---
title: "Z-transform of k² a^(k−1)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-k2-a-k-1"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-k2-a-k-1
site: BCT Engineering
---

# Z-transform of k² a^(k−1)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2082 Chaitra (new course), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-k2-a-k-1
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the Z-transform of x(k)=k²aᵏ⁻¹.

## Short answer

Z-transform of k²a^(k-1) using Z[a^k]=z/(z-a) and multiplication by k property. Result: Z[k²a^(k-1)] = z(z+a)/(z-a)³ for |z|>|a|.

## Answer

Use the known pair $\mathcal{Z}[a^k] = \frac{z}{z-a}$ and the **multiplication by $k$** property $\mathcal{Z}[k\,x(k)] = -z\frac{d}{dz}X(z)$.

#### Step 1: $\mathcal{Z}[k a^{k-1}]$

$$
\mathcal{Z}[k a^k] = -z\frac{d}{dz}\left(\frac{z}{z-a}\right) = -z\cdot\frac{(z-a) - z}{(z-a)^2} = \frac{az}{(z-a)^2}
$$

so $\mathcal{Z}[k a^{k-1}] = \frac{1}{a}\cdot\frac{az}{(z-a)^2} = \frac{z}{(z-a)^2}$.

#### Step 2: $\mathcal{Z}[k^2 a^{k-1}]$

$$
\begin{aligned}
\mathcal{Z}[k\cdot k a^{k-1}] &= -z\frac{d}{dz}\left[\frac{z}{(z-a)^2}\right] \\
&= -z\cdot\frac{(z-a)^2 - 2z(z-a)}{(z-a)^4} \\
&= -z\cdot\frac{(z-a) - 2z}{(z-a)^3} = \frac{z(z+a)}{(z-a)^3}
\end{aligned}
$$

#### Check

Series of $\frac{z(z+a)}{(z-a)^3} = z^{-1} + 4a z^{-2} + 9a^2 z^{-3} + \cdots$, i.e. coefficients $k^2a^{k-1}$ ✔.

**Answer:** $\mathcal{Z}[k^2a^{k-1}] = \dfrac{z(z+a)}{(z-a)^3} = \dfrac{z^{-1}(1+az^{-1})}{(1-az^{-1})^3}$, $|z| > |a|$.

## Check yourself

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

1. Which property turns Z[x(k)] into Z[k x(k)]?
   A. Multiply X(z) by z
   B. -z d/dz of X(z)
   C. d/dz of X(z) divided by z
   D. Replace z by z/k
   Answer: B. Multiplying by k in time corresponds to -z dX/dz in the z-domain. (Key point: Multiplication by k property)

2. What is Z[k a^k] obtained from Z[a^k] = z/(z-a)?
   A. z(z+a)/(z-a)^3
   B. z/(z-a)^2
   C. az/(z-a)^2
   D. a/(z-a)^2
   Answer: C. -z d/dz [z/(z-a)] = -z(-a)/(z-a)^2 = az/(z-a)^2. (Key point: Z[k a^k])

3. How is Z[k a^(k-1)] obtained from Z[k a^k]?
   A. Replace z by az
   B. Differentiate again
   C. Multiply by a
   D. Divide by a, giving z/(z-a)^2
   Answer: D. a^(k-1) = a^k / a, so divide by a: z/(z-a)^2. (Key point: Getting k a^(k-1))

4. What is Z[k^2 a^(k-1)]?
   A. z(z-a)/(z+a)^3
   B. z(z+a)/(z-a)^3
   C. az(z+a)/(z-a)^3
   D. z/(z-a)^3
   Answer: B. Applying -z d/dz to z/(z-a)^2 gives z(z+a)/(z-a)^3. (Key point: Final result)

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