Embed Size (px)
DESCRIPTION
Reed-Solomon Codes. Reed-Solomon Codes. [1] Codes over GF(2 r ) [2] Reed-Solomon codes [3] Decoding Reed-Solomon codes [4] Transform approach to Reed-Solomon codes (Omitted) [5] Berlekamp-Massey algorithm [6] Erasures (Omitted). Reed-Solomon Codes. [1] Codes over GF(2 r ) - PowerPoint PPT Presentation
Citation preview
Reed-Solomon Codes
p2.
Reed-Solomon Codes
[1] Codes over GF(2r) [2] Reed-Solomon codes [3] Decoding Reed-Solomon codes [4] Transform approach to Reed-Solomon
codes (Omitted) [5] Berlekamp-Massey algorithm [6] Erasures (Omitted)
p3.
Reed-Solomon Codes
[1] Codes over GF(2r)
1. GF(2r)[x]
Polynomials with coefficients from GF(2r) K[x], K= GF(2)={0,1}, polynomials with binary
coefficients
2. Recall BCH code of length n=2r-1
c(x)C iff 1, 2, 3, 4 are all roots of c(x), c(x)K[x]
1, 2, 4 share same minimal polynomial m1(x) and 3
has minimal polynomial m3(x) so the generator polynomial is gK(x)=m1(x)m3(x)
p4.
Reed-Solomon Codes
3. Generalization of BCH codes
Choosing c(x)GF(2r)[x] c(x)C iff g(x)=(x+)(x+2)(x+3)(x+4) divides c(x)
4. Subfield subcode
CK is a BCH code C over GF(2r) is an example of Reed-Solomon code CK is said to be a subfield subcode of C:
CKC All words in CK have all their digits in the subfield K
of GF(2r) CK=C Kn
Both CK and C are cyclic
p5.
Reed-Solomon Codes
5. If g(x) generates a linear cyclic code of length 2r-1 over GF(2r), the generator of the binary subfield subcode is the polynomial gK(x)
Let 1, 2,…, t be distinct non-zero elements of GF(2r). Then g(x)=(1+x) (2+x)…(t+x) generates a linear cyclic code of length 2r-1 over GF(2r)(Theorem 6.1.1)
Example:
Let F= GF(24) constructed using 1+x+x4. g(x)=(+x) (2+x)= 3+5x+x2 generates a linear cyclic code over F of length 15. The codeword corresponding to g(x) is of course 351000000000000
p6.
Reed-Solomon Codes
6. Cyclic codes over GF(2r)
Let C be a linear cyclic code of length n over GF(2r). Every codeword c(x) can be written uniquely as m(x)g(x) for some m(x) in GF(2r)[x] of degree less than n-deg(g(x)). Also, g(x) divides f(x) iff f(x) is a codeword, and g(x) divides 1+xn
(Theorem 6.1.3)
p7.
Reed-Solomon Codes
Let g(x) have degree n-k. If g(x) generates a linear cyclic code C over GF(2r) of length n=2r-1 then
is a generating matrix for C.
|C|=2rk . There are 2rk polynomials m(x), since each of the k coefficients in m(x) can be any one of the 2r field elements
)(
)(
)(
1 xgx
xxg
xg
G
k
p8.
Reed-Solomon Codes Example:
Construct GF(23) using 1+x+x3 with as the primitive element.
Let g(x)=(+x)(2+x)=3+4x+x2. Then g(x) generates a linear cyclic code C over GF(23) of length n=7 then
C has 85 codewords. The codeword corresponding to m(x)=1+x+3x41003=m, for example, is
m(x)g(x) mG= 304613
10000
01000
00100
00010
00001
43
43
43
43
43
G
p9.
Reed-Solomon Codes
[2] Reed-Solomon codes
1. Lemma 6.2.1
Let 1, 2,…,t be non-zero elements of GF(2r). Then
tijji
tttt
t
t
1
12
12
222
11
211
)(
1
1
1
det
p10.
Reed-Solomon Codes
Example 6.2.2
Using lemma 6.2.1 and GF(24) constructed using 1+x+x4,
we find that:
510
147
42
1
1
1
det
=(7+2)(10+2)(10+7)
= 12 4 6
= 7
p11.
Reed-Solomon Codes
2. Theorem 6.2.4
Let g(x)=(m+1+x)(m+2+x)…(m+-1+x) be the generator of a linear cyclic code C over GF(2r) of length n=2r-1, where is a primitive element in GF(2r) and m is some integer. Then d(C)
<Proof> any -1 rows in the parity check matrix H (see p.131) is linearly independent so d(C)
p12.
Reed-Solomon Codes
3. Definition of RS(2r, )
A binary Reed-Solomon code RS(2r, ) is a cyclic code
over GF(2r) with generator g(x)=(m+1+x)(m+2+x)… (m+-1+x) for some integer m and some primitive element of GF(2r)
4. Theorem 6.2.5
If C is an RS(2r, ) then (a) n = 2r-1 (b) k = 2r- (c) d = (d) |C| = 2rk
p13.
Reed-Solomon Codes
5. Binary representation of RS(2r, )
Any RS(2r, ) code C, can be represented as a binary code simply by replacing each digit in each codeword by the binary word
Example:
Let C be the RS(4,2) with g(x)= +x and where GF(22) is constructed using 1+x+x2. From Theorem 6.2.5, C has n=3, k=2, d=2 and |C|=16, a generating matrix for C is
10
01
G
p14.
Reed-Solomon Codes
The 16 messages u with their binary representation
along with the corresponding codewords c = uG of C
and their binary representations
u
c
1001111111110101011111110
11111101111100100110110
0100110110110111101111010
001011010001100011010010010
1000011011011011000101100
11100110101110100000100
0101011100101100010101000
001101000001000000000000000
2222
22222
222
22
222
22
2
uuGcuucuGcuu
p15.
Reed-Solomon Codes
6. Shortened RS(2r, ) code C(s)
Example
Let C be the RS(4,2) code of previous example. The
shortened RS(4,2) code C(1) is formed by taking all
codewords that have 0’s in the last s=1 position,
namely:
000,10, 20 and 120
then deleting the last s positions. So
C(1)={ 00, 1, 2, 12 }
p16.
Reed-Solomon Codes
The distance d(s) of C(s):
d(C(s)) d(C) =
d(s) n(s) – k(s) + 1
= 2r-1-s-(2r--s)+1
=
d(s) =
Theorem 6.2.12
Let C be an RS(2r, ) code and let C(s) be the shortened RS(2r, ) code with parameters n(s),k(s),and d(s), then
n(s) = 2r-1-sk(s) = 2r--sd(s) =
p17.
Reed-Solomon Codes
Example In example 6.1.5 we constructed an RS(23,3) code C
with generator polynomial g(x)= 3+4x+x2. The shortened RS(23,3) code C(2) has generating matrix
and has parameters n(2)=5, k(2)=3 and d(2)=3. G(2) is formed by deleting the last s=2 rows columns of the original matrix G
100
100
100
)2(43
43
43
G
p18.
Reed-Solomon Codes
[3] Decoding Reed-Solomon codes 1. Error locations and error magnitudes
The error locations of a received word are the coordinates i in which the error pattern is non-zero, referred by an error location number I
The error magnitude of an error location i is the element of GF(2r) that occurs in coordinate i of the error pattern
Example 6.3.1: Using RS(8, 3) constructed in Example 6.1.5, if c= 3400000 is transmitted and w=3450000 is received then the most likely error pattern is c+w=e=0040000. So the error location number is 2 and the corresponding error magnitude is 4
p19.
Reed-Solomon Codes
2. Decoding algorithm
Notation:
g(x)=(m+1+x)(m+2+x)…(m+-1+x) t=(-1)/2a1, …, ae = the error location numbers
b1, …, be = the error magnitudes
p20.
Reed-Solomon Codes
Symdromes: s = wH
Let w(x)=w0+w1x+ … + wn-1xn-1
c(x)=c0+c1x+ … + cn-1xn-1
e(x)=e0+e1x+ … + en-1xn-1
)1)(1()2)(1()1)(1(
121
110
121
111
...
...
mnmnmn
mmm
n
mmm
www
sss
p21.
Reed-Solomon Codes Syndrome: sm+1,…,sm+-1
sj=w(j) for m+1 j m+-1
w(j)=c(j)+e(j)=e(j)= (6.1)
Error locator polynomial: A(x), let A={a1,…,ae}
A(x)=(a1-x) (a2-x) … (ae-x) (6.2)
Define j to be the coefficient of xj in A(x)
A(x)= 0+ 1x+ … + e-1xe-1+xe (6.3)
(6.4)
which may be written as
(6.5)
t
i
jiiab
1
t
i
ejii
t
i
jii
t
i
jii ababab
11
1
10
1
)()(0
ejjj sss 110
11110 eejjjej ssss
p22.
Reed-Solomon Codes
Substitute j=m+1,…,m+e in turn to obtain e linear equations
(6.6)
The exe matrix above be called M, which does have full rank
em
em
em
eememem
emmm
emmm
s
s
s
sss
sss
sss
2
2
1
1
1
0
121
132
21
1
111
1
111
111
1
1
1
0
011
eee
e
mee
m
ee
e
e
aa
aa
ab
ab
aa
aaM
p23.
Reed-Solomon Codes
Equations (6.1) can most easily be represented as follows:
(6.7)
Define the M’ be the extended matrix:
em
m
m
eem
eemem
me
mm
me
mm
s
s
s
b
b
b
aaa
aaa
aaa
2
1
2
1
21
222
21
112
11
ememem
emmm
emmm
sss
sss
sss
M
21
232
121
'
p24.
Reed-Solomon Codes
Algorithm 6.3.2
1. Calculate sj=w(j) for m+1 j m+2t
2. Setting e=t, find the rank of the extended matrix M’
3. Let e be the rank of M’ and solve the linear system
(6.6) for 0,…,e-1
4. Find the roots of A(x)=0+ 1x+…+ e-1xe-1+xe
These roots are the error location numbers a1,…,ae
5. Solve the linear system (6.7) for b1,…,be
There are the error magnitudes corresponding to
a1,…,ae
p25.
Reed-Solomon Codes
Example 6.3.3
Let g(x)=(1+x)(+x)(2+x)(3+x)=6+5x+ 5x2+2x3+x4
be the generator of an RS(23,5) code (m=-1, t=2),
where GF(23) is constructed using 1+x+x3. Suppose
that the received word is w= 6 5 210 2
Using algorithm 6.3.2:
1. Since m=-1 and =5, we calculate the 4 syndromes
s0=w(0)=6++5+2+1+0+2=1
s1=w(1)=6+2+7+5+4+0+8=3
s2=w(2)=6+3+9+8+8+0+14=3
s3=w(3)=6+4+11+11+12+0+20=1
p26.
Reed-Solomon Codes
2. Setting e=t=2, the extended matrix M’ is
Row reducing M’ yields the matrix
which has rank 2
1
1'
33
33
M
24
33
0
1
p27.
Reed-Solomon Codes
3. Since M’ has full rank. e=2 and so we now solve the linear system
However, as observed above, we have already row reduced M in Step 2, so we have to solve
Then 41+ 2=0 so 1= 5, and 0+ 35 + 3 = 0 so 0= 1
3
2
1
0
s
sM
2
3
1
0
4
3
0
1
p28.
Reed-Solomon Codes
4. We now know the error locator polynomial
A(x) = 0+1x+x2=1+5x+x2. On substituting
field elements into A(x), we find that A()=0
and A(6)=0. Therefore
A(x)=1+5x+x2 =(+x)(6+x)
So the error location numbers are a1= and
a2= 6
p29.
Reed-Solomon Codes5. Now we solve the following linear system:
or
Then 5b2=1 so b2=2, and b1+b2=1 so b1=6.
Therefore the most likely error pattern is and
the most likely codeword is
c = w+e = 65 52100
3
2
1
6
111
b
b
1
1
0
11
2
1
5 b
b
p30.
Example 6.3.4 Let
be the generator of an RS(24,7) code (so m=-1
and t=3), where GF(24) is constructed using
(see Table 5.1)
Suppose that the received word is
Reed-Solomon Codes
))()()()()(1()g( 5432 xxxxxxx 65941232241 xxxxxx
41 xx
659341)( xxxxxw
p31.
1.
Reed-Solomon Codes
7303416955
9242913844
12182410733
912197622
614451
79400
1)(
1)(
1)(
1)(
11)(
11)(
dws
ws
ws
ws
ws
ws
p32.
2.
So M has rank 2 and therefore the most likely
error pattern has weight e=2
Reed-Solomon Codes
27
6712
1297
79129
9129
1297
0
0
1
1
1
'M
0000
0
16712
1297
p33.
3. With e=2, the linear system (6,7) becomes
but in Step 2 we row reduced this matrix:
Then 12 1+7=0 so 1=10, and 7 0+1+9=0
so 0=6.
Reed-Solomon Codes
12
9
1
0
9
7
1
1
7
9
1
0
12
7
0
1
p34.
Reed-Solomon Codes
4. A(x)=6+ 10x+x2=(2+x)(4+x). Therefore a1= 2
and a2=4
5.
so
Therefore b2= 12 and b1= 2. So the most likely error
pattern is e=0020120…0 and the most likely codeword is
c=w+e=142129100…0
1
11 7
2
1
42
b
b
7
7
2
14
0
11
b
b
p35.
Reed-Solomon Codes
[4] Transform approach to Reed-Solomon codes (Omitted)
[5] Berlekamp-Massey algorithm
1. Notation
Reverse error locator polynomialR(x)=1+t-1x+t-2x2+…+0xt
Syndrome polynomials(x)=1+sm+1x+sm+2x2+…+sm+2tx2t
we can write R(x)s(x)=q(x)x2t+1+r(x) with deg(r(x)) t
p36.
