---
title: "CRC detection of burst errors"
subject: "Data Communication"
chapter: 8
key: "semester_5/data_communication:crc-burst-errors"
source: https://www.bctengineer.com/q/semester_5/data_communication/crc-burst-errors
site: BCT Engineering
---

# CRC detection of burst errors

Data Communication · Chapter 8: Information Theory and Coding

- Asked in: 2076 Asoj, 10 marks; 2075 Chaitra, 10 marks (asked 2 times)
- Video: [Lec-30: Cyclic Redundancy Check(CRC)  for Error Detection and Correction  | Computer Networks](https://www.youtube.com/watch?v=5Q-Yv6_0Qcw) — Gate Smashers
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/crc-burst-errors
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-8

## Question

Explain with example how CRC-5 works to detect 3 burst errors.

## Short answer

CRC-5 uses generator G(x)=x^5+x^2+1 (100101). Data 11010011101100 is appended with 5 zeros, divided modulo-2 to give remainder 01110, transmitted as 1101001110110001110. A 3-bit burst error flipping bits 5-7 gives received 1101110110110001110. Dividing by 100101 yields nonzero remainder 01111, so the error is detected; any burst of length 5 or less is always detected.

## Answer

A **burst error** of length *b* is a run of corrupted bits where the first and last bits are in error (the bits in between may or may not be). A CRC with an r-bit remainder detects every burst of length ≤ r.

#### CRC-5 setup
Generator G(x) = x⁵ + x² + 1 → **100101** (degree 5, so 5 check bits). Data = 11010011101100 (14 bits).

#### Sender
Append 5 zeros: 11010011101100**00000** and divide by 100101 (modulo-2). The remainder is **01110**.
Transmitted codeword = 1101001110110001110 (19 bits).

#### Receiver, no error
Dividing the codeword by 100101 gives remainder 00000, so no error is found.

#### Burst of 3 errors
Take a burst error of length 3: bits 5, 6 and 7 are inverted.

```text
 sent     1101001110110001110
 error    0000111000000000000   (E(x), burst length 3)
 received 1101110110110001110
```
Dividing the received word by 100101 gives remainder **01111 ≠ 0**, so the error is detected.

#### Why a burst of 3 is always detected
The received word is T(x) + E(x). T(x) is divisible by G(x), so the remainder equals E(x) mod G(x). A burst of length 3 has the form E(x) = xᵏ·B(x), where B(x) has degree 2 (less than 5). G(x) has a constant term 1, so it shares no factor with xᵏ. Therefore G(x) can divide E(x) only if it divides B(x), which is impossible because B(x) has a lower degree than G(x). So E(x) is never divisible by G(x) and the remainder is non-zero.

General rule for an r-bit CRC: all bursts of length ≤ r are detected; a burst of length r+1 is missed with probability 2^−(r−1); longer bursts are missed with probability 2^−r.

**Answer: the 3-bit burst gives the non-zero remainder 01111, so it is detected.**

## Check yourself

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

1. For CRC-5 with G(x) = x^5 + x^2 + 1, how many bits are appended to the data before division?
   A. 6 zeros
   B. 4 zeros
   C. 5 zeros
   D. 14 zeros
   Answer: C. The degree of G(x) is 5, so 5 zeros are appended and the remainder gives 5 check bits. (Key point: Generator polynomial)

2. With no error, what remainder does the receiver get when dividing the received codeword by 100101?
   A. 00000
   B. 11111
   C. The original data
   D. The generator itself
   Answer: A. The sender makes the codeword exactly divisible by G, so an error-free word leaves remainder zero. (Key point: Receiver check)

3. Why is a burst of length 3 always detected by this CRC-5?
   A. The data is only 14 bits long
   B. E(x) = x^k B(x) with B of degree 2, and G(x) (constant term 1, degree 5) cannot divide it
   C. The remainder is always 01111
   D. Because the burst has an odd length
   Answer: B. G cannot divide x^k because it has a constant term 1, nor B(x) because B has lower degree, so the remainder is non-zero. (Key point: Why bursts are detected)

4. For an r-bit CRC, which bursts are always detected?
   A. Only bursts of length exactly r
   B. Bursts of length up to r
   C. Bursts of any length
   D. Only bursts of length greater than r
   Answer: B. Every burst of length r or less is detected; longer bursts may escape with small probability. (Key point: Burst length rule)

## Related in Data Communication

- [CRC error detection with example](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acrc-explain) — Data Communication, chapter 8, asked 5×
- [Checksum error detection](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Achecksum-12-15-15-10-5-2) — Data Communication, chapter 8

## Related in other subjects

- [Back-face detection algorithm](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Aback-face-detection-method) — Computer Graphics, chapter 7, asked 8×
- [Deadlock conditions, detection, action](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Adeadlock-conditions-detection-recovery-action) — Operating System, chapter 7
- [Deadlock conditions; detection with example](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Adeadlock-conditions-detection-example) — Operating System, chapter 7
- [Synchronous detection of DSB-SC; phase and frequency error](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asynchronous-dsbsc-phase-freq-error) — Communication Systems, chapter 2