73172633 giao trinh ly thuyet thong tin

7/28/2019 73172633 Giao Trinh Ly Thuyet Thong Tin

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I HC THI NGUYNKHOA CNG NGH THNG TIN

V Vinh Quang Ch bin

Nguyn nh Dng, Nguyn Hin Trinh, Dng Th Mai Thng

GIO TRNH

L THUYT THNG TIN

THI NGUYN NM 2010


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LI MU

Gio trnh l thuyt thng tin c bin son da trn cc bi ging c ging dy nhiu nm cho i tng l sinh vin chnh quy ngnh Cngngh thng tin ti khoa Cng ngh thng tin i hc Thi Nguyn cng vivic tham kho mt s gio trnh ca cc trng i hc khc cng nh ccti liu nc ngoi. c gio trnh ny, ngi c cn phi c trang by cc kin thc v ton cao cp, xc sut thng k, l thuyt thut ton vmt ngn ng lp trnh cbn (C hoc Pascal).

Gio trnh c cu trc gm 5 chngChng 1 trnh by mt s khi nim cbn v l thuyt thng tin nh

cu trc ca h thng truyn tin, phn loi mi trng truyn tin, vn rirc ha cc ngun tin lin tc v cc khi nim viu ch v gii iu ch.

Chng 2 a ra cc khi nim cbn v tn hiu v cc cch phn tchph cho tn hiu, khi nim v nhiu trong qu trnh truyn tin.

Chng 3 trnh by cc khi nim cbn vo thng tin, lng tin,entropi v mi quan h gia lng tin v entropi, cc cng thc xc nhlng tin v entropi da trn csca l thuyt xc sut, khi nim v tc

lp tin v thng lng knh trong qu trnh truyn tin.Chng 4 gii thiu cc khi nim chung v m ha, iu kin thit lp,cc phng php biu din, cc thut ton m ha cbn, khi nim v mchng nhiu v m tuyn tnh.

Chng 5 ca gio trnh gii thiu v mt s h mt m ni ting trn thgii ngi c tham kho.

Trong qu trnh son tho gio trnh chc chn khng trnh khi nhngthiu xt v ni dung cng nh hnh thc, nhm bin son trn trng cm nnhng kin qu bu ca cc bn c gio trnh c hon thin hn.

Thi Nguyn, thng 01 nm 2010Thay mt nhm bin son

V Vinh Quang


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CHNG 1. NHNG KHI NIM CBN

1.1 Gii thiu v l thuyt thng tin

Trong th gii ngy nay, chng ta hng ngy phi tip xc vi rt nhiucc h thng chuyn ti thng tin khc nhau nh: Cc h thng truyn hnhpht thanh, h thng in thoi c nh v di ng, h thng mng LAN,Internet, cc h thng ny u vi mc ch l chuyn thng tin t ni phtn ni thu vi nhng mc ch khc nhau. nghin cu v cc h thngny, chng ta cn phi nghin cu v bn cht thng tin, bn cht ca qutrnh truyn tin theo quan im ton hc, cu trc vt l ca mi trng truyn

tin v cc vn lin quan n tnh cht bo mt, ti u ha qu trnh.Khi nim u tin cn nghin cu l thng tin: thng tin c hiu l

tp hp cc tri thc m con ngi thu c qua cc con ng tip nhn khcnhau, thng tin c mang di dng nng lng khc nhau gi l vt mang,vt mang c cha thng tin gi l tn hiu.

L thuyt v nng lng gii quyt tt vn xy dng mch, tn hiu.Nhng vn v tc , hin tng nhiu, mi lin h gia cc dng nnglng khc nhau ca thng tin cha gii quyt c m phi cn c mt lthuyt khc l l thuyt thng tin.

L thuyt thng tin l l thuyt nhm gii quyt vn cbn ca qu

trnh truyn tin nh vn v ri rc ha ngun, m hnh phn phi xc sutca ngun v ch, cc vn v m ha v gii m, kh nng chng nhiuca h thng...

Cn ch rng l thuyt thng tin khng i su vo vic phn tch cccu trc vt l ca h thng truyn tin m ch yu nghin cu v cc m hnhton hc m t qu trnh truyn tin trn quan im ca l thuyt xc sut thngk, ng thi nghin cu v cc nguyn tc v cc thut ton m ha cbn,cc nguyn tc m chng nhiu...

1.2 H thng truyn tin

Trong thc t, chng ta gp rt nhiu cc h thng truyn thng tin tim ny ti im khc, trong thc t nhng h thng truyn tin c th m con


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ngi s dng v khai thc c rt nhiu dng, khi phn loi chng ngi ta

c th da trn nhiu cskhc nhau.

1.2.1 Cc quan im phn loi cc h thng truyn tin

Theo nng lng- Nng lng mt chiu (in tn)- V tuyn in (sng in t)- Quang nng (cp quang)- Sng siu m (la-de)

Theo biu hin bn ngoi- H thng truyn s liu

- H thng truyn hnh pht thanh- H thng thng tin thoi

Theo dng tn hiu- H thng truyn tin ri rc- H thng truyn tin lin tc

Xut pht t cc quan im , trong thc t trong nhiu lnh vc cbit l lnh vc truyn thng tn ti cc khi nim nh: H pht thanh truynhnh, h truyn tn hiu s, ...

1.2.2 S truyn tin v mt skhi nim trong h thng truyn tin

nh ngha: Truyn tin(transmission): L qu trnh dch chuyn thng tin tim ny sang im khc trong mt mi trng xc nh. Hai im ny sc gi l im ngun tin (information source) v im nhn tin (informationdestination). Mi trng truyn tin cn c gi l knh tin (chanel).

S khi chc nng ca mt h thng truyn tin tng qut gm c 3thnh phn chnh: Ngun tin, knh tin v nhn tin.

Trong :

NGUN TIN KNH TIN NHN TIN


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Ngun tin:l ni sn sinh ra hay cha cc tin cn truyn i,haynguntin l tp hp cc tin m h thng truyn tin dng to cc bn tin khc nhau truyn tin.

Knh tin: l mi trng lan truyn thng tin. c th lan truyn c thng tin trong mt mi trng vt l xc

nh, thng tin phi c chuyn thnh tn hiu thch hp vi mi trngtruyn lan. Nh vy ta c thnh ngha knh tin:

Knh tin l ni hnh thnh v truyn tn hiu mang tin ng thi ysinh ra cc tp nhiu ph hu thng tin.

Trong l thuyt truyn tin, knh l mt khi nim tru tng i din cho

s hn hp gia tn hiu v tp nhiu. T khi nim ny, s phn loi knh sd dng hn, mc d trong thc t cc knh tin c rt nhiu dng khc nhau.

V d:- Truyn tin theo dy song hnh, cp ng trc.- Tn hiu truyn lan qua cc tng in ly.- Tn hiu truyn lan qua cc tng i lu.- Tn hiu truyn lan trn mt t, trong t.- Tn hiu truyn lan trong nc..

Nhn tin: L ccu khi phc thng tin ban u t tn hiu thu ctu ra ca knh

tm hiu chi tit hn ta i su vo cc khi chc nng ca s truyn tin v xt n nhim v ca tng khi.

1.3 Ngun tin nguyn thu

1.3.1 Khi nim chung

nh ngha: Ngun tin nguyn thu l tp hp nhng tin ban u m h thngthu nhn c cha qua mt php bin i nhn to no.

V mt ton hc, cc tin nguyn thu l nhng hm lin tc theo thi

gian )(tf hoc l nhng hm bin i theo thi gian v mt s thng s khc

nh hnh nh en trng ),,( tyxh trong yx, l cc to khng gian, hoc

nh cc thng tin kh tng ),( tg i trong i l cc thng s kh tng nh

nhit , m, tc gi,..


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Thng tin nguyn thu cng c th l cc h hm theo thi gian v cc

thng s nh trng hp thng tin hnh nh mu:

( , , )

( , , ) ( , , ) .

( , , )

f x y z

K x y z g x y z

h x y z

=

Cc tin nguyn thu phn ln l hm lin tc ca thi gian trong mtngng ngha l c th biu din mt thng tin no di dng mt hm

( )S t tn ti trong qung thi gian T v ly cc gi tr bt k trong phm vi

( maxmin , SS ) trong min max,S S l ngng nh nht v ln nht m h thng

c th thu nhn c.

Smax

Smin

Tin nguyn thu c th trc tip a vo h thng truyn tin nhng cn

phi qua cc php bin i sao cho ph hp vi h thng tng ng. Nh vyxt v quan im truyn tin th c hai loi tin v hai loi h thng tng ng:

Tin ri rc ng vi- Ngun ri rc- Knh ri rc

Tin lin tc ng vi- Ngun lin tc- Knh lin tc

S phn bit v bn cht ca ngun ri rc vi ngun lin tc c hiul s lng cc tin trong ngun ri rc l hu hn v s lng cc tin trongngun lin tc l khng m c.


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Ni chung cc tin ri rc, hoc nguyn thu ri rc, hoc nguyn thu

lin tc c ri rc ho trc khi a vo knh thng thng u qua thitb m ho. Thit b m ho bin i tp tin nguyn thu thnh tp hp nhngtin thch hp vi c im c bn ca knh nh kh nng cho qua (thnglng), tnh cht tn hiu v tp nhiu.

1.3.2 Bn cht ca thng tin theo quan im truyn tin

Ch c qu trnh ngu nhin mi to ra thng tin. Mt hm gi l ngunhin nu vi mt gi tr bt k ca i s, gi tr ca hm l mt i lngngu nhin (cc i lng vt l trong thin nhin nh nhit mi trng, psut khng kh l hm ngu nhin ca thi gian).

Mt qu trnh ngu nhin c quan st bng mt tp cc gi tr ngunhin. Qu trnh ngu nhin c coi l bit r khi thu nhn v x lc mttp nhiu cc gi trc trng ca n.

Gi s qu trnh ngu nhin X(t) c mt tp cc gi tr mu (hay cn

c gi l cc bin) ( )x t , khi ta biu din qu trnh ngu nhin bi:

{ }( ) ( )x X

X t x t

=

V d: Quan st thi gian vo mng ca cc sinh vin trong 1 ngy,

ngi ta tin hnh phng vn 10 sinh vin, gi X l thi gian vo mng,k

x

l thi gian vo mng ca sinh vin th , ( 1, 2,...,10)k k= ta thu c mu

nh sau:

{ }10, 50, 20,150,180, 30, 30, 5, 60, 0X = n v tnh (pht)

Vic on trc mt gi tr ngu nhin l kh khn. Ta ch c th tmc quy lut phn b ca cc bin thng qua vic p dng cc qui lut caton thng k x l cc gi tr ca cc bin ngu nhin m ta thu c tcc tn hiu.

Qu trnh ngu nhin c th l cc hm trong khng gian 1 chiu, khi ta c quy lut phn phi xc sut 1 chiu v hm mt phn phi xc sutc xc nh bi cc cng thc

( )( ) ( ); ( )

dF xF x p X x w x

dx= < =


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Trong :

x l bin ngu nhinp(x) xc sut xut hin X x= trong qu trnh ngu nhin, thng

c vit l ( ) ( )p x p X x= = .

Nu qu trnh ngu nhin l cc hm trong khng gian 2 chiu khi quy lut ngu nhin c biu hin bi cc cng thc

2

( , ) ( ; ); ( , ) .xyF

F x y p X x Y y w x yx y

= < < =

Tng t, ta cng c cc quy lut phn phi xc sut trong khng giannhiu chiu.

Cc c trng quan trng ca bin ngu nhin

1. Tr trung bnh (k vng ton hc) ca mt qu trnh ngu nhin ( )X t ( ) ( ) ( ) ( )E X X t x t w x dx

+

= =

2. Tr trung bnh bnh phng

2 2 2( ) ( ) ( ) ( )E X X t x t w x dx+

= =

3. Phng sai( ) ( )

2 2( ) ( ) ( ) ( )D X X X x t E x w x dx

+

= =

4. Hm tng quanM t mi quan h thng k gia cc gi tr ca 1 qu trnh ngu nhin

cc thi im t1, t2

( )1 2 1 2 1 2 1 2 1 2( , ) ( ), ( ) ( ( ), ( ))xB t t E X t X t x x w x t x t dx dx+ +

= =

Nu hai qu trnh ,X Y khc nhau hai thi im khc nhau, khi

( )1 2 1 2 1 2( , ) ( ), ( ) ( ( ), ( ))xyB t t E X t Y t xyw x t y t dxdy+ +

= =


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gii quyt bi ton mt cch thc t, ngi ta khng th xc nh tc

thi m thng ly tr trung bnh ca qu trnh ngu nhin. C hai loi trtrung bnh theo tp hp v tr trung bnh theo thi gian. Ta cn nghin cu trtrung bnh theo tp hp, tuy vy s gp nhiu kh khn khi tip nhn v x ltc thi cc bin ngu nhin v cc bin ngu nhin lun bin i theo thigian. tnh tr trung bnh theo thi gian, ta chn thi gian ln quan stcc bin ngu nhin d rng hn v c iu kin quan st v s dng cc cngthc thng k, khi vic tnh cc gi tr trung bnh theo thi gian c xcnh bi cc cng thc:

0

1( ) ( )

T

T

X t Lim x t dt

T+=

Tr trung bnh bnh phng:

2 2

0

1( ) ( )

T

TX t Lim x t dt

T+=

Khi thi gian quan st T dn n v cng th tr trung bnh tp hp bngtr trung bnh thi gian. Trong thc t ta thng chn thi gian quan st lnch khng phi v cng nh vy vn tho mn cc iu kin cn nhng ngin hn, khi ta c tr trung bnh theo tp hp bng tr trung bnh theo thigian. Ta c:

0

1( ) ( ) ( ) ( )

T

TX t x t w x dx Lim x t dt

T

+

+

= =

Tng t:

2 2 2

0

1( ) ( ) ( ) ( )

T

TX t x t w x dx Lim x t dt

T

+

+

= =

Trng hp ny gi chung l qu trnh ngu nhin dng theo hai ngha:

Theo ngha hp: Tr trung bnh ch ph thuc khong thi gian quan st2 1t t = m khng ph thuc gc thi gian quan st.


