Band-pass Butterworth filter

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Guglielmo Giambartolomei
评论： Star Strider 2017-3-2
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!
  2 个评论
Jan
Jan 2017-3-1
Please post, what you have tried and explain "doesn't work" with details. Then suggestion an improvement is much easier.
Guglielmo Giambartolomei
Guglielmo Giambartolomei 2017-3-1
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!

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采纳的回答

Star Strider
Star Strider 2017-3-1
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.
  2 个评论
Guglielmo Giambartolomei
Guglielmo Giambartolomei 2017-3-2
编辑：Guglielmo Giambartolomei 2017-3-2
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
Star Strider 2017-3-2
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.

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