chng 2

1 GIO TRNH

L THUYT THNG TIN

2 CHNG 1: NHNG KHI NIM C BN

2.1 Gii thiu v l thuyt thng tin

Trong th gii ngy nay, chng ta hng ngy phi tip xc vi rt nhiu cc h thng chuyn ti thng tin khc nhau nh: Cc h thng truyn hnh pht 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 pht n ni thu vi nhng mc ch khc nhau. nghin cu v cc h thng ny, chng ta cn phi nghin cu v bn cht thng tin, bn cht ca qu trnh 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. Cc vn thng c gi l cc l thuyt thng tin, l thuyt nng lng.

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 khc nhau, 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 nng lng khc nhau ca thng tin cha gii quyt c m phi cn c mt l thuyt khc l l thuyt thng tin.

L thuyt thng tin l l thuyt nhm gii quyt vn c bn ca qu trnh truyn tin nh vn v ri rc ha ngun, m hnh phn phi xc sut ca ngun v ch, cc vn v m ha v gii m, kh nng chng nhiu ca h thng...

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

2.2 H thng truyn tin

Trong thc t, chng ta gp rt nhiu cc h thng truyn thng tin t im ny ti im khc, trong thc t nhng h thng truyn tin c th m con ngi s dng v khai thc c rt nhiu dng, khi phn loi chng ngi ta c th da trn nhiu c s khc nhau.

2.2.1 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

2.2.2 S truyn tin v mt s khi nim trong h thng truyn tin

nh ngha: Truyn tin(transmission): L qu trnh dch chuyn thng tin t im ny sang im khc trong mt mi trng xc nh. Hai im ny s c gi l im ngun tin (information source) v im nhn tin (information destination). 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 3 khu chnh: Ngun tin, knh tin v nhn tin.

Trong :

Ngun tin: l ni sn sinh ra hay cha cc tin cn truyn i.Khi mt ng truyn c thit lp truyn tin t ngun tin n nhn tin, mt dy cc phn t c s (cc tin) ca ngun s c truyn i vi mt phn b xc sut no . Dy ny c gi l mt bn tin (Message). Nh vy ta cn c th nh ngha ngun tin:

Ngun tin l tp hp cc tin m h thng truyn tin dng lp 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 trng truyn lan. Nh vy ta c th nh ngha knh tin:Knh tin l ni hnh thnh v truyn tn hiu mang tin ng thi y sinh 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 s d 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, ng dn sng.

-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 c cu khi phc thng tin ban u t tn hiu ly c t u 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.

2.3 Ngun tin nguyn thu

2.3.1 Khi nim chung

nh ngha: Ngun tin nguyn thu l tp hp nhng tin nguyn thu m h thng thu 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 hoc l nhng hm bin i theo thi gian v mt hoc nhiu thng s khc nh hnh nh en trng trong l cc to khng gian ca hnh, hoc nh cc thng tin kh tng: trong l cc thng s kh tng nh nhit , m, tc gi,..

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:

Cc tin nguyn thu phn ln l hm lin tc ca thi gian v mc: Ngha l c th biu din mt thng tin no di dng mt hm tn ti trong qung thi gian v ly cc gi tr bt k trong phm vi () trong 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 vy xt 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 l s lng cc tin trong ngun ri rc l hu hn v s lng cc tin trong ngun lin tc l khng m c.

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 thit b m ho. Thit b m ho bin i tp tin nguyn thu thnh tp hp nhng tin thch hp vi c im c bn ca knh nh kh nng cho qua (thng lng), tnh cht tn hiu v tp nhiu.

2.3.2 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 ngu nhin nu vi mt gi tr bt k ca i s gi tr ca mt hm l mt i lng ngu nhin (cc i lng vt l trong thin nhin nh nhit mi trng, p sut khng kh l hm ngu nhin ca thi gian)

Mt qu trnh ngu nhin c quan st bng mt tp cc gi tr ngu nhin. Qu trnh ngu nhin c coi l bit r khi thu nhn v x l c mt tp nhiu cc gi tr c 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) , khi ta biu din qu trnh nh sau:

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 l thi gian vo mng, l thi gian vo mng ca sinh vin th ta thu c mu nh sau:

n v tnh (pht)Vic on trc mt gi tr ngu nhin l kh khn. Ta ch c th tm c quy lut phn b ca cc bin thng qua vic p dng cc qui lut ca ton thng k x l cc gi tr ca cc bin ngu nhin m ta thu c t cc 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 sut c xc nh bi cc cng thc

Trong :

l bin ngu nhin

p(x) xc sut xut hin trong qu trnh ngu nhin, thng c vit l .

