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Convolutional codes

A convolutional encoder:

Encoding is done as a continuous process

Data stream may be shifted into the encoder as fixed

size blocks but are not block coded Encoder is a machine with memory

Block codes are based on algebraic/combinatorialtechniques.

Convolutional codes are based on constructiontechniques.

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Convolutional codes

A Convolutional code is specified by three parameters

or

is the coding rate, determining thenumber of data bits per coded word.

Kis the constraint length of the encoder, where theencoder has K-1 memory elements each of size k.

),,( Kkn),/( Knk

nkRc /

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1u

2u

A Rate Convolutional encoder

(representations)

1u

2u

k = 1 bit , n = 2 bits , Rate = k/n =

L = no. of memory elements

K = L + 1 = 2 + 1 = 3 or (L+1) x n = 6

k = 2 bits , n = 4 bits , Rate = k/n =

L = no. of memory elements

K = L + 1 = 2 + 1 = 3 or (L+1) x n = 12

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Input databits

Output codedbitsm

First coded bit1u

2u

Second coded bit

21,uu

(Branch word)

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A Rate Convolutional encoder)101(mMessage sequence:

1 0 01t

1u

2u

11

21 uu

0 1 02t

1u

2u

01

21 uu

1 0 13t

1u

2u

0021

uu

0 1 04t

1u

2u

0121

uu

Time Output OutputTime(Branch word) (Branch word)

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Encoder)101(m )1110001011(U

0 0 15t

1u

2u

11

21 uu

0 0 06t

1u

2u

00

21 uu

Time Output Time Output(Branch word) (Branch word)

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Effective code rate

Initialize the memory before encoding the first bit(all-zero)

Clear out the memory after encoding the last bit(all-zero), a tail of zero-bits is appended to data bits.

Effective code rate : M is the number of data bits and k=1 is assumed:

Data Encoder CodewordTail

ceff RKMn

MR

)1(

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Encoder representation

Vector representation:

m

1u

2u

21 uu

)101(

)111(

2

1

g

g

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Impulse response representation:

1110001011

1110111

0000000

1110111

OutputInput m

Modulo-2 sum:

11001

0101011100

111011:sequenceOutput

001:sequenceInput

21 uu

Branch wordRegistercontents

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Polynomial representation:

The output sequence is found as follows:

22)2(

2

)2(

1

)2(

02

22)1(

2

)1(

1

)1(

01

1..)(

1..)(

XXgXggX

XXXgXggX

g

g

)()(withinterlaced)()()( 21 XXXXX gmgmU

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In more detail:

1110001011

)1,1()0,1()0,0()0,1()1,1()(.0.0.01)()(

.01)()(

1)1)(1()()(

1)1)(1()()(

432

432

2

432

1

422

2

4322

1

U

Ugm

gm

gm

gm

XXXXXXXXXXX

XXXXXX

XXXXX

XXXXXXXX

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State diagram

A finite-state machine only encounters a finitenumber of states.

State of a machine: the smallest amount ofinformation that, together with a current inputto the machine, can predict the output of themachine.

In a Convolutional encoder, the state isrepresented by the content of the memory.

Hence, there are states, K is theconstraint length.

12

K

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State diagram

A state diagram is a way to represent theencoder.

A state diagram contains all the states and allpossible transitions between them.

Only two transitions initiate from a state

Only two transitions end up in a state

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Input data

bits

Output coded

bitsm

1u

2u

First coded bit

Second coded bit

21,uu

(Branch word)

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State diagram

CurrentState

Input NextState

Output

S0

00

0 S0 00

1 S2 11S1

01

0 S0 11

1 S2 00

S2

10

0 S1 10

1 S3 01S3

11

0 S1 01

1 S3 10

10 01

00

11

0S

1S2S

3S

1/11

1/00

1/01

1/10

0/11

0/00

0/01

0/10

Input

Output

(Branch word)

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Trellis Representation

Trellis diagram is an extension of the state diagramthat shows the passage of time.

Timeit 1it

State 000 S

011 S

102 S

113 S

0/00

1/10

0/11

0/10

0/01

1/11

1/01

1/00

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Trellis

6t

1t

2t 3t 4t 5

t

Input bits

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

1 0 1 0 0

11 10 00 10 11

Output bits

Tail bits

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Trellis

1/11

0/00

0/10

1/11

1/01

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

0/00

0/11

0/00

6t

1t

2t

3t

4t 5t

1 0 1 0 0

11 10 00 10 11

Input bits

Output bits

Tail bits

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Block diagram of the DCS

sequencereceived

321 ,...),...,,,( iZZZZZ

Information

source

Rate 1/n

Conv. encoderModulator

Information

sink

Rate 1/n

Conv. decoder Demodulator

sequenceInput

,...),...,( mi,m2m1m

sequenceCodeword

321 ,...),...,,,( iUUUU

G(m)U

,...),...,,( 21 immmm

Channel

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Soft and hard decision

decoding In hard decision: The demodulator makes a firm or hard decision

whether 1 or 0 is transmitted and provides no other

information for the decoder such as the reliability of thedecision.

In Soft decision:

The demodulator provides the decoder with some sideinformation together with the decision. The sideinformation provides the decoder with a measure ofconfidence for the decision.

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Soft and hard decision

decoding ML hard-decisions decoding rule:

Choose the path in the trellis with minimum

Hamming distance from the received sequence

ML soft-decisions decoding rule:

Choose the path in the trellis with minimumEuclidean distance from the received sequence

MLMaximum Likelihood

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Decoding of Convolutional Codes

Maximum likelihood decoding of convolutional codes

Finding the code branch in the trellis that was most likely

transmitted

Based on calculating code Hamming distances for each branchforming encoded word

Assume that the information symbols applied into an AWGN

channel are equally alike and independent,

0 1 2... ...jx x x xx

0 1... ...

jy y yy

non -erroneous code:

Decoder(=DistanceCalculation &Comparison)x

y

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Decoding of Convolutional Codes

Probability to decode the symbols is then

The most likely path through the trellis will maximize this

metric.

Since probabilities are often small numbers, often ln( ) is

taken from both sides, yielding,

0

( , ) ( | )j j

j

p p y x

y x

received code:

non -erroneous code:

1

ln ( , ) ln ( )j mj

j

p p y x

y x

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Example of Exhaustive

Maximal Likelihood Detection

Assume a three bit message is transmitted [encoded by

(2,1,2) convolutional encoder]. To clear the decoder,

two zero-bits are appended after message. Thus 5 bits

are encoded resulting in 10 bits of code. Assume

channel error probability is p = 0.1. After the channel

10,01,10,11,00 is produced (including some errors).

What comes after the decoder, e.g. what was most likely

the transmitted code and what were the respective

message bits?

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a

b

c

d

states

decoder outputsif this path is selected

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Received Sequence 10 01 10 11 00

All Zero Sequence 00 00 00 00 00

Hamming Distance 5

Path metric 5 (-2.3) + 5 (-0.11) = -12.05

ln [ p(0/0)] = ln [p(1/1)] = ln (0.9) = - 0.11

ln [ p(1/0)] = ln [p(0/1)] = ln (0.1) = - 2.30



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correct:1+1+2+2+2=8;8 ( 0.11) 0.88

false:1+1+0+0+0=2;2 ( 2.30) 4.6

total path metric: 5.48



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The Viterbi Algorithm

Viterbi algorithm performs ML decoding

finds a path through trellis with the largest metric,

processes demodulator outputs in an iterative manner,

At each step in the trellis, it compares the metric of

all paths entering each state, and keeps only the path

with the largest metric, called the survivor, together

with its metric,

It proceeds in the trellis by eliminating the least likely

paths.

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Example of Hard decision

Viterbi decoding

1/11

0/00

0/10

1/11

1/01

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

0/00

0/11

0/00

6t

1t

2t 3t 4t 5

t

)101(m

)1110001011(U )1110000011(Z

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Viterbi decoding-contd Label all branches with the branch metric (Hamming distance)

0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0

i =1

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Viterbi decoding-contd

0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0 2

0

State metricPath metric

i=2

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0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0 2

0

2

4

1

1

i=3

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0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0 2

0

2

4

1

1 2

2

1

2

XX

XX

i=4

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0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0 2

0

2

4

1

1 2

2

1

2 3

1

XX

XX

X

i=5

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0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0 2

0

2

4

1

1 2

2

1

2 3

1

XX

XX

X

i=6

X

1

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0

2

1

2

1

0

2

1

1

2

1

0

0

1

0

2

1

0

2

6t

1t

2t 3t 4t 5

t

1

0 2

0

2

4

1

1 2

2

1

2 3

1

XX

XX

X

X

1

)101(m)101(m

)1110001011(U)1110000011(Z

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Viterbi decoding

1/11

0/00

0/10

1/11

1/01

0/00

0/11

0/10

0/01

1/11

1/01

1/00

0/00

0/11

0/10

0/01

0/00

0/11

0/00

6t

1t

2t 3t 4t 5

t

)101(m

)1110001011(U )0110111011(Z

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0

2

0

1

2

1

0

1

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0

i=1

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0

2

0

1

2

1

0

1

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0 2

0

i=2

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0

2

0

1

2

1

0

1

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0 2 3

0

2

30

i=3

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0

2

0

1

2

1

01

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0 2 3 0

3

2

3

0

2

30

i=4

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0

2

0

1

2

1

01

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0 2 3 0 1

3

2

3

20

2

30

i=5

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0

2

0

1

2

1

01

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0 2 3 0 1 2

3

2

3

20

2

30

i=6

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Example of Hard decisionViterbi decoding-contd

0

2

0

1

2

1

01

1

0

1

2

2

1

0

2

1

1

1

6t

1t

2t 3t 4t 5

t

1

0 2 3 0 1 2

3

2

3

20

2

30

)100( m

)101(m

)1110001011(U

)0110111011(Z

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Free distance of Convolutional

Codes Since the code is linear, the minimum distance of thecode is the minimum distance between each of thecodewords and the all-zero codeword.

This is the minimum distance in the set of all arbitrarylong paths along the trellis that diverge and remergeto the all-zero path.

It is called the minimum free distance or the free

distance of the code, denoted by

ffree dd or

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Free distance

2

0

1

2

1

0

2

1

1

2

1

0

0

2

1

1

0

2

0

6t

1t

2t

3t

4t 5t

Hamming weight

of the branchAll-zero path

The path diverging and remerging to

all-zero path with minimum weight

5fd

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Performance bounds

Error correction capability of Convolutional codes isgiven by,

The coding gain is upper bounded by,2/)1(

fdt

)(log10gaincoding 10 fcdR

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How to end-up decoding?

In the previous example it was assumed that theregister was finally filled with zeros thus finding theminimum distance path

In practice with long code words zeroing requiresfeeding of long sequence of zeros to the end of themessage bits: this wastes channel capacity &introduces delay

To avoid this path memory truncationis applied:

It has been experimentally tested that negligibleerror rate increase occurs by truncating at 5L

Note this also introduces a delay of 5L

dct lecture 24 2011

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