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Plot transfer function of band pass filter

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MIRK
MIRK am 24 Feb. 2015
Beantwortet: Rick Rosson am 24 Feb. 2015
I want to compare between BPF with diffrent order using butterworth the figured produced doesn't look like the transfer function of band pass what is the problem in my code??
fs = 100*7000;
ts = 1/fs ;
t = -2 : ts : 2- ts ;
NO = length(t);
f=(-NO/2 : NO/2 -1)/(NO * ts);
imp = zeros(1,NO);
imp(1+2*fs)=NO;
cutoff_PPF=[6000,10000] ;
[a11,b11]=butter(1,2*cutoff_PPF*ts); % PBF of Order 2
[a22,b22]=butter(2,2*cutoff_PPF*ts); % PBF of Order 3
[a33,b33]=butter(3,2*cutoff_PPF*ts); % PBF of Order 6
zP_O1 = filter(a11,b11,imp);
zP_O4 = filter(a22,b22,imp);
zP_O6 = filter(a33,b33,imp);
ZP_O2 = abs(fftshift(fft(zL_O1)/NO));
ZP_O4 = abs(fftshift(fft(zL_O4)/NO));
ZP_O6 = abs(fftshift(fft(zL_O6)/NO));
figure(2);
subplot(221);
plot(f,ZP_O2);
xlabel('Frequency');
ylabel('Magnitude');
legend('ZP_O2');
title('The Transfer Function of PPF _order 2)');
%xlim([-100000,100000])
subplot(222);
plot(f,ZP_O4);
xlabel('Frequency');
ylabel('Magnitude');
legend('ZP_O4');
title('The Transfer Function of PPF _order 4)');
%xlim([-,])
subplot(223);
plot(f,ZP_O6);
xlabel('Frequency');
ylabel('Magnitude');
legend('ZP_O6');
title('The Transfer Function of PPF _order 6)');
%xlim([-,])

Antworten (1)

Rick Rosson
Rick Rosson am 24 Feb. 2015
doc fvtool
doc freqz

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