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

# Z-transform of k² e^(−ak) from unit step

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2067 Mangsir, 8 marks
- Video: [Z transform of Step , Ramp , Exponential, cosine , sine , hyperbolic cosine and sine functions.](https://www.youtube.com/watch?v=01-C5Oqbyb4) — Seema Prasad
- 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-e-ak
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Starting from the z transform of unit step function, determine the z transform of k²e⁻ᵃᵏ.

## Short answer

Multiply unit step by e^{-ak} using scaling Z[b^k x(k)]=X(z/b), then apply Z[k x(k)]=-z dX/dz twice. Result: Z[k^2 e^{-ak}] = e^{-a} z (z+e^{-a}) / (z-e^{-a})^3.

## Answer

Start from the unit step and build the result in three steps: **multiply by $e^{-ak}$** (scaling in the z-domain), then **multiply by $k$ twice** (differentiation in the z-domain).

#### Properties used

1. Unit step: $\mathcal{Z}[1(k)] = \displaystyle\sum_{k=0}^{\infty}z^{-k} = \frac{z}{z-1}$, $|z| > 1$
2. Scaling (multiplication by $b^k$): $\mathcal{Z}[b^k x(k)] = X\left(\dfrac{z}{b}\right)$
3. Multiplication by $k$: $\mathcal{Z}[k\,x(k)] = -z\dfrac{d}{dz}X(z)$

Proof of property 3: $-z\dfrac{d}{dz}\sum x(k)z^{-k} = -z\sum(-k)x(k)z^{-k-1} = \sum k\,x(k)z^{-k}$.

#### Step 1: $e^{-ak}$

Use property 2 with $b = e^{-a}$, writing $E = e^{-a}$ for short:

$$
\mathcal{Z}[e^{-ak}] = \frac{z/E}{z/E - 1} = \frac{z}{z - E} = \frac{z}{z - e^{-a}}
$$

#### Step 2: $k\,e^{-ak}$

$$
\mathcal{Z}[k\,e^{-ak}] = -z\frac{d}{dz}\left(\frac{z}{z-E}\right) = -z\cdot\frac{(z-E) - z}{(z-E)^2} = \frac{Ez}{(z-E)^2}
$$

#### Step 3: $k^2e^{-ak}$

$$
\begin{aligned}
\mathcal{Z}[k^2e^{-ak}] &= -z\frac{d}{dz}\left[\frac{Ez}{(z-E)^2}\right] \\
&= -z\cdot\frac{E(z-E)^2 - Ez\cdot2(z-E)}{(z-E)^4} \\
&= -z\cdot\frac{E(z-E) - 2Ez}{(z-E)^3} = -z\cdot\frac{-E(z+E)}{(z-E)^3} \\
&= \frac{Ez(z+E)}{(z-E)^3}
\end{aligned}
$$

#### Result

$$
\mathcal{Z}[k^2e^{-ak}] = \frac{e^{-a}z\,(z + e^{-a})}{(z - e^{-a})^3} = \frac{e^{-a}z^{-1}(1 + e^{-a}z^{-1})}{(1 - e^{-a}z^{-1})^3}, \quad |z| > e^{-a}
$$

#### Check

Take $e^{-a} = 0.5$. Then $X(z) = \dfrac{0.5z(z+0.5)}{(z-0.5)^3}$, and its series is $0.5z^{-1} + z^{-2} + 1.125z^{-3} + z^{-4} + \dots$. Direct values of $k^2(0.5)^k$ for $k = 0$ to $4$ are $0, 0.5, 1, 1.125, 1$, which match.

As a special case, $a = 0$ gives $\mathcal{Z}[k^2] = \dfrac{z(z+1)}{(z-1)^3}$, the known result.

**Answer:** $\mathcal{Z}[k^2e^{-ak}] = \dfrac{e^{-a}z(z + e^{-a})}{(z - e^{-a})^3}$.

## Check yourself

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

1. Z[k x(k)] equals:
   A. -z dX/dz
   B. z dX/dz
   C. X(z)/z
   D. -dX/dz
   Answer: A. Differentiating the sum of x(k)z^{-k} brings down -k z^{-k-1}, so multiplying by -z gives k x(k). (Key point: Multiplication by k)

2. Z[e^{-ak}] obtained from the unit step is:
   A. z/(z - e^{-a})
   B. z/(z - e^{a})
   C. z/(z - 1)
   D. e^{-a} z/(z-1)
   Answer: A. Replace z by z/b with b = e^{-a} in z/(z-1). (Key point: Scaling step)

3. Z[k e^{-ak}] is:
   A. E z/(z-E)^2
   B. z/(z-E)^2
   C. E/(z-E)^2
   D. -E z/(z-E)^2
   Answer: A. -z d/dz [z/(z-E)] = Ez/(z-E)^2, with E = e^{-a}. (Key point: Transform of k e^{-ak})

4. Z[k^2 e^{-ak}] is:
   A. E z (z+E)/(z-E)^3
   B. E z (z-E)/(z-E)^3
   C. E z/(z-E)^3
   D. E^2 z (z+E)/(z-E)^2
   Answer: A. Differentiating once more gives Ez(z+E)/(z-E)^3; with a = 0 this becomes z(z+1)/(z-1)^3. (Key point: Final result)

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