Inverse z-transform with repeated pole

X(z) = z/((z − 0.4)(z + 1.5)²), ROC |z| < 0.4. Poles: 0.4 and −1.5 (double). The ROC is inside all the poles, so x[n] is left-sided (anticausal).

Partial fractions of X(z)/z = 1/((z − 0.4)(z + 1.5)²):

1/((z − 0.4)(z + 1.5)²) = A/(z − 0.4) + B/(z + 1.5) + C/(z + 1.5)²

• A = 1/(0.4 + 1.5)² = 1/3.61 = 100/361
• C = 1/(−1.5 − 0.4) = −1/1.9 = −10/19
• B = −A = −100/361 (from the large-z behaviour: the coefficients of the 1/z terms must add to zero)

X(z) = (100/361) z/(z − 0.4) − (100/361) z/(z + 1.5) − (10/19) z/(z + 1.5)²

Inverse transforms, left-sided (ROC inside the poles):

• z/(z − a) = 1/(1 − az⁻¹) ↔ −aⁿ u[−n−1]
• z/(z − a)² ↔ −n a^(n−1) u[−n−1]

With a = 0.4 and a = −1.5:

x[n] = [ −(100/361)(0.4)ⁿ + (100/361)(−1.5)ⁿ + (10/19) n (−1.5)^(n−1) ] u[−n−1]

Check (series in z): expanding X(z) about z = 0 gives X(z) = −1.1111z − 1.2963z² − 4.7222z³ − …, so x[−1] = −1.1111, x[−2] = −1.2963, x[−3] = −4.7222 and x[n] = 0 for n ≥ 0. The formula gives x[−1] = −1.1111, x[−2] = −1.2963, x[−3] = −4.7222. ✓

Answer: x[n] = [ (100/361)((−1.5)ⁿ − (0.4)ⁿ) + (10/19) n (−1.5)^(n−1) ] u[−n−1]