Computer Networks · TEE · mocks

Q2A · Error detection and correction

TEE mock B · full paper

TEE · 100 attempted marks · solutions for every choice · current syllabus 702AI0C023

TEE practice mock · Q2A. For message bits 101101 and generator 10011, calculate the CRC and transmitted frame. Check the frame at the receiver and show a detected one-bit error. [10 mock marks]

Answer

Given

Message=101101; generator=10011; generator degree r=4. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.

To find

Exactly r CRC bits, the transmitted frame and receiver syndromes.

Formula and why

R=(M\,2^r)\bmod_2 G,\qquad T=(M\,2^r)\oplus RLaTeX: R=(M\,2^r)\bmod_2 G,\qquad T=(M\,2^r)\oplus R

Modulo-2 division uses aligned XOR, not ordinary decimal division.

Step-by-step solution

Step 1. Preserve all polynomial coefficients

Message bits=101101; generator bits=10011. Leading exponent is 4, so CRC length is 4.

Step 2. Append zeros

101101 followed by 4 zeros = 1011010000.

Step 3. Perform each nonzero XOR subtraction

BeforeAligned generatorAfter XOR
101101000010011000000010110000
001011000000100110000000101000
000010100000001001100000001110

Remainder padded to 4 bits = 1110. Leading zero columns need no subtraction.

Step 4. Construct the transmitted frame

Message + CRC = 101101 + 1110 = 1011011110.

Step 5. Check the full frame

BeforeAligned generatorAfter XOR
101101111010011000000010111110
001011111000100110000000100110
000010011000001001100000000000

Receiver remainder = 0000; the uncorrupted frame passes.

Step 6. Demonstrate detection

Flip the last bit: 1011011111. Dividing by 10011 leaves 0001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.

Final answer: CRC=1110; transmitted frame=1011011110; uncorrupted receiver remainder=0000.