---
title: "Convolve {3, 1, 5} and {1, 4, −2, 3}"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-315"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-315
site: BCT Engineering
---

# Convolve {3, 1, 5} and {1, 4, −2, 3}

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2079 Chaitra, 5 marks
- Video: [Method to Find Discrete Convolution](https://www.youtube.com/watch?v=Nj8no2GMm8s) — TutorialsPoint
- 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-315
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Convolve the signal x[n] = {3, 1, 5} and h[n] = {1, 4, −2, 3}

## Short answer

Convolution sum yields y[n] = {3, 13, 3, 27, -7, 15} for n = 0 to 5, with the first sample at 3.

## Answer

The convolution sum is $y[n]=\sum_k x[k]h[n-k]$. Since the origin is not marked, take the **first sample of each sequence at $n=0$**:
$x[n]=\{3,1,5\}$ for $n=0,1,2$ and $h[n]=\{1,4,-2,3\}$ for $n=0,\dots,3$.

Length of $y$ = $3+4-1=6$, running from $n=0+0=0$ to $n=2+3=5$.

#### Tabular (multiplication) method
Each row is $x[k]$ times the whole of $h$, shifted by $k$:

| | $n=0$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $3\cdot h$ | 3 | 12 | −6 | 9 | | |
| $1\cdot h$ | | 1 | 4 | −2 | 3 | |
| $5\cdot h$ | | | 5 | 20 | −10 | 15 |
| $y[n]$ | **3** | **13** | **3** | **27** | **−7** | **15** |

#### Check of each value
$$
\begin{aligned}
y[0]&=3(1)=3\\
y[1]&=3(4)+1(1)=13\\
y[2]&=3(-2)+1(4)+5(1)=3\\
y[3]&=3(3)+1(-2)+5(4)=27\\
y[4]&=1(3)+5(-2)=-7\\
y[5]&=5(3)=15
\end{aligned}
$$

Check: $\sum y = 54 = (\sum x)(\sum h) = 9\times 6$. Correct.

**Answer: $y[n]=\{3,\ 13,\ 3,\ 27,\ -7,\ 15\}$ for $n=0$ to $5$** (↑ at 3).

## Check yourself

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

1. x = {3,1,5} and h = {1,4,-2,3}. The output has:
   A. 7 samples
   B. 6 samples, n = 0 to 5
   C. 12 samples
   D. 4 samples
   Answer: B. Length = 3 + 4 - 1 = 6. (Key point: Output length)

2. y[2] equals:
   A. 3(-2) + 1(4) + 5(1) = 3
   B. 3(1) + 1(4) + 5(-2) = -3
   C. 3(4) + 1(1) = 13
   D. 5(1) = 5
   Answer: A. y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0]. (Key point: Value y[2])

3. The sum of y should be:
   A. 15
   B. 54
   C. 9
   D. 6
   Answer: B. (3+1+5)(1+4-2+3) = 9 x 6 = 54. (Key point: Sum check)

4. y[5] equals:
   A. 15
   B. 5
   C. 3
   D. -7
   Answer: A. y[5] = x[2]h[3] = 5 x 3 = 15. (Key point: Last sample)

## Related in Signals and Systems

- [Convolve {1, 2, 0, −1} and {2, 0, 2}](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-1-2-0-minus1) — Signals and Systems, chapter 6, asked 2×
- [Convolve {3, 5, 6, 9} and {−1, 3, 5}](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-3569) — Signals and Systems, chapter 6
- [Convolve δ-sum with u[n] − u[n−1]](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-delta-first-difference) — Signals and Systems, chapter 6