TTS - Ch4

7/23/2019 TTS - Ch4

1/48

25/3/2014

1

Hthng thng tin sin hnhHthng thng tin sin hnh

Tn hiu tng tvo

Knh

thng

A/D

M

ha

ngun

Mtm

ha

M

ha

knh

Ghp

knh

iuch

atruy

cp

Gii Gii Gii Gi Gii

Khi m ha knh: lm nhim va thm cc bit dvo

tn hiu stheo mt quy lut no y, nhm gip cho bn

thu c thpht hin v thm ch sa c cli xy ra

trn knh truyn.

D/A mngun mtm mknh

c

knh iuch truycp

Tn hiu tng tra

7/23/2019 TTS - Ch4

2/48

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2

NI DUNGNI DUNG

CCss ll thuy tthuy t mm haha knhknh

GiiGii thiuthiu vv iuiu khinkhin lili

MM khikhi

MM khikhi tuyntuyn tnhtnh

MM vngvng

MM chpchp

CSL THUYT MCSL THUYT MHA KNHHA KNH

7/23/2019 TTS - Ch4

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3

Smt mt tin do nhiuSmt mt tin do nhiu

Lngtincakt nguniTX:

1log)i(I

2TXTX [bit]

Knhcnhiu:lngtinnhnthnlngtintruynmtlng,

do khngchcchncaquytnh

TX

)i(I)vi(p

log)i:v(I TXTXRXTX

2TXRXRX [bit]

Knhkhngnhiu:lngtincbotonkhitruynquaknh

1)vi(p RXTX )i(I

)i(p

1log)i:v(I TXTX

TX

2TXRXRX

TX

[bit]

Smt mt tin do nhiuSmt mt tin do nhiu (tt)(tt)

Entropyhiuqu:entropynhn

i

RXRXRXeff

)i(p

)ii(plog)i(I

TX

RXTX

2RXRX

.

C nghing ldokhngchckt nhnccgingvikt phthaykhng

effHHE

7/23/2019 TTS - Ch4

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4

Dung lng knhDung lng knh

nhngha:lngtintiaknhchophptruynquatrongmtnv thigianmkhnggyli

Khiu:C

Biuthctnh:tyvoknh

KnhnhiuGaussetrng:

n n ng ruy n c qua n

S/N = 0

)N/S1(logBC2

Truyn tin trong knh c nhiuTruyn tin trong knh c nhiu

nh l Shannonnh l Shannon Lngtintctruynquaknhb haohtdonhiu

.

tinb nhiuphhytrongmtnv thigian

Shannon:

)EH(nEnHnC max00max0

u < c m a ruy n n qua n v x c

sut li b ty

Phng php m ha gim xc sut li gi l m ho knh

7/23/2019 TTS - Ch4

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5

GII THIU VIUGII THIU VIUKHIN LIKHIN LI

Cc yu cu thit ktuyn truyn dn thngCc yu cu thit ktuyn truyn dn thng

tintin

1. Truynthngtinvimttc bityucutytheodchv

2. Truynthngtintrongmtbngthnghnchcamtknh

truynsnc

3. Truynthngtinvimtcngsuthnchtyngdngc th

4. Truynthngtintrongmtkhongthigiantr hnch

5. Chtlngtruyndn mcchpnhnciukhinli

7/23/2019 TTS - Ch4

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6

nhnh gigi chtcht lnglng truyntruyn dndn

Thams nhgi:BERhayPb

BER(BitErrorRate):t s litrungbnh,ctnhltchPbRb(Rbl

tc bittruyntrongknh)

Gitr Pbinhnh:

H thngPCMtuyntnh:107

H th ngPCMnnphituy n:10

H thngADPCM:104

Cc phng php iu khin liCc phng php iu khin li

KhiBERhayPblnqumcchophp:5phngphp:

1. Tngcngsutpht

2. S dngphntp(diversity)

3. Kimtraecho(truynsongcng)

4. Yuculplit ng(AutomaticRepeatreQuest)

5. Mhasalikhngphnhi(ForwardErrorCorrection

Coding)

Phngphp4v5:Yucudngmckh nngphthinv

sali (M ha knh)

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7

1. Tngcngsutpht: D thchin

Khngphilcnocngthchinc,vd trongthitb didng

CcCc phngphng phpphpiuiu khinkhin lili ((tttt))

ngc pn n c t cp n n.

2. S dngphntp: C3kiu:phntpkhnggian,tns,thigian

athm d vod liubngcchtruyngpi(qua2antenna,2tns hay2thiimkhcnhau)

3. Truynsongcng: P tt nngc p ttr n n t pr ng.

Yucugpibngthng=>khngthchhpkhicntndngph.


4. Yuculplit ng(AutomaticRepeatreQuest)

Bnpht:mhakhitinphtbngmphthinli.

Bnthu:kimtralitrongkhitinthutr lichobnphtACK

(thung)hocNAK(thusai)trnmtknhhitipringbn

phts phtlikhitinsaikhinhnNAK.

C2k thut:StopandwaitARQvContinuousARQ

ngdngtrongcch thngthngtinmytnh(vcsnknhsong

cn

Khngphhpvicch thngthigianthcvcch thngctrtruyndnln,vd thngtinv tinh

7/23/2019 TTS - Ch4

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8


5. Mhasalikhngphnhi(ForwardErrorCorrection

Coding)

Bnpht:mhakh itinphtb ngms al i

Bnthu:kimtralitrongkhitinthuvt salinuphthinc

li

Gimxcsutlinh lidngs khcnhaugiatc truyndnv

dunglngknh;tr gibngvictngthigiantruyndotng

d mcth phthinvsali.

ngdng trongcch thngthigianthcvcch thngc

khongcchthuphtln

SoSo snhsnh chtcht lnglng knhknh cc vv khngkhng mm haha

KhiEb/Nothp:s dngmhaknhkhnglmtngchtlngknhtruyn.

KhiEb/Nocao:s dngmhaknhlmgimt l li,tngchtlngknh.

7/23/2019 TTS - Ch4

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9

Chtlngknhvbngthng: S dngmhaknhlmtngchtlngknh(BERA>BERC).

Phiathmccbitd vobntin.


Nuh thngthigianthc(khngchpnhntr)>phitngtctruyn>tngbngthng

Cngsutvbngthng: Gimcngsutpht>gimEb(t B>AhocC)

S dngmhaknh:gimcngsutphtkhnglmgimchtlng


knh.

Tngbngthngdoccbitd camhaknh

7/23/2019 TTS - Ch4

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10

Phn loi m iu khin liPhn loi m iu khin li

M iu khin li

M khi M chp

M khng

tuyn tnhM tuyn tnh

(M nhm)

M khng M vng

v ng

MGolay

RS BCH nhphn

Hamming

(e=1)e>1

MM khikhi

Thams camkhi:n,kvmatrnsinh/athcsinh

Bm ha khik bittin

n bit

m ha

Cmtngthibittinvbitd trongt m=>mh thng.

Tm khin bit

(n-k) bitk bit

Phn tin Phn d Tlm R = k/n

(thng t -1)

nhnghanghimngt:kbittinlintcthnh1khi,(nk)bitd lintcthnh1khi.

nhnghatnghimngt:ch cncmtccbittinvbitd.

7/23/2019 TTS - Ch4

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11

PhnPhn loiloi mm khikhi MM khikhi tuyntuyn tnhtnh

M khi tuyn tnh:

C cha t m ton bit 0;

C tnh chtng: vi hai t m Ci, Cj bt k, ta c:

Ci + Cj = Ck

Vi Ck cng l t m

a = 00 00000b = 01 00111

c = 10 11100d = 11 11011

cdb,bdc,dbc

PhnPhn loiloi mm khikhi MM vngvng

M vng:

M khi tuyn tnh khng c t m ton 0;

C tnh cht: dch vng mt t m th cngc mt t

m trong cng b m

, , , ,

1000110, 0100011, 1010001

7/23/2019 TTS - Ch4

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12

MM chpchp

Thams camchp:n,k,Kvathcsinh

T mph thucvo:

v o;

