Convolution codes （卷积码）

Chapter 8

李莉 . 信息机电学院 . SHNU

Linearcodes: 1 infor. bit 1 codeword;

(n,k)or [n,k,d] ， k-bit infor. one n-bit codeword

Convolution code:. 1 infor. bit m codewords; (n,k,m)

k-bit infor.+m previous infor.bits 1 n-bit codeword

Representation to convolution code:

Algebraically, discrete convolution, generator matrices,

polynomials;

解析法：离散卷积法、生成矩阵法、码多项式法；Graphically, tree, trellis, state diagrams.

图形法：树图法、格图法、状态图法 .

SHNU. 信息机电学院

8.1 Convolution ( 卷积 ) The discrete convolution.

1)Linear feed-forward shift register

the links: g0, g1, g2, g3, g4 = 0, the link is absent

or 1, the link is present.

Input sequence: u=( u0 u1, u2,u3,…) ( 输入码序列 ).

output sequence: v=( v0 v1, v2,v3,…) ( 输出码序列 ).

Assuming the initial states of the shift registers =0, output:

1st tick: v0=u0g0.

2nd tick: v1=u1g0+u0g1.

3rd, v2=u2g0+u1g1+u0g2.

4th , v3=u3g0+u2g1+u1g2+u0g3. (8.1)

5th , v4=u4g0+u3g1+u2g2+u1g3+u0g4.

6th , v5=u5g0+u4g1+u3g2+u2g3+u1g4.

u vg4

g3g2

g1g0

A register with m stages each input ~ m+1 outputs.


2) Outputs for a finite input sequence

The last input should appear at the output,

an addition m zero inputs are required.

The input: u=(u0 u1 u2 … u7,0000), the last outputs:

v7= u7g0+u6g1+u5g2+u4g3+u3g4. m=4

v8= u8g0+u7g1+u6g2+u5g3+u4g4.

v9= u9g0+u8g1+u7g2+u6g3+u5g4.

v10= u10g0+u9g1+u8g2+u7g3+u6g4.

v11= u11g0+u10g1+u9g2+u8g3+u7g4.

Setting u8=u9=u10=u11=0,

All the 7 inputs appear at the output, the output sequence is completed.

Any further zero inputs a zero output. The additional m 0 bits returns reg. to zero state. The m 0 should not be confused with the information bits.

v8=u7g1+u6g2+u5g3+u4g4.

v9= u7g2+u6g3+u5g4.

v10= u7g3+u6g4.

v11= u7g4.


3) Discrete convolution

For a LFFSR with m stages and links gi, i=0,1,2,…,m feedin

g into a modulo-2 adder, the output from the adder is :

vj= uj-rgr.= u*g (8.2, 8.3)

where uj-r=0, if j<r. operation * denotes convolution.

Each output vj is a convolution of (m+1) input with g0, g1,…,gm.

Generator sequence g=(g0, g1,…,gm) ,生成序列 Discrete convolution Physical vision ( 物理含义 ):

10 A window of (m+1) bits are multiplied by g0, g1,…,gm and a

dded together.

20 The window is then slid along the input sequence by 1 bit

and multiplication and summation repeated.

30 The process continues until the window has covered the e

ntire input sequence.

m

0r


4) Determination of generator sequence

10 by inspection of the shift register circuit or

20 by considering the impulse response when the input u=

(1 000…0), 1 followed by m 0s.

The impulse response gives the generator sequence g.

N.B. The generator sequence is not the output of the shift re

gister, But the output from the modulo-2 adder.

I.e A line of stages feeding into more than one modulo-

2 adder will have a generator sequence for each adder

( 同一列移存器送入二个以上的模 2 加法器，则每个加法器输出有自己的生成序列 ).

by inspection ： v1 and v2 g sequences

g1=(g0 g1 g2 g3)=(0011)

g2=(g0 g1 g2 g3)=(1011)

v2u g3

g2g0

v1


8.2 Encoding convolution codes (n,k,m) Convolution Code (CC).

1) Structure

10 k LFSRs, num. of stages may different.

(in Space)ith LFSR’s input sequence:

u(i)=(u0(i)u1

(i)u2(i) …) , i=1,2,…,k. (8.4)

(in Time)Bth tick, k-bit input block

uB=(uB(1)uB

(2) …uB(k)) , B=0,1,2,…,L,....

The whole u=(u0(1)u0

(2) …u0(k),

u1(1)u1

(2) …u1(k),

u2(1) u2

(2) …u2(k) ,…) (8.5)

Fig.8.4 Encoder for an (n,k,m) convolution code

v

v(1)

v(2)

v(n)

v(n-1)

v(3)

v(4)

u

u(1)

u(2)

u(k)

20 Memory order ( 存储器级数，编码存储 ):

The max. num. of stages m in any shift register.

