Assuming we are using the hamming algorithm presented in your text and even parity to design an error-correcting code, find the code word to represent the 8-bit information word 10011011

Before redundancy :100101011100

After checking parity redundancy: 100101010011

Explanation:

First calculate redundant bits ,

We know that the number of redundant bits can be calculated using the following formula:

2^r ≥ m + r + 1

r = redundant bit, m = data bit

total data bits = 8

so ,

2^4 ≥ 8 + 4 + 1

16 ≥ 13

so, redundant bits is 4

Now let they denoted by r1, r2, r4, and r8, as the redundant bits are placed at positions corresponding to power of 2:    1, 2, 4, and 8.  

All the redundant bits are initialized by zero.

For the data word 10011011, we can let  the bits of the data word as  w8,w7, w6, w5, w4, w3, w2, w1

Now , put data bits and redundancy bits  as follows.

d12  d11  d10  d9  d8   d7  d6  d5  d4  d3  d2  d1

w8   w7  w6  w5   r8  w4  w3  w2 r4   w1   r2  r1

 1      0    0      1      0     1     0     1    0    0    0   0

r1 = d1  xor  d3  xor d5  xor d7  xor  d9  xor d11

r1 =  0  xor  0  xor 1  xor 1  xor  1  xor 0

r1 =  1

r2 = d2  xor  d3  xor d6  xor d7  xor  d10  xor d11

r2 = 0  xor  0  xor 0  xor 1  xor  0  xor 0

r2 = 1

r4 = d4  xor  d5  xor d6  xor d7  

r4 = 0  xor  1  xor 0  xor 1

r4 = 0

r8 = d8  xor  d9  xor d10  xor d11   xor d12

r8 = 0  xor  1  xor 0  xor 0 xor  1

r8 = 0

So, the data transferred is

d12  d11  d10  d9  d8   d7  d6  d5  d4  d3  d2  d1

w8   w7  w6  w5   r8  w4  w3  w2 r4   w1   r2  r1

 1      0    0      1      0     1     0     1    0    0    1   1

The bits give the binary number as 0011 whose decimal representation is 3. Thus, the bit 3 contains an error. To correct the error the 3th bit is changed from 1 to 0.