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Principles of Communications

Faculty of Electronics and Telecommunications

Ho Chi Minh University of Natural Sciences

1

Convolutional codes

09/2008


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Introduction In block coding, the encoder accepts k-

bit message block and generates n-bitcodewordBlock-by-block basis

2

block before generating the codeword

When the message bits come in serially

rather than in large blocks, using bufferis undesirable

Convolutional coding


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Definitions An convolutional encoder: a finite-state machine that

consists of an M-stage shift register, n modulo-2 adders

L-bit message sequence produces an output sequencewith n(L+M) bits

Code rate:

3

L>>M, so

ol)(bits/symb)( MLn

Lr+

=

ol)(bits/symb1

nr=


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Definitions

Constraint length (K): the number ofshifts over which a single message bit

4

M-stage shift register: needs M+1 shiftsfor a message to enter the shift register

and come out K=M+1


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Example Convolutional code (2,1,2)

n=2: 2 modulo-2 adders or 2 outputs

k=1: 1 input

M=2: 2 stages of shift register (K=M+1=2+1=3)

5

Path 2

OutputInput


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Example Convolutional code (3,2,1)

n=3: 3 modulo-2 adders or 3 outputs

k=2: 2 input

M=1: 1 stages of each shift register (K=2 each)

6

OutputInput


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Generations Convolutional code is nonsystematiccode

Each path connecting the output to the input can becharacterized by impulse responseor generator

polynomial

denoting the impulse response of the),,,...,()(

0

)(

1

)(

2

)( iiii

M gggg

7

pa Generator polynomial of the ith path:

D denotes the unit-delay variabledifferent from X of

cyclic codes A complete convilutional code described by a set of

polynomials { }

)(

0

)(

1

2)(

2

)()( ...)( iiiMiMi

gDgDgDgDg ++++=

)(),...,(),( )()2()1( DgDgDg n


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Example(1/8)

Consider the case of (2,1,2) Impulse response of path 1 is (1,1,1)

The corresponding generator polynomial is2)1(

8

Impulse response of path 2 is (1,0,1)

The corresponding generator polynomial is

Message sequence (11001)

Polynomial representation:

1)( 2)2( += DDg

1)( 34 ++= DDDm


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Example(2/8) Output polynomial of path 1:

1

1

)1)(1(

)()()(

236

2345456

234

)1()1(

++++=

++++++++=

++++=

=

DDDD

DDDDDDDD

DDDD

DgDmDc

9

Output sequence of path 1 (1001111)

Output polynomial of path 2:

Output sequence of path 2 (1111101)

1

)1)(1(

)()()(

23546

234

)2()2(

+++++=

+++=

=

DDDDD

DDD

DgDmDc


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Example(3/8)

m= (11001) c(1)=(1001111)

c(2)=(1111101)

10

Encoded sequence c=(11,01,01,11,11,10,11) Message length L=5bits

Output length n(L+K-1)=14bits

A terminating sequence of K-1=2 zeros isappended to the last input bit for the shiftregister to be restored to its zero initial state


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Example(4/8) Another way to calculate the output:

Path 1:

11001001

output111m

11

11110000

10110100

10011001

11001001

00110010

01100100

c(1)=(1001111)


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Example(5/8)

Path 2m 101 output

001001 1 1

00100 11 1

12

001 001 1 1

00 100 11 1

0 010 011 0001 0011 1

c(2)=(1111101)


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Example(6/8) Consider the case of (3,2,1)

13

denoting the impulseresponse of the jth path corresponding to ith

input

Output

Input

),,...,,( )( 0,)(

1,

)(

1,

)(

,

)( j

i

j

i

j

Mi

j

Mi

j

i ggggg =


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Example(7/8)

OutputInput

14DDgg

DDgg

DDgg

Dgg

Dgg

DDgg

==

+==

==

====

+==

)()10(

1)()11(

)()10(

1)()01(

1)()01(

1)()11(

)1(

1

)3(

2

)1(

1

)3(

1

)2(

2

)2(

2

)2(

1

)2(

1

)1(1)1(2

)1(

1

)1(

1


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Example(8/8) Assume that:

m(1)=(101)m(1)(D)=D2+1

m(2)=(011)m(1)(D)=D+1 Outputs are:

c(1)=m(1)*g1(1)+m(2)*g2

(1)