(k-bits input 在编码器中需存储的单位时间数。 )SHNU. 信息机电学院

30 Output: n modulo-2 adders

whose inputs depends on the particular code.

jth adders output sequence:

v(j)=(v0(j)v1

(j)v2(j) …) , j=1,2,…,n. (8.5)

Bth tick, n-bit output block

vB=(vB(1) vB

(2) …vB(n)) , B=0,1,2,…,L+m,….

the whole v=(v0(1)v0

(2) …v0(n),

v1(1)v1

(2)…v1(n), v2

(1)v2(2)…v2

(n) , …) (8.7)

40 Constraint length ( 编码约束长度 ): NAn(m+1)

( 相互约束的码元个数 )

**Constraint ( 约束度 ): N=m+1. ( 相互约束的码段个数 ).

m:the parameter of the complexity of an encoder.

1 in-bit affects the n out-bits, 1 in-bit stays in the encoder for up to m+1 inputs, 1 input bit affects n(m+1) output bits.


The n-bit vL block leaving the encoder depends on

the k-bit entering the encoder

up to m previous k-bit input blocks.

2) Finite Input and output : If u(i) is a finite sequence

u(i)=(u0(i)u1

(i)u2(i) … uL-1

(i) ) , i=1,2,…,k. L=length(u(i))

uB=(uB(1)uB

(2) …uB(k)) , B=0,1,2,…,L-1.

u is a total of kL bits input.

An output sequence v(j) of length L+m,

v(j)=(v0(j)v1

(j)v2(j) … vL+m-1

(j) ) , j=1,2,…,n.

vB=(vB(1) vB

(2) …vB(n)) , B=0,1,2,3…,L+m-1

v=(v0(1)v0

(2) …v0(n), v1

(1)v1(2)…v1

(n), v2(1)v2

(2)…v2(n) , …

v(1)Lv

(2)L…v(n)

L, …, v(1)L+m-1v

(2)L+m-1…v(n)

L+m-1)

(a total of n(L+m) bits output) the last m blocks of v the additional m zero inputs required to

return the encoder to its zero state.SHNU. 信息机电学院

3) The generator sequences :

A LFSR n modulo-2 adders n generator sequences.

& the (n,k,m) CC has k LFSR nk generator sequences.Eg8.1a An encoder for the (4,3,2) CC.

Input k=3, outputs n=4 ,

memory order m=2.

A total of 4*3=12 g sequences to

determine the output.

Let g(i,j)=(g0(i,j) g1

(i,j) g2(i,j) …gm

(i,j)) An (4,3,2) CC Encoder

v(1)

v(2)

v(3)

v(4)u

u(1)

u(2)

u(3)

the g sequence connecting the ith input to the jth output.

g(1,1)=(g0(1,1)g1

(1,1)g2(1,1))=(100); g(1,2)=(100); g(1,3)=(100); g(1,4)=(100);

g(2,1)=(g0(2,1)g1

(2,1)g2(2,1))=(000); g(2,2)=(110); g(2,3)=(010); g(2,4)=(100);

g(3,1)=(g0(3,1)g1

(3,1) g2(3,1))=(000);g(3,2)=(010); g(3,3)=(101); g(3,4)=(101);

(8.8) SHNU. 信息机电学院

Example 8.1b

Determine the output sequence from the (4,3,2) CC encoder s

hown in Eg.8.1a, given the input sequences u=(110,011,1

01)

Solution:

The input blocks is L=3 , each in-block length =3-bit.

The register order m=2.

The output blocks are (L+m)=5, each output block=4-bit.

The total number of output bits are

(L+m)n=(3+2)*4=20bits.

ur=0, if r<0.

First the inputs should be transformed from sequence to pa

rallel blocks of 3 bits

( 输入序列要进行串并转换，一组组地同时送入 3个 LFSR).


∵ v(i,j)=u(i)*g(i,j), & the 3 LFSR’s input sequences

u(1)=(u0(1) u1

(1) u2(1)00) =(10100),

u(2)= (11000),

u(3)= (01100)

Then the output sequence to each adder,

Ticks: r = 0, 1, 2, 3, 4, 5 …

1st adder: vr(1)=ur

(1)= 1, 0, 1, 0, 0, 0,…

2nd adder : vr(2)=ur

(1)+ur(2)+ur-1

(2)+ur-1(3)= 0, 0, 1, 1, 0, 0,…

3rd adder : vr(3)=ur

(1)+ur-1(2)+ur

(3)+ur-2(3)= 1, 0, 1, 1, 1, 0,…

4th adder : vr(4)=ur

(1)+ur(2) +ur

(3)+ur-2(3) = 0, 0, 0, 1, 1, 0,…

Finally transform from the 4 bits parallel block into the

final output sequence

v=(1010 0000 1110 0111 0011 0000…) SHNU. 信息机电学院

4) Error-control properties ( 各种距离度量 )

The error-control properties of CC the distance character

istics of the resulting encoded sequences.

