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Ammar Abh-Hhdrohss Islamic University -Gaza ١

Majority Logic Decoding

Slide ٢Channel Coding Theory

One Step majority logic decoding

Consider C as an (n, k) cyclic code with parity check matrix H. His an (n, n - k) which generate the dual code ,denoted by Cd.

For any code v in C and w in Cd, the inner product of v and w is zero; that is ,

Now, suppose that a codeword v in c is transmitted. Let e = (e0, e1, …., en-1) and r = (r0, r1, …., rn-1) be the error vector and the received vector, respectively.

Then, r = v + e

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Slide ٣Channel Coding Theory

One Step majority logic decoding

For any vector w in the dual code Cd, we can form the following

Which is called a parity check sum. A must be zero for no errors. If there is an error then A is equal to

Suppose that there is J vectors in the dual code Cd,

Slide ٤Channel Coding Theory

One Step majority logic decoding

That have the following properties1) the (n-1) component of each vector is 1; that is

2) For i ≠ n – 1 , there is at most one vector whose i th component is a 1. For example if w1,i = 1, then w2,i = w3,i = … = wJ,i = 0.

I.E these vector are orthogonal on the n -1 digital position.

So we can form the following check-sums

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Slide ٥Channel Coding Theory

One Step majority logic decoding

Using the first properties, and w.r = w.e, we can rewrite previous equation as

Using 2nd property of w then

Slide ٦Channel Coding Theory

From previous equation en -1 can be decoded correctly based on majority if there is j/2 errors has happened.

Example: Consider a (15, 7) cyclic code generated by the polynomial

The parity check on systematic form is found to as follows

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Slide ٧Channel Coding Theory

Consider the following linear combinations of rows of H,

if the received error vector is r, then

It is clear that if two errors happen or less e14 can be decoded correctly

Slide ٨Channel Coding Theory

Steps for majority decoding

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Slide ٩Channel Coding Theory

For our example

Slide ١٠Channel Coding Theory

TYPE I

The parity check sum on error digit can be formed from the syndrome digits:

H is the parity check matrix on the systematic form.

As the orthogonal vectors w1, w2,…, wj are vectors in the row space of H, they are linear combination of rows of H.

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Slide ١١Channel Coding Theory

This implies:

Because of the systematic structure of H, it can be seen

Let r = (r0, r1, …, rn-1) be the received vector. Then the syndrome of r is :

Where the ith syndrome digit

Slide ١٢Channel Coding Theory

Now, consider the parity-check sum

or

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Slide ١٣Channel Coding Theory

Steps of type – I majority logic decoding

Step1 : the syndrome is computed using the shift of the received polynomial.

Step 2: the J-parity check sums orthogonal on en-1 are form by taking the proper sum of the syndrome digits. These check-sum are fed to the majority logic circuit.

Step 3: the first received digit is corrected. And the syndrome is shifted to correct the next diigt.

Step 4: step 2 and 3 is applied on the new syndrome to correct rn-2

Step 5: the process is repeated until all received digits are corrected.

Slide ١٤Channel Coding Theory

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Slide ١٥Channel Coding Theory

Example

Consider the (15, 7) BCH code given in Example 8.1 from w1, w2, w3, and w4 that are orthogonal on the digit position 14.

We find that the parity check orthogonal on e14 are equal to

Based on these sums, we construct the type I one-step majority logic decoder shown at the next slide.

If (0, 0, …. , 0) is transmitted and r (x) = x 13 + x 14 is received.

By shifting the received sequence the syndrome register contains (0 0 1 1 1 0 0 1).

Slide ١٦Channel Coding Theory

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Slide ١٧Channel Coding Theory

The four parity check sums orthogonal

So the majority decoder decide e14 = 1;

The buffer register is shifted once and the new parity check sums are

So the majority decoder decide e13 = 1;

The next thirteen bits are error free (you can check by shifting again and again)

Slide ١٨Channel Coding Theory

Class of one step majority logic decodable codes

Let C be a (n , k) cyclic code generated by g (x), where n = 2m – 1. we extend the vector v = (v0, v1, … , v n-1) by adding overall parity bit v∞ where,

And the codeword of the extended code, Ce ,become

Now if we assign 0 to v∞ and to v0 , 2 to v1 , Or the component vi is numbered i.

