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Problem with design of iir lpf filter.
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Question

Respected Sir / Madam I am developing scilab for iir butterworth LPF. Once after designing the filter I am trying to check the filter output. The output does not what I was looking for. Please help me in this regard. I am posting the code as well. Your help will be priceless. Yours sincerely T.V.Chandra Shekar

//Low Pass IIR Butterworth Filter
clc;
clear;
close;

fs = 8000;
omegap = 500;
omegas = 2000;
wp = omegap/fs;
ws = omegas/fs;
A1_dB = -3;
A2_dB = -40;
A1 = (10)^(A1_dB/20);
A2 = (10)^(A2_dB/20);
N = 0.5*(log10(1/A1^2 -1)-log10(1/A2^2 -1))/(log10(500)-log10(2000));
disp(ceil(N));
hz = iir(ceil(N),'lp','butt',[wp ws],[A1 A2]);
[hzm fr] = frmag(hz,1024);
subplot(2,1,1);
plot2d(fr',hzm');xgrid(color("green"));
xtitle('Magnitude Spectrum','Normalized Frequency','Magnitude in dB');
disp(hz);
n = coeff(hz.num);
n = flipdim(n,2);
disp(n);
d = coeff(hz.den);
d = flipdim(d,2);
disp(d);
n = 0:1023;
x = sin(2*%pi*500*n/fs) + cos(2*%pi*2000*n/fs);
y = filter(n,d,x);
Y = abs(fft(y))/1024;
disp(Y);
f = (0:1023)*fs/1024;
subplot(2,1,2);
plot2d(f,Y);


Scilab 15-12-17, 8:01 p.m. chandrashekar

Answers:

0

If you can tell us what you are expecting as the output, we can help.

05-02-18, 4:48 p.m. Sid11235


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