ch3 ma khoi tuyen tinh
TRANSCRIPT
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Csl thuytthng tin
Chng3: M ha knhM Khituyntnh
TS. Phm Hing
02/12/2013 Slice 1Trng H Bch Khoa H Ni
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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.
02/12/2013 Slice 2Trng H Bch Khoa H Ni
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Phn1: Khi nimcbn
Phn loili
Liclpthngk: Lixuthintrong qu trnh truyntin trnknh truyn, xuthinclpkhng lin quan tinhau. V d:
nhiuGaussian.
Lichm: Lic phn blin hvinhau.
02/12/2013 Slice 3Trng H Bch Khoa H Ni
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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)
02/12/2013 Slice 4Trng H Bch Khoa H Ni
1-p
IN OUT
0 0
1 1
1-p
p
p
p l xc sutliBERBit Error Rate (BER)
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Phn1: Khi nimcbn
V d: Knh truyntin ang
02/12/2013 Slice 5Trng H Bch Khoa H Ni
receiving signal
time
strength
0
sending signal
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Phn1: Khi nimcbn
V d: Knh truyntin ang
02/12/2013 Slice 6Trng H Bch Khoa H Ni
RX power
TX power
Channel Fading
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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
02/12/2013 Slice 7Trng H Bch Khoa H Ni
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Phn2: Cc khi nimcbncam ha sali
Tcm
Khongcch Hamming (Hamming distance)
Khongcch tithiu(minimum distance)
Ma trnsinh, m trnkimtra chnl.
02/12/2013 Slice 8Trng H Bch Khoa H Ni
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Phn2: Cc khi nimcbncam ha sali
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
02/12/2013 Slice 9Trng H Bch Khoa H Ni
2
2
n
2
( , )n k 2k 2n
( , )n k k n
( , )n k
k
Rn
( ,1)n
1R
n
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Phn2: Cc khi nimcbncam ha sali
Ma trnsinh
Vi biudinthng tin (message).l tm (codeword) cam lp
Qu trnh m ha cbiudindidngma trn. Ma trn
sinh cam lpl
02/12/2013 Slice 10Trng H Bch Khoa H Ni
m
C ( ,1)n
, , , ...C m m m m
C mG
1,1,1,...,1G
G
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Phn2: Cc khi nimcbncam ha sali
Biudinm lp(3,1) trong khng gian
Khongcch Hamming trong m binary ctnh bngs
cc imkhc bittrong 2 tm.
02/12/2013 Slice 11Trng H Bch Khoa H Ni
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
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Phn2: Cc khi nimcbncam ha sali
nhngha: Khongcch tithiu(min distance) l
khongcch Hamming nhnhtgia2 tm btk.
Phngphp giim ML: tm kimtm c khongcchgnnhtvitm thu c.
02/12/2013 Slice 12Trng H Bch Khoa H Ni
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Phn2: Cc khi nimcbncam ha sali
Lin hgiakhongcch Hamming tithiuv khnng
pht hinv sali.
Vim binary
Khnngpht hinliKhnngsali
02/12/2013 Slice 13Trng H Bch Khoa H Ni
( , )n k
min 1d
min 1
2
dt
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Phn2: Cc khi nimcbncam ha sali
02/12/2013 Slice 15Trng H Bch Khoa H Ni
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
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Phn2: Cc khi nimcbncam ha sali
02/12/2013 Slice 16Trng H Bch Khoa H Ni
V d1: M kimtra chnl(6,5)
Bntin m = (10110) => tm c=(101101)
Bntin m = (11011) => tm c=(110110)
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Phn2: Cc khi nimcbncam ha sali
02/12/2013 Slice 17Trng H Bch Khoa H Ni
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
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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
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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
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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
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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
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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
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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.
Phn3: M khituyntnh Giim sali
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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)
Phn3: M khituyntnh Giim sali
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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
Phn3: M khituyntnh Giim sali
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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
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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
Phn3: M khituyntnh M hthng
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Tm vector syndrome trong v d
Tm thu ccr= [1101001]
000
100
010
001
111
011
101
110
1001011HcsT
r
Phn3: M khituyntnh M hthng
h i
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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
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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)
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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
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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
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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
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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
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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