Computer Networks · TEE · papers
Q6(b) · CRC
2025-02-19 · 2023-24 / 2024-25
Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.
Q6(b). Given the data bit sequence 1010011110 and the CRC polynomial of x^3+x+1. Determine the transmitted message from the sender. Illustrate the process of checking the received message for any errors at the receiver side. [10 marks]
Answer
Given
Data = 1010011110; generator bits = 1011; degree r = 3.
To find
CRC remainder, complete transmitted frame and receiver check.
Formula and why
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
Step 1. Append zeros
1010011110 → 1010011110000
Step 2. Perform XOR division
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 1010011110000 | 1011000000000 | 0001011110000 |
| 4 | 0001011110000 | 0001011000000 | 0000000110000 |
| 8 | 0000000110000 | 0000000101100 | 0000000011100 |
| 9 | 0000000011100 | 0000000010110 | 0000000001010 |
| 10 | 0000000001010 | 0000000001011 | 0000000000001 |
Step 3. Build the frame
Remainder = 001. Transmit 1010011110001.
Step 4. Divide the received frame
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 1010011110001 | 1011000000000 | 0001011110001 |
| 4 | 0001011110001 | 0001011000000 | 0000000110001 |
| 8 | 0000000110001 | 0000000101100 | 0000000011101 |
| 9 | 0000000011101 | 0000000010110 | 0000000001011 |
| 10 | 0000000001011 | 0000000001011 | 0000000000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.