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

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

Cc c trng quan trng ca bin ngu nhin

1. Tr trung bnh (k vng ton hc) ca mt qu trnh ngu nhin

2. Tr trung bnh bnh phng

3. Phng sai

4. Hm tng quan

M t mi quan h thng k gia cc gi tr ca 1 qu trnh ngu nhin cc thi im t1, t2

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

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 tr trung bnh theo tp hp v tr trung bnh theo thi gian. Ta cn nghin cu tr trung bnh theo tp hp, tuy vy s gp nhiu kh khn khi tip nhn v x l tc thi cc bin ngu nhin v cc bin ngu nhin lun bin i theo thi gian. tnh tr trung bnh theo thi gian, ta chn thi gian ln quan st cc bin ngu nhin d rng hn v c iu kin quan st v s dng cc thut ton thng k, khi vic tnh cc gi tr trung bnh theo thi gian c xc nh bi cc cng thc:

Tr trung bnh bnh phng:

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

Tng 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 st m khng ph thuc gc thi gian quan st.

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

Tm li: nghin cu nh lng ngun tin, h thng truyn tin m hnh ho 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. Trong cc h thng thng tin thoi, truyn thanh, truyn hnh vi cc tn hiu iu bin, iu tn thng thng ta gp cc ngun nh vy.

2. Qu trnh ngu nhin ri rc: l qu trnh ngu nhin lin tc sau khi c lng t ho theo mc tr thnh 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 ho theo thi gian hoc trong h thng thng tin c xung lng t ho.

2.4 H thng knh tin

2.4.1 Khi nim

Ta bit rng, cho n nay khoa hc tha nhn rng: Vt cht ch c th dch chuyn t im ny n mt im khc trong mt mi trng thch hp v di tc ng ca mt lc thch hp. Trong qu trnh dch chuyn ca mt ht vt cht, nhng thng tin v n hay cha trong n s c dch chuyn theo. y chnh l bn cht ca s lan truyn thng tin.

Vy c th ni rng vic truyn tin chnh l s dch chuyn ca dng cc ht 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 tn hiu v lan truyn n i.

Vic tn hiu lan truyn trong mt mi trng xc nh chnh l dng cc ht vt cht chu tc ng ca lc, lan truyn trong mt cu trc xc nh ca mi trng. Dng vt cht mang tin ny ngoi tc ng dch chuyn, cn chu tc ng ca cc lc khng mong mun sn c trong cng nh ngoi mi trng v chu va p vi cc ht ca mi trng. y cng chnh l nguyn nhn lm bin i dng vt cht khng mong mun hay l nguyn nhn gy ra nhiu trong qu trnh truyn tin.

Nh vy: Knh tin l mi trng hnh thnh v truyn lan tn hiu mang tin, trong knh din ra s truyn lan ca tn hiu mang tin v chu tc ng ca tp nhiu.2.4.2 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, c bn c th tng qut ho v knh.

Khi tn hiu i qua mi trng do tc ng ca tp nhiu trong mi trng s lm bin i nng lng, dng ca tn hiu. Mi mi trng c mt dng 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 ngun nhiu cng nghip, v tr sinh ra, lun lun tn ti trong cc mi trng truyn lan tn 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 b phn mi trng truyn lan khi tn hiu i qua. N lm nhanh, chm tn hiu (thng sng ngn) lm tng gim bin tn hiu (lc to, lc nh, c lc tt hn). Mi trng gm c nhiu cng v nhiu nhn

2.4.3 M t s truyn tin qua knh:

Ta c biu thc m t nhiu trong trng hp tng qut

Trong thc t, ngoi cc nhiu cng v nhiu nhn, tn hiu cn chu tc ng ca h s c tnh xung ca knh do :

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

Tn hiu vo ca knh truyn hin nay l nhng dao ng cao tn vi nhng thng s bin i theo quy lut ca thng tin. Cc thng s c th l bin , 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 theo thng tin nh bin xung, tn s lp li, thi im xut hin. Trong trng hp dao ng lin tc, biu thc tng qut ca tn hiu c dng sau:

Trong l bin , : tn s gc, : gc pha, cc thng s ny bin i theo quy lut ca thng tin mang tin v nhiu tc ng s lm thay i cc thng s ny lm sai lc thng tin trong qu trnh truyn.

Theo m hnh mng 2 ca ca knh tin, k hiu l xc sut nhn c tin khi pht i tin , nu u vo ta a vo tin vi xc sut xut hin l ta nhn c u ra mt tin vi xc sut xut hin ng vi . Vi yu cu truyn tin chnh xc, ta cn phi m bo phi l tin nhn c t tc l . iu ny ch c c khi knh khng c nhiu. Khi knh c nhiu, c th trn u ra ca knh chng ta nhn c mt tin khc vi tin c pht, c ngha l 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 l mt tham s c trng cho c tnh truyn tin ca knh.

