---
title: "Convolve {2, 0, −1, 2} and {1, 2, −1, 3}"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-20-12"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-20-12
site: BCT Engineering
---

# Convolve {2, 0, −1, 2} and {1, 2, −1, 3}

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2077 Chaitra, 6 marks
- Video: [#110 Solved problems -2 on Convolution Sum || EC Academy](https://www.youtube.com/watch?v=hl2T2adZWFI) — EC 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-20-12
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Find the convolution between the following sequences. h[n] = {2, 0, −1, 2}, x[n] = {1, 2, −1, 3}. Choose origin as per your choice.

## Short answer

y[n] = {2, 4, -3, 6, 5, -5, 6} starting at n = 0 (origin at first sample).

## Answer

Choose the origin at the **first sample** of each sequence:
$x[n]=\{1,2,-1,3\}$ and $h[n]=\{2,0,-1,2\}$ for $n=0,1,2,3$.

The output has $4+4-1=7$ samples, from $n=0$ to $n=6$, with $y[n]=\sum_k x[k]h[n-k]$.

#### Tabular method

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

#### Step-by-step values
$$
\begin{aligned}
y[0]&=1(2)=2\\
y[1]&=1(0)+2(2)=4\\
y[2]&=1(-1)+2(0)+(-1)(2)=-3\\
y[3]&=1(2)+2(-1)+(-1)(0)+3(2)=6\\
y[4]&=2(2)+(-1)(-1)+3(0)=5\\
y[5]&=(-1)(2)+3(-1)=-5\\
y[6]&=3(2)=6
\end{aligned}
$$

Check: $\sum y=15=(\sum x)(\sum h)=5\times3$.

**Answer: $y[n]=\{2,\ 4,\ -3,\ 6,\ 5,\ -5,\ 6\}$, with $y[0]=2$ (↑ at the first value).**

If a different origin is chosen, the values stay the same; only the starting index changes (start index of $y$ = start of $x$ + start of $h$).

## Check yourself

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

1. Both sequences have 4 samples starting at n = 0. The output has:
   A. 7 samples, n = 0 to 6
   B. 8 samples
   C. 4 samples
   D. 16 samples
   Answer: A. Length = 4 + 4 - 1 = 7. (Key point: Output length)

2. y[2] for x = {1,2,-1,3}, h = {2,0,-1,2}:
   A. -3
   B. 3
   C. -1
   D. 0
   Answer: A. y[2] = 1(-1) + 2(0) + (-1)(2) = -3. (Key point: Value y[2])

3. sum y =
   A. 8
   B. 15
   C. 5
   D. 3
   Answer: B. (1+2-1+3)(2+0-1+2) = 5 x 3 = 15. (Key point: Sum check)

4. If a different origin is chosen:
   A. The values change
   B. The values stay the same, only the start index shifts (start x + start h)
   C. The output length changes
   D. The sum check fails
   Answer: B. Origin only relabels the index; the sequence values are unaffected. (Key point: Change of origin)

## Close variants set in other exams

- [Convolve {2, 1, 5, 7} and {1, 3, 5, 9, 6}](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-2157) — Signals and Systems, chapter 6, asked 2×

## Related in Signals and Systems

- [Convolve |n| and 2n sequences](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-abs-n-ramp) — Signals and Systems, chapter 6
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- [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 {1,2,1,−1} and {1,2,3,1}](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Aconvolution-origin-given) — Signals and Systems, chapter 6, asked 2×

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