The code words at the output of channel encoder of a (6, 2) linear block code are given as $C_{0} = [000000]; C_{1} = [010010]; c_{2} = [101101]; C_{3} = [111111]$
The minimum distance of the given linear block code is equal to _________.
2

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Solution

The correct option is A 2

$C_{j}$ $Code word$ $Hamming Weight$

$C_{0}$ $000000$ $0$

$C_{1}$ $010010$ $2$

$C_{2}$ $101101$ $4$

$C_{3}$ $111111$ $6$

Thus $d_{m i n} = 2$

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Similar questions

c_{0} = [000000]; c_{1} = [010010]; c_{2} = [101101]; c_{3} = [111111]

(d_{m i n})

T h e p a r i t y c h e c k m a t r i x o f a (7, 4) l i n e a r b l o c k c o d e s i s g i v e n b y, I f t h e c o d e w o r d r e c e i v e d a t t h e o u t p u t o f t h e c h a n n e l i s [1110100] . I f t h e r e i s a p o s s i b i l i t y f o r a t m o s t o n e - b i t e r r o r d u e t o c h a n n e l n o i s e, t h e n t h e c o r r e c t c o d e w o r d t r a n s m i t t e d a t t h e i n p u t o f t h e c h a n n e l w i l l b e . H = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 & 0 & 1 & 0 & 0 1 & 0 & 1 & 1 & 0 & 1 & 0 1 & 1 & 1 & 1 & 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

(d_{m i n})

d_{m i n} = 4

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