Download The Engineer’s Error Coding Handbook by A. Houghton PDF

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By A. Houghton

Error coding is an engaging topic as a lot, if no more so, because it is an vital a part of modem engineering platforms. regrettably, in a bid to stay normal and to create an outstanding starting place upon which to construct, many books in this topic are out of the succeed in of these with extra engineering-based, or non-mathematical backgrounds. this can be a pity simply because in lots of instances the mathematics is tractable with few and straightforward principles. If we're content material to think that it really works, and allow others fear in regards to the deeper mysteries of the way or why, then with a bit perform the layout and implementation of useful errors coding platforms turns into elementary. during this textual content i've got tried to bare the important kernel of the topic, removal the shell of phrases and proofs that typically surrounds it. Being a little empirical in nature ( an empiricist), and sometimes heard to cite the adage, 'if it really works two times it is a law', my causes take this way. for plenty of, together with myself, summary principles are frequently larger grasped by way of functional representation than from yards of theory.

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Extra info for The Engineer’s Error Coding Handbook

Example text

For a 6 we will have to clock the circuit nine times. Since we are counting forwards, when we ought to be counting backwards, this number represents 15 - k where 15 is the sequence length, found from 2m - 1 where the eRe is m bits. So in this case, k = 15 - clocks. If the remainder is 1, then we need not perform all 15 clocks only to produce the value 15 to subtract it from 15. To illustrate the principle, consider the second example above where the remainder is 0101 (a 9 ). 4. ) Successive clocks are then applied to the circuit until the register contents are 0001.

In fact, our distinction between the two solutions, aa and a b, is quite notional. One would initially think that we must first solve for c, d and e, then use these to solve f, g and h. However, because this distinction is notional, c, d and e actually reveal both solutions as wouldf, g and h if we had chosen to solve them instead. We know that the answers are 011 and 001. e. c = 0 and e = 1. Because d shares both states (0 and 1) in the two solutions, we arrive at no fixed result. Our two solutions must therefore be 001 and 011.

E. the LSB is in error. As with normal maths, aD is 1, or a 15 . If the remainder is 1, then we know that k = 0. 2 A HARDWARE ERROR LOCATOR As far as correction is concerned we could simply use a look-up table to translate the remainder into its corresponding power of a (k), and invert bit k. This is fme for small fields but could be a little unwieldy if the purpose is to fit the error detection/correction circuits onto a single low-cost Ie. If time is not a problem we can actually use the CRC circuit itself to perform the error location.

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