Error Detect and Correct.ppt
- 2. 10.2 Data can be corrupted during transmission. Some applications require that errors be detected and corrected. Note
- 3. 10.3 10-1 INTRODUCTION Let us first discuss some issues related, directly or indirectly, to error detection and correction. Types of Errors Redundancy Topics discussed in this section:
- 4. 10.4 Figure 10.1 Single-bit error
- 5. 10.5 Figure 10.2 Burst error of length 8
- 6. 10.6 Figure 10.3 The structure of encoder and decoder To detect or correct errors, we need to send redundant bits
- 7. 10.7 Figure 10.11 Two-dimensional parity-check code
- 8. 10.8 Figure 10.11 Two-dimensional parity-check code
- 9. 10.9 Figure 10.11 Two-dimensional parity-check code
- 10. 10.10 10-4 CYCLIC CODES Cyclic codes are special linear block codes with one extra property. In a cyclic code, if a codeword is cyclically shifted (rotated), the result is another codeword. Cyclic Redundancy Check Hardware Implementation Polynomials Cyclic Code Analysis Advantages of Cyclic Codes Other Cyclic Codes Topics discussed in this section:
- 11. 10.11 Cyclic Redundancy Checksum The CRC error detection method treats the packet of data to be transmitted as a large polynomial. The transmitter takes the message polynomial and using polynomial arithmetic, divides it by a given generating polynomial. The quotient is discarded but the remainder is “attached” to the end of the message.
- 12. 10.12 Cyclic Redundancy Checksum The message (with the remainder) is transmitted to the receiver. The receiver divides the message and remainder by the same generating polynomial. If a remainder not equal to zero results, there was an error during transmission. If a remainder of zero results, there was no error during transmission.
- 13. 10.13 More Formally M(x) - original message treated as a polynomial To prepare for transmission: Add r 0s to end of the message (where r = degree of generating polynomial) Divide M(x)xr by generating polynomial P(x) yielding a quotient and a remainder Q(x)+R(x)/P(x).
- 14. 10.14 More Formally Add (XOR) remainder R(x) to M(x)xr giving M(x)xr+R(x) and transmit. Receiver receives message (M(x)xr+R(x)) and divides by same P(x). If remainder is 0, then there were no errors during transmission. (Any expression which has exactly P(x) as a term is evenly divisible by P(x).)
- 15. Data word to be sent - 100100 Key - 1101 10.15
- 16. Sender Side Therefore, the remainder is 001 and hence the encoded data sent is 100100001. 10.16
- 17. Receiver Side Code word received at the receiver side 100100001 10.17 Therefore, the remainder is all zeros. Hence, the data received has no error
- 18. Error in transmission Sender’s Side Data word to be sent - 100100 Key - 1101 10.18 Therefore, the remainder is 001 and hence the code word sent is 100100001.
- 19. Receiver’s Side Let there be error in transmission media Code word received at the receiver side - 100000001 10.19 Since the remainder is not all zeroes, the error is detected at the receiver side.
- 20. 10.20 Common CRC Polynomials CRC-12: x12 + x11 + x3 + x2 + x + 1 CRC-16: x16 + x15 + x2 + 1 CRC-CCITT: x16 + x15 + x5 + 1 CRC-32: x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1 ATM CRC: x8 + x2 + x + 1
- 21. 10.21 CRC Example Given a pretend P(x) = x5 + x4 + x2 + 1 and a message M(x) = 1010011010, calculate the remainder using long hand division and a shift register.
- 22. 10.22 10-5 CHECKSUM The last error detection method we discuss here is called the checksum, or arithmetic checksum. The checksum is used in the Internet by several protocols although not at the data link layer. However, we briefly discuss it here to complete our discussion on error checking Idea One’s Complement Internet Checksum Topics discussed in this section:
- 23. 10.23 Suppose our data is a list of five 4-bit numbers that we want to send to a destination. In addition to sending these numbers, we send the sum of the numbers. For example, if the set of numbers is (7, 11, 12, 0, 6), we send (7, 11, 12, 0, 6, 36), where 36 is the sum of the original numbers. The receiver adds the five numbers and compares the result with the sum. If the two are the same, the receiver assumes no error, accepts the five numbers, and discards the sum. Otherwise, there is an error somewhere and the data are not accepted. Example 10.18
- 24. 10.24 We can make the job of the receiver easier if we send the negative (complement) of the sum, called the checksum. In this case, we send (7, 11, 12, 0, 6, −36). The receiver can add all the numbers received (including the checksum). If the result is 0, it assumes no error; otherwise, there is an error. Example 10.19
- 25. 10.25 How can we represent the number 21 in one’s complement arithmetic using only four bits? Solution The number 21 in binary is 10101 (it needs five bits). We can wrap the leftmost bit and add it to the four rightmost bits. We have (0101 + 1) = 0110 or 6. Example 10.20
- 26. 10.26 How can we represent the number −6 in one’s complement arithmetic using only four bits? Solution In one’s complement arithmetic, the negative or complement of a number is found by inverting all bits. Positive 6 is 0110; negative 6 is 1001. If we consider only unsigned numbers, this is 9. In other words, the complement of 6 is 9. Another way to find the complement of a number in one’s complement arithmetic is to subtract the number from 2n − 1 (16 − 1 in this case). Example 10.21
- 27. 10.27 Let us redo Exercise 10.19 using one’s complement arithmetic. Figure 10.24 shows the process at the sender and at the receiver. The sender initializes the checksum to 0 and adds all data items and the checksum (the checksum is considered as one data item and is shown in color). The result is 36. However, 36 cannot be expressed in 4 bits. The extra two bits are wrapped and added with the sum to create the wrapped sum value 6. In the figure, we have shown the details in binary. The sum is then complemented, resulting in the checksum value 9 (15 − 6 = 9). The sender now sends six data items to the receiver including the checksum 9. Example 10.22
- 28. 10.28 The receiver follows the same procedure as the sender. It adds all data items (including the checksum); the result is 45. The sum is wrapped and becomes 15. The wrapped sum is complemented and becomes 0. Since the value of the checksum is 0, this means that the data is not corrupted. The receiver drops the checksum and keeps the other data items. If the checksum is not zero, the entire packet is dropped. Example 10.22 (continued)
- 29. 10.29 Figure 10.24 Example 10.22 1 1 1 1 1 0 0 0 0
- 30. 10.30 Sender site: 1. The message is divided into 16-bit words. 2. The value of the checksum word is set to 0. 3. All words including the checksum are added using one’s complement addition. 4. The sum is complemented and becomes the checksum. 5. The checksum is sent with the data. Note
- 31. 10.31 Receiver site: 1. The message (including checksum) is divided into 16-bit words. 2. All words are added using one’s complement addition. 3. The sum is complemented and becomes the new checksum. 4. If the value of checksum is 0, the message is accepted; otherwise, it is rejected. Note
- 32. 10.32 Let us calculate the checksum for a text of 8 characters (“Forouzan”). The text needs to be divided into 2-byte (16-bit) words. We use ASCII (see Appendix A) to change each byte to a 2-digit hexadecimal number. For example, F is represented as 0x46 and o is represented as 0x6F. Figure 10.25 shows how the checksum is calculated at the sender and receiver sites. In part a of the figure, the value of partial sum for the first column is 0x36. We keep the rightmost digit (6) and insert the leftmost digit (3) as the carry in the second column. The process is repeated for each column. Note that if there is any corruption, the checksum recalculated by the receiver is not all 0s. We leave this an exercise. Example 10.23
- 33. 10.33 Figure 10.25 Example 10.23 F F F F F F