2.5 H thng nhn tin

Nhn tin l u cui ca h thng truyn tin. Nhn tin thng gm c b nhn bit thng tin c pht v x l thng tin. Nu b phn x l thng tin l thit b t ng ta c mt h thng truyn tin t ng.

V tn hiu nhn c u ra ca knh l mt hn hp tn hiu v tp nhiu xy ra trong knh, nn ni chung tn hiu ra khng ging vi tn hiu a vo knh. Nhim v chnh cn thc hin ti nhn tin l t tn hiu nhn c phi xc nh c no c a vo u vo ca knh. Bi ton ny c gi l bi ton thu hay phc hi tn hiu ti im thu.

2.6 Mt s vn c bn ca h thng truyn tin

Cc vn l thuyt thng tin cn gii quyt trong qu trnh truyn tin l: hiu sut, chnh xc ca qu trnh truyn tin trong .

2.6.1 Hiu sut ( tc truyn tin)

L lng tin tc cho php truyn i trong mt n v thi gian vi sai st cho php.

2.6.2 chnh xc: (hay kh nng chng nhiu ca h thng)

L kh nng gim ti a sai nhm thng tin trn ng truyn, yu cu ti a vi bt k mt h thng truyn tin no l thc hin c s truyn tin nhanh chng v chnh xc. Nhng khi nim v l thuyt thng tin cho bit gii hn tc truyn tin trong mt knh tin, ngha l khi lng thng tin ln nht m knh cho truyn qua vi mt sai nhm nh ty .

Trong nhiu trng hp ngun tin nguyn thy l lin tc nhng dng knh ri rc truyn tin. Vy ngun tin lin tc trc khi m ha phi c ri rc ha. xc minh php bin i ngun lin tc thnh ngun ri rc l mt php bin i tng ng 1 1 v mt thng tin, trc ht ta kho st c s l thuyt ca php ri rc ha gm cc nh l ly mu v quy lut lng t ha.

2.7 Ri rc ha mt ngun tin lin tc

Trong cc h thng truyn tin m thit b u v cui l nhng thit b x l thng tin ri rc (v d my tnh s) nh cc h thng truyn s liu th khng cho php truyn trc tip tin lin tc. Do vy nu cc ngun tin l lin tc, nht thit trc khi a tin vo knh phi thng qua mt php bin i lin tc thnh ri rc. Sau s p dng cc phng php m ha p ng c cc ch tiu k thut ca h thng truyn tin c th.

Php bin i ngun tin lin tc thnh ri rc gm hai khu c bn:

Khu ri rc ha theo thi gian hay l khu ly mu.

Khu lng t ha.

C s l thuyt ca php bin i ny gm cc nh l ly mu v lut lng t ha nh sau.

2.7.1 Khu ly mu

Gi s ngun tin lin tc dng tn hiu c biu din bng hm tin ph thuc thi gian

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 thay th tng ng. Tng ng y l v ngha thng tin, ngha l hm thay 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 m di tc ng ca in p no . Thi gian ng mch ca r le l thi gian ly mu , chu k ly mu l , tn sut ly mu l . T lin tc, ta thu c theo ngha ri rc (Hnh 1.1)

u(t) T ( S(t)

S *(t)

Hnh 1.1

Trong k thut, vic ly mu phi tha mn mt s iu kin ca nh l ly mu trong khng gian thi gian cho qu trnh ngu nhin c bng tn hn ch. Sau y chng ta xt mt s khi nim Bin i Fourier: hm c gi l c bin i Fourier l nu:

Gi s c tn hiu lin tc c bin i Fourier l c gi l c bng tn hn ch nu vi , trong l tn s cao nht ca tn hiu . Mt tn hiu nh th c biu din mt cch duy nht bi nhng mu ca vi tn s ly mu l vi . Ta thy ngoi min tn s nng lng coi nh bng 0 nn:

Tn hiu c bng tn hn ch c ly mu vi tn s ly mu l c th khi phc li t cc mu ca n theo cng thc ni suy sau:

trong : l cc mu ca ly ti vi

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 kt qu trn c pht biu bi nh l sau y:

nh l ly mu Shanon: Hm trong khong hon ton c xc nh bng cch ly mu vi tn s ly mu .

2.7.2 Khu lng t ho

Gi thit hm tin bin thin lin tc vi bin ca n thay i trong khong . Ta chia khong thnh khong:

Nh vy hm tin lin tc qua phng php ri rc s bin i thnh c dng bin i bc thang gi l hm lng t ho vi mi mc lng t S la chn cc mc thch hp s gim s sai khc gia v .

