0
1.5kviews
Let H= $\begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 0&1&1 \\ 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix}$ be parity check matrix.Determine the group code $e_H: B^2->B^5$

Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 8 Marks

Year: Dec 2015

1 Answer
0
31views

we have

$$N=\begin{bmatrix} 1&0&0 \\ 1&1&0 \end{bmatrix}$$

And

$$B^2= \{00, 01, 10, 11\}.$$

Then,

$$00 * \begin{bmatrix} 1&0&0 \\ 1&1&0 \end{bmatrix} =000, \ \ \ \ \ \ \ \ \ \ 01 * \begin{bmatrix} 1&0&0 \\ 1&1&0 \end{bmatrix} =110 \\ 10 * \begin{bmatrix} 1&0&0 \\ 1&1&0 \end{bmatrix} =100, \ \ \ \ \ \ \ \ \ \ 11 * \begin{bmatrix} 1&0&0 \\ 1&1&0 \end{bmatrix} =010$$

Hence, the required group code is

$$e_H= (00) =00000, \ \ \ \ \ \ \ \ \ \ e_H= (01) =01110 \\ e_H= (10) =10100 \ \ \ \ \ \ \ \ \ \ e_H= (11) =11010$$

Please log in to add an answer.