(Error Detection and Correction) Chapter 10 오류 발견과 교정 (Error Detection and Correction)

10 장 오류 검출 및 수정 10.1 개요 10.2 블록 코딩 10.3 선형 블록 코드 10.4 순환 코드 10.5 검사합 10.6 요약

Data can be corrupted during transmission. 오류 검출 및 수정 데이터는 전송 중에 변경될 수 있다. 신뢰성 있는 통신을 위해 오류들은 검출·정정되어야 한다. Data can be corrupted during transmission. Some applications require that errors be detected and corrected.

Topics discussed in this section: 10.1 개요 Let us first discuss some issues related, directly or indirectly, to error detection and correction. Topics discussed in this section: Types of Errors Redundancy Detection Versus Correction Forward Error Correction Versus Retransmission Coding Modular Arithmetic

단일-비트 오류(Single-Bit Error) 데이터 부분의 한 비트만 변경

In a single-bit error, only 1 bit in the data unit has changed. 단일 비트 오류는 데이터 단위 중 하나의 비트만이 변경되었을 때를 말한다. In a single-bit error, only 1 bit in the data unit has changed.

폭주 오류(Burst Error) 데이터 부분의 2개 또는 그 이상의 비트가 변경

A burst error means that 2 or more bits in the data unit have changed. 폭주 오류는 데이터 단위에서 2개 이상의 비트들이 변경되는 경우를 일컫는다. A burst error means that 2 or more bits in the data unit have changed.

To detect or correct errors, we need to send 오류 검출은 목적지에서 오류를 검출하기 위해서 여분의 비트를 추가하는 중복(잉여)개념을 이용 To detect or correct errors, we need to send extra (redundant) bits with data.

중복(redundancy)

오류 검출은 목적지에서 오류를 검출하기 위해 여분의 비트들을 추가하는 중복의 개념을 사용한다. 오류 검출은 목적지에서 오류를 검출하기 위해 여분의 비트들을 추가하는 중복의 개념을 사용한다.

In this book, we concentrate on block codes; we leave convolution codes to advanced texts.

모듈 연산 법(modulus) N 이라는 제한된 정수를 사용 0 부터 N-1까지 정수 사용 예: 모듈로-2 연산에서 덧셈과 뺄셈 덧셈: 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 1 뺄셈: 0 – 1 = 0 0 – 1 = 1 1 – 0 = 1 1 – 1 =0 In modulo-N arithmetic, we use only the integers in the range 0 to N −1, inclusive.

Figure 10.4 모듈로-2 연산(XORing of two single bits or two words)

Topics discussed in this section: 10.2 블록 코딩(BLOCK CODING) In block coding, we divide our message into blocks, each of k bits, called datawords. We add r redundant bits to each block to make the length n = k + r. The resulting n-bit blocks are called codewords. Topics discussed in this section: Error Detection Error Correction Hamming Distance Minimum Hamming Distance

그림 10.5 블록 코딩에서 데이터워드 (Dataword와 코드워드 codeword)

Example 10.1 The 4B/5B block coding discussed in Chapter 4 is a good example of this type of coding. In this coding scheme, k = 4 and n = 5. As we saw, we have 2k = 16 datawords and 2n = 32 codewords. We saw that 16 out of 32 codewords are used for message transfer and the rest are either used for other purposes or unused.

Figure 10.6 Process of error detection in block coding

Example 10.2 Let us assume that k = 2 and n = 3. Table 10.1 shows the list of datawords and codewords. Later, we will see how to derive a codeword from a dataword. Assume the sender encodes the dataword 01 as 011 and sends it to the receiver. Consider the following cases: 1. The receiver receives 011. It is a valid codeword. The receiver extracts the dataword 01 from it.

Example 10.2(continued) 2. The codeword is corrupted during transmission, and 111 is received. This is not a valid codeword and is discarded. 3. The codeword is corrupted during transmission, and 000 is received. This is a valid codeword. The receiver incorrectly extracts the dataword 00. Two corrupted bits have made the error undetectable.

