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Error detection and Correction
- 1. Gandhinagar Institute ofTechnology MCWC ALA Topic: Error Correction and Error Detection Prepared By: Tarj Mehta (170120107074) Guided By: Prof. Parita Shah
- 2. Introduction • For reliablecommunication, errors must be detected and corrected. • The data may get corrupted during transmission. • The data link layer uses some error control mechanism to ensure that frames (stream of data bits) are transmitted with accuracy. • In OSI model, error detection and correction are implemented either at data link layer or transport layer.
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- 4. Error Detection • Ituses concept of redundancy, means adding extra bits for detecting errors at receiving-end. • There are 3 methods: 1. Parity Check 2. Cyclic redundancy check 3. Checksum
- 5. Parity Check • Anextra bit is added at the end of data and number of 1’s remain even. • If block has even number of 1’s, 0 is added at the end. • If block has odd number of 1’s, 1 is added at the end. • On the receiver side parity bit is computed and check with received data.
- 6. • Sender: Data: 11 0 1 0 0 As it has odd number of 1’s, 1 will be added With parity: 1 1 0 1 0 0 1 • This will be transmitted via transmitting media. • Receiver: Parity bit will be computed. If parity bit are correct (1 1 0 1 0 0 1) then only data will be accepted otherwise it will be rejected. • Disadvantage: Error in more than one bit cannot be detected.
- 7. Cyclic Redundancy check •In this method parity bits are calculated for each row and column and are sent along with data to the receiver. • At receiving end these are compared with the parity bits calculated on the received data. • Suppose, the given data: 1101101 and divisor: 10101 • Append 4 zeros at the end of data (divisor – 1) • Division is done by X – OR.
- 8. _1110111______ 10101 | 11011010000 |10101 011100 10101 010011 10101 001100 00000 011000 10101 011010 The remainder is not 0, so there is some 10101 error. Now, Append the remainder to data 011110 10101 01011
- 9. 1110111______ 10101| 11011011011 | 10101 011100 10101 010011 10101 001101 00000 011010 10101 011111As the remainder is 0, the received data is error free 10101 010101 10101 00000
- 10. Checksum • The givendata is divided into data segments of k size. • Each segment is added using 1’s complement method and at the end they are complemented and the check sum is sent along with data segments. • At the receiving end all the segments are added with checksum. • If the answer is 0, the data is accepted. • If there is any tampering with the bits, the answer will not be 0 and error can be detected.
- 11. Example • Question: 1011001110101011 01011010 11010101 10110011 10101011 1| 01011110 1 01011111 01011010 10111001 11010101 1| 10001111 1 10001111 Now 1’s complement is: 0111000 (checksum)
- 12. • Performing theaddition again with check sum block… 10110011 10101011 1| 01011110 1 01011111 01011010 10111001 11010101 1| 10001110 1 10001111 01110000 11111111 Complementing this we get 00000000, hence data will be accepted.
- 13. Error correction • Ithas 1 technique: 1. Hamming code
- 14. Hamming Code • Itis a set of error correction codes that can be used to detect and correct the errors encountered during transmission. • It has various steps to detect and correct the error. • Example: A 7bit hamming code is received as 1011011. Assume even parity and state whether the received code is correct or wrong, if wrong locate the bit in error.
- 15. • Solution: ReceivedHamming code is • Step1: Detecting error Analyzing bits P1 D3 D5 D7 (skip one take one) 1 0 1 1 = odd parity = error exists Now, keeping P1=1 Step2: Analyzing bits P2 D3 D6 D7 1 0 0 1 = even 1’s = no error D7 D6 D5 P4 D3 P2 P1 1 0 1 1 0 1 1
- 16. • Step3: Analyzingbits P4 D5 D6 D7 1 1 0 1 = odd 1’s = error exists Keeping P4=1 Here, P4 and P1 are not equal to zero. Correcting Error: E = Decimal value for 101 is 5, which shows that 5th bit is error bit. So, we write correct word by simply inverting 5th bit. Final answer: 1001011 P4 P2 P1 1 0 1
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