15

= + + + +

=D3+D2+D+1+D+1=D3+D2c(1)=(1100)

c(2)=m(1)*g1(2)+m(2)*g2

(2)

= (D2+1)(1)+(D+1)(D)

=D2

+1+D2

+D=D+1 c(2)

=(0011) c(3)=m(1)*g1

(3)+m(2)*g2(3)

= (D2+1)(D+1)+(D+1)(D)

=D3+D2+D+1+D2+D=1=D3+1c(3)=(1001)

Output c=(101,100,010,011)


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State diagramstate Binary

description

a 00

b 10

c 01

11d

1/10

Consider convolutional code (2,1,2)

16

4 possible states Each node has 2 incoming

branches, 2 outgoing branches A transition from on state to

another in case of input 0 isrepresented by a solid line andof input 1 is represented bydashed line

Output is labeled over the

transition line

00

10 01

a

b c

0/00

1/11

0/10

1/00

0/11


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Example Message 11001

Start at state a Walk through the

state diagram in

10 01

11

b

d

c

0/10

1/01

1/10

0/01

1/00

17

message sequence

00

a

0/00

1/110/11

00

a

10 01 00

b c a

State

Input 1 0 0

Output

10

b

1

11

01

c

0

01

00

a

0

11 11 10 11

11

d

01

1


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Trellis(1/2)

a=00

b=10

0/00 0/00 0/00 0/00 0/00 0/00 0/00 0/00

18

c=01

d=111/10 1/10 1/10 1/10

Level j=0 1 5432 L-1 L L+1 L+2


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Trellis(1/2) The trellis contains (L+K) levels

Labeled as j=0,1,,L,,L+K-1

The first (K-1) levels correspond to the

19

enco er s epar ure rom e n a s a e a The last (K-1) levels correspond to the

encoders return to state a

For the level j lies in the range K-1jL, allthe states are reachable


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Example Message 11001

a=00

0/00 0/00 0/00 0/00 0/00 0/00 0/00

Input 1 1 0 0 1 0 0

20

=10

c=01

d=111/10 1/10 1/10

Level j=0 1 5432

Output

6 7

11 01 01 11 11 10 11


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Maximum Likelihood Decoding

of Convolutional codes mdenotes a message vector

cdenotes the corresponding code vector

rdenotes the received vector

With a given r, decoder is required to make estimateof messa e vector e uivalentl roduce an estimate

mc

21

of the code vector otherwise, a decoding error

happens

Decoding ruleis said to be optimum when the

propability ofdecoding erroris minimized

The maximum likelihood decoder or decision rule isdescribed as follows:

Choose the estimate for which the log-likelihoodfunction logp(r/c)is maximum

ccmm == ifonly

c


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of Convolutional codes Binary symmetric channel: both c and r are binary

sequences of length N

==N

i

ii crpcrp1

)|()|(

=

=N

i

ii crpcrp1

)|(log)|(log

22

rdiffers from cin dpositions, or dis the Hamming

distance between rand c

=

=

ii

ii

iicr

cr

p

pcrpif

if1

)|(with

)1log(1

log

)1log()(log)|(log

pNp

pd

pdNpdcrp

+

=

+=

k l h d d


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of Convolutional codes

Decoding rule is restated as follows: Choose the estimate that minimizes the

Hamming distance between the received vector r

c

23

The received vector ris compared with eachpossible code vector c, and the one closestto ris chosen as the correct transmitted

code vector


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The Viterbi algorithm

Choose a path in the trellis whosecoded sequence differs from the

24

number of positions


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The Viterbi algorithm

The algorithm operates by computing ametric for every possible path in the trellis Metric is Hamming distance between coded

