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STATISTIC & INFORMATION THEORY (CSNB134) MODULE 12 ERROR DETECTION & CORRECTION

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Noise in Channels With real channels being imperfect, we have to consider the noise introduced: Noise is one of the reasons for transmission errors. It may be caused by electrical interference, thermal noise, etc. A transmission error occurs when a bit sent with a specific value reaches the drain in a different state (e.g. sent ‘1’, received ‘0’ and vice versa)

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Error Detection All network technologies employ some form of error detection. The basic idea is to use extra bits added to message to determine if errors have been introduced. In the extreme case of error detection – we can transmit additional copy of message. If the receiver has received 2 identical messages, then no error has occurred. However this is not practical since it results in 100% redundancy. Thus, it is best to adopt error detection mechanism which is effective enough to detect most errors yet which yield low level of redundancy (i.e. that require the least extra bits for error detection purposes)

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Error Correction There are also techniques available to correct errors in addition to detect errors (i.e. error correcting codes). However, error correction techniques often requires many more redundant bits. Sometimes, for less crucial security application, it is ‘cheaper’ to use error detection and throw away bad messages and / or request for retransmission of erroneous messages (i.e. avoid having to correct any erroneous messages).

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Error Detection & Error Correction Here, we shall study about two methods of error correction and detection, namely:- (i) Parity checking (ii) Hamming Coding

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Parity Checking Parity Bit was inserted in the early days per Byte: 7 bits plus 1 parity bit (sometimes still used today – remember ASCII code, 7 data bits + 1 parity bit that make up a byte). The parity bit identifies an odd / even number of '1' per group of data bits (e.g. 7 data bits) Example (even parity) Data BitsData + Parity Bits Example (odd parity) Data BitsData + Parity Bits Note: Number of ‘1’ is even Note: Number of ‘1’ is odd

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Parity Checking (cont.) Limitations of Party checking error detection: - Detects any odd number of errors. - Cannot detect even number of errors. (Note: See example below based on even parity)

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2-Dimensional Parity Checking Extension of 1-D parity checking. Example (even parity): parity bit parity byte Detects any odd number of errors and all 2-bit errors

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Exercise 1 Derive the parity bit P 1, P 2, P 3 and P 4 and the parity byte P 5 P 6 P 7 P 8 P 9 P 10 P 11 P 12 assuming even parity checking is being deployed. parity bit P P P P 4 parity byte P 5 P 6 P 7 P 8 P 9 P 10 P 11 P

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Exercise 2 Identify erroneous bit/s and erroneous codeword/s assuming even parity is being deployed. parity bit parity byte Therefore possible error could be either: the pair of bit A and bit D or the pair of bit B and bit C Thus, it is capable of detecting any 2 bits error of 2 different code- words, HOWEVER incapable of correcting them. AB CD

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Exercise 3 Identify erroneous bit/s and erroneous codeword/s assuming even parity is being deployed. parity bit parity byte Therefore possible error is bit A! Thus, this 2-D parity checking only allows for 1 bit error correction!! A

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Hamming Coding Hamming Coding is introduced mainly to increase the capability of error detection and error correction without having too many redundant bits. Before understanding Hamming Coding, we shall need to learn two other parameters related to this: (1) Hamming Weight - The number of non-zero elements in a codeword c. It is written as w(c). (2) Hamming Distance - The number of places in which two codewords c i and c j differ. It is written as d(c i, c j )

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Exercise 4 Let code C consists of 4 codewords: {110101, , , } Derive the Hamming Weights Derive the Hamming Distances cic1c2c3c4 Codeword Hamming Weight w(ci)4334 cicjd(ci,cj)cicjd(ci,cj)

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Hamming Coding (cont.) (n, k) Hamming Code consists of: - n total number of bits per codeword (i.e. data bits + parity bits) - k total number of data bits per codeword Suppose m is the total number of parity bits per codeword, n and k can be derived based on the following formulae: (n, k) Hamming Code = (2 m – 1, 2 m – 1 – m) Hamming Code Note: n – k = m Where m = total number of parity bits / control bits

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Exercise 5 Suppose that a Hamming Code of n bits consists of k data bits – i.e. (n, k) Hamming Code. (a) If the total control bits m = 3, derive n and k. (b) If the total control bits m = 4, derive n and k. Compare the percentage of the redundancy in (a) and (b). n = 2 m – 1 = 2 3 – 1 = 7 where as k = n – m = 7 – 3 = 4 Percentage of redundancy in (a) = (3/4)x100% = 0.75% Percentage of redundancy in (b) = (4/11)x100% = 0.36% n = 2 m – 1 = 2 4 – 1 = 15 where as k = n – m = 15 – 4 = 11

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Hamming Coding (cont.) Previously we have learned how to calculate n and k of an (n, k) Hamming Code. We also know that n - k = m total number of parity check bits in Hamming code. Unlike in normal parity checking, where parity bit is always placed at the first bit, in Hamming code, the parity check bits are placed at 2 x position. For example for (7,4) Hamming code, where total number of parity bits = 7 – 4 = 3, the parity check bits are placed at: (1) 2 0 position = 1 st bit (2) 2 1 position = 2 nd bit (3) 2 2 position = 4 th bit

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Hamming Coding (cont.) Suppose that a (7, 4) Hamming Code is decoded based as the following formulas: B7B6B5B4B3B2B1Remarks Data ParityDataParity Code- word Data- - -ParityB1 Even Parity Data -- ParityB2 Even parity Data Parity---B4 Even Parity Note: The parity bits B1, B2 and B4 are related to the data bits B3, B5, B6 AND B7 B7 affecting B1, B2 & B4 B6 affecting B2 & B4 B5 affecting B1 & B4 B3 affecting B1 & B2

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Exercise 6 Based on the previous formulae, derive the Hamming codeword if we want to transmit 1101: B7B6B5B4B3B2B1Remarks Data 1 Data 0 Parity 0 Data 1 Parity 1 Parity 0 Code- word B1 Even Parity B2 Even parity B4 Even Parity Thus, the Hamming codeword is

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Exercise 7 Based on the previous formulae, derive the Hamming codeword if we want to transmit 0100: B7B6B5B4B3B2B1Remarks Data 0 Data 1 Data 0 Parity 1 Data 0 Parity 1 Parity 0 Code- word B1 Even Parity B2 Even parity B4 Even Parity Thus, the Hamming codeword is

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Exercise 8 Detect the erroneous data bit, assuming a single error has occurred in the received Hamming Code (which follows previous formulae) B7B6B5B4B3B2B1Remarks Data 1 Parity 0 Data 1 Parity 1 Parity 0 Code- word ERROR! OK! ERROR! B1 & B4 indicate an error!! Common data bits effecting B1 & B4 - {B7, B5} - potential erroneous bits However, B2 is free of error. Data bits affecting B2 – {B7, B6 and B3} - eliminate B7 from potential erroneous bit list! Therefore, the erroneous bit is B5

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Exercise 9 Detect the erroneous data bit, assuming a single error has occurred in the received Hamming Code (which follows previous formulae) B7B6B5B4B3B2B1Remarks Data 1 Data 0 Parity 1 Data 0 Parity 1 Parity 0 Code- word ERROR! B1, B2 & B4 indicate an error!! Common data bits effecting B1, B2 & B4 - {B7} - potential erroneous bit Therefore, the erroneous bit is B7

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STATISTIC & INFORMATION THEORY (CSNB134) ERROR DETECTION & CORRECTION --END--

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