---
title: "Inverse Z-transform of z(z+2)/(z−1)²"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:inverse-integral-z-z-plus-2"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-integral-z-z-plus-2
site: BCT Engineering
---

# Inverse Z-transform of z(z+2)/(z−1)²

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2080 Chaitra, 4 marks; 2072 Magh, 4 marks; 2070 Magh, 4 marks (asked 3 times)
- Video: [Inverse Z- Transform/ using Residue Method/Problem#1](https://www.youtube.com/watch?v=MR2Ma25QB74) — Dr.Jayaudhaya ,Simple and Easy Way
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/inverse-integral-z-z-plus-2
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the inverse Z-transform of X(z)=z(z+2)/(z-1)² by inversion integral method.

## Short answer

X(z)z^(k-1) = z^k(z+2)/(z-1)² has a double pole at z=1. The residue is lim_{z→1} d/dz[z^(k+1)+2z^k] = (k+1)+2k = 3k+1. Hence x(k) = 3k + 1.

## Answer

By the inversion integral method,

$$
x(k) = \sum \text{residues of } X(z)z^{k-1} \text{ at the poles of } X(z)z^{k-1}
$$

$$
X(z)z^{k-1} = \frac{z(z+2)z^{k-1}}{(z-1)^2} = \frac{z^k(z+2)}{(z-1)^2}
$$

For $k \ge 0$ the only pole is a **double pole at $z = 1$** (no pole at the origin).

#### Residue at a double pole

$$
\begin{aligned}
x(k) &= \frac{1}{(2-1)!}\lim_{z\to1}\frac{d}{dz}\left[(z-1)^2\frac{z^k(z+2)}{(z-1)^2}\right] \\
&= \lim_{z\to1}\frac{d}{dz}\left[z^{k+1} + 2z^k\right] \\
&= \lim_{z\to1}\left[(k+1)z^k + 2kz^{k-1}\right] \\
&= (k+1) + 2k = 3k + 1
\end{aligned}
$$

#### Check

| k | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $3k+1$ | 1 | 4 | 7 | 10 |

Long division: $X(z) = \frac{z^2+2z}{z^2-2z+1} = 1 + 4z^{-1} + 7z^{-2} + 10z^{-3} + \cdots$, which agrees.

**Answer:** $x(k) = 3k + 1$, $k = 0, 1, 2, \dots$

## Check yourself

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

1. X(z)z^(k-1) has:
   A. A double pole at z = 1
   B. Two simple poles
   C. A pole at the origin
   D. No poles
   Answer: A. (z-1)^2 in denominator. (Key point: Pole type)

2. Double-pole residue uses:
   A. The function value at z=1
   B. d/dz of (z-1)^2 X(z) z^(k-1) at z=1
   C. Second derivative at z=0
   D. Product of poles
   Answer: B. 1/(2-1)! times the first derivative. (Key point: Residue formula)

3. x(k) =
   A. k + 3
   B. 2k + 1
   C. 3k + 1
   D. 3k
   Answer: C. (k+1) + 2k. (Key point: Result)

4. Long division gives the first values:
   A. 1, 3, 5, 7
   B. 0, 3, 6, 9
   C. 1, 2, 4, 8
   D. 1, 4, 7, 10
   Answer: D. 3k+1 at k=0..3. (Key point: Check)

## Close variants set in other exams

- [Inverse Z-transform by inversion integral](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-step-exponential) — Digital Control System, chapter 2, asked 10×
- [Inverse Z-transform of z⁻¹(1−z⁻²)/(1+z⁻¹)²](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-minus-cube) — Digital Control System, chapter 2, asked 10×
- [Inverse Z-transform: double pole at z=1](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z2-double-pole) — Digital Control System, chapter 2, asked 10×
- [Inverse Z-transform by inversion integral (cubic)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Ainverse-integral-z-1-cubic) — Digital Control System, chapter 2, asked 10×

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