---
title: "DT unit impulse, step and ramp relationship"
subject: "Signals and Systems"
chapter: 1
key: "semester_4/signals_and_systems:dt-impulse-step-ramp"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-impulse-step-ramp
site: BCT Engineering
---

# DT unit impulse, step and ramp relationship

Signals and Systems · Chapter 1: Signals

- Asked in: 2082 Bhadra (new course), 3+2 marks
- Video: [Discrete Time Unit Impulse and Step Signal](https://www.youtube.com/watch?v=DQAeINlf_i8) — PMRF Teachings
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-impulse-step-ramp
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-1

## Question

Briefly explain discrete time unit impulse, unit step and unit ramp signal. What is the relationship between unit impulse, unit step and unit ramp signal?

## Short answer

Unit impulse δ[n] equals 1 at n=0 else 0. Unit step u[n] is 1 for n≥0. Unit ramp r[n]=n u[n]. Summation links: u[n]=∑δ[k], r[n]=∑u[k-1]; differencing: δ[n]=u[n]-u[n-1], u[n]=r[n+1]-r[n].

## Answer

#### Discrete-time unit impulse
$$
\delta[n] = \begin{cases} 1, & n = 0 \\ 0, & n \ne 0 \end{cases}
$$

A single sample of height 1 at $n = 0$. It has the sampling property $x[n]\delta[n-k] = x[k]\delta[n-k]$, and any sequence can be written as $x[n] = \sum_k x[k]\delta[n-k]$. The response of a system to $\delta[n]$ is its impulse response $h[n]$.

#### Discrete-time unit step
$$
u[n] = \begin{cases} 1, & n \ge 0 \\ 0, & n < 0 \end{cases}
$$

Samples of height 1 from $n = 0$ onwards. Used to switch signals on at $n = 0$ (make them causal).

#### Discrete-time unit ramp
$$
r[n] = n\,u[n] = \begin{cases} n, & n \ge 0 \\ 0, & n < 0 \end{cases}
$$

Sample values $0, 1, 2, 3, \dots$ growing linearly for $n \ge 0$.

```text
 delta[n]         u[n]              r[n]
                                         o
   o          o  o  o  o ..           o  |
   |          |  |  |  |           o  |  |
 --o--> n   --o--o--o--o--> n   o--o--o--o--> n
   0          0  1  2  3        0  1  2  3
```

#### Relationship
- Step from impulse (running sum) and impulse from step (first difference):

$$
u[n] = \sum_{k=-\infty}^{n}\delta[k], \qquad \delta[n] = u[n] - u[n-1]
$$

- Ramp from step and step from ramp:

$$
r[n] = \sum_{k=-\infty}^{n-1}u[k] = \sum_{k=-\infty}^{n} u[k-1], \qquad u[n-1] = r[n] - r[n-1]
$$

  (equivalently $u[n] = r[n+1] - r[n]$).

So summation moves impulse to step to ramp, and the first difference moves back the other way, just as integration and differentiation link $\delta(t)$, $u(t)$ and $r(t)$ in continuous time.

## Check yourself

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

1. Which describes the discrete-time unit impulse delta[n]?
   A. 1 for all n >= 0
   B. 1 only at n = 0 and 0 elsewhere
   C. n for n >= 0
   D. 1 at n = 1 and 0 elsewhere
   Answer: B. delta[n] is a single sample of height 1 at n = 0 and zero for every other n. (Key point: Unit impulse definition)

2. How is the unit step obtained from the unit impulse?
   A. u[n] = delta[n] - delta[n-1]
   B. u[n] = n delta[n]
   C. u[n] = running sum of delta[k] from k = -infinity to n
   D. u[n] = delta[n+1] - delta[n]
   Answer: C. Accumulating the impulse gives 1 for all n >= 0, which is the step. (Key point: Impulse to step)

3. Which expression recovers delta[n] from u[n]?
   A. delta[n] = u[n] + u[n-1]
   B. delta[n] = u[n] - u[n-1]
   C. delta[n] = n u[n]
   D. delta[n] = u[n] u[n-1]
   Answer: B. The first difference u[n] - u[n-1] is 1 only at n = 0. (Key point: Step to impulse)

4. What is the discrete unit ramp r[n]?
   A. u[n] - u[n-1]
   B. n for all n
   C. n u[n], i.e. 0, 1, 2, 3, ... for n >= 0
   D. u[n] u[n]
   Answer: C. The ramp is the step multiplied by n, giving values growing linearly from n = 0. (Key point: Unit ramp)

## Close variants set in other exams

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