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If the original size of data is $40$ then after adding error detection redundancy bit the size of data length is

1. $26$
2. $36$
3. $46$
4. $56$

### 1 comment

C ??

2r >= m+r+1

for r=6 it satisfies i.e  64>= 40+6+1

thus m+r = 46

### 1 comment

Can you point to a video resource if possible ? I am unable to understand it
→ Imagine that we want to design a code with m message bits and r check bits that will allow all single errors to be corrected.
→ Each of the 2​ m​ legal messages has n illegal codewords at a distance of 11 from it.
→ These are formed by systematically inverting each of the n bits in the n-bit codeword formed from it. Thus, each of the 2​ m​ legal messages requires n+1 bit patterns dedicated to it.
→ Since the total number of bit patterns is 2​^n​ ,​ We must have (n+1)2​^m​ ≤ 2​^n​ .
→ Using n=m+r, this requirement becomes
= (m+r+1) ≤ 2​^r
= 40+6+1 ≤ 2^6
= 47 ≤ 2
r=6
Message size will be 6+40=46
by
As for adding redundant bits it must follow

2^r >= m + r +1

Where r is number of redundant bits and m is number of data bits.

So,  for r=6

64 > 40 + 6 +1

Therefore size of codeword becomes m+r == 40 +6 ==46

option c