slides - Fei Hu

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Transcript slides - Fei Hu

Channel Coding (II)
Cyclic Codes and Convolutional Codes
1
Topics today


Cyclic codes
– presenting codes: code polynomials
– systematic and non-systematic codes
– generating codes: generator polynomials
– encoding/decoding circuits realized by shift registers
Convolutional codes
– presenting codes
 convolutional encoder
 code trees and state diagram
 generator sequences
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Defining cyclic codes: code polynomial


An (n,k) linear code X is called a cyclic code when every cyclic shift of
a code X, as for instance X’, is also a code, e.g.
X  ( x 0 x1
x n  2 x n 1 )
X '  ( x n 1 x 0
x n 3 x n  2 )
Each cyclic code has the associated code vector with the polynomial
X ( p )  x 0  x1 p 



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 xn2 p
n2
 x n 1 p
n 1
Note that the (n,k) code vector has the polynomial of degree of n-1 or
less. Mapping between code vector and code polynomial is one-to-one,
e.g. they specify the same code uniquely
Manipulation of the associated polynomial is done in a Galois field (for
instance GF(2)) having elements {0,1}, where operations are performed
mod-2
For each cyclic code, there exist only one generator polynomial whose
degree equals the number of check bits in the encoded word
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The common factor of cyclic codes

GF(2) operations (XOR and AND):
Modulo-2 Addition
+
0
1
0
0
1
1
1
0

Modulo-2 Multiplication
*
0
1
0
0
0
1
0
1
Cyclic codes have a common factor pn+1. In order to see this we
consider summing two (unity shifted) cyclic code vectors:
X ( p )  x 0  x1 p 
 x n2 p
X '( p )  x n 1  x 0 p  x1 p 
2
p X ( p )  x 0 p  x1 p 
2

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n2
 x n 1 p
 x n2 p
 x n2 p
n 1
n 1
Unshifted
Right rotated
n 1
 x n 1 p
n
Right shifted by multiplication
Question is how to make the cyclic code from the multiplied code?
Adding the last two equations together reveals the common factor:
p X ( p )  X '( p )  x n 1 p  x n 1  x n 1 ( p  1)
n
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n
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Factoring cyclic code generator polynomial


Any factor of pn+1 (Note: decompose it into factors) with the degree
q=n-k generates an (n,k) cyclic code
Example: Consider the polynomial p7+1. This can be factored as
p  1  (1  p )(1  p  p )(1  p  p )
7

3
2
3
For instance the factors 1+p+p3 or 1+p2+p3, can be used to generate an
unique cyclic code. For a message polynomial 1+p2 (I.e. 110), the
following encoded word is generated:
(1  p )(1  p  p )  1  p  p  p
2
3
2
5
and the respective code vector (of degree n-1, n=7, in this case) is
(1 1 1 0 0 1 0)
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Obtaining a cyclic code from another cyclic code

Therefore unity cyclic shift is obtained by (1) multiplication by p where
after (2) division by the common factor yields a cyclic code
1
X '( p )  p X ( p ) m od( p  1)
n
and by induction, any cyclic shift is obtained by
X ( p )  p X ( p ) m od( p  1)
(i)


Example:
right shift 101
(n=3)
(i)
101  X ( p )  1  p
1 p
3
1
3
p p
1 p
3
3
1 p
2
pX ( p )  p  p
pX ( p )

n
1 p
1
1 p
3
not a three-bit code,
divide by the common factor
 110
Important point is that division by mod pn+1 and multiplication by the
generator polynomial is enabled by tapped shift register.
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Using shift registers for multiplication
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
Figure shows a shift register to realize multiplication by 1+p2+p3

In practice, multiplication can be realized by two equivalent topologies:
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Example: multiplication by using a shift register
adding dashed line would
enable division by 1+pn
determined by
the tapped connections
word to be
encoded
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
(1  p )(1  p  p )
2
3
1 p  p  p  p  p
2
3
3
 1  p  p  p  11101
2
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4
4
Encoded word
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1
0
0
0
0
0
0
0
0
1
out
0
0
0
1
1
1
0
1
0
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Examples of cyclic code generator polynomials

The generator polynomial for a (n,k) cyclic code is defined by
G ( p )  1  g1 p 
 g q 1 p
q 1
 p ,q  n  k
q
and G(p) is a factor of pn+1. Any factor of pn+1 that has the degree q
may serve as the generator polynomial. We noticed that a code is
generated by the multiplication
X ( p )  M ( p )G ( p )

where M(p) is a block of k message bits. Hence this gives a criterion to
select the generator polynomial, e.g. it must be a factor of pn+1.
Only few of the possible generating polynomials yield high quality
codes (in terms of their minimum Hamming distance)
G ( p)  p  0  p  1
3
Some cyclic codes:
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Systematic cyclic codes

Define the length q=n-k check vector C and the length-k message
vector M by
M ( p)  m  m p 
m p
k 1
0
C ( p )  c 0  c1 p 

k 1
1
 c q 1 p
q 1
Thus the systematic n:th degree codeword polynomial is
X( p)  p
nk
( m 0  m1 p 
 c 0  c1 p 
 m k 1 p
 c q 1 p
 p M ( p)  C( p)
q 1
k 1
)
message bits
q
Check bits determined by:
C ( p )  m od  p
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nk
M ( p) / G ( p)
check bits
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Determining check-bits

Prove that the check-bits C ( p ) can be calculated from the message bits
M(p) by
C ( p )  m od  p
p
p
nk
nk
nk
M ( p) / G ( p)
10 / 4  2  2 / 4
M ( p) / G ( p)  Q ( p)  C( p) / G ( p)
M ( p) / G ( p)  C( p) / G ( p)  Q ( p)  C( p) / G ( p)  C( p) / G ( p)
0
p
nk
M ( p )  C ( p )  G ( p )Q ( p )
message
check
must be a systematic code
based on its definition
(previous slide)
X ( p )
Example: (7,4) Cyclic code:
G ( p )  p  p  1

3
M
(
p
)

p
 p

 p 74 M ( p )  p 6  p 4

3
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2
 p M ( p) / G ( p)  p  p  1  1
C( p)

Q( p)
 nk
3
3
6
4
 p M ( p)  C( p)  p ( p  p)  1  p  p  1
 Q ( p ) G ( p )  ( p 3  p 2  1)( p 3  p 2  1)  p 6  p 4  1


nk
3
2
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Example: Encoding of systematic cyclic codes
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Decoding cyclic codes


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Every valid, received code word R(p) must be a multiple of G(p),
otherwise an error has occurred. (Assume that the probability for noise
to convert code words to other code words is very small.)
Therefore dividing the R(p)/G(p) and considering the remainder as a
syndrome can reveal if the error has happened and sometimes also to
reveal in which bit (depending on code strength)

The error syndrome of n-k-1 degree is therefore

This can be expressed also in terms of error E(p) and the
code word X(p)
S ( p )  m od  R ( p ) / G ( p ) 
R ( p)  X( p)  E( p)
error syndrome
S(p) is:
S ( p )  m od  X ( p )  E ( p )  / G ( p )
S ( p )  m od  E ( p ) / G ( p ) 
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Decoding cyclic codes: example
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Using denotation of this example:
16.20 s ( x )  m od  e ( x ) / g ( x ) 
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Decoding cyclic codes (cont.)
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Table 16.6
error
syndrome
error
s ( x )  m od  r ( x ) / g ( x ) 
code
g (x)
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msg
Part II. Convolutional coding






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Block codes are memoryless
Convolution codes have memory that utilizes previous bits to encode or
decode following bits
Convolutional codes are specified by n, k and constraint length that is
the maximum number of information symbols upon which the symbol
may depend
Thus they are denoted by (n,k,L), where L is the code memory depth
Convolutional codes are commonly used in applications that require
relatively good performance with low implementation cost
Convolutional codes are encoded by circuits based on shift registers and
decoded by several methods as
 Viterbi decoding that is a maximum likelihood method
 Sequential decoding (performance depends on decoder
complexity)
 Feedback decoding (simplified hardware, lower performance)
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Example: convolutional encoder
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x ' j  m j  2  m j 1  m j
x '' j  m j  2  m j
X out  x '1 x ''1 x ' 2 x '' 2 x '3 x ''3 ...




(n,k,L) = (2,1,2) encoder
Convolutional encoder is a finite state machine processing information
bits in a serial manner
Thus the generated code word is a function of input and the state of the
machine at that time instant
In this (n,k,L)=(2,1,2) encoder, each message bit influences a span of
n(L+1)=6 successive output bits that is the code constraint length
Thus (n,k,L) convolutional code is produced that is a 2n(L-1) state finitestate machine
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(3,2,1) Convolutional encoder
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x ' j  m j3  m j 2  m j
x '' j  m j  3  m j 1  m j
x ''' j  m j  2  m j
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Here each message bit influences
a span of n(L+1)=3(1+1)=6
successive output bits
Representing convolutional code: code tree
x ' j  m j  2  m j 1  m j
x '' j  m j  2  m j
X out  x '1 x ''1 x ' 2 x '' 2 x '3 x ''3 ...
Tells how one input bit
is transformed into two output bits
(initially register is all zero)
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Representing convolutional codes compactly:
code trellis and state diagram
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Input state ‘1’
indicated by dashed line
Shift register states
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