Inverse z of z(3z²-6z+4)/((z-1)²(z-2))

Method: partial fractions of X(z)/z.

X(z)/z = (3z² − 6z + 4) / [(z − 1)²(z − 2)] = A/(z − 1) + B/(z − 1)² + C/(z − 2)

Constants

C = (3·4 − 12 + 4)/(2 − 1)² = 4
B = (3 − 6 + 4)/(1 − 2)     = −1
Compare z² coefficients: A + C = 3  →  A = −1

So

X(z) = −z/(z − 1) − z/(z − 1)² + 4z/(z − 2)

Inverse using standard pairs

• Z⁻¹[z/(z − 1)] = 1
• Z⁻¹[z/(z − 1)²] = k
• Z⁻¹[z/(z − 2)] = 2ᵏ

Answer: x(k) = 4·2ᵏ − k − 1, k ≥ 0.

Check: x(0) = 3 (equals lim X(z) = 3), x(1) = 6, x(2) = 13, x(3) = 28 — the series expansion of X(z) gives the same values.