raw the filter of the equation x(t) given below. Design the filter using a Butterworth filter. Verify the filter design and adjust n as needed. - Test the filter and explain what you get? x(t) = 3 sin(51130t)+ 15cos(61250t)
Script:
clc;close all;clear all;
%plot x(t)
t=0:2*(10^(-4)):0.05;
x=(cos(2*pi*100*t))+(cos(2*pi*500*t));
figure;
subplot(311)
plot(t,x);xlabel('t');ylabel('y(t)');;grid;
title('Input x(t) ')
%design Butterworth Lpf for N=20
[b,a]=butter(20,2*pi*300,'s');
Hsys=tf(b,a);
%Magntiude response of butterworth lpf
w=2*pi.*(0:800);
H=freqs(b,a,w);
subplot(312)
plot([0:800],abs(H));grid;
xlabel('f in Hz');ylabel('|H|')
title('Magntitude response of butterworth LPF for N=20')
%filtering
y=lsim(Hsys,x,t);
subplot(313)
plot(t,y);grid;
xlabel('t');ylabel('y(t)');
title('Output y(t) after filtering')
Observation:
The output y(t) has only one frequency component f=1/0.01s=100hz. The high-frequency component is not available at the output.
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