(K1)b kbitvotrc

=>mchp:mcnh

K: dirngbuc(constraitlength)

Giimmchp:thuttonViterbi

2

1

T0 T0 T0 k = 1, n = 2, K = 3

Khnng pht hin v sa li caKhnng pht hin v sa li ca

m khim khi

Khong cch Hamming gia cc tm trong mt b

m c lin quan n khnng pht hin li v sa li ca b

m :

d l khong cch Hamming

1srd

r s p n c

s l sli sa c

r >= s

7/23/2019 TTS - Ch4

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13

V dminh ha khnng pht hinV dminh ha khnng pht hin

v sa li ca m khiv sa li ca m khiK t A B C D E F G H

M1

m

K t B C E H

Tm 001 010 100 111

K t B G

M2

M3

Bm M1 (d = 1) khng c khnng pht hin li

Bm M2 (d = 2) pht hin c 1 li, khng sa c li

Bm M3 (d = 3) pht hin v sa c 1 li

Tm 001 110

Quan hgia n v k (sai 1 li)Quan hgia n v k (sai 1 li)

Sbit tin: k Sbit tng cng: n

Stm tng cng: 2n

Stm khng dng (cm): 2n 2k

Strng hp sai 1 li i vi 1 tm: n

knk 222.n

S tr ng hp sai 1 l i t ng cng: n.

c thpht hin v sa ht cc li trn, yu cu:

1n

22

nk

k/n = 1/3k/n = 4/7k/n = 7/11

7/23/2019 TTS - Ch4

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14

M KHIM KHI

M kim tra chn l(parity)M kim tra chn l(parity)

M khi n gin, dng trong truyn sliu dng ASCII;

C 2 loi:

Loi chn (Even parity): tng sbit 1 trong k t(kcbit P) l

schn

Loi l(Odd parity): tng sbit 1 trong k t(kcbit P) l s

.

7/23/2019 TTS - Ch4

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15


R = k/n = 7/8B0 B1 B2 B3 B4 B5 B6 P

7 bit tin 1bit kim tra

Nguyn tc m ha/gii m:

- Bn pht: tnh bit P gn vo cui k t7 bit phti

- Bn thu: tnh bit P so snh vi bit P thu. N u k t qu

ging nhau khng c li, nu khc nhau c li


Khong cch Hamming ca m: 2

Khnng pht hin li theo l thuyt: 1

Khnng pht hin li trn thc t: pht hin c s

li l

Khnng sa li: 0

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16

Mch tnh bit PMch tnh bit P

B6 B5 B4 B3 B2 B1 B0

P lP chn

M kim traM kim tra tng khi BCCtng khi BCC(B(Blocklock sumsum CCheckheck CCharacterharacter))

Truyn khi k t: 1 k tbli -> ckhi bli

n, bsung thm tp cc bit parity tnh trn ckhi

Nguyn tc m ha/gii m: tng tparity n

Bn pht: tnh v gn bit P cho tng k t(P hng) tnh v gn

thm tp cc bit P cho ckhi k t(BCC)

Bn thu: tnh v ki m tra cbit P n v BCC

ng dng: truyn sliu l mng k tASCII

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17

M kim traM kim tra tng khi BCCtng khi BCC(B(Blocklock sumsum CCheckheck CCharacterharacter))

Khnng pht hin li:

C th pht hinccc li chn trong tngk t

Khng th pht hin li chn xy racng ct vo cng thi

im

Kh nng sa li:

- C th sac lin

V dmV dm kim tra kim tra tng khi BCCtng khi BCC

1 0 0 1 1 1 0

B0 B1 B2 B3 B4 B5 B6 PC0

0 1 1 0 0 0 1

1 1 0 0 0 0 0

1 1 1 1 0 0 1

cbitparityhn

0

1

1

BCC = Cc bit parityct

Bit PchoBCC

0 1 1 1 10 0 0

7/23/2019 TTS - Ch4

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M kim traM kim tra tng khi BCC b 1tng khi BCC b 1

Nguyn tc m ha/gii m: tnh BCC bng tng b 1

Bn pht:

