GATE CSE 2007 | Question: 68, ISRO2016-73

Question

The message $11001001$ is to be transmitted using the CRC polynomial $x^3 +1$ to protect it from errors. The message that should be transmitted is:

• A. $11001001000$

• B. $11001001011$

• C. $11001010$

• D. $110010010011$

Comments

CRC Calculation

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How come 1100-1001 is 101? Should it not be 11?

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https://www.geeksforgeeks.org/error-detection-in-computer-networks/

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We are not doing division of binary numbers, we are doing XOR operation.

XOR of 1100 and 1001 is 101

Answer 1

Answer - B.

Degree of generator polynomial is $3$ hence $3\text{-bits}$ are appended before performing division

After performing division using $2's$ complement arithmetic remainder is $011$

The remainder is appended to original data bits and we get  $M' = 11001001\bf{011}$ from $M = 11001001.$

Courtesy, Anurag Pandey

Comments

Polynomial function is $x^{3}$ +1 

It can be interpreted as 1$x^{3}$ + 0$x^{2}$ + 0$x^{1} + 1$$x^{0}$

Considering all the coefficients: 1001 is the required divisor.

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@Sumaiya23 Since the degree of CRC polynomial is 3 that's why 3 bits are appended.

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In polynmial method , We continue to divide until degree of rmainder is less than degree of divisor => x3 degree 3 so remainder of the form x2 + x + 1 , hence 3 bits.

Answer 2

110011 011 where 011 is crc

Answer 3

Answer is d the no. Of bits u add for n degree poly is n+ 1

Answer 4

ans a)

Comments

Can you please give detailed explanation?

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seriously she does nt explain anything....

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how are you xoring 1011 with 1001 such that you are taking in the next step 1001?