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solution4

# solution4 - University of California San Diego ECE 259A...

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University of California San Diego ECE 259A: Solutions to Problem Set #4 1 . A binary polynomial g x generates a linear cyclic code of length 15 if and only if it is a divisor of x 15 1 over GF 2 . Using the factorization x 15 1 x 1 x 2 x 1 x 4 x 1 x 4 x 3 1 x 4 x 3 x 2 x 1 we conclude that there are 2 5 32 binary linear cyclic codes of length 15, including the trivial cyclic codes 0 and IF n 2 . Note that the factorization of x 15 1 itself is not essential to answer this question. All one needs to know is the total number of factors, which may be easily found by writing out the cyclotomic cosets. 2 . (a) If x 1 is a factor of g x then g 1 0 , and thereforee for all c x , we have c 1 c 0 c 1 c n 1 f 1 g 1 0 . Since g x , we have g 0 g 1 g r g 1 0 which means that 1 α 0 is a root of g x . Therefore, the minimal polynomial of α 0 div- ides g x . Yet the minimal polynomial of α 0 is x 1 over any field. (b) The all-one vector is described by the polynomial 1 x x n 1 x n 2 x 1 . In any field we have x n 1 x 1 1 x . Since n and q are relatively prime, all the roots of x n 1 are distinct and therefore 1 α 0 is not a root of 1 x . Thus if 1 x then α 0 cannot be a root of g x . On the other hand, since all the nonzero powers of α are roots of 1 x , if α 0 is not a root of g x then necessarily g x 1 x and hence 1 x .

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