Inverse z-transform, ROC |z| < 1

X(z) = z/((z − 1)(z − 2)²), ROC |z| < 1. Poles: z = 1, and z = 2 (double). The ROC lies inside all poles, so x[n] is left-sided.

Partial fractions of X(z)/z = 1/((z − 1)(z − 2)²):

1/((z − 1)(z − 2)²) = A/(z − 1) + B/(z − 2) + C/(z − 2)²

• A = 1/(1 − 2)² = 1
• C = 1/(2 − 1) = 1
• B = −A = −1

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

Inverse transform for ROC inside the poles (left-sided):

• z/(z − a) ↔ −aⁿ u[−n−1]

• z/(z − a)² ↔ −n a^(n−1) u[−n−1]

• z/(z − 1) ↔ −u[−n−1]

• −z/(z − 2) ↔ +2ⁿ u[−n−1]

• z/(z − 2)² ↔ −n 2^(n−1) u[−n−1]

Answer: x[n] = [ 2ⁿ − 1 − n 2^(n−1) ] u[−n−1]

Check: expanding X(z) about z = 0 gives X(z) = −0.25z − 0.5z² − …, so x[−1] = −0.25, x[−2] = −0.5. The formula gives x[−1] = 0.5 − 1 + 0.25 = −0.25 and x[−2] = 0.25 − 1 + 0.25 = −0.5. ✓ Also x[n] = 0 for n ≥ 0.