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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:

1. $11001001000$

2. $11001001011$

3. $11001010$

4. $110010010011$

edited | 7.1k views
+12

CRC Calculation

0
How come 1100-1001 is 101? Should it not be 11?

Degree of generator polynomial is $3$ hence $3-bits$ 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

by Loyal (8.6k points)
edited
0

m + r + 1 < 2r

messege, m = 8 bits

8 + 1 + 1 > 21

8 + 2 + 1 > 22

8 + 3 + 1 > 23

8 + 4 + 1 < 24

Hence r = 4

then sending mesg is 8 + 4 = 12 bits

ANS : d

+1
This is for hamming code.
0
In the end you appended 011 and not 0011, why like that?
+1
How did you find the divisor ? Pl explain thax
0
Since the degree of G(x) is 3, the transmitted code must have only 3 more bits. Hence B,
0

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.

+1

@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.
110011 011 where 011 is crc
by (49 points)
Answer is d the no. Of bits u add for n degree poly is n+ 1
by Active (3.3k points)
ans a)
by Loyal (5.2k points)
0
Can you please give detailed explanation?
+1
seriously she does nt explain anything....
0
how are you xoring 1011 with 1001 such that you are taking in the next step 1001?