ISI CSB 2017

Question

Consider a simple code in which each codeword consists of $2$ data bits $[d1, d0]$ and $3$ check bits $[c2, c1, c0]$.

The check bits are computed as follows:

$c2 = d1 \oplus d0$, where $\oplus$ is the modulo – $2$ sum

$c1 = d1$, and

$c0 = d0$.

(i) Determine the minimum Hamming distance between any two distinct codewords of this code.

(ii) How many errors in a codeword can be detected by this code? Justify your answer.

(iii) How many errors in a codeword can be corrected by this code? Justify your answer.

Comments

suppose , codeword is in  format :- c₀d₀c₁d₁c₂

now , if d₀ = d₁ = 0 then codeword will be 00000

 and  if  d₀ = 1 , d₁ = 0 then codeword will be 11001

so , minimum distance will be 3 b/w these codewords.

since , for "d" bit error detection , minimum distance >= d+1 ...so d = 2

and  for "d" bit error correction , minimum distance >= 2d+1 ...so d = 1

now , we can take any codeword format and put any combination of d₀ and d₁ like 00,01,10,11 and find the 4 possible codeword for each codeword format..minimum distance will be 3...

because when we put d₀ and d₁ as 00 then codeword = 00000 for each codeword format. now we can't  get any codeword in the same format when hamming distance will be 2 because when we put 01 as data bits then we get check bits as 011 . so we always get one extra 1 due to exor. like this we can analyze how we are not getting minimum distance as 2 for any codeword format..