp,Ws,Rp,Rs); 
[b,a]=butter(n,Wn);
[sos,g]=tf2sos(b,a);
%fvtool(sos,'Analysis','freq')

deltap1filtfilt=filtfilt(sos,g,deltap_1);

This doesn't work. Instead I tried:

Fcp_low=1;  %lower cutoff frequency
Fcp_high=50;  %higher cutoff frequency

[z,p,k]=butter(8,[Fcp_low Fcp_high]/(Fsp/2),'bandpass');
[sos,g]=zp2sos(z,p,k);
%fvtool(sos,'Analysis','freq')

deltap1bpfiltfilt=filtfilt(sos,g,deltap_1);

This last code seems to work!

Answer 1

[Star Strider]

There is an error in your code that prevents you from calculating the passbands and stopbands correctly.

This calculation:

Wp = [1 49]/Fsp/2;

gives you these passbands:

Wp =
       0.0005       0.0245

However the passbands you want are:

Wp = [1 49]/(Fsp/2);

Wp =
        0.002        0.098

Those will give you the correct result. The parentheses around ‘(Fsp/2)’ make all the difference.

If you want to design a filter with rolloffs as steep as you want, a Butterworth filter is not the best option. I would use a Chebyshev Type II design.

The Code —

Fsp = 1000;                                     % Create Data
Fn = Fsp/2;
Wp = [1.0 49]/Fn;
Ws = [0.5 50]/Fn;

Rp=10;
Rs=30; 
[n,Ws] = cheb2ord(Wp,Ws,Rp,Rs); 
[z,p,k] = cheby2(n,Rs,Ws);
[sos,g] = zp2sos(z,p,k);

figure(1)
freqz(sos, 2^16, Fsp)
set(subplot(2,1,1), 'XLim',[0 100])             % ‘Zoom’ X-Axis
set(subplot(2,1,2), 'XLim',[0 100])             % ‘Zoom’ X-Axis

The passband ripple in a Chebyshev Type II filter are irrelevant, since the filter has a flat passband. Allowing them to be 10 dB allows you to design a stable filter.

[Guglielmo Giambartolomei]

Thank you Star Strider! What about the code:

Fcp_low=1;  %lower cutoff frequency
Fcp_high=50;  %higher cutoff frequency

[z,p,k]=butter(8,[Fcp_low Fcp_high]/(Fsp/2),'bandpass');
[sos,g]=zp2sos(z,p,k);
%fvtool(sos,'Analysis','freq')

deltap1bpfiltfilt=filtfilt(sos,g,deltap_1);

Do you think it is correct?

[Star Strider]

My pleasure.

It looks like it should work. It depends on the filter characteristic you want.

The filter I posted in my Answer is the one you originally described in your Question. A Butterworth filter cannot do that effectively.

I would compare the two filters to see which design works best in your application.