---
title: "Inverse z of z(3z²-6z+4)/((z-1)²(z-2))"
subject: "Applied Mathematics"
chapter: 2
key: "semester_4/applied_mathematics:inverse-z-3z2-6z-4"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/inverse-z-3z2-6z-4
site: BCT Engineering
---

# Inverse z of z(3z²-6z+4)/((z-1)²(z-2))

Applied Mathematics · Chapter 2: The Z-Transform

- Asked in: 2074 Bhadra, 5 marks
- 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-3z2-6z-4
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-2

## Question

Obtain inverse Z-transform of z(3z² - 6z + 4)/((z - 1)²(z - 2)).

## Short answer

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

## Answer

**Method: partial fractions of X(z)/z.**

X(z)/z = (3z² − 6z + 4) / [(z − 1)²(z − 2)] = A/(z − 1) + B/(z − 1)² + C/(z − 2)

**Constants**

```text
C = (3·4 − 12 + 4)/(2 − 1)² = 4
B = (3 − 6 + 4)/(1 − 2)     = −1
Compare z² coefficients: A + C = 3  →  A = −1
```

So

X(z) = −z/(z − 1) − z/(z − 1)² + 4z/(z − 2)

**Inverse using standard pairs**

- Z⁻¹[z/(z − 1)] = 1
- Z⁻¹[z/(z − 1)²] = k
- Z⁻¹[z/(z − 2)] = 2ᵏ

**Answer:** x(k) = 4·2ᵏ − k − 1, k ≥ 0.

Check: x(0) = 3 (equals lim X(z) = 3), x(1) = 6, x(2) = 13, x(3) = 28 — the series expansion of X(z) gives the same values.

## Check yourself

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

1. Why divide X(z) by z first?
   A. Standard pairs have a factor z in the numerator
   B. To remove poles
   C. To get a polynomial
   D. To apply initial value theorem
   Answer: A. z/(z-a) form. (Key point: Method)

2. C (for 1/(z-2)) is
   A. -1
   B. 4
   C. 3
   D. 2
   Answer: B. (3*4 - 12 + 4)/(1)^2 = 4. (Key point: Constant C)

3. The (z-1)^2 factor gives
   A. only A/(z-1)^2
   B. only one term
   C. terms A/(z-1) and B/(z-1)^2
   D. terms in z^2
   Answer: C. Repeated roots need both powers. (Key point: Repeated pole)

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

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