# An 8 bit data 01101101 after transmission is received as 01001101. Explain how SEC code will detect and correct – IGNOU MCA Assignment 2015 – 16

By | August 31, 2015

MASTER OF COMPUTER APPLICATIONS

Course Code : MCS-012
Course Title : Computer Organisation and Assembly Language Programming
Assignment Number : MCA(I)/012/Assignment/15-16
Maximum Marks : 100
Weightage : 25%

An 8 bit data 01101101 after transmission is received as 01001101. 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  2                    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 ? ? 0 ? 1 1 0 ? 1 1 0 1

P1     =      ?        0        1        0        1        0

(Positions)  1        3        5        7        9        11

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

We get 01010

It’s even parity will be 0

P2     =      ?        0        1        0        1        0

(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 01010

It’s even parity will be 0

P3     =      ?        1        1        0        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 1101

It’s even parity will be 1

P3     =      ?        1        1        0        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 1101

It’s even parity will be 1

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

Assuming all parity sent correctly

The 8-bit Sent Data = 0110 1101

The 8-bit Sent Data = 0100 1101

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

P1=   D1     D2     D4     D5     D7 =   01010         even parity will be 0   – correct

P2=   D1     D3     D4     D6     D7=   00010         even parity will be 1   – incorrect

P3=   D2     D3     D4     D8           =    1001         even parity will be 0   – incorrect

P4=   D5     D6     D7     D8           =    1101         even parity will be 1   – 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.