CH3 Ma Khoi Tuyen Tinh

8/10/2019 CH3 Ma Khoi Tuyen Tinh

1/36

Csl thuytthng tin

Chng3: M ha knhM Khituyntnh

TS. Phm Hing

02/12/2013 Slice 1Trng H Bch Khoa H Ni


2/36

Phn1: Khi nimcbn

M knh/M sali

M ha knh (channel Coding) hay cn gil m sali(ErrorCorrection coding) l kthutkhngch, pht hinv sali

trong qu trnh truyndliuqua knh c nhiu.

M salisdngthng tin dtha(redundancy) cm ha

thm vo dliupha bn pht. Thng tin dthascphathu sdngsali- m khng cnyu cupht litin.



3/36

Phn1: Khi nimcbn

Phn loili

Liclpthngk: Lixuthintrong qu trnh truyntin trnknh truyn, xuthinclpkhng lin quan tinhau. V d:

nhiuGaussian.

Lichm: Lic phn blin hvinhau.



4/36

Phn1: Khi nimcbn

V d: Knh truyntin khng nh

(Binary Symmetric Memoryless Channel).

Lixyra vibit 0 v 1 vicng xc sutp (symmetric)

Lixyra ngunhin v clpgiacc bit (memoryless)


1-p

IN OUT

0 0

1 1

1-p

p

p

p l xc sutliBERBit Error Rate (BER)


5/36

Phn1: Khi nimcbn

V d: Knh truyntin ang


receiving signal

time

strength

0

sending signal


6/36

Phn1: Khi nimcbn

V d: Knh truyntin ang


RX power

TX power

Channel Fading


7/36

Phn1: Khi nimcbn

Phn loim sali:

M khi(block codes): thng tin cm ha v chnthm phndthatheo tngkhi.

M khi

M khituyntnh

M vng CRC

M BCH, Reed-Solomon, LDPC

M chp(Convolutional codes): thng tin cbini

theo cc hm truynt(php tch chp). Khng c giihnr rng giathng tin v phndtha.

M chp(covolutional codes)

M Turbo



8/36

Phn2: Cc khi nimcbncam ha sali

Tcm

Khongcch Hamming (Hamming distance)

Khongcch tithiu(minimum distance)

Ma trnsinh, m trnkimtra chnl.



9/36


Tcm

Githit l tphp2 phnt0 v 1.biudinvector n phntca

Sbinary l tphp imtrong khng gian

M l m chpnhn bit uvo v tora bit ura.

nhngha: tcm cam l

V d: M lp(repetition code) nhn1 bit uvo v to

ra n bit lpli ura. Tcm l


2

2

n

2

( , )n k 2k 2n

( , )n k k n

( , )n k

k

Rn

( ,1)n

1R

n


10/36


Ma trnsinh

Vi biudinthng tin (message).l tm (codeword) cam lp

Qu trnh m ha cbiudindidngma trn. Ma trn

sinh cam lpl


m

C ( ,1)n

, , , ...C m m m m

C mG

1,1,1,...,1G

G


11/36


Biudinm lp(3,1) trong khng gian

Khongcch Hamming trong m binary ctnh bngs

cc imkhc bittrong 2 tm.


1 1,1,1C

0 0, 0, 0C

1

2

12

1,0,1,1,1,0

1,1,0,1,1,1

3

C

C

d


12/36


nhngha: Khongcch tithiu(min distance) l

khongcch Hamming nhnhtgia2 tm btk.

Phngphp giim ML: tm kimtm c khongcchgnnhtvitm thu c.



13/36


Lin hgiakhongcch Hamming tithiuv khnng

pht hinv sali.

Vim binary

Khnngpht hinliKhnngsali


( , )n k

min 1d

min 1

2

dt


14/36


15/36



V dm kimtra chn

Trong trnghp n = k+1, bntin cbsung thm 1

bit kimtra chnl

Trong trnghpschncc bit 1, Bit kimtra chnl

c gi tr

Trong trnghpslcc bit 1, bit kimtra c gi tr

Bit kimtra chnlcthm vo mboschncc

bit 1 trong tm.

M kimtra chnlchpht hinc(tia) 1 li,

khng c khnngsali.

1 (mod 2)k iiq m

1 q


16/36



V d1: M kimtra chnl(6,5)

Bntin m = (10110) => tm c=(101101)

Bntin m = (11011) => tm c=(110110)


17/36



V d2: Bngm kimtra chnl(4,3)

Dataword Codeword

111

011

101

001

110

010

100

000

1111

0011

0101

1001

0110

1010

1100

0000


18/36

M khi(n,k) cbiudindngvector

Bntin d=(d1d2.dk)Tm c=(c1c2..cn)

M khicxy dng

c=dG

ViG l ma trnsinh

k

2

1

21

22221

11211

a

.

a

a

...

......

...

...

G

knkk

n

n

aaa

aaa

aaa

Phn3: M khituyntnh


19/36

Phn3: M khituyntnh

mbo2 bntin khng c chung 1 tm

(khng thgiim/sali), cc hs phic

lptuyntnh. Nu l 2 tm btk

cngl 1 tm

Hqu: khigmton bit 0 cngl 1 tm

1

c a

k

i i

i

d

ai

ic

kc

i kc c c


20/36

Khnngsalicam khituyntnh

Khongcch Hamming cam khituyntnhl khongcch Hamming nhnhtcacc tm khc 0

tm khongcch Hamming nhnht, cntm kimtra 2ktm tm khongcchHamming nhnht.

Phn3: M khituyntnh


21/36

Phn3: M khituyntnh

Vim (4,2), c ma trnsinh

Vid=[1,1]

1010

1101G

0111

____1010

1101

c

a1= [1011]

a2= [0101]

V d1: m khituyntnh


22/36

Error Syndrome salim khituyntnh, sdngphngphp

Error Syndrome Nucr ltm thu c pha thu, vector pht hin

li(error syndrome) scacrs = crH

T

Nucrbli, gie l vector licr= c + e

do s = (c + e) HT= cHT+ eHT

s = 0 + eHT

Vector syndrome c gi trchphthucvo vectorlie.

Phn3: M khituyntnh Giim sali


23/36

Error Syndrome

Nhnxt: nucngvector e vicc tmkhc th vnthu c1 vector syndrome. Tngcngc 2(n- k)syndromes, 2nvector li. V d: vim (3,2), c 2 syndromes v 8 vector li

e. R rng khng thsattccc litrongtrnghpny.

V d: vim (7,4) c 8 syndromes v 128 vectorlie.

V vy, vi8 syndromes, ta cnbtr cc gi tr

khc nhau sac7 li(li1 bit) v 1trnghpkhng lis=0.



24/36

Bnglitk li Bnglitk licxy dngnhsau:

c1(all zero)

e1

e2

e3eN

c2+e1

c2+e2

c2+e3

c2+eN

c2

cM+e1

cM+e2

cM+e3

cM+eN

cM

s0s1

s2

s3sN

Cc hng ucchung gi tr

syndromeCc hng khcnhau c vectorsyndrome khcnhau

Bngc 2kct(tngngvicc tm hpl)v 2n-khng (slngcc syndrome)



25/36

Bnglitk licxy dngtheo cch Litk ttccc li 1 bit Litk ttccc li2 bit

Kimtra mbc cc licbsung vo bngc vectorsyndrome khc nhau. Vicxy dngbngktthc khi s

dnghtcc vector syndrome.

giim: Tnh vector syndrome theo cng thcs = 0 + eHT

Tra bngtngng, tm vector lie tngng. Cngmodulo vector e v tm thu cgiim

c = cr+ e



26/36

Phn3: M khituyntnh M hthng

M hthngc phnthng tin v phnkimtra

cphn tch trong tm, phnthng tin lkhng thay iso viban u

Ma trnsinh cam hthngc dng

P|I

..1..00................

..0..10

..0..01

G

21

22221

11211

kRkk

R

R

ppp

ppp

ppp

k R R= n- k


27/36

Vim hthng, ma trnkimtra cxy dng

G = [ I | P] and so H = [-PT| I]

V d: m (7,4) vikhongcch Hamming dmin= 3

1111000

0110100

1010010

1100001

P|IG

1001011

0101101

0011110

I|P-HT



28/36

Tm vector syndrome trong v d

Tm thu ccr= [1101001]

000

100

010

001

111

011

101

110

1001011HcsT

r


h i


29/36

M Hamming l dngcbitcam khituyntnh

Vimi Tm c di

Bntin c di

Tcm

Khongcch Hamming , khnngsa1 li

M Hamming l m c tcm R lnnhtvicng

khongcch Hamming

Phn4: M Hamming

2r

2 1r

n

2 1r

k r

1

2 1r

k rR

n

min 3d

min 3d

h i


30/36

M Hamming (7,4) dnghthng

1111000

0110100

10100101100001

P|IG

1001011

0101101

0011110

I|P-HT

1001011

0101010

0011001

0000111

G

1010101

1100110

1111000

H

Phn4: M Hamming (7,4)

M Hamming (7,4) dngkhng hthng

Ph 4 M H i (7 4)


31/36

V d: m Hamming (7,4) khng hthngd = 1011

c = 1110000

+ 0101010

+ 1101001

= 0110011

e = 0010000

cr= 0100011

s = crHT= eHT= 011

Ch : gi trvector syndrome l vtr li.

Phn4: M Hamming (7,4)

Ph 4 M H i (7 4) BER


32/36

Phn4: M Hamming (7,4) BER

For a given channel bit error rate (BER), what is the BER

after correction (assuming a memoryless channel, i.e., noburst errors)?

To do this we will compute the probability of receiving 0,

1, 2, 3, . errors

And then compute their effect

Bit E R t ft D di


33/36

Bit Error Rates after Decoding

Example A (7,4) Hamming code with a channel BER of 1%,

i.e.,p= 0.01

P(0 errors received) = (1 p)7

= 0.9321P(1 error received) = 7p(1 p)6= 0.0659

P(3 or more errors) = 1 P(0) P(1) P(2) = 0.000034

002.0)1(

2

67received)errorsP(2 52

pp

Bit E R t ft D di


34/36

Bit Error Rates after Decoding

Single errors are corrected, so,

0.9321+ 0.0659 = 0.998 codewords are correctlydetected

Double errors cause 3 bit errors in a 7 bit codeword, i.e.,(3/7)*4 bit errors per 4 bit dataword, that is 3/7 bit errorsper bit.

Therefore the double error contribution is 0.002*3/7 =0.000856


35/36


36/36

Perfect CodesSo,

errors3toupcorrectto6

2)-1)(n-n(n

2

1)-n(n

n1

errors2toupcorrectto2

1)-n(nn1

error1toupcorrectto12

nR

If equality then code is Perfect

Only known perfect codes are SEC Hamming

codes and TEC Golay (23,12) code (dmin=7).Using previous equation yields

)1223(11 2220486

2)-1)(23-23(23

2

1)-23(23231

CH3 Ma Khoi Tuyen Tinh

Documents