---
title: "Z-transforms of t²e^(−at) and t·x(t−nT)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-t2eat-shifted"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-t2eat-shifted
site: BCT Engineering
---

# Z-transforms of t²e^(−at) and t·x(t−nT)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2082 Chaitra, 5 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-t2eat-shifted
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Determine the Z-transform for (i) x(t)=t²e⁻ᵃᵗ (ii) x(t)=t· x(t-nT).

## Short answer

(i) For x(t)=t²e^(-at), X(z)=T²e^(-aT)z(z+e^(-aT))/(z-e^(-aT))³. (ii) For y(t)=t·x(t-nT), Y(z)=z^(-n)[nTX(z)-Tz dX/dz] using translation and multiplication by t properties.

## Answer

#### (i) $x(t) = t^2e^{-at}$

Sampled: $x(kT) = T^2k^2(e^{-aT})^k$. Let $r = e^{-aT}$.

From $\mathcal{Z}[1] = \frac{z}{z-1}$ and $\mathcal{Z}[k\,x(k)] = -z\frac{dX}{dz}$:

$$
\mathcal{Z}[k] = \frac{z}{(z-1)^2}, \qquad \mathcal{Z}[k^2] = \frac{z(z+1)}{(z-1)^3}
$$

Multiplying by $r^k$ replaces $z$ by $z/r$:

$$
\mathcal{Z}[k^2r^k] = \frac{(z/r)(z/r+1)}{(z/r-1)^3} = \frac{rz(z+r)}{(z-r)^3}
$$

$$
X(z) = \frac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}
$$

#### (ii) $y(t) = t\cdot x(t-nT)$

Assume $x(t) = 0$ for $t < 0$ and $\mathcal{Z}[x(t)] = X(z)$.

**Step 1 (real translation):** $\mathcal{Z}[x(t-nT)] = z^{-n}X(z)$.

**Step 2 (multiplication by $t$):** for sampled signals $t = kT$, and $\mathcal{Z}[kT\,f(kT)] = -Tz\frac{d}{dz}F(z)$.

$$
\begin{aligned}
Y(z) &= -Tz\frac{d}{dz}\left[z^{-n}X(z)\right] \\
&= -Tz\left[-nz^{-n-1}X(z) + z^{-n}\frac{dX(z)}{dz}\right] \\
&= z^{-n}\left[nT\,X(z) - Tz\frac{dX(z)}{dz}\right]
\end{aligned}
$$

**Example:** if $x(t) = 1(t)$, $X(z) = \frac{z}{z-1}$ and $\frac{dX}{dz} = \frac{-1}{(z-1)^2}$, so $Y(z) = z^{-n}\left[\frac{nTz}{z-1} + \frac{Tz}{(z-1)^2}\right]$, the transform of $kT$ for $k \ge n$.

**Answers:** (i) $\dfrac{T^2e^{-aT}z(z+e^{-aT})}{(z-e^{-aT})^3}$; (ii) $z^{-n}\left[nTX(z) - Tz\dfrac{dX(z)}{dz}\right]$.

## Check yourself

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

1. After sampling, x(kT) for t^2 e^(-at) equals?
   A. T k^2 e^(-aT k)
   B. T^2 k^2 (e^-aT)^k
   C. k^2 e^(-aT)
   D. T^2 k (e^-aT)^k
   Answer: B. Substituting t = kT gives T^2 k^2 r^k with r = e^(-aT). (Key point: Sampling t^2 e^-at)

2. Multiplying a sequence by r^k has what effect on its z-transform?
   A. Shifts X(z) by r samples
   B. Multiplies X(z) by r
   C. Replaces z by z/r
   D. Replaces z by rz
   Answer: C. Z[r^k x(k)] = X(z/r). (Key point: Exponential multiplication)

3. Which is the Z-transform of t^2 e^(-at)?
   A. T e^(-aT) z/(z-e^(-aT))^2
   B. T^2 e^(-aT) z(z-e^(-aT))/(z+e^(-aT))^3
   C. T^2 e^(-aT) z(z+e^(-aT))/(z-e^(-aT))^3
   D. T^2 z(z+1)/(z-1)^3
   Answer: C. Replace z by z/r in z(z+1)/(z-1)^3 and multiply by T^2. (Key point: Result of (i))

4. For y(t) = t x(t-nT), what are the two steps?
   A. Apply delay z^-n X(z), then use -Tz d/dz on the result
   B. Differentiate X(z), then multiply by z^n
   C. Use initial value theorem then shift
   D. Multiply by t using -Tz d/dz, then apply the delay z^-n only to the derivative
   Answer: A. First Z[x(t-nT)] = z^-n X(z), then apply the multiplication-by-t rule to it. (Key point: Part (ii) method)

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