NHĐT Ly Thuyet Thong Tin

8/14/2019 NHT Ly Thuyet Thong Tin

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Chn cu ng sau : d Tin l dng vt cht biu din hoc thhin thng tinChn pht biu ng trong nhng cu sau : a Thng tin l nhng tnhcht xc nh ca vt cht m con ngi nhn c t th gii vtcht bn ngoi

Mn l thuyt thng tin bao gm vic nghin cu: b Cc qu trnhtruyn tin v L thuyt m haChn cu ng nht v ngun tin a Ngun tin l ni sn ra tinChn cu ng nht v ng truyn tin a L mi trng Vt l,trong tn hiu truyn i t my pht sang my thu bin i mt tn hiu lin tc theo bin v theo thi gian thnh tn hius, chng ta cnthc hin qu trnh no sau y: a Ri rc ha theo trc thi gianv lng t ha theo trc bin Chn cu ng v tn hiu: d Tn hiu l qu trnh ngu nhin .Chn pht biu ng nht v c trng thng k : a c trng cho

cc qu trnh ngu nhin chnh l cc quy lut thng k v cc ctrng thng kChn cu ng nht v hm t tng quan : c Hm t tngquan 1 2( , )x R t t c trng cho s ph thuc thng k gia hai gi tr hai thi im thuc cng mtVic biu din mt tn hiu gii hp thnh tng ca hai tn hiu iu binbin thin chm s lmcho vic phn tch mch v tuyn in di tc ng ca n tr nn phc tp

b SaiNgi ta gi tn hiu gii rng nu b rng ph ca n tho mn bt ng

thc sau: 10

. Cc tn hiu iu tn, iu xung, iu xung ct, manip tn

s, manip pha, l cc tn hiu gii rng. b ngChn cu ng v cng thc xc nh mt ph cng sut c

2

( )

( ) ( ) limT

xT

S

G M G M T

l cng thc xc nh mt ph cng

sut ca cc qutrnh ngu nhin.Chn cu ng v cng thc quan h gia mt ph cng sut v hm t

tng quan d ( ) ( )j

G R e d

Trong trng hp h thng tuyn tnh th ng c suy gim th nhng thiim t >> t0= 0 (thi im t tc ng vo), qu trnh ngu nhin u ras c coi l dng. Khi hm t tng quan v mt ph cng sut caqu trnh ngu nhin u ra s lin h vi nhau theo biu thc sau

1( ) ( )

2

j

ra ra R G e d

b ng


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*( )aS t l hm lin hp phc ca ( ) : ( ) ( ) ( )a aS t S t x t j x t

l tn hiu gii tch. ng

bao ca tn hiu gii tch c th biu din bng cng thc sau:*( ) ( ). ( )a aA t S t S t b ng

Mt mch v tuyn in tuyn tnh c tham s khng i v c tnh truyn

t dng ch nht(hnh di) chu tc ng ca tp m trng dng. Tm hm t tng quan ca

tp m ra theo cng thc2

0

0

( ) ( ) cos2

ra

G R K d

ta c kt qu no ?

Hnh chng 2 cu 10 a 20

sin2( ) cos/ 2

ra raR

Cho qu trnh ngu nhin dng c biu thc sau: 0( ) cos(2 ) X t A f t Trong A = const, f0= const, l i lng ngu nhin c phn b u trong

khong ( , ) . Tnh k vng ( )M X t theo cng thc ( ) ( ) ( )M X t X t w d

ta c gi tr no di y a ( ) 0M x t

Cho qu trnh ngu nhin dng c biu thc sau: 0( ) cos(2 ) X t A f t Trong A = const, f0= const, l i lng ngu nhin c phn b u trongkhong ( , ) . Tnh hm t tng quan 1 2( , )R t t theo biu thc

1 2( , ) ( ). (R t t M X t X t ta c gi tr no di y: c

2

1 2 0

1( , ) cos 2

2 R t t A f

Tn hiu in bo ngu nhin X(t) nhn cc gi tr + a; - a vi xc sut nhnhau v bng 1/2. Cn xc sut trong khong c N bc nhy l:

( )( , )

!

N

P N eN

>0 (theo phn b Poisson). T cc gi thit trn tnh c

hm t tng quan 2 2( )x R a e . Khi mt ph cng sut ( )xG ca X(t)

c tnh theo cng thc0

( ) 2 ( ) cosx xG R d

ta c gi

tr no sau y a2

2 2

4( )

4x

aG

Mt qu trnh ngu nhin dng c hm t tng quan: 2 0( ) cosxR e Khi mt ph cng sut ca cc qu trnh ngu nhin trn l

d 2 2 20 0

2 2( )

( ) ( )xG

Khi nim lng tin c nh ngha da trn: b bt nh ca tinChn pht biu ng nht v Entropy ca ngun tin, H(X): a Li lng c trng cho bt nh trung bnh ca ngun tin


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Chn pht biu sai v bt nh c bt nh ca mt phn t cgi tr 1 bit khi xc sut chn phn t l 1Entropy ca ngun ri rc nh phn H(A) =plog p(1p)log(1p).Khi p=1/2th H(A) t max Chn cu ng v H(A)max c H(A)max= 1 btTrong mt trn thi u bng Quc t, i tuyn Vit Nam thng i tuyn

Brazin, thng tin ny c bt nh l : b V cng lnHc sinh A c thnh tch 12 nm lin t danh hiu hc sinh gii, hc sinh Blc hc km Thi ttnghip ph thng trung hc, hc sinh A trt, hc sinh B th khoa Thngtin v hc sinh B th khoa, hc sinh A trt c bt nh l: b

V cng lnChn ngu nhin mt trong cc s t 0 n 7 c xc sut nh nhau Khi xc sut ca s cchn ngu nhin l: b 1/8Mt thit b v tuyn in gm 16 khi c tin cy nh nhau v c mcni tip Gi s c mt khi no b hng, khi xc sut ca mt khi

hng l: a 1/16B t l kh 52 qun (khng k fng teo), A rt ra mt qun bi bt k Xcsut v qun bi m A rt l: a Bng 1/52Chn cu sai trong cc cu sau : c Lng thng tin = bt nh tinnghim + bt nh hu nghimChn cu ng sau : a Lng thng tin = thng tin tin nghim -thng tin hu nghimChn cu sai trong cc cu sau : a Thng tin tin nghim (k hiuI(Xk)) c xc nh theo cng thc sau:I(Xk) = log P(Xk)I(Xk,Yl) l lng thng tin cho vXkdo Ylmang li c tnh bng cng thc

no sau y a

1 1

log log( ) ( / )k k l P x P x y

I(Xk/Yl)= - log ( / )k l P x y l thng tin hu nghim v kx ( / )k lp x y = 1 khi victruyn tin khngb nhiu. Chn cu sai trong nhng cu v I(Xk/Yl) di y: d I(Xk/Yl) =1/2 khi knh khng c nhiuI(Xk/Yl) l thng tin hu nghim v kx ( / )k l p x y = 1 khi vic truyn tin khng bnhiu Chn cusai trong nhng cu v I(Xk/Yl) di y: c I(Xk/Yl) >1/2 khi knhkhng c nhiu

I(Xk/Yl) l lng thng tin ring ca kx khi bit ly v I(Xk/Yl) = 0 khi khngc nhiu Cuny ng hay sai ? a ngChn cu sai trong nhng cu sau: b I(Xk,Yl) l lng tin ring ca kx

v ly

Cho tin ( )ix c xc sut l ( )iP x 0,5 lng tin ring I ( )ix ca tin ny bng cci lng no


