Construct the Hamming code for the bit sequence10011101

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