S(t)

S(t)

(i

Hnh 1.2

Php bin i thnh c gi l php lng t ho. (i gi l mc lng t ho.

Nu , ta c qui lut lng t ho u ngc li ta gi l lng t ha khng ng u. Do s bin thin trong thc t thng l khng u nn ngi ta thng dng qui lut lng t khng u. Vic chia li lng t khng u ny ph thuc vo mt xc sut cc gi tr tc thi ca . Ta thng chn sao cho cc gi tr tc thi ca trong phm vi l hng s. V mt thng k, php lng t ha chnh l vic to mu phn khong vi di khong l v ng vi mi khong xc nh tn s xut hin ca tn hiu trong khong, khi ta nhn c bng phn khong ca tn hiu tng ng sau khi ri rc ha.

Tm li: Vic bin mt ngun lin tc thnh mt ngun ri rc cn hai php bin i: ly mu v lng t ho. Th t thc hin hai php bin i ny ph thuc vo iu kin c th ca h thng:

Lng t ho sau ly mu: BX, TX

Ly mu sau lng t ho.

Thc hin ng thi hai php trn.

2.8 iu ch v gii iu ch2.8.1 iu ch

Trong cc h thng truyn tin lin tc, cc tin hnh thnh t ngun tin lin tc c bin i thnh cc i lng in (p, dng) v chuyn vo knh. Khi mun chuyn cc tin y qua mt c ly ln, phi cho qua mt php bin i khc gi l iu ch.

nh ngha: iu ch l php bin i nhm chuyn thng tin ban u thnh mt dng nng lng thch hp vi mi trng truyn lan sao cho nng lng t b tn hao, t b nhiu trn ng truyn tin.Cc phng php iu ch

Cc phng php iu ch cao tn thng dng vi tn hiu lin tc

iu ch bin AM (Amplitude Modulation)

iu ch n bin SSB (Single Side Bande)

iu tn FM (Frequency Modulation)

iu pha PM (Phase Modulation)

Vi tn hiu ri rc, cc phng php iu ch cao tn cng ging nh trng hp thng tin lin tc, nhng lm vic gin on theo thi gian, gi l manip hay kha dch. Gm cc phng php sau. Manip bin ASK (Amplitude Shift Key)

Manip tn s FSK (Frequency Shift Key)

Manip pha PSK (Phase Shift Key)

2.8.2 Gii iu ch

nh ngha: Gii iu ch l nhim v thu nhn lc tch thng tin nhn c di dng mt in p lin tc hay mt dy xung in ri rc ging nh u vo, vi mt sai s cho php.Cc phng php gii iu ch

V 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 c m chng ta c cc phng php lc tin thng thng nh:

Tch sng bin ,

Tch sng tn s

Tch sng pha

3 CHNG 2: TN HIU

3.1 Mt s khi nim c bn

Tn hiu l cc thng tin m con ngi thu nhn c t mi trng bn ngoi thng qua cc gic quan hay cc h thng o lng. V d nh: Sng a chn, nhp tim ca bnh nhn, lu lng ca cc dng sng 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 tng qut . Sau y chng ta s nghin cu cc dng tn hiu c bn.

3.1.1 Tn hiu duy tr:

Th hin s duy tr ca tn hiu vi cng khng thay i c biu hin bng hm s

(2.1)

trong l cng ca tn hiu. Tn hiu duy tr th loi tn hiu khng thay i trong sut qung thi gian, v d ting ca m thanh, nhp pht manip vi gi tr khng i, nh sng vi cng mt cng ,

3.1.2 Tn hiu xung (t ngt)

Biu hin tn hiu xut hin t ngt trong khong thi gian cc nh (xung) vi mt cng cc k ln sau khng xut hin

(2.2)

Tn hiu xung thng rt hay gp trong cc tn hiu o ca cc thit b vt l hay c hc.

3.1.3 Tn hiu iu ho

Biu hin cc loi tn hiu tun hon trong mt khong chu k no , c biu din bng cng thc tng qut

(2.3)

Trong : l bin dao ng, l tn s, l chu k ca dao ng c bn. Dao ng c bn cn c th biu din bng cng thc tng qut hn

(2.4)

Khi ta c th biu din dao ng c bn nh mt vect trong h trc ta cc hay di dng s phc tng qut

vi l n v o.

3.2 Phn tch ph cho tn hiu

Trong 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 nhin bt k, chng ta phi tm cch tch t tn hiu ngu nhin thnh tng tn hiu n sc, vic phn tch gi l php phn tch ph.

Nu tn hiu iu ho c dng phng trnh sau:

,

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

Khi chng ta c dng ph phc

A/2 A/2

(-( 0 (+(Hnh 2.2

3.2.1 Chui Fourier v ph ri rc

nh ngha 1:

Cho 2 hm s lin tc kh tch trn , nh ngha

(2.5)

c gi l tch v hng ca 2 hm trn khng gian . K hiu

(2.6)

c gi l chun ca trn .

nh ngha 2

Cho h hm xc nh lin tc trn .

H c gi l h trc giao nu tha mn iu kin

(2.7)

H c gi l h trc chun nu tha mn iu kin

(2.8)

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

(2.9)

nh ngha 3:

Cho h l mt h trc giao v l mt hm s bt k xc nh lin tc trn , khi khai trin

(2.10)

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

xc nh cc h s khai trin, ta nhn hai v vi v ly tch phn trn on , ta c

Do tnh cht trc giao ca h ta thu c

Hay

Tc l

(2.11)

Cng thc (2.7) l cng thc xc nh h s khai trin Fourier trong trng 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 h trc giao khc nhau

V d 1: Xt h trn on

Ta c

,

Tc l

Hay ni cch khc, h l trc giao trn on . Khi xt hm bt k, ta lun c khai trin trong

Hon ton tng t, ta cng chng minh c cc h hm , l cc h trc giao trn cc on tng ng.

Tng qut, c th chng minh rng h trc giao trn on

V d 2: Gi s quan st tn hiu tun hon vi chu k trong khong thi gian , xt h hm

Ta c th chng minh rng h hm trc giao trn on tc l

Khi s dng phng php khai trin Fourier, ta khai trn hm thng qua h hm trc giao

Trong :

Hay

Trong

Trong trng hp tn hiu l hm s chn tc l hm s l hm s l, khi h s Khi

.

Hon ton tng t, nu tn hiu l hm s l tc l hm s l hm s l, khi h s Khi

.

Nhn xt:

+ Vi mt tn hiu tun hon vi chu k th h hm c chn l h trc giao tng qut trn on trong nu tn hiu l chn th h trc giao c xc nh l cn nu tn hiu l l th h trc giao c xc nh l

+ i vi mt tn hiu bt k th chng ta cn phi xc nh chu k ca tn hiu cng nh tnh cht chn l ca tn hiu trc khi khai trin.

V d 3: Phn tch ph cho tn hiu l dy xung sau:

A -( -(/2 (/2 (

Hnh 2.3

Ta c chu k ca tn hiu l . Xt trn on , khi

Tn hiu l hm chn. S dng cc cng thc khai trin vi h trc giao ta c trong

Hay

Nh vy ta c khai trin

2A/ (

A/2

2A/3 ( 0

0 2(/T 4(/T 6(/T 8(/T

Hnh 2.4

Ph ca tn hiu c m t bi hnh 1.6

3.2.2 Tch phn Fourier v ph lin tc

Vi tn hiu lin tc ta c hm trong ph thi gian tng ng vi trong ph tn s. S dng cng thc khai trin Fourier trong trng hp tng qut, ta c:

(2.12)

Ngc li ta c:

(2.13)

Tng t nh xt vi ta c ph ca nh sau

Ph phc:

.

Ph bin :

.

Ph pha:

.

V d:

Xt mt xung vung sau:

A

S(j() =

-(/2 (/2

Ta c:

Hnh 2.5

Nh vy ph

A(

0 2(/( 4(/( 6(/( Hnh 2.6

3.2.3 Ph cc 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 theo qui lut ca tn hiu thng tin. Tn hiu cao tn c dng:

(2.14)

Ta c th iu ch 2 thng s bin v gc . Vi gc ta c th iu ch theo tn s (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.

3.2.3.1 Phng php iu bin

Trong phng php iu bin, ta bin i bin ca tn hiu cao tn theo qui lut ca thng tin tc l bin i c cha lng tin cn truyn, cn tn s v gc pha khng i.

Gi s lng tin cn truyn l , khi ta c cng thc bin i:

(2.15)

Trong : c gi l h s iu ch, trong k thut iu ch, thng tin iu ch m bo chnh xc, ta cn chn . Hm s 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 php phn tch ph nghin cu chng trc.

Gi s l hm n sc c dng mt dao ng iu ho

Khi ta c

Nh vy tn hiu qua qu trnh iu bin s gm ba thnh phn, Mt thnh phn dao ng vi tn s mang v 2 thnh phn dao ng vi tn s bin . Bin ca tn s bin bng nhau v bng .

a0

M0/2

M0/2

(0 - ( (0 (0 + (Hnh 2.7

Trong trng hp tn hiu l khng n sc th tn hiu iu bin l mt min, khng phi l ph vch (Hnh 1.9)

th vc t ca tn hiu iu bin nh sau:

C

D

(

A -(B

a0

(0

(

Hnh 2.8

Trong

OA: Tn hiu mang

AB, AC: Tn s bin

OD: Tn hiu iu ch

Nhn xt:

OD max khi AD=AB + AC = Ma0 ( OD = a0 + Ma0 = a0 (1+M). khng nhiu th AD ( a0 hay M ( 1.

OD // OA: Th tn hiu hm tin l n sc v ph l ph vch nu hm tin khng n sc th ph l mt min.

Theo th th bin sng mang (a0) ln chim nhiu hn 70% nng lng nn thng nn tit kim nng lng gim hao ph.

3.2.3.2 Phng php iu tn

Trong phng php iu tn, ngi ta bin i tn s ca sng mang theo tn hiu ca hm tin, tc l

.

Trong h s

Xut pht t cng thc tch phn

Qua qu trnh iu tn, ta nhn c

Gi s sng mang c dng v hm tin l hm n sc

. Khi qua phng php iu bin

Nh vy qua qu trnh iu tn, pha ca sng mang c tch thnh 2 thnh phn cha tn s ca sng mang v thnh phn cha thnh phn tin .

3.2.3.3 Phng php iu pha

Tng t nh phng php iu tn, phng php iu pha bin i gc pha c cha hm tin cn bin v tn s khng i. Ta c cng thc bin i trong trng hp tn hiu n sc:

Tc l

Nhn xt: V hnh thc th c th coi tn hiu iu tn, iu pha ging nhau trong cng thc tng qut sau y:

(2.16)

Tn hiu iu tn th , tn hiu iu pha th

.

3.2.4 Phn tch tn hiu ngu nhin

Do 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 tren c s phn tch mi tng quan gia cc i lng ngu nhin ca l thuyt xc sut thng k.

3.2.4.1 Phng php phn tch tng quan

Nh chng trc gii thiu, tn hiu ngu nhin c thi gian tn ti hu hn ph thuc vo . Hm tng quan c tnh theo cng thc:

(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 xt trong qu trnh dng theo ngha rng th hm c tnh nh gi tr trung bnh ca v tc l

(2.18)

Hm tng quan c mt s tnh cht nh sau:

1. Hm tng quan l mt hm chn

2. Tr s hm tng quan khi trng vi cng sut trung bnh ca qu trnh:

.

3. Gi tr hm tng quan khi t gi tr cc i

4. Nu hm tng quan tha mn iu kin

Th gia v khng tn ti tng quan thng k

5. Khi th gia v s c lp vi nhau khi hm tng quan s dn ti 0.

th m t hm tng quan c dng nh hnh v

Bx(0)

Hnh 2.9 (3.2.4.2 Phng php phn tch ph

Quan st cc qu trnh ngu nhin ta ch c th xc lp c ph chy

(2.18)

Hm tng quan:

(2.19)

Trong

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

(2.20)

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

(2.21)

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

;

(2.22)

Nu ( = 0 th:

3.3 Nhiu trng

Cc 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 ph rt rng gi l nhiu trng. Nhiu l thnh phn khng th b qua khi nghin cu v cc knh, nhiu trng cng l mt loi tn hiu ngu nhin. Qua o c nghin cu ta tm c cng thc tnh mt phn b xc sut ca nhiu theo quy lut ca phn phi chun Gauss

(2.23)

Trong ( c gi l cng sut trung bnh ca nhiu. T ta thy quy lut phn b xc sut ca nhiu c xc nh bi hm phn phi xc sut

(2.24)

Trong gi l tr s tng i ca nhiu,

i vi hm phn phi, ta c cc tnh cht sau y

l hm l

, nh vy c tnh hi t

Dng phng php phn tch ph kho st nhiu, ta coi nhiu trng nh mt hm ngu nhin trong khong . Xt trong mt on di c xung. Ngi ta phn tch v thu c kt qu:

trong

Khi ta gi l ph nng lng ca nhiu c xc nh theo ca tng xung, trong thc t nhiu n mt gi tr no s gim nh khi

G(()

G(0)

(Hnh 2.10

4 CHNG 3: LNG TIN, ENTROPI NGUN RI RC

4.1 o thng tin

4.1.1 Khi nim o:

i 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 l o. 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 tr o c phi cng cao.

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

4.1.2 o thng tin.

Khi nghin cu v thng tin, hin nhin y cng l mt i lng vt l, v vy chng ta cng phi xc nh mt o cho thng tin. xy dng o 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 khi n 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 xut hin ca tin tc.

K hiu tin l vi xc sut xut hin l . Khi hm o k hiu l l hm t l nghch vi xc sut .

Mt tin s l khng c ngha nu chng ta bit v n hay xc sut xut hin . Trong trng hp ny o phi bng khng tc l .

Xt 2 tin l c lp thng k vi xc sut xut hin tng ng l khi tin l tin khi xut hin ng thi 2 tin cng mt thi im. Do theo tnh cht tuyn tnh, chng ta phi c

.

Nh vy xy dng hm o thng tin, ta thy hm phi l hm khng m v tha mn ng thi c 3 iu kin nu. D thy trong tt c cc hm ton hc bit th nu chn

th tt c cc iu kin u c tha mn bi v

l hm s nghch bin vi xc sut.

vi c lp.

Xut pht t nhng l do trn, trong l thuyt thng tin, hm s

(3.1)

c chn lm o thng tin hay lng o thng tin ca mt tin ca ngun.

Trong cng thc xc nh o thng tin ny, c s ca hm logarit c th chn ty tha mn tuy nhin ngi ta thng dng cc n v o nh sau:

Bit hay n v nh phn khi c s l 2.

Nat hay n v t nhin khi c s l e.

Hartley hay n v thp phn khi c s l 10.

4.2 Lng tin ca ngun ri rc

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

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

Mt tin i vi ngi nhn c hai phn, hay hai ni dung.

bt ng ca 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 l tnh hm ca tin, khng nh hng n cc vn c bn ca h thng truyn tin nh tc hay chnh xc. N chnh l ngha ca nhng tin m con ngi mun trao i vi nhau thng qua vic truyn tin.

bt ng ca tin li rt lin quan n cc vn c bn ca h thng truyn tin. V d: mt tin cng bt ng, s xut hin ca n cng him, th r rng thi 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 ng ca 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 t c 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 th c m t bi cc quy lut thng k.

V mt truyn tin ta ch quan tm n bt ng ca tin hay xc sut xut hin cc k hiu. nghin cu vn ny ta dng cc quy lut thng k. Php bin i tng qut trong h thng truyn tin l php bin i cu trc thng k ca ngun

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 hnh thnh bi nhng dy k hiu hu hn l mt k hiu bt k thuc ngun c gi i thi im . Tin c dng: vi xc sut xut hin .

V mt ton hc ngun tin cng ng ngha vi mt trng xc sut hu hn gm cc im . trong khng gian chiu. l tng s cc im c tnh bng .

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 tc no 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.

(

Hnh 3.1

Trong

l ngun u vo vi b ch v phn b xc sut cc k hiu .

l ngun u ra vi b ch v phn b xc sut cc k hiu .

Nu ( l quy lut bin i th ta c mi quan h ( =.

Chng ta c th m t ngun tin u vo bng tp tin v quy lut phn b xc sut cc tin . Trong , l cc tin thuc xy ra cc thi im .

Tng t ngun tin u ra c m t bng tp tin vi quy lut phn b xc sut . Trong , l mt k hiu thuc b ch xy ra tun t thi im .

Cc tin hay c xem nh nhng phn t ca tp hay ; hoc nhng b ca tp tch ct ca tp.

Nh vy v ln lt l phn t ca tp:

vi

vi

Ngun tin c xem nh khng gian im ri rc nhiu chiu, mi mt im i din cho mt tin. Php bin i ngun chuyn mt khng gian tin ny sang mt khng gian tin khc. V d php ri rc ha, chuyn mt khng gian tin lin tc thnh khng gian tin ri rc.

Php m ha chuyn mt khng gian tin ri rc chiu thnh mt khng gian tin chiu vi mt nh x mt i mt gia cc tin.

Php bin i trong knh cng c th c xem nh nhng php bin i ngun khc, tuy nhin v c tc ng ca nhiu nn s chuyn i gia cc tin thng thng khng phi l mt mt

V d 1: Php bin i trong knh nh phn i xng. Tp vo gm hai k hiu , tp ra gm hai k hiu . Php bin i trong knh c th gy ra nhng kt qu sau:

Nu vo l thu c ta c

Nu vo l thu c ta c

Nu vo l thu c ta c

Nu vo l thu c ta c

Kt qu bin i cc tin trong knh c th c xem nh cc phn t ca tp tch . Quy lut phn b xc sut cc tin ca tp tch ty thuc vo quy lut phn b xc sut ca tp vo v tnh cht thng k ca knh ngha l xc sut chuyn i t tin thnh tin : . .

Nu c ngun tin vi s k hiu bt k . u ra thu c ngun . Ta xt nh sau:

x1 y1

yn

x2 y2

(xi,yj)

yj

xi yj

y2

y1

xn yn

x1 x2 xi xn

Hnh 3.2

Php bin i trong knh to ra mt ngun mi , vi cc tin l cc cp , trong , , theo quy lut phn b xc sut . Cc tin l cc im ri rc trn mt phng .

Theo l thuyt xc sut, s lin h gia cc xc sut ca cc phn t trong tp v c th tnh nh sau:

V d 2: Php m ha nh phn; cho mt ngun tin dng m nh phn m ha ngun tin, vi php m ha nh sau:

Trong ; ; cc m hiu thit lp nh trn l cc phn t ca mt tp tch v c i biu bng nhng im ri rc trong 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:

p dng cc biu thc trn trong vic xc nh xc sut ca m hiu, khi nhn c u ra ca b m ha ln lt cc k hiu ca mt dy no . Gi s u ra nhn c dy . Hy tnh xc sut ca tin sau khi nhn c ln lt cc k hiu ca dy.

Xc sut ca tin sau khi nhn c k hin c tnh theo xc sut c iu kin

Trong

Xc sut ca tin sau khi nhn c k hiu , tnh theo xc sut c iu kin sau:

Trong

Xc sut ca tin sau khi nhn c k hiu , , ch c kh nng xy ra l : , cn li cc tin khc u c xc sut bng 0. Kt qu tnh ton c cho trong bng 3.1

Xc sut ca tin sau khi nhn c k hiu

1/4

000

1/4

1/24/50

1/8

000

1/8

1/400

1/16

000

1/16

1/81/51

1/16

000

1/16

1/800

Bng 3.1

4.2.2 Lng tin ring, lng tin tng h, lng tin c iu kin

Nh trong phn trc ta cp v o thng tin, hm loga c chn nh gi, nh lng cc lng tin. i vi mi tin ca ngun u c lng tin ring nh trn ta bit:

Error! Objects cannot be created from editing field codes.Nu ngun Error! Objects cannot be created from editing field codes. thng qua mt php bin i tr thnh ngun v d thng qua s truyn lan trong knh th php bin i c th khng phi l 1-1.

u vo ca knh l cc tin , cc tin trong qu trnh truyn lan trong knh b nhiu ph hoi, lm cho s chuyn i t ngun sang ngun khng phi l 1-1 . Mt tin c th chuyn thnh mt tin u ra ca knh vi nhng xc sut chuyn i khc nhau ty thuc theo tnh cht nhiu trong knh.

Bi ton truyn tin trong trng hp ny t ra l: Cho bit cu trc thng k ca ngun Error! Objects cannot be created from editing field codes., tnh cht tp nhiu ca knh biu th di dng cc xc sut chuyn i ca tin, khi nhn c mt tin , hy xc nh tin tng ng ca ngun Error! Objects cannot be created from editing field codes..

y l bi ton thng k, li gii khng nh l khng c c. Li gii tm c s c dng: Vi tin nhn c, tin no ca ngun X c nhiu kh nng c pht i nht.

Mun gii quyt vn ny ta ln lt qua hai bc(1) Tnh cc lng tin v mt tin bt k cha trong tin nhn c, lng tin gi l lng tin tng h gia v .

Mun xc nh lng tin tng h ta phi tm lng tin ban u c trong , sau khi thc hin qu trnh truyn tin ta tm lng tin cn li trong , hiu hai lng tin ny cho ta thy lng tin truyn t sang .

Lng tin ban u l lng tin ring c xc nh bng xc sut tin nghim ca tin: Error! Objects cannot be created from editing field codes.Lng tin cn li ca sau khi nhn c c xc nh bng xc sut hu nghim: , lng tin ny cn gi l lng tin c iu kin, trong qu trnh truyn tin, lng tin chnh l lng tin b tp nhiu ph hy khng n u thu c.

Nh vy lng tin tng h c tnh theo cng thc sau:

=

(2) em so snh cc lng tin tng h vi nhau, v lng tin no cc i s cho bit tin c kh nng nhiu nht chuyn thnh trong qu trnh truyn tin.

Trong trng hp phc tpM rng khi nim lng tin tng h trong trng hp m ha hay php bin i phc tp hn. Lc lng tin tng h cng c xc nh theo cc cng thc xc sut tin nghim v xc sut hu nghim ca tin ang xt. V d mt php bin i kp .

Lng tin ban u ca c xc nh theo xc sut ban u hay xc sut tin nghim. Sau khi nhn c tin , xc sut ca tin tr thnh xc sut c iu kin , v cui cng khi nhn c v th xc sut ca l .

Nu ta xem l xