CN106656215A - Low-complexity (47,24,11) square residual code decoding method - Google Patents
Low-complexity (47,24,11) square residual code decoding method Download PDFInfo
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Abstract
A low-complexity (47,24,11) square residual code decoding algorithm of the invention is based on the Galois field characteristic of cyclic codes. A complex algebraic operation process needed for the traditional (47,24,11) square residual code decoding algorithm is avoided. By converting complex algebraic operation into simple binary shift and exclusive-or operation, the complexity of the decoding algorithm is reduced effectively. In computer simulation, the operation speed of the algorithm is improved nearly one hundred times compared with the traditional algebraic decoding algorithm.
Description
Technical field
The present invention relates to channel error correction encoding field, more specifically to a kind of decoding suitable for cyclic code
Method, its can using (47,24,11) quadratic residue code.
Background technology
As the important guarantee of reliability transmission, from Shannon in 1948 in " mathematical principle of communication " this
Since proposing channel coding theorem in initiative paper, channel error correction encoding is always all by the extensive pass of scholars
Note.Cyclic code is one of most important error correcting code, including Golay codes, Hamming code, Bose Chaudhuri
Hocquenghem (BCH) code, quadratic residue code etc..(47,24,11) quadratic residue code is one of weight
The pattern wanted, but due to being limited to the decoding algorithm of complexity, the pattern fails to apply in actual applications always.
Galois field characteristic of the present invention based on cyclic code, a kind of low complex degree of proposition (47,24,11) quadratic residue
Code decoding algorithm.
The content of the invention
Present invention is primarily targeted at providing (47,24,11) quadratic residue that a kind of error correcting capability is 5
Code decoding algorithm, its can be applicable to (47,24,11) quadratic residue code decoding on, can effectively reduce (47,24,
11) quadratic residue code complexity in the product.
In order to reach above-mentioned purpose, the method that the present invention is provided is
A kind of low complex degree (47,24,11) quadratic residue code coding method, it is characterised in that:Order generates many
Xiang Shiwei g=(11110111011011100011000),1≤i≤46;0≤j≤46;Reception code word is r=(r0,r1,…,r46), include the step of decoding,
Step 1:Calculate syndrome s=rp+r23g(23)+r24g(24)+…+r46g(46), wherein
rp=(r0,r1,…,r22) and rm=(r23,r24,…,r46)。
Step 2:If 1 quantity in the weight of syndrome, i.e. syndrome s, w (s)≤5, then wrong mould
Formula be e=(s | 01×(n-k));Now, the reception code word after correction is c '=r+e, successfully decoded.
Step 3:If w (s) > 5, it is meant that at least 1 mistake occurs in rm;Now successively by rmIn
A bit riNegate, and recalculate syndrome s+=g(i).If w (s)≤4, error pattern is
E=(s | 01×(n-k))+1i;Now, the reception code word after correction is c '=r+e, successfully decoded.
Step 4:If w (s) > 4, represent that at least two mistakes occur in rm;Now successively by rmIn
Two bit riAnd rjNegate, and recalculate syndrome s=s+g(i)+g(j).It is wrong if w (s)≤3
By mistake pattern be e=(s | 01×(n-k))+1i+1j;Now, the reception code word after correction is c '=r+e, successfully decoded.
A kind of low complex degree (47,24,11) quadratic residue code coding method, it is characterised in that:If
W (s) > 3, then mean that most two mistakes occur in rp;Code word will now be received and moves to left or move to right 23 bits,
Obtain and receive code word after a displacement, there are two kinds of situations for code word is received after displacement:1) most two mistakes
Occur in r by mistakem;2) three mistakes occur in rm, it is one of to occur in r23;Now decoding procedure includes,
Step 1:Calculate and receive after displacement code word syndrome s=rp+r23g(23)+r24g(24)+…+r46g(46),
Wherein rp=(r0,r1,…,r22) and rm=(r23,r24,…,r46)。
Step 2:If 1 quantity in the weight of syndrome, i.e. syndrome s, w (s)≤5, then wrong mould
Formula be e=(s | 01×(n-k));Now, the reception code word after correction is c '=r+e;If receiving a code word left side (right side)
23 bits are moved, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
Step 3:If w (s) > 5, it is meant that at least 1 mistake occurs in rm;Now successively by rmIn
A bit riNegate, and recalculate syndrome s+=g(i).If w (s)≤4, error pattern is
E=(s | 01×(n-k))+1i;Now, the reception code word after correction is c '=r+e;If receiving a code word left side (right side)
23 bits are moved, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
Step 4:If w (s) > 4, represent that at least two mistakes occur in rm;Now successively by rmIn
Two bit riAnd rjNegate, and recalculate syndrome s=s+g(i)+g(j).It is wrong if w (s)≤3
By mistake pattern be e=(s | 01×(n-k))+1i+1j;Now, the reception code word after correction is c '=r+e;If received
A code word left side (right side) moves 23 bits, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
Step 5:If w (s) > 3, now successively by rmIn three bit r23、riAnd rjNegate, and
Recalculate syndrome s=s+g(23)+g(i)+g(j).If w (s)≤2, error pattern is
E=(s | 01×(n-k))+123+1i+1j;Now, the reception code word after correction is c '=r+e;If receiving code word
Left (right side) moves 23 bits, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
A kind of low complex degree (47,24,11) quadratic residue code coding method, it is characterised in that:g(i)Calculation procedure be,
Step 1:g(i-1)> > 1 are by g(i-1)Move to right 1 bit;
Step 2:Represent g(i-1)The 22nd bit;IfFor 1, then g(i)=(g(i-1)> > 1)+g;
IfFor 0, then g(i)=(g(i-1)> > 1).
Description of the drawings
Accompanying drawing 1 decodes flow chart for (47,24,11) quadratic residue code of low complex degree.
Specific implementation method
Above-mentioned functions realize technical scheme combine accompanying drawing conduct further description it is as follows:
Low complex degree according to Fig. 1 (47,24,11) quadratic residue code decoding flow chart, it is characterised in that:
Step 1:Generator polynomial is made to be g=(11110111011011100011000),1≤i≤46;g(i-1)> > 1 are by g(i-1)Move to right 1 bit;Table
Show g(i-1)The 22nd bit;IfFor 1, then g(i)=(g(i-1)> > 1)+g;IfFor 0, then
g(i)=(g(i-1)> > 1);0≤j≤46;Reception code word is r=(r0,r1,…,r46),
Iter=0.
Step 2:Calculate syndrome s=rp+r23g(23)+r24g(24)+…+r46g(46), wherein
rp=(r0,r1,…,r22) represent message part, rm=(r23,r24,…,r46) represent check part;
Step 3:If 1 quantity in the weight of syndrome, i.e. syndrome s, w (s)≤5, then wrong mould
Formula be e=(s | 01×(n-k));Now, into step 8.
Step 4:If w (s) > 5, it is meant that at least 1 mistake occurs in rm;Now successively by rmIn
A bit riNegate, and recalculate syndrome s+=g(i).If w (s)≤4, error pattern is
E=(s | 01×(n-k))+1i;Now, into step 8.
Step 5:If w (s) > 4, represent that at least two mistakes occur in rm;Now successively by rmIn
Two bit riAnd rjNegate, and recalculate syndrome s=s+g(i)+g(j).It is wrong if w (s)≤3
By mistake pattern be e=(s | 01×(n-k))+1i+1j;Now, into step 8.
Step 6:If w (s) > 3 and iter=1, now successively by rmIn three bit r23、riAnd rjTake
Instead, and syndrome s=s+g is recalculated(23)+g(i)+g(j).If w (s)≤2, error pattern is
E=(s | 01×(n-k))+123+1i+1j;Now, into step 8.
Step 7:If w (s) > 3 and iter=0, mean that most two mistakes occur in rp;Now
Code word will be received and move to left or move to right 23 bits, to be obtained and receive code word after a displacement, for receiving code after displacement
There are two kinds of situations in word:1) most two mistakes occur in rm;2) three mistakes occur in rm, one of them
Occur in r23;Now, into step 2.
Step 8:Reception code word after correction is c '=r+e;If iter=1, represent and receive a code word left side (right side)
23 bits are moved, then the reception code word right side (left side) after correction is moved into 23 bits, it is successfully decoded.
Using example:
Example 1:
Order sends code word, c=(00110100000100101001110011110101111001000000000);
Order receives code word, r=(11110100000100101001111101110101111001000000000);
The error pattern is made to be, e=(11000000000000000000001110000000000000000000000);
5 mistakes of generation altogether are we have found that, wherein 2 in message part, 3 in check part, tool
The decoding procedure of body is for example following:
1. syndrome is calculated:
S=(01001100110110010010101).Because w (s)=11 > 5, into step 4.
2. because all of w (s) > 4, into step 5.
3. i=23, j=24 are worked as, then s=(11000000000000000000001).Because w (s)=3≤3,
Then e=(11000000000000000000001110000000000000000000000), into step 8.
4., because iter=0, the reception code word after now decoding is:
C '=r+e=(00110100000100101001110011110101111001000000000).It is successfully decoded.
Example 2:
Order sends code word,
C=(00101000111010110110111011110000110010000000000).
Order receives code word,
R=(11101000111010110110111101110000110010000000001)
The error pattern is made to be,
E=(11000000000000000000000110000000000000000000001)
5 mistakes of generation altogether are we have found that, wherein 3 in message part, 2 in check part, tool
The decoding procedure of body is for example following:
1. syndrome is calculated:
S=(10100010000001010100101).Because w (s)=8 > 5, into step 4.
2. because all of w (s) > 4, into step 5.
3. because all of w (s) > 3, into step 7.
4. because w (s) > 3 and iter=0, it is meant that most 2 mistakes occur in rp;According to cyclic code
Definition, will receive code word 23 bits of cyclic shift, obtain new reception code word,
R=(10111000011001000000000111101000111010110110111).Iter=1 is set, is entered
Step 2.