110 .... nvvvv

110 ,,,, nvvvv

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Slide ١٩Channel Coding Theory

We can do permutation of the location using Z = aY + b Where Y is location of component ve , and a, b are constants.

This permutation is called affine permutation Example

Slide ٢٠Channel Coding Theory

An extended cod Ce of length 2m is said to be invariant under the group of affine permutation if every affine permutation carries the codeword in Ce into another codeword in Ce.

Let h be a nonnegative integer less than 2m. The radix-2 expansion of h is

Where i = 0 or 1. let h’ be another nonnegative integer less than 2m whose radix 2 expansion is

The integer h ’ is said to be a descent of h if

11

2210 222

mmh

11

2210 222

mmh

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Slide ٢١Channel Coding Theory

For example for m = 5, all the numbers 0, 1, 4, 5, 16, 17, 20 are descendants of 21.

(h) denotes the set of all nonzero proper descendants of h,

The extended code Ce is invariant under affine permutation if and only if for every h that is a root of the generator polynomial g (x) of C and for every h ’ in (h), h’ is also a root of g (x) and 0 is not a root of g (x).

The generator polynomial is said to have doubly transitive invariant (DTI) property.

If Ce is invariant under affine permutation, then C is a cyclic code (HOW?)

Slide ٢٢Channel Coding Theory

Now, it is the time to present a class of one-step majority-logic decodable codes whose dual codes have the DTI property.

Let J and L be two factors of 2m – 1 such that 2m – 1 = JL, then

Let

But

Which yields the polynomial has roots

112 mJL

1JX LJLL 120 ,,,,

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Slide ٢٣Channel Coding Theory

To form H (x) we notice that H (x) has h as a root only if 1) h is a root of (x) 2) for every h’ in (h) h’ is also a root of (x) .

Let i be a root of h(x) and i(x) is the minimal polynomial of i

.

Or

The generator polynomial of C is

Slide ٢٤Channel Coding Theory

So first we need to determine J vectors from C’ that are orthogonal on the digit location

(x) is a code polynomial in C’ generated by H (x).

This imply that x(x) , x2(x) , …., xJ-1(x) are also code polynomials of C’.

But for i ≠ j implies xi(x) and xj(x) do not have any common terms.

So by adding an overall parity bit to each of these vectors, we obtain J vectors u0, u1, …, uj-1

12 m

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Slide ٢٥Channel Coding Theory

The J vectors are orthogonal on the digit location ∞. Now, we apply the affine permutation

To the vectors u0, u1, …, uj-1 to achieve z0, z1, …, zj-1 which are also in C’e.

But the permutation carries the component ui at location ∞ to the location

The final step is to delete the locations ∞ from z0, z1, …, zj-1and we can form the J check sums from the resultant vectors.

Therefore the cyclic code generated by Can correct up to J/2. this is called type-0

22 m

YZ

22 m

Slide ٢٦Channel Coding Theory

Example

Let m = 5 , the factorization applies as

Thus J = 5, L = 3 and

10551512 11114

XXXXX

1051 XXx

129635 ,,,1, rootshas1 X

1413111087542105 ,,,,,,,,, rootshas1 XX

8542 ,,,, rootshas XH

8421 ,,, forpolynomialminimaltheis X

1055 , forpolynomialminimaltheis X

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Slide ٢٧Channel Coding Theory

Then H (x) can be calculated as

It can be easily checked that H (x) divides (x) and divides x15 + 1.

Thus

H(x) generates C’ where (x) ,x(x) , x2(x) , …., xJ-1(x) are code polynomials in C’.

2451 11 xxxxxxxH

65431 xxxxxH

985431 xxxxxxG

Slide ٢٨Channel Coding Theory

The dual code C is generated by

The vectors (x) ,x(x) , x2(x) , …., xJ-1(x) are given as

Adding the overall parity bit we get

965419 1)( xxxxxxGxxg

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Slide ٢٩Channel Coding Theory

Applying the affine permutation Z = Y+ 14

Slide ٣٠Channel Coding Theory

Deleting the overall parity bit

These vector are orthogonal on the digit location 14. Let r = (r0, r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14) .

The parity check sums orthogonal on e14 are

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Slide ٣١Channel Coding Theory

The C is capable of correcting 2 errors