Inverse z of z/((z+1)²(z-1))

Partial fractions of X(z)/z.

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

A = 1/(1 + 1)²   = 1/4
B = 1/(−1 − 1)   = −1/2
A + C = 0 (no z² term in the numerator)  →  C = −1/4

X(z) = ¼ z/(z − 1) − ¼ z/(z + 1) − ½ z/(z + 1)²

Standard pairs

• Z⁻¹[z/(z − 1)] = 1
• Z⁻¹[z/(z + 1)] = (−1)ᵏ
• Z[k aᵏ] = az/(z − a)²; for a = −1: Z[k(−1)ᵏ] = −z/(z + 1)², so Z⁻¹[z/(z + 1)²] = −k(−1)ᵏ

Answer:

x(k) = ¼ − ¼(−1)ᵏ + ½ k (−1)ᵏ, k ≥ 0.

Check: x(0) = 0, x(1) = 0, x(2) = 1, x(3) = −1, x(4) = 2, … The expansion X(z) = z⁻² − z⁻³ + 2z⁻⁴ − … agrees.