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Full Version: SOLUTION OF DIFFERENCE EQUATION WITH INITIAL CONDITIONS
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Code:
clear
close all
clc
% SOLVE THE GIVEN DIFFERENCE EQUATION Y(N) - 1.5Y(N-1) + 0.5Y(N-2) = X(N), N >= 0
% X(N) = (1/4)^n * U(N) SUBJECT TO Y(-1) = 4 AND Y(-2) = 10
% CALCULATE AND PLOT THE OUTPUT OF THE SYSTEM
% FIND THE TRANSFER FUNCTION OF THE GIVEN DIFFERENCE EQUATION H(Z) = Y(Z)/X(Z)
% NUMERATOR POLYNOMIAL IS a AND DENOMINATOR POLYNOMIAL IS b
% FOR THE ABOVE EXAMPLE b = [1] AND a = [1 -1.5 0.5]

b = [1];
a = [1 -1.5 0.5];
% FIND THE INITIAL CONDITIONS FOR THE INPUT X SUCH THAT THE GIVEN INITIAL CONDITIONS
% FOR Y ARE OBTAINED
yic = [4 10];
xic = filtic(b,a,yic);
xic
% GENERATE THE DISCRETE TIME INDEX
n = [0:7];
% GENERATE THE GIVEN INPUT X
x = (1/4).^n;
% FIND THE OUTPUT OF THE SYSTEM FOR THE INPUT SEQUENCE X
y = filter(b,a,x,xic);

% PLOT THE INPUT AND THE OUTPUT
figure(1)
subplot(2,1,1);
stem(n,x);
grid on;
xlabel('n --> ');
ylabel('x --> ');
title ('Input Sequence');
subplot(2,1,2);
stem(n,y);
xlabel('n --> ');
ylabel('y --> ');
title ('Output Sequence');
grid on;

RESULT:- DIFF EQN WITH INITIAL COND