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Cyclic Code

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Given $ \mbox{$g := X^3+3X^2+2X+4 \subseteq\mathbb{F}_5[X]$}$.

  1. Find $ \mbox{$h\in\mathbb{F}_5[X]$}$ with $ \mbox{$X^6-1=gh$}$.
  2. Find a generator matrix and a check matrix for the cylic code $ \mbox{$C_g$}$ of the length $ \mbox{$6$}$ that is generated by $ \mbox{$g$}$.
  3. Prove, that $ \mbox{$Y^2+2$}$ is irreducible in $ \mbox{$\mathbb{F}_5[Y]$}$.
  4. Prove, that $ \mbox{$\alpha=Y+3\in\mathbb{F}_5[Y]/(Y^2+2)$}$ is a primitive $ \mbox{$6$}$-th root of unity over $ \mbox{$\mathbb{F}_5$}$.
  5. Prove, that $ \mbox{$d(C_g) = 4$}$ holds.
  6. Code $ \mbox{$(4,2,1)$}$.

Solution

  1. Polynomial division of $ \mbox{$X^6-1$}$ by $ \mbox{$g = X^3 + 3X^2 + 2X + 4$}$ gives $ \mbox{$h = X^3 + 2X^2 + 2X + 1$}$.
  2. Generator matrix and check matrix are
    $ \mbox{$\displaystyle \begin{array}{rl} G := \begin{pmatrix}4&2&3&1&0&0\\ 0... ...&2&2\\ 0&1&2\\ 0&0&1\end{pmatrix}\in\mathbb{F}_5^{6\times 3}. \end{array} $}$
  3. The polynomial $ \mbox{$Y^2+2$}$ of degree $ \mbox{$2$}$ has no roots within $ \mbox{$\mathbb{F}_5$}$. If it would be irreducible we could seperate a linear factor corresponding to a root. Therefore the polynomial is irreducible.
  4. We get
    $ \mbox{$k$}$ $ \mbox{$1$}$ $ \mbox{$2$}$ $ \mbox{$3$}$ $ \mbox{$4$}$ $ \mbox{$5$}$ $ \mbox{$6$}$
    $ \mbox{$\alpha^k$}$ $ \mbox{$Y+3$}$ $ \mbox{$Y+2$}$ $ \mbox{$4$}$ $ \mbox{$4Y+2$}$ $ \mbox{$4Y+3$}$ $ \mbox{$1$}$
  5. From $ \mbox{$G$}$ we can see, that $ \mbox{$d(C_g)\leq 4$}$ must hold. We have $ \mbox{$g(\alpha^{-1}) = g(\alpha^0) = g(\alpha^1) = 0$}$. The well known estimate for the minimal distance of cyclic codes gives us $ \mbox{$d(C_g)\geq 4$}$. So $ \mbox{$d(C_g) = 4$}$ must hold.
  6. It is $ \mbox{$u(X) = X^5 + 2 X^4 + 4 X^3$}$. Polynomial division by $ \mbox{$g(X)$}$ gives the residue $ \mbox{$r(X) = 3 X^2 + 4 X$}$. The code word to use is $ \mbox{$c = (0,1,2,4,2,1)$}$.

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  automatisch erstellt am 6.  7. 2005