1. 7/2/2015Errors1 Transmission errors are a way of life. In the digital world an error means that a bit value is flipped. An error can be isolated to a single bit. Errors on some media come in bursts –Harder to detect and correct than isolated errors.

2. 7/2/2015Errors2 Dealing with Errors Error detecting codes –provide enough redundant information to enable the receiver to deduce that an error occurred Error correcting codes –provide enough redundant information to enable the receiver to deduce that an error occurred AND how to fix it So a message consists of m data bits and r redundant or check bits.

3. 7/2/2015Errors3 Hamming Distance Hamming distance: –the number of bit positions in which two codewords differ Simple to calculate find the XOR If two codewords are a Hamming distance d apart, it will require d single-bit errors to convert one into the other. The Hamming distance of a code is the minimum Hamming distance between any two codewords.

4. 7/2/2015Errors4 Hamming Distance 2 Code 000 011 101 110 Note that not all of the 8 different bit patterns are included in the code Any single error will convert a valid codeword into an invalid codeword. If we know the valid codewords we can detect the error

5. 7/2/2015Errors5 How Error Detection Works Invalid Code Words Valid codeword Valid codeword 2e2e

6. 7/2/2015Errors6 Parity A simple single error detecting code could be constructed by counting bits. –Any codeword with an even number of bits is consider valid (you could also make it the other way around). The minimum distance of this code is 2, so it is capable of detecting single errors. This code can be created by adding a parity bit: –chose the parity bit so that the number of ones in the codeword is even (or odd).

7. 7/2/2015Errors7 Parity in Action Want to send: 10 Data Link Sends: 110 Receive: 111 (ERROR)

8. 7/2/2015Errors8 Protecting Blocks The probability of detecting a burst error on a block using a single parity bit is 50%. This can be improved by viewing the block as a n by k bit matrix. A parity bit is then computed for each column. The check bits are placed in a k-bit row and affixed to the matrix as the last row. Bursts of length n can be detected.

9. 7/2/2015Errors9 Detecting Burst Errors 10010000 11000011 11011011 11010010 11011101 11001111 01000001 11000110 11011110 11001001 11001010 11001001 1001000 1100001 1101101 1101001 1101110 1100111 0100000 1100011 1101111 1100100 1100101 ? Data VRC (Vertical Redundancy Check) LRC (Longitudinal Redundancy Check) n

10. 7/2/2015Errors10 What About Error Correction? How do we get error correction? –Must increase the minimum distance of the code The key to error correction is that it must be possible to detect and locate the error. The minimum distance must be at least 2e+1

11. 7/2/2015Errors11 Error Correction Valid codeword Valid codeword Invalid codewords The +1 ensures the circles will not overlap

12. 7/2/2015Errors12 A Simple Single Error Correcting Code

13. 7/2/2015Errors13 Hamming Codes Hamming codes are n-bit codes that can correct single errors. The basic idea is to split the codeword into two portions –information or message bits (m) –parity bits (k) The result are codewords that consist of m+k bits

14. 7/2/2015Errors14 Choosing m and k Selecting m is easy, you are usually told what it is. How do you pick k? The parity bits are used to generate a k-bit word that identifies where in the codeword the error, if any, occurred. Consequently, k must satisfy the following:

15. 7/2/2015Errors15 Constructing the Codeword The codeword consists of m+k bits. The location of each of the m+k bits is assigned a decimal value, 1 is assigned to the MSB, and m+k to the LSB. Parity bits go in positions 1, 2, 4, …, 2 k-1 12435876...m+k9 p0p0 p1p1 m0m0 p2p2 m1m1 m2m2 m3m3 p3p3 m4m4...m m+k

16. 7/2/2015Errors16 Parity Checks The parity checks must be specified so that when an error occurs, the position number will take on the the value assigned to to location of the error

17. 7/2/2015Errors17 Putting It Together

18. 7/2/2015Errors18 Example Send the message 0010 using a hamming code Step 1: Find k. Here k=3 Step 2: Determine where things go Step 3: Figure out the parity bits p1 will cover 1,3,5,7,9,11,… p2 will cover 2,3,6,7,10,11,… p3 will cover 4,5,6,7,12,13,14,15,...

19. 7/2/2015Errors19 Correcting Burst Errors Hamming codes can be used to correct burst errors A sequence of s consecutive codewords are arranged as a matrix, one codeword per row. Transmit data one column (s bits) at a time. The matrix is reconstructed by the receiver one column at a time. If a burst error of size s occurs, only a single column will be affected.

20. 7/2/2015Errors20 Correcting Burst Errors 00110010000 10111001001 11101010101 01101011001 01101010110 01111001111 10011000000 11111000011 10101011111 11111001100 00111000101 1001000 1100001 1101101 1101001 1101110 1100111 0100000 1100011 1101111 1100100 1100101 HammIngcodeHammIngcode CharacterASCIICheck Bits s

21. 7/2/2015Errors21 Correcting vs. Detecting Most often error detection followed by retransmission is more efficient. Consider a channel with an error rate is 10 -6 per bit (one error per million bits sent) –Block size 1000 == 10 check bits ( k == 10 ) –For parity one check bit will suffice Overhead for sending 1MB –Hamming == 10,000 bits –Parity == 2001 bits (since 1 block will be retransmitted)

22. 7/2/2015Errors22 Checksums Both sides agree on a checksum function Sender –Computes checksum while sending message –Attaches result to the end of the message Receiver –Computes checksum while reading message –Compares result to checksum at end of message

23. 7/2/2015Errors23 Error Detection

24. 7/2/2015Errors24 Basic Idea Treat the entire message as a binary number To calculate the checksum –Divide message by another fixed number –Use the remainder as the checksum CRC treats bit strings as representations of polynomials with coefficients of 0 and 1. –110001 == x 5 + x 4 + x 0

25. 7/2/2015Errors25 The Generator Polynomial Both the sender and the receiver must agree upon a generator polynomial, G(x). –Both the high and low order bits of the generator must be 1. –The length of the generator is one bit longer than the FCS. –Finally the frame must be longer than the generator. This is what we use mathematicians for

26. 7/2/2015Errors26 Standard Polynomials CRC-12 (x 12 +x 11 +x 3 +x 2 +x 1 +1) –used when the character length is 6 CRC-16 (x 16 +x 15 +x 2 +1) CRC-CCITT (x 16 +x 12 +x 5 +1) –used for 8 bit characters –catches all single and double errors –all errors of an odd length –all bursts of 16-bits or less, 99.997% of 17-bits, and 99.998% of 18-bits and longer.

27. 7/2/2015Errors27 The Algorithm To compute the checksum –Append n 0s to the end of the message, where n is the number of bits in the checksum –The resulting value is divided by the generator polynomial –Each division step is carried out in the conventional manner, except that we use polynomial arithmetic

28. 7/2/2015Errors28 Polynomial Arithmetic 1 1 0 0 -1-0-1-0 -------- 0 1 1 0 1 1 0 0 +1+0+1+0 -------- 0 1 1 0 Subtraction and addition as usual but no borrows or carries Both operations are identical to XOR

29. 7/2/2015Errors29 Polynomial Arithmetic Addition and subtraction, are a single operation, that is its own inverse By collapsing addition and subtraction, the arithmetic discards any notion of magnitude –Beyond the power of the highest bit 1010 is clearly greater than 10 1010 is no longer greater than 1001 –1010 = 1001 + 0011 –1010 = 1001 - 0011

30. 7/2/2015Errors30 Polynomial Multiplication 1101 x 1011 ---- 1101 11010 000000 1101000 ------- 1111111

31. 7/2/2015Errors31 Polynomial Division 1101 1011 1111111 1011 1001 1011 0

32. 7/2/2015Errors32 The Algorithm (continued) The division produces a quotient which is discarded. The remainder replaces the 0s appended to the frame (subtracted from the frame modulo 2). The resulting frame is now evenly divisible by the generator polynomial. The receiver performs the same division, a non- zero remainder indicates that an error occurred.

33. 7/2/2015Errors33 CRC Example (transmit) Frame contents:111011 Polynomial:11101 (x 4 + x 3 +x 2 + x 0 ) Frame with 0s:1110110000 100001 11101 1110110000 11101 ----- 10000 11101 ----- 1101 Frame to send:1110111101

34. 7/2/2015Errors34 CRC Example (receive) Frame contents:1110111101 Polynomial:11101 (x 4 + x 3 +x 2 + x 0 ) 100001 11101 1110111101 11101 ----- 11101 ----- 0

35. 7/2/2015Errors35 Fast Polynomial Division Start 0000000 1111111 0000000 Shift 0000001 1111110 0000000 Shift 0000011 1111100 0000000 Shift 0000111 1111000 0000000 Shift 0001111 1110000 0000000 XOR/Inc 0000100 1110000 0000001 Shift 0001001 1100000 0000010 XOR/Inc 0000010 1100000 0000011 Shift 0000101 1000000 0000110 Shift 0001011 0000000 0001100 XOR/Inc 0000000 0000000 0001101

36. 7/2/2015Errors36 Optimize For CRC –We do not need the quotient –If the divisor is W bits long The remainder will be at most W-1 bits long Only need 1 register

37. 7/2/2015Errors37 Faster Polynomial Division Start 000 1111111 Shift 001 111111 Shift 011 11111 Shift 111 1111 Shift 1 111 111 XOR 100 111 Shift 1 001 11 XOR 010 11 Shift 101 1 Shift 1 011 XOR 000

38. 7/2/2015Errors38 CRC Simple Version Consider the polynomial 10111 with a CRC of size W=4 To perform the division perform the following: –Load the register with zero bits. –Augment the message by appending W zero bits to the end of it. –While (more message bits) Shift the register left by one bit, reading the next bit of the augmented message into register bit position 0. If (a 1 bit popped out of the register during step 3) –Register = Register XOR Poly. The register contains the remainder. Source: http://www.repairfaq.org/filipg/LINK/F_crc_v33.html