tts - ch4
TRANSCRIPT
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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
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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
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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
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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
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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
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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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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.
CcCc phngphng phpphpiuiu khinkhin lili ((tttt))
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
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CcCc phngphng phpphpiuiu khinkhin lili ((tttt))
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.
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Chtlngknhvbngthng: S dngmhaknhlmtngchtlngknh(BERA>BERC).
Phiathmccbitd vobntin.
SoSo snhsnh chtcht lnglng knhknh cc vv khngkhng mm haha
Nuh thngthigianthc(khngchpnhntr)>phitngtctruyn>tngbngthng
Cngsutvbngthng: Gimcngsutpht>gimEb(t B>AhocC)
S dngmhaknh:gimcngsutphtkhnglmgimchtlng
SoSo snhsnh chtcht lnglng knhknh cc vv khngkhng mm haha
knh.
Tngbngthngdoccbitd camhaknh
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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.
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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
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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
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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
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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
.
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15
M kim tra chn l(parity)M kim tra chn l(parity)
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
M kim tra chn l(parity)M kim tra chn l(parity)
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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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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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
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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
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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
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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
MM khi tuyn tnhkhi tuyn tnh (4,7)(4,7)
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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MM khi tuyn tnhkhi tuyn tnh (4,7)(4,7)
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
MM khi tuyn tnhkhi tuyn tnh (4,7)(4,7)
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
MM khi tuyn tnhkhi tuyn tnh (4,7)(4,7)
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
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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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25
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
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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
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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
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Mch m ha CRC (tt)Mch m ha CRC (tt)
1
a thc sinh G(x) = x4 + x3 + 1
TxC
0
1 0 0 10 1 0 0
0 1 1 0 0 1 1
1
Mch m ha CRC (tt)Mch m ha CRC (tt)
1
a thc sinh G(x) = x4 + x3 + 1
TxC
0
0 1 0 01 0 1 1
0 1 1 0 0 1
1 1
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30
Mch m ha CRC (tt)Mch m ha CRC (tt)
1
a thc sinh G(x) = x4 + x3 + 1
TxC
0
1 0 1 11 1 0 0
0 1 1 0 0
1 1 1
Mch m ha CRC (tt)Mch m ha CRC (tt)
1
a thc sinh G(x) = x4 + x3 + 1
TxC
0
0 1 1 01 1 0 0
0 1 1 0
0 1 1 1
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Mch m ha CRC (tt)Mch m ha CRC (tt)
1
a thc sinh G(x) = x4 + x3 + 1
TxC
0
1 0 1 00 1 1 0
0 1 1
0 0 1 1 1
Mch m ha CRC (tt)Mch m ha CRC (tt)
1
a thc sinh G(x) = x4 + x3 + 1
TxC
0
1 1 0 01 0 1 0
0 1
1 0 0 1 1 1
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Mch m ha CRC (tt)Mch m ha CRC (tt)
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
Mch m ha CRC (tt)Mch m ha CRC (tt)
a thc sinh G(x) = x4 + x3 + 1
0
TxC
0
1
0 1 1
0 1 1 0 0 1 1 1
-
7/23/2019 TTS - Ch4
33/48
25/3/2014
33
Mch m ha CRC (tt)Mch m ha CRC (tt)
a thc sinh G(x) = x4 + x3 + 1
0
TxC
1
0 1 1
0 0 1 1 0 0 1 1 1
Mch m ha CRC (tt)Mch m ha CRC (tt)
a thc sinh G(x) = x4 + x3 + 1
0
TxC
1
0 1
0 0 1 1 0 0 1 1 11
-
7/23/2019 TTS - Ch4
34/48
25/3/2014
34
Mch m ha CRC (tt)Mch m ha CRC (tt)
a thc sinh G(x) = x4 + x3 + 1
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
35/48
25/3/2014
35
Mch gii m CRCMch gii m CRC
0
01
10
RxD
0 0 0 0
111001
1 0 0 0
RxC
c song song byte (xN)
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
c song song byte (xN)
SIPO
1
-
7/23/2019 TTS - Ch4
36/48
25/3/2014
36
Mch gii m CRC (tt)Mch gii m CRC (tt)
0
01
10
RxD
1 1 1 01001
1 1 0 0
RxC
c song song byte (xN)
SIPO
1 1
Mch gii m CRC (tt)Mch gii m CRC (tt)
0
01
10
RxD
0 1 1 1001
1 1 1 0
RxC
c song song byte (xN)
SIPO
1 1 1
-
7/23/2019 TTS - Ch4
37/48
25/3/2014
37
Mch gii m CRC (tt)Mch gii m CRC (tt)
0
01
10
RxD
1 0 1 001
0 1 1 1
RxC
c song song byte (xN)
SIPO
11 10
Mch gii m CRC (tt)Mch gii m CRC (tt)
0
01
10
RxD
1 1 0 1
1
1 0 1 0
RxC
c song song byte (xN)
SIPO
1 110 0
-
7/23/2019 TTS - Ch4
38/48
25/3/2014
38
Mch gii m CRC (tt)Mch gii m CRC (tt)
0
01
10
RxD
0 1 1 11 1 0 1
RxC
c song song byte (xN)
SIPO
1 1 1001
Mch gii m CRC (tt)Mch gii m CRC (tt)
0
01
10
RxD
1 0 1 00 1 1 1
RxC
c song song byte (xN)
SIPO
11 10 01 1
-
7/23/2019 TTS - Ch4
39/48
25/3/2014
39
Mch gii m CRC (tt)Mch gii m CRC (tt)
01
10
RxC
RxD
SIPO
c song song byte (xN)
Mch gii m CRC (tt)Mch gii m CRC (tt)
1
10
RxC
RxD
SIPO
0 0 1 10 1 0 1
c song song byte (xN)
1 1100110
-
7/23/2019 TTS - Ch4
40/48
25/3/2014
40
Mch gii m CRC (tt)Mch gii m CRC (tt)10
RxC
RxD
SIPO
0 0 0 00 0 1 1
c song song byte (xN)
1 1100110
Mch gii m CRC (tt)Mch gii m CRC (tt)0
RxC
RxD
SIPO
0 0 0 00 0 0 0
c song song byte (xN)
1 1100110
-
7/23/2019 TTS - Ch4
41/48
25/3/2014
41
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
-
7/23/2019 TTS - Ch4
42/48
25/3/2014
42
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
- Dch vng tri ln 2 c: 1010111, chia cho 1101, d110
- Dch vng tri ln 3 c: 0101111, chia cho 1101, d001
- 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
-
7/23/2019 TTS - Ch4
43/48
25/3/2014
43
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
-
7/23/2019 TTS - Ch4
44/48
25/3/2014
44
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
-
7/23/2019 TTS - Ch4
45/48
25/3/2014
45
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
-
7/23/2019 TTS - Ch4
46/48
25/3/2014
46
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
-
7/23/2019 TTS - Ch4
47/48
25/3/2014
47
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
-
7/23/2019 TTS - Ch4
48/48
25/3/2014
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: