Error detection and correction
codes
Objective of the lecture
• (i)Undersatnd concept of parity
• (ii)Apply parity method of error detection
• (iii)Make use of various error detection codes
• (iv)Use Hamming code
• When digital signals (group of 0’s and 1’s) are transmitted from one
circuit or system to another circuit or system,error may occur during
transmission. The change of a 1 to 0 or a 0 to 1 during transmission
is known as error. It is necessary to detect and correct the error to
obtain a correct message.
• A number of codes exist for detection and correction of error in
digital transmission.
• The concept of parity of a group of bits or digital word is the key for
error detection and correction.
•
• Parity method for error detection:-• Many system use a parity bit as a means for bit error detection.
• Any group of bits contain either even or an odd no. of 1’s.
• A parity bit is attached to a group of bit to make the total no. of 1’s in
a group always even or always odd.
• An even parity bit makes the total no.of 1’s even, and an odd parity
bit makes the total odd.
Even parity
P
BCD
0
0000
1
0001
1
0010
0
0011
1
0100
0
0101
0
0110
1
0111
1
1000
0
1001
Odd
P
1
0
0
1
0
1
1
0
0
1
parity
BCD
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
•
•
•
•
•
•
The parity bit can be attached to the code at either the beginning or the
end,depending on system design.
Notice that the total no.of 1’s including the parity bit,is always even for even
parity and always odd for odd parity.
Detecting an error : A parity bit provides for the detection of a single bit
error
Let us assume BCD code 0101
(parity can be used with any no.of bits:we are using four for illustration)
The total code transmitted including parity bit,is
• Let us assume the error occurs in the third
• bit from the left
• Redundancy bit:To calculate the no. of redundancy bits(r) required to
include in data bits(m).
• 2r >= m+r+1
• For example for m=1 r= 2
• For m=2
r= 3
• For m=3
r=3
• For m=4
r=3
• For m=5 r=4
Hamming codes
•
Q-Determine the single error correcting code for the BCD
•
•
•
number 1001 using even parity.
Sol:- Step-1 Find the no of parity bit required :-that is = 3
Step-2 Construct a bit position table,and enter the information
bits.Parity bits are determined in the following steps.
Bit
P1
Designation
P2
MI
P3
M2 M3
M4
Bit position
1
2
3
4
5
7
Binary
position no
001
010 011
100
101 110
111
0
1
Information
bit
Parity bit
1
6
0
•
•
•
•
Step-3:Determine the parity bit as follows
Bit P1 checks bit position 1,3,5 and 7 and put 0 for an even no. of 1s in this
group.
Bit P2 checks bit position 2,3,6,and 7 and put 0 for even no of 1s in this
group
Bit P3 checks bit position 4,5,6 and7 and put 1 for even no of 1s in this
group
Bit Designation
P1
P2
MI
P3
M2
M3
M4
Bit position
1
2
3
4
5
6
7
Binary position no
001
010
011
100
101
110
111
0
0
1
Information bit
Parity bit
1
0
0
1
Q-2 Determine the single error correcting code for the
information code 10110 for odd parity.
• Sol:---- Step-1 Find the no of parity bit required :-- that is = 5
•
Step-2 Construct a bit position table,and enter the information
bits.Parity bits are determined in the following steps.
Bit Designation
P1
P2
MI
P3
M2
M3
M4
P4
M5
Bit position
1
2
3
4
5
6
7
8
9
Binary position
no
0001
0010
0011
0100
0101
0110
0111
1000
1001
0
1
1
Information bit
Parity bit
1
0
•
•
•
•
•
Step-3:Determine the parity bit as follows
Bit P1 checks bit position 1,3,5 ,7and 9 and put 1 for an odd no. of 1s in this group.
Bit P2 checks bit position 2,3,6,and 7 and put 0 for odd no of 1s in this group
Bit P3 checks bit position 4,5,6 and7 and put 1 for odd no of 1s in this group.
Bit P4 checks bit position 8 and 9 and must be 1 for there to be an odd no of 1s in
the group
Bit Designation
P1
P2
MI
P3
M2
M3
M4
P4
M5
Bit position
1
2
3
4
5
6
7
8
9
Binary position
no
0001
0010
0011
0100
0101
0110
0111
1000
1001
0
1
1
Information bit
Parity bit
1
1
0
1
0
1
• Q-3 Assume that the code word in (0011001) is transmitted and
0010001 is received. Receiver does not know what is transmitted
and must look for proper parities to determine if the code is correct.
Designate any error that has occurred in transmission if even parity
is used.
Bit Designation
P1
P2
MI
P3
M2
M3
M4
Bit position
1
2
3
4
5
6
7
Binary position no
001
010
011
100
101
110
111
Received code
0
0
1
0
0
0
1
•
•
•
•
•
•
•
•
•
First parity check
P1 for position1,3,5,7
Parity check is good-------------------0(LSB)
Second parity check
P2 for position2,3,6,7
Parity check is good-------------------0
Third parity check
P3 for position4,5,6,7
Parity check is bad-------------------1(MSB)
• RESULT: The error position code is
100.This says that bit in position 4 is in
error.It is a zero and should be 1.The
corrected code is 0011001,which agrees
with the transmitted code.
• The code 101101010 is received.Correct any
errors.There are four parity bits,and odd parity is used.
Bit
Designation
P1
P2
MI
P3
M2
M3
M4
P4
M5
Bit position
1
2
3
4
5
6
7
8
9
0
1
0
Information
bit
Parity bit
1
1
0
1
0
1
•
•
•
•
•
•
•
•
•
•
•
•
•
•
First parity check
P1 for position1,3,5,7 and 9
Parity check is bad-------------------1(LSB)
Second parity check
P2 for position 2,3,6,7
Parity check is bad-------------------1
Third parity check
P3 for position 4,5,6,7
Parity check is bad-------------------1
Fourth parity check:
Bit 4 checks positions 8 and 9.
Parity check is good-------------------0(MSB)
RESULT:
The error position code is 0111.This says that the bit in position 7 is in
error.The corrected code is therefore 101101110
Bit Designation
P1
P2
MI
P3
M2
M3
M4
Bit position
1
2
3
4
5
6
7
Binary position no
001
010
011
100
101
110
111
Information bit
Parity bit