Respuesta :
Answer:
Bit sequence 10011101
For finding Redundant bits ,we check-
2^r >= r+m+1
m is no of bits.
We check from 1,m=8
2>10 which is wrong.
We check from 2,m=8
4>11 which is wrong.
We check from 3,m=8
8>12 which is wrong.
We check from 4,m=8
16>13 which is right
So r=4, there are 4 redundant bits;
r1,r2,r4,r8
Inserting redundant bits in bit sequence-
1 0 0 1 r8 1 1 0 r4 1 r2 r1
r1 is the first parity bit
r2 is the second parity bit
r4 is the third parity bit
r8 is the forth parity pit
We will use even parity bit-we have to make no of 1's = even , if already no of ones is even, then simply r = 0 else r=1.
for r1, we will check at the positions which consist one's at the first position(1,3,5,7,9,11) .
for r2, we will check at the positions which consist one's at the second position (2,3,6,7,10,11) .
for r4, we will check at the positions which consist one's at the forth position(4,5,6,7,12) .
for r8, we will check at the positions which consist one's at the eighth position(8,9,10,11,12) .
Calculating r1
as bit no. 3 = 1
bit no. 5 = 0
bit no. 7 = 1
bit no. 9 = 1
bit no. 11 = 0
to make no. of 1's even we have r1 = 1
For r2
count no of 1's at bit (2,3,6,7,10,11)
No. of 1's are = 3 (on bit 3,6,7)
As 3 is odd
for even parity,we have to make it 1
Similarly ,
r4 =1
r8 =0
Hence Following are the redundancy bits.
r1 = 1
r2 = 1
r4 = 1
r8 = 0
So the Hamming code is 100101101111
