# Recent questions tagged crc-polynomial

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1
Let $G(x)$ be generator polynomial used for CRC checking. The condition that should be satisfied by $G(x)$ to correct odd numbered error bits, will be: $(1+x)$ is factor of $G(x)$ $(1-x)$ is factor of $G(x)$ $(1+x^{2})$ is factor of $G(x)$ $x$ is factor of $G(x)$
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The goal of this lab exercise is to implement an error-detection mechanism using the standard CRC algorithm described in the text. Write two programs, generator and verifier. The generator program reads from standard input a line of ASCII text containing an n-bit message ... see that the message is correct, but by typing generator <file | alter arg | verifier you should get the error message.
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A 1024-bit message is sent that contains 992 data bits and 32 CRC bits. CRC is computed using the IEEE 802 standardized, 32-degree CRC polynomial. For each of the following, explain whether the errors during message transmission will be detected by the receiver: (a) There was a ... were 47 isolated bit errors. (e) There was a 24-bit long burst error. (f) There was a 35-bit long burst error.
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A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x 3 + 1. Show the actual bit string transmitted. Suppose that the third bit from the left is inverted during transmission. Show that this error is detected at the receiver’s end. Give an example of bit errors in the bit string transmitted that will not be detected by the receiver.
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What is the remainder obtained by dividing $x^7 + x ^5 + 1$ by the generator polynomial $x^ 3 + 1?$
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Can anyone please explain why do we need x+1 to be a factor of G(x).ie generator polynomial to detect odd no of bit errors ??
2 votes
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7
For detecting a single bit error using CRC, it is needed that $x^{i}$ should not be divisible by g(x). So, we make g(x) of at least 2 terms, which renders a single term of e(x) indivisible. But then what is the logic behind keeping MSB as 1. Isn't just keeping ... to make any single bit indivisible? For example, $x^{3}+x^{2}$ is guarantees to detect a single bit error at any position. Is it not?
1 vote
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I think for (C), it should be The probability of "not detecting" a burst error of size 9 is $\frac{1}{2^7}$ And for (D), the probability of detecting burst error of size 15 should be $1-\frac{1}{2^8}$ Correct me if I am wrong.
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CRC can detect any odd number of errors. CRC can detect all burst errors of less than the degree of the polynomial. Please explain and if possible give proof
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10
A bit stream 10011101 is transmitted using the standard CRC method described in the text. The generator polynomial is x ^3 + 1. Suppose that the third bit from the left is inverted during transmission.. Give an example of bit errors in the bit string transmitted that will not be detected by the receiver.
2 votes
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11
________ can detect burst error of length less than or equal to degree of the polynomial and detects burst errors that affect odd number of bits. Hamming Code CRC VRC None of the above
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12
In CRC, if the degree of Generator polynomial is n, then the number of bits in divisor is?
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13
Let G(x) be the generator polynomial used for CRC checking. What is the condition that should be satisfied by G(x) to detect all isolated double errors? G(x) should have at least two terms. G(x) should have the factor x+1 G(X) shouldn’t divide xt+1 (for t less than frame length The coefficient of the term x0 should be 1.
4 votes
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From forouzan : If the generator has more than one term and coefficient of x^0 is 1 then all single bit error can be caught. So if e(x)=x^i // means some power of 2 number g(x) = at least two terms and coefficient of x^0 is always 1, means some odd number. So is it saying that the number some 2^i will never be divisible by odd number?
26 votes
3 answers
15
A computer network uses polynomials over $GF(2)$ for error checking with $8$ bits as information bits and uses $x^{3}+x+1$ as the generator polynomial to generate the check bits. In this network, the message $01011011$ is transmitted as: $01011011010$ $01011011011$ $01011011101$ $01011011100$
4 votes
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16
In CRC checksum method, assume that given frame for transmission is 1101011011 and the generator polynomial is $G(x) = x^{4}+ x + 1$. After implementing $CRC$ encoder, the encoded word sent from sender side is _____. 11010110111110 11101101011011 110101111100111 110101111001111
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In CRC based design, a CRC Team consists of one or two users representatives several programmers project co-ordinators one or two system analysts a and c a, b, c and d a, c, and d a, b, and d
3 votes
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18
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 110010011001 11001001 110010011001001 11001001011
1 vote
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19
If the data unit is 111111 and the divisor is 1010. In CRC method, what is the dividend at the transmission before division? 1111110000 1111111010 111111000 111111
1 vote
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20
Why don't we use CRC in IP header instead of Checksum?
10 votes
1 answer
21
In CRC if the data unit is 100111001 and the divisor is 1011 then what is dividend at the receiver? 100111001101 100111001011 100111001 100111001110
3 votes
4 answers
22
Consider the use of Cyclic Redundancy Code (CRC) with generator polynomial $G(x)$ for error detection. Recall that error detection with a CRC works by appending the CRC value to the bit sequence to make it a multiple of $G(x)$ ... a burst error of length $5$ in such a way that the error cannot be detected by the CRC with the $G(x)$ given above.
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23
5 votes
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24
If the frame to be transmitted is $1101011011$ and the CRC polynomial to be used for generating checksum is $x^{4}+x+1$, than what is the transmitted frame? $11010110111011$ $11010110111101$ $11010110111110$ $11010110111001$
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18. Suppose we want to transmit the message 11100011 and protect it from errors using the CRC polynomial x3 + 1. (a) Use polynomial long division to determine the message that should be transmitted. (b) Suppose the leftmost bit of the message is inverted due ... the transmission link. What is the result of the receiver&rsquo;s CRC calculation? How does the receiver know that an error has occurred?
0 votes
3 answers
26
19. Suppose we want to transmit the message 1011 0010 0100 1011 and protect it from errors using the CRC8 polynomial x8+ x2+ x1 + 1. (a) Use polynomial long division to determine the message that should be transmitted. (b) Suppose the leftmost bit of the ... on the transmission link. What is the result of the receiver&rsquo;s CRC calculation? How does the receiver know that an error has occurred?
1 vote
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27
24 votes
4 answers
28
Consider the following message $M = 1010001101$. The cyclic redundancy check (CRC) for this message using the divisor polynomial $x^5+x^4+x^2+1$ is : $01110$ $01011$ $10101$ $10110$
22 votes
4 answers
29
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: $11001001000$ $11001001011$ $11001010$ $110010010011$
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