---
title: "Z-transform of ramp minus delayed ramp"
subject: "Digital Control System"
chapter: 2
key: "semester_6/digital_control_system:z-transform-ramp-minus-delayed-ramp"
source: https://www.bctengineer.com/q/semester_6/digital_control_system/z-transform-ramp-minus-delayed-ramp
site: BCT Engineering
---

# Z-transform of ramp minus delayed ramp

Digital Control System · Chapter 2: The Z-Transform

- Asked in: 2075 Baisakh, 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-ramp-minus-delayed-ramp
- Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-2

## Question

Find the z-transform of x(t)=1/4t-1/4(t-4)1(t-4).

## Short answer

x(t)=(1/4)t–(1/4)(t–4)1(t–4). Z-transform with T=1 s: X(z)=(1/4)·z^{-1}(1–z^{-4})/(1–z^{-1})^2=(z^3+z^2+z+1)/[4z^3(z–1)].

## Answer

$x(t)$ is a ramp of slope $\tfrac14$ minus the same ramp delayed by 4 s. It rises from 0 to 1 over 4 s and then stays at 1.

#### Step 1: Standard pairs

- $\mathcal{Z}[t\cdot1(t)] = \dfrac{Tz^{-1}}{(1-z^{-1})^2}$
- Real translation (shift) theorem: if $4 = nT$, then $\mathcal{Z}[x(t - nT)1(t-nT)] = z^{-n}X(z)$

#### Step 2: General sampling period $T$ (4 an integer multiple of $T$)

$$
X(z) = \frac{1}{4}\cdot\frac{Tz^{-1}}{(1-z^{-1})^2}\left(1 - z^{-4/T}\right)
$$

#### Step 3: With $T = 1$ s (usual case, $n = 4$)

$$
\begin{aligned}
X(z) &= \frac{1}{4}\cdot\frac{z^{-1}(1-z^{-4})}{(1-z^{-1})^2} \\
&= \frac{z^{-1}(1+z^{-1}+z^{-2}+z^{-3})}{4(1-z^{-1})} \\
&= \frac{z^3+z^2+z+1}{4z^3(z-1)}
\end{aligned}
$$

Here $1 - z^{-4} = (1-z^{-1})(1+z^{-1}+z^{-2}+z^{-3})$ was used.

#### Step 4: Check with the samples

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

Long division gives $X(z) = 0.25z^{-1} + 0.5z^{-2} + 0.75z^{-3} + z^{-4} + z^{-5} + \dots$, which matches. The final value from the theorem, $\lim_{z\to1}(1-z^{-1})X(z) = 1$, is also correct.

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

## Check yourself

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

1. Z[t 1(t)] is?
   A. T/(1-z^-1)
   B. Tz/(z-1)
   C. Tz^-1/(1-z^-1)^2
   D. z^-1/(1-z^-1)
   Answer: C. Standard ramp pair. (Key point: Standard pair)

2. Ramp delayed by 4 s with T = 1 multiplies by?
   A. z^4
   B. z^-1
   C. 4z^-1
   D. z^-4
   Answer: D. n = 4 samples. (Key point: Delay)

3. 1 - z^-4 factors as?
   A. (1-z^-1)(1+z^-1+z^-2+z^-3)
   B. (1-z^-1)^4
   C. (1+z^-1)(1-z^-3)
   D. (1-z^-2)^2
   Answer: A. Cancels one factor of (1-z^-1). (Key point: Factoring)

4. X(z) for T = 1?
   A. (z-1)/(4z^3)
   B. (z^3+z^2+z+1)/(4z^3(z-1))
   C. (z^3+z^2+z+1)/(z^3(z-1))
   D. 1/(4z^4(z-1))
   Answer: B. Check: samples 0, .25, .5, .75, 1, 1, ... (Key point: Result)

## Related in Digital Control System

- [Z-transform of a step delayed by 4](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Az-transform-delayed-step) — 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
- [Z-transform of ramp-and-hold signal (T = 1 s)](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Awaveform-ramp-to-1-at-4) — Digital Control System, chapter 2
- [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

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