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di y : b 1 btCho tin ( )ix c xc sut l ( )iP x 1/ 4 lng tin ring I ( )ix ca tin ny bng cci lng nodi y : a 2 btCho tin ( )ix c xc sut l ( )iP x 1/ 8 lng tin ring I ( )ix ca tin ny bng cc

i lng nodi y : b 3 btCho tin ( )ix c xc sut l ( )iP x 1/16 lng tin ring I ( )ix ca tin ny bng cci lng nodi y: d 4 btCho tin ( )ix c xc sut l ( )iP x 1/ 27 lng tin ring I ( )ix ca tin ny bngcc i lng nodi y: c Log27 bt;Cho tin ( )ix c xc sut l ( )iP x 1/ 9 lng tin ring I ( )ix ca tin ny bng cci lng no di y : c 2log3 bt

Cho tin ( )ix c xc sut l ( )iP x 1/ 25 lng tin ring I ( )ix ca tin ny bngcc i lng no di y : b 2log5 btTm cu sai trong nhng cu di y c bt nh ca tin vlng tin c ngha nh nhau nhng gi tr khc nhauLng thng tin ring ( bt nh) ca mt bin ngu nhin kx l ( )kI x . Chn

biu thc sai trong cc biu thc di y a ( ) ln ( )k k I x p x n vo l bitLng thng tin ring ( bt nh) ca mt bin ngu nhin kx l ( )kI x c

tnh bng biu thc no di y : a ( ) ln ( )k k I x k p x

Lng thng tin ring ( bt nh) ca mt bin ngu nhin kx l ( )kI x ctnh nh sau ( ) ln ( )k k I x k p x , trong k l h s t l. Tm cu sai v cchchn k trong cc cu di y : a Chn k = 1 ta c ( ) ln ( )k k I x p xEntropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin

ring ca cc tin thuc A. K hiu: 1( );H A 1( ) ( )iH A M I a

vi1

1( )

aA

p a

2

2( )

a

p a

....

.... ( )s

s

a

p a

0p(ai) 1 ;

1

( ) 1s

i

i

p a

; 1( )H A c tnh bng

biu thc no di y d 1 1( ) ( ) log ( )

s

i ii

H A p a p a (bt)

Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin


vi1

1( )

aA

p a


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2

2( )

a

p a

....

.... ( )s

s

a

p a

0p(ai) 1 ;

1

( ) 1s

i

i

p a

; 1( )H A c tnh bng

biu thc no : c 11

1( ) ( ) log

( )

s

i

i i

H A p ap a

(bt)

Entropy ca ngun tin ri rc A l trung bnh thng k ca lng thng tin


vi0

0( )

aA

p a

1

1( )

a

p a

....

....1

1( )

s

s

a

p a

0p(ai) 1 ;

1

1

( ) 1s

i

i

p a

; 1( )H A c tnh nh sau: d

1

1

0

( ) ( ) log ( )s

i i

i

H A p a p a

(bt)A v B l hai trng bin c bt k, Entropy ca 2 trng bin c ng thiC=AB l H(AB),

trong cc tnh cht ca H(AB) di y, tnh cht no sai: c1

1

( ) ( ) log ( )s

i i

i

H A p a p a

(bt)Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B). Trongcc tnh cht ca H(A/B) di y, tnh cht no sai a H(A / B)H(B / A)Entropy c iu kin v 1 trng tin B khi r trng tin A l H(B/A),Tnhcht no ca H(B/A)di y l ng b 0H(B/A)Entropy ca trng bin c ng thi H(AB) c tnh bng cng thc nosau y d H(B) + H(A/B)Lng thng tin cho trung bnh (k hiu l I(A,B) c cc tnh cht no sauy b 0I(A,B)H(A)Lng thng tin cho trung bnh (k hiu l I(A,B) Trong cc tnh cht diy, tnh cht nosai: b I(A,B)=H(A) khi knh c nhiu;Lng thng tin cho trung bnh (k hiu l I(A,B) c cc tnh cht no sauy b 0I(A,B) v I(A,B)=0 khi knh b tLng thng tin cho trung bnh (k hiu l I(A,B) ), trong cc tnh cht diy ca I(A,B), tnh cht no sai d I(A,B)1 khi knh b t;Lng thng tin cho trung bnh (k hiu l I(A,B)) , tm biu thc sai trongcc biu thc di y c I(A,B) = H(A) - H(B/A)Mnh no sau y sai c I(A,B) = H(A) + H(B) + H(AB)Chn ngu nhin mt trong cc s t 0 n 7 c xc sut nh nhau btnh ca s c chn ngu nhin l c 3 btMt thit b v tuyn in gm 16 khi c tin cy nh nhau v c mcni tip Gi s c mt khi no b hng, bt nh ca khi hng l:

d 4 btB t l kh 52 qun (khng k fng teo), A rt ra mt qun bi bt k bt nh v qun bi


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m A rt l: b Ln hn 5 nh hn 6 btMt hp c 8 ng tin kim loi , trong c 02 ng tin 500 ng; 02 ngtin 1000 ng, 2ng tin 2000 v 2 ng tin 5000 ng Chn ngu nhin 1 trong 8 ngtin Khi xc sut ca ng tin c chn ngu nhin l: d

1/4 btMt hp c 8 ng tin kim loi , trong c 02 ng tin 500; 02 ng tin1000, 2 ng tin2000 v 2 ng tin 5000 Chn ngu nhin 1 trong 8 ng tin Khi bt nh ca ng tin c chn ngu nhin l: d 2 btCho ngun tin X = {x1, x2, x3} vi cc xc sut ln lt l {1/2, 1/4, 1/4},

Entropy ca ngun tin H(X) c tnh l c1 1 1

log 2 log 4 log 42 4 4

Cho ngun tin X = {x1, x2, x3, x4, x5, x6, x7, x8, x9} vi cc xc sut lnlt l {1/4, 1/8, 1/8,1/8, 1/16, 1/16, 1/8, 1/16, 1/16}. Trong cc kt qu tnh Entropy di y, kt

qu no sai: a 1 1 1log4 4 log8 4 log164 8 8

Entropy 1( )H A ca ngun ri rc1

1( )

aA

p a

2

2( )

a

p a

....

.... ( )s

s

a

p a

0p(ai)

1 ;1

( ) 1s

i

i

p a

Trong cc tnh cht ca H(A) di y tnh cht no l saia Khi ( ) 1, ( ) 0k i p a p a ,vi Vi kth 1 1 min( ) ( ) 1H A H A

Cho ngun ri rc A1

1( )

aA

p a

2

2( )

a

p a

....

.... ( )s

s

a

p a

0p(ai) 1 ;

1

( ) 1s

i

i

p a

Gi entropy ca ngun A l 1( )H A , trong cc biu thc tnh 1 max( )H A

logs di y, biu thc no sai: a 1 max1

1( ) log ( ) log log

( )

s

i

i i

H A s p a sp a

Cho ngun ri rc A1

1( )

aA

p a

2

2( )

a

p a

....