25

received sequence For each node, two paths enter the node, the

lower metric is survived. The other isdiscarded

Computation is repeated every level j in therange K-1jL

Number of survivors at each level 2K-1=4


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The Viterbi algorithm c=(11,01,01,11,11,10,11),r=(11,00,01,11,10,10,11)

a=00 0/00 0/00 0/00 0/00 0/00 0/00 0/00

Input 11 00 01 11 10 10 11

2 2 3 2 41 24 325

26

b=10

c=01

d=111/10 1/10 1/10

11 01 01 11 11 10 11

0 4

1

1

3 2

6

1

4 3

32

4 3

4 3

24

2 5

43

25

Code

Output 1 1 0 0 1 0 0


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Free distance of a conv. code Performance of a conv. code depends on

decoding algorithm and distance properties ofthe code.

Free distance, denoted by dfree, is a measure of

27

Free distance: minimum Hamming distancebetween any two codewords in the code

dfree>2t

Since a convolutional code doesn't use blocks,processing instead a continuous bitstream, thevalue oftapplies to a quantity of errors located

relatively near to each other


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Free distance of a conv. code Conv. code has linear property

So, free distance also defined:

....000)]([ min =

XXwdfree

28

ufree

y r u

Generating function viewed the transferfunction of encoder

Relating input and output by convolution

Generation func relating initial and final stateby multiplication Free distance

Decoding error probability


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Free distance of a conv. code Modify state diagram

11d

1/10

d

DL

29

00

10 01

a

b c

0/00

1/11

0/10

1 01 0/01

1/00

0/11

a0 b c a1

D2L

DL DD

L

D2

Signal-flow graph

Exponent of D: Hamming weight of encoderoutput on that branch.

Exponent of L: number of nonzero messagebits


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Free distance of a conv. code State equations:

cDa

DLdDLbd

DdDbc

LcLaDb

2

1

0

2

=

+=

+=

+=

DL

30

d

a0

b c a1

D2L

DL D

D

L

D2

a0,b,c,d,a1: node signals of the graphSolve the equation set for a1/a0. Sothe generating func is:

=

=

==0

55

0

1 )2(21

),(i

iDLLD

DL

LD

a

aLDT

4

5

537265 2...42),(

=

=+++=d

d

dd LDLDLDLDLDT


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Free distance of a conv. code

T(D,L) represents all possible transmittedsequences that terminate with c-e

4

5

5372652...42),(

=

=+++= dd

ddLDLDLDLDLDT

31

r

For any d5, there are 2d-5 paths withweight w(X)=dthat terminate with c-etransition, those paths are generated bymessages containing d-4 nonzero bits

The free distance is the smallest of w(X),

so dfree=5


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Systematic conv. code The message elements appear explicitly in the

output sequence together with the redundantelements

32

Path 2

OutputInput


33/37

Systematic conv. code

Impulse response of path 1 is (1,0,0) The corresponding generator polynomial is

2)1()( DDg =

33

Impulse response of path 2 is (1,0,1) The corresponding generator polynomial is

Message sequence (11001)

1)( 2)2( += DDg


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Output sequence of path 1 (1100100) Output sequence of path 2 (1111101)

34

c(1)=(1100100)

c(2)=(1111101)

Encoded sequencec=(11,11,01,01,11,00,01)


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Another example of systemmatic conv.code

35

Path 2

OutputInput


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Systematic vs nonsystematic

Assumption: T(D,L) is convergent

When T(D,L) is nonconvergent, an finite number

4

5

537265 2...42),(

=

=+++= dd

dd LDLDLDLDLDT

36

of transmission errors can cause an infinitenumber of decoding errors

The code is called catastrophic code

Systematic conv. code cannot be catastrophic

But, for the same constraint length, free distanceof systematic code is smaller than that ofnonsystematic code

Table 10.8


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Systematic vs nonsystematic Maximum free distance with systematic and

nonsystematic conv codes of rate 1/2

K Systematic Nonsystematic

2 3 3

37

3 4 5

4 4 6

5 5 7

6 6 8

7 6 10

8 7 10

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