hatilima jasper 11042005 hw03

7/31/2019 Hatilima Jasper 11042005 HW03

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0 ( )

O --- 0

11042005 D31207

HatilimaJasper V.

1. Given a (2,1,3) convolutional code with generator polynomials (13,15)8=(1+x+x , 1+x +x ),

1). Draw the corresponding encoder diagram, state diagram, and trellis.

2). Determine its transfer function by using Masson signal flow analysis.

3). List the first three Hamming weight terms.

4). If the received sequence over binary symmetric channel (BMC) is given by

1 0, 0 0, 1 1, 1 0, 0 1, 1 1, 0 0, 1 0, 0 0.

If Hamming correlation is utilized as the incremental path metric (namely, H(0,1)=H(1,0)=1,H(1,1)=H(0,0)=0), please determine the restored message sequence by following the maximumlikelihood decoding algorithm steps.

2. Given a (2,1,3) recursive systematic convolutional code with generator polynomials(15/13)8=(1+x+x

3/1+x

2+x

3),

1). Draw the corresponding encoder diagram, state diagram, and trellis.

2). Determine its transfer function by using Masson signal flow analysis.

3). List the first three Hamming weight terms.

Notes:

1. The homework can be written either in English or in Chinese.2. Format and print your homework report on 176x250mm A4 white paper, using 10pt Times New

Roman (or 5) for single-spaced text and 16pt Times New Roman (or 3 ) for title.The left, right, top and bottom margins should all be set to 20mm.

3. Print out your report according to the required page layout requirement and hand in it together withthis sheet as cover.


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QUESTION 13

1 813 1g X X and2 3

2 815 1g X X

Code= ( , , )n k m = ( , , 1)cn k L = (2, 1, 3)Part 1: Encoder diagram, State diagram, Trellis diagram

Encoder Diagram:

a(l-1) a(l-2) a(l-3)

X1

X1

u(l)

State Diagram:This encoder can assume one of the eight possible states which we will designate by:

0

1

2

3

000

100

010

110

s

s

s

s

4

5

6

7

001

101

011

111

s

s

s

s

S1

100

S4

001

S0

000

S2

010

S3

110

S6

011

S5

101

S7

111

1/01

1/10

1/11

1/00

1/001/100/10

0/00

1/111/01

0/00

0/10

0/11 0/01

0/01

0/11


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Trellis Diagram:

S0=000

S4=001

S2=010

S6=011

S1=100

S5=101

S3=110

S7=111

1/11

1/01

1/00

1/01

1/10

1/10

1/00

1/11

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

0/10

0/01

0/11

0/10

0/11

0/00

0/01

0/00 0/00T=0 T=1 T=2 T=3 T=4 T=5 T=6

Part 2:Transfer function from Masons formula:

We first transform the state diagram by splitting the zero state as show below and using the following notations for the

transitions:

L : The number of nodes traversed

W: the hamming weight of the input vector (uncoded input).D: the hamming weight of output vector (coded output).

The transformed state diagram becomes:


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4

S1

100

S4

001

S0

000

S2

010

S3110

S6011

S5

101

S7

111

WDL

WDL

WL

WL

DL

WDL

L

DL

DL

DL

D^2 L

S0

000

WD^2 L

WDL

WD^2 LD^2 L

Masons formula:

( , , )i i

i

F

T W D L

Where:1 . .

i t m j v h

i t m j v h

L L L L L L

C C C C C C

Andi

is calculated in a similar manner but deletes the loops touchingi

K from the summation.

