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 21 22 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.