- Tnh tng b 1 ca cc k ttrong khi

- o ngc kt qutng b 1 to thnh BCC

Bn thu:

- Tnh tng b 1 ca tt ck t. Kt qubng 0 l khng c

li v ngc li

Khnng pht hin li: tt hn so vi phng php BCC

tng modulo-2

Thc hin: bng phn mm

Bn pht Bn thu

0 1 1 1 0 0 1 0 1 1 1 0 0 1

V dmV dm BCC tng b 1BCC tng b 1

1 0 0 0 1 1 0 1 0 0 0 1 1 00 0 0 0 0 1 1 0 0 0 0 0 1 11 0 0 1 1 1 1 1 0 0 1 1 1 1

1 1 0 1 0 0 0 1

1

0 1 0 1 1 0 1

1 1 1 1 1 1 1 0

1 1 1 1 1 1 1 = s0trong sb

1

1 0 1 0 0 1 0 = tng b 1

0 1 0 1 1 0 1 = BCC

o

bit

7/23/2019 TTS - Ch4

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19

Hthng thng tin sin hnhHthng thng tin sin hnh

Tn hiu tng tvo

Knh

thng

A/D

M

ha

ngun

Mtm

ha

M

ha

knh

Ghp

knh

iuch

atruy

cp

Gii Gii Gii Gi Gii

Khi m ha knh: lm nhim va thm cc bit dvo

tn hiu stheo mt quy lut no y, nhm gip cho bn

thu c thpht hin v thm ch sa c cli xy ra

trn knh truyn.D/A m

ngun

mt

m

m

knh

c

knh

iu

ch

truy

cp

Tn hiu tng tra

M KHI TUYN TNHM KHI TUYN TNH

7/23/2019 TTS - Ch4

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MM khi tuyn tnhkhi tuyn tnh (4,7)(4,7)

M haM ha

4243232221212

4143132121111

IaIaIaIaP

IaIaIaIaP

aij: 0 hoc 1

Bit tin: I1, I2 I3, I4; Bit kim tra (d): P1, P2, P3


M ha (tt)M ha (tt)

Bc 2: Lp ma trn kim tra H:

4343332321313

4243232221212

4143132121111

IaIaIaIaP

IaIaIaIaP

IaIaIaIaP

100:aaaa

010:aaaa

001:aaaa

H

34333231

24232221

14131211

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M ha (tt)M ha (tt)Bc 3: Lp ma trn sinh G:

100:aaaa

010:aaaa

001:aaaa

H

34333231

24232221

14131211

342414

332313

322212

312111

aaa:1000

aaa:0100

aaa:0010

aaa:0001

G


M ha (tt)M ha (tt)

Bc 4: Tnh tm khi tuyn tnh bng cch nhn m

]PPPIIII[c 3214321

332313

322212

312111

4321aaa:0100

aaa:0010

aaa:0001

xIIIImxGc

v , rong m vec or n:

342414

aaa:1000

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22


Gii mGii mSyndrome: tm c lp vi tm pht, phthuc vo

tm thu b li

Bng syndrome: tp hp tt ccc syndrome c thc

Gii m: tnh syndrome, da vo bng syndrome suy

ra vtr bit li v sa

TTTTTT

s: vector syndrome

r: vector tm thu

e: vector li

V dmV dm khi tuyn tnhkhi tuyn tnh (4,7)(4,7)

I1 I2 I3 I4 P1 P2 P3

43213

43212

43211

xI0xI1xI1xI1P

xI1xI0xI1xI1PxI1xI1xI0xI1P

100:0111

010:1011001:1101

H

011:1000

101:0100

110:0010G

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V dmV dm ha m khi tuyn tnhha m khi tuyn tnh (4,7)(4,7)

I1 I2 I3 I4 P1 P2 P3

111:0001

011:1000

101:0100

110:0010

G

Cc bit tin l: 1 0 1 1

Cc bit dl: 1 0 0

Tm khi pht i: 1 0 1 1 1 0 0

V dgii mV dgii m m khi tuyn tnhm khi tuyn tnh (4,7)(4,7)

