How can I plot Low Pass Butterworth Filter's response graph when its input wave is square wave?

Hi everyone,

How can I plot Low Pass Butterworth Filter's response graph when its input wave is square wave?

Could you inform me?

Answer 1

[Star Strider]

Plot it however you like.

Perhaps —

[z,p,k] = butter(5, 0.05);
[sos,g] = zp2sos(z,p,k);

t = linspace(0, 10, 1E+3);
s = sin(2*pi*t*0.75) > 0;

sf = filtfilt(sos,g,+s);

figure
plot(t, s)
hold on
plot(t,sf)
hold off

See the documentation on filtfilt for details.

.

[Erkam Aydar]

Thank you for your answer Star Strider.

I want to design a butterworth filter as fallows and want to see how my filter will respond when I apply a 90 kHz square wave. I cannot do when add your code to my code?. Could you help me?

%Butterworth filter
    %   Sample freq = 1000 Hz
    %   Low Pass Filter
    %   3 dB ripple at Passband 
    %   Passband is between 0 and 40 Hz 
    %   Rs stopband attenuation is 60 dB
    %   Stopband stars at 150 Hz
    %   Nyquist freq = 500 Hz (Sample freq/2).

 Wp = 40/500;  %passband edge frequency (normalized)
 Ws = 150/500; %stopband edge frequency (normalized)

%[n,Wn] = buttord(Wp,Ws,Rp,Rs);  % finds the minimum order,
%Rp passband ripple, Rs stopband attenuation.
[n,Wn] = buttord(Wp,Ws,3,60);


%[Z,P,K] = butter(N,Wn,'low')% design butterworth filter
%n:order,Wn:cutoff frequency

% Gives zeroes, poles and gain as a vector. 
[z,p,k] = butter(n,Wn);


%Each row of the SOS matrix describes a 2nd order transfer function:
SOS = zp2sos(z,p,k);  

    figure 
    freqz(SOS,1024,1000);           % Plots frequency response
   
    figure 
    zplane(z,p);                    % Plots Z-plane zero-pole 
    grid on
    
    figure
    [B,A] = butter(n,Wn); 
    freqz(B,A)

[Star Strider]

As always, my pleasure!

The stated requirements are going to be difficult, because it is not possible to create a 90 kHz square wave with a 1 kHz sampling frequency. The sampling frequency has to be more that two times the square wave frequency (so >180 kHz) for this to work. For that reason, I chose a 1 MHz sampling frequency. I kept the filter specifications the same, otherwise. Because the filter passband is so low compared to the signal frequency, essentially nothing (only a D-C signal) is passed after some initial transients.

It should be straigtforward to change the sampling frequency and the square wave frequency in this code if you want to use other values (such as a 1 kHz sampling frequency and a 90 Hz square wave). It will also be necessary to change the subplot xlim values in the figure that plots the Fourier transforms.

%Butterworth filter
    %   Sample freq = 1000 Hz
    %   Low Pass Filter
    %   3 dB ripple at Passband 
    %   Passband is between 0 and 40 Hz 
    %   Rs stopband attenuation is 60 dB
    %   Stopband stars at 150 Hz
    %   Nyquist freq = 500 Hz (Sample freq/2).

Fs = 1E+6;
Fn = Fs/2;
Wp = 40/Fn;  %passband edge frequency (normalized)
Ws = 150/Fn; %stopband edge frequency (normalized)

%[n,Wn] = buttord(Wp,Ws,Rp,Rs);  % finds the minimum order,
%Rp passband ripple, Rs stopband attenuation.
[n,Wn] = buttord(Wp,Ws,3,60);


%[Z,P,K] = butter(N,Wn,'low')% design butterworth filter
%n:order,Wn:cutoff frequency

% Gives zeroes, poles and gain as a vector. 
[z,p,k] = butter(n,Wn);


%Each row of the SOS matrix describes a 2nd order transfer function:
[SOS,g] = zp2sos(z,p,k);  

    figure 
    freqz(SOS,2^16,Fs);           % Plots frequency response

   
    figure 
    zplane(z,p);                    % Plots Z-plane zero-pole 
    grid on

    
%     figure
%     [B,A] = butter(n,Wn); 
%     freqz(B,A)

t = linspace(0, 1E+6-1, 1E+6)/Fs;
Fsqwv = 90E+3;                                  % Square Wave Frequency
s = sin(2*pi*t*Fsqwv) > 0;
Q = 1/t(2)

Q = 1000000

sf = filtfilt(SOS,g,+s);

figure
plot(t, s, 'DisplayName','Original Signal')
hold on
plot(t,sf, 'DisplayName','Filtered Signal')
hold off
grid

L = numel(t);
NFFT = 2^nextpow2(L);
FTsfs = fft([s(:) sf(:)]-mean([s(:) sf(:)]), NFFT)/L;
Fv = linspace(0, 1, NFFT/2+1)*Fn;
Iv = 1:numel(Fv);

trfn = FTsfs(:,2) ./ FTsfs(:,1);

figure
subplot(3,1,1)
plot(Fv, abs(FTsfs(Iv,1))*2, 'DisplayName','Original Signal')
grid
legend('Location','best')
xlim([0 1.5E+5])
subplot(3,1,2)
plot(Fv, abs(FTsfs(Iv,2))*2, 'DisplayName','Filtered Signal')
grid
legend('Location','best')
xlim([0 1.5E+5])
subplot(3,1,3)
plot(Fv, abs(trfn(Iv)), 'DisplayName','Filter Transfer Function')
grid
xlim([0 1.5E+5])
legend('Location','best')
sgtitle('Fourier Transforms')
xlabel('Frequency (Hz)')

.