---
title: "Convolve {1, 3, 4, 3, 1} and {2, 2, −2}"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-13431"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-13431
site: BCT Engineering
---

# Convolve {1, 3, 4, 3, 1} and {2, 2, −2}

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2076 Baisakh, 6 marks
- Video: [Discrete Time Convolution (Tabular Method)](https://www.youtube.com/watch?v=C_kwjJLepLg) — Neso Academy
- 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-13431
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Find the convolution sum of x[n] = {1, 3, 4, 3, 1} and h[n] = {2, 2, −2}

## Short answer

The convolution sum gives y[n] = {2, 8, 12, 8, 0, -4, -2} for n = 0 to 6 (origin at first sample).

## Answer

Take the origin at the first sample of each sequence (none is marked):
$x[n]=\{1,3,4,3,1\}$, $n=0..4$ and $h[n]=\{2,2,-2\}$, $n=0..2$.

Output length $=5+3-1=7$, for $n=0$ to $6$; $y[n]=\sum_k x[k]h[n-k]$.

#### Tabular method

| | $n=0$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| $1\cdot h$ | 2 | 2 | −2 | | | | |
| $3\cdot h$ | | 6 | 6 | −6 | | | |
| $4\cdot h$ | | | 8 | 8 | −8 | | |
| $3\cdot h$ | | | | 6 | 6 | −6 | |
| $1\cdot h$ | | | | | 2 | 2 | −2 |
| $y[n]$ | **2** | **8** | **12** | **8** | **0** | **−4** | **−2** |

#### Check by formula
$$
\begin{aligned}
y[2]&=x[0]h[2]+x[1]h[1]+x[2]h[0]=-2+6+8=12\\
y[4]&=x[2]h[2]+x[3]h[1]+x[4]h[0]=-8+6+2=0
\end{aligned}
$$

Sum check: $\sum y=24=(\sum x)(\sum h)=12\times2$.

**Answer: $y[n]=\{2,\ 8,\ 12,\ 8,\ 0,\ -4,\ -2\}$ for $n=0,\dots,6$** (↑ at 2).

## Check yourself

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

1. x has 5 samples and h has 3. The output has:
   A. 7 samples, n = 0 to 6
   B. 8 samples
   C. 15 samples
   D. 5 samples
   Answer: A. Length = 5 + 3 - 1 = 7. (Key point: Output length)

2. y[2] =
   A. 12
   B. 8
   C. 6
   D. 10
   Answer: A. y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = -2 + 6 + 8 = 12. (Key point: Value y[2])

3. y[4] =
   A. 0
   B. 2
   C. -2
   D. 4
   Answer: A. y[4] = x[2]h[2] + x[3]h[1] + x[4]h[0] = -8 + 6 + 2 = 0. (Key point: Value y[4])

4. sum y =
   A. 24
   B. 14
   C. 8
   D. 12
   Answer: A. (1+3+4+3+1)(2+2-2) = 12 x 2 = 24. (Key point: Sum check)

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