Let f(r)= x- 1 and the irreducible polynomial p(r) r +x+ 1. Then the powers of a in GF(2") Zz[r]/(p(x)) are 1, a, a2,a a+1.04 a + a, a a +a+1,a=a + 1 Let C be the two-error correcting Reed-Solomon code that results from the polynomial p(x). Given that the quotient polynomial resulting from dividing r by the syndrome polynomial S(r) is Q(x) x2 +r+ a', then the errors occur at the positions O a. a, a" O b. a, a" O c.a, a O d. 2, r

icon
Related questions
Question
Let f(r) a?
1 and the irreducible polynomial p(x) = x +x + 1. Then the powers of a in GF(2°) = Z2[x|/(p(x)) are
1, a, a2, a
= a + 1, q4
a + a, a =
a + a+ 1, a°
a² + 1
Let C be the two-error correcting Reed-Solomon code that results from the polynomial p(x). Given that the quotient polynomial resulting from dividing r by the syndrome polynomial S(r) is
Q(r) x tr+a, then the errors occur at the positions
O a. a, a
O b. x4, r
Oca, 5
O d. z, r7
Transcribed Image Text:Let f(r) a? 1 and the irreducible polynomial p(x) = x +x + 1. Then the powers of a in GF(2°) = Z2[x|/(p(x)) are 1, a, a2, a = a + 1, q4 a + a, a = a + a+ 1, a° a² + 1 Let C be the two-error correcting Reed-Solomon code that results from the polynomial p(x). Given that the quotient polynomial resulting from dividing r by the syndrome polynomial S(r) is Q(r) x tr+a, then the errors occur at the positions O a. a, a O b. x4, r Oca, 5 O d. z, r7
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer