Question

A (7, 4) block code has a generator matrix as shown.

\(G = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1&1&0\\ 0&1&0&0&0&1&1\\ 0&0&1&0&1&1&{1}\\ 0&0&0&1&1&0&1 \end{array}} \right]\)

If there is error in the 7^th Bit then syndrome for the same will be
1. 001
2. 010
3. 100
4. 011

Answer 1

Correct Answer - Option 1 : 001

The generator Matrix is given by

\(G = \left[ {{I_K}{P^T}} \right]\)

\({P^T} = \left[ {\begin{array}{*{20}{c}} 1&1&0\\ 0&1&1\\ 1&1&1\\ 1&0&1 \end{array}} \right]\)

The parity check matrix is given by:

H = [P I_{kn– K}]

Syndrome

S = eH^T

\({H^T} = \left[ {\begin{array}{*{20}{c}} {{P^T}}\\ {{I_{n - k}}} \end{array}} \right]\)

\({H^T} = \left[ {\begin{array}{*{20}{c}} 1&1&0\\ 0&1&1\\ 1&1&1\\ 1&0&1\\ 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

S = eH^T

For error in 7^th Bit

E = [000 0001]

\(S = \left[ {000\;000\;1} \right]\left[ {\begin{array}{*{20}{c}} 1&1&0\\ 0&1&1\\ 1&1&1\\ 1&0&1\\ 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

S = [ 0 0 1]

Extra information:

Syndrome for all possible errors

Error Pattern	Syndrome
0000000	000
0000001	001
0000010	010
0000100	100
0001000	101
0010000	111
0100000	011
1000000	110