Forward Loops [K]:3 6 6

1 0 1 3 7 6 4 0 1K S S S S S S S F W D L 2 8 5

2 0 1 3 6 4 0 2K S S S S S S F W D L 3 10 7

3 0 1 3 6 5 2 4 0 3K S S S S S S S S F W D L 6 4

4 0 1 2 4 0 4K S S S S S F WD L 4 8 8

5 0 1 3 7 6 5 2 4 0 5K S S S S S S S S S F W D L 4 8 8

6 0 1 2 5 3 7 6 4 0 6K S S S S S S S S S F W D L

3 10 7

7 0 1 2 5 3 6 4 0 7K S S S S S S S S F W D L Other loops [L]:


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5

2 2

1 2 5 2 1

2 3

2 1 2 4 1 2

2 4 4

3 1 3 6 4 1 3

3 2 5

4 1 3 7 6 4 1 4

L S S S C WD L

L S S S S C WD L

L S S S S S C W D L

L S S S S S S C W D L

2

5 7 7 5

3 6 6

6 1 3 6 5 2 4 1 6

4 4 7

7 1 3 7 6 5 2 4 1 7

4 4 7

8 1 2 5 3 7 6 4 1 8

3 6 6

9 1 2 5 3 6 4 1 9

2 4 3

10 5 3 6 5 10

3 2 4

11 5 3 7 6 5 11

L S S C WD L

L S S S S S S S C W D L

L S S S S S S S S C W D L



L S S S S C W D L

L S S S S S C W D L

Product of non-touching pairs of loops :2 2 2 2 4 31 5

3 2 4 2 3 4 4 7

11 2

2 3 6 6 4 8 7

5 6

2 4 4 3 6 5

5 3

2 3 6 6 4 8 7

5 9

2 4 3 2 3 3 6 6

10 2

2 4 3 2 3 6 4

10 5

1 3

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

L L WD L WD L W D L

L L W D L WD L W D L

L L WD L W D L W D L

L L WD L WD L W D L




L L

2 2 2 4 4 3 6 6

2 3 2 2 4 4

2 5

2 2 3 2 5 4 4 7

1 4

( )( ) ( )

( ) ( )

WD L W D L W D L

L L WD L WD L W D L


Product of non-touching triplets of loops :2 2 3 6 5 4 8 7

1 3 5

2 4 3 2 4 4 4 8 7

10 2 5

( ) ( )

( ) ( )

L L L WD L W D L W D L

L L L W D L W D L W D L

There are no sets of four or more non-touching loops.

1i t m j v h

i t m j v h

L L L L L L

C C C C C C

2 2 2 3 2 4 4 3 2 5 2 3 6 6

4 4 7 4 4 7 3 6 6 2 4 3 3 2 4 2 4 3

4 4 7 4 8 7 3 6 5 4 8 7 3 6 6 3 6 4

3 6 6 2 4 4 4 4 7 4 8 7 4 8 7

1 (

)

WD L WD L W D L W D L WD L W D L

W D L W D L W D L W D L W D L W D L


W D L W D L W D L W D L W D L

2 2 3 3 2 5 4 3 6 5 41 ( ) ( ) ( )WD L L L W D L L W D L L And we have


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6

2 2

1 1

2 2 2 2 4 3

2 1 5 1 5

2

3 5

4 10 11 5 10 5

2 4 3 3 2 4 2 3 6 4

5

6

2

7 5

1 1

1 1

1 1

1

.... 1

1

1

1 1

L WD L

L L L L WD L WD L W D L

L WD L

L L L L L

W D L W D L WD L W D L

L WD L

1 1 2 2 3 3 4 4 5 5 6 6 7 7

( , , )i i

i

F

T W D L

F F F F F F F

3 6 6 2 2 2 8 5 2 2 2 2 4 3

3 10 7 2

6 4 2 4 3 3 2 4 2 3 6 4

4 8 8 4 8 8 3 10 7 2

2 2 3 3 2 5 4 3 6 5 4

(1 ) (1 )(1 )

(1 )

(1 )

1 ( ) ( ) ( )

W D L WD L W D L WD L WD L W D LW D L WD L

WD L W D L W D L WD L W D L

W D L W D L W D L WD L

WD L L L W D L L W D L L

Part 3: Listing the first three hamming weight terms

0

( , , )P

P

T W D L aS b

2

0 0 0 0 0 0a WD L 2[ 0 0 0 0 0 Tb D L

2

2

0 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

WL

DL WDL

DL WDL

S DL WDL

DL WDL

D L WL

L WD L

2P L For 4,5,6L , we have 2,3,4P . We therefore obtain the first three hamming weight terms as follows:

