---
title: "Inverse z of (3z³+2z)/((z-3)²(z-2)), inversion integral"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:inverse-z-3z3-2z-inversion"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-3z3-2z-inversion
site: BCT Engineering
---

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

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2073 Bhadra, 5 marks; 2071 Magh, 5 marks (asked 2 times)
- Video: [12. Inverse Z-Transform by Inversion Integral Method | Problem #2 | Complete Concept](https://www.youtube.com/watch?v=6ID2ujXivfU) — MKS TUTORIALS by Manoj Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-3z3-2z-inversion
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Obtain the inverse z-transform of X(z) = (3z³ + 2z)/((z - 3)²(z - 2)) by using inversion integral method.

## Short answer

Compute residues of X(z)z^(k−1) = (3z²+2)z^k/(( z−3)²(z−2)). Residue at z=2: 14·2^k. Double pole at z=3 gives −11·3^k+29k·3^(k−1). Sum: x(k)=14·2^k−11·3^k+29k·3^(k−1), k≥0.

## Answer

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

```text
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 + …)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. x(k) is?
   A. Residue at z=0 only
   B. Sum of residues of X(z) z^{k-1} at all poles
   C. Residues of X(z) only
   D. Sum of zeros
   Answer: B. Contour inside the ROC. (Key point: Inversion integral)

2. Why no pole at z=0?
   A. z=0 is outside
   B. X(z) is zero
   C. X(z) has a factor z cancelling z^{k-1}'s negative power
   D. k is positive
   Answer: C. X z^{k-1} has numerator z^k. (Key point: Pole at 0)

3. What is the residue at z=2?
   A. 3 · 2^k
   B. 29 · 2^k
   C. −11 · 2^k
   D. 14 · 2^k
   Answer: D. (3z²+2) z^k/(z-3)² at 2. (Key point: Residue at 2)

4. What is x(k)?
   A. 14·2^k − 11·3^k + 29k·3^{k−1}
   B. 14·2^k + 11·3^k
   C. 14·2^k − 11·3^k + 29k·3^k
   D. 3·2^k
   Answer: A. x(0)=3. (Key point: Result)

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