Question

Instruction: For every question you answer, please show the calculation steps, if you do not show the calculation step and show only the final result, you will not receive any score. You can use online resources for conversion, if the conversion is not a part of the question. There are a total of eight questions.

Question 5: Suppose we are working with an error-correcting code that will allow all single-bit errors to be corrected for memory words of length 7. We have already calculated that we need 4 check bits, and the length of all code words will be 11. Code words are created according to the Hamming Algorithm presented in the text. We now receive the following code word:

1 0 1 0 1 0 1 1 1 1 0

Assuming even parity, is this a legal code word? If not, according to our error-correcting, where is the error?

Answer 1

Solution

We know at sender end parity bits are at positions 1 , 2 , 4 and 8.

D11

D10

D9

P8

D7

D6

D5

P4

D3

P2

P1

Parity bits are calculated at receiver end . The decimal equivalent of the parity bits (c1c2c3c4) is calculated. If it is 0, there is no error. Otherwise, the decimal value gives the bit position which has error.

c1 = EvenParity(1, 3, 5, 7, 9, 11)

c2 = EvenParity(2, 3, 6, 7, 10, 11)

c3 = EvenParity(4,5,6,7)

c4= EvenParity(8,9,10,11)

Code word received is :

1

0

1

0

1

0

1

1

1

1

0

Bit numbers for above codeword are:

11

10

9

8

7

6

5

4

3

2

1

c1= 0 XOR 1 XOR 1 XOR 1 XOR 1 XOR 1 = 1

c2= 1 XOR 1 XOR 0 XOR 1 XOR 0 XOR 1 = 0

c3 = 1 XOR 1 XOR 0 XOR 1 = 1

c4 = 0 XOR 1 XOR 0 XOR 1 = 0

Bit error position will be C1C2C3C4 = 1010 = 10 , since this is non zero so this code word is not legal.

So the error is in bit D10 so we can invert this bit from 0 to 1, , so correct pattern was 1 1 1 0 1 0 1 1 1 1 0