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Q6(a) · CRC

2023-11-22 · Final 2023-24 / Re-exam 2022-23

Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.

Q6(a). Consider data generated at sender 110101 and CRC polynomial x^3+x+1. Calculate i) CRC code generated at the sender side ii) Final message transmitted from sender to receiver iii) Show message is received without error at receiver. [10 marks]

Answer

Given

Data = 110101; 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

110101 → 110101000

Step 2. Perform XOR division

Leading position	Before XOR	Aligned generator	After XOR
1	110101000	101100000	011001000
2	011001000	010110000	001111000
3	001111000	001011000	000100000
4	000100000	000101100	000001100
6	000001100	000001011	000000111

Step 3. Build the frame

Remainder = 111. Transmit 110101111.

Step 4. Divide the received frame

Position	Before	Generator	After
1	110101111	101100000	011001111
2	011001111	010110000	001111111
3	001111111	001011000	000100111
4	000100111	000101100	000001011
6	000001011	000001011	000000000

Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.