🏠 Home > Data Communication & Networks > 10. Error Detection and Correction > XOR Error Detecting Technique
← Back to 10. Error Detection and Correction

📝 XOR Error Detecting Technique

📖 From Data Communication & Networks • 10. Error Detection and Correction • 11 questions available

Practice MCQs for XOR Error Detecting Technique. Test your knowledge with carefully crafted questions across easy, medium, and hard difficulty levels.

🔄 Last updated: 2026-07-31

3
Easy Questions
5
Medium Questions
3
Hard Questions

📝 Sample Questions

Q1. If a data block \(D\) is XOR‑ed with a checksum \(C\) to produce the transmitted word \(T = D \oplus C\), and the received word is \(R = T \oplus E\) where \(E\) is a single‑bit error, which of the following statements is true about the computed checksum at the receiver?

🔹 A. The receiver will recover the original checksum \(C\) regardless of the error.
🔹 B. The receiver will detect the error because the recomputed checksum will differ from \(C\).
🔹 C. The error will be corrected automatically by the XOR operation.
🔹 D. The error will be invisible to the checksum, so it will go undetected.

💡 Difficulty: easy | ✅ Correct: B

Q2. A generator polynomial \(G(x)=x^5 + x^2 + 1\) can detect all burst errors up to length 5. If a burst error of length 6 occurs, what is the logical implication regarding detection probability?

🔹 A. The error will always be detected because any longer burst contains a shorter detectable pattern.
🔹 B. The error will be detected with probability 31/32.
🔹 C. The error will be detected with certainty only if the error pattern matches the generator.
🔹 D. Detection is impossible for bursts longer than the generator degree.

💡 Difficulty: medium | ✅ Correct: B

Q3. A sender uses generator polynomial \(G(x)=x^6 + x + 1\) to compute a CRC. An adversary flips bits at positions 2 and 5 of the transmitted frame. Assuming the frame length is 20 bits, determine whether the CRC will detect the error. Choose the correct statement.

🔹 A. The error will always be detected because the error pattern has odd weight.
🔹 B. The error will be detected unless the flipped bits happen to correspond to a multiple of the generator polynomial.
🔹 C. The error will be detected with probability 1−1/2^6.
🔹 D. The error may go undetected because the error polynomial could be divisible by \(G(x)\).

💡 Difficulty: hard | ✅ Correct: D

⬆️ View all questions in the quiz below

🔗 Related Topics

📝 Types of Errors in Data Link Layer📝 Redundancy error detecting and correcting📝 Error Detection versus Correction in Data Link Layer📝 Redundancy hrough Coding📝 Block Coding in Data Link Layer
🚀 Start Quiz 📝 Practice Mode