.... ( )s

s

a

p a

0p(ai) 1 ;

1

( ) 1s

i

i

p a

Nu ngun A c s du ng xc sut , khi biu thc no di

y l sai a1 1

1 ( ) 0s s

i

i i

p as

Kh nng thng qua ca knh ri rc C l gi tr cc i ca lng thng tincho trung bnh truyn qua knh trong mt n v thi gian ly theo mi kh

nng c th c ca ngun tin A ' ' ' 'max ( , ) max ( , )( ); .k kA A

C I A B v I A B bps C v C

vi

max ( , ); kA

C I A B v biu th s du m knh truyn c (c truyn qua


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knh) trong mt n v thi gian I(A,B) l lng thng tin truyn qua knhtrong mt n v thi gian C c gi l kh nng thng qua ca knh i vimi du C c cc tnh cht no di y : a C 0, C = 0 khi Av B c lp ; C vklogs , = vklogs khi knh khng nhiuI(A,B) l lng thng tin trung bnh c truyn qua knh ri rc c tnh cht

: I(A,B) H(A)V mt s nh ngha : ' ' ' 'max ( , ) max ( , )( ); .k k

A AC I A B v I A B bps C v C

vi

max ( , ); kA

C I A B v biu th s du m knh truyn c trong mt n v thigian. T cc tnh cht v nh ngha trn cho bit cc biu thc di y,

biu thc no sai a( )

( , )H AK K

v I A B v

Cho ngun ri rc ch c hai du:1

1( )

aA

p a

2

2( )

a

p a

Ngun ri rc nh phn l

ngun A trn tho mn iu kin sau1

2

"0"

"1"

a

a

vi xc sut

1

2

( )

( ) 1

p a p

p a p

Khi

ngun ri rc nh phn A c th vit biu thc no d1

aA

p

2

1

a

p

1a

Ap

2

1

a

p

l ngun ri rc nh phn Tho mn iu kin

1

2

"0"

"1"

a

a

vi xc

sut1

2

( )

( ) 1

p a p

p a p

Khi Entropy 1( )H A c tnh bng cng thc no sau y

d p log p(1p)log(1p)1( / ) H A b l lng thng tin tn hao trung bnh ca mi tin u pht khi u

thu thu c jb 1 1 11

( / ) ( / ) log ( / ); ( / )s

i i i

i

H A b p a b p a b H B a

l lng thng tinring trung bnh cha trong mi tin u thu khi u pht pht i mt tin

ia c tnh theo cng thc1

( / ) ( / ) log ( / )t

i j i j i

j

H B a p b a p b a

Trong trng hpknh b t (b nhiu tuyt i) ta c biu thc no di y l sai : a

( / ) ( ) ( )jH A b H A H B 1( / ) H A b l lng thng tin tn hao trung bnh ca mi tin u pht khi u

thu thu c jb 1 1 11

( / ) ( / ) log ( / ); ( / )s

i i i

i




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c tnh theo cng thc:1

( / ) ( / ) log ( / )t

i j i j i

j

H B a p b a p b a

Trong trng hpknh khng nhiu biu thc no di y l ng : d

( / ) ( / ) 0kH A b H A B

1( / ) H A b l lng thng tin tn hao trung bnh ca mi tin u pht khi uthu thu c jb 1 1 1

1

( / ) ( / ) log ( / ); ( / )s

i i i

i



ia c tnh theo cng thc1

( / ) ( / ) log ( / )t

i j i j i

j

H B a p b a p b a

Trong trng hpb nhiu tuyt i, A v B l c lp nhau suy ra :

( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi ta c biu thc no

sau y l ng: a1

( / ) ( ) log ( )s

j i i

i

H A b p a p a

v ( / )i H B a

1

( ) log ( )s

j j

i

p b p b

Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B), cxc nh theo

cng thc sau:1 1

( / ) ( ) log ( / )s t

i j i j

i j

H A B p a b p a b

Trong trng hp b nhiutuyt i, A v B l c lp nhau, suy ra :

( / ) ( ); ( / ) ( ) ( ) ( ) ( )i j i j i j i j i jp a b p a p b a p b p a b p a p b khi biu thc no sau y

l ng: d 1 1( / ) ( ) ( ) log ( )

t s

j i ij iH A B p b p a p a

Entropy c iu kin v 1 trng tin B khi r trng tin A l H(B/A), cxc nh theo

cng thc sau:1 1

( / ) ( ) log ( / )s t

j i j i

i j

H B A p b a p b a



l ng: a1 1

( / ) ( ) ( ) log ( )s t

i j j

i j

H B A p a p b p b

Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A/B), cxc nh theo

cng thc sau:1 1

( / ) ( ) log ( / )s t

i j i j

i j

H A B p a b p a b



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l ng: a1 1

( / ) ( ) ( ) log ( ) ( )t s

j i i

j i

H A B p b p a p a H A

Entropy c iu kin v 1 trng tin B khi r trng tin A l H(B/A), c

xc nh theocng thc sau:

1 1

( / ) ( ) log ( / )s t

j i j i

i j

H B A p b a p b a



l ng: c1 1

( / ) ( ) ( ) log ( ) ( )s t

i j j

i j

H B A p a p b p b H B

Entropy c iu kin v 1 trng tin A khi r trng tin B l H(A B), cxc nh theo

cng thc sau: 1 1( / ) ( ) log ( / )

s t

i j i ji j

H A B p a b p a b vi

( )

( / ) ( )

i j

i j

j

p ab

a b p b T cngthc ny c th khai trin ( / ) H A B thnh cng thc no sau y: c

1 1

( ) ( / ) log ( / )t s

j i j i j

j i

p b p a b p a b

Lng thng tin cho trung bnh (k hiu l I(A,B)) ( , ) ( , )i jI A B M I a b

vi

( / )( , ) log

( )

i j

i j

i

p a b I a b

p a Xc sut c thng tin ( , )i j I a b l ( )i j p ab Do c th

vit I(A,B)) bng cng thc no sau y: b 1 1

( / )

( ) log ( )

s ti j

i ji j i

p a b

p ab p a Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :

1 1

( / )( , ) ( ) log

( )

s ti j

i j

i j i

p a b I A B p a b

p a Khi c th khai trin I(A,B) thnh

1 1

( , ) ( ) log ( / ) ( )s t

i j i j i

i j

I A B p a b p a b p a

b ngLng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :

1 1

( / )( , ) ( ) log

( )

s ti j

i j

i j i

p a b I A B p a b


1 1 1 1

( , ) ( ) log ( / ) ( ) log ( ) s t s t

i j i j i j i

i j i j

I A B p a b p a b p a b p a

a ng


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Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :

1 1

( / )( , ) ( ) log

( )

s ti j

i j

i j i

p a b I A B p a b


1 1

( , ) ( ) log ( / ) ( )s t

i j i j j

i j

I A B p a b p a b p b

a SaiLng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :

1 1

( / )( , ) ( ) log

( )

s ti j

i j

i j i

p a b I A B p a b

p a Trong cc biu thc khai trin I(A,B) di y,

biu thc

no sai b1 1

( , ) ( ) log ( / ) ( )s t

i j i j j

i j

I A B p a b p a b p b

Lng thng tin cho trung bnh (k hiu l I(A,B)) c vit thnh :

1 1

( / )( , ) ( ) log

( )

s ti j

i j

i j i

p a b I A B p a b

p a Entropy c iu kin v 1 trng tin A khi r

trng tin B l ( / ) H A B , c xc nh theo cng thc sau:

1 1

( / ) ( ) log ( / )s t

i j i j

i j

H A B p a b p a b

Khi trong cc kt qu tnh I(A,B), kt quno sai b I(A,B) = H(A)+ H(A/B)

Xt 2 trng s kin A v B sau : 1, ; 1,( ) ( )

ji

i j

ba A i s B j t

p a p b

Khi ,

trng s kin ng thi C=A.B Nu A v B l c lp th C c th vit thnh

biu thc no di y: d( ) ( )

i j

i j

a bC

p a p b

Xt 2 trng s kin A v B sau : 1, ; 1,( ) ( )

ji

i j

ba A i s B j t

p a p b

Trng s

kin ng thi C=A.B c entropy H(C) c tnh bng cng thc no di

y: d1 1

( ) ( ) log ( )s t

i j i j

i j

H C p a b p a b

Chn cu sai : a Xc sut xut hin cng ln, lng tin thuc cng ln cng s cng nhChn cu sai sau : a Xc sut xut p(x) cng ln th lng tin khi

nhn c tin ny cng s cng lnLng tin c iu kin hu nghim v Kx ( thng tin ring v Kx sau khi c ly )c nh nghal ( / ) log ( / )k l k l I x y p x y Chn cu sai trong cc cu sau b Khi

( / )k lp x y 1 th ( / )k l I x y 1 v ngc li


11/30

Lng thng tin v Kx khi r tin ly l( / )

( / ) log( )

k lk l

k

p x yI x y

p x Chn cu sai

sau : b Nu ( / )k l p x y = 0 suy ra1

( / ) log( )

k l

k

I x yp x

Lng thng tin hu nghim v Kx ( thng tin ring v Kx sau khi c ly ) cvit l :

1( / ) log

( )k l

k

I x yp x

Lng thng tin ring v Kx l1

( ) log( )

k

k

I xp x

Lng thng tin cho v Kx do ly mang li l :1 1

( / ) log log( ) ( / )

k l

k k l

I x y p x p x y

Tm cu sai sau : b Lng thng tin ring c th m

Cho 2 ngun tin A v B c cc xc sut tng ng l :1

1/ 2

aA

2

1/ 4

a 3

1/ 8

a

4 1;

1/ 8 0,5

a bB

2

0,25

b 3

0,125

b 4

0,125

b

Entropy ca ngun A ( k hiu l H(A)), entropy

ca ngun B ( k hiu l H(B)) c quan h theo cc h thc no di y a H(A)=H(B)

Cho ngun tin A c cc xc sut tng ng l :1

1/ 2

aA

2

1/ 4

a 3

1/ 8

a 4

1/ 8

a Khi

Entropy ca ngun A ( k hiu l H(A)) bng cc i lng no di ybH(A) = 1,75 bt


0,45

aA

2

0,2

a 3

0,15

a 4

0,1

a 5

0,1

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yb H(A) = 2,06 bt


0,4

aA

2

0,25

a 3

0,15

a 4

0,1

a 5

0,1

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yd H(A) = 2,1 bt


0,4

aA

2

0,25

a 3

0,15

a 4

0,15

a 5

0,05

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no diy d H(A) = 2,07 bt


0,4

aA

2

0,25

a 3

0,2

a 4

0,1

a 5

0,05

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yc H(A) = 2,04 bt


12/30


0,4

aA

2

0,3

a 3

0,15

a 4

0,1

a 5

0,05

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di yd H(A) = 2,01 bt

Gi s ngun tin A c cc xc sut tng ng l : 10,4aA

2

0,3a 3

0,2a 4

0,05a 5

0,05a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no diy a H(A) = 1,95 bt

Gi s ngun tin A c cc xc sut tng ng l :1

0,35

aA

2

0,35

a 3

0,2

a 4

0,05

a 5

0,05

a

Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng nodi y d H(A) = 1,96 bt


0,35

aA

2

0,3

a 3

0,25

a 4

0,05

a 5

0,05

a

Khi Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng nodi y c H(A) = 1,98 bt


0,35

aA

2

0,3

a3

0,2

a4

0,1

a 5

0,05

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no diy a H(A) = 2,06 bt


0,3

aA

2

0,3

a 3

0,2

a 4

0,1

a 5

0,1

a

Khi

Entropy ca ngun A ( k hiu l H(A)) gn bng cc i lng no di y

c H(A) = 2,17 bt

Gi s ngun tin A v B c cc xc sut tng ng l :1

1/ 4

aA

2

1/ 8

a 3

1/16

a 4

1/16

a

5

1/ 4

a 6

1/ 8

a 7

1/16

a 8

1/16

a

1

0,25

bB

2

0,125

b3

0,0625

b4

0,0625

b5

0,25

b6

0,125

b7

0,0625

b 8

0,0625

b

Entropy

ca ngun A ( k hiu l H(A)), entropy ca ngun B ( k hiu l H(B)) cquan h theo cc h thc no di y d H(A) > H(B)


1/ 4

aA

2

1/ 8

a 3

1/16

a 4

1/16

a 5

1/ 4

a 6

1/ 8

a

7

1/16

a 8

1/16

a

khi Entropy ca ngun A ( k hiu l H(A)) bng cc i lng

no di y b H(A) = 2,75 bt


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Gi s ngun tin B c cc xc sut tng ng l :1

0,25

bB

2

0,125

b 3

0,0625

b 4

0,0625

b

5

0,25

b 6

0,125

b 7

0,0625

b 8

0,0625

b

Khi Entropy ca ngun B ( k hiu l H(B)) bng

cc i lng no di y c H(B) = 2,75 btCho mt knh nh phn nh hnh bn: hnh chng 3 cu 92 Trong :Phn b xc sut ca tin u ra 1( )p b c tnh theo cng thc sau :

2

1 1

1

( ) ( ) ( / )i ii

p b p a p b a

T cng thc ny, c th khai trin 1( )p b thnh cng thcno di y: c 1 1 1 1 2 1 2 p(b ) p(a ).p(b / a ) p(a ).p(b / a ) Cho mt knh nh phn nh hnh bn: hnh chng 3 cu 93 Trong :Phn b xc sut ca tin u ra 2( )p b c tnh theo cng thc sau :

2

2 2

1

( ) ( ) ( / )i ii

p b p a p b a

T cng thc ny, c th khai trin 1( )p b thnh cng

thc no di y: b 2 1 2 1 2 2 2p(b ) p(a ).p(b / a ) p(a ).p(b / a ) Cho knh i xng nh phn nh hnh bn: hnh chng 3 cu 94 Bit

1 2( ) ; ( ) 1p a p p a p 1 2 2 1( / ) ( / ) 1s dp b a p b a p p 1 1 2 2( / ) ( / ) dp b a p b a p 2 2

1 1

( / ) ( ) ( / ) log ( / )i j i j ii j

H B A p a p b a p b a

Khai trin H(B/A) theo pd v ps s ckt qu no di y b ( / ) log (1 ) log(1 ) s s s s H B A p p p p Cho knh i xng nh phn nh hnh bn: hnh chng 3 cu 95 Bit

1 2( ) ; ( ) 1p a p p a p 1 2 2 1( / ) ( / ) 1s dp b a p b a p p 1 1 2 2( / ) ( / ) dp b a p b a p

( / ) log (1 ) log(1 ) s s s s H B A p p p p T truyn tn hiu cho knhvK= 1/T

v '1 1

max ( , ) max ( ) ( / )A A

C I A B H B H B AT T

Khi '

'

max

C

Cc tnh theo biu

thc no di y c'

'

max

1 log (1 ) log(1 ) s s s sC

p p p pC

Chn cu ng v m ha c M ha l mt nh x 1- 1 t tp cctin ri rc ia ln tp cc t m ; :i i

n n

i i if a Chn cu ng v m : d M (hay b m) l sn phm ca php mha

Chn cu ng v m : b M l mt tp cc t m c lp nntheo mt lut nh di t m ni l s cc du m cn thit dng m ha cho tin ai . Chncu ng v di t m b Nu ni = constvi mi i th mi tm u c cng di B m tng ng c gi l b m uChn cu sai v di t m d di t m ni l mt snguyn c th m


14/30

di t m ni l s cc du m cn thit dng m ha cho tin ai . Chncu ng v di t m a Nu ninjth b m tng ng cgi l b m khng uS cc du m khc nhau (v gi tr) c s dng trong b m c gi lc s m Ta k hiu

gi tr ny l m . Chn cu sai v cc du m m: c Nu m = 0 thb m tng ng c gi l m uGi s c t m 7i =0 1 1 0 1 0 1Chn cu ng nht v t m

7

i b

T m 7i trong b m nh phn c m=2 ( tc l c 2 du m l0 v 1) v c di l 7Gi s c t m 7i =0 1 1 0 1 0 1 . Chn cu ng v t m

7

i c T

m 7i c di bng 7

Chn cu ng v di t m a di trung bnh ca t m n

l k vng ca i lng ngu nhin nic xc nh nh1

( )s

i i

i

n p a n

Ni dung ca nh l m ho th nht ca Shannon (i vi m nh phn)c pht biu nh sau: Lun lun c th xy dng c mt php m hocc tin ri rc c hiu qu m di trung bnh ca t m c th nh tu ,nhng khng nh hn entropie xc nh bi cc c tnh thng k cangun . Chn cu ng v di t m : d Chiu di trung bnhcc t m nh nht trong tt c cc cch m haNi dung ca nh l m ho th nht ca Shannon (i vi m nh phn)c pht biu nh sau: Lun lun c th xy dng c mt php m hocc tin ri rc c hiu qu m di trung bnh ca t m c th nh tu ,nhng khng nh hn entropie xc nh bi cc c tnh thng k ca

ngun. Chn cu sai v di t m : d Chiu di trung bnh

cc t m tho mn h thc 11

( ) ( )s

i i

i

n p a n H A

Khong cch gia hai t m bt k ni v

n

j l s cc du m khc nhau tnh

theo cng mt v

tr gia hai t m ny, k hiu d( ni ,n

j ) Gi sn

i =0 1 1 0 1 0 1;n

j =1 0 0

1 1 1 0 . Khong cch gia 2 t m ni vn

j l d((7

i ,7

j ) bng cc i lng

no di y b 6Khong cch gia hai t m bt k ni v

n


theo cng mt vtr gia hai t m ny, k hiu d( ni ,

n

j ) . Tm biu thc sai v khong cch

m d trong cc biu thc sau: c d( ni ,n

j ) d(n

i ,n

j )

Khong cch gia hai t m bt k ni vn


theo cng mt v


15/30

tr gia hai t m ny, k hiu d( ni ,n

j ) Tm biu thc sai v khong cch

m d trong cc biu

thc sau: c d( ni ,n

j ) 0

Khong cch gia hai t m bt k ni vn


theo cng mt vtr gia hai t m ny, k hiu d( ni ,

n

j ) . Chn cu ng sau : d d(n

i ,n

j ) 0 ; d(n

i ,n

j ) = 0 khin

i n

j

Trng s ca mt t m W( ni ) l s cc du m khc khng trong t m V

d: ni = 0 1 1 0 1 0 1 th W(n

i ) = 4. Chn cu ng v cc tnh cht ca

trng s W( ni ) d d(n

i ,n

j ) = W(n

i +n

j ) v 0 W(n

i ) 1

Trng s ca mt t m W( ni ) l s cc du m khc khng trong t m .

Gi s ni = 0 1 1 0 1 0 1 th trng s W(n

i ) bng s no di y. a

4Coi mi t m ni l mt vct n chiu trong mt khng gian tuyn tnh nchiu Vn, khi phpcng c thc hin gia hai t m tng t nh php cng gia hai vcttng ng c thc hin trn trng nh phn GF(2) Php cng theo modulo2 ny c m t nh sau: hnh cu 19 chng 4 Cho 7i = 0 1 1 0 1 0 1

(0, 1, 1, 0, 1, 0, 1) 7j = 1 0 0 1 1 1 0 (1, 0, 0, 1, 1, 1, 0) Khi 7 7 7

k i j bng gi tr no di y b (1 1, 1, 1, 0, 1, 1)Coi mi t m ni l mt vct n chiu trong mt khng gian tuyn tnh nchiu Vn, khi phpcng c thc hin gia hai t m tng t nh php cng gia hai vcttng ng c thc hin trn trng nh phn GF(2) Php cng theo modulo2 ny c m t nh sau: hnh cu 20 chng 4 Cho 8i =1 1 1 0 1 0 1 0

(1, 1, 1, 0, 1, 0, 1, 0) 7j = 1 0 0 1 1 1 0 1 (1, 0, 0, 1, 1, 1, 0, 1). Khi 8 8 8

k i j bng gi tr no di y c (0 1, 1, 1, 0, 1, 1, 1 )Chn cu ng v m tuyn tnh b M tuyn tnh di n l mm t m ca n c cc du m l cc dng tuyn tnhChn cu ng v m h thng tuyn tnh : b M h thng tuyntnh (n,k) l m tuyn tnh di n trong ta c th ch ra c v

tr ca k du thng tin trong t mChn cu ng v m tuyn tnh ngu nhin : b M tuyn tnh ngunhin l m tuyn tnh c cc du m c chn ngu nhin t ccdng tuyn tnh c th c m t m tuyn tnh, c th s dng ma trn sinh Gk,n Trong i s tuyntnh ta bit rng


16/30

vi mi Gk,n s tn ti ma trn Hrxn tha mn: G.HT= 0 Chn cu ng sau :a H c gi l ma trn kim tra ca m tuyn tnh (n,k) v

H cha r vc t hng trc giao vi cc vc t hng ca GGi s sau khi thc hin m ha, tin aic m ha thnh 1001, khi di t m nica tin

ny l : a 4Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,5 ) c thtnh c di trung bnh n Vi mi cch m ha khc nhau s tnh c n

khc nhau Trong cc kt qu m ha, gi tr n no di y c gi l ti u:

b n =1,51Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,4 ) c thtnh c di trung bnh n Vi mi cch m ha khc nhau s tnh c n


b n =1,45Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)=1,9 ) c th

tnh c di trung bnh n Vi mi cch m ha khc nhau s tnh c n khc nhau Trong cc kt qu m ha, gi tr n no di y c gi l ti u:

c n =1,905Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)= 1,905) c thtnh c di trung bnh n Vi mi cch m ha khc nhau s tnh c n


a n =1,906Sau khi thc hin m ha ngun ri rc A ( c entropy l H(A)= 2,01 bt) cth tnh c di trung bnh n Vi mi cch m ha khc nhau s tnh

c n khc nhau Trong cc kt qu m ha, gi tr n no di y c gil ti u: d n =202Cho m Cyclic C(n, k) = C(7,4) c a thc sinh l g(x) th d T g(x) cth xc nh c ma trn sinh h thng G v ma trn kim tra Hcho b mCho m Cyclic C(n, k) = C(7,4), c a thc sinh g(x) l c g(x) = 1 +x2 +x3

Cho m Cyclic C(n, k) = C(7,4), c a thc sinh g(x) l b g(x) = 1 +x+x3

Cho m Cyclic C(n, k) = C(7,3) , tm cu sai v g(x) a g(x)khng l c cax7 + 1Cho m Cyclic C(n, k) = C(7,3), c a thc sinh g(x) l b g(x) = 1 +x+x2 +x4

Cho m Cyclic C(n, k) = C(7,3), c a thc sinh g(x) l d g(x) = 1 +x+x2 +x4

Pht biu sau ng hay sai : Cho m Cyclic C(n, k) = C(7,3) , g(x) l a thcbc 4 b ngCho m tuyn tnh (7,4) S nh cn thit nh b m l b 28


17/30

Cho m Xyclic (7,4) S cc nh cn thit nh b m l a 7Cho m tuyn tnh (7,3) S nh cn thit nh b m l b 21Cho m Xyclic (7,3) S cc nh cn thit nh b m l d 7Cho m tuyn tnh (n,k) S nh cn thit nh b m l b knCho m Xyclic (n,k) S cc nh cn thit nh b m l d n7

Chn cu ng sau: b Cc dng tuyn tnh ca k bin c lpx1,x2,,xkl cc biu thc c dng: 1

1

( ,..., ) ( )k

k i i

i

f x x a x

Trong :

ia F , F l mt trngChn cu sai v m tuyn tnh : c Trong i s tuyn tnh ta bitrng vi mi G s tn ti ma trn Hrn tha mn m t m tuyn tnh, c th s dng ma trn sinh Gk,nTrong i s tuyntnh ta bit rng vi mi Gk,ns tn ti ma trn Hrn tha mn: G.HT = 0Chncu sai sau : b HTc gi l ma trn k hng, n ctKhi xy dng mt m tuyn tnh (n,k,d0)ngi ta phi tm c cc m c

tha nh nhngli c kh nng khng ch sai ln. Ngi ta thng xy dng m ny datrn cc bi ton ti u tm cu sai trong cc cu di y: d Vi n vs sai khi sa t xc nh, ta phi tm c m c s du thng tin kl ln nht (hay s du tha l nh nht) Tng ng vi bi ton

ny ta c gii hn Hamming sau:0

2t

n k i

n

i

C

Chn nh ngha sai v m xyclic trong cc nh ngha sau b M xyclic(n, k) l mt b m m a thc sinh c bc r = n+kChn cu ng ca nh l v kh nng sa sai c M u nh phn c tha (D > 0) vi khong cch Hamming d0 3c kh nng sa

c t sai tho mn iu kin: 0 12

dt v c kh nng pht hin t

sai tho mn iu kintd01T nh l m ho th 1 ca Shannon i vi m nh phn ta c :

1

1

( ) ( )s

i i

i

n p a n H A

T biu thc trn tm cu ng nht trong cc biu thc

sau : c1 1

( ) ( ) log ( )s s

i i i i

i i

p a n p a p a

Theo nh l m ho th 1 ca Shannon i vi m nh phn ta c:

1

1

( ) ( )

s

i i

i

n p a n H A

= 1 ( ) log ( )s

i i

i

p a p a

T biu thc ny suy ra di t m ni

v xc sut p(ai) lin h vi nhau:1

log (*)( )

i

i

np a

T (*) tm cu ng sau v

nguyn tc lp m tit kim: c Cc tin c xc sut xut hin lnc m ha bng cc t m c di nh v ngc li cc tin cxc sut xut hin nh c m ha bng cc t m c di ln


18/30

Cho ngun tin 1 2 3 4 5, , , ,X x x x x x vi cc xc sut ln lt l {1/2, 1/4, 1/8,1/16, 1/16} Bit x1c m ha thnh 0, x2c m ha thnh 10, x3cm ha thnh 110, x4c m ha thnh 1110, x5c m ha thnh 1111B m ti u cho ngun trn c chiu di trung bnh tnh theo cng thc :

5

1( )i i

in p x n

l : c 1,875Yu cu ca php m ha: nhng t m c di nh hn khng trng viphn u ca t m c di ln hn Cc tin c xc sut xut hin ln hnc m ha bng cc t m c di nh v ngc li. Cho ngun tin

1 2 3 4 5, , , ,X x x x x x vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16} Bitx1c m ha thnh 0, x2c m ha thnh 10 . Chn cu ng di y m ha cho x3 c 110Yu cu ca php m ha: nhng t m c di nh hn khng trng viphn u ca t m c di ln hn Cc tin c xc sut xut hin ln hnc m ha bng cc t m c di nh v ngc li. Cho ngun tin

1 2 3 4 5, , , ,X x x x x x vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16} Bitx1c m ha thnh 0, x2c m ha thnh 11, x3 c m ha thnh 100Chn cu ng di y m ha cho x4 d 1010Yu cu ca php m ha: nhng t m c di nh hn khng trng viphn u ca t m c di ln hn Cc tin c xc sut xut hin ln hnc m ha bng cc t m c di nh v ngc li. Cho ngun tin

1 2 3 4 5, , , ,X x x x x x vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16} Bitx1c m ha thnh 0, x2c m ha thnh 10, x3 c m ha thnh110, x4 c m ha thnh 1110. Chn cu ng di y m ha cho x5

b 1111

Cho m Cyclic C(7, 4) c a thc sinh l g(x) =1 +x+x3v a thc sinh G

sau :

3

2 4

2 3 5

3 4 6

1 x x

x x xG

x x x

x x x

Ma trn no sau y l mt ma trn sinh G ng vi m

Cyclic C(7,4) trn a

1101000

0110100

0011010

0001101


19/30

Cho m Cyclic C(7, 4) c a thc sinh l g(x) = 1 +x2 +x3v a thc sinh G

sau

2 3

3 4

2 4 5

3 5 6

1 x x

x x xG

x x x

x x x

Ma trn no sau y l mt ma trn sinh G ng vi m

Cyclic C(7,4) trn b

1011000

0101100

0010110

0001011

Cho m Cyclic C(7, 4) c a thc sinh l : g(x) = 1 +x+x3v ma trn sinh

G. T ma trn sinh G tnh c ma trn kim tra H bn:

2 3 4

3 4 5

2 4 5 6

1 x x x

H x x x x

x x x x

Chuyn ma trn kim tra H sang dng khng gian tuyn tnh. Ma trn nosau y l mt ma trn kim tra H ( dng khng gian tuyn tnh) a1011100

0101110

0010111

Cho m Cyclic C(7, 4) c a thc sinh l : (x) = 1 +x2 +x3 v ma trn sinh GT ma trn sinh G

tnh c ma trn kim tra H bn :

2 4

2 3 5

2 3 4 6

1 x x x

H x x x x x x x x

Chuyn ma trn kim

tra H sang dng khng gian tuyn tnh. Ma trn no sau y l mt ma trn

kim tra H ( dng khng gian tuyn tnh) a

1110100

0111010

0011101

Tin ri rc ai A; RA: T m fi (x)tng ng vi ai Thut ton xy dng t mxyclic gmtheo 4 bc: M,N,P,Q c sp xp ngu nhin M: M t tin ai trong tp tin

cn m ha (gm 2

k

tin) bng mt a thc ai (X)vi deg ai (X) k

1. N: Chiaai (X).xn-k cho a thc sinh g(X) tm phn d ri (X). P: Nng bc ai (X) bngcch nhn n vi xn-k. . Q : Xy dng t m xyclic: ( ) ( ). ( )n ki i if X a X x r X

.Chn cc sp xp th t ng thut ton xy dng t m xyclic: a

M-P-N-Q


20/30

Gi s sau khi thc hin m ha ngun ri rc A. Ta c kt qu m ho sau :A1 A4 A5 A2 A3

00 10 110 1111 1110

Gii m cho dy bt nhn c c dng

11100011010111100 s ckt qu no sau y: b A3-A1-A5-A4-A2-A1Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :

A1 A4 A5 A2 A3

00 10 110 1111 1110


11110011010111100 s c. kt qu no sau y: d A2-A1-A5-A4-A2-A1Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :

A1 A4 A5 A2 A3

00 10 110 1111 1110


1111001101011110010 s c kt qu no sau y: c A2-A1-

A5-A4-A2-A1-A4Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :A1 A4 A5 A2 A3

00 10 110 1111 1110


11110011010111100110 s c kt qu no sau y: b A2-A1-A5-A4-A2-A1-A5Gi s sau khi thc hin m ha ngun ri rc A . Ta c kt qu m ho sau :

A1 A4 A5 A2 A3

00 10 110 1111 1110


1111001101011110011000 s c kt qu no sau y: b A2-A1-

A5-A4-A2-A1-A5-A1Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 66 chng 4

Tnh di m trung bnh theo biu thc5

1

( )i ii

n n p a

ta c gi tr no :c 2,1

Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 67 chng 4

Tnh di m trung bnh theo biu thc

5

1( )i i

in n p a

ta c gi tr no :

a 1,8Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh cc


21/30

m nh phn c di m nitng ng nh bng sau: hnh cu 68 chng 4


1

( )i ii

n n p a

ta c gi tr no :d 2,75

Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )c

m ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 69 chng 4


1

( )i ii

n n p a

ta c gi tr no :d 2,45 du m

Gi s sau khi thc hin m ha, cc tin aivi xc sut tng ng P(ai )cm ha thnh ccm nh phn c di m nitng ng nh bng sau: hnh cu 70 chng 4


1

( )i ii

n n p a

ta c gi tr no :

a 2,5Mt dy tin 1 2, , ... n X x x x vi ; 1i x X i n Lng tin I(X) cha trong dy tin

X s l:1 2

1 1 1( ) log log ... log

( ) ( ) ( )nI X

p x p x p x Gi s cho ngun

1 2 3 4 5, , , ,X x x x x x vi cc xc sut ln lt l {1/2, 1/4, 1/8, 1/16, 1/16}.

Lng tin I(X) cha trong dy tin 1 2 1 1 3 4 1 1 5X x x x x x x x x x l: c 18 btKhi xy dng mt m tuyn tnh (n,k,d0 ) ngi ta phi tm c cc m c tha nh nhng li c kh nng khng ch sai ln Ngi ta thng xydng m ny da trn cc bi ton ti u Tm cu sai trong cc cu di y:

a Vi k v d0 xc nh, ta phi tm c m c di vi tm l ln nht. Tng ng vi bi ton ny ta c gii hn n = kChn cu ng ca nh l v kh nng pht hin sai sau c M unh phn c tha (D > 0) vi khong cch Hamming d0>1c khnng pht hin t sai tho mn iu kintd0-1kh nng sa c t

sai tho mn iu kin: 01

2

dt

Thc hin m ha Shannon - Fano ngun ri rc A sau :1 1

1( ) 1/ 2

a aA

p a

21/ 4

a

3

1/ 8

a

4

1/16

a

5

1/16

a s c kt qu no sau y: d

1a

0

2

10

a

3

110

a

4

1110

a

5

1111

a


1( ) 1/16

a aA

p a

21/16

a

3

1/ 8

a 41/ 4

a

5

1/ 2

a

s c kt qu no sau y: d1a

1111

21110

a 3

110

a 4

10

a

5

0

a


22/30


1( ) 1/ 4

a aA

p a

21/ 2

a

3

1/ 8

a 41/16

a

5

1/16

a

s c kt qu no sau y: a1a

10

20

a 3

110

a 4

1110

a

5

1111

a


1( ) 1/ 8

a aA

p a

21/ 4

a

3

1/ 2

a 41/16

a

5

1/16

a

s c kt qu no sau y: b1a

110

210

a 3

0

a 4

1110

a

5

1111

a


1( ) 1/16

a aA

p a

21/ 4

a

3

1/ 8

a 41/ 2

a

5

1/16

a


1110

210

a 3

110

a 4

0

a

5

1111

a


1( ) 1/ 2

a aA

p a

21/ 4

a

3

1/16

a 41/ 8

a

5

1/16

a


0

210

a 3

1110

a 4

110

a

5

1111

a


1( ) 1/ 2

a aA

p a

21/ 8

a

3

1/ 4

a 4

1/16

a

5

1/16

a


0

2

11

a 3

01

a 4

111

a

5

1111

a


1( ) 1/ 2

a aA

p a

21/16

a

3

1/ 8

a 41/ 4

a

5

1/16

a

s c kt qu no sau y: c1a

0

21110

a 3

110

a 4

10

a

5

1111

a


1( ) 1/ 2

a aA

p a

21/ 4

a

3

1/16

a 4

1/16

a

5

1/ 8

a


0

2

10

a 3

1111

a 4

1110

a

5

110

a


1( ) 1/16

a aA

p a

21/ 4

a

3

1/ 8

a 41/16

a

5

1/ 2

a


1111

210

a 3

110

a 4

1110

a

5

0

a


23/30

Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :

1 1

1( ) 1/ 2

a aA

p a

2

1/ 4

a 3

1/ 8

a 4

1/16

a 5

1/16

a

Ta c kt qu m ho sau :1a

0

210

a 3

110

a 4

1110

a

5

1111

a di t m trung bnh

n

v entropy H(A) c tnh theo biu thc5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

s c kt qu no sau y : b n=H(A)=1,875


1( ) 0,5

a aA

p a

2

0,25

a

3

0,125

a 4

0,625

a 5

0,625

a


0

210

a 3

110

a 4

1110

a

5

1111

a


1( ) 0,25

a a

A p a

2

0,5

a

3

0,125

a 4

0,625

a 5

0,625

a


10

20

a 3

110

a 4

1110

a

5

1111

a


1( ) 0,125

a aA

p a

2

0,25

a 3

0,5

a 4

0,625

a 5

0,625

a


110

210

a 3

0

a 4

1110

a

5

1111

a


1( ) 0,625

a aA

p a

2

0,25

a3

0,125

a4

0,5

a 5

0,625

a


1110

210

a 3

110

a 4

0

a

5

1111

a


1( ) 0,625

a a

A p a

2

0,25

a3

0,125

a4

0,625

a 5

0

a


1111

210

a 3

110

a 4

1110

a

5

0

a


24/30


1( ) 0,5

a aA

p a

2

0,125

a

3

0,25

a 4

0,625

a 5

0,625

a


0

2110

a 3

10

a 4

1110

a

5

1111

a


1( ) 0,5

a aA

p a

2

0,625

a

3

0,125

a 4

0,25

a 5

0,625

a


0

21110

a 3

110

a 4

10

a

5

1111

a


1( ) 0,5

a aA

p a

2

0,625

a

3

0,125

a 4

0,625

a 5

0,25

a


0

21111

a 3110

a 4

1110

a

5

10

a


1( ) 0,5

a aA

p a

2

0,25

a

3

0,625

a4

0,125

a 5

0,625

a


0

210

a 3

1110

a 4

110

a

5

1111

a


1 1

1( ) 0,5

a aA

p a

2

0,25

a3

0,125

a4

0,625

a 5

0,625

a


0

210

a 3

110

a

4

1111

a

5

1110

a

di t m trung bnh n v entropy H(A) c tnh theo biu

thc

5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : an =H(A)=1,875


1 1

1( ) 0,25

a aA

p a

2

0,125

a 3

0,625

a 4

0,625

a 5

0,25

a 6

0,125

a 7

0,625

a 8

0,625

a

Ta c kt qu m ho

sau :1a

00

5

01

a

2

100

a

6

101

a

3

1100

a

4

1101

a

7

1110

a

8

1111

a di t m trung bnh

n v

entropy H(A) c tnh theo biu thc8

1

( )i ii

n p a n

v8

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : a n =H(A)=2,75Gi s sau khi thc hin m ha Shannon - Fano ngun ri rc A :

1 1

1( ) 1/ 4

a aA

p a

2

1/ 8

a 3

1/16

a 4

1/16

a 5

1/ 4

a 6

1/ 8

a 7

1/16

a 8

1/16

a

Ta c kt qu m ho sau : 1a

00


25/30

5

01

a 2

100

a 6

101

a 3

1100

a 4

1101

a 7

1110

a

8

1111

a

di t m trung bnh n v entropy

H(A) c tnh theo biu thc8

1

( )i ii

n p a n

v8

1

( ) ( ) log ( )i ii

H A p a p a

c kt

qu no sau y :b n =H(A)=2,75

Gi s sau khi thc hin m ha ngun ri rc A :1 1

1( ) 0,5

a aA

p a

4

0,25

a 5

0,125

a

2

0,625

a 3

0,625

a


0

410

a 5

110

a 2

1111

a

3

11100

a

di t

m trung bnh n v entropy H(A) c tnh theo biu thc

5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : d n =1,9375 vH(A)=1,875


1( ) 1/ 2

a aA

p a

4

1/ 4

a 5

1/ 8

a 2

1/16

a

3

1/16

a


0

410

a 5

110

a 2

1111

a

3

11100

a

di t m trung

bnh n v entropy H(A) c tnh theo biu thc5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : c n =1,9375 vH(A)=1,875


1( ) 1/ 2

a aA

p a

4

1/ 4

a 5

1/ 8

a 2

1/16

a

3

1/16

a


00

410

a 5

110

a 2

1111

a

3

1100

a

di t m trung


1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : b

n=2,375 v

H(A)=1,875


1( ) 1/ 2

a aA

p a

4

1/ 4

a 5

1/ 8

a 2

1/16

a

3

1/16

a


0

410

a 5

1100

a 2

1111

a

3

1100

a

di t m trung


26/30


1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : c n =2 v H(A)=1,875

Gi s sau khi thc hin m ha ngun ri rc A : 1 11( ) 0,5

a aAp a

4

0,25a 5

0,125a

2

0,625

a 3

0,625

a


0

410

a 5

1100

a 2

1111

a

3

1100

a

di t


5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : d n =2 v H(A)=1,875


1( ) 0,5

a a

A p a

4

0,25

a5

0,125

a

2

0,625

a 3

0,625

a


00

410

a 5

110

a 2

1111

a

3

1110

a

di t m

trung bnh n v entropy H(A) c tnh theo biu thc

5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : c n =2,375 vH(A)=1,875

Gi s sau khi thc hin m ha ngun ri rc A : 1 11( ) 0,5

a aA

p a 4

0,25

a5

0,125

a

2

0,625

a 3

0,625

a


00

410

a 5

1100

a 2

1111

a

3

1110

a

di t


5

1

( )i ii

n p a n

v5

1

( ) ( ) log ( )i ii

H A p a p a

c kt qu no sau y : a n =2,5 vH(A)=1,875

Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3

tng ng a thcthng tin a(x) = x + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : a 1100101Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + x2 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y: c 0011010


27/30

Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x + x2 S dng thut ton 4 bc thit lp t m h thngta c kt qu no di y: c 1000110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + x3 S dng thut ton 4 bc thit lp t m h

thng ta c kt qu no di y : d 0111001Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x2 + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 0100011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + x2 + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 1001011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = 1 + xS dng thut ton 4 bc thit lp t m h thngta c kt qu no di y: c 1011100Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thc

thng tin a(x) = 1 + x

+ x2

S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y: b 0101110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x + x2 + x3 S dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 0010111Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x3 S dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : c 1010001Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3 tng ng a thcthng tin a(x) = x S dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : b 0110100

Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+x+x3

tng ng a thcthng tin a(x) = x2 S dng thut ton 4 bc thit lp t m h thng tac kt qu no di y: a 1110010Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x3 s dng thut ton 4 bc thit lp t m h thngta c kt qu no di y : c 1000101Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x2 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : d 0111010Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x2 s dng thut ton 4 bc thit lp t m h thng

ta c kt qu no di y : d 0010110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x3 s dng thut ton 4 bc thit lp t m h thngta c kt qu no di y : a 1101001Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x2 + x3 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : a 1010011


28/30

Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x2 + x3 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : a 0001011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x s dng thut ton 4 bc thit lp t m h thng

ta c kt qu no di y : c 0101100 Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = 1 + x + x2 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : b 1001110Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x + x2 + x3 s dng thut ton 4 bc thit lp t m hthng ta c kt qu no di y : b 0100111Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x3 s dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : b 0111011Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thc

thng tin a(x) = x s dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : a 1110100Cho m Cyclic C(7,4) c a thc sinh l g(x) =1+ x2+x3 tng ng a thcthng tin a(x) = x2 s dng thut ton 4 bc thit lp t m h thng tac kt qu no di y : c 1110110Thu tn hiu khi c nhiu l mt bi ton thng k: a ngNhim v ca my thu l phi chn li gii, do my thu cn c gi l s gii Yu cu ca s gii l phi tm li gii ng (pht i ta phi tmc i ) Trong thc t c rt nhiu s gii Trong tt c cc s gii cth c th ti mt s bo m xc sut nhn ln phi ng l ln nht(xc sut gii sai l b nht) S ny c gi l s gii ti u. Khi

ngi ta gi My thu xy dng theo s gii ti u c gi l my thu tiu Kt lun ny ng hay sai ? b ng

/

( )( )

( )

il i

l

pu

p

vi

1,

1

i m

i

l quy tc gii ti u vit di dng hm hp l.

Chn cung trong cc cu sau: b Nu mi tn hiu gi i u ng xcsut th / ( ) 1l i u vi i l nh ngha b lc phi hp tuyn tnh th ng : i vi mt tn hiu xc

nh, mt mch tuyn tnh th ng m bo t s rara

N

S

cc i mt

thi im quan st no y s c gi l mch lc phi hp tuyn tnh thng ca tn hiu Trong ra l t s gia cng sut trung bnh ca nhiu u ra b lc y v cng sut nh ca tn hiu nh ngha ny ng haysai ? a SaiQu trnh x l tn hiu trong my thu ti u c gi l x l ti u tn hiuX l nhn li gii c xc sut sai b nht Chn cu ng nht v x l tiu cc tn hiu: a Da vo cc tiu chun ti u, bng cng c


29/30

thng k ton hc ngi ta xc nh c quy tc gii ti u, do ngy nay l thuyt truyn tin cho php bng ton hc tng hpc my thu ti u

Gi s i l tn hiu gi i, c xc sut p( i ))- c gi l xc sut tinnghim my thu ta nhn c u(t), t u(t)qua s gii ta s c li gii 1

no Nh vy l c gi i vi mt xc sut p( /l u )- c gi l xcsut hu nghim. Do xc sut gii sai s l:

p(sai/u, l ) =1p( /l u )Xt hai s gii: - T u(t)cho ta 1 - gi l s (1) - Tu(t)cho ta 2 - gi l s (2). Nu p(sai/u, 1 ) < p(sai/u, 2 ) ta rtra kt lun s (2) ti u hn (1) Khng nh ny ng hay sai ? a

SaiTn hiu tng qut c dng: 0 0( ) ( ) cos( ( ) )i iC t C t t t Chn cu sai trong cccu sau: a Vic x l ti u tn hiu khng ph thuc ng

bao 0 ( )iC t v tn s tc thi( )

( )id t

t dt

Cho u vo mch tuyn tnh th ng mt dao ng c dng:( ) ( ) ( )iy t C t n t Trong ( )iC t l th hin ca tn hiu pht i (cn c gi l

tn hiu ti) n(t) l nhiu cng, trng, chun bi ton tng hp mch l tmbiu thc gii tch ca hm truyn phc Ki()ca mch tuyn tnh th ngsao cho mt thi im quan st (dao ng nhn c) no ra tmax, p dng cng thc bin i ngc Fourier tnh c

2

0

1(2 )ra ivS f df

N

Trong ( )ivS l mt ph (bin) phc ca th hin

tn hiu u vo mch tuyn tnh Theo nh l Parseval, ta c: max0

(*)iraEN

trong 2(2 )i iv E S f df

l nng lng ca tn hiu ti d T (*) chng

t t s rara

S

N

hon ton ph thuc vo dng ca n ca tn hiu m

hon ton khng ph thuc vo dng ca n.Cho knh nh phn, i xng, khng nh c nhiu cng, trng, chun theom hnh sau: hnh cu 9 chng 5 Xc sut sai ton phn sp ( xc sut sai

khng iu kin) l 1 2 1 2 1 2( ). ( / ) ( ). ( / )s p p p p p Gi s tnh c

04 2

2 1

0

1( / ) exp

2 22

P T

NP T

p dG

khi sp s bng biu thc no

di y: d0

12

s

P Tp

G


30/30

Ti u vo b lc tuyn tnh tc ng tn hiu: x(t) = s(t) + n(t) Trong n(t) l tp m trng, chun, dng Cn s(t) l xung th tn c lp vi n(t) v

c dng:( ).

( )0

A t T A es t

t T

t T

Hm truyn *0 ( ) ( )

j T K j kS j e Hm truyn

ca b lc (

*

0 ( ) ( )

j T

K j kS j e

) sao cho t s tn trn tp u ra ca b lct cc i s l biu thc no di y b 0 ( )

kAK j

A j

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