However with CCs there are several minimum distance mea

sures. The most important distance measure is

Minimum free distance: it is defined as the minimum distan

ce between any two encoded sequences.

As CCs are linear codes

v1, v2 v3 =v1+v2, WH(v3)=d(v1,v2) . & v30 if v1v2.

the free distance of a CC = the minimum-weight sequenc

e of any length produced by a nonzero input sequence.

The definition of free distance includes all sequences an

d not just sequence with the same length.


8.3 Generator matrices for convolutional codes

( 卷积码的生成矩阵 )

If we define a matrix G, its elements are the g sequence,

Given an input sequence u, the output sequence v can be obt

ained by matrix multiplication:

v =uG (8.14)

G0 G1 G2 … Gm 0 0 …0 G0 G1 G2 … Gm 0 …0 0 G0 G1 G2 … Gm …. …

G= (8.15)

gr(1,1) gr

(1,2) … gr(1,n)

gr(2,1) gr

(2,2) … gr(2,n)

┇ ┇gr

(k,1) gr(k,2) … gr

(k,n)

Gr= (8.16)


8.4 Generator polynomials for convolution codes 1) Definition of input, output, and generator

polynomials of CC

** The input sequence u=(u0u1u2…) , the input poly. u(D)

u(D)=u0+u1D+u2D2+… (8.18)

D is the unit-delay operator , a delay of 1 bit. E.g.

u1 is 1 bit delay relative to u0.

Bit u2 is 2-bit delay relative to u0,

1-bit delay relative to u1.

** An output sequence v=(v0v1v2…) , output poly. v(D)

v(D)=v0+v1D+v2D2+… (8.19)

** If v(D) is the output arising from u(D) then

v(D)=u(D)g(D) (8.20) the generator polynomial. SHNU. 信息机电学院

2) Determination of g(D)

The g(D) from the encoder.

** Each link to a modulo-2 adder contributes a Dr term to g

(D)

r is the number of stages that a bit has to pass through to

arrive at the adder.

** The rth component of

the generator sequence g=(g0, g1,…,gm)

gr= 0 , exist a link to the adder, after the rth stage

1 , no link to the adder, after the rth stage

rth component in the g(D) is grDr.

** For m stages, the generator polynomial g(D)=g0+g

1D+g2D2+…+gmDm (8.21)


3) The generator polynomials for an arbitrary (n,k,m) CC

** An (n,k,m) CC has k generator polynomials for each of the

n output sequences, a total of nk generator polynomia

ls. g(i,j)(D) gr(i,j)Dr (8.22)

g(i,j)(D): for the jth output arising from the ith input,

gr(i,j): the components of the generator sequence g(i,j).

** The jth output polynomial for (n,k,m) CC is

v(j)(D)= u(i)(D)g(i,j)(D) (8.23)

j=1,2,…,n.

The total output v(D):

v(D)= Dj-1v(j)(Dn) (8.24)

The components of v given by the coefficients of v(D).

m

0r

n

1j

k

1i


8.5 Graphical representation of convolution codes

( 卷积码的图形表示 , 树图 , 格图，状态图 ) CCs can be represented graphically using tree, trellis and s

tate diagrams ( 树图，格图，状态图 ),

show the state of a register and

the encoder output for all possible inputs.

The state of a register: the contents of the stages at a give

n point during encoding.

The trellis diagram plays an important role when decoding

using the Viterbi algorithm

Take (2,1,2) as an example to illustrate the graphical represen

tations.


Example 8.2

An encoder for (2,1,2) CC. Structure.

at input point j, the encoder can be in one of the 4 states

Qj-1Qj-2=Si=00,01,10,11, i=0,1,2,3.

The output: as sequence vj(1)vj

(2).

vj(1) = Qj+Qj-1+Qj-2

vj(2) = Qj +Qj-2

j=0,1,2,3,…Specifications:

The encoder has memory M=2.

The code's rate is k/n=1/2.

The encoder has three flip-flops;

the code has constraint 3

A (2,1,2) convolution encoder

Qj-2

v

u v(2)v(1)

Qj-1Qj


1) (2,1,2) CC. State diagrams ( 状态图 ):

Branches: transitions from one state Si

to another Sj.

The output: next to the relevant branch.

Given an arbitrary input u=(u0u1u2

…)

The output,next state following the t

ransitions via the state diagram, v

=(v0(1)v0

(2)v1(1)v1

(2)…) is nonsystematic

CC.

a) the input sequence u=(1101 00).

Find the output of the (2,1,2) CC encod

er.

11

10 01

0011 11

10

1001 01

00

00

States Qj-1Qj-2

S0=00S1=01S2=10S3=11

Outvj

(1)vj(2)

0 input1 input

State diagram for the (2,1,2) CC.

