---
title: "Z-transform of 9k·2^(k−1) − 2^k + 3"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-9k-2k"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-9k-2k
site: BCT Engineering
---

# Z-transform of 9k·2^(k−1) − 2^k + 3

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2079 Chaitra, 4 marks
- Video: [Calculating Z transform of given discrete signals.](https://www.youtube.com/watch?v=3rHRevJhvIc) — Engg-Course-Made-Easy
- 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-9k-2k
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the Z-transform of {x(k)=9k(2ᵏ⁻¹)-2ᵏ+3}, for k≥ 0.

## Short answer

By linearity and pairs for a^k, k a^{k-1}, and constant: X(z)=9z/(z-2)^2 - z/(z-2) + 3z/(z-1) = z(2z^2+1)/[(z-2)^2(z-1)], ROC |z|>2.

## Answer

Use linearity and three standard pairs.

#### Standard pairs used

| $x(k)$ | $X(z)$ |
|---|---|
| $1$ | $\dfrac{z}{z-1}$ |
| $a^k$ | $\dfrac{z}{z-a}$ |
| $k\,a^{k-1}$ | $\dfrac{z}{(z-a)^2}$ |

The last pair comes from $\mathcal{Z}[k\,x(k)] = -z\dfrac{d}{dz}X(z)$:

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

Dividing by $a$ gives $\mathcal{Z}[k a^{k-1}] = \dfrac{z}{(z-a)^2}$.

#### Transform each term ($a = 2$)

$$
\begin{aligned}
\mathcal{Z}[9k\,2^{k-1}] &= \frac{9z}{(z-2)^2} \\
\mathcal{Z}[-2^k] &= -\frac{z}{z-2} \\
\mathcal{Z}[3] &= \frac{3z}{z-1}
\end{aligned}
$$

$$
X(z) = \frac{9z}{(z-2)^2} - \frac{z}{z-2} + \frac{3z}{z-1}
$$

#### Combine over a common denominator $(z-2)^2(z-1)$

$$
\begin{aligned}
\text{Numerator} &= 9z(z-1) - z(z-2)(z-1) + 3z(z-2)^2 \\
&= z\left[9z - 9 - (z^2 - 3z + 2) + 3(z^2 - 4z + 4)\right] \\
&= z\left[2z^2 + 0\cdot z + 1\right]
\end{aligned}
$$

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

#### Check

From the formula: $x(0) = 0 - 1 + 3 = 2$, $x(1) = 9 - 2 + 3 = 10$, $x(2) = 36 - 4 + 3 = 35$, $x(3) = 108 - 8 + 3 = 103$.

Long division of $X(z)$ gives $2 + 10z^{-1} + 35z^{-2} + 103z^{-3} + \dots$, which matches.

**Answer:** $X(z) = \dfrac{9z}{(z-2)^2} - \dfrac{z}{z-2} + \dfrac{3z}{z-1} = \dfrac{z(2z^2+1)}{(z-2)^2(z-1)}$, ROC $|z| > 2$.

## Check yourself

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

1. How is Z of a sum of terms found?
   A. Only the largest term counts
   B. Differentiate
   C. Transform each term and add (linearity)
   D. Multiply the transforms
   Answer: C. The z-transform is linear. (Key point: Linearity)

2. Z[k a^(k-1)] with a = 2 is?
   A. 2z/(z-2)^2
   B. z/(z-2)
   C. z/(2z-1)^2
   D. z/(z-2)^2
   Answer: D. Z[k a^(k-1)] = z/(z-a)^2. (Key point: Pair)

3. What is Z[9k 2^(k-1)]?
   A. 9z/(z-2)^2
   B. 9z/(z-2)
   C. 18z/(z-2)^2
   D. 9z/(z-1)^2
   Answer: A. Nine times z/(z-2)^2. (Key point: Term 9k 2^(k-1))

4. What is X(z)?
   A. 9z/(z-2)^2
   B. z(2z^2+1)/((z-2)^2(z-1))
   C. z(2z^2-1)/((z-2)^2(z-1))
   D. z/((z-2)^2(z-1))
   Answer: B. Numerator combines to z(2z^2 + 1). (Key point: Result)

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