Inverse z of (3z³+2z)/((z-3)²(z-2)), inversion integral

X(z) = (3z³ + 2z)/((z − 3)²(z − 2)).

Inversion integral: x(k) = (1/2πi) ∮ X(z) zᵏ⁻¹ dz = sum of residues of X(z) zᵏ⁻¹ at all poles.

X(z) zᵏ⁻¹ = (3z² + 2) zᵏ / ((z − 3)²(z − 2)).

Poles: z = 3 (double) and z = 2 (simple). (There is no pole at z = 0 since X(z) has a factor z.)

Residue at z = 2 (simple):

R₂ = (3z² + 2) zᵏ/(z − 3)² at z = 2 = (14) 2ᵏ/1 = 14 · 2ᵏ

Residue at z = 3 (double):

R₃ = d/dz [ (3z² + 2) zᵏ/(z − 2) ] at z = 3

Let N = (3z² + 2) zᵏ. Then N′ = 6z zᵏ + k (3z² + 2) zᵏ⁻¹, and

R₃ = [ N′(z − 2) − N ]/(z − 2)² at z = 3 = N′(3) − N(3)

• N(3) = 29 · 3ᵏ
• N′(3) = 18 · 3ᵏ + 29 k 3ᵏ⁻¹

R₃ = 18 · 3ᵏ + 29 k 3ᵏ⁻¹ − 29 · 3ᵏ = −11 · 3ᵏ + 29 k 3ᵏ⁻¹

Answer:

x(k) = 14 · 2ᵏ − 11 · 3ᵏ + 29 k · 3ᵏ⁻¹, k ≥ 0

k    : 0   1    2    3    4
x(k) : 3  24  131  598  2465

Check: x(0) = 14 − 11 = 3 = lim X(z) as z → ∞. ✓ (Long division gives 3 + 24z⁻¹ + 131z⁻² + …)