In information theory and coding theory with applications in computer science and telecommunication, error detection and correction (EDAC) or error control are techniques that enable reliable delivery of digital data over unreliable communication channels. Many communication channels are subject to channel noise, and thus errors may be introduced during transmission from the source to a receiver. Error detection techniques allow detecting such errors, while error correction enables reconstruction of the original data in many cases.
Error Control Coding By Shu Lin Pdf Free 14
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Good error control performance requires the scheme to be selected based on the characteristics of the communication channel. Common channel models include memoryless models where errors occur randomly and with a certain probability, and dynamic models where errors occur primarily in bursts. Consequently, error-detecting and correcting codes can be generally distinguished between random-error-detecting/correcting and burst-error-detecting/correcting. Some codes can also be suitable for a mixture of random errors and burst errors.
If the channel characteristics cannot be determined, or are highly variable, an error-detection scheme may be combined with a system for retransmissions of erroneous data. This is known as automatic repeat request (ARQ), and is most notably used in the Internet. An alternate approach for error control is hybrid automatic repeat request (HARQ), which is a combination of ARQ and error-correction coding.
Automatic repeat request (ARQ) is an error control method for data transmission that makes use of error-detection codes, acknowledgment and/or negative acknowledgment messages, and timeouts to achieve reliable data transmission. An acknowledgment is a message sent by the receiver to indicate that it has correctly received a data frame.
Shannon's theorem is an important theorem in forward error correction, and describes the maximum information rate at which reliable communication is possible over a channel that has a certain error probability or signal-to-noise ratio (SNR). This strict upper limit is expressed in terms of the channel capacity. More specifically, the theorem says that there exist codes such that with increasing encoding length the probability of error on a discrete memoryless channel can be made arbitrarily small, provided that the code rate is smaller than the channel capacity. The code rate is defined as the fraction k/n of k source symbols and n encoded symbols.
The actual maximum code rate allowed depends on the error-correcting code used, and may be lower. This is because Shannon's proof was only of existential nature, and did not show how to construct codes which are both optimal and have efficient encoding and decoding algorithms.
Error-correcting memory controllers traditionally use Hamming codes, although some use triple modular redundancy. Interleaving allows distributing the effect of a single cosmic ray potentially upsetting multiple physically neighboring bits across multiple words by associating neighboring bits to different words. As long as a single-event upset (SEU) does not exceed the error threshold (e.g., a single error) in any particular word between accesses, it can be corrected (e.g., by a single-bit error-correcting code), and the illusion of an error-free memory system may be maintained.[22]
From problem 3.17, the rst inequality garantees the existence ofa systematic linear codewith minimum distance dmin.14Chapter 44.1 Aparity-check matrix for the (15, 11) Hamming code isH =1 0 0 0 1 00 1 1 0 1 0 1 1 10 1 0 0 1 1 0 0 0 1 1 1 0 1 10 0 1 0 0 1 1 0 1 0 11 1 0 10 0 0 1 0 0 1 1 0 1 0 1 1 1 1Let r = (r0, r1, . . . , r14)be the received vector. The syndrome of r is (s0, s1, s2, s3)withs0 = r0 +r4 +r7 +r8 +r10 +r12 +r13 +r14,s1 = r1 +r4 +r5 +r9+r10 +r11 +r13 +r14,s2 = r2 +r5 +r6 +r8 +r10 +r11 +r12 +r14,s3 = r3+r6 +r7 +r9 +r11 +r12 +r13 +r14.Set up the decoding table as Table4.1. From the decoding table, we nd thate0 = s0 s1 s2 s3, e1 = s0s1s2 s3, e2 = s0 s1s2 s3,e3 = s0 s1 s2s3, e4 = s0s1 s2 s3, e5 =s0s1s2 s3,. . . , e13 = s0s1 s2s3, e14 = s0s1s2s3.15Table 4.1:Decoding Tables0 s1 s2 s3 e0 e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12e13 e140 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 0 0 0 1 0 0 0 0 0 0 00 0 0 0 0 0 00 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 1 0 00 0 0 0 0 0 0 0 0 00 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 01 1 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 00 0 11 0 0 0 0 0 0 1 0 0 0 0 0 0 0 01 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 001 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 00 1 0 1 0 0 0 0 0 0 0 0 0 1 00 0 0 01 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 00 1 1 1 0 0 0 0 0 0 0 00 0 0 1 0 0 01 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 01 1 0 1 0 0 0 0 00 0 0 0 0 0 0 0 1 01 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1r0r1BufferRegisterr14+...s0+s1+s2+s3...s0s1s2s3s0s1s2s3s0s1s2s3s0s1s2s3+e0r0+e1r1+e2r2+e14r14Decodedbits... ... ...164.3 From (4.3), the probability of an undetectederror for a Hamming code isPu(E) = 2m1 + (2m1)(1 2p)2m1 (1 p)2m1=2m+ (1 2m)(1 2p)2m1(1 p)2m1. (1)Note that(1 p)2 1 2p, (2)and(1 2m)0. (3)Using (2) and (3) in (1), we obtain the followinginequality:Pu(E) 2m+ (1 2m)
p3(1 p)20.The probability of a decoding error isP(E) = 1P(C).5.29(a) Consider two single-error patterns, e1(X) = Xiande2(X) = Xj, where j > i. Supposethat these two error patternsare in the same coset. Then Xi+ Xjmust be divisible byg(X) = (X3+1)p(X). This implies that Xji+ 1 must be divisible by p(X). Thisisimpossible since j i i. Suppose that e1(X) and e2(X) are in the samecoset. ThenXi+Xj+Xj+1must be divisible by g(X) = (X3+1)p(X). Thisis not possible since g(X)has X + 1 as a factor, however Xi+ Xj+Xj+1does not have X + 1 as a factor. Hence nosingle-error patternand a double-adjacent-error pattern can be in the samecoset.Consider two double-adjacent-error patterns,Xi+Xi+1andXj+Xj+1where j > i. Supposethat these two error patterns are inthe same cosets. Then Xi+ Xi+1+ Xj+ Xj+1must bedivisible by (X3+1)p(X). Note thatXi+ Xi+1+Xj+ Xj+1= Xi(X + 1)(Xji+ 1).We see thatfor Xi(X + 1)(Xji+ 1) to be divisible by p(X), Xji+ 1 must bedivisibleby p(X). This is again not possible since j i
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