Question

4. A 3rd-order lowpass filter is described by the difference equation y(n) = 0.01 x(n) + 0.05 x(n − 1) + 0.05 x(n − 2) + 0.01 x(n − 3) + 1.75 y(n − 1) − 1.18 y(n − 2) + 0.28 y(n − 3). Plot the magnitude and the phase responses of this filter and verify that it is a lowpass filter.

how do I code on matlab

Answer 1

clear;
clc;
close all;
syms n z y(n) x(n);
eqn1 = y(n) == 0.01*x(n) + 0.05*x(n-1) + 0.05*x(n-2) +0.01*x(n-3) + 1.75*y(n-1) - 1.18*y(n-2) + 0.28*y(n-3);
z_eqn1 = ztrans(eqn1);
syms Y X;
z_eqn1 = subs(z_eqn1,[ztrans(y(n), n, z) ztrans(x(n), n, z)],[Y X]);
z_eqn1 = subs(z_eqn1,[y(-1) y(-2) y(-3) x(-1) x(-2) x(-3)],[0 0 0 0 0 0]); %placing the initial conditions as 0
z_eqn1 = subs(z_eqn1,X,1); % to find the mag/phase response considering X(z) = 1 i.e x(n) = kroneckerDelta.
Y = solve(z_eqn1,Y); %Solving for Y and get the z-domain response of the kroneckerDelta input
w_range = -pi:0.1:pi;
%for finding the mag/phase plot, we need to substitute z = exp(j*w) and
%then w varies from 0 to 2*pi.
syms w;
response_w = subs(Y,z,exp(1i*w));
fft_response = subs(response_w,w,w_range);
mag_response = abs(fft_response);
phase_response = angle(fft_response);
subplot 211
plot(w_range/(2*pi),mag_response);
title("Magnitude Response)");
xlabel("f");
ylabel("magnitude");
subplot 212
plot(w_range/(2*pi),phase_response);
title("Phase Response)");
xlabel("f");
ylabel("magnitude");

Fro the above plot, it is clear that the given filter is a low pass filter, since it allows the lower frequencies and stops higher frequencies.

If you have any doubts, please feel free to ask in the comments.