Bch and reed solomon codes generation in frequency domain

GENERATION OF ERROR CONTROL

CODES – BCH AND REED-

SOLOMON CODES IN FREQUENCY

DOMAIN

Fourier Transforms in Galois Field, Linear

complexity, LFSR, Conjugacy Constraints

FOURIER TRANSFORMS IN GALOIS FIELDS

Discrete Fourier transforms reveal periodicities in input data

and the relative strengths of any periodic components.

Discrete Fourier transform of a real sequence of numbers

will be a sequence of complex numbers of the same length.

In complex field,

Discrete Fourier transform of p = {pi, i=0,… N-1}, vector of

complex number, is a vector P = {Pk, k=0,…N-1}, given by

where j = √(-1).

The Fourier kernel exp(-j2πN-1

) is Nth root of unity in field of complex numbers.

In finite field GF(qm), element α of order n is an nth root of unity.

FOURIER TRANSFORMS IN GALOIS FIELDS Drawing analogy between exp(-j2πN

-1) and α, Fourier

transform in Galois field can be given by—

Let v = {vi, i=0,… n-1}, be a vector over GF(q), where n

divides qm-1 for some m, and let α be an element of GF(qm)

of order n.

The Galois field Fourier transform of the vector v is the

vector V = {Vj, j=0,… n-1}, given by -

Discrete index i is called time and v time domain function or

signal .

Discrete index j is called frequency and V frequency

domain function or spectrum.

FOURIER TRANSFORMS IN GALOIS FIELDS

In DFT, though function p is real, transform P is complex.

Similarly in GFFT, though time domain function v is over

GF(q), the spectrum V is over GF(qm).

In DFT, exp(-j2πN-1

) exists for all values of n.

In Galois fields, α exists only if n divides qm-1.

To ensure v over GF(q) and V over GF(qm), n = qm-1.

Inverse Fourier transform :

EXAMPLE Let 1011000 be a vector over GF(2), and n = 23-1 for m=3,

and let α be an element of GF(23) of order 7.Find its FT in

Galois field GF(23).

Solution: Vj = ∑ αij vi , i = 0…n-1

V0 = ∑ αo i vi = 1+1+1 = 1

V1 = ∑ αi vi = 1+1. α2

+ 1. α3 = α4

.

V2 = ∑ α2i vi = 1+1. α4

+ 1. α6 = α .

V3 = ∑ α3i vi = 1+1. α6

+ 1. α9 = 0 .

V4 = ∑ α4i vi = 1+1. α8

+ 1. α12 = α2

.

V5 = ∑ α5i vi = 1+1. α10

+ 1. α15 = 0 .

V6= ∑ α6i vi = 1+1. α12

+ 1. α18 = 0 .

V = 1 α4 α 0 α2

0 0 .

Confirm by finding inverse Fourier transform.

EXAMPLE Confirm by finding inverse Fourier transform.

Solution: vi = (1/n)∑ α-ij Vj , j = 0…n-1

v0 = (1/7) ∑ Vj = (1+ α4 + α + α2

)/7 = 1 mod-7

v1 = (1/7) ∑ α-j Vj = (1+ α4 α-1

+ α α-2+ α2

α-4)/7

= (1+ α4 α6+ α α5

+ α2 α3

)/7 = 0

v2 = (1/7) ∑ α-2jVj = (1+ α4 α-2

+ α α-4+ α2

α-8)/7 = 1

v3 = (1/n) ∑ α-3jVj = (1+ α4 α-3

+ α α-6+ α2

α-12 )/7 = 1

v4 = (1/n) ∑ α-4jVj = (1+ α4 α-4

+ α α-8+ α2

α-16 )/7 = 0

v5 = (1/n) ∑ α-5jVj = (1+ α4 α-5

+ α α-10+ α2

α-20 )/7 = 0

v6 = (1/n) ∑ α-6jVj = (1+ α4 α-6

+ α α-12+ α2

α-24 )/7 = 0

v = 1011000

FOURIER TRANSFORMS IN GALOIS FIELDS-

PROPERTIES

If vi ↔ Vj then--

v + v’ ↔ V + V’ Additivity

αil vi ↔ V((j+l)) Modulation

v((i-l)) ↔ αjl Vj Translation

Proof

Modulation ∑ αij (αil

vi ) = ∑ αi(j+l) vi = V((j+l))

OR (1/n)∑ α-ij V ((j+l)) j+l=k or j=k-l

(1/n)∑ α-i(k-l) Vk = (1/n)∑ α-ik

αil V ((j+l)) = αil

(1/n)∑ α-ik V ((j+l)) =

= αil vi

Translation (1/n)∑ α-ij αjl

Vj= (1/n)∑ α-(i-l)j Vj = v((i-l))

OR ∑ αij v((i-l)) i-l=k or i=k+l

∑ α(k+l)j vk= ∑ αjl

αkj vk = αjl

∑ αkj vk = αjl

Vj

= αil vi

FOURIER TRANSFORMS IN GALOIS FIELDS-

PROPERTIES

If Ej = FjGj then ei =∑n-1

l=0 f((i-l)) gl. Convolution

If ei = fi gi then Ej =(1/n)∑n-1

k=0 F((j-k)) Gk.

CONVOLUTION

Let ei = figi i=0,… n-1.

Then Fourier transform Ej, where double parenthesis

denotes modulo-n arithmetic on the indices, is given as-

Proof: Taking FT of ei = figi

FOURIER TRANSFORMS IN GALOIS FIELDS -

THEOREMS The polynomial v(x) has a zero (root) at αj

iff the jth

frequency component Vj equals zero.

The polynomial V(x) has a zero (root) at α-i iff the i

th time

component vi equals zero. where

v(x) = vn-1xn-1

+…+ v1x + v0

V(x) = Vn-1xn-1

+…+ V1x + V0

Proof:

v(αj) = ∑viαij = Vj = 0 From FT formula

Same for other statement.

Code can be generated in time domain as well as frequency

domain.

EXAMPLE Example v = 1011000 V = 1 α4 α

0 α2 0 0.

v(x) = 1+x2+x

3. V(x) = 1+α4

x+αx2

+ α2

x4

.

V3, V5, and V6 = 0

Check if α3 , α5

, and α6

are roots of v(x) and other elements are not.

v(α3) = 1+ α6

+ α9 = 0 .

v(α5) = 1+ α10

+ α15 = 0 .

v(α6) = 1+ α12

+ α18 = 0 .

Similarly--

v1, v4, v5, and v6 = 0

Check if α-1 , α-4

, α-5

, and α-6 are roots of V(x) and other elements

are not.

V(α-1) = 1+ α4 α-1

+ α α-2+ α2

α-4 = 0 .

V(α-4) = 1+ α4 α-4

+ α α-8+ α2

α-16 = 0

V(α-5) = 1+ α4 α-5

+ α α-10+ α2

α-20 = 0

V(α-6) = 1+ α4 α-6

+ α α-12+ α2

α-24 )/7 = 0

Proved

EXAMPLE

Consider vector v of block length n and weight 1 with vm as

single non-zero component.

Vj =∑n-1

i=0 αij

vi = αjm vm j=1,…n-1.

Vj = vm+ αm vm+ α2m

vm+… α(n-1)m vm.

Above spectrum V can be generated by Recursion

(generating future values using past values of any length).

By multiplying αm to previous one value starting from v0=vm.

Spectrum of a vector of weight 1 can be generated by a

linear recursion of length 1.

Special case of linear complexity.

IMPORTANCE OF REED-SOLOMON CODES

RS codes are important for several reasons:

(i) They are the natural codes to use when a code is

required of length less than the size of the field. For, being

MDS, they have the highest possible minimum distance.

(ii) They are convenient for building other codes, as they

can be mapped into binary codes with high minimum

distance.

(iii) They are also used in constructing concatenated and

Justesen codes.

(iii) They are useful for correcting bursts of errors.

LINEAR COMPLEXITY LA

La = Degree of its minimum polynomial m

= Length of shortest Linear Feedback Shift Register (LFSR)

capable of producing a sequence which agrees with ‘a’ for first k entries a0…ak-1.

= maximum number of linearly independent vectors among

state vectors produced by LFSR.

La = n-k = d-1

LINEAR FEEDBACK SHIFT REGISTER (LFSR)

Machine consisting of registers R0 ..Rm-1, arranged in a row

with X-OR gates.

Feedback connections provided from Rm-1 to shift registers

through X-OR based on constant Ci , 1≤i ≤m, being 0 or 1. c(x) = 1+c1x+c2x

2+…. cm-1xm-1

.

Example: LFSR of length 10.

c(x) = 1+x+x3+x

4+x

7+x

10.

Initial state s0…s9 =1001001

Generatesμ 100 100 …

+

1 0 0 1 0 0 1 0 0 1

+ + +

LINEAR FEEDBACK SHIFT REGISTER (LFSR)

Any sequence generated by LFSR of length L with feedback polynomial C is also generated by LFSR whose feedback polynomial is a multiple of C.

A sequence generated by LFSR of length L with feedback polynomial C may also be generated by ‘shorter’ LFSR with feedback polynomial C’ iff C is not irreducible over GF(q).

c(x) = 1+x+x3+x

4+x

7+x

10= (1+x+x

7)(1+x

3)

Hence same sequence will be generated by (1+x3).

Hence linear complexity = 3

0 0 1

LINEAR COMPLEXITY - THEOREMS

Linear complexity of a finite length vector V, cyclically

repeated, is equal to hamming weight to its inverse FT v.

Hamming weight of a vector V is equal to the linear

complexity of its inverse FT v.

Example: v = 1011000 V = 1 α4 α 0 α2

0 0.

La of V = 3

La of v = 4

SPECTRAL PROPERTIES OF CYCLIC CODES –

REED-SOLOMON CODES

A R-S code C of block length n over GF(q), with n a divisor

of q-1, is defined as the set of all codewords whose FT is

equal to zeros in a specified block of d-1 consecutive

components j0, j0+1,… j0+d - 2 .

As n divides q-1, codewords and spectrum components are

symbols in same field GF(q).

If codeword C is cyclically shifted by one place, then its

spectral component cj is multiplied by αj .

If cj = zero, cjαj = zero, cyclic shift of C does not alter

spectrum and hence a valid codeword.

Coded using g(x)= (x- α j0) (x- α j0+1) (x- α j0+2) …(x- α j0+d-2

)

dmin = d2t+1

Polynomial of degree n-k = d-1

FREQUENCY /TRANSFORM DOMAIN ENCODER

Every spectrum consistent with the spectral constraints

yields a code word.

(d-1) cyclically consecutive frequencies j= j0, j0+1,… j0+d - 2

chosen as set of spectral components constrained to be

zero.

(n-(d-1)) unconstrained components filled with data symbols

from GF(q).

Inverse FT produces non-systematic codewords.

d=2t+1, (n,n-d+1) codeword.

Roots of g(x) = roots of c(x)

If αi is root of c(x), Vi = 0.

d-1=n-k roots, and d-1 components zero.

EXAMPLE

(7,5) cyclic code over GF(8) with t=1, j0= 1, (common choice)

g(x)= (x- α) (x- α2) (x- α3) …(x- α2t

) d=2t+1

g(x)= (x- α) (x- α2) = x

2 + α4

x + α3 .

Let data d(x) = α4x

4 + x

3 + α3

x2 + αx +α6

.

c(x) = d(x)g(x)=α4x

6 + α3

x5 + α2

x4 + 0.x

3+ α5

x2 + α6

x +α2.

(7,3) RS code over GF(8) with t=2, j0= 4,

g(x)= x4 + α6

x3 + α6

x2 + α3

x +α.

Let data d(x) = α4x

2 + x +α3

.

c(x) = α4x

6 + αx

5 + α6

x4 + 0.x

3+ 0.x

2 + α5

x +α4.

(15,11) RS code over GF(16) with t=2, j0= 1,

d(x) = x4 + α13

x3+ α6

x2 + α3

x + α10.

c(x) = ?

CONJUGACY CONSTRAINT

Any vector in GF(q) gives FT in GF(qm)

But any vector in GF(qm) does NOT always give Ŧ-1 in

GF(q).

As in complex field, Ŧ-1(real) = real iff V(-f) = V(f).

Conjugacy Constraint Theorem: Let V be a vector of

length n of elements of GF(qm) where n is a divisor of qm-1.

Then the inverse Fourier transform v is a vector of length n

of elements of GF(q) if and only if following is satisfied:-

Vj

q = V((qj)) j=0,…n-1.

Proof: In GF, (a+b)pr = apr

+bpr for any integer r.

If vi is an element of GF(q), then vi

q= vi. Hence..

Vj

q = (∑ αij

vi )q = (∑ αqij

vi

q ) = (∑ αqij

vi) = V((qj))

CONJUGACY CLASSES

To apply Conjugacy theorem, the integers modulo-n are divided into a collection of sets known as Conjugacy classes as—

Aj = { j, jq, jq2, …jqmj-1}

where m is smallest positive integer satisfying jqmj =j modulo n. Example: q=2, n=7, ..

A0 = {0}

A1 = {1,2,4}

A2 = {2,4,1} A1 = A2

A3 = {3,6,5} Test each set on previous example.

Conjugacy class Aj specifies a set of frequencies in the spectrum called a Chord.

If the time domain signal is in GF(q), then the value of the spectrum at any frequency in a conjugate class specifies the value of the spectrum at all other frequencies in that class.

Find conjugate classes for modulo-15, 21, 31.

Q-ARY TRACE

The q-ary trace of an element β of GF(qm) is the sum-

trace(β)= ∑m-1

i=0 βqi

= β+βq+ βq2 +…+ βqm-1 .

Q-ary trace is an element of GF(q).

Example: if q=2, m=3, GF(23)

trace(β)= (β+β2+ β22 )

(trace(β))q = (∑ βqi )q = ∑ βqi+1

= βq+ βq2 +…+ βqm-1 +βqm

= trace(β) βqm = β

If β has m elements in its Conjugacy class, then trace(β) is

sum of all elements in class.

If β has r elements in its Conjugacy class, r divides m, then

each element added into trace m/r times.

Check above with modulo-15 chord.

GENERATION OF CYCLIC CODES IN FREQUENCY DOMAIN

Non-systematic cyclic code c(x) = g(x)d(x).

This is cyclic convolution in time domain as-

Ci =∑n-1

k=0 g((i-k)) dk.

In frequency domain Cj = GjDj.

Any spectrum satisfying above is frequency-domain

codeword over GF(qm) provided all components in time

domain are GF(q)-valued.

Given a set of spectral components j1,…jn-k called parity

frequencies, the cyclic code C is the set of words over GF(q)

whose spectrum is constrained to zero in consecutive

components j1,…jn-k.

Only few of spectra designed as above give codeword in

GF(q) if they satisfy Conjugacy constraint theorem.

GENERATION OF CYCLIC CODES IN FREQUENCY DOMAIN-

REED-SOLOMON CODES

n= q-1 or a factor.

Set of 2t consecutive frequencies of the spectrum are

chosen symbols, to be constrained to zero. 2t = d-1 = n-k

The n-2t unconstrained components of spectrum are filled

with information symbols from GF(q).

Inverse FT then produces nonsystematic (n,n-2t) codeword.

GENERATION OF CYCLIC CODES IN FREQUENCY

DOMAIN- REED-SOLOMON CODES

GENERATION OF CYCLIC CODES: EXAMPLE

SIMPLE HAMMING CODES

Encode SEC (7,4) hamming(Cyclic) code in frequency domain in GF(23).

Components C1 and C2 are chosen as parity frequencies to correct single error. t=1

k=4, 16 combinations. Hence information is contained in two information frequency components C0 and C3.

Using Conjugacy constrain theorem, obligatory frequency-

C1

4 = C2

2 = C4 = 0

C3

4 = C6

2 = C5 = 0

C0

2 = C0. Hence C0 can be either 0 or 1.

Hence equivalent bit content of C0 is one bit.

Equivalent bit content of C3 is three bits. GF(23).

Hence 4 information bits uniquely specify the spectrum.

Find 16 codewords after inverse FT.

GENERATION

OF CYCLIC

CODES:

EXAMPLE

SIMPLE

HAMMING

CODES

Frequency Domain codeword

C0 C

1 C

2 C

3 C

4 C

5 C

6

0 0 0 0 0 0 0

0 0 0 α0 0 α0 α0

0 0 0 α1 0 α4 α2

0 0 0 α2 0 α1 α4

0 0 0 α3 0 α5 α6

0 0 0 α4 0 α2 α1

0 0 0 α5 0 α6 α3

0 0 0 α6 0 α3 α5

1 0 0 0 0 0 0

1 0 0 α0 0 α0 α0

1 0 0 α1 0 α4 α2

1 0 0 α2 0 α1 α4

1 0 0 α3 0 α5 α6

1 0 0 α4 0 α2 α1

1 0 0 α5 0 α6 α3

1 0 0 α6 0 α3 α5

GENERATION

OF CYCLIC

CODES:

EXAMPLE

SIMPLE

HAMMING

CODES

Time Domain codeword

c0 c

1 c

2 c

3 c

4 c

5 c

6

0 0 0 0 0 0 0

1 1 1 0 1 0 0

0 0 1 1 1 0 1

0 1 0 0 1 1 1

1 1 0 1 0 0 1

0 1 1 1 0 1 0

1 0 0 1 1 1 0

1 0 1 0 0 1 1

1 1 1 1 1 1 1

0 0 0 1 0 1 1

1 1 0 0 0 1 0

1 0 1 1 0 0 0

0 0 1 0 1 1 0

1 0 0 0 1 0 1

0 1 1 0 0 0 1

0 1 0 1 1 0 0

GENERATION OF CYCLIC CODES: BCH CODES Example: Design TEC BCH code over GF(64). t=3

Choosing C1, C2, C3, C4, C5, and C6 to be parity frequencies and constrained to be zero. (2t symbols)

Then C0, C7, C9, C11, C13, C15, C21, C23, C27 andC31 are information symbols. Check for validity. How?

Find smallest n for which Cjn = Cjn mod-63 = Cj (for n=64,32, 16,

8, 4, 2). Always mod-63 as GF(64).

C31 –

n=64– (C31)64

= C((1984)) = C31

n=32 -- (C31)32

= C((992)) ≠ C31

n=16 -- (C31)16

= C((496)) ≠ C31

n=8 -- (C31)8= C((248)) ≠ C31

n=4 -- (C31)4= C((124)) ≠ C31

n=2 -- (C31)2= C62 ≠ C31

Hence for C31 n=64. 64 symbols possible for position C31.

Needs 6 bits to encode in time domain. GF(26).

GENERATION OF CYCLIC CODES: BCH CODES

C27 –

n=64– (C27)64

= C((1728)) = C27 mod-63

n=32 -- (C27)32

≠ C27 ,

n=16 -- (C27)16

≠ C27

n=8 -- (C27)8= C((216)) = C27

n=4 -- (C27)4= C((108)) ≠ C27

n=2 -- (C27)2= C54 ≠ C27

Hence for C27 n=8. Needs 3 bits to encode. GF(23).

8 symbols possible for position C27.

Needs 3 bits to encode in time domain. GF(23).


Similarly C23, C15, C13, C11, C7, n=64. Each position can

have any of 64 symbols.

Needs 6 bits to encode each in time domain. GF(26).

For C21 n=4. Needs 2 bits to encode. GF(22). Solve

For C9 n=8. Needs 3 bits to encode. GF(23).

For C0 n=2. Needs 1 bit to encode. GF(2 ).

Hence total information symbols are --

2.64.8.64.64.64.4.64.8.64 = 35184372088832 = 245

Hence total 45 information bits to specify these symbols in

time domain. Data = 245

Gives TEC (63,45) BCH code.

Remaining symbols (63-16=46)?


Remaining symbols (63-16=46)?

Remaining symbols are called obligatory symbols and are

obtained by property of Conjugacy.


Complete spectrum of 64 symbols are transformed into time

domain 245 codewords.

Operation of decoder is same as earlier irrespective of

time/frequency domain encoding.

At last step, to extract information, however, decoder must

know how information is stored in codeword.

If information bits are encoded in the frequency domain,

they must be read out from frequency domain.

Engineering

Bch and reed solomon codes generation in frequency domain