---
title: "Z-transform of ramp-and-hold signal (T = 1 s)"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:waveform-ramp-to-1-at-4"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/waveform-ramp-to-1-at-4
site: BCT Engineering
---

# Z-transform of ramp-and-hold signal (T = 1 s)

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2078 Chaitra, 4 marks
- Video: [Z Transform of common sequences ( delta, unit ramp, unit step, exponential, sinusoidal sequence)](https://www.youtube.com/watch?v=LDqAPyCjph4) — EC Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/digital_control_system/waveform-ramp-to-1-at-4
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Obtain the z-transform of x(t) for which time response is given by (assume sampling period T = 1 s). [Figure: continuous time signal x(t) rising linearly from 0 at t = 0 to 1 at t = 4 (0.5 marked on the axis), then constant at 1; time axis marked 1 to 13]

## Short answer

x(t) = (1/4)t – (1/4)(t–4)1(t–4). Sampling at T=1 s yields X(z) = (1/4)·z^{-1}(1–z^{-4})/(1–z^{-1})^2 = (z^3+z^2+z+1)/(4z^3(z–1)).

## Answer

#### Step 1: Express $x(t)$ as ramp minus delayed ramp

The ramp has slope $\dfrac{1}{4}$ and reaches 1 at $t = 4$ s, after which it holds at 1:

$$
x(t) = \frac{1}{4}t\cdot1(t) - \frac{1}{4}(t-4)\cdot1(t-4)
$$

#### Step 2: Sampled values ($T = 1$ s)

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | ... |
|---|---|---|---|---|---|---|---|---|
| $x(k)$ | 0 | 0.25 | 0.5 | 0.75 | 1 | 1 | 1 | ... |

#### Step 3: Z-transform

With $\mathcal{Z}[t] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$ ($T = 1$) and delay of 4 samples $\to z^{-4}$:

$$
\begin{aligned}
X(z) &= \frac{1}{4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} - \frac{1}{4}z^{-4}\cdot\frac{z^{-1}}{(1-z^{-1})^2} \\
&= \frac{1}{4}\cdot\frac{z^{-1}(1 - z^{-4})}{(1-z^{-1})^2}
\end{aligned}
$$

Using $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$:

$$
X(z) = \frac{z^{-1}(1+z^{-1}+z^{-2}+z^{-3})}{4(1-z^{-1})} = \frac{z^3+z^2+z+1}{4z^3(z-1)}
$$

#### Step 4: Check

Series: $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$, which matches the table.

Final value: $\lim_{z\to1}(1-z^{-1})X(z) = \dfrac{1\cdot 4}{4} = 1$, which is correct.

**Answer:** $X(z) = \dfrac{z^{-1}(1-z^{-4})}{4(1-z^{-1})^2} = \dfrac{z^3+z^2+z+1}{4z^3(z-1)}$.

## Check yourself

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

1. What is the slope of the ramp reaching 1 at t = 4?
   A. 4
   B. 1/2
   C. 1/4
   D. 1
   Answer: C. Rise 1 over 4 s. (Key point: Slope)

2. How is x(t) written?
   A. (1/4)t 1(t) + (1/4)(t-4)1(t-4)
   B. t 1(t) - (t-4)1(t-4)
   C. (1/4)(t-4)1(t-4)
   D. (1/4)t 1(t) - (1/4)(t-4)1(t-4)
   Answer: D. A delayed ramp of equal slope cancels the rise. (Key point: Decomposition)

3. X(z) for T = 1?
   A. (z^3+z^2+z+1)/(4z^3(z-1))
   B. (z^3+z^2+z+1)/(z^3(z-1))
   C. z/(4(z-1)^2)
   D. (z^2+z+1)/(4z^2(z-1))
   Answer: A. z^-1(1+z^-1+z^-2+z^-3)/(4(1-z^-1)). (Key point: Result)

4. Final value of x(k)?
   A. 0
   B. 1
   C. 4
   D. 0.25
   Answer: B. 4 terms times 1/4 = 1. (Key point: Final value)

## Related in Digital Control System

- [Z-transform of ramp-and-hold continuous signal](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Awaveform-ramp-to-1-at-6) — Digital Control System, chapter 2
- [Z-transform of ramp-and-hold curve (to 1 at t=2)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Acurve-ramp-to-1-at-2) — Digital Control System, chapter 2
- [Z-transform of a ramp-and-hold input (to 3)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Awaveform-ramp-to-3) — Digital Control System, chapter 2