5. the new syndrome for receiving code word is calculated:
S=(01110001000000101010010).Because w (s)=8 > 5, into step 4.
6. because all of w (s) > 4, into step 5.
7. because all of w (s) > 3, into step 6.
8. i=24, j=25, thens=(11000000000000000000000) are worked as.Because w (s)=2≤2,
Then e=(11000000000000000000000111000000000000000000000), into step 8.
9. c '=r+e=(01111000011001000000000000101000111010110110111) is had.
Because iter=1, to the right the bits of cyclic shift c ' 23, obtain
C '=(00101000111010110110111011110000110010000000000), it is now successfully decoded.
It will appreciated by the skilled person that realizing that all or part of step in above-described embodiment is
Related hardware can be instructed by program to complete, described program can be stored in embodied on computer readable
In storage medium, described storage medium, such as ROM/RAM, disk, CD.
Presently preferred embodiments of the present invention is the foregoing is only, it is all at this not to limit the present invention
Any modification, equivalent and improvement made within the spirit and principle of invention etc., should be included in the present invention
Protection domain within.
Claims (3)
1. a kind of low complex degree (47,24,11) quadratic residue code coding method, it is characterised in that:Make generator polynomial
For g=(11110111011011100011000),1≤i≤46;0≤j≤46;Reception code word is r=(r0,r1,…,r46), include the step of decoding,
Step 1:Calculate syndrome s=rp+r23g(23)+r24g(24)+…+r46g(46), wherein rp=(r0,r1,…,r22) and
rm=(r23,r24,…,r46);
Step 2:If 1 quantity in the weight of syndrome, i.e. syndrome s, w (s)≤5, then error pattern be
E=(s | 01×(n-k));Now, the reception code word after correction is c '=r+e, successfully decoded.
Step 3:If w (s) > 5, it is meant that at least 1 mistake occurs in rm;Now successively by rmIn one
Individual bit riNegate, and recalculate syndrome s+=g(i).If w (s)≤4, error pattern is
E=(s | 01×(n-k))+1i;Now, the reception code word after correction is c '=r+e, successfully decoded.
Step 4:If w (s) > 4, represent that at least two mistakes occur in rm;Now successively by rmIn two
Bit riAnd rjNegate, and recalculate syndrome s=s+g(i)+g(j).If w (s)≤3, error pattern
For e=(s | 01×(n-k))+1i+1j;Now, the reception code word after correction is c '=r+e, successfully decoded.
2. a kind of low complex degree according to claim 1 (47,24,11) quadratic residue code coding method, it is special
Levy and be:If w (s) > 3, mean that most two mistakes occur in rp;Code word will now be received to move to left
Or 23 bits are moved to right, and to obtain and receive code word after a displacement, there are two kinds of situations for code word is received after displacement:
1) most two mistakes occur in rm;2) three mistakes occur in rm, it is one of to occur in r23;Now translate
Code step include,
Step 1:Calculate and receive after displacement code word syndrome s=rp+r23g(23)+r24g(24)+…+r46g(46), wherein
rp=(r0,r1,…,r22) and rm=(r23,r24,…,r46);
Step 2:If 1 quantity in the weight of syndrome, i.e. syndrome s, w (s)≤5, then error pattern be
E=(s | 01×(n-k));Now, the reception code word after correction is c '=r+e;If receiving a code word left side (right side) to move
23 bits, the then reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
Step 3:If w (s) > 5, it is meant that at least 1 mistake occurs in rm;Now successively by rmIn one
Individual bit riNegate, and recalculate syndrome s+=g(i).If w (s)≤4, error pattern is
E=(s | 01×(n-k))+1i;Now, the reception code word after correction is c '=r+e;If receiving a code word left side (right side)
23 bits are moved, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
Step 4:If w (s) > 4, represent that at least two mistakes occur in rm;Now successively by rmIn two
Bit riAnd rjNegate, and recalculate syndrome s=s+g(i)+g(j).If w (s)≤3, error pattern
For e=(s | 01×(n-k))+1i+1j;Now, the reception code word after correction is c '=r+e;If it is left to receive code word
(right side) moves 23 bits, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
Step 5:If w (s) > 3, now successively by rmIn three bit r23、riAnd rjNegate, and count again
Calculate syndrome s=s+g(23)+g(i)+g(j).If w (s)≤2, error pattern is
E=(s | 01×(n-k))+123+1i+1j;Now, the reception code word after correction is c '=r+e;If receiving code word
Left (right side) moves 23 bits, then the reception code word right side (left side) after correcting moves 23 bits, successfully decoded.
3. a kind of low complex degree according to claim 1 (47,24,11) quadratic residue code coding method, it is special
Levy and be:g(i)Calculation procedure be,
Step 1:g(i-1)> > 1 are by g(i-1)Move to right 1 bit;
Step 2:Represent g(i-1)The 22nd bit;IfFor 1, then g(i)=(g(i-1)> > 1)+g;Such as
ReallyFor 0, then g(i)=(g(i-1)> > 1).
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