2

4 ( , , )LT W D L aS b


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2 2

2

2

2

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

WL D L

DL WDL

DL WDL

WD L DL WDL

DL WDL

D L WL

L WD L

2

4 3 2 5 3 2 4 3 3 3 3

0

0

[ 0 0 0 ] 0

0

0

0

D L

WD L W D L W D L W D L

6 4WD L

3

5 ( , , )LT W D L aS b 2 2

0 0 0 0 0 0WD L S S b

4 3 2 5 3 2 4 3 3 3 3[ 0 0 0 ]WD L W D L W D L W D L Sb 2 8 5W D L

4

6 ( , , )LT W D L aS b 2 3

0 0 0 0 0 0WD L S S b 2 6 4 2 5 4 3 3 4 2 4 4 3 6 4 3 5 4 4 5 4[ ]W D L W D L W D L W D L W D L W D L W D L S b

2 8 6 3 6 6W D L W D L

6 4

4LT WD L 2 8 5

5

2 8 6 3 6 6

6

L

L

T W D L

T W D L W D L

Part 4: Decoding the received sequence

The maximum likelihood decoder maximizes the likelihood function over the entire sequence( )arg max ( )m

m

L P Z U

Where Z is the received code sequence, U is the possible code sequence.

So the procedure will be a special case of maximum likelihood that has a Markov property, it is the Viterbi algorithm and

with require the calculation of incremental path metric

{ ( ) ( )} { ( ), ( )}H

z l u l d z l u l

Then adding the incremental path metric to the old cumulative state metric, i.e.


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( ) ( 1) { ( ), ( )}j j HM l M l d z l u l where 0,1,...,2 1m

j

At 0t :

Timei

t Received bits,

Z Cumulative State Metric ( )

jM l

0t -0 0 0( ) (000) 0M S M

1t 1 01 0 1 0 0( ) (000) ( ) (10, 00) 1

HM S M M S d

1 1 1 0 0( ) (100) ( ) (10,11) 1

HM S M M S d

2t 0 02 0 1 0( ) ( ) (00, 00) 1HM S M S d

2 1 1 0( ) ( ) (00,11) 3

HM S M S d

2 2 1 1( ) ( ) (00,10) 2HM S M S d

2 3 1 1( ) ( ) (00,01) 2

HM S M S d

3t 1 13 0 2 0

( ) ( ) (11, 00) 3HM S M S d

3 1 2 0( ) ( ) (11,11) 1HM S M S d

3 2 2 1( ) ( ) (11,10) 4HM S M S d

3 3 2 1( ) ( ) (11,01) 4HM S M S d

3 4 2 2( ) ( ) (11, 01) 3HM S M S d

3 5 2 2( ) ( ) (11,10) 3HM S M S d

3 6 2 3( ) ( ) (11,11) 2

HM S M S d

3 7 2 3( ) ( ) (11, 00) 4HM S M S d From here, the states will be entered by two paths and the one that gives a larger

state metric is eliminated.

4t 1 03 0

4 0

3 4

( ) (10,00) 4( ) {

( ) (10,11) 4..

H

H

M S dM S

M XS d

3 0

4 1

3 4

( ) (10,11) 4( ) {( ) (10, 00) 4..

H

H

M S dM SM XS d

3 1

4 2

3 5

( ) (10,10) 1( ) {

( ) (10, 01) 5..

H

H

M S dM S

M XS d

3 1

4 3

3 5

( ) (10,01) 3( ) {

( ) (10,10) 3..

H

H

M S dM S

M XS d

3 2

4 4

3 6

( ) (10, 01) 6..( ) {

( ) (10,10) 2

H

H

M S dM S

M

X

S d

3 24 5

3 6

( ) (10,10) 4..( ) {

( ) (10,01) 4H

H

M S dM S

M

X

S d

3 3

4 6

3 7

( ) (10,11) 5( ) {

( ) (10, 00) 5..

H

H

M S dM S

M XS d

3 3

4 7

3 7

( ) (10, 00) 5( ) {

( ) (10,11) 5..

H

H

M S dM S

M XS d

So from the above, the elimination is done. For the paths entering a node and with

equal values of the cumulative state metric, one is arbitrarily chosen. The same


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algorithm is done for the remaining stages and only the surviving paths are shown in

this table from 5t to 9t 5t 0 1

5 0 4 0( ) ( ) (01,11) 3

HM S M S d

5 1 4 4( ) ( ) (01, 00) 3HM S M S d

5 2 4 5( ) ( ) (01, 01) 4HM S M S d

5 3 4 1

( ) ( ) (01, 01) 4H

M S M S d

5 4 4 2( ) ( ) (01,01) 1HM S M S d

5 5 4 2( ) ( ) (01, 01) 3HM S M S d

5 6 4 3( ) ( ) (01,11) 4

HM S M S d

5 7 4 3( ) ( ) (01, 00) 4HM S M S d

6t 1 16 0 5 4

( ) ( ) (11,11) 1HM S M S d

6 1 5 4( ) ( ) (11, 00) 3HM S M S d

6 2 5 5( ) ( ) (11, 01) 4HM S M S d

6 3 5 5( ) ( ) (11,10) 4

HM S M S d

6 4 5 4( ) ( ) (11,10) 5HM S M S d

6 5 5 6( ) ( ) (11, 01) 5HM S M S d

6 6 5 3( ) ( ) (11,00) 3

HM S M S d

6 7 5 7( ) ( ) (11,11) 5HM S M S d

7t 0 07 0 6 0( ) ( ) (00, 00) 1HM S M S d

7 1 6 0( ) ( ) (00,11) 3HM S M S d

7 2 6 1( ) ( ) (00,10) 4

HM S M S d

7 3 6 1( ) ( ) (00, 01) 4HM S M S d

7 4 6 6( ) ( ) (00,10) 4HM S M S d

7 5 6 6( ) ( ) (00, 01) 4

HM S M S d

7 6 6 7( ) ( ) (00, 00) 5HM S M S d

7 7 6 3( ) ( ) (00, 00) 4HM S M S d

8t 1 08 0 7 0( ) ( ) (10,00) 2HM S M S d

8 1 7 0( ) ( ) (10,11) 2

HM S M S d

8 2 7 1( ) ( ) (10,10) 3HM S M S d

8 3 7 5( ) ( ) (10,10) 4

HM S M S d

8 4 7 6( ) ( ) (10,10) 5

HM S M S d

8 5 7 2( ) ( ) (10,10) 4HM S M S d

8 6 7 7( ) ( ) (10, 00) 5HM S M S d

8 7 7 7( ) ( ) (10,11) 5

HM S M S d

9t 0 09 0 8 0( ) ( ) (00,00) 2HM S M S d

9 1 8 0( ) ( ) (00,11) 4

HM S M S d

9 2 8 1( ) ( ) (00,10) 3HM S M S d

9 3 8 1( ) ( ) (00, 01) 3

HM S M S d


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9 4 8 2( ) ( ) (00,01) 4HM S M S d

9 5 8 2( ) ( ) (00,10) 4HM S M S d

9 6 8 7( ) ( ) (00,00) 5

HM S M S d

9 7 8 3( ) ( ) (00, 00) 4HM S M S d

From the state metrics at 9t , we trace the surviving paths back and have the following trellis:

S0=000

S4=001

S2=010

S6=011

S1=100

S5=101

S3=110

S7=111

0/00 0/00 0/00T=0 T=1 T=2 T=3 T=4 T=5 T=6

1/11

T=9

2

4

4

3

5

4

4

3

0/11

0/01

0/10

0/000/000/00

We see that state0

S has the lowest metric of value 2 and so we trace it back into the trellis to produce the following code

sequence:

