10.2 Data can becorrupted during transmission. Some applications require that errors be detected and corrected. Note
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10.3 10-1 INTRODUCTION 10-1 INTRODUCTION Letus first discuss some issues related, directly or Let us first discuss some issues related, directly or indirectly, to error detection and correction. indirectly, to error detection and correction. Types of Errors Redundancy Detection Versus Correction Forward Error Correction Versus Retransmission Coding Modular Arithmetic Topics discussed in this section: Topics discussed in this section:
10.12 10-2 BLOCK CODING 10-2BLOCK CODING In block coding, we divide our message into blocks, In block coding, we divide our message into blocks, each of k bits, called each of k bits, called datawords datawords. We add r redundant . We add r redundant bits to each block to make the length n = k + r. The bits to each block to make the length n = k + r. The resulting n-bit blocks are called resulting n-bit blocks are called codewords codewords. . Error Detection Error Correction Hamming Distance Minimum Hamming Distance Topics discussed in this section: Topics discussed in this section:
10.14 Let us assumethat k = 2 and n = 3. Table 10.1 shows the list of datawords and codewords (even parity). It is only good for detecting one bit error. Example 10.2 Table 10.1 A code for error detection (Example 10.2)
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10.15 An error-detecting codecan detect only the types of errors for which it is designed; other types of errors may remain undetected. Note
10.17 The Hamming distancebetween two words is the number of differences between corresponding bits. Note
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10.18 Let us findthe Hamming distance between two pairs of words. 1. The Hamming distance d(000, 011) is 2 because Example 10.4 2. The Hamming distance d(10101, 11110) is 3 because
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10.19 The minimum Hammingdistance is the smallest Hamming distance between all possible pairs in a set of words. Note
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10.20 Find the minimumHamming distance of the coding scheme in Table 10.1. Solution We first find all Hamming distances. Example 10.5 The dmin in this case is 2.
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10.21 To guarantee thedetection of up to s errors in all cases, the minimum Hamming distance in a block code must be dmin = s + 1. Note Why? More than s-bit error is possible to detect, but not guaranteed.
10.24 To guarantee correctionof up to t errors in all cases, the minimum Hamming distance in a block code must be dmin = 2t + 1. Note
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10.25 A code schemehas a Hamming distance dmin = 4. What is the error detection and correction capability of this scheme? Solution This code guarantees the detection of up to three errors (s = 3), but it can correct up to one error. In other words, if this code is used for error correction, part of its capability is wasted. Error correction codes should have an odd minimum distance (3, 5, 7, . . . ). Example 10.9
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10.26 10-3 LINEAR BLOCKCODES 10-3 LINEAR BLOCK CODES Almost all block codes used today belong to a subset Almost all block codes used today belong to a subset called called linear block codes linear block codes. A linear block code is a code . A linear block code is a code in which the XOR (addition modulo-2) of two valid in which the XOR (addition modulo-2) of two valid codewords creates another valid codeword. codewords creates another valid codeword. Minimum Distance for Linear Block Codes Some Linear Block Codes Topics discussed in this section: Topics discussed in this section:
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10.27 In a linearblock code, the exclusive OR (XOR) of any two valid codewords creates another valid codeword. Note
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10.28 Let us seeif the two codes we defined in Table 10.1 belong to the class of linear block codes. The scheme in Table 10.1 is a linear block code because the result of XORing any codeword with any other codeword is a valid codeword. For example, the XORing of the second and third codewords creates the fourth one. Example 10.10
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10.29 In our firstcode (Table 10.1), the numbers of 1s in the nonzero codewords are 2, 2, and 2. So the minimum Hamming distance is dmin = 2. Example 10.11 In a linear block code, the minimum Hamming distance is the number of 1s in the nonzero valid codeword with the smallest number of 1s. Note
Hamming Code Thechecker in the decoder creates a 3-bit syndrome (s2s1s0). In which each bit is the parity check for 4 out of the 7 bits in the received codeword: s0 = b2 + b1 + b0 + q0 s1 = b3 + b2 + b1 + q1 s2 = b1 + b0 + b3 + q2 The equations used by the checker are the same as those used by the generator with the parity-check bits added to the right-hand side of the equation. 10.41
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10.42 Table 10.5 Logicaldecision made by the correction logic analyzer Hamming code C(7, 4) can : • detect up to 2-bit error (dmin -1) • can correct up to 1 bit error (dmin-1)/2
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10.43 Figure 10.13 Bursterror correction using Hamming code Split burst error between multiple codewords
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10.44 10-4 CYCLIC CODES 10-4CYCLIC CODES Cyclic codes Cyclic codes are special linear block codes with one are special linear block codes with one extra property. In a cyclic code, if a codeword is extra property. In a cyclic code, if a codeword is cyclically shifted (rotated), the result is another cyclically shifted (rotated), the result is another codeword. codeword. Cyclic Redundancy Check Hardware Implementation Polynomials Cyclic Code Analysis Advantages of Cyclic Codes Other Cyclic Codes Topics discussed in this section: Topics discussed in this section:
10.53 10-5 CHECKSUM 10-5 CHECKSUM Thelast error detection method we discuss here is The last error detection method we discuss here is called the checksum. The checksum is used in the called the checksum. The checksum is used in the Internet by several protocols although not at the data Internet by several protocols although not at the data link layer. However, we briefly discuss it here to link layer. However, we briefly discuss it here to complete our discussion on error checking complete our discussion on error checking Idea One’s Complement Internet Checksum Topics discussed in this section: Topics discussed in this section:
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10.54 Suppose our datais 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
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10.55 We can makethe 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
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10.56 How can werepresent 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
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10.57 How can werepresent 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. Example 10.21
10.59 Sender site: 1. Themessage 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
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10.60 Receiver site: 1. Themessage (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
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Internet Checksum Example Note When adding numbers, a carryout from the most significant bit needs to be added to the result Example: add two 16-bit integers 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 1 wraparound sum checksum
Editor's Notes
#61 Kurose and Ross forgot to say anything about wrapping the carry and adding it to low order bit