Question

use ztransform to solve the difference equation

$y(k)-1.4y(k-1)+0.45y(k-2)=x(k)$

for k greater than or equal to 0 with initial conditions y(-1)=3, y(-2)=-1, and $x(k)=(0.5)^{k}u(k)$

Answer 1

y (k) = 1.4 y)k - 1) + 0.45 y (k - 1) = x (k) x (k) = (0.5)^k u (k) y (-1) = 3 y (-2) = -1 Apply z-transform y (z) - 1.4 [z^-1 y (z) - y (-1)] + 0.45 [z^-2 y (z) - z^-1 cy (-1) -y (-2)] = z/z - 0.5 y (z) [1 - 1.4 z^-1 + 0.45 z^-2] = z/z - 0.5 + 1.4 y (-1) + 0.45 z^-1 y () + y (-2) Y (z) [z^2 - 1.4 z + 0.45/z^2] = z/z - 0.5 + 1.4 (z) + 0.45 z^-1 (-1) + 3 = z + 1.4 z (z - 0.5) + 0.45 z^-1 (z - 0.5) + (z - 0.5)/(z - 0.5) Y (z) [z^2 - 1.4z + 0.5/z^2] = z + 1.4z^2 - 0.7z + (-1) 0.45 + 0.225z^-1 + 3z - 1.5/(z - 0.5) y (z) [(z - 0.9) (z - 0.5)/2] = 1.4z^2 + 1.3z + 0.225 z^-1 - 1.95/(z - 0.5) Y (z) (z - 0.9) (z - 0.5)/z^2 = 1.4 z^3 + 1.3 z^