Table 10.1 A code for error detection (Example 10.2)

An error-detecting code can detect only the types of errors for which it is designed; other types of errors may remain undetected.

Figure 10.7 Structure of encoder and decoder in error correction

Example 10.3 Let us add more redundant bits to Example 10.2 to see if the receiver can correct an error without knowing what was actually sent. We add 3 redundant bits to the 2-bit dataword to make 5-bit codewords. Table 10.2 shows the datawords and codewords. Assume the dataword is 01. The sender creates the codeword 01011. The codeword is corrupted during transmission, and 01001 is received. First, the receiver finds that the received codeword is not in the table. This means an error has occurred. The receiver, assuming that there is only 1 bit corrupted, uses the following strategy to guess the correct dataword.

Example 10.3(continued) 1. Comparing the received codeword with the first codeword in the table (01001 versus 00000), the receiver decides that the first codeword is not the one that was sent because there are two different bits. 2. By the same reasoning, the original codeword cannot be the third or fourth one in the table. 3. The original codeword must be the second one in the table because this is the only one that differs from the received codeword by 1 bit. The receiver replaces 01001 with 01011 and consults the table to find the dataword 01.

Table 10.2 A code for error correction (Example 10.3)

해밍 코드(Hamming Code) 오류 제어를 위해 해밍 거리(hamming distance) 이용 두 개의 같은 크기의 워드간에 차이가 나는 비트의 갯수 두 워드에 XOR 연산을 해서 얻은 결과 값의 1의 갯수 The Hamming distance between two words is the number of differences between corresponding bits.

Example 10.4 Let us find the Hamming distance between two pairs of words. 1. The Hamming distance d(000, 011) is 2 because 2. The Hamming distance d(10101, 11110) is 3 because

The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.

Example 10.5 Find the minimum Hamming distance of the coding scheme in Table 10.1. Solution We first find all Hamming distances. The dmin in this case is 2.

Example 10.6 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.

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.

Example 10.7 The minimum Hamming distance for 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 and 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 and the errors are not detected.

Example 10.8 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 and the errors are undetected.

Figure 10.8 Geometric concept for finding dmin in error detection

Figure 10.9 Geometric concept for finding dmin in error correction

To guarantee correction of up to t errors in all cases, the minimum Hamming distance in a block code must be dmin = 2t + 1.

Example 10.9 A code scheme has 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 need to have an odd minimum distance (3, 5, 7, . . . ).

Topics discussed in this section: 10.3 선형 블록 코드(LINEAR BLOCK CODE) 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. Topics discussed in this section: Minimum Distance for Linear Block Codes Some Linear Block Codes

In a linear block code, the exclusive OR (XOR) of any two valid codewords creates another valid codeword.

Example 10.10 Let us see if the two codes we defined in Table 10.1 and 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 and 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 10.11 In our first code (Table 10.1), the numbers of 1s in the nonzero codewords are 2, 2, and 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, and 4. So in this code we have dmin = 3.

A simple parity-check code is a single-bit error-detecting code in which n = k + 1 with dmin = 2.

Table 10.3 Simple parity-check code C(5, 4)

Figure 10.10 Encoder and decoder for simple parity-check code

Example 10.12 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(continued) 4. An error changes r0 and 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 bits—a3, a2, and a1—are 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.

A simple parity-check code can detect an odd number of errors.

All Hamming codes discussed in this book have dmin = 3. The relationship between m and n in these codes is n = 2m − 1.

Figure 10.11 Two-dimensional parity-check code

Figure 10.11 Two-dimensional parity-check code

Figure 10.11 Two-dimensional parity-check code

Table 10.4 Hamming code C(7, 4)

Figure 10.12 The structure of the encoder and decoder for a Hamming code

Table 10.5 Logical decision made by the correction logic analyzer

Example 10.13 Let us trace the path of three datawords from the sender to the destination: 1. The dataword 0100 becomes the codeword 0100011. The codeword 0100011 is received. The syndrome is 000, the final dataword is 0100. 2. The dataword 0111 becomes the codeword 0111001. The syndrome is 011. After flipping b2 (changing the 1 to 0), the final dataword is 0111. 3. The dataword 1101 becomes the codeword 1101000. The syndrome is 101. After flipping b0, we get 0000, the wrong dataword. This shows that our code cannot correct two errors.

