Received codeword

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= c(x) + e(x)

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g(x) g(x)

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g(x)

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The first term at the right-hand side of the equality does not have a remainder (according to the definition of codeword) So the syndrome is actually the remainder of the second term on the right-hand side If this term does not have a remainder (syndrome = 0), either e(x) is 0 or e(x) is divisible by g(x) We do not have to worry about the first case (there is no error); the second case is very important Those errors that are divisible by g(x) are not caught

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

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Let us show some specific errors and see how they can be caught by a welldesigned g(x) Single-Bit Error What should be the structure of g(x) to guarantee the detection of a single-bit error A single-bit error is e(x) =xi, where i is the position of the bit If a single-bit error is caught, then xi is not divisible by g(x) (Note that when we say not divisible, we mean that there is a remainder) If g(x) ha~ at least two terms (which is normally the case) and the coefficient of xO is not zero (the rightmost bit is 1), then e(x) cannot be divided by g(x)

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H the generator has more than one term and the coefficient of xO is 1,

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all single errors can be caught

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Example 1015

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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 x 3

c 1

SECTION 104

CYCLIC CODES

Solution

a No xi can be divisible by x + 1 In other words, xi/ex + 1) always has a remainder So the syndrome is nonzero Any single-bit error can be caught b If i is equal to or greater than 3, xi is divisible by g(x) The remainder of x i /x 3 is zero, and the receiver is fooled into believing that there is no error, although there might be one Note that in this case, the corrupted bit must be in position 4 or above All single-bit errors in positions I to 3 are caught c All values of i make j divisible by g(x) No single-bit error can be caught In addition, this g(x) is useless because it means the codeword is just the dataword augmented withn - k zeros

Two Isolated Single-Bit Errors

Now imagine there are two single-bit isolated errors Under what conditions can this type of error be caught We can show this type of error as e(x) =: xl + xi The values of i and j define the positions of the errors, and the difference j - i defines the distance between the two errors, as shown in Figure 1023

Figure 1023 Representation o/two isolated single-bit errors using polynomials

Difference: j - i

We can write e(x) = 1(xj - i + 1) If g(x) has more than one term and one term is xo, it cannot divide 1, as we saw in the previous section So if g(x) is to divide e(x), it must divide x j - i + 1 In other words, g(x) must not divide Y! + 1, where t is between 0 and n - 1 However, t=:O is meaningless and t = I is needed as we will see later This means t should be between

2andn-1

H a generator cannot divide

r + 1 (t between 0 and n - 1),

then all isolated double errors can be detected

Example 1016

Find the status of the following generators related to two isolated, single-bit errors a x+ 1

b x 4 + I c x7 + x 6 + 1

d x ls +x I4 +1

Solution

a 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 The two errors can be anywhere, but if their distance is 4, they remain undetected c This is a good choice for this purpose d This polynomial cannot divide any error of typext + 1 if t is less than 32,768 This means that a codeword with two isolated errors that are next to each other or up to 32,768 bits apart can be detected by this generator