2bits: return to zero state


For (2,1,2) CC, & input u: The register order m=2. L=4 ,

The output blocks are (L+m)=6, length(output block)=2-bit.

The total number of output bits: (L+m)n=6*2=12bits.

starting from S0=00, Ticks j Input

bit Qj

Qj-1Qj-2 output vj

(1)vj(2)

Transition to Qj-1Qj-2

1 1 00 10=S2

2 1 10 11=S3

3 0 11 01=S1

4 1 01 10=S2

5 0 10 01=S1

6 0 01

11010100

1011 00=S0

0 00 00 S0

11

10 01

0011 11

10

1001 01

00

00

0 input1 input

The final output:

v=(11,01,01,00,10,11).

return to zero state


States Qj-1Qj-2

S0=00S1=01S2=10S3=11

Outvj

(1)vj(2)

1 input0 input

2) Tree diagrams ( 树图 ) for the (2,1,2) CC.

StartS0

S0

S0

S0

S2

S2

S3

S3

S2

S1

S1

S0

S2

S3

S1

11

00

0000

11

01

10

10

01

00

11

01

10

11

10010011011011001001001101101100

S3

S1S2

S0S3

S1S2

S0

S3

S1S2

S0S3

S1S2

S0

input sequence u=(1101 00)

output:

v=(11,01,01,00,10,11).

output

TimeSHNU. 信息机电学院

3) Trellis diagrams ( 格图 ) for the (2,1,2) C:

States Qj-1Qj-2

S0=00S1=01S2=10S3=11

Outvj

(1)vj(2)

0 input1 input

input sequence u=(1101 00)

output: v=(11,01,01,00,10,11). 6 ticks needed

Time

S0

S1

S2

S3

00

11

10

01

10

11

10

01

00

11

01

10

11

00 00 00

11

01

10

00

01

00

11

CurrentState

NextState

Output

10

01

11

10

00

01

00

11

10

11

00

01

11

00StartTime j = 1 2 3 4 5 6

7


8.6 The Viterbi decoder

1) The Viterbi decoder is a ML decoder.

find the path that is the least distance away from the deco

der input w. The distance metrics.

2) Branch metric ( 分支度量 ): For a BSC,

the branch metric is the Hamming distance between the br

anch output and the corresponding decoder input ( 译码器输入 w 的子码。 ).

3) Path metric ( 部分路径度量 )

Each path metric = (branch metrics forming the path).

Survivor branch and path ( 留存支路和部分路径 ): The branc

h and the path with lower metric. If the path metrics a

re the same then either path can be taken as the survivo

r.

Note that metric values depend on the decoder input and ar

e not fixed attributes of a trellis. SHNU. 信息机电学院

4) Viterbi's decoding algorithm for (2,1,2) CC

S= {a,b,c,d: S0,S1,S2,S3}. If s, t S ,

B(s,t) = 0 transition by a 0 input.

1 transition by a 1 input.

not defined. No links

lj-1,j(s,t) = the Hamming distance between the enc

oder 2-bit output and the received 2-bit subc

ode, on the trellis edge joining sj-1 to tj.

l j-1,j(s,t)= + . no such edge

Viterbi's algorithm computes 2 metrics:

The metrics j(s), sS,: the length of the shortest

path from a0 to sj;

the survivor Bj(s) is a binary string of length j whi

ch represents a shortest path from a0 to sj.

s

t a 0 - 1 -b 0 - 1 -

c - 0 - 1d

a b c d

- 0 - 1

10

00

01

00

11

S0

S1

S2

S3


10. Initially set 0(a) 0, and 0(a)=+ for all sa.

Set B0(a)= ( 空 ), j=1.

20. For each sS, find a tS for which j-1(t)+lj-1,j(s,t) is a minim

um. Then set

j(s) j-1(t)+ lj-1,j(s,t),

Bj(s) Bj-1(t)*B(s,t).

The operation * : concatenation( 串联 ), e.g. 1101*0=11010

30. If j=L+m, output the first L bits of Bj(a) and stop;

otherwise set j j+1, go to step 2.

L: the information length,

m: the degree of the encoder.


Key points to chapter 8:1. Concept of (n,k,m) convolution code, 2. (n,k,m) CC Encoder input k-bit infor block. output n-bit codeword block. Each LFSR input: length L The LFSR Total Output : (L+m)*n bits

3. Representations to conv. code G g(D) Graphical: tree, trellis, state diagram

4. Viterbi decoder Understand.


Homework:

1. http://xxjd.shnu.edu.cn/ 下载中心 /more...

/problems.pdf

P255, 8.1 8.2

8.5 8.6


END & THANK YOU ！


Documents

Convolution codes （ 卷积码）

Convolution codes （卷积码）