---
title: "Convolve δ-sum with u[n] − u[n−1]"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-delta-first-difference"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-delta-first-difference
site: BCT Engineering
---

# Convolve δ-sum with u[n] − u[n−1]

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2082 Bhadra (new course), 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-delta-first-difference
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Find the convolution between the signal x[n] = δ[n−2] − δ[n−1] + 2δ[n] + δ[n+1] + δ[n+2] and h[n] = u[n] − u[n−1]

## Short answer

h[n] = u[n] - u[n-1] simplifies to δ[n]. Convolution with δ[n] is the identity operation, so y[n] = x[n] = δ[n+2] + δ[n+1] + 2δ[n] - δ[n-1] + δ[n-2]. The sequence values are {1, 1, 2, -1, 1} for n = -2 to 2, zero elsewhere.

## Answer

First simplify $h[n]$:

$$
h[n]=u[n]-u[n-1]
$$

$u[n]-u[n-1]$ is 1 only at $n=0$ (both are 1 for $n\ge1$ and both are 0 for $n<0$), so

$$
h[n]=\delta[n]
$$

Convolution with the unit impulse leaves a signal unchanged ($x[n]*\delta[n]=x[n]$, the identity property). Hence

$$
\begin{aligned}
y[n]&=x[n]*h[n]=x[n]*\delta[n]=x[n]\\
&=\delta[n+2]+\delta[n+1]+2\delta[n]-\delta[n-1]+\delta[n-2]
\end{aligned}
$$

| $n$ | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| $y[n]$ | 1 | 1 | 2 | −1 | 1 |

**Answer: $y[n]=\{1,\ 1,\ 2,\ -1,\ 1\}$ for $n=-2,\dots,2$** (↑ at 2), and zero elsewhere.

## Check yourself

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

1. What does h[n]=u[n]-u[n-1] simplify to?
   A. u[n]
   B. delta[n-1]
   C. delta[n]
   D. 1 for all n
   Answer: C. It is 1 only at n=0. (Key point: Simplifying h)

2. Why is u[n]-u[n-1] zero for n>=1?
   A. u[n] is 0 there
   B. u[n-1] is infinite there
   C. Convolution forces it to 0
   D. Both are 1 there, so they cancel
   Answer: D. For n>=1 both steps equal 1, and for n<0 both equal 0. (Key point: Why)

3. What does x[n]*delta[n] equal?
   A. 0
   B. x[n]
   C. x[n-1]
   D. 2x[n]
   Answer: B. Convolution with the unit impulse leaves the signal unchanged. (Key point: Identity property)

4. Which is y[n] for n=-2..2?
   A. {1, -1, 2, 1, 1}
   B. {1, 1, 2, 1, -1}
   C. {1, 1, 2, -1, 1}
   D. {-1, 1, 2, 1, 1}
   Answer: C. y=x, and delta[n+2] has weight 1 while delta[n-1] has weight -1. (Key point: Result)

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