64 votes 64 votes Let $G(x)$ be the generator polynomial used for CRC checking. What is the condition that should be satisfied by $G(x)$ to detect odd number of bits in error? $G(x)$ contains more than two terms $G(x)$ does not divide $1+x^k$, for any $k$ not exceeding the frame length $1+x$ is a factor of $G(x)$ $G(x)$ has an odd number of terms. Computer Networks gatecse-2009 computer-networks error-detection normal + – Kathleen asked Sep 22, 2014 edited Jun 20, 2018 by Milicevic3306 Kathleen 29.0k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments KUSHAGRA गुप्ता commented Jan 23, 2020 reply Follow Share $Ans: C$ $*$ Error should not be divisible by your generator $a)$ $\dfrac{E(x)}{G(x)}=\dfrac{1+x+x^2}{1+x+x^2}: Wrong$ $b)$ $G(x)$ does not divide $1+x^k,$ so let's take $G(x)=x$ $\dfrac{E(x)}{G(x)}=\dfrac{x+x^2+x^3}{x}: Wrong$ $d)$ $\dfrac{E(x)}{G(x)}=\dfrac{1+x+x^2}{1+x+x^2}: Wrong$ $c)$ $\dfrac{E(x)}{G(x)}=\dfrac{E(x)}{(1+x)G(x)}=\dfrac{odd \#\ of\ terms}{even \#\ of\ terms}:Not\ possible: \checkmark$ 19 votes 19 votes rupesh17 commented Nov 18, 2020 reply Follow Share Reference :-Tanenbaum answer to question is marked by red. Rest marked by yellow are also properties of G(x) to catch different type of errors. 21 votes 21 votes Abhrajyoti00 commented Aug 21, 2022 i edited by Kabir5454 Aug 21, 2022 reply Follow Share @KUSHAGRA गुप्ता Why did you take ${E(x)}$$={x+x^2+x^3}$ when it says $G(x)$ does not divide $1+x^k$ ? How is $x+x^{2}+x^{3}$ the expansion of $1+x^{k}$ ? 2 votes 2 votes Please log in or register to add a comment.
1 votes 1 votes ans c) Aditi Dan answered Dec 23, 2014 Aditi Dan comment Share Follow See all 6 Comments See all 6 6 Comments reply saurabhrk commented Feb 1, 2015 reply Follow Share Please elaborate 2 votes 2 votes Gate Mm commented Dec 1, 2015 reply Follow Share Please elaborate the reason for the answer 2 votes 2 votes minal commented Dec 16, 2015 reply Follow Share http://www.cs.jhu.edu/~scheideler/courses/600.344_S02/CRC.html 2 votes 2 votes aditya dhanraj commented Jun 30, 2016 reply Follow Share please explain. 1 votes 1 votes tusharp commented May 11, 2018 reply Follow Share In most of the answers answered by you, you just tell the options. Please elaborate so that it can b helpful for others. 6 votes 6 votes Radha mohan commented Sep 15, 2018 reply Follow Share everyone knows the answer either right or wrng explaination matters 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes G(x) contains more than two terms = No meaning here If 1+x is factor of generator polynomial then we can find all odd number of error. So B is true. To detect 1 bit error we need xk +1. where k is any constant. Prashant. answered Oct 22, 2016 Prashant. comment Share Follow See all 0 reply Please log in or register to add a comment.