---
title: "Inverse z of (2z³+z)/((z-2)²(z-1))"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:inverse-z-2z3-z"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-2z3-z
site: BCT Engineering
---

# Inverse z of (2z³+z)/((z-2)²(z-1))

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2075 Baisakh, 5 marks; 2068 Bhadra (asked 2 times)
- 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-2z3-z
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Obtain the inverse z-transform of X(z) = (2z³ + z)/((z - 2)²(z - 1)) by using partial fraction method.

## Short answer

Using partial fractions, X(z)/z = 3/(z-1) - 1/(z-2) + 9/(z-2)^2, so X(z) = 3z/(z-1) - z/(z-2) + 9z/(z-2)^2. Inverting gives x(k) = 3 - 2^k + 9k 2^{k-1} for k≥0.

## Answer

X(z) = (2z³ + z)/((z − 2)²(z − 1)).

**Step 1.** Divide by z and use partial fractions (a double pole at 2 and a simple pole at 1):

X(z)/z = (2z² + 1)/((z − 2)²(z − 1)) = A/(z − 1) + B/(z − 2) + C/(z − 2)²

- A = (2z² + 1)/(z − 2)² at z = 1 = 3/1 = 3
- C = (2z² + 1)/(z − 1) at z = 2 = 9/1 = 9
- B: for large z, X/z ≈ 2/z, so A + B = 2, which gives B = −1

(Check at z = 0: the left side is 1/((4)(−1)) = −1/4 and the right side is −3 + 1/2 + 9/4 = −1/4. ✓)

**Step 2.** Multiply by z:

X(z) = 3z/(z − 1) − z/(z − 2) + 9z/(z − 2)²

**Step 3.** Invert with the standard pairs:

- Z⁻¹[z/(z − 1)] = 1
- Z⁻¹[z/(z − 2)] = 2ᵏ
- Z⁻¹[z/(z − a)²] = k aᵏ⁻¹, so Z⁻¹[z/(z − 2)²] = k 2ᵏ⁻¹

**Answer:**

x(k) = 3 − 2ᵏ + 9k · 2ᵏ⁻¹,  k ≥ 0

```text
k    : 0   1   2    3    4
x(k) : 2  10  35  103  275
```

(Checked by long division: X(z) = 2 + 10z⁻¹ + 35z⁻² + 103z⁻³ + …)

## Check yourself

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

1. Why divide X(z) by z?
   A. To remove poles
   B. So the partial fractions give terms z/(z-a) with known inverses
   C. To make it proper
   D. To find x(0)
   Answer: B. Multiplying back by z. (Key point: Divide by z)

2. What is the coefficient of 1/(z-1)?
   A. 9
   B. -1
   C. 3
   D. 2
   Answer: C. (2z²+1)/(z-2)² at 1 = 3. (Key point: Coefficient)

3. What is Z⁻¹[z/(z-2)²]?
   A. 2^k
   B. k 2^k
   C. (k-1)2^k
   D. k 2^{k-1}
   Answer: D. Standard pair. (Key point: Inverse of double pole)

4. What is x(k)?
   A. 3 − 2^k + 9k 2^{k−1}
   B. 3 + 2^k + 9k 2^k
   C. 3 − 2^k + 9k 2^k
   D. 2^k + 9
   Answer: A. x(0)=2 matches. (Key point: Result)

## Close variants set in other exams

- [Inverse z of (z-4)/((z-1)(z-2)²)](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-4-over-z-1-z-2-sq) — Applied Mathematics, chapter 2, asked 5×
- [Inverse z of (2z²-5z)/((z-2)(z-3))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-2z2-5z) — Applied Mathematics, chapter 2, asked 5×

## Related in Applied Mathematics

- [Inverse z of (3z³+2z)/((z-3)²(z-2)), inversion integral](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-3z3-2z-inversion) — Applied Mathematics, chapter 2, asked 3×
- [Inverse z of z(3z²-6z+4)/((z-1)²(z-2))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-3z2-6z-4) — Applied Mathematics, chapter 2
- [Inverse z of (z+2)/((z-2)(z-3))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-plus-2-over-z-2-z-3) — Applied Mathematics, chapter 2
- [Inverse z of z/((z-2)(z-1))](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-over-z-2-z-1) — Applied Mathematics, chapter 2
- [Inverse z of (z-4)/((z-1)(z-2)²), inversion integral](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-4-inversion-integral) — Applied Mathematics, chapter 2, asked 3×
- [Inverse z of z/(z²-3z+2)](https://www.bctengineer.com/question.md?key=semester_4%2Fapplied_mathematics%3Ainverse-z-z-over-z2-3z-2) — Applied Mathematics, chapter 2, asked 2×

## Related in other subjects

- [Inverse Z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-pfe-2z3-z) — Digital Control System, chapter 2, asked 5×
- [Inverse z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Ainverse-z-partial-fraction) — Digital Signal Analysis and Processing, chapter 2
- [Inverse Z-transform by partial fractions](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-pfe-exponential-as-printed) — Digital Control System, chapter 2, asked 5×
- [Inverse Z-transform of z⁻³/((1−z⁻¹)(1−0.2z⁻¹))](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-z3-over-product) — Digital Control System, chapter 2