Y.S.Han RS Codes 2 degree less than or equal to k-1. The minimum distance of an (n,k)RS code is dmin =n-k+1.It can be proved by follows. Since u(x)has at most k-1 roots,there are at most k-1 zero positions in each nonzero codeword.Hence, dmin >n-k+1.By the Singleton bound, dmin <n-k+1.So dmin =n-k+1. School of Electrical Engineering Intelligentization,Dongguan University of TechnologyY. S. Han RS Codes 2 degree less than or equal to k − 1. • The minimum distance of an (n, k) RS code is dmin = n − k + 1. It can be proved by follows. • Since u(x) has at most k − 1 roots, there are at most k − 1 zero positions in each nonzero codeword. Hence, dmin ≥ n − k + 1. By the Singleton bound, dmin ≤ n − k + 1. So dmin = n − k + 1. School of Electrical Engineering & Intelligentization, Dongguan University of Technology