0
1.2kviews
We know that systematic form of $G = [I_k | P]$

Number of bits in codeword n = 6

No. of bits in each message vector K = 3

No. of extra bits added q = n - k = 3

Given

$\begin{bmatrix} \ 1 \hspace{0.5cm} 0 \hspace{0.5cm} 0 & 1 \hspace{0.5cm} 1 \hspace{0.5cm} 1 \\ \ 0 \hspace{0.5cm} 1 \hspace{0.5cm} 0 & 1 \hspace{0.5cm} 1 \hspace{0.5cm} 0\\ \ 0 \hspace{0.5cm} 0 \hspace{0.5cm} 1 & 0 \hspace{0.5cm} 1 \hspace{0.5cm} 1\end{bmatrix}$

0
1views

$\therefore P = \begin{bmatrix} \ 1 & 1 & 1 \\ \ 1 & 1 & 0 \\ \ 0 & 1 & 1 \\ \end{bmatrix} H = \begin{bmatrix} \ 1 1 0 & 1 0 0 \\ \ 1 1 1 & 0 1 0 \\ \ 1 0 1 & 0 0 1 \\ \end{bmatrix}$

Number of different message vector $2^k = 2^3 = 8$

$C4 = d_1 \oplus d_2 \oplus d_3\\ C5 = d_1 \oplus d_2\\ C6 = d_2 \oplus d_3$

Where [C] = [M] [P]

Let $d_1, d_2, d_3$ are message bits

M = [$d_1 \hspace{1cm} d_2 \hspace{1cm} d_3$]

C = [$C_4 \hspace{1cm} C_5 \hspace{1cm} C_6$]

$d_1$ $d_2$ $d_3$ $C_4$ $C-5$ $C_6$ codeword Weight
0 0 0 0 0 0 000 000 -
0 0 1 1 0 1 001 101 3
0 1 0 1 1 1 010 111 4
0 1 1 0 1 0 011 010 3
1 0 0 1 1 0 100 110 3
1 0 1 0 1 1 101 011 4
1 1 0 0 0 1 110 001 3
1 1 1 1 0 0 111 100 4

$d_{min} = MIN [w(x)]\\ \therefore d_{min} = 3$

Error deletion capability:

$d_{min} \geq S + 1\\ 3 \geq S + 1\\ \therefore S \leq 2$

Error correction Capability:

$d_{min} \geq 2t + 1\\ \therefore t \leq 1$

It can correct up to 1 bit in error.

Number of single bit error pattern = $n_c1 = 6_C1 = 6$

Decoding table is:-

Error pattern Syndrome
$E_1$ 1 0 0 0 0 0 1 1 1 $S_1$
$E_2$ 0 1 0 0 0 0 1 1 0 $S_2$
$E_3$ 0 0 1 0 0 0 0 1 1 $S_3$
$E_4$ 0 0 0 1 0 0 1 0 0 $S_4$
$E_5$ 0 0 0 0 1 0 0 1 0 $s_5$

$H^T = \begin{bmatrix} \ 1 & 1 & 1 \\ \ 1 & 1 & 0 \\ \ 0 & 1 & 1 \\ \ 1 & 0 & 0 \\ \ 0 & 1 & 0 \\ \ 0 & 0 & 1 \\ \end{bmatrix}$

Let $y_1$ = 1 0 1 1 0 1

$S_1 = y_1 . H^T = [1 0 1 1 0 1] \begin{bmatrix} \ 1 & 1 & 1 \\ \ 1 & 1 & 0 \\ \ 0 & 1 & 1 \\ \ 1 & 0 & 0 \\ \ 0 & 1 & 0 \\ \ 0 & 0 & 1 \\ \end{bmatrix} \\ = \begin{bmatrix} 1 \oplus 1 & 1 \oplus 1 & 1 \oplus 1 \oplus 1 \end{bmatrix} \\ = \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \hspace{2cm} S_1 \neq 0 \Rightarrow error$

Corresponding error pattern is 0 0 0 0 0 1

Corrected codeword will be 1 0 1 1 1 0