Bng syndrome:

e s

010:1011

001:1101

H

101

011

111

1010010000

0110100000

1111000000

0000000000 100:0111

001

010

100110H

T

THxes0010000001

0100000010

1000000100

7/23/2019 TTS - Ch4

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24

1011101r

V dgii mV dgii m m khi tuyn tnhm khi tuyn tnh (4,7)(4,7)

0000000000

1011100c001110

101

011

1011101s

1000000100

1100001000

1010010000

0110100000

1111000000

001010

0010000001

0100000010

M VNGM VNG

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c im ca mc im ca m vngvng

L mt lp con ca m khi tuyn tnh

sa li cao

Thc hin m vng bng phn mm hoc phn cng

Dch vng mt tm cng c mt tm thuc cng b

m.

C thbiu din m vng bng a thc

C thto ra tm vng bng cch nhn modulo-2

vector mang tin vi a thc sinh m vng khng hthng

MM kim tra dvng CRCkim tra dvng CRC

(C(Cyclicyclic RRedundancyedundancy CCheckheck)) Nguyn tc m ha/ gii m:

- Bn pht: da vo ni dung khung tin tnh ton cc bit

kim tra gn thm vo cui khung pht i

- Bn thu: bn thu tnh ton tng tnhbn pht pht

hin/ sa li

- Pht hin hu ht cc loi li, c thsa c 1 li

Thc hin m CRC:

- Bng phn cng hoc phn mm

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MM ha v gii m CRCha v gii m CRC

)x(R)x(Q

x)x(Mr

Thc hin php chia:

)x(Rx)x(M)x(Tr

Q(x) l thng s, R(x) l sd

t:

M(x) l a thc tin bc k-1, G(x) l a thc sinh bc r

biu din cho tm CRCa thc sinh: a thc c trng cho m CRCTa thc sinh, to ra tt ccc tm trong bm

)x(Q)x(G

)x(Rx)x(M

r

Nu khng c li xut hin th bn thu, sau khi chia tmthu cho a thc sinh ta sc phn dl 0

V dmV dm ha v gii m CRCha v gii m CRC

Truyn khung tin 11100110 qua ng truyn sliu dngm CRC, a thc sinh 11001

11100110 0000

a: c n p p c a:

11001

Vy, khung tin m ha CRC l: 11100110 0110

11100110 1111

Gii m: gisthu 11100110 1111Thc hin php chia:

11001

D1001, kt lun c li

7/23/2019 TTS - Ch4

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27

1 1 1 0 0 1 1 0 0 0 0 0 1 1 0 0 1

1 1 0 0 1 1 0 1 1 0 1 1 0

0 0 1 0 1 1 1

1 1 0 0 1

1 1 1 0 0

1 1 0 0 1

0 0 1 0 1 0 0

0 1 1 0 1 0 1 1 0 0 1

0 0 0 1 1 0

Chn a thc sinh G(x)Chn a thc sinh G(x)

Khnng pht hin li ca G(x) bc r c t nht 3 s1:

- Tt ccc li n, hu ht cc li i

- Tt ccc li xy ra vi sl

- Tt ccc li chm ngn hn r

- Hu ht cc li chm di bng hoc hn r

Cc a thc sinh hay gp:

CRC 16 : G(x) = x16

+ x15

+ x2

+ 1CRC ITU: G(x) = x16 + x12 + x5 + 1

CRC 32 : G(x) = x32 + x26 + x23 + x16 + x12 + x11 + x10 + x8

+ x7+ x5 + x4 + x2 + x + 1

7/23/2019 TTS - Ch4

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Thc hin m CRCThc hin m CRC

a thc sinh G(x) = xr+ ar-1xr-1 + . . . + a2x2 + a1x + 1

x0 x1 x2 xr-1

a1 a2 ar-1

Thanh ghi CRC

Mch m ha CRCMch m ha CRC

1

a thc sinh G(x) = x4 + x3 + 1

TxC

0 0 0 0

0

1 0 0 1

PISO

Np song song tng byte trong Nbyte tin

0 1 1 0 0 1 1 1

7/23/2019 TTS - Ch4

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Mch m ha CRC (tt)Mch m ha CRC (tt)