Example 10.14 We need a dataword of at least 7 bits. Calculate values of k and n that satisfy this requirement. Solution We need to make k = n − m greater than or equal to 7, or 2m − 1 − m ≥ 7. 1. If we set m = 3, the result is n = 23 − 1 and k = 7 − 3, or 4, which is not acceptable. 2. If we set m = 4, then n = 24 − 1 = 15 and k = 15 − 4 =11, which satisfies the condition. So the code is C(15, 11)

Figure 10.13 Burst error correction using Hamming code

Topics discussed in this section: 10.4 순환 코드(CYCLIC CODE) 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. Topics discussed in this section: Cyclic Redundancy Check Hardware Implementation Polynomials Cyclic Code Analysis Advantages of Cyclic Codes Other Cyclic Codes

Table 10.6 A CRC code with C(7, 4) 순환 중복 확인 LAN이나 WAN에서 많이 사용 Table 10.6 A CRC code with C(7, 4)

Figure 10.14 CRC encoder and decoder

Figure 10.15 Division in CRC encoder

Figure 10.16 Division in the CRC decoder for two cases

Figure 10.17 Hardwired design of the divisor in CRC

Figure 10.18 Simulation of division in CRC encoder

Figure 10.19 The CRC encoder design using shift registers

Figure 10.20 General design of encoder and decoder of a CRC code

Figure 10.21 A polynomial to represent a binary word

Figure 10.22 CRC division using polynomials

The divisor in a cyclic code is normally called the generator polynomial or simply the generator.

In a cyclic code, If s(x) ≠ 0, one or more bits is corrupted In a cyclic code, If s(x) ≠ 0, one or more bits is corrupted. If s(x) = 0, either a. No bit is corrupted. or b. Some bits are corrupted, but the decoder failed to detect them.

In a cyclic code, those e(x) errors that are divisible by g(x) are not caught.

If the generator has more than one term and the coefficient of x0 is 1, all single errors can be caught.

Example 10.15 Which of the following g(x) values guarantees that a single-bit error is caught? For each case, what is the error that cannot be caught? a. x + 1 b. x3 c. 1 Solution a. No xi can be divisible by x + 1. Any single-bit error can be caught. b. If i is equal to or greater than 3, xi is divisible by g(x). All single-bit errors in positions 1 to 3 are caught. c. All values of i make xi divisible by g(x). No single-bit error can be caught. This g(x) is useless.

Figure 10.23 Representation of two isolated single-bit errors using polynomials

If a generator cannot divide xt + 1 (t between 0 and n – 1), then all isolated double errors can be detected.

Example 10.16 Find the status of the following generators related to two isolated, single-bit errors. a. x + 1 b. x4 + 1 c. x7 + x6 + 1 d. x15 + x14 + 1 Solution This is a very poor choice for a generator. Any two errors next to each other cannot be detected. b. This generator cannot detect two errors that are four positions apart. c. This is a good choice for this purpose. d. This polynomial cannot divide xt + 1 if t is less than 32,768. A codeword with two isolated errors up to 32,768 bits apart can be detected by this generator.

A generator that contains a factor of x + 1 can detect all odd-numbered errors.

❏ All burst errors with L ≤ r will be detected. ❏ All burst errors with L = r + 1 will be detected with probability 1 – (1/2)r–1. ❏ All burst errors with L > r + 1 will be detected with probability 1 – (1/2)r.

Example 10.17 Find the suitability of the following generators in relation to burst errors of different lengths. a. x6 + 1 b. x18 + x7 + x + 1 c. x32 + x23 + x7 + 1 Solution a. This generator can detect all burst errors with a length less than or equal to 6 bits; 3 out of 100 burst errors with length 7 will slip by; 16 out of 1000 burst errors of length 8 or more will slip by.