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Theo ngha rng: Gi l qu trnh ngu nhin dng khi tr trung bnh lmt hng s v hm tng quan ch ph thuc vo hiu hai thi gian quan st

2 1t t = . Khi ta c mi tng quan

1 2 2 1

1 2 1 2 1 2

( , ) ( ) ( ) ( ). ( )

1( , ) ( ) ( )

x

T

T

B t t B t t B X t X t

x x w x x dx dx Lim x t x t dtT

+ +

+

= = = = +

= = +

Tm li: nghin cu nh lng ngun tin, h thng truyn tin m hnhho ngun tin bng 4 qu trnh sau:

1. Qu trnh ngu nhin lin tc: Ngun ting ni, m nhc, hnh nh l tiu

biu cho qu trnh ny.2. Qu trnh ngu nhin ri rc: l qu trnh ngu nhin lin tc sau khi

c ri rc ha theo mc trthnh qu trnh ngu nhin ri rc.3. Dy ngu nhin lin tc: y l trng hp mt ngun lin tc c

gin on ha theo thi gian, nh thng gp trong cc h thng tin xung nh:iu bin xung, iu tn xung ... khng b lng t ha.

4. Dy ngu nhin ri rc: Ngun lin tc c gin on theo thi gianhoc trong h thng thng tin c xung lng t ho.

1.4 H thng knh tin

1.4.1 Khi nim

Nh chng ta bit: vt cht ch c th dch chuyn tim ny nmt im khc trong mt mi trng thch hp v di tc ng ca mt lcthch hp. Trong qu trnh dch chuyn ca mt dng vt cht, nhng thng tinv n hay cha trong n sc dch chuyn theo. y chnh l bn cht cas lan truyn thng tin.

Vy c th ni rng vic truyn tin chnh l s dch chuyn ca dng ccht vt cht mang tin (tn hiu) trong mi trng. Trong qu trnh truyn tin,h thng truyn tin phi gn c thng tin ln cc dng vt cht to thnh tnhiu v lan truyn i.

Vic tn hiu lan truyn trong mt mi trng xc nh chnh l dng ccht vt cht chu tc ng ca lc, lan truyn trong mt cu trc xc nh ca


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mi trng. Dng vt cht mang tin ny ngoi tc ng dch chuyn, cn

chu tc ng ca cc lc khng mong mun trong cng nh ngoi mitrng. y cng chnh l nguyn nhn lm bin i dng vt cht khngmong mun hay l nguyn nhn gy ra nhiu trong qu trnh truyn tin.Nhvy: Knh tin l mi trng hnh thnh v truyn lan tn hiu mang tinng thi sinh ra cc tp nhiu ph hy thng tin.

1.4.2 Phn loi mi trng truyn tin

Knh tin l mi trng hnh thnh v truyn lan tn hiu mang tin. m t v knh chng ta phi xc nh c nhng c im chung, cbn c th tng qut ho v knh.

Khi tn hiu i qua mi trng do tc ng ca tp nhiu trong mitrng s lm bin i nng lng, dng ca tn hiu. Mi mi trng c mtdng tp nhiu khc nhau. Vy ta c th ly s phn tch, phn loi tp nhiu phn tch, phn loi cho mi trng (knh)

Mi trng trong tc ng nhiu cng l ch yu Nc(t):Nhiu cng l nhiu sinh ra mt tn hiu ngu nhin khng mong mun

v tc ng cng thm vo tn hiu u ra. Nhiu cng l do cc ngunnhiu cng nghip sinh ra, lun lun tn ti trong cc mi trng truyn lantn hiu.

Mi trng trong tc ng nhiu nhn l ch yu Nn(t):Nhiu nhn l nhiu c tc ng nhn vo tn hiu, nhiu ny gy ra do

phng thc truyn lan ca tn hiu, hay l s thay i thng s vt l ca bphn mi trng truyn lan khi tn hiu i qua. N lm nhanh, chm tn hiu(thng sng ngn) lm tng gim bin tn hiu.

Mi trng gm c nhiu cng v nhiu nhn1.4.3 M t struyn tin qua knh:

Xt h thng truyn tin trong VS ( t ) l thng tin truyn, RS ( t ) l

thng tin thu

( )VS t ( )RS t Ta c biu thc m t nhiu trong trng hp tng qut

Knh tin


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( ) ( ) ( ) ( )R n V cS t N t S t N t = + Trong thc t, ngoi cc nhiu cng v nhiu nhn, tn hiu cng chu tc

ng ca h sc tnh xung ca knh ( )H t do :

( ) ( ). ( ). ( ) ( )R n V cS t N t H t S t N t = +

c tnh knh khng l tng ny s gy ra mt s bin dng ca tnhiu ra so vi tn hiu vo gi l mo tn hiu v l mt ngun nhiu trong qutrnh truyn tin.

Tn hiu vo ca knh truyn hin nay l nhng dao ng cao tn vinhng thng s bin i theo quy lut ca thng tin. Cc thng s c th lbin , tn s hoc gc pha, dao ng c th lin tc hoc gin on, nu l

gin on s c nhng dy xung cao tn vi cc thng s xung thay i theothng tin nh bin xung, tn s lp li, thi im xut hin. Trong trnghp dao ng lin tc, biu thc tng qut ca tn hiu c dng sau:

( ) ( )cos( ( ))V

S t a t t t = +

trong ( )a t l bin , : tn s gc, ( )t : gc pha, cc thng s ny

bin i theo quy lut ca thng tin mang tin v nhiu tc ng s lm thayi cc thng s ny lm sai lch thng tin trong qu trnh truyn.

Theo m hnh mng ca knh tin, k hiu ( / )p y x l xc sut nhn

c tin ( )y t khi pht i tin ( )x t , nu u vo ta a vo tin ( )x t vi

xc sut xut hin l ( )p x ta nhn c u ra mt tin ( )y t vi xc sut

xut hin ( )p y ng vi ( )x t . Vi yu cu truyn tin chnh xc, ta cn phi

m bo ( )y t phi l tin nhn c t ( )x t tc l ( / ) 1p y x = . iu ny

ch c c khi knh khng c nhiu. Khi knh c nhiu, c th trn u raca knh chng ta nhn c mt tin khc vi tin c pht, c ngha l

( / ) 1p y x < v nu nhiu cng ln th xc sut ny cng nh. Nh vy v

mt ton hc, chng ta c th s dng xc sut ( / )p y x l mt tham sctrng cho c tnh truyn tin ca knh.

1.5 H thng nhn tin


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Php bin i ngun tin lin tc thnh ri rc gm hai qu trnh cbn:

Qu trnh ri rc ha theo thi gian hay l khu ly mu.Qu trnh lng t ha.

Csl thuyt ca php bin i ny gm cc nh l ly mu v lut lngt ha nh sau.

1.7.1 Qu trnh ly muGi s ngun tin lin tc dng tn hiu c biu din bng hm tin ph

thuc thi gian ( ) ( )cos( )S t a t t = + Vic ly mu mt hm tin c ngha l trch t hm ra cc mu ti

nhng thi im nht nh. Ni mt cch khc l thay hm tin lin tc bng

mt hm ri rc l nhng mu ca hm trn ly ti nhng thi im gin an.Vn t ra y l xt cc iu kin cho s thay th l mt s thayth tng ng. Tng ng y l v ngha thng tin, ngha l hmthay th khng b mt mt thng tin so vi hm c thay th.

Vic ly mu c th thc hin bng mt r le in, in t bt k ng

mdi tc ng ca in p ( )u t no . Thi gian ng mch ca rle l

thi gian ly mu , chu k ly mu l T, tn sut ly mu l1

fT

= . T

( )S t lin tc, ta thu c *( )S t theo ngha ri rc (Hnh 1.1)

u(t) T

S(t)

S *(t)

Hnh 1.1


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Nh vy nu thi gian ly mu di v s mu ln th nng lng

ca tn hiu ly mu tng ng vi nng lng ca tn hiu gc. Cc ktqu trn c pht biu bi nh l sau y:

nh l ly mu Shanon: Hm ( )S t trong khong ( )max max,f f hon ton

c xc nh bng cch ly mu vi tn s ly mu max2Sf f= .

1.7.2 Khu lng tho

Gi thit hm tin ( )S t bin thin lin tc vi bin ca n thay i

trong khong ( )min max,S S . Ta chia khong ( )min max,S S thnh n khong:

min 0 1 2 max... nS S S S S S = < < < < =

Nh vy hm tin lin tc ( )S t qua phng php ri rc s bin i

thnh '( )S t c dng bin i bc thang gi l hm lng t ho vi mi mc

lng t 1 , ( 0.. 1).i i iS S i n+ = = S la chn cc mc i thch hp s

gim s sai khc gia ( )S t v '( )S t .

S(t) S(t)

i

Hnh 1.2

Php bin i ( )S t thnh '( )S t c gi l php lng t ho.

,( 0,..., 1)i i n = gi l mc lng t ho.

Nu max min , 0,..., 1iS S

i nn

= = , ta c qui lut lng t ho u

ngc li ta gi l lng t ha khng ng u. Do s bin thin ( )S t trong

thc t thng l khng u nn ngi ta thng dng qui lut lng t khng


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Manip tn s FSK (Frequency Shift Key)Manip pha PSK (Phase Shift Key)

1.8.2 Gii iu ch

nh ngha: Gii iu ch l nhim v thu nhn, lc, tch thng tin nhn cdi dng mt in p lin tc hay mt dy xung in ri rc ging nhuvo vi mt sai s cho php.

Cc phng php gii iu chV phng php gii iu ch ni cch khc l php lc tin, ty theo hn

hp tn hiu nhiu v cc ch tiu ti u v sai s ( chnh xc) phi t cm chng ta c cc phng php lc tin thng thng nh:

Tch sng bin ,Tch sng tn sTch sng pha


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CHNG 2. TN HIU

2.1 Mt s khi nim cbn

Tn hiu l cc thng tin m con ngi thu nhn c t mi trng bnngoi thng qua cc gic quan hay cc h thng o lng. V d nh: Sng achn, nhp tim ca bnh nhn, lu lng ca cc dng chy hay m thanh,sng in t, tn hiu s,. V mt ton hc, tn hiu c hiu nh mt hm

s ph thuc vo thi gian ( )S t . Sau y chng ta s nghin cu cc dng tnhiu cbn.

2.1.1 Tn hiu duy trTh hin s duy tr ca tn hiu vi cng khng thay i theo thi

gian, tn hiu c biu hin bng hm s

, 0,( )

0, 0

a tI t

t

=


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Trong : A l bin dao ng, 2f

= l tn s,2

T

= l chu k ca

dao ng cbn. Dao ng cbn cn c th biu din bng cng thc tngqut hn

( ) cos sinS t a t b t = + (2.4)Khi ta c th biu din dao ng cbn nh mt vecttrong h trc ta

cc hay di dng s phc tng qut ( ) j tS t re = vi j l n vo.

2.2Phn tch ph cho tn hiuTrong thc t, mt tn hiu ngu nhin gm hu hn hay v hn cc tn

hiu n sc (nguyn t), khi nghin cu v x l tn hiu ngu nhinbt k, chng ta phi tm cch tch t tn hiu ngu nhin thnh tng tn hiun sc, vic phn tch gi l php phn tch ph.

Nu tn hiu iu ho c dng:

( ) cos( )S t A t = + ,khi chng ta c cc khi nim ph bin , ph pha v ph thc nh sau:

A A

Ph bin Ph pha Ph thc

Hnh 2.1

Trong cc loi ph trn, nng lng tp trung ch yu .

Nu tn hiu cho di dng phc ( )( ) ( )( )2

j t j tAS t e e + = +

Khi chng ta c dng ph phcA/2 A/2

- 0 + Hnh 2.2


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21

2.2.1 Chui Fourier v phri rc

nh ngha 1Cho 2 hm s ( ), ( )x x lin tc kh tch trn [ , ]a b , nh ngha

, ( ) ( )b

a

x x dx < >= (2.5)

c gi l tch v hng ca 2 hm trn khng gian [ , ]a bC . K hiu

2

[ , ]( )

b

a b

a

x dx = (2.6)

c gi l chun ca ( )x trn [ , ]a bC .nh ngha 2

Cho h hm 1 2( ), ( ),..., ( ),....nx x x xc nh lin tc trn [ , ]a b .