1


TxC

0

1 0 0 10 1 0 0

0 1 1 0 0 1 1

1


1


TxC

0

0 1 0 01 0 1 1

0 1 1 0 0 1

1 1

7/23/2019 TTS - Ch4

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1


TxC

0

1 0 1 11 1 0 0

0 1 1 0 0

1 1 1


1


TxC

0

0 1 1 01 1 0 0

0 1 1 0

0 1 1 1

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1


TxC

0

1 0 1 00 1 1 0

0 1 1

0 0 1 1 1


1


TxC

0

1 1 0 01 0 1 0

0 1

1 0 0 1 1 1

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1a thc sinh G(x) = x4 + x3 + 1

0

TxC

0

0 1 1 01 1 0 0

1

0

1 1 0 0 1 1 1



0

TxC

0

1

0 1 1

0 1 1 0 0 1 1 1

7/23/2019 TTS - Ch4

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0

TxC

1

0 1 1

0 0 1 1 0 0 1 1 1



0

TxC

1

0 1

0 0 1 1 0 0 1 1 11

7/23/2019 TTS - Ch4

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0

TxC

1

0

0 0 1 1 0 0 1 1 11 1

Mch gii m CRCMch gii m CRC

0

01

10

RxD

0 0 0 0

111001

RxC

c song song byte (xN)

SIPO

0 0 0 0 0 0 0 0

7/23/2019 TTS - Ch4

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35

Mch gii m CRCMch gii m CRC

0

01

10

RxD

0 0 0 0

111001

1 0 0 0

RxC


SIPO

Mch gii m CRC (tt)Mch gii m CRC (tt)

0

01

10

RxD

1 1 0 0

11001

1 0 0 0

RxC


SIPO

1

7/23/2019 TTS - Ch4

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0

01

10

RxD

1 1 1 01001

1 1 0 0

RxC


SIPO

1 1


0

01

10

RxD

0 1 1 1001

1 1 1 0

RxC


SIPO

1 1 1

7/23/2019 TTS - Ch4

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0

01

10

RxD

1 0 1 001

0 1 1 1

RxC


SIPO

11 10


0

01

10

RxD

1 1 0 1

1

1 0 1 0

RxC


SIPO

1 110 0

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0

01

10

RxD

0 1 1 11 1 0 1

RxC


SIPO

1 1 1001


0

01

10

RxD

1 0 1 00 1 1 1

RxC


SIPO

11 10 01 1

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01

10

RxC

RxD

SIPO



1

10

RxC

RxD

SIPO

0 0 1 10 1 0 1


1 1100110

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Mch gii m CRC (tt)Mch gii m CRC (tt)10

RxC

RxD

SIPO

0 0 0 00 0 1 1


1 1100110

Mch gii m CRC (tt)Mch gii m CRC (tt)0

RxC

RxD

SIPO

0 0 0 00 0 0 0


1 1100110

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M CRC sa 1 liM CRC sa 1 li

Quan hgia n v k phi tha yu cu:

n

Chn a thc sinh da vo vic phn tch xn + 1 ra

tha snguyn t, a thc sinh l a thc c bc r

1n2

k

)1xx)(1xx)(1x(1x3237

G(x) G(x)

V d: n = 7, k = 4, r = 3

M = 1

Dch vng T'(x) sang tri

BeginThutThut

tonton

sasa

N

Chia T'(x) cho G(x) - dl R'(x)

Tnh trng lng dW

W 1 ?

