---
title: "Inverse z of z/((z+1)²(z-1))"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:inverse-z-z-over-z1sq-z-1"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-z-over-z1sq-z-1
site: BCT Engineering
---

# Inverse z of z/((z+1)²(z-1))

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2069 Bhadra
- Video: [Find the inverse Z-transform of  z(z+1) by (z-1)^3](https://www.youtube.com/watch?v=1tYCaIvgxII) — Simha's Classes
- 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-z-over-z1sq-z-1
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Find the inverse z-transform of z/((z + 1)²(z - 1)).

## Short answer

Use partial fractions of X(z)/z: X(z)=1/4 z/(z-1)-1/4 z/(z+1)-1/2 z/(z+1)^2. Inverse gives x(k)=1/4 -1/4(-1)^k +1/2 k(-1)^k for k≥0.

## Answer

**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)

```text
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.

## Check yourself

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

1. Partial fractions of X(z)/z use
   A. A/(z-1) + B/(z+1)^2 + C/(z+1)
   B. A/(z-1) + B/(z+1)^2 only
   C. A/(z-1) + C/(z+1)
   D. A/(z-1)^3
   Answer: A. Repeated factor needs both. (Key point: Form)

2. A and B are
   A. A = 1/2, B = 1/4
   B. A = 1/4, B = -1/2
   C. A = -1/4, B = 1/2
   D. A = 1/4, B = 1/2
   Answer: B. A = 1/(2)^2; B = 1/(-2). (Key point: A and B)

3. C comes from
   A. B + C = 0
   B. C = A
   C. A + C = 0 (no z^2 term in numerator)
   D. C = 1
   Answer: C. Compare z^2 coefficients. (Key point: C)

4. Z inverse of z/(z+1)^2 is
   A. k(-1)^k
   B. (-1)^k
   C. k
   D. -k(-1)^k
   Answer: D. Z[k a^k] = az/(z-a)^2 with a = -1 gives -z/(z+1)^2. (Key point: Repeated pole pair)

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