GATE CS 2017 Set 1 — Question 32
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Computer Networks → Data Link Layer → Error Detection (Parity, CRC)
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Question
A computer network uses polynomials over for error checking with 8 bits as information bits and uses as the generator polynomial to generate the check bits. In this network, the message 01011011 is transmitted as
Correct answer
(C) 01011011101
Solution
The message is . Ignoring the leading zero, the bit sequence is . The generator polynomial is , which corresponds to the bit sequence .To find the transmitted message (CRC), we append 3 zeros (degree of ) to the message: .
We perform modulo-2 division (XOR division) of by .Dividend:
Divisor:
(1 time)
(shift divisor? No, align MSB)
Actually, let's do step-by-step:
Message polynomial (from 1011011).
Multiply by : .
Divisor .
Remainder:
Remainder:
Remainder: corresponds to binary .
So the check bits are .
The transmitted message is the original message appended with : .
We perform modulo-2 division (XOR division) of by .Dividend:
Divisor:
1.Take first 4 bits . .
2.Bring down next bits until we have a leading 1. The sequence becomes . The first '1' is at the 6th position (from left). We align the divisor with this '1'.
Current remainder part: (1 time)
(shift divisor? No, align MSB)
Actually, let's do step-by-step:
- vs -> 0
- Next significant block starts at the 2nd '1' in the original message? No, the result of XOR was 0. We bring down the rest: .
- Leading bit is 0, shift. .
- (Remainder )
- (Remainder )
- (Remainder )
Message polynomial (from 1011011).
Multiply by : .
Divisor .
Remainder:
Remainder:
Remainder: corresponds to binary .
So the check bits are .
The transmitted message is the original message appended with : .
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