---
title: "Convolve {1, 2, 0, −1} and {2, 0, 2}"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:convolution-1-2-0-minus1"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/convolution-1-2-0-minus1
site: BCT Engineering
---

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

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2080 Asoj, 6 marks; 2074 Bhadra, 7 marks (asked 2 times)
- 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-1-2-0-minus1
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Perform convolution sum of signal x[n] = {1, 2, 0, −1} and h[n] = {2, 0, 2}

## Short answer

Assuming both sequences start at n=0, convolution yields y[n] = {2, 4, 2, 2, 0, −2} for n = 0 to 5, obtained by tabular or direct summation.

## Answer

No origin is marked, so take the first sample of each sequence at $n = 0$: $x[n] = \{\underset{\uparrow}{1}, 2, 0, -1\}$ for $0 \le n \le 3$ and $h[n] = \{\underset{\uparrow}{2}, 0, 2\}$ for $0 \le n \le 2$.

**Convolution sum:**

$$
y[n] = \sum_{k} x[k]\,h[n-k]
$$

**Output range:** starts at $0 + 0 = 0$; length $= 4 + 3 - 1 = 6$, so $0 \le n \le 5$.

#### Tabular method

Each row is one input sample times the shifted $h[n]$; add the columns:

| Row | $n=0$ | $n=1$ | $n=2$ | $n=3$ | $n=4$ | $n=5$ |
|---|---|---|---|---|---|---|
| $x[0]\,h[n]$ | 2 | 0 | 2 |  |  |  |
| $x[1]\,h[n-1]$ |  | 4 | 0 | 4 |  |  |
| $x[2]\,h[n-2]$ |  |  | 0 | 0 | 0 |  |
| $x[3]\,h[n-3]$ |  |  |  | -2 | 0 | -2 |
| $y[n]$ | **2** | **4** | **2** | **2** | **0** | **-2** |

#### Check by direct calculation

- $y[0] = x[0]h[0] = 1\times2 = 2$
- $y[1] = x[0]h[1] + x[1]h[0] = 0 + 4 = 4$
- $y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = 2 + 0 + 0 = 2$
- $y[3] = x[1]h[2] + x[2]h[1] + x[3]h[0] = 4 + 0 - 2 = 2$
- $y[4] = x[2]h[2] + x[3]h[1] = 0 + 0 = 0$
- $y[5] = x[3]h[2] = -1\times2 = -2$

**Sum check:** $\sum y = (\sum x)(\sum h) = 2 \times 4 = 8$, and $2 + 4 + 2 + 2 + 0 - 2 = 8$.

**Answer:**

$$
y[n] = \{\underset{\uparrow}{2},\ 4,\ 2,\ 2,\ 0,\ -2\}
$$

i.e. $y[n] = 2\delta[n] + 4\delta[n-1] + 2\delta[n-2] + 2\delta[n-3] - 2\delta[n-5]$.

(If $x$ and $h$ had different origins, the values stay the same and only the starting index shifts: start 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. For x[n] = {1,2,0,-1} (4 samples) and h[n] = {2,0,2} (3 samples), how many samples does y[n] have and over which range?
   A. 7 samples, 0 to 6
   B. 6 samples, 0 to 5
   C. 5 samples, 0 to 4
   D. 12 samples, 0 to 11
   Answer: B. Length = 4 + 3 - 1 = 6, starting at 0 + 0 = 0, so n = 0 to 5. (Key point: Output length)

2. Which expression gives y[3]?
   A. x[0]h[3] + x[1]h[2] + x[2]h[1]
   B. x[1]h[2] + x[2]h[1] + x[3]h[0]
   C. x[0]h[0] + x[1]h[1] + x[2]h[2]
   D. x[3]h[2] only
   Answer: B. y[3] = sum of x[k]h[3-k] with valid k = 1,2,3: 4 + 0 - 2 = 2. (Key point: Direct convolution value)

3. What is the quick check on the convolution result?
   A. Sum of y equals sum of x plus sum of h
   B. Sum of y equals (sum of x)(sum of h) = 8
   C. Sum of y equals zero
   D. Sum of y equals the number of samples
   Answer: B. Sum of convolution = product of sums: 2 x 4 = 8, matching 2+4+2+2+0-2. (Key point: Sum check)

4. What is the full output y[n]?
   A. {2, 4, 2, 2, 0, -2}
   B. {2, 4, 2, 0, 2, -2}
   C. {2, 4, 2, 2, 0, 2}
   D. {1, 4, 2, 2, 0, -2}
   Answer: A. The tabular method (rows x[k]h shifted by k) added by columns gives 2, 4, 2, 2, 0, -2. (Key point: Final result)

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