Inverse z-transform by partial fractions

X(z) = (1 + 2z⁻¹ + z⁻²)/(1 + 1.5z⁻¹ + 0.5z⁻²), ROC |z| > 1 (right-sided, causal).

Step 1. Factor.

• Numerator: 1 + 2z⁻¹ + z⁻² = (1 + z⁻¹)²
• Denominator: 1 + 1.5z⁻¹ + 0.5z⁻² = (1 + z⁻¹)(1 + 0.5z⁻¹)

The factor (1 + z⁻¹) cancels, so X(z) = (1 + z⁻¹)/(1 + 0.5z⁻¹).

Step 2. Partial fractions. Numerator and denominator have the same degree, so there is a constant term. Write in z:

X(z) = (z + 1)/(z + 0.5) = 1 + 0.5/(z + 0.5)

Equivalently, in the form A + B/(1 + 0.5z⁻¹): X(z) = 2 − 1/(1 + 0.5z⁻¹). (Check: 2 + z⁻¹ − 1 = 1 + z⁻¹ over (1 + 0.5z⁻¹). ✓)

Step 3. Invert (causal). Using aⁿu[n] ↔ 1/(1 − az⁻¹) with a = −0.5:

x[n] = 2δ[n] − (−0.5)ⁿ u[n]

Answer: x[n] = 2δ[n] − (−0.5)ⁿ u[n]

Check: x[0] = 2 − 1 = 1; x[1] = 0.5; x[2] = −0.25; x[3] = 0.125. Long division of (1 + z⁻¹) by (1 + 0.5z⁻¹) gives 1 + 0.5z⁻¹ − 0.25z⁻² + 0.125z⁻³ …. ✓

The pole at z = −1 given by the denominator is cancelled by the zero, so the effective ROC is |z| > 0.5, which contains the stated ROC |z| > 1.