Reed-Solomon Codes
Produce a sequence of polynomials Pi(x) and integer Di
If Pi(x)s(x)=qi(x)xi+1+ri(x) with deg(ri(x)) i ,then deg(Pi(x)) i - Di /2 and deg(ri(x)) i - (1+Di)/2
Pi(x) is a combination of Pi-1(x) and some specific previous polynomial Pz(i-1)(x)
2. Algorithm 6.5.1
Calculate sj=w(j) for m+1 j m+2t
Define: q-1(x)=1+sm+1x+sm+2x2+…+sm+2tx2t
q0(x)=sm+1+sm+2x+…+sm+2tx2t-1
p-1(x)=x2t+1
p0(x)=x2t
Let D-1=-1 and D0=0 and let z0=-1
p37.
Reed-Solomon Codes
For 1 i 2t, recursively define qi(x), pi(x), Di, and zi
as follows
(a) If qi-1,0=0, then let qi(x)= qi-1(x)/x pi(x)= pi-1(x)/x Di= 2+Di-1
zi= zi-1
(b) If qi,0 0, then letxxq
q
qxqxq
i
i
zz
iii /))()(()(
1
1 0,
0,11
p38.
Reed-Solomon Codes
which can be truncated to have degree at most 2t-1-i; and let
If e t errors have occurred during transmission then p2t(x)= R(x) has degree e; the error locator polynomial
is:
which has e distinct roots
xxpq
qxpxp
i
i
zz
iii /))()(()(
1
1 0,
0,11
},min{211
izii DDD
otherwisez
DDifiz
i
zii
i
,
)(,1
1
1 1
eetetet xxpxppx
1
1,21,2,2)(
p39.
Reed-Solomon Codes
Example 6.5.2:
Consider example-6.3.4 with the syndrome s0=7, s1=0, s2=9, s3=12, s4=9, s5=7. Working step by step through Algorithm 6.5.1, we obtain the
following. We begin by setting
q-1(x) = 1+7x+x2+9x3+12x4+9x5+7x6
q0(x) = 7+x+9x2+12x3+9x4+7x5
p-1(x) = x7
p0(x) = x6
D-1 = -1, D0= 0, z0= -1
p40.
Reed-Solomon Codes
Let i=1. Since q0,0=70, we use step 3(b):
(q0(x)+7q-1(x))/x =3+x+13x2+14x3+14x4+14x5
which is truncated to degree 2t-i-1=4 to give
q1(x) =3+x+13x2+14x3+14x4
p1(x) =1+7x D1 = 2+min{D-1, D0}=0
and since D0D-1, zi=i-1=0
Before proceeding, we shall adopt a more concise format by representing the polynomials by their corresponding words. Then the information we have found so far is the following table:
p41.
Reed-Solomon Codes
i qi - pi Di zi
-1 0 7 0 9 12 9 7 - 0 -1 0 7 0 9 12 9 7 - 0 0 -1 1 3 0 13 14 14 - 0 7 1 0
Proceeding from i=2 to i=2t=6 we obtain the following table i qi - pi Di zi
-1 0 7 0 9 12 9 7 - 0 -1 0 7 0 9 12 9 7 - 0 0 -1 1 3 0 13 14 14 - 0 7 1 0 2 12 7 6 12 - 0 8 2 1 3 0 10 9 - 0 12 1 3 2 4 0 0 - 0 10 6 4 3 5 0 - 0 10 6 6 3 6 - 0 10 6 8 3
So finally we obtain (x) by reading p2t(x)= p6(x)= backwards:
(x) = 6 + 10x+ x2
p42.
Reed-Solomon Codes
Example 6.5.3: Let C be the RS(24,9) code with generator g(x)=(1+x)(+x)… (7+x) and GF(24) constructed using 1+x+x4. Suppose that w is the received word and the syndromes of w are:
s0=12, s1=9, s2=6, s3=3, s4=5, s5=12,s6=6, s7=6
Using algorithm 6.5.1 and the notation describe in Example 6.5.2, we obtain the error locator polynomial as follows:
i qi - pi Di zi
-1 0 12 9 6 3 5 12 6 6 - 0 -1 0 12 9 6 3 5 12 6 6 - 0 0
-1 1 0 0 0 10 7 5 2 - 0 12 1 0 2 0 0 10 7 5 2 - 0 12 3 0 3 0 10 7 5 2 - 0 12 5 0 4 10 7 5 2 - 0 12 7 0
5 0 8 5 - 0 12 0 0 13 2 46 8 5 - 0 12 0 0 13 4 47 0 - 0 12 13 10 13 6 48 - 0 12 13 10 13 8 4
Therefore the error locator polynomial is: (x) = 13 + 10x+ 13x2+12x3+x4
p43.
Reed-Solomon Codes
3. Error evaluator polynomial
Since r2t(x)=p2t(x)q-1(x) mod x2t+1
Either r(x)=r2t(x)or (x)=r2t(x)-p2t(x)
called the error evaluator polynomial, used to calculate the error magnitudes bj, given the error locations aj
Example 6.5.2
[6] Erasures (Omitted)