Example 10.17(continued) b. This generator can detect all burst errors with a length less than or equal to 18 bits; 8 out of 1 million burst errors with length 19 will slip by; 4 out of 1 million burst errors of length 20 or more will slip by. c. This generator can detect all burst errors with a length less than or equal to 32 bits; 5 out of 10 billion burst errors with length 33 will slip by; 3 out of 10 billion burst errors of length 34 or more will slip by.

A good polynomial generator needs to have the following characteristics: 1. It should have at least two terms. 2. The coefficient of the term x0 should be 1. 3. It should not divide xt + 1, for t between 2 and n − 1. 4. It should have the factor x + 1.

Table 10.7 표준 다항식

Topics discussed in this section: 10-5 검사합(CHECKSUM) 여기서 논의하는 마지막 오류 검출 방법은 검사합이다. 검사합은 데이터 링크층 뿐만 아니라 인터넷에서 여러 프로토콜에 의해 사용되고 있다.그렇지만, 여기서는 오류 확인에 대한 논의를 마치기 위해 간단하게 살펴본다 Topics discussed in this section: Idea One’s Complement Internet Checksum

Example 10.18 목적지에 보내고자 하는 5개의 4 비트 숫자 데이터가 있다고 가정하자. 이들 숫자에 덧붙여 숫자의 합도 보낸다. 예를 들면, 숫자 (7, 11, 12, 0, 6)에 대해 (7, 11, 12, 0, 6, 36)를 보낸다, 여기서 36은 숫자의 합이다. 수신자는 5개의 숫자를 더해서 그 결과를 합과 비교한다. 만약 두 값이 같으면 수신자는 오류가 없다고 판단하고, 5개의 숫자를 받아들인다. 그렇지 않으면 어딘가 오류가 있다고 판단하고 데이터를 받아들이지 않는다.

Example 10.19 검사합(checksum)이라는 합의 음수(보수)를 보내면 수신자가 하는 일을 쉽게 할 수 있다. 이 경우에 (7, 11, 12, 0, 6, −36)를 보내게 된다. 수신자는 검사합을 포함해서 모든 숫자를 더한다. 결과가 0 이면 오류가 없고 그렇지 않으면 오류가 있다고 간주한다.

수 21을 단 4 비트만 사용하여 one’s complement arithmetic 으로 표시할 수 있는가? Example 10.20 1의 보수 0 부터 2n-1 사이의 부호 없는 수를 n 비트만 사용하여 나타냄 수가 n 비트 보다 많으면 왼편의 남은 비트를 n개의 오른편 비트에 더해짐 수 21을 단 4 비트만 사용하여 one’s complement arithmetic 으로 표시할 수 있는가? 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 10.21 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 10.22 예제 10.19를 1의 보수 연산을 이용하여 다시 해보자. 그림 10.24 송신자와 수신자에서 처리 과정을 보여준다. 송신자는 검사합을 0 으로 초기화한 다음 모든 데이터 항목과 검사합을 더한다(검사합도 하나의 데이터 항목으로 간주한다. 결과는 36이다. 그렇지만 36은 4 비트로 나타낼 수 없다. 여분의 2 비트는 오른쪽으로 돌려져서 결국에 합은 6이 된다. 그림에서 2진수로 자세하게 보여주고 있다. 합은 보수를 취하고 결과가 검사합이 된다. 9 (15 − 6 = 9). 송신자는 검사합 9를 포함해서 수신자에게 6 개의 데이터 아이템을 보낸다.

Example 10.22(continued) 수신자는 송신자와 같은 절차를 수행한다. 모든 데이터 아이템을 더한다(검사합 포함). 결과는 45이다. 합은 돌려지고 15가 된다. 돌려진 합은 보수를 취하고 결과는 0이 된다. 검사합이 0 이기 때문에 이것은 데이터가 손상되지 않았음을 의미한다. 수신자는 검사합을 버리고 다른 아이템은 보관한다. 만약 검사합이 0이 아니면 전체 패킷은 버려진다.

Figure 10.24 Example 10.22

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.

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.

Example 10.23 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.

Figure 10.25 Example 10.23

10.6 요약