Due: Monday, April 30, 2001
Problem 1. Let V
= ( g(x) ) be the binary cyclic code
of length 7 with generator polynomial
g(x) = x4 + x2 +
x + 1
Problem 2. Let a be the primitive
element of GF(24) which is the zero of the primitive polynomial
p(x) = x4 + x + 1 .
Let g(x) be the binary polynomial of smallest degree having
a and a5
as roots. Let V = ( g(x) )
be the cyclic code of smallest length having g(x) as a
generator polynomial. Use a and a5 to construct
a parity check matrix H of V . (Do not explicitly compute g(x) . )
Hint: Let K denote the matrix:
[ 1 a a2
a3
a4
... a14 ]
K=
[
]
[ 1 (a5)1 (a5)2 (a5)3 (a5)4
... (a5)]14 ]
Then
f(x) = f0+f1x+f2x2+ ... f14x14
e V
iff
(f0, f1, f2,
... , f14)KT = [ f(a), f(a5) ] = [ 0,
0 ]
Next let H' denote the
binary matrix formed by replacing each element of GF(24)
in K' with the corresponding binary 4-tuple written as a
column vector. Then the row space of H' is VPerp.
Unfortunately, the rows of H' may be linearly dependent. Form the parity
check matrix H by putting H' in row
echelon canonical form and deleting the all zero rows.
Problem 3.
Let x be the
primitive element of GF(26) which is the zero of the primitive polynomial:
p(x) = x6 + x + 1 .
Let g(x) be the polynomial of smallest degree having the
following zeros:
x, x2, x3, x4, x5, x6, x7, x8
Let V
= ( g(x) ) be the corresponding cyclic
code of shortest length.
Note: If you have not installed the symbolic fonts on your web browser, then the above ksi's will look like x's.
Last Modified:
April 23, 2001