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Channel Coding (II)

Cyclic Codes and Convolutional Codes

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2 of 20Topics 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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3 of 20Defining 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.

Each cyclic code has the associated code vector with the polynomial

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

2 1

0 1 2 1( ) n n

n np x x p x p x p X

0 1 2 1( )

n nx x x x X

1 0 3 2' ( )

n n nx x x x X

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4 of 20The common factor of cyclic codes

GF(2) operations (XOR and AND):

Cyclic codes have a common factor pn+1. In order to see this we consider summing two (unity shifted) cyclic code vectors:

Question is how to make the cyclic code from the multiplied code? Adding the last two equations together reveals the common factor:

Modulo-2 Addition+ 0 10 0 11 1 0

Modulo-2 Multiplication* 0 10 0 01 0 1

2 1

1 0 1 2

2 1

0 1 2 1

2 1

0 1 2 1( )

'( )

( )

n

n n

n n

n n

n n

n np x x p x p x p

p x x p x p x p

p p x p x p x p x p

X

X

X

1 1 1( ) '( ) ( 1)n n

n n np p p x p x x p X X

Right rotated

Right shifted by multiplication

Unshifted

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5 of 20Factoring 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

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:

and the respective code vector (of degree n-1, n=7, in this case) is

7 3 2 31 (1 )(1 )(1 )p p p p p p

2 3 2 5(1 )(1 ) 1p p p p p p

(111 0 0 1 0)

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6 of 20Obtaining 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

and by induction, any cyclic shift is obtained by

Example:right shift 101

(n=3)

Important point is that division by mod pn+1 and multiplication by the generator polynomial is enabled by tapped shift register.

'( ) ( )mod( 1)np p p p X X

( ) ( )( ) ( )mod( 1)i i np p p p X X

2101 ( ) 1p p X3( )p p p p X

3

( ) 11 110

1 1 pp p

p

X

not a three-bit code,divide by the common factor

33

3

1

1

1

1

p pp

p

p

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7 of 20Using shift registers for multiplication

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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8 of 20Example: multiplication by using a shift register

2 3

2 3

(1 )(1 )

1

p p p

p p

3p p 4

2 41 11101

p

p p p

out1 1 0 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 0 0 00 0 1 1 0 0 0 0 0 0 00 0 0 1 1 0 0 0 0 0 10 0 0 0 1 1 0 0 0 0 10 0 0 0 0 1 1 0 0 0 10 0 0 0 0 0 1 1 0 0 00 0 0 0 0 0 0 1 1 0 10 0 0 0 0 0 0 0 1 1 0

determined by the tapped connectionsword to be

encoded

adding dashed line would enable division by 1+pn

Encoded word

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9 of 20Examples of cyclic code generator polynomials

The generator polynomial for a (n,k) cyclic code is defined by

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

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)

1

1 1( ) 1 ,q q

qp g p g p p q n k

G

( ) ( ) ( )p p pX M G

Some cyclic codes:

3( ) 0 1p p p G

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10 of 20Systematic cyclic codes

Define the length q=n-k check vector C and the length-k message vector M by

Thus the systematic n:th degree codeword polynomial is

1

0 1 1( ) k

kp m m p m p

M 1

0 1 1( ) q

qp c c p c p

C

1

0 1 1

1

0 1 1

( ) ( )

( ) ( )

n k k

k

q

q

q

p p m m p m p

c c p c p

p p p

X

M C

Check bits determined by:

( ) mod ( ) / ( )n kp p p pC M G

check bits

message bits

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11 of 20Determining check-bits

Prove that the check-bits can be calculated from the message bits M(p) by ( ) mod ( ) / ( )n kp p p pC M G

( ) / ( ) ( ) ( ) / ( )n kp p p p p p M G Q C G

0

( ) / ( ) ( ) / ( ) ( ) ( ) / ( ) ( ) / ( )n kp p p p p p p p p p

M G C G Q C G C G

( )

( ) ( ) ( ) ( )n k

p

p p p p p

X

M C G Q checkmessage

3 2

3

7 4 6 4

( ) 1

( )

( )

p p p

p p p

p p p p

G

M

M

3 2

3 3 6 4

3 2 3 2 6 4

( )( )

( ) / ( ) 1 1

( ) ( ) ( ) 1 1

( ) ( ) ( 1)( 1) 1

n k

n k

pp

p p p p p

p p p p p p p p

p p p p p p p p

CQ

M G

M C

Q G

Example: (7,4) Cyclic code:

must be a systematic codebased on its definition (previous slide)

10 / 2 /4 42

( )pC

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12 of 20Example: Encoding of systematic cyclic codes

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13 of 20Decoding cyclic codes

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)

( ) mod ( ) / ( )p p pS R G

( ) ( ) ( )p p p R X E

( ) mod ( ) ( ) / ( )

( ) mod ( ) / ( )

p p p p

p p p

S X E G

S E G

error syndrome S(p) is:

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14 of 20Decoding cyclic codes: example

16.20 ( ) mod ( ) / ( )s x e x g xUsing denotation of this example:

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15 of 20

( )g x

( ) mod ( ) / ( )s x r x g x

Table 16.6Decoding cyclic codes (cont.)

msgcode

error

error

syndrome

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16 of 20Part II. Convolutional coding

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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17 of 20Example: convolutional 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 finite-state machine

2 1'

j j j jx m m m

2''

j j jx m m

1 1 2 2 3 3' '' ' '' ' '' ...

outX x x x x x x (n,k,L) = (2,1,2) encoder

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18 of 20(3,2,1) Convolutional encoder

3 2'

j j j jx m m m

3 1''

j j j jx m m m

2'''

j j jx m m

Here each message bit influences a span of n(L+1)=3(1+1)=6 successive output bits

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2 1'

j j j jx m m m

2''

j j jx m m

1 1 2 2 3 3' '' ' '' ' '' ...

outX x x x x x x

Tells how one input bitis transformed into two output bits(initially register is all zero)

Representing convolutional code: code tree

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20 of 20Representing convolutional codes compactly: code trellis and state diagram

Shift register states

Input state ‘1’ indicated by dashed line