Question

I want to get inverse Z transform of the following matrix,

Want to get it in k variable.

G = [z(z-0.6) / (z^2-1.2z+1) 0.8z / (z^2-1.2z+1) ; 0.8z / (z^2-1.2z+1) z(z-0.6) / (z^2-1.2z+1)] --->2 x 2 matrix

Answer 1

%Matlab code to generate inverse Z transform of given matrix

clc

clear all

syms z k

G=[z*(z-0.6)/(z^2-1.2*z+1) 0.8*z/(z^2-1.2*z+1);0.8*z/(z^2-1.2*z+1) z*(z-0.6)/(z^2-1.2*z+1)];

t=iztrans(G,k);

% After executing above script on matlab, we will get

t =

[ (-1)^k*cos(k*(pi - acos(3/5))), (4*(-1)^k*cos(k*(pi - acos(3/5))))/3 + ((-1)^k*5^(1 - k)*(- 3 - 4i)^(k - 1)*5i)/6 - ((-1)^k*5^(1 - k)*(- 3 + 4i)^(k - 1)*5i)/6]

[ (4*(-1)^k*cos(k*(pi - acos(3/5))))/3 + ((-1)^k*5^(1 - k)*(- 3 - 4i)^(k - 1)*5i)/6 - ((-1)^k*5^(1 - k)*(- 3 + 4i)^(k - 1)*5i)/6, (-1)^k*cos(k*(pi - acos(3/5)))]