Decode the following words relative to a maximum likelyhood decoding function. i)11001 ii) 01010 iii) 00111

95views

written 7.9 years ago by

teamques10 ★ 68k

To decode the word 11001 we first arrange the given code as

$00000 \ \ \ \ \ 00110 \ \ \ \ \ 01001 \ \ \ \ \ 01111 \ \ \ \ \ 10011 \ \ \ \ \ 10101 \ \ \ \ \ 11010 \ \ \ \ \ 11000$

Then

$d(11001, 00000)= 3 \ \ \ \ \ \ \ \ \ \ d(11001, 00110)=5 \\ d(11001, 01001)=2 \ \ \ \ \ \ \ \ \ \ d(11001, 01111)=2 \\ d(11001, 10011)=3 \ \ \ \ \ \ \ \ \ \ d(11001, 10101)=2 \\ d(11001, 11010)=4 \ \ \ \ \ \ \ \ \ \ d(11001, 11000)=3$

We observe that 01001, 01111 and 10101 are equidistance from 11001. But 01001 comes first in our list of codes. Since e(010)=01001, we decode 11001 as 010.
To decode the word 01010 we first arrange the given code as

$00000 \ \ \ \ \ 00110 \ \ \ \ \ 01001 \ \ \ \ \ 01111 \ \ \ \ \ 10011 \ \ \ \ \ 10101 \ \ \ \ \ 11010 \ \ \ \ \ 11000$

Then

$d(01010, 00000)= 2 \ \ \ \ \ \ \ \ \ \ d(01010, 00110)=1 \\ d(01010, 01001)=3 \ \ \ \ \ \ \ \ \ \ d(01010, 01111)=3 \\ d(01010, 10011)=4 \ \ \ \ \ \ \ \ \ \ d(01010, 10101)=5 \\ d(01010, 11010)=2 \ \ \ \ \ \ \ \ \ \ d(01010, 11000)=2$

We observe that 00110 are equidistance from 01010. Since e (001) = 00110, we decode 01010 as 001.
To decode the word 00111 we first arrange the given code as

$00000 \ \ \ \ \ 00110 \ \ \ \ \ 01001 \ \ \ \ \ 01111 \ \ \ \ \ 10011 \ \ \ \ \ 10101 \ \ \ \ \ 11010 \ \ \ \ \ 11000$

Then

$d(00111, 00000)= 3 \ \ \ \ \ \ \ \ \ \ d(00111, 00110)=3 \\ d(00111, 01001)=2 \ \ \ \ \ \ \ \ \ \ d(00111, 01111)=3 \\ d(00111, 10011)=2 \ \ \ \ \ \ \ \ \ \ d(00111, 10101)=3 \\ d(00111, 11010)=2 \ \ \ \ \ \ \ \ \ \ d(00111, 11000)=5$

We observe that 01001, 10011and 11010 are equidistance from 00111. But 01001 comes first in our list of codes. Since e (010) =01001, we decode 00111 as 010.

ADD COMMENT EDIT