00 00 11 10 01 00 00

Which is then decoded to the following information bits:0 0 1 0 0 0 0 0, but the last three zeros are for zero termination. Therefore the final sequence of bits is:

0 0 1 0 0

QUESTION 2: RECURSIVE SYSTEMATIC CONVOLUTION ENCODER3

1 1g X X and2 3

2 1g X X Part 1: Encoder diagram, State diagram, Trellis diagram:

1 1( ) ( ) ( ) ( )X D U D x l u l

(1)

32 1 1

2

( )( ) ( ) ( ) ( ) ( ){1 }

( )

U DX D g D A D g D A D D D

g D

32 ( ) ( ) ( ) ( )X D A D A D D A D D

(2)Where

2 3

2

( ) ( )( )

( ) 1

U D U DA D

g D D D


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2 3( ) ( ) ( ) ( )A D A D D A D D U D 2 3( ) ( ) ( ) ( )A D U D A D D A D D (3)

Transforming equations (2) and (3) into time domain:

32 2( ) ( ) ( ) ( ) ( ) ( ) ( 1) (1 3)X D A D A D D A D D x l a l a l a

2 3( ) ( ) ( ) ( ) ( ) ( ) ( 2) ( 3)A D U D A D D A D D a l u l a l a l

Note that ( )u l is the input message bit sequence and ( )a l are the register contents.

Encoder Diagram:

a(l-1) a(l-2) a(l-3)

X1

X2

u(l) a(l)

State Diagram:

S1

100

S4

001

S0

000

S2

010

S3

110

S6

011

S5

101

S7

111

1/10

0/01

1/11

0/00

0/000/010/01

0/001/11

1/10

0/00

0/01

1/11 1/10

1/10

1/11

Trellis Diagram:


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S0=000

S4=001

S2=010

S6=011

S1=100

S5=101

S3=110

S7=111

1/11

1/10

0/00

1/10

0/01

0/01

0/00

1/11

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

0/01

1/10

1/11

0/01

1/11

0/00

1/10

0/00T=0 T=1 T=2 T=3 T=4 T=5

Part 2: Transfer function from Masons formula:

To determine the transfer function we first transform the state diagram by splitting the zero state as show below and usingthe following notations for the transitions:

L : The number of nodes traversed

W: the hamming weight of the input vector (uncoded input).

D: the hamming weight of output vector (coded output).

The transformed state diagram becomes:


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S1

100

S4

001

S0

000

S2

010

S3

110

S6

011

S5

101

S7

111L

WDL

S0

000

WD^2 L

WD^2 L

L

L

WDLDL

DL DL

WD^2 L

DL

WDLDL

WD^2 L

Masons formula:

( , , )

i i

i

F

T W D L

Forward Loops [K]:

3 6 6

1 0 1 3 7 6 4 0 1K S S S S S S S F W D L 4 8 5

2 0 1 3 6 4 0 2K S S S S S S F W D L 7 10 7

3 0 1 3 6 5 2 4 0 3K S S S S S S S S F W D L 3 6 4

4 0 1 2 4 0 4K S S S S S F W D L 6 8 8

5 0 1 3 7 6 5 2 4 0 5K S S S S S S S S S F W D L 2 8 8

6 0 1 2 5 3 7 6 4 0 6K S S S S S S S S S F W D L 3 10 7

7 0 1 2 5 3 6 4 0 7K S S S S S S S S F W D L Other loops [L]:

