Chapter 10 Error Detection in addition to Correction
Burns, Julie, Midday Host;News Director has reference to this Academic Journal, PHwiki organized this Journal 10. Chapter 10 Error Detection in addition to Correction Copyright © The McGraw-Hill Companies, Inc. Permission required as long as reproduction or display. 10. The Hamming distance between two words is the number of differences between corresponding bits. 10. Let us find the 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. The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words. 10. Find the minimum Hamming 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. 10. Find the minimum Hamming distance of the coding scheme in Table 10.2. Solution We first find all the Hamming distances. The dmin in this case is 3. Example 10.6
10. To guarantee the detection of up to s errors in all cases, the minimum Hamming distance in a block code must be dmin = s + 1. 10. The minimum Hamming distance as long as our first code scheme (Table 10.1) is 2. This code guarantees detection of only a single error. For example, if the third codeword (101) is sent in addition to one error occurs, the received codeword does not match any valid codeword. If two errors occur, however, the received codeword may match a valid codeword in addition to the errors are not detected. Example 10.7 10. Our second block code scheme (Table 10.2) has dmin = 3. This code can detect up to two errors. Again, we see that when any of the valid codewords is sent, two errors create a codeword which is not in the table of valid codewords. The receiver cannot be fooled. However, some combinations of three errors change a valid codeword to another valid codeword. The receiver accepts the received codeword in addition to the errors are undetected. Example 10.8
10. Figure 10.8 Geometric concept as long as finding dmin in error detection 10. Figure 10.9 Geometric concept as long as finding dmin in error correction 10. To guarantee correction of up to t errors in all cases, the minimum Hamming distance in a block code must be dmin = 2t + 1.
10. A code scheme has a Hamming distance dmin = 4. What is the error detection in addition to 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 as long as error correction, part of its capability is wasted. Error correction codes need to have an odd minimum distance (3, 5, 7, ). Example 10.9 10. 10-3 LINEAR BLOCK CODES Almost all block codes used today belong to a subset called linear block codes. A linear block code is a code in which the exclusive OR (addition modulo-2) of two valid codewords creates another valid codeword. Minimum Distance as long as Linear Block Codes Some Linear Block Codes Topics discussed in this section: 10. In a linear block code, the exclusive OR (XOR) of any two valid codewords creates another valid codeword.
10. Let us see if the two codes we defined in Table 10.1 in addition to Table 10.2 belong to the class of linear block codes. 1. 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 in addition to third codewords creates the fourth one. 2. The scheme in Table 10.2 is also a linear block code. We can create all four codewords by XORing two other codewords. Example 10.10 10. In our first code (Table 10.1), the numbers of 1s in the nonzero codewords are 2, 2, in addition to 2. So the minimum Hamming distance is dmin = 2. In our second code (Table 10.2), the numbers of 1s in the nonzero codewords are 3, 3, in addition to 4. So in this code we have dmin = 3. Example 10.11 10. A simple parity-check code is a single-bit error-detecting code in which n = k + 1 with dmin = 2. Even parity (ensures that a codeword has an even number of 1s) in addition to odd parity (ensures that there are an odd number of 1s in the codeword)
10. Table 10.3 Simple parity-check code C(5, 4) 10. Figure 10.10 Encoder in addition to decoder as long as simple parity-check code 10. Let us look at some transmission scenarios. Assume the sender sends the dataword 1011. The codeword created from this dataword is 10111, which is sent to the receiver. We examine five cases: 1. No error occurs; the received codeword is 10111. The syndrome is 0. The dataword 1011 is created. 2. One single-bit error changes a1 . The received codeword is 10011. The syndrome is 1. No dataword is created. 3. One single-bit error changes r0 . The received codeword is 10110. The syndrome is 1. No dataword is created. Example 10.12
10. 4. An error changes r0 in addition to a second error changes a3 . The received codeword is 00110. The syndrome is 0. The dataword 0011 is created at the receiver. Note that here the dataword is wrongly created due to the syndrome value. 5. Three bitsa3, a2, in addition to a1are changed by errors. The received codeword is 01011. The syndrome is 1. The dataword is not created. This shows that the simple parity check, guaranteed to detect one single error, can also find any odd number of errors. Example 10.12 (continued) 10. A simple parity-check code can detect an odd number of errors. 10. All Hamming codes discussed in this book have dmin = 3 (2 bit error detection in addition to single bit error correction). A codeword consists of n bits of which k are data bits in addition to r are check bits. Let m = r, then we have: n = 2m -1 in addition to k = n-m
10. Figure 10.11 Two-dimensional parity-check code 10. Figure 10.11 Two-dimensional parity-check code 10. Figure 10.11 Two-dimensional parity-check code
10. Table 10.4 Hamming code C(7, 4) – n=7, k = 4 10. Modulo 2 arithmetic: r0 = a2 + a1 + a0 r1 = a3 + a2 + a1 r2 = a1 + a0 + a3 Calculating the parity bits at the transmitter : Calculating the syndrome at the receiver: s0 = b2 + b1 + b0 s1 = b3 + b2 + b1 s2 = b1 + b0 + b3 10. Figure 10.12 The structure of the encoder in addition to decoder as long as a Hamming code
10. Figure 10.25 Example 10.23
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