Question

rn] h[n], let rn] (3)un] and h[n ()"un]: Problem 1. For a system described by yln] _ Compute yn] analytically. Plot yln] in MATLAB for -5n 15 a. b. Compute yn] = rn] . h[n ] using the conv command in MATLAB. Plot yln] in MATLAB for -5 and compare it to your results from part (a.) n15

Digital Signal Processing

Answer 1

a. x(n) = 1/3^n u(n): h(n) = (4/5)^n u (n) apply z transform to x(n) & h(n) apply z transform to x(n) & h(n) rightarrow x(z) = z/ z - 1/3 & h(z) = z/z -4/5 y(n) = x(n) * h(n) = x(n) * h(n) rightarrow y(n) = x(z) . h(z) = z^2/(z - 1/3 (z - 4/5) rightarrow y(z)/z = z/(z - 1/3)(z - 4/5) = A/z - 1/3 + B/z -4/5 A = lt_z - 1/3 (z - 1/3) y(z)/z = -0.71743 B = -07143Z/z - 1/3 + 1.7143 z/z - 4/5 apply increase z-Transform to y(z) rightarrow y[n] = [-0.7143 (1/3)^n + 1.7143 (4/5)^n] u(n) lets plot y(n) from (1) using Matlab

MATLAB code is given below in bold letters.

clc;
close all;
clear all;

% define the range of n
n = -5:15;

% define the signal y[n] from analytical solution
y = (-0.7143*(1/3).^n + 1.7143*(4/5).^n).*(n>=0);

% plot the signal y[n]
figure;
stem(n,y);grid on;
xlabel('n');title('y[n] analytical solution');

% Question b

% define x and h as follows
x = ((1/3).^n) .*(n>=0);
h = ((4/5).^n) .*(n>=0);

% now compute y[n] using conv command as follows
y = conv(x,h);

% Now plot the signal y[n]
figure;
stem(2*min(n):2*max(n),y);grid on;
xlabel('n');title('y[n] using MATLAB conv function ');
xlim([-5 15]);

It is observed that both the analytical solution and the MATLAB conv function solution are the same.