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

# Inverse z of z/((z-2)(z-1))

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2070 Magh, 5 marks
- Video: [Inverse Z Transform - Part 1 | Most Important | Engineering Mathematics](https://www.youtube.com/watch?v=yEAE9O_aGD8) — Mathematics Tutor
- 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-z-2-z-1
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Obtain inverse Z-transform of z/((z - 2)(z - 1)).

## Short answer

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

## Answer

**Partial fractions of X(z)/z.**

X(z)/z = 1/[(z − 2)(z − 1)] = A/(z − 2) + B/(z − 1)

```text
A = 1/(2 − 1) = 1
B = 1/(1 − 2) = −1
```

X(z) = z/(z − 2) − z/(z − 1)

Using Z⁻¹[z/(z − a)] = aᵏ:

**Answer:** x(k) = 2ᵏ − 1, k ≥ 0.

Check: x(0) = 0, x(1) = 1, x(2) = 3, x(3) = 7, x(4) = 15. The series expansion of X(z) = z⁻¹/[(1 − 2z⁻¹)(1 − z⁻¹)] gives the same values.

## Check yourself

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

1. X(z)/z =
   A. 1/[(z-2)(z-1)]
   B. z/[(z-2)(z-1)]
   C. 1/z
   D. (z-2)/(z-1)
   Answer: A. Cancel z. (Key point: Divide by z)

2. A (for z-2) and B (for z-1) are
   A. 1 and 1
   B. 1 and -1
   C. -1 and 1
   D. 2 and 1
   Answer: B. A = 1/(2-1), B = 1/(1-2). (Key point: Coefficients)

3. X(z) =
   A. z/(z-2) + z/(z-1)
   B. 1/(z-2) - 1/(z-1)
   C. z/(z-2) - z/(z-1)
   D. z/(z-1) - z/(z-2)
   Answer: C. Multiply back by z. (Key point: Transform)

4. x(k) =
   A. 2^k + 1
   B. 1 - 2^k
   C. 2^(k-1)
   D. 2^k - 1
   Answer: D. x(0)=0, x(1)=1, x(2)=3. (Key point: Result)

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