Inverse z-transform for three ROCs

X(z) = 1/((z − 0.5)(z + 2)). Poles at z = 0.5 and z = −2.

Partial fractions: 1/((z − 0.5)(z + 2)) = 0.4/(z − 0.5) − 0.4/(z + 2)

(A at z = 0.5: 1/2.5 = 0.4; B at z = −2: 1/(−2.5) = −0.4.)

Write each term in z⁻¹ form: 1/(z − a) = z⁻¹/(1 − az⁻¹):

X(z) = 0.4 z⁻¹/(1 − 0.5z⁻¹) − 0.4 z⁻¹/(1 + 2z⁻¹)

Standard pairs (with the one-sample delay):

• Causal (ROC outside pole a): z⁻¹/(1 − az⁻¹) ↔ a^(n−1) u[n−1]
• Anticausal (ROC inside pole a): z⁻¹/(1 − az⁻¹) ↔ −a^(n−1) u[−n]

i) ROC 0.5 < |z| < 2

Pole 0.5 is inside the ring (causal part), pole −2 is outside (anticausal part).

x[n] = 0.4 (0.5)^(n−1) u[n−1] + 0.4 (−2)^(n−1) u[−n]

ii) ROC |z| < 0.5

Both poles are outside the ROC: both terms left-sided.

x[n] = −0.4 (0.5)^(n−1) u[−n] + 0.4 (−2)^(n−1) u[−n] = 0.4 [(−2)^(n−1) − (0.5)^(n−1)] u[−n]

iii) ROC |z| > 2

Both poles are inside the ROC: both terms right-sided.

x[n] = 0.4 [(0.5)^(n−1) − (−2)^(n−1)] u[n−1]

Checks: (iii) X(z) = 1/(z² + 1.5z − 1) = z⁻² − 1.5z⁻³ + …, so x[2] = 1 and x[3] = −1.5; the formula gives 0.4(0.5 + 2) = 1 and 0.4(0.25 − 4) = −1.5. ✓ (ii) X(0) = 1/((−0.5)(2)) = −1, and the formula gives x[0] = 0.4[(−2)⁻¹ − (0.5)⁻¹] = −1. ✓

Only case (i) includes the unit circle, so it is the stable (but non-causal) system.