 An Error occurs when the information received by

the receiver does not match with the sender’s

information.
 During transmission, digital signals suffer from
noise that can introduce errors in the binary bits
travelling from sender to receiver.
 That means a 0 bit may change to 1 or a 1 bit may
change to 0.
 The various types of errors are:

https://www.faceprep.in/computer-networks/computer-networks-error-detection/
a) Single bit error

b) Multiple bit error

c) Burst error

https://www.faceprep.in/computer-networks/computer-networks-error-detection/
https://www.youtube.com/watch?v=EMrY-8m8D1E
Error detection techniques Error Correction
techniques
1. Simple Parity check Hamming Code
2. Two-dimensional Parity
check
ARQ (retransmission of loss
3. Checksum packets)
4. Cyclic redundancy
check

https://www.faceprep.in/computer-networks/computer-networks-error-detection/
 Error detecting codes are either implemented at the data link layer or
transport layer of OSI model.
 Whenever a message is transmitted, it may get scrambled by noise or
data may get corrupted.
 To avoid this, we use error-detecting codes which are additional data
added to a given digital message to help us detect if an error has
occurred during transmission of the message.

Basic approach used for error detection is the use of redundancy bits,
where additional bits are added to facilitate detection of errors.

 Some popular techniques for error detection are:

 1. Simple Parity check

 2. Two-dimensional Parity check
 3. Checksum
 4. Cyclic redundancy check

https://www.faceprep.in/computer-networks/computer-networks-error-detection/
1. Simple Parity check: Blocks of data from the source are subjected to a
check bit or parity bit generator form, where a parity of :
 1 is added to the block if it contains an odd number of 1’s, and
 0 is added if it contains an even number of 1’s

This scheme makes the total number of 1’s even, that is why it is called
even parity checking.

https://www.geeksforgeeks.org/error-detection-in-computer-networks/
2. Two-dimensional Parity check
Parity check bits are calculated for each row, which is equivalent
to a simple parity check bit. Parity check bits are also calculated
for all columns, then both are sent along with the data. At the
receiving end these are compared with the parity bits calculated
on the received data..

https://www.geeksforgeeks.org/error-detection-in-computer-networks/
2. Checksum

 In checksum error detection scheme, the data is

divided into k segments each of m bits.
 In the sender’s end the segments are added using
1’s complement arithmetic to get the sum. The sum
is complemented to get the checksum.
 The checksum segment is sent along with the data
segments.
 At the receiver’s end, all received segments are
added using 1’s complement arithmetic to get the
sum. The sum is complemented.
 If the result is zero, the received data is accepted;
otherwise discarded.

https://www.geeksforgeeks.org/error-detection-in-computer-networks/
Sender Side

Receiver Side

https://www.youtube.com/watch?v=AtVWnyDDaDI
3. Checksum

https://www.geeksforgeeks.org/error-detection-in-computer-networks/
0 + 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 = 10 (which is 0 carry 1)
4. Cyclic redundancy check (CRC)

• Unlike checksum scheme, which is based on

addition, CRC is based on binary division.
• In CRC, a sequence of redundant bits, called cyclic
redundancy check bits, are appended to the end of
data unit so that the resulting data unit becomes
exactly divisible by a second, predetermined binary
number.
• At the destination, the incoming data unit is divided
by the same number. If at this step there is no
remainder, the data unit is assumed to be correct
and is therefore accepted.
• A remainder indicates that the data unit has been
damaged in transit and therefore must be rejected.

https://www.geeksforgeeks.org/error-detection-in-computer-networks/
https://www.youtube.com/watch?v=A9g6rTMblz4
4. Cyclic redundancy check (CRC)

https://www.geeksforgeeks.org/error-detection-in-computer-networks/
Find the CRC of Data Block 100100 with the divisor 1101 ?

Sender Receiver

https://www.youtube.com/watch?v=A9g6rTMblz4
 Message Bits = M
Total Bits = M + R
 Redundant Bits = R
M M
Y
Correct
(Y/N)?
Encode / R Bits
Extractor /
Generator Checker
N

M+R

Discard
 Hamming codes are linear block codes
 It is an error correcting codes
 Parity bits are used here
 They are inserted in between the data bits
 The most commonly used is 7-bit hamming
code
 Structure (7-bit Hamming code)
7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

22 21 20
D – Data Bit
P – Parity Bit 2m = 0,1,2,3,….

https://www.youtube.com/watch?v=QSiuqhb7sw8
https://www.youtube.com/watch?v=m2dFjkZVlVw
 7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

111 110 101 100 011 010 001

P1 - > (1,3,5,7) These bits are adjusted to 0 or

1 depending upon the
P2 - > (2,3,6,7) condition
P3 - > (4,5,6,7)

https://www.youtube.com/watch?v=QSiuqhb7sw8
 Problem: A bit word 1011 is to be
transmitted. Construct even parity 7-bit
hamming code.
7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

1 0 1 1

P1 - > (1,3,5,7) P1, 1,1,1 P1 = 1

P2 - > (2,3,6,7) P2, 1,0,1 P2 = 0

P3 - > (4,5,6,7) P3, 1,0,1 P3 = 0

Hamming Code 1010101

https://www.youtube.com/watch?v=JeFT_XA-ryY
 Problem: A bit word 1010 is to be
transmitted. Construct even parity 7-bit
hamming code.
7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

1 0 1 0

P1 - > (1,3,5,7) P1, 0,1,1 P1 = 0

P2 - > (2,3,6,7) P2, 0,0,1 P2 = 1

P3 - > (4,5,6,7) P3, 1,0,1 P3 = 0

Hamming Code 1010010

https://www.youtube.com/watch?v=JeFT_XA-ryY
 Problem: Code received =1011011. Assume even
parity. State whether received code is correct or
wrong. If wrong, locate the bit in error.

7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

1 0 1 1 0 1 1

P1 - > (1,3,5,7) (1,0,1,1) – Odd (Error) Error Word P3, P2, P1

P2 - > (2,3,6,7) (1,0,0,1) - Even Error Word 1, 0, 1

P3 - > (4,5,6,7) (1,1,0,1) – Odd (Error) Error Word (5)

Error is in 5th bit
Correct Code 1001011
https://www.youtube.com/watch?v=JeFT_XA-ryY
 Problem: A bit word 1101 is to be
transmitted. Construct even parity 7-bit
hamming code.
7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

1 1 0 1
4th
1100110
P1 - > (1,3,5,7) P1, 1,0,1 P1 = 0
Error
P2 - > (2,3,6,7) P2, 1,1,1 P2 = 1
1101110
P3 - > (4,5,6,7) P3, 0,1,1 P3 = 0

Hamming Code 1100110

https://www.youtube.com/watch?v=JeFT_XA-ryY
 Problem: Code received =1101110. Assume even
parity. State whether received code is correct or
wrong. If wrong, locate the bit in error.

7 6 5 4 3 2 1

D4 D3 D2 P3 D1 P2 P1

1 1 0 1 1 1 0

P1 - > (1,3,5,7) (0,1,0,1) – Even Error Word P3, P2, P1

P2 - > (2,3,6,7) (1,1,1,1) - Even Error Word 1, 0, 0

P3 - > (4,5,6,7) (1,1,0,1) – Odd (Error) Error Word (4)

Error is in 4th bit
Correct Code 1100110
https://www.youtube.com/watch?v=JeFT_XA-ryY