Tng M ln1

lili

bngbng

phngphng

Y

End

T'(x) = T'(x) + R'(x)

Dch vng T'(x) sang phi M ln

phpphpbyby

lili

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V dm CRC sa 1 liV dm CRC sa 1 li

Khung tin:1100

a thc sinh:G(x) = x3 + x2 + 1

Chia 1100 000 cho 1101, sdl:

Tm CRC (4,7) pht:1100 101

V dm CRC sa 1 li (tt)V dm CRC sa 1 li (tt)Gistm CRC thu: 1110 101

Kim tra li: chia 1110 101 cho 1101, c d

Sa li:

- Dch vng tri ln 1 c: 1101011, chia cho 1101, d011



- Cng 0101111 vi 001, c: 0101110

- Dch vng phi ln 1 c: 0010111

- Dch vng phi ln 2, c: 1001011

- Dch vng phi ln 3, c: 1100101.y l tm c sa li ng

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M HAMMINGM HAMMING

M Hamming (1950)M Hamming (1950)

Do Richard Hamming pht minh (1950)

M c khnng pht hin v sa c 1 li

M hnh c a m Hamming

(4,7) bao gm 4 bit dliu

(3,5,6,7) v 3 bit chn l

(1,2,4) (dng P chn)

Bit 1 kim tra bit (3, 5, 7)

Bit 2 kim tra bit (3, 6, 7) Bit 4 kim tra bit (5, 6, 7)

Lu : cc vtr(1,2,4 ...)

thc ra l 20, 21, 22

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V dm Hamming (1950)V dm Hamming (1950)

Tm Hamming:

0 1 1 0 0 1 1

M ha/gii m HammingM ha/gii m Hamming

M ha:

...

- Tnh XOR cc snhphn chvtr cc bit 1

- Kt quphp XOR chnh l cc bit P

Gii m:

-

- Nu kt quphp XOR l 0 th tm thu khng c li

Nu kt quphp XOR khc 0 th tm thu c li bit

l kt quphp XOR

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V dm ha/gii m HammingV dm ha/gii m Hamming

Tm mang tin: 1 0 1 1 1 0 0

Tm Hamming: P1 P2 1 P4 0 1 1 P8 1 0 0

M ha:

- Tnh

- Tm Hamming: 1 1 1 0 0 1 1 1 1 0 0

Gii m:

1248 PPPP10111001011101100011

Gisthu 1 1 1 0 0 1 1 1 1 0 1 pht hin v sa c

li

Gisthu 11101010101 pht hin c li nhng

khng sa c

Kthut sa li chmKthut sa li chm

sa li chm, dng kthut to lon (an xen)

(interleaving)

To lon/gii to lon:

M ha Gii mTo lonGii

to lon

thay i trt tcc bit: biti vo theo hng/ct v i ra

theo ct/hng

ngha: bin li chm thnh cc li n

Trgi: tng thi gian tr

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V dto lonV dto lon M = 4, N = 6M = 4, N = 6

11 55 99 1313 1717 2121

22 66 1010 1414 1818 2222

33 77 1111 1515 1919 2323

44 88 1212 1616 2020 2424

Vo: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

Ra: 1 5 9 13 17 21 2 6 10 14 18 22 3 7 11 15 19 23 4 8 12 16 20 24

HTCHNG4HTCHNG4

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Gisxc sut mt li bit l p v cc li xut hin c lpnhau, xc sut k li xut hin trong khi n bit c thtnh theo

cn thc x xPoisson vi n ln v nh:

npandkwhere

ek

ppCnkPk

knkk

n

...,2,1,0

!)1(),(

Phn loiPhn loi ARQARQ

1. ARQdngvi:

Bnphtphttngkhitindnglich

Nunhntr lilACKphtkhitinmi

Nunhntr lilNAKphtlikhitinc

2. ARQlintc:

Bnphtlintcphtcckhitin

ARQliliN:khinobnphtnhntr liNAK(+s th t)phtliNkhik t khili

ARQchnlc:khinobnphtnhntr liNAKch phtli

khili

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Gistt ccc loi li u xy ra vi xc sut nhnhau

v cc li xut hin c lp nhau, xc sut j li xut hin

trong khi n bit l:

)p1(pC)n,j(P jnjjn

)!jn(!j

!nCjn

y, p l xc sut mt k hiu nhn c bli.

Suy ra, trong m parity, xc sut li khng pht hin c

j2n)oddn(,2/n

)evenn(,2/)1n(

1j

j2j2

nund )p1(pCP

trong mt khi n bit l:

TTS - Ch4

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Transcript of TTS - Ch4