2 2

1 2 5 2 1

2 3

2 1 2 4 1 2

2 4 4

3 1 3 6 4 1 3

2 5

4 1 3 7 6 4 1 4

L S S S C WD L

L S S S S C WD L

L S S S S S C W D L

L S S S S S S C WD L


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14

2

5 7 7 5

5 6 6

6 1 3 6 5 2 4 1 6

4 4 7

7 1 3 7 6 5 2 4 1 7

4 7

8 1 2 5 3 7 6 4 1 8

6 6

9 1 2 5 3 6 4 1 92 4 3

10 5 3 6 5 10

2 4

11 5 3 7 6 5 11

L S S C WD L



L S S S S S S S S C D L

L S S S S S S S C WD L

L S S S S C W D L

L S S S S S C WD L

Product of non-touching pairs of loops :2 2 2 2 4 3

1 5( ) ( )L L WD L WD L W D L 2 4 2 3 2 4 7

11 2( ) ( )L L WD L WD L W D L 2 5 6 6 6 8 7

5 6( ) ( )L L WD L W D L W D L 2 2 4 4 3 6 5

5 3( ) ( )L L WD L W D L W D L 2 6 6 2 8 7

5 9( ) ( )L L WD L WD L W D L 2 4 3 2 3 3 6 610 2( ) ( )L L W D L WD L W D L

2 4 3 2 3 6 4

10 5( ) ( )L L W D L WD L W D L 2 2 2 4 4 3 6 6

1 3( ) ( )L L WD L W D L W D L 2 3 2 2 4 4

2 5( ) ( )L L WD L WD L W D L 2 2 2 5 2 4 7

1 4( ) ( )L L WD L WD L W D L

Product of non-touching triplets of loops :2 2 3 6 5 4 8 7

1 3 5

2 4 3 2 4 4 4 8 7

10 2 5

( ) ( )

( ) ( )

L L L WD L W D L W D L

L L L W D L W D L W D L

There are no sets of four or more non-touching loops.

1i t m j v h

i t m j v h

L L L L L L

C C C C C C

2 2 2 3 2 4 4 2 5 2 5 6 6

4 4 7 4 7 6 6 2 4 3 2 4 2 4 3

2 4 7 6 8 7 3 6 5 2 8 7 3 6 6 3 6 4

3 6 6 2 4 4 2 4 7 4 8 7 4 8 7

1 (

)

WD L WD L W D L WD L WD L W D L

W D L D L WD L W D L WD L W D L


W D L W D L W D L W D L W D L

2 2 2 3 2 5 2 5 6 6 4 4 7

4 7 6 6 2 4 2 4 7 6 8 7 3 6 5 2 8 7

3 6 6 3 6 4 4 8 7

1

2

2 2

WD L WD L WD L WD L W D L W D L

D L WD L WD L W D L W D L W D L W D L

W D L W D L W D L

And we have2 2

1 11 1L WD L 2 2 2 2 4 3

2 1 5 1 51 1L L L L WD L WD L W D L 2

3 51 1L WD L


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15

4 10 11 5 10 5

2 4 3 2 4 2 3 6

1

.... 1

L L L L L

W D L WD L WD L W D L

51

61

2

7 51 1L WD L

1 1 2 2 3 3 4 4 5 5 6 6 7 7

( , , )i i

i

F

T W D L

F F F F F F F

3 6 6 2 2 4 8 5 2 2 2 2 4 3

7 10 7 2

3 6 4 2 4 3 2 4 2 3 6 4

6 8 8 2 8 8 3 10 7 2

(1 ) (1 )

(1 )

(1 )

(1 )

W D L WD L W D L WD L WD L W D L

W D L WD L

W D L W D L WD L WD L W D L

W D L W D L W D L WD L

3 6 6 4 8 8 4 8 5 5 10 7 5 10 6 6 12 8 7 10 7

8 12 8 3 6 4 4 8 5 6 8 8 2 8 8 3 10 7 4 12 8

2 2 2W D L W D L W D L W D L W D L W D L W D L

W D L W D L W D L W D L W D L W D L W D L

Part 3: Listing the first three hamming weight terms

0

( , , )P

P

T W D L aS b

2

0 0 0 0 0 0a WD L 2[ 0 0 0 0 0 Tb W D L

2

2

0 0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

L

WDL DL

DL WDL

S DL WDL

WDL DL

WD L L

L WD L

2P L For 4,5,6L , we have 2,3,4P . We therefore obtain the first three hamming weight terms as follows:

2

4 ( , , )LT W D L aS b


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2 2

2

2

2

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0

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hatilima jasper 11042005 hw03

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