An 8 bit data 10110011 after transmission is received as 10010011. Explain how SEC code will detect and correct – IGNOU MCA Assignment 2014 – 15

By | September 27, 2014

MASTER OF COMPUTER APPLICATIONS

Course Code : MCS-012

Course Title : Computer Organisation and Assembly Language Programming

Assignment Number : MCA (2)/012/Assign /2014-15

Maximum Marks : 100

Weightage : 25%

 

An 8 bit data 10110011 after transmission is received as 10010011. Explain how SEC code will detect and correct this problem.

 

SEC means Single Error Correction

This if found in Hamming Error Correction Code

First find out the number parity bits in SEC Code

Formula:

2i – 1 >= N + i

where

i = number of parity bits in SEC Code

N = number of  bits in Data Word

 

In this case

N=8

i=?

So,

2i – 1 >= N + I                at  i=3

 

23 – 1 >= 8 + 3

7 >= 11 (Not True)

So,

2i – 1 >= N + I                at  i=4

 

24 – 1 >= 8 + 4

15 >= 12 (True)

Condition Satisfied.

 

 

Coorection bits (parity bits) are 4.

 

Parity bits are placed at 20  222  23                     i.e.   1 2 4 8          respectively

 

P1

P2

D1

P3

D2

D3

D4

P4

D5

D6

D7

D8

1

2

3

4

5

6

7

8

9

10

11

12

?

?

?

?

 

D1 D2 D3 D4 D5 D6 D7 D8 are the data sent and recd.

 

Before sending calculate Parity of  Data to be sent.

 

P1

P2

D1

P3

D2

D3

D4

P4

D5

D6

D7

D8

1

2

3

4

5

6

7

8

9

10

11

12

?

?

1

?

0

1

1

?

0

0

1

1

 

P1     =      ?        1        0        1        0        1

(Positions)  1        3        5        7        9        11

 

Starting Point will be after Parity position P1 i.e 2 and Take 1 Skip 1till the end of table data.

We get 10101

It’s even parity will be 1

 

P2     =      ?        1        1        1        0        1

(Positions)  2        3        6        7        10      11

 

Starting Point will be after Parity position P2 i.e 3 and Take 2 Skip 2 till the end of table data.

We get 11101

It’s even parity will be 0

 

P3     =      ?        0        1        1        1

(Positions)  4        5        6        7        12

 

Starting Point will be after Parity position P3 i.e 4 and Take 4 Skip 4 till the end of table data.

We get 0111

It’s even parity will be 1

 

P3     =      ?        0        0        1        1

(Positions)  8        9        10      11      12

 

Starting Point will be after Parity position P3 i.e 4 and Take 8 Skip 8 till the end of table data. (Note:- data ends before taking 8 elements.)

We get 0011

It’s even parity will be 0

 

We found all parity bits, we will fill in the table:-

 

P1

P2

D1

P3

D2

D3

D4

P4

D5

D6

D7

D8

1

2

3

4

5

6

7

8

9

10

11

12

1

0

1

1

0

1

1

0

0

0

1

1

 

Assuming all parity sent correctly

The 8-bit Sent Data = 1011 0011

The 8-bit Sent Data = 1001 0011

 

Check with parity bits, before that Create a new parity bit with new data

 

P1=   D1     D2     D4     D5     D7=   10101         even parity will be 1   - correct

P2=   D1     D3     D4     D6     D7=   10101         even parity will be 1   - incorrect

P3=   D2     D3     D4     D8=             10101         even parity will be 0   - incorrect

P4=   D5     D6     D7     D8=             10101         even parity will be 0   - correct

 

After Checking

We find that common data bit numbers are D3 & D4

But D4 is also present in P1that means error has been occurred in D3 only.

 By this we dectect error in D3 and Correct it by replacing it by 0 to 1.