error control coding2

8/2/2019 Error Control Coding2

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Linear block codes: Cyclic codes

For practical applications rather large n and kmust be used. This is

because in order to correct up to terrors lhs and rhs of the followingshould be as close as possible:

Hence for large n and kmust be used

Advantages of cyclic codes: Encoding and syndrome computation easy by shift registers

Handy decoding schemes exist due to inherent algebraic structure

For most of the linear codes there exist cyclic codes, as for instance for

Hamming, BCH and Golay codes

12 1 ...

1 2

n k

n

t

i

n

i

n n n

t

=

=

+ + + =

number of syndromes

(or check bit error patterns)

number of error patters

that can be corrected

in an encoded word

2 1

11 log note: (1 )

C C

t

i

n

i R q n k n R

n =

= =

1C

R

min

( 1) / 2t d=


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Example

Consider a relatively high SNR channel such that only 1 or 2 bit errors

are likely to happen. Consider the ration

Take a constant code rate ofRc=0.8 and consider with some values of

larger n and k:

This demonstrates that long codes are more advantages when a high

code rate and high error correction capability is required

2

2

1

11 log

1 /( , )

1log

1 2

C

t

i

n

iR

n

k nn k

n n

n

=

=

+

(10,8) 0.35, (32,24) 0.89, (50,40) 0.97 = = =


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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.

Each cyclic code has the associated code vector with the polynomial

Note that the (n,k) code vector has the polynomial of degree ofn-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(forinstance 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 n x x x x

=X

1 0 3 2' ( )

n n nx x x x

=X


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An example of a (7,4) cyclic code,

generator polynomial G(p)=1+p+p3( )pM ( )pX ( ) ( )p pM G

( )( )31 1

1

X X X

X

+ + +

= + 3X X+ + 2 4X X+ +


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Factoring cyclic code generator polynomial

Any factor ofpn+1 with the degree q=n-kgenerates an (n,k) cyclic code

Example: Consider the polynomialp7+1. This can be factored as

For instance the factors 1+p+p3 or 1+p2+p3, can be used to generate an

unique cycliccode. For a message polynomial 1+p2 the following

encoded word is generated:

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

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

2 3 2 5(1 )(1 ) 1 p p p p p p+ + + = + + +

(111 0 0 1 0)


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Obtaining a cyclic code from another cyclic code

Therefore unity cyclic shift is obtained by multiplication byp where

after division by the common factor yields a cyclic code

and by induction, any cyclic shift is obtained by

Example:

right shift 101

Important point is that division by modpn+1 and multiplication by the

generator polynomial is enabled by tapped shift register.

'( ) ( )mod( 1)n p p p p= +X X

( ) ( )( ) ( )mod( 1)i i n p p p p= +X X

2101 ( ) 1p p = +X3( ) p p p p= +X

3 3

( )

1 1101 1

1p p

p 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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Using 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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Example: 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

+

= + + +

out

1 1 0 0 0 0 0 1

0 1 1 0 0 0 0 1

0 0 1 1 0 0 0 1

0 0 0 1 1 0 0 0

0 0 0 0 1 1 0 1

0 0 0 0 0 1 1 0determined by

the tapped connectionsword to be

encoded

Encoded word


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Examples of cyclic code generator polynomials

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

and G(p) is a factor ofpn+1. Any factor ofpn+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 ofkmessage bits. Hence this gives a criterion to

select the generating polynomial, e.g. it must be a factor ofpn+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 p=X M G

Some cyclic codes:

3( ) 0 1 p p p= + + +G


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Systematic cyclic codes

Define the length q=n-kcheck vector C and the length-kmessage

vector M by

Thus the systematic n:th degree codeword polynomial is

1

0 1 1( )k

k p 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 pc c p c p

p p p

= + + ++ + + +

= +

X

M C

Check bits determined by:

Question: Why these denote the message bits still

the message bits are M(p) ???

[ ]( ) mod ( ) / ( )n k p p p p=C M G

check bits

message bits


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Determining check-bits

Prove that the check-bits can be calculated from the message bits

M(p) by [ ]( ) mod ( ) / ( )n k p p p p=C M G

( ) / ( ) ( ) ( ) / ( )n k p p p p p p = +M G Q C G

0

( ) / ( ) ( ) / ( ) ( ) ( ) / ( ) ( ) / ( )n k p 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

= + += +

= +

GM

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 code

based on its definition

(previous slide)

10 / 2 / 4 42= +

( )pC


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Example: Encoding of systematic cyclic codes


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Circuit for encoding systematic cyclic codes

We noticed earlier that cyclic codes can be generated by using shift

registers whose feedback coefficients are determined directly by the

generating polynomial

For cyclic codes the generator polynomial is of the form

In the circuit, first the message flows to the transmitter, and feedback

switch is set to 1, where after check-bit-switch is turned on, and the

feedback switch to 0, enabling the check bits to be outputted

2 1

1 2 1( ) 1 q q q

q qp g p g p g p p

= + + + + +G

1

0


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Decoding cyclic codes

Every valid, received code word R(p) must be a multiple ofG(p),

otherwise an error has occurred. (Assume that the probability for noiseto 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)

Division is accomplished by a shift registers with reversedtap order (why?)

The error syndrome ofn-k-1 degree is therefore

This can be expressed also in terms of error E(p) and the

code wordX(p)

[ ]( ) mod ( ) / ( ) p p p=S R G

( ) ( ) ( ) p p p= +R X E

[ ]{ }

[ ]

( ) mod ( ) ( ) / ( )

( ) mod ( ) / ( )

p p p p

p p p

= +

=

S X E G

S E G

hence


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Decoding cyclic codes: example

[ ]16.20 ( ) mod ( ) / ( )s x e x g x=Using denotation of this example:


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( )g x

[ ]( ) mod ( ) / ( )s x r x g x=

Table 16.6Decoding cyclic codes (cont.)


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Decoding circuit for (7,4) code

syndrome computation

While first receiving the code, switch is set to 0

The shift register is stepped until all the received code bits have entered

the register

This results 3-bit syndrome (n - k= 3 ) that is then left to the register

Then the switch is turned to the direction 1 that drives the syndrome

out of the register

3( ) 1 p p p= + +G

1

0

received code syndrome


19/27

Convolutional coding

Block codes are memoryless

Convolutional codes have memory that utilizes previous bits to encodeor decode following bits

Convolutional codes are specified by n, kand constraint length that is

the maximum number of information symbols upon which the symbol

may depend

Thus they are denoted by (n,k,L), whereL 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

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)=6successive 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 j

x m m m

=

2'' j j j x 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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(3,2,1) Convolutional encoder

3 2' j j j j

x m m m

=

3 1''

j j j j x m m m

=

2'''

j j j x m m

=

Here each message bit influences

a span ofn(L+1)=3(1+1)=6

successive output bits


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Generator sequences

(n,k,L) Convolutional code can be described by the generator sequences

that are the impulse responses for each coder output branch:

Generator sequences specify convolutional code completely by thegenerator matrix

Encoded convolutional code is produced by matrix multiplication of the

input and the generator matrix

1 2, ,... ng g g

1

2

[1 0 11]

[1111]

=

=

g

g

(2,1,3) encoder

Note that the generator sequence length

exceeds register depth by 1


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Encoding equations

Encoder outputs are formed by modulo-2 discrete convolutions:

Therefore the l:th bit of thej:th output branch is

Input for extraction of generator sequences is

Hence for this circuit the following equations result:

(1) (1)

( 2) ( 2)

**

==

v u gv u g

( ) ( ) ( ) ( ) ( )

0 0 1 1... , 1,2... , 2 j j j j jmil l i i l l l m mv u g u g u g u g j m m L= = = + + + = = +

,

0,otherwisel i

i l iu

=

(1)

2 3

( 2)

1 2 3

l l l l

l l l l l

v u u u

v u u u u

= + +

= + + +

1

2

[1 0 11]

[111 1]

= =

g

g

(1) ( 2 ) (1) ( 2 ) (1) ( 2)

0 0 1 1 2 2[ ...]v v v v v v=v

Encoder output:


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25/27

Example of using generator matrix

1

2

[1 0 11]

[111 1]

= =

g

g

Verify that you can obtain the result shown!

11 10

01

11 00 01 11 01 =


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2 1' j j j j

x m m m

=

2''

j j j x m m

=

1 1 2 2 3 3' '' ' '' ' '' ...outX x x x x x x=Tells how one input bit

is transformed into two output bits

(initially register is all zero)

Representing convolutional code: code tree


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Representing convolutional codes compactly:

code trellis and state diagram

Shift register states

Input state 1

indicated by dashed line

error control coding2

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