H{ }1( )k x

c gi l h trc giao nu tha mn iu kin

2

0 ,,

, .k l

k

k l

k l

< >=

=(2.7)

H{ }1( )k x

c gi l h trc chun nu tha mn iu kin

0 ,,

1 , .k lk l

k l

< >=

=(2.8)

Nhn xt: Vi mi h trc giao bt k, lun lun tn ti php bin i v htrc chun bng

( )( ) : .kk

k

xx

= (2.9)

nh ngha 3

Cho h { }1( )k x

l mt h trc giao v ( )f x l mt hm s bt k

xc nh lin tc trn [ ],a b , khi khai trin


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1( ) ( )k k

kf x A x

+

== (2.10)

c gi l khai trin Fourier tng qut thng qua h trc giao trong kA

c gi l h s khai trin.

xc nh cc h s khai trin, ta nhn hai v vi ( )n

x v ly tch

phn trn on [ ],a b , ta c

1

( ) ( ) ( ) ( )b b

n k k n

ka a

f x x dx A x x dx +

=

=

Do tnh cht trc giao ca h{ }1( )k x ta thu c

2

1

( ) ( ) ( ) ( ) ( )b b b

n k k n n n

ka a a

f x x dx A x x dx A x dx +

=

= =

Hay2

, n n nf A =

Tc l

2

2

( ) ( ),

.

( )

b

n

n an b

nn

a

f x x dxf

A

x dx

= =

(2.11)

Cng thc (2.7) l cng thc xc nh h s khai trin Fourier trongtrng hp tng qut vi mt h trc giao bt k.

Sau y ta xt mt s v d p dng phng php khai trin vi cc htrc giao khc nhau

V d 1: Xt h{ }1sin kx+

trn on [ ]0,2

Ta c

( )

2 2

0 0

1sin sin cos( ) cos( ) 02kx lxdx k l x k l x dx

= + = ,


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28

Nh vy ph

, 0,( ) 2

0, .

A t

S j kt

==

=

A

0 2/ 4/ 6/Hnh 2.6

2.2.3 Phcc tn hiu iu ch

Tn hiu thng tin mun truyn i xa phi nh tn hiu cao tn. tn

hiu cao tn mang thng tin ta phi lm cho tn hiu cao tn bin thin theoqui lut ca tn hiu thng tin. Tn hiu cao tn c dng:

0 0 0( ) cos( ) cos ( )S t a t a t = + = (2.14)

Ta c thiu ch 2 thng s bin 0a v gc ( )t . Vi gc ( )t ta c

thiu ch theo tn s 0 (gi l tn hiu iu tn) theo gc pha (gi l

iu pha). Sau y chng ta s xt chi tit cc phng php iu ch.

Phng php iu bin

Trong phng php iu bin, ta bin i bin ca tn hiu cao tn

theo qui lut ca thng tin ( )u t tc l bin i c cha lng tin cn truyn,cn tn s v gc pha khng i.

Gi s tin cn truyn l ( )u t , khi ta c cng thc bin i:

0 0 0( ) [ ( )]cos( )S t a M u t t = + + (2.15)

trong : 00

aMa

= c gi l h siu ch, trong k thut iu ch,

thng tin iu chm bo chnh xc, ta cn chn 0 1M . Hm s ( )u t c gi l hm tin, hm tin thng chn l hm n sc, nu hm tin l cc

thng tin phc tp, ta phi tch thnh cc tn hiu n sc bng phng phpphn tch ph nghin cu chng trc.

Gi s ( )u t l hm n sc c dng mt dao ng iu ho


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29

( ) cos( )u t t = +

Khi ta c

0 0 0

0 0 0 0

0 00 0 0 0

1 2 3

( ) [ cos( )]cos( )

cos( ) cos( )cos( )

cos( ) cos[( ) ] cos[( ) ]2 2

( ) ( ) ( )

S t a M t t

a t M t t

M Ma t t t

S t S t S t

= + + +

= + + + +

= + + + + + + +

= + +Nh vy tn hiu qua qu trnh iu bin s gm ba thnh phn, Mt

thnh phn 1( )S t dao ng vi tn s mang 0 v 2 thnh phn 2 3( ), ( )S t S t

dao ng vi tn s bin 0( ) . Bin ca tn s bin bng nhau v

bng 02

M.

a0

M0/2 M0/2

0 - 0 0 +

Hnh 2.7Trong trng hp tn hiu l khng n sc th tn hiu iu bin l mt

min, khng phi l ph vch, th vc tca tn hiu iu bin nh sauC D

A - Ba0

0

Hnh 2.8Trong

OA: Tn hiu mang

O


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31

Phng php iu pha

Tng t nh phng php iu tn, phng php iu pha bin i gc

pha c cha hm tin ( )u t cn bin v tn s khng i. Ta c cng thcbin i trong trng hp tn hiu n sc:

0 0( ) ( ) cos( )t t u t t t = + + = + + +

Tc l

( )0 0( ) cos cos( )S t a t t = + + + Nhn xt: V hnh thc th c th coi tn hiu iu tn, iu pha ging nhautrong cng thc tng qut sau y:

( )0 0 1 1( ) cos cos( )mS t a t t = + + + (2.16)

Tn hiu iu tn th 1 1, , 2m

= = =

, tn hiu iu pha th

1 1, ,m = = = .

2.2.4 Phn tch tn hiu ngu nhinDo cc tn hiu ngu nhin l cc i lng ngu nhin tun theo cc quy

lut phn phi xc nh nn vic phn tch cc tn hiu ngu nhin da trn csphn tch mi tng quan gia cc i lng ngu nhin ca l thuyt xcsut thng k.

Phng php phn tch tng quan

Nh chng trc gii thiu, tn hiu ngu nhin ( )x t c thi gian

tn ti hu hn ph thuc vo . Hm tng quan ( )B c tnh theo cngthc:

( ) ( ) ( )xB x t x t dt +

= + (2.17)

Hm tng quan phn nh mi lin h gia tn hiu v bn thn n sau

khi dch chuyn mt qung thi gian . Thc ra do c s bin thin nn ta xttrong qu trnh dng theo ngha rng th hm ( )xB c tnh nh gi tr

trung bnh ca ( )x t v ( )x t + tc l


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33

0( ) ( )

T

j tTS S t e dt = (2.18)

Hm tng quan:

( ) ( ) ( )T

j tB S t e dF

= (2.19)

Trong )(2

1

G

d

dF=

Ngi ta gi G() l ph nng lng, khi _1

( ) ( )2j

B G e d

= (2.20)

Trong trng hp ny, G() c xem nh l bin i Fourier ca B()_

( ) ( ) jG B e d

= (2.21)

Do B() v G() l cc hm chn nn ch ly gi trcos( ) tc l

0

1( ) ( )cosB G d

+

= ;

0

( ) 2 ( )cosG B d +

= (2.22)

Nu = 0 th:

1(0) ( )

2B G d

+

=

2.3Nhiu trngCc hin tng xo ng nhit trong cc phn t ca mch in hay dy

dn, hoc bc x trong kh quyn u gy ra mt loi tn hiu nhiu c di phrt rng gi l nhiu trng. Nhiu l thnh phn khng th b qua khi nghincu v cc knh, nhiu trng cng l mt loi tn hiu ngu nhin. Qua o c


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35

CHNG 3. LNG TIN, ENTROPI NGUN RI RC

3.1 o thng tin

3.1.1 Khi nim oi vi mt i lng vt l bt k, nghin cu v i lng

chng ta phi trang b mt n v xc nh ln ca i lng c gi lo. Mi o phi tha mn 3 tnh cht sau:

o l mt i lng khng m.o phi cho php ta xc nh c ln ca i lng . i

lng cng ln, gi tro c phi cng cao.

o phi tuyn tnh: tc l gi tro c ca i lng tng cngphi bng tng gi tr ca cc i lng ring phn khi s dng o ny o chng.

3.1.2o thng tin.

Khi nghin cu v thng tin, hin nhin y cng l mt i lng vtl, v vy chng ta cng phi xc nh mt o cho thng tin. xy dngo cho thng tin chng ta cn ch mt s vn sau y:

Theo bn cht ca thng tin th hin nhin thng tin cng c ngha khin cng t xut hin, nn o ca n phi t l nghch vi xc sut xut hin

ca tin hay ni cch khc hm o phi l hm t l nghch vi xc sut xuthin ca tin tc.

K hiu x l tin vi xc sut xut hin l ( )p x . Khi hm o k

hiu l1

( )( )

I x fp x

=

l hm t l nghch vi xc sut ( )p x .

Mt tin x s l khng c ngha nu chng ta bit v n hay xc sut

xut hin ( ) 1p x = . Trong trng hp ny o phi bng khng tc l

( ) 0I x = .

Xt 2 tin ,x y l c lp thng k vi xc sut xut hin tng ng l( ), ( )p x p y khi tin z xy= l tin khi xut hin ng thi 2 tin ,x y cng

mt thi im. Do theo tnh cht tuyn tnh, chng ta phi c


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37

3.2 Lng tin ca ngun ri rc

3.2.1 Mi lin h ca lng tin v l thuyt xc sut

Khi nim thng tin l mt khi nim c hnh thnh t lu trong tduy ca con ngi. din t khi nim ny, ta gi thit rng trong mt tnhhung no , c th xy ra nhiu s kin khc nhau v vic xy ra mt skin no trong tp hp cc s kin c th lm cho ta thu nhn c thngtin.

Mt tin i vi ngi nhn c hai phn

bt ngca tin. ngha ca tin. so snh cc tin vi nhau, ta c th ly mt trong hai hoc c hai ni

dung trn lm thc o. Nhng ni dung hay ngha ca tin m ta cn gi ltnh hm ca tin, khng nh hng n cc vn c bn ca h thngtruyn tin nh tc hay chnh xc. N chnh l ngha ca nhng tin mcon ngi mun trao i vi nhau thng qua vic truyn tin.

bt ngca tin lin quan n cc vn cbn ca h thng truyntin. V d: mt tin cng bt ng, s xut hin ca n cng him, th r rngthi gian n chim trong mt h thng truyn tin cng t.

Nh vy, mun cho vic truyn tin c hiu sut cao th khng th coi cc

tin nh nhau nu chng xut hin t nhiu khc nhau.nh lng thng tin trong cc h thng truyn tin, ta ly bt ng

ca tin so snh cc tin vi nhau. Ta quy c rng lng tin cng ln nu bt ngca tin cng cao. iu ny l hp l v khi ta nhn c mt tin bit trc th xem nh khng nhn c g, v vic nhn c mt tin m ta tc hy vng nhn c th li rt qu i vi chng ta.

Mi tin tc c th hin qua mi s kin. Cc s kin l cc hin tng

ngu nhin c thc m t bi cc quy lut thng k.

V mt truyn tin ta ch quan tm n bt ngca tin hay xc sut

xut hin cc k hiu. nghin cu vn ny ta dng cc quy lut thngk. Php bin i tng qut trong h thng truyn tin l php bin i cu trc

thng k ca ngun


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38

By gi chng ta xem xt mi lin h gia khi nim tin tc vi l

thuyt xc sut. Mt ngun tin ri rc c xem nh mt tp hp cc tin )(kx

hnh thnh bi nhng dy k hiu hu hn ix l mt k hiu ia bt k thuc

ngun A c gi i thi imj

t . Tin )(kx c dng: )...,,,( 21)(

n

k xxxx =

vi xc sut xut hin )( )(kxp .

V mt ton hc ngun tin X cng ng ngha vi mt trng xc sut

hu hn gm cc im )(kx . ),...,2,1( Mk= trong khng gian n chiu. M l

tng s cc im c tnh bng nmM = .

Php bin i tng qut trong mt h thng truyn tin l php bin i c

cu trc thng k ca ngun. Chng ta c th ly bt k mt khu x l tin tcno trong h thng nh ri rc ha, m ha, iu ch, truyn lan, gii iu

ch, gii m u c th xem nh mt php bin i ngun. Ni cch khc

php x l bin i cu trc thng k ca tp tin u vo khu h

thng tr thnh mt tp tin mi vi mt cu trc thng k mong mun u

ra.

{ }, ( )A p a { }, ( )B p b

Hnh 3.1Trong

{ })(, apA l ngun u vo vi b chA v phn b xc sut cc khiu )(ap .

{ })(, bpB l ngun u ra vi b chB v phn b xc sut cc khiu )(bp .

Nu l quy lut bin i th ta c mi quan h ={ })(, bpB .

Chng ta c th m t ngun tin u vo bng tp tin { })(iuU = v quy

lut phn b xc sut cc tin )(

)(i

up . Trong ),...,,( 21)(

n

i

uuuu = , ku l cctin thuc A xy ra cc thi im kt .


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40

Php bin i trong knh to ra mt ngun mi YXU .= , vi cc tin l cc

cp ),( ji yx , trong Xxi , Yyj , theo quy lut phn b xc sut

),( ji yxp . Cc tin ),( ji yx l cc im ri rc trn mt phng XY.

Theo l thuyt xc sut, s lin h gia cc xc sut ca cc phn t

trong tp YX, v YXU .= c th tnh nh sau:

==XxYy

yxpypyxpxp ),()(;),()(

)/()()/()(),( yxpypxypxpyxp ==

=

Yy

xypxp

xypxpyxp

)/()(

)/()()/(

V d 1: Php m ha nh phn, cho mt ngun tin { }0 1 7U u ,u ,...,u= dng

m nh phn m ha ngun tin, vi php m ha nh sau:

0000 zyxu

0011 zyxu

0102 zyxu

0113 zyxu

1004 zyxu

1015 zyxu

1106 zyxu

1117 zyxu

trong 0000 === zyx ; 1111 === zyx ; cc m hiu thit lp nh trn l

cc phn t ca mt tp tch ZYX .. v c i biu bng nhng im ri rctrong mt khng gian 3 chiu.

S lin h gia quy lut phn b xc sut trong cc tp hp v tp tch cho trong l thuyt xc sut nh sau:

==

ZzYyYy

zyxpyxpxp

;

),,(),()(

==ZzXxXx

zyxpyxpyp;

),,(),()(


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44

Trong trng hp phc tp

M rng khi nim lng tin tng h trong trng hp m ha hayphp bin i phc tp hn. Lc lng tin tng h cng c xc nhtheo cc cng thc xc sut tin nghim v xc sut hu nghim ca tin ang

xt. V d mt php bin i kp ZYX .

Lng tin ban u ca ix c xc nh theo xc sut ban u hay xc

sut tin nghim. Sau khi nhn c tin jy , xc sut ca tin ix trthnh xc

sut c iu kin )|( ji yxp , v cui cng khi nhn c jy v kz th xc

sut ca ix l )|( kji zyxp .

Nu ta xem )( ixp l xc sut tin nghim v )|( ji yxp l xc sut hunghim th ta li trli trng hp bin i n gin ni trn v c lng

tin tng h giai

x v jy : ),( ji yxI .

Nu ta xem )|(ji

yxp l xc sut tin nghim v )|(kji

zyxp l xc

sut hu nghim, ta s xc nh c lng tin tng h gia ix v kz vi

iu kin bit jy : )|(

)|(log)|,(

ji

kji

jkiyxp

zyxpyzxI =

Nu ta xem )( ixp l xc sut tin nghim v )|( kji zyxp l xc sut

hu nghim th lng tin tng h gia ix v cp ),( kj zy l:

)(

)|(log),(

i

kji

kjixp

zyxpzyxI =

Mt cch trc gic c th nhn thy lng tin v ix cha trong cp

),( kj zy phi bng lng tin v ix cha trong jy cng vi lng tin tng h

vi

x cha trongk

z khi bitj

y . iu ny c thc xc minh mt cch

d dng nh sau:

)|,(),(

)|(

)|(

)(

)|(log),( jkiji

ji

ji

i

kji

kji yzxIyxI

yxp

yxp

xp

zyxpzyxI +==

Nu thay th t cc tp ZYX ,, th s c cc biu thc mi nhng

ngha hon ton khng thay i.


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46

)|,()|(log)|( kjikiki zyxIzxpzxI = .

Gii thch lng tin ring c iu kin )|( ki zxI , )|( kj zyI cng tng

t nh gii thch lng tin ring. Lng tin ring c iu kin chnh l lngtin tng h vi cng mt iu kin xc nh mt cch n tr gia cc tinvi nhau. Lng tin tng h c th phn thnh tng ca nhng lng tintng h khc:

)|,(),(),( jkijikji yzxIyxIzyxI +=

3.2.4 Lng tin trung bnh

Lng tin ring ch c ngha i vi mt tin no , nhng khng phn

nh c gi tr tin tc ca ngun. Ni mt cch khc ( )iI x chnh gic v mt tin tc ca mt tin khi n ng ring r, nhng khng th dng

nh gi v mt tin tc ca tp hp trong ix tham gia. Trong thc t

iu ta quan tm l gi tr tin tc ca mt tp hp ch khng phi gi tr tin tcmt phn t no trong tp hp. nh gi hon chnh gi tr tin tc ca

mt tin ix trong c bng tin ta dng khi nim lng tin trung bnh.

nh ngha: Lng tin trung bnh l lng tin tc trung bnh cha trong mtk hiu bt k ca ngun cho.

( ) ( ) log ( )x X

I X p x p x=

(3.6)

V d: Mt ngun tin c hai k hiu l 0x , 1x vi xc sut %99)( 0 =xp v

%1)( 1 =xp . Nh vy khi nhn mt tin ta bit gn nh chc chn l l 0x ,

tin ny khng cn bt ngnn gi tr tin tc rt nh. Th nhng xt lng tin

ring ca 1x :

1( ) log 0,01 6,64I x = (bit/k hiu)

l gi tr rt ln, iu khng phn nh ng gi tr ca tin nhxt trn. Nu xt lng tin trung bnh:

08,0014,0066,0)(log)()(log)()( 1100 =+== xpxpxpxpXI (Bit/k hiu)Nh vy lng tin trung bnh rt nh phn nh ng thc t gi tr ca

ngun tin.


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53

Nu c mt tin gm cc k hiu : 31212411 xxxxxxxx . c )(XH cc i phi

c xc sut cc k hiu bng nhau bng 1/8Mun vy ta m ho ngun X trn thnh ngun Ynh sau :

1114

0113

012

01

yyyx

yyyx

yyx

yx

=

=

=

=

Ta c dy cc k hiu ca tin : 01100100111100 yyyyyyyyyyyyyy

Ngun ny c 1)( =YH

M ho ngun ny thnh ngun Z vi cc k hiu sau :

114

013

102

001

yyz

yyz

yyz

yyz

=

=

=

=

Ta c dy cc k hiu ca tin : 3231441 zzzzzzz

Xc sut cc k hiu ca ngun Z bng nhau v bng 1/4 nn :

24log)( ==ZH

Vy bng php m ho ngun X thnh ngun Z ta c th nng Entropi

ca ngun X l )(XH = 7/4 thnh Entropi 2)( =ZH m vn m bo lngtin trong cc bn tin c bo ton v c cng gi tr l 14 (bit)

dca ngun ch ra s chnh lch gia entropi ca ngun v gi tr cc i c th

c ca n ta dng d ca ngun:

max( ) ( )R H X H X= (3.17)Ngoi ra cng c th dng d tng i ca ngun nh gi:

max

max max

( ) ( ) ( )1

( ) ( )

H X H X H Xr

H X H X

= = (3.18)


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56

CHNG 4. L THUYT M

4.1 Khi nim m v iu kin thit lp m

Trong cc h thng truyn tin ri rc hoc truyn cc tn hiu lin tcnhng c ri rc ha, bn tin thng phi thng qua mt s php bini hay cn gi l m ha pha ngun pht, pha thu tin cn phi thng quanhng php bin i ngc li l gii m.

S m ha thng tin cho php ta k hiu ha thng tin hay s dng cck hiu quy c biu din bn tin dng ph hp cho ni s dng. Chnhnhm ha, ta c th hin thc thng tin, thng tin c bn cht l cc khi

nim. i vi mt h thng truyn tin, vic m ha cho php tng tnh huhiu v tin cy ca h thng truyn tin, ngha l tng tc truyn tin vtng kh nng chng nhiu ca h thng.

Khi tc lp tin R ca ngun cn cch xa thng lng C ca knh,nhim v ca m ha l bin i tnh thng k ca ngun lm cho tc lptin tip cn vi kh nng truyn ca knh. Trong trng hp truyn tin trongknh c nhiu, iu cn quan tm nhiu l chnh xc ca s truyn tin, haycc tin truyn i t b sai nhm. y chnh l nhim v th hai ca m ha.

Trong chng ny, trc tin ta cp ti cc khi nim v nh nghav m: th no l m hiu? cc thng s cbn ca m hiu l g? cc iu

kin v yu cu i vi m hiu l g?

4.1.1 M hiu v cc thng scbn

nh ngha: M hiu l mt ngun tin vi mt s thng k c xy dngnhm tho mn mt s yu cu do h thng truyn tin t ra nh tng tc lp tin, tng chnh xc cho cc tin

Nh vy m hiu chnh l mt tp hu hn cc k hiu ring hay bngch ring c phn b xc sut tha mn mt s yu cu quy nh.

- Vic m ho l php bin i 1 1 gia cc tin ca ngun c m hovi cc t m do cc du m to thnh.

Cho ngun S (A,P) vi A- tp k hiu ngun, P-xc sut tng ng. Khi php m ha l song nh f: A -> M, M l tp cc t m tng ng.


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59

a=00

b=01c=10d=11

Mt tin aaabcdb c m ho nh sau: 00000001101101 v truyn i .Khi nhn c tin v gii m, nu xc nh c gc ca dy k hiu

trn, chng ta ch c th tch mt cch duy nht thnh dy tin ban u bngcch t gc tri chia thnh tng nhm hai k hiu m tng ng. Nh vyb m trn cho php phn tch cc t m mt cch duy nht v c gi l mphn tch c.

Cng tin trn nu ta m bng b m khc:a=0b=01c=101d=1

Vn ngun trn m theo b m ny ta c: 00001101101. Khi nhn tin vgii m ta c th gii m nh sau: aaaaddbdb hoc aaabcdb. Nh vy tinnhn c s sai lc so vi ngun v vy b m ny khng dng c. Vykhi nim m phn tch c nh ngha nh sau:

S tn ti quy lut cho php tch c mt cch duy nht dy cc k

hiu m thnh cc tmc gi l iu kin thit lp m chung cho b m.B m tha mn iu kin thit lp m cn c gi l b m phn tch c.

iu kin ring cho tng loai m (m u v khng u)i vi mi b m cn tn ti nhng iu kin ring phi c tha mn

khi thit lp n.- Vi m khng u (m thng k ti u) ta phi chn b m sao cho

t c di trung bnh m ti thiu.- Vi m u (m sa sai) th b m c kh nng pht hin v sa sai

cng nhiu cng tt.Cc iu kin ring cho mi b m chnh l nhng iu kin v hnh

thc, v yu cu k thut hoc ch tiu k thut ring m b m cn t c.Cc iu kin ny l khc nhau vi mi loi m c th.


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Nu nhn c dy: 1000101001011101101

Ta ch c th tch c duy nht thnh: 100-01-01-00-1011-1011-01.Nh vy b m trn l loi b m phn tch c

Khi nim chm gii m l s k hiu nhn c cn thit phi cmi phn tch c mt t m. chm gii m c th l hu hn nhngcng c th l v hn. xc nh tnh phn tch c ca mt b m v chm gii m hu hn hay v hn ta xy dng bng th nh sau:Bc 1: Sp xp cc t m vo ct u tin ca bng (ct 1)Bc 2: So snh cc t m ngn vi cc t m di hn trong ct 1, nu t mngn ging phn u t m di th ghi phn cn li trong t m di sang ct 2.Bc 3: i chiu cc t hp m trong ct 2 vi cc t m trong ct 1 lyphn cn li ghi vo ct tip theo (ct 3).Bc 4: i chiu cc t hp m trong ct 3 vi cc t m trong ct 1 thchin ging nh trn cho n khi c mt ct trng th dng.

iu kin cn v m c thphn tch l trong ct j 2 khng cthp no trng vi mt tm trong ct 1.

r hn v thut ton ta quan st cc v d sau:V d 1:

1 2 3

00 - -

01 - -100 - -

1010 - -

1011 - -

Bng th c cc ct t th 2 l rng nn b m ny phn tch c, chm gii m ca b m ny bng di t m.

Nh vy c th ni cch khc l c tnh phn tch c iu kin cnv l bt k t m no cng khng c trng vi phn u ca t m khctrong cng b m.


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if not (St[i] in S) then

Beginn:=n+1;A[n]:= St[i];S:=S+[St[i]];

end;for i:=1 to n do P[i]:=0;for i:=1 to n dobegin

for k:=1 to length(St) doif A[i]=St[k] then P[i]:= p[i]+1;p[k]:=p[k]/length(st);

end;end;Procedure Sorting;var i,j:integer;tgp:real;TgA: char;Begin

For i:=1 to n-1 doFor j:=i+1 to n do

If P[i]


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begin

nn[i]:=0;S:=1;repeat nn[i]:=nn[i]+1;S:=S*0.5;until S1 thenbegin

ma[i]:=ma[i]+1;pp[i]:=pp[i]-1;endelse ma[i]:=ma[i]+0;

until length(ma[i])=nn[i];End;BEGIN

Input;create;Sorting;Mahoa_shanon;Output;Readln;END.

C th chng minh rng nhng b m c xy dng theo hai phngphp trn u l m prefix v hai phng php trn l tng ng nhau.


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(3.3) Gn cc gi tr ring 00,1, 2,..., 1n cho cc nhnh trn cy( )iT

tng ng cc nhnh ni vi cc cy con0

( ) ( ) ( )1 2, ,...,

i i i

nT T T .

End;Cy cui cng thu c l cy m gm cc nt l l cc t m tng ng

vi cc tin ca ngun. Mi t m l dy cc k hiu m { }00,1,..., 1n c

nh du trn ng i t gc ti nt l.Thut ton m ha trn c th m t mt cch khc nh sau:

Bc 1: Sp xp ngun X theo xc sut ( )i

p u gim dn

Bc 2: Chn n0 l s nguyn ln nht tho mn iu kin: 2 n0 m v

0

11

Nn

l mt s nguyn.

Bc 3: Lp, thay th 0n tin cui cng c xc sut nh nht thnh 1 tin ph c

xc sut bng tng xc sut trong nhm sau sp xp cc tin c li cng tinph theo xc sut gim dn. Qu trnh lp li cho n khi tin ph c xc sut

bng 1 (S bc lp l 1N ).Bc 4: Lp m: Xut pht t tin ph c xc sut bng 1, nh cc tin trong

nhm cc tr t 00... 1n , t cc tin ph sau tri, m ca cc tin trong nhm

chnh bng m ca tin ph cng vi cc tr t 00... 1n . Qu trnh s dng li

khi tt c cc tin c gn m.Bc 5: Loi b tt c cc tin ph, ta thu c b m cn tm.

r hn v thut ton ta xt v d sau:

Cho ngun tin U gm 7 tin: { }721 ,....,, uuuU = vi cu trc thng k

iu

1u 2u 3u 4u 5u 6u 7u

)( iup

0,34 0,23 0,19 0,10 0,07 0,06 0,01

Ta ly cs m m=2, nn chn n0=2

Sau khi thc hin thut ton ta c kt qu sau:


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00 10 11

011

0100

1010 01011Thut ton trn c m phng bi chng trnh sauProgram huffman;

uses wincrt;var st:string;A:array[1..255] of char;Ma:array[1..255] of string[20];

mma:array[1..255] of string[20];P:array[1..255] of real;

aa:array[1..255] of char;pp:array[1..255] of real;

n,i,j,k:integer;Procedure Input;

Beginwrite(cho xau can ma hoa:);readln(St);

end;

1

0

0 0

0

0

0

1

1

1

1

1

0,24

0,58

0,34

0,42

0,23 0,19

0,14 0,10

0,07 0,07

0,06 0,01

Mc 2

Mc 1

Mc 3

Mc 4

Mc 5

T(2)

T(1)

T)

T(4)T

(5)

T

(6)

1


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Procedure Output;

var k:integer;xauma:string;Begin

xauma:=;for i:=1 to length(st) dofor k:=1 to n doif st[i]=a[k] then xauma:=xauma+ma[k]+ ;writeln(Ket qua ma hoa);writeln(xauma);

end;Procedure Create;var s:set of char;Begin

n:=0;s:=[];for i:=1 to length(st) do

if not (St[i] in S) thenBegin

n:=n+1;A[n]:= St[i];S:=S+[St[i]];

end;for i:=1 to n do P[i]:=0;for i:=1 to n dobegin

for k:=1 to length(St) doif A[i]=St[k] then P[i]:= p[i]+1;p[i]:=p[i]/length(st);

end;for i:=1 to n do Ma[i]:= ;end;

Procedure Sorting(n1:integer);var i,j,tgp:integer;TgA: char;


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if aa[i]=# then

begink:=i;repeat k:=k+1;until b[k]=true;mma[k]:=mma[i]+0;mma[k+1]:=mma[i]+1;b[k]:=false;b[k+1]:=false;

end;until i=2*n-1;for i:=1 to n dofor j:=1 to 2*n-1 doif a[i]=aa[j] then ma[i]:=mma[j];

End;BEGINInput;create;Mahoa_huffman;Output;Readln;END.

4.5 Thut ton m ho Lempel-Ziv

Ta thy rng thut ton m ho ngun Huffman l thut ton m hongun ti u theo ngha b m to ra c tnh prefix v c di trung bnh tithiu. m ho cho ngun ri rc khng nh, ta phi bit xc sut xut hinca tt c cc k hiu, v i vi ngun c nh ta phi bit hm mt xcsut ca cc khi k hiu. Tuy nhin trong thc t tnh cht thng k cangun thng khng bit trc m ta thng c lng cc gi tr xc sutca ngun ri rc bng cch quan st mt chui di cc k hiu. Ngoi tr

vic c lng cc xc sut kp tng ng vi tn sut xut hin cc k hiu

ring r ca ngun, phc tp tnh ton trong vic c lng cc xc sut

ng thi l rt cao. Nh vy s dng thut ton m ho ngun Huffman chocc ngun c nhtrong thc t ni chung rt phc tp. Tri li vi thut ton


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Bc 1: Phn on m theo di li bit.

Bc 2: Biu din di dng thp phn cho cc phn on mBc 3: Biu din di dng cp s(ri, si).Bc 4: Khi phc dy ngun.V d : Cho dy m 101000000010100010110010000

Tp k hiu ngun {0,1,2}Bc 1 : Phn on dy m.k=3

l1 = )1*3log( = 2 , l2 = )2*3log( = 3, l3 = 4, l4 = 4, l5 = 4, l6 = 5, l7 = 5

Vy ta c : B1 = 10 ; B2 = 100 ; B3 = 0 ; B4 = 101 ;B5 = 1 ; B6 = 1100 ; B7 =

10000.Bc 2 : i v thp phn cho cc phn on m.D1 = 2 ;D2 = 4 ; D3 = 0 ; D4 = 5 ; D5 = 1 ; D6= 12 ; D7= 16Bc 3 : Biu din di dng cp s (ri, si).(r1, s1) = (0,2); (r2, s2) = (1,1); (r3, s3) = (0,0); (r4, s4) = (1,2); (r5, s5) = (0,1);(r6, s6) = (4,0) (r7, s7) = (5,1)Bc 4: Vit li phn on ngun tng ng viBi.

I1 = 2, I2 = 21, I3 =0, I4 = 22, I5 = 1, I6= 220, I7= 11.Vy dy ngun cho l: u = 221022122011.

4.6 M chng nhiu

Phng php m ho thng k lm cho cu trc thng k ca ngun trnn hp l. Trong phn trc (tc lp tin) c mt v d ni n vicnng cao entropi ca ngun bng cch m ho hai ln, mun nh vy c sca m cui cng t nht cng phi bng s tin ca ngun c. iu ny khngthun tin cho vic truyn tin. Thng thng ngi ta ch dng cc loi m ccs b (m=2 hoc m=3). V vy phng php m ho thng k lm cho cutrc thng k ca ngun tin trnn hp l c ngha l lm cho entropi ca b

m c dng c tr s cc i v di trung bnh ca t mm

Hn

log= .

S p dng cc loi m thng k trong cc h thng truyn tin cho phpt c nhng ch tiu kinh t tt, ngha l vi lng tin cn truyn cho th


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Tng s cc t m n k hiu c tk hiu sai l Cnt= n!/((n-t)!t!).

Xc sut xut hin mt t m n k hiu c tk hiu sai l:p(n,t)= Cn

tps

t(1-ps)

n-t.

Khi tr sps nh th xc sut nhn cc t m sai t k hiu ln hn xc sutnhn cc t m sai nhiu k hiu:p(n,t) > p(n,t) nu t < t.iu ny cho php ta xy dng cc m hiu chng nhiu hu hiu trong knhnh phn i xng. Cc m hiu ny c kh nng pht hin v sa sai cc thp m c s k hiu sai b, nhiu hn l i vi cc t hp m c s k hiusai ln.

4.6.2 Cchpht hin sai

Nu m hiu l tp hp cc t m n k hiu th s t m c chn phib hn tng s cc t hp c th c tn k hiu. Nhng t hp khng dnglm t m gi l t hp cm. Khi nhn tin nhn c t hp cm ta pht hin

ra sai. Vi m nh phn c n k hiu th c s t hp l: 0 2nN = .

M mun pht hin sai c th s t hp dng 0N N< v ta c s t hp cm

l: 0N N .

V d: 2n = dng 4 t hp m ho tin 1 2 3 4, , ,a a a a nh sau:

1 2 3 400, 01, 10, 11a a a a= = = =

Khi nhn c bt k t hp no ta cng thy hp l v vy ta khng pht hinra sai nhm. Nhng nu ta ch dng 3 t hp nh sau: 1 2 300, 01, 10a a a= = =

th khi nhn c t hp 11 ta bit l c s sai nhm. T hp 11 chnh l thp cm

Cchsa saiCch sa sai ca m hiu ng thi cng l nguyn l gii m phi da

vo tinh cht thng k ca knh m bo mc tiu sai nhm ti thiu. Munvy cn phi da vo tnh cht nhiu trn knh phn nhm cc t hp cm,mi nhm tng ng vi mt t m m chng c kh nng b chuyn i sangnhiu nht.

Cn c vo c tnh thng k s t m sai t k hiu xy ra nhiu hnnn ta u tin x l trc. Coi mi t m l mt vc t, mi k hiu l mt


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to ca vc t. Nhiu cng c coi nh l mt vc tgi l vc tsai. T

m sai coi nh l tng hp ca vc tm v vc tsai.V d 1:

Vc tm: 0101, vc tsai: 0100, t m sai: 0001T m trn b sai k hiu th 2 (bit 1 trthnh bit 0)V d 2: Vi mt t m c 4 k hiu vc tsai cng c 4 k hiu, c ccphng n sai 1 k hiu, 2 k hiu, 3 k hiu .. Ta c bng sau:

Vec tm a1 a2 a3 a4

Vec tsai 0001 0101 1110 1111 S Kh sai

0001 0000 0100 - -

0010 0011 0111 1100 11010100 - - 1010 10111000 1001 1101 0110 0111

1

0011 0010 0110 1101 11000101 0100 0000 1011 10101001 1000 1100 0111 01101010 1011 - 0100 -1100 1101 1001 0010 0011

2

0111 0110 0010 1001 10001011 1010 - - 0100

1101 1100 1000 0011 00101110 - 1011 0000 -

3

1111 - 1010 - 0000 4

Vi nhn xt trn ta c cc t hp cm c phn thnh nhm, cc t m cmtrong nhmBi tng ng vi t m ng ai. Ta c bng sau:

B1 B2 B3 B4

0000 0100 1100 1101

0011 0111 1010 10111001 1101 0110 0111


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Ta thy nhn c t hp 0111 hoc 1101 do sai 1 k hiu ta c th

gii m v sa thnh 0101 ( 2a ), nh vy theo nguyn tc trn ta c th sa

sai cho b m. Tuy nhin nh b m trn cng cho thy c th sa khng

chnh xc v t hp 1101 c th sa thnh 2a cng c th sa thnh 4a .

gii quyt ngi ta khc phc bng cch dng bng chn hoc tnh khongcch m ca b m.

4.6.3 Xy dng b m sa sai bng bng chn

Bng cch lp bng chn ta chn c t hp t m dng, hp l mbo sa sai c.

V d: Xydng bng m c 4 t m mi t m c 4 k hiu.Xy dng bng c 42 2n = ct nh sau:

sct

1 2 3 4 5 6 7 12 13 14 15 16

u

e 0000 0001 0010 0011 0100 0101 0110 ... 1011 1100 1101 1110 1111

0011 0011 0010 0001 0000 0111 0110 0101 ... 1000 1111 1110 1101 1100

0110 0110 0111 0100 0101 0010 0011 0000 ... 1101 1010 1001 1000 1001

1100 1100 1101 1110 1111 1000 1001 1010 ... 0111 0000 0001 0010 0011v : vec tm, e : vec tsai.

T bng trn ta chn cc t m s dng bng cch:

Chn mt t bt k v d chn t m 2a = 0001 ta xem bn thn n v

cc t hp cm ca n nu trng vi cc t hp ca t mia no th nh

du ct i loi. Nh trn loi ct 3, 5, 8, 12, 14, 15. Chn t m th 2 trong

cc ct cn li gi s chn 6a loi cc ct c cc t hp trng vi cc t hp

cm ca 6a ta s loi ct 1, 4, 7, 10, 11, 16.. Chn tip tng t ta chn c

4 t m l:

1 0001a = (ct 2)

2 0101a = (ct 6)


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V d: Vc tV(v) = 01011 th W(v) = 3

Khong cch Hamming

Trong l thuyt thng tin, khong cch Hamming (Hamming distance)gia hai dy k t c chiu di bng nhau l s cc k hiu v tr tngng c gi tr khc nhau. Ni mt cch khc, khong cch Hamming o slng thay th cn phi c i gi tr ca mt dy k t sang mt dy k tkhc hay l s lng li xy ra khi bin i mt dy k t sang mt dy k tkhc.

V d: Khong cch Hamming gia 1011101 v 1001001 l 2.Khong cch Hamming gia 2143896 v 2233796 l 3.

Khong cch Hamming gia "toned" v "roses" l 3.nh ngha: Khong cch Hamming gia hai vc tU(u1,u2,u3,..,un),V(v1,v2,v3,..,vn) l s cc to khc nhau gia chng. K hiu lD(U,V).V d: Tnh khong cch hai vc t 11001 v 10110 l 4

Tnh ngha khong cch Hamming gia hai vec tta c kt qu sau:D(U,V)= W(U+V) trong U+V l mt vec tc c t php cng modul 2gia hai vec tUv vec tV. U= 11001, V= 10110: U+ V= 01111. Nhvy khong cch ca Uv Vl W(U+ V) = 4.

Cchpht hin sai bng khong cch Hamming

Ta thy khong cch gia hai t m s thay i 1 n v nu thay i 1k hiu no trong mt t m. Khong cch bng 0 nu 2 t m trng nhau.Vy nu 2 m hiu c khong cch D = 1 th khng pht hin c sai 1 khiu, mun pht hin sai 1 k hiu gia hai t m phi c khong cch mDti thiu bng 2.

Vy mun pht hin t m sai tk hiu th khong cch m phi tho

mn iu kin D t+1.

V d1: T m 1110 di tc dng ca t m sai 0001 s thnh 1111, nh vynu ta chn a1 = 1110, a2 = 1111 (D =1) th khi a1 sai mt k hiu thnh1111 khi nhn ta khng pht hin ra sai c.

V d 2: Gi s c b m: a1 = 0011; a2 = 0101; a3 = 1100; a4 = 1111


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B m ny c D = 2 khi mt t m bt k sai i 1 k hiu ta nhn ra

ngay. Gi s t m a1 sai 1 k hiu trthnh 0111, 1011, 0001, 0010 ta nhnthy cc t hp ny l t hp khng dng vy c hin tng sai.

Cchsa sai bng khng cch Haming

By gihy xt iu kin m khng nhng ch pht hin sai m cn sacha c sai nhm. iu kin ny c tho mn nu s sai nhm lmchuyn i t m ban u thnh nhng t hp m gn t m hn l bt cmt t m no khc.

Ta thy khi t m sai i tk hiu th khong cch ca n so vi mt tm bt k khc cng lch i t n v. Vy mt t m sai lch i so vi t m

gc ca n khong cch nh hn D/2 th ta s quy v t m gc ca n c.Do t m sai tk hiu, ta mun xc nh c t hp sai thuc t m no

th khong cch m phi tho mn iu kin:D 2t +1.

V d: Cho b m: a1 = 00000, a2 = 01101, a3 = 10110, a4 = 11011B m ny cD = 3 nn c th sa sai cho 1 k hiu. Gi s t m a2 b

sai 1 k hiu thnh a2 = 01111. Ta tnh khong cch m ca a2 vi cc ttrong b m trn c:D(a1,a2) = 4,D(a2,a2) = 1,D(a2,a3) = 3, D(a2,a4) = 2.Nh vy a2 gn a2 nht ta sa a2 thnh a2 = 01101. R rng vic sa ny lchnh xc.

4.6.5 Mt sbin php xy dng b m pht hin sai v sa sai

Dng ParityKhi xy dng bng tin pht i, ta thm vo mt bit vo cui ni dung

thng tin to ra tng s cc bit mang tr 1 l chn (Parity chn) hay l(Parity l). Khi nhn tin ta kim tra tng cc bit 1 xem c b thay i khng(chn-l) nu dng Parity chn m tng s bit 1 l th ta bit nhn tin sai.Mun khi phc tin ta tch cc bit parity ra khi tin.


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V d: Ta c parity chn cc bit parity thm vo nh sau:

Thng tin Parity0 1 1 0 1

1 1 1 0 1

0 1 1 0 0

M khiPhng php ny ta lm ging nh dng Parity nhng mc cao hn.

Ta to tin thnh tng khi v trn cc dng, ct ta u thm cc bit parity theocch trn. Nh vy ta kim tra hai ln theo dng v theo ct nn kh nng phthin cao hn nhiu, ngoi ra ta cn c kh nng sa sai nu xc nh c to sai chnh xc.

V d:

1 1 0 0 1

0 0 1 0 1

1 1 0 1 0

0 0 1 1 0

1 0 1 0 1

M i xng (M thun nghch)Gi s ta c b m vi cc t m 1 2, ,..., na a a . Xut pht t b m trn

ta xy dng b m mi nh sau:

1 1 1 2 2 2 3 3 3; ; ;...,R R R R

n n nb a a b a a b a a b a a= = = = trong R

ka l xu nghch o

ca xu ( 1,2,..., )ka k n= . Do tnh cht i xng nn b m m bo vicpht hin sai v sa sai tng i d dng v chnh xc.

1

0

0

1

0

1

0

1

10101


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V d:

B m gc: 1 2 3 40100; 100; 1101; 001a a a a= = = = B m i xng:

1 2 3 401000010; 100001; 11011011; 001100b b b b= = = =

M tlK hiu p l s bit c tr (0), q l s bit c tr (1) trong t m, ta a ra

quy tc xy dng mt b m tha mn tnh chtp

rq

= trong r l mt s

cho trc c gi l h s t l, tnh cht trn cn ng vi mi t m. Vitnh cht t l, vic pht hin sai v sa sai cng tng i d dng.

V d: B m vi h s t l 2r=

1 2 3 4 5001; 010; 100; 011000; 110000a a a a a= = = = =

4.7 M tuyn tnh

Phng php biu din m tuyn tnh s dng cc php ton i stuyn tnh nn cch biu din m l thun tin

4.7.1 Phng php xy dng m tuyn tnhNgi ta s dng cc php ton ca i s tuyn tnh biu din m.

Gi s b m V gm cc t m c di n , mi t m c gi l mt vectm gm n thnh phn. Cs m l m (trong trng hp nh phn th 2m = ).

Biu din bng ma trn sinhXy dng ma trn G gm k hng, mi hng l mt t m thuc V

c gi l vectnn ca m V , khi G c gi l ma trn sinh ca m

V khi v ch khi bt k mt t m no thuc V cng l mt t hp tuyn tnh

ca cc hng ca ma trn V.

Vc gi l khng gian hng ca ma trn sinh G , nu s chiu ca

khng gian vectV l k th ma trn G s c k hng.

Nh vy cc t hp tuyn tnh khc nhau tng ng vi nhng vectmkhc nhau, Nu V c k vectnn th h gm k vect s l hc lptuyn tnh.


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Gi sG =

ka

a

a

..2

1

th khi mt vectm 1 1 2 1 ... k kv a a a = + + + . Trong

i l cc k hiu ca m. M ( , )V n k s ckm t m.

V d : Ta c ma trn sinh sau : 1 01101a = ; 2 10110a =

=

10110

01101G

M (5,2)V c 5n = v 2k= , cc t m nh sau :

1 1 2 2v a a = +

V= {00000; 01101; 10110; 11011}

V d: Xc nh m (5,3)V t ma trn sinh G :

=

00101

01010

10011

G

Ta thy s t hp l 25 s c 32 t hp nhng ta ch dng 8 t hp (8 vectm) cn li l nhng t hp khng dng, m ny cng cho php ta pht hin

v sa sai c. Cch biu din ny rt gn v thun tin, nht l khi cc s n v k ln trong khi nu ta biu din bng bng m th rt cng knh v c thkhng thc hin c. Ngoi ra, ma trn sinh cn c gi l ma trn sinh mkim tra chn l.

Biu din bng ma trn thXy dng mt ma trn H gm n-khng c lp tuyn tnh (trong k l

s hng ca ma trn G ), H l ma trn m bt k mt vec tm v V th

0TvH = (0 l vec tkhng, TH l chuyn v ca H). Khng gian vec t 'V

to bi ma trn Hc gi l khng gian khng ca khng gian hng V(hay

'V l khng gian hng ca ma trn H, khng gian khng ca G ). Hcgi l ma trn th ca m V .

Mi vec tm ca Vu trc giao vi mi vectthuc '( , )V n n k .


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V d :

=

00101

01010

10011

G c ma trn th

=

10101

11010H

M h thngXt b m nh phn. Nu bng mt php bin i ta chuyn ma trn G

v dng sau :

=

knkkk

kn

kn

ppp

ppp

ppp

G

,2,1,

,22,21,2

,11211

..1...00

..0...01

...0...10

Vi mi b 1 2( , ,..., )k = ta c mt vect m 'v G=

1 2 1( , ,..., , ,..., )k n kv C C = trong =

=k

i

jiij pC1

,

k thnh phn u ca v c gi l nhng k hiu mang tin, n k thnhphn sau c gi l cc k hiu d hay k hiu th (k hiu l r).

Ta thy n k thnh phn sau l t hp ca cc thnh phn u, v vyqu trnh m ho n gin. B m ny c gi l m h thng.

4.7.2 Nguyn l gii m

Trong ni dung trn, vic gii m thc hin i chiu vec tnhn ctrong cc lp k ca bng gii m, tuy nhin cch lm ny khng hiu qu khi

,n k ln. khc phc hn ch ta thc hin theo thut ton da trn cstnh

Syndrom. Cho V l m tuyn tnh ( , )n k bao gm cc t m l cc vec t: 1h

(vec t khng) v nhng vec t m cn li 2 3 2, ,..., kh h h . Cc vec t sai

1 2, ,..., pe e e . thun tin trong qu trnh lp thut ton gii m ta nh ngha

khi nim lp k nh sau :

nh ngha 1: jV l lp k th j nu { }: , 1, 2,..., 2k

j i jV v v h e i= = + = , je vv c gi l vec tto lp k.


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nh l 1: Nu bng gii m theo cch xp lp k, khi nhn c vec t v s

gii m ng thnh vec t u khi v ch khi vec tsai v u+ l vec t to lpk.

nh ngha 2: Syndrom ca vec t v gm r thnh phn c xc nhTS vH= .

Khi v l t m ng th Syndrom s bng khng, ngc li Syndromkhc 0.

nh l 2: Hai vec t 1 2,v v cng nm trong mt lp k khi v ch khi Syndrom

ca chng bng nhau.Trn cscc nh l trn, ta c thut ton gii m :

Bc 1: Thit lp bng gii m bng cch xp cc lp k vi nhng Syndromtng ng cho tng lp.

Bc 2: Khi nhn c mt dy 'v di n th tnh Syndrom ca 'v v tm lp

kj

V ng vi Syndrom tnh.

Bc 3: T m gc ' jv v e= + . je l vec tto lp k jV .

V d: Cho ma trn sinh G

=

00101

01010

10011

G c ma trn th

=

10101

11010H

Bc 1 : Xp lp k :

v

e 00000 00101 01010 01111 10011 10110 11001 11100

1V 00001 00100 01011 01110 10010 10111 11000 11101

2V 00010 00111 01000 01101 10001 10100 11011 11110

3V 10000 10101 11010 11111 00011 00110 01001 01100

Tnh Syndrom ca cc lp 1, 2, 3 ln lt bng : 01, 10, 11.


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98

Cc vec t to lp k : 00001, 00010, 10000 l nhng vec t sai. M

ny khng sa c ht tt c cc t m sai mt bit, v khong cch m tithiu 2D = . Mun sa c th 3D = .

Bc 2: Gi thit nhn c ' 10111v = , Syndrom ' 01TS v H= = ng vi vectto lp k 00001.

Bc 3: T m gc: 10111 00001 10110v = + = .

4.7.3 Mt sgii hn ca m tuyn tnh

Trong mt m tuyn tnh, kh nng sa sai ca m l mt tiu chunhng u. Kh nng ny c biu th bng s k hiu sai ti a c th sac trong mt t hp m v c lin quan n trng lng ti thiu ca m(khong cch m). Cho nn khi xy dng m sa sai ti u, cn phi xc nhtrc mt s gii hn ca m tuyn tnh ca m nh gii hn trn v trnglng ti thiu, gii hn di v s k hiu thi vi m tuyn tnh. Diy l mt s gii hn i vi m nh phn.

Gii hn trn ca khong cch m ti thiuTng trng s ca m tuyn tnh (n, k) nh sau :

Mi t m c n k hiu, b m c 2k t m. Nh vy c tng s k hiu

l n . 2k. Trong c 1 k hiu khc 0. Gi thit1

(0) (1)2

p p= = . Tng trng

s l: 11

.2 . .22

k kn n = .

Ta thy khong cch m ca t m 0 n t m c trng s thp nhtchnh l trng s ca t m gi thit l trng s ti thiu D . R rngtrng s ti thiu th phi nh hn trng s trung bnh. T ta c th rt ra

iu kin cho khong cch m D phi tho mn iu kin:12

2 1

k

knD

Nh vy D phi nh hn hoc ln nht l bng t s trn, y chnh lgii hn trn ca khong cch m ti thiu.


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100/136


101/136

101

0 0....0 1

0 0....1 0

..............

0 1....0 0

1 0....0 0

kI

=

to thnh ma trn sinh *, ,,n k k n k k G I C =

V d : Xy dng mt m vng (7,4) .

Trc ht xc nh a thc sinh ( )P x ly trong nhng tha s chung ca

nh thc:7 3 2 31 ( 1)( 1)( 1)x x x x x x+ = + + + + + .

3 23 ( ) 1r n k P x x x= = = + + , vit di dng t hp nh phn l: 1101.

Xc nh ma trn cc s tha 3,4C ca php chia7x cho ( )P x . Sau khi

thc hin php chia nh phn ta c ma trn:

3,4

1 0 1

1 1 1

0 1 1

1 1 0

C

=

=> *7,4

0 0 0 11 0 1

0 0 1 0 1 1 1

0 1 0 0 0 1 1

1 0 0 0 1 1 0

G

=

c th thc hin mt cch thun tin cc thit b to m v gii m,

cn c thut ton lp v gii m. Cc thut ton ny u da trn tnh cht cbn l cc t m chia chn cho a thc sinh. T m tm c:

( ) ( ). ( ) ( ) ( )n kF x P x Q x x A x R x= = + trong ( ) ( )

( )( ) ( )

n kx A x R xQ x

P x P x

= + .

V d : T hp m n gin 2 2( ) 1, n kA x x x x= + = , 5 3. ( )n kx A x x x = + . em

chia 5 3x x+ cho 3 2( ) 1P x x x= + + ta c:5 3 2

23 2 3 21 1

x x x xx x

x x x x

+ += + +

+ + + +.

V vy t hp m s l: 5 3 2x x x x+ + + , vit di dng t hp m nh phn l101110.


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102

4.8.3 Nguyn l gii m

em chia t m nhn c cho a thc ( )P x , nu php chia chn chng

t t m thu c l ng, trong trng hp php chia cn s tha th chngt t m thu c c k hiu sai. S tha cho php xc nh v tr ca k hiu

sai. Vec tgc u v e= + . Trong v l vec tthu c, e l vec tsai.Phng php thc hin sa sai c th da trn ma trn pht hin v sa

sai

2 ,

,

n k

nn k n

n k k

I

M I

C

=

V d: i vi m vng (7,4) ta c cc ma trn sau:

3,4

1 0 1

1 1 1

0 1 1

1 1 0

C

=

3

0 0 1

0 1 0

1 0 0

I

=

7

0 0 0 0 0 0 1

0 0 0 0 0 1 0

0 0 0 0 1 0 0

0 0 0 1 0 0 0

0 0 1 0 0 0 0

0 1 0 0 0 0 0

1 0 0 0 0 0 0

I

=

Cc hng ca ma trn 7I i biu cho cc vec tsai, v tr ca sn v chobit v tr ca k hiu sai trong t hp m tuyn tnh t tri sang phi theo bc

t cao gim xung thp: 6 5 4 3 2 1 0, , , , , ,x x x x x x x .

V d: hng th ba ca ma trn sn vv tr th 5 tng ng vi k hiu

sai 2x . Ma trn pht hin sai v sa sai ca m (7,4) l:


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103

10,7

0 0 0 0 0 0 1 0 0 1

0 0 0 0 0 10 0 1 0

0 0 0 0 1 0 0 1 0 0

0 0 0 1 0 0 0 1 0 1

0 0 1 0 0 0 0 1 1 1

0 1 0 0 0 0 0 0 11

1 0 0 0 0 0 0 1 1 0

M

=

Gi s thu c t m: 0111100, trong khi u pht gi i tm: 0110100. Nh vy trong qu trnh truyn tin gy sai k hiu th 4 k ttri. Qu trnh pht hin sai v sa sai c thc hin nh sau:

a thc sinh c chn nh trc 3 2( ) 1P x x x= + + . Tin hnh kim tra

t hp m thu c l ng hay sai bng cch em chia t m cho a thc

sinh 3 2( ) 1P x x x= + + . Sau khi thc hin php chia c s tha l: 101, em

i chiu trong ma trn 10,3M thy s tha thuc hng th t tng ng vi

vec tsai: 0001000, iu ny cho thy bit th t ca t m sai. Vec tgc= 0111100+0001000 =0110100.

thc hin cc s to m v gii m vng, s dng cc mch nhn,chia a thc vi a thc. Cc mch ny c xy dng trn cscc phn tcbn ca mch nh sau:

: Cng moun 2 ia : Nhn vi mt h s bng ai : Mch ghi chuynS nhn hoc chia mt a thc cho mt a thc l nhng dng t hp

khc nhau ca cc phn t cbn trn c m t nh sau:

S nhn mt a thc vi mt a thc: ( ). ( )A x P x

0 1

0 1

( ) ...

( ) ...

i

i

r

r

A x a a x a x

P x b b x b x

= + + +

= + + +

20 0 0 1 1 0 0 2 1 1 2 0

1 11 1

( ) ( ) ( ) ( ) ...

( ) ( )i r ri r i r i r

A x P x a b a b a b x a b a b a b x

a b a b x a b x+ +

= + + + + + + +

+ + +


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104

S chia mt a thc cho mt a thc:

Output

Inpu

aj aj+1 aj+2 aj+r- aj+r-

br br- br- b1 b0

Output

Input 1 2 3 r-1 r

-b0 -b1 -b2 -br-

-br


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105

CHNG 5. GII THIU V H MT M

5.1 Khi nim h mt m

i tng cbn ca mt m l to ra kh nng lin lc trn mt knh

cho hai ngi s dng (Gi s l ngi A v ngi B) sao cho i phng

khng th hiu c thng tin c truyn i. Knh ny c th l mt ng

dy in thoi hoc mt mng my tnh. Thng tin m B mun gi cho A (bn

r) c th l mt vn bn ting Anh, cc d liu bng s hoc bt c ti liu

no c cu trc tu . Khi B s m ho bn r bng mt kho c xc

nh trc v gi bn m kt qu trn knh. i phng c bn m thu c

trn knh song khng th xc nh ni dung ca bn r, nhng A (ngi

bit kho m) c th gii m v thu c bn r.

Ta s m t m hnh ton hc tng qut nh sau:

Mt h mt l mt b 5 (P, C, K, E, D) tho mn cc iu kin sau:

+ P (Plaintext): L mt tp hu hn cc bn r c th.

+ C (Ciphertext): L mt tp hu hn cc bn m c th.

+ K (Key): L tp hu hn cc kho c th.+ E (Encryption): L tp cc hm m ho.

+ D (Decryption): L tp cc hm gii m.

i vi mi k K c mt quy tc m ek: P C v mt quy tc gii m

tng ng dk D.

Mi ek: P C v dk: C P l nhng hm m dk (ek (x)) = x vi mi

bn r x P.

5.2 Phn loi cc h thng mt m

Theo cch m ho v gii m:H m cin (h m i xng): dng 1 kho m ho v gii m.


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106

H m hin i (h m bt i xng): dng mt kho m ho v 1

kho gii m. Kho m ho c th cng khai, kho gii m phi gi b mt.

Theo cch m ho:M khi: m ho s dng cc thut ton khi, d liu c chia thnh

khi trc khi m vi kch thc tu nhng phi cnh.

M dng: l cc thut ton m ho v gii m thc hin theo tng bit ti

mi thi im.

5.3 H m cin (Symmetric-key encryption)

5.3.1 Khi nim

H m cin l loi m c thc hin thng qua hm f c tnh thun

nghch, s dng f m ho, bit f c th suy ra hm gii m f-1. y l h m

dng cng mt kho m ho v gii m, kho phi c gi b mt.

5.3.2 Mt sh m cin

M hon v(MHV) tng ca MHV l gi cc k t ca bn r khng thay i nhng s

thay i v tr ca chng bng cch sp xp li cc k t ny. MHV (cn cgi l m chuyn v) c dng t hng trm nm nay. Gi s m l mt s

nguyn dng xc nh no , k hiuP = C = (Z26 )m v cho tt c cc hon

v ca {1, . . ., m}. i vi mt kho (tc l mt hon v) ta xc nh:

e(x1, . . . , xm ) = (x(1), . . . , x(m)),

d(x1, . . . , xm ) = (y-1

(1), . . . , y-1

(m))

trong -1 l hon v ngc ca .

Gi s ta c bn r l : khoacongnghethongtindaihocthainguyen.

Trc tin ta nhm bn r thnh cc nhm 6 k t :

khoaco | ngheth | ongtin | daihoc | thaing | uyenzz


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107

Mi nhm 6 ch ci c sp xp li theo php hon v, ta c:

AOKOHC| EHNHGT | HIDCAO | IATGHN | NEUZYZ

Nh vy bn m l :

AOKOHCEHNHGTHIDCAOIATGHNNEUZYZ

Vic thc hin gii m c thc hin thng qua hon vo -1

M dch vng (shift cipher)Trong phn ny s m t m dch (MD) da trn l thuyt s hc theo

modulo, chng ta s xt mt snh ngha cbn ca s hc lin quan n

vn ny.

nh ngha: Gi s a v b l cc s nguyn v m l mt s nguyn dng, k

hiu a b (mod m) nu m chia ht cho b-a. Mnh a b (mod m) c gi

l "a ng d vi b theo modulo m".

Gi s chia a v b cho m v ta thu c thng nguyn v phn d, cc

phn d nm gia 0 v m-1, ngha l a = q1m + r1 v b = q2m + r2 trong 0

r1 m-1 v 0 r2 m-1. Khi c th d dng thy rng a b (mod m) khi

v ch khi r1 = r2 . Ta s dng k hiu a mod m xc nh phn d khi a c

chia cho m. Nh vy a b (mod m) khi v ch khi a mod m = b mod m. Nu

thay a bng a mod m th ta ni rng a c rt gn theo modulo m.

Nhn xt: Nhiu ngn ng lp trnh ca my tnh xc nh a mod m l phn

d trong khong (-m+1),..., (m-1) c cng du vi a, v d -18 mod 7 s l -4,

gi tr ny khc vi gi tr 3 l gi trc xc nh theo cng thc trn. Tuy

nhin, thun tin ta s xc nh a mod m lun l mt s khng m.

By gi ta nh ngha modulo m k hiu l Zm l tp hp {0,1,. . .,m-1}

c trang b hai php ton cng v nhn, vic cng v nhn trong Zmc thchin ging nh cng v nhn cc s thc vi cc kt quc rt gn theo

modulo m.


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108

V d tnh 11 13 trong Z16 . Tng t nh vi cc s nguyn ta c 11

13 = 143. rt gn 143 theo modulo 16, ta thc hin php chia bnh

thng: 143 = 8 16 + 15, bi vy 143 mod 16 = 15 trong Z16 .

nh ngha m dch vng

Gi s P = C = K = Z26 vi 0 k 25, nh ngha m dch vng dng

ton hc nh sau:

ek(x) = x + k ( mod 26 )

dk(x) = y - k ( mod 26 ) (vi x,y Z26)

Nhn xt: Trong trng hp K = 3, h mt thng c gi l m Caesar tng c Julius Caesar s dng.

Ta s s dng MDV (vi modulo 26) m ho mt vn bn ting Anh

thng thng bng cch thit lp s tng ng gia cc k t v cc thng d

theo modulo 26 nh sau: A 0,B 1, . . ., Z 25.

V d:Gi s kho cho MDV l K = 5 v bn r l: Khoacongnghethongtin.

Trc tin bin i bn r thnh dy cc s nguyn nhdng php tng ng

trn:


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109

Cui cng bin i dy s nguyn ny thnh cc k t thu c bn m sau:

PMTFHTSLSLMJYMTSLYNS gii m bn m ny, ta s bin i bn m thnh dy cc s nguyn

ri tri gi tr cho 11 (rt gn theo modulo 26) v cui cng bin i li dy

ny thnh cc k t.

Ch : Trong v d trn, ta dng cc ch in hoa cho bn m, cc ch

thng cho bn r tin phn bit. Quy tc ny cn tip tc s dng sau ny.

Nhn xt: MDV (theo modulo 26) l khng an ton v n c th b thm theo

phng php vt cn. Do ch c 26 kho nn d dng th mi kho dK c th

cho ti khi nhn c bn r c ngha.

M thay th(MTT)Cho P =C = Z26, K cha mi hon v ca 26 k hiu 0,1,...,25. Vi mi

php hon vK, k hiu

e(x) = (x)

d(y) = -1(y)

trong -1 l hon v ngc ca .

Sau y l mt v d v php hon v ngu nhin to nn mt hm mho:


110/136


111/136

111

Cc hm ny c gi l cc hm Affine (ch rng khi a = 1, ta c MDV).

vic gii m c th thc hin c, yu cu cn thit l hm Affine phi l

n nh. Ni cch khc, vi bt k y Z26, phng trnh

ax + b y (mod 26)

c nghim x duy nht. iu ny tng ng vi:

ax y-b (mod 26).

V y thay i trn Z26 nn y-b cng thay i trn Z26. V vy ta ch cn nghin

cu phng trnh ng d:

ax y (mod 26), (y Z26 ).nh l: ng d thc ax b mod m ch c mt nghim duy nht x Zm vi

mi b Zm khi v ch khi UCLN(a,m) = 1.

V 26=213 nn cc gi tr a Z26 tho mn UCLN(a,26)=1 l a=1, 3, 5,

7, 9, 11, 13, 15, 17, 19, 21, 23 v 25. Tham s b c th l mt phn t bt k

trong Z26. Nh vy, m Affine c 1226 = 312 kho c th. By gita s xt

bi ton chung vi modulo m.

nh ngha 1: Gi s a 1 v m 2 l cc s nguyn. UCLN(a,m) = 1 th ta

ni rng a v m l nguyn t cng nhau. S cc s nguyn trong Zm nguyn t

cng nhau vi m thng c k hiu l (m) (hm ny c gi l hm

Euler).

Mt kt qu quan trng trong l thuyt s cho ta gi tr ca (m) theo cc

tha s trong php phn tch theo lu tha cc s nguyn t ca m. Mt s

nguyn p >1 l s nguyn t nu n khng c c dng no khc ngoi 1 v

p. Mi s nguyn m >1 c th phn tch c thnh tch ca cc lu tha cc

s nguyn t theo cch duy nht. V d 60 = 23

3 5 v 98 = 2 72

.By gi ta s xt xem cc php ton gii m trong mt m Affine vi

modulo m=26. Gi s UCLN(a,26)=1. gii m cn gii phng trnh ng


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112

d y ax+b (mod 26) theo x. T trn ta thy rng phng trnh ny c mt

nghim duy nht trong Z26. iu quan trng l cn xc nh mt thut ton

xc nh nghim ca phng trnh.

nh ngha 2: Gi s a Zm . Phn t nghch o (theo php nhn) ca a l

phn t a-1 Zm sao cho aa-1 a-1a 1 (mod m).

Bng cc l lun tng t nh trn, c th chng t rng a c nghch o

theo modulo m khi v ch khi UCLN(a,m) =1, v nu nghch o ny tn ti

th n phi l duy nht. Ta cng thy rng, nu b = a-1 th a = b-1 .

Sau y chng ta s m t mt thut ton hu hiu tnh cc nghcho caZm vi m tu .

Xt phng trnh ng d y ax+b (mod 26). Phng trnh ny tng

ng vi phng trnh axy-b ( mod 26). V UCLN(a,26) =1 nn a c nghch

o theo modulo 26. Nhn c hai v ca ng d thc vi a-1 ta c

a-1(ax) a-1(y-b) (mod 26) .

S dng tnh cht kt hp ca php nhn modulo, ta c

a-1(ax) (a-1a)x 1x x.

T x a-1(y-b) (mod 26) hay hm gii m l:

d(y) = a-1(y-b) mod 26

M VigenreTrong c hai h MDV v MTT, mi k t sc nh x vo mt k t

duy nht. V l do , cc h mt cn c gi h thay thn biu. Sau y

ta s trnh by mt h mt khng phi l b chn, l h m Vigenre.

Mt m ny do Blaise de Vigenre xut vo th k XVI.

S dng php tng ng A 0, B 1, . . . , Z 25 trn, ta c thgn cho mi kho K vi mt chui k t c di m c gi l t kho. Mt


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113

m Vigenre s m ho ng thi m k t: Mi phn t ca bn r tng

ng vi m k t.

nh ngha: Cho m l mt s nguyn dng cnh no , vi kho K = (k1,

k2, . . . , km) k hiu

ek(x1, x2, . . . ,xm) = (x1+k1, x2+k2, . . . , xm+km)

dk(y1, y2, . . . ,ym) = (y1-k1, y2-k2, . . . , ym-km)

trong tt c cc php ton c thc hin trong Z26.

V d: Gi s m =6 v t kho l CIPHER. T kho ny tng ng vi dy s

K = (2,8,15,4,17).

Gi s bn r l xu: thiscryptosystemisnotsecure

Ta s bin i cc phn t ca bn r thnh cc thng d theo modulo

26, vit chng thnh cc nhm 6 ri cng vi t kho theo modulo 26 nh sau:

Do dy k t ca xu bn m s l

VPXZGIAXIVWPUBTTMJPWIZITWZT

gii m ta c th dng cng t kho nhng thay cho cng, ta tr cho

n theo modulo 26.


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114

Nhn xt: Ta thy rng cc t kho c th vi s di m trong mt m

Vigenre l 26m, vi cc gi tr m kh nh, phng php tm kim vt cn

cng yu cu thi gian kh ln. Trong h mt Vigenre c t kho di m,

mi k t c thc nh x vo trong m k t c th c, mt h mt nh vy

c gi l h mt thay tha biu (polyalphabetic). Ni chung, vic thm m

h thay tha biu s kh khn hn so vic thm m hn biu.

Mt m Hill (Do Lester S.Hilla ra nm 1929).Gi s m l mt s nguyn dng, t P = C = (Z26)

m, tng y l

ly m t hp tuyn tnh ca m k t trong mt phn t ca bn r to ra mk tmt phn t ca bn m. V d nu m = 2 ta c th vit mt phn t

ca bn r l x=(x1,x2) v mt phn t ca bn m l y=(y1,y2). y, y1 v y2

u l cc t hp tuyn tnh ca x1 v x2. Chng hn, c th ly

y1 = 11x1+ 3x2

y2 = 8x1+ 7x2

Hay biu din di dng ma trn

(y1, y2) = (x1, x2)

3

11

7

8

Tng qut: ta c th ly mt ma trn K kch thc mm lm kho, vi

x=(x1, x2, . . . ,xm) P v K K, ta tnh y = ek(x) = (y1, y2, . . . ,ym) theo cng

thc y = xK, hin nhin x=yK-1.

Cho ti lc ny ta ch ra rng c th thc hin php gii m khi v ch

khi K l ma trn kh nghch. Tnh kh nghch ca mt ma trn vung ph

thuc vo gi trnh thc ca n.

Nhn xt: nh thc ca mt ma trn vung cp mm c thc tnh theo

cc php ton scp. Trn Z26 , ta c ma trn K c nghch o theo modulo 26


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khi v ch khi UCLN(det K,26) = 1. Sau y s chng minh ngn gn kt qu

ny.

Trc tin, gi s rng UCLN(det K,26) = 1. Khi det K c nghch o

trong Z26 . Vi 1 i m, 1 j m, nh ngha Kij l ma trn thu c t ma

trn K bng cch loi b hng th i v ct th j, ma trn K* c phn t (i,j)

ca n nhn gi tr bng (-1)det Kij, (K*c gi l ma trn b i s ca K).

Khi c th chng t rng: K-1 = (det K)-1K* , Nh vy K l kh nghch.

Ngc li K c nghch o K-1 . Theo quy tc nhn ca nh thc

1 = det I = det (KK-1) = det K det K-1

Nh vy det K c nghch o trong Z26 .

Cc h m dngTrong cc h mt nghin cu trn, cc phn t lin tip ca bn r u

c m ho bng cng mt kho K tc l xu bn m y nhn c c dng:

y = y1y2. . . = eK(x1) eK(x2 ) . . .

Cc h mt thuc dng ny thng c gi l cc m khi. Mt quan

im khc l mt m dng, mc ch chnh l to ra mt dng kho z = z1z2 . .

v dng n m ho mt xu bn r x = x1x2 . . . theo quy tc:

y = y1y2. . . = ez1(x1) ez2(x1). . .

M dng hot ng nh sau: Gi s K K l kho v x = x1x2 . . . l xu

bn r. Hm fic dng to zi (zi l phn t th i ca dng kho) trong

fi l mt hm ca kho K v i-1 l k tu tin ca bn r:

zi = fi (K, x1 , . . ., xi-1 )

Phn t zi ca dng kho c dng m xi to ra yi = eiz(xi) . Bi vy,

m ho xu bn r x1 x2 . . . ta phi tnh lin tip: z1, y1, z2 , y2 ...

Vic gii m xu bn m y1y2. . . c thc thc hin bng cch tnh

lin tip: z1, x1, z2 , x2 ...


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nh ngha: Mt m dng l mt b (P,C,K,L,F,E,D) tho mn cc iu kin

sau:

1. P l mt tp hu hn cc bn r c th.

2. C l tp hu hn cc bn m c th.

3. K l tp hu hn cc kho c th (khng gian kho)

4. L l tp hu hn cc b ch ca dng kho.

5. F = (f1 f2...) l b to dng kho. Vi i 1

6. fi : K Pi -1L

Vi mi z L c mt quy tc m ez E v mt quy tc gii m tngng dzD. ez : PC v dz: CP l cc hm tho mn dz(ez(x))= x vi mi

bn r x P.

Ta c th coi m khi l mt trng hp c bit ca m dng trong

dng kho khng i: Zi = K vi mi i 1.

Sau y l mt s dng c bit ca m dng:

+ M dng c gi l ng b nu dng kho khng ph thuc vo xu

bn r, tc l nu dng kho c to ra ch l hm ca kho K.

+ Mt h m dng c gi l tun hon vi chu k d nu zi+d= zi vi s

nguyn i 1. M Vigenre vi di t kho m c th coi l m dng tun

hon vi chu k m. Trong trng hp ny, kho l K = (k1, . . . km ). Bn thn

K s to m phn tu tin ca dng kho: zi = ki, 1 i m, sau dng

kho s t lp li. Nhn thy rng, trong m dng tng ng vi mt m

Vigenre, cc hm m v gii m c dng ging nh cc hm m v gii

m c dng trong MDV

ez(x) = x+z v dz(y) = y-z


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Cc m dng thng c m t trong cc b ch nh phn tc l P=

C=L= Z2. Trong trng hp ny, cc php ton m v gii m l php cng

theo modulo 2.

ez(x) = x +z mod 2 v dz(x) = y +z mod 2.

Nu k hiu "0" k hiu gi tr "sai" v "1" k hiu gi tr "ng" th php

cng theo moulo 2 sng vi php hoc c loi tr. Bi vy php m (v gii

m ) d dng thc hin bng vic thit k cc vi mch.

5.3.3 H m DES

tng ca h mt Des l to ra mt thut ton bin i d liu thtphc tp i phng khng th tm ra mi lin quan ca on tin m vibn r cng nh khng th thit lp c mi quan h no gia on tin cm ha v kha.

Thut ton tin hnh theo 3 giai on:

1. Bn r cho trc x (64 bit), mt xu bit x0 sc xy dng bng

cch hon v cc bit ca x theo php hon v cnh ban u IP, khi d liu

chia thnh 2 na (na tri v na phi). Ta vit: x0= IP(X) = L0R0, (L0 gm 32

bit u v R0 l 32 bit cui).2. Tnh ton 16 ln lp theo mt hm xc nh, xc nh LiRi, 1 i 16

theo quy tc sau:

Li = Ri-1 , Ri = Li-1 f(Ri-1,Ki) trong k hiu l php hoc loi tr

ca hai xu bit (cng theo modulo 2), f l mt hm m ta s m tsau.

K1,K2, . . . ,K16 l cc xu bit di 48 c tnh nh hm ca kho K.

( trn thc t mi Ki l mt php chn hon v bit trong K). K1, . . ., K16 s to

thnh bng kho.

3. p dng php hon v ngc IP -1 cho xu bit R16L16, ta thu c

bn m y = IP -1 (R16L16). (Cn ch th t o ca L16 v R16).


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Nh vy thc hin m ha xu bn r x th ta phi xc nh:

- Php hon v IP. (cho sn)

- Tnh c Li v Ri. (Tnh hm f v tnh kha Ki)

- Php hon vo: IP-1 (cho sn)

Tnh bng kha

Kha K l mt xu bit di 64, trong 56 bit l kho v 8 bit kim

tra tnh chn l nhm pht hin sai. Cc bit cc v tr 8,16, . . ., 64 c xc

nh sao cho mi byte cha mt s l cc s "1". Bi vy mt sai st n l c

th pht hin c trong mi nhm 8 bit. Cc bit kim tra b b qua trong qutrnh tnh ton bng kho.

Cc bc tnh kha:

1. Vi mt kho K 64 bit cho trc, ta loi b cc bit kim tra tnhchn l bng cch p dng hon v cc bit ca K theo php hon v cnh

PC-1, sau chia kha thnh 2 phn C0: 28 bit u, D0: 28 bit cui.

PC-1(K) = C0D0

2.

Vi i thay i t 1 n 16: Thc hin dch tri 1 hoc 2 bit phthuc vo s vng lp.

3. Cc bit ca kha c chn ra theo hon v nn PC-2 (hon v la

chn): 56 bit -> 48 bit.

Ci = LSi(Ci-1), Di = LSi(Di-1)


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Tnh bng kho DES.

Tnh hm f

Hm f c hai bin vo: bin th nht A (thnh phn Ri-1) l xu bit

di 32, bin th hai J (kha Ki) l mt xu bit di 48. u ra ca f l mt

xu bit di 32.

Cc bc sau c thc hin:

1. A (na phi ca d liu Ri-1c mrng thnh xu bit di 48

theo mt hm mrng cnh), E(A) gm 32 bit c mrng thnh 48 bittheo hon v m rng E (nhm mc ch to ra d liu c cng kch cvi

dng kha thc hin php ton XOR).

K

PC-1

C0 D0

LS1 LS1

C1 D1 PC-2 K1

.

.

LS16 LS16

C16 D16 PC-2 K16


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2. Tnh E(A)J v vit kt qu thnh mt chui 8 khi 6 bit

B1B2B3B4B5B6B7B8.

3. Mi khi Bj sau c a vo mt hm Sj (S-box): Cj=Sj (Bj) tr

v mt khi 4 bit.

Mi khi c thc hin trn mt hp S ring (B1S1, B8S8,).

Hp S l bng gm 4 hng v 16 dng, vi khi bit c di 6, k hiu

Bi = b1b2b3b4b5b6.

Tnh Sj(Bj) nh sau:

+ Hai bit b1b6 xc nh biu din nh phn ca hng r ca Sj (0r 3)v bn bit (b2b3b4b5) xc nh biu din nh phn ca ct c ca Sj (0 c 15).

+ Sj(Bj)=Sj(r,c); (rhng, c-ct) trong hp S). Phn t ny vit di

dng nh phn l mt xu bit c di 4.

+ Bng cch tng t tnh cc Cj = Sj(Bj), 1 j 8.

4. Ghp cc xu bit C = C1C2... C8 c di 32 c hon v theo php

hon v cnh P, xu kt qu l P(C) c xc nh l f(A,J).

f(Ri-1, Ki) = P(S1(B1) . S8(B8))

Hm f ca DES


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5.4 Mt s h m ho cng khai5.4.1 Khi nim chung

Trong phn ny, chng ta s cp n nhng php bin i ca m

kho cng khai hay m khng i xng, trong h m kho cng khai, mi thc

th A c mt kho cng khai e v tng ng l mt kho ring d. H thng

ny bo m vic tnh d t e l khng th lm c. Kho cng khai xc nh

mt php m ho Ee, trong khi kho ring xc nh php gii m Dd. Bt

k thc th B no mun gi mt vn bn ti A thu c mt bn sao xc thc

t kho cng khai ca A l e, s dng php m ho thu c bn mc=Ee(m) v truyn ti A. gii m c, A p dng php gii m, thu c vn

bn gc m=Dd(c).

Kho cng khai khng cn gi b mt, mt thun li chnh ca nhng h

thng ny l cung cp tnh xc thc kho cng khai, thng d hn so vi vic

bo m phn phi kho b mt. Mc ch chnh ca m kho cng khai l

cung cp s b

73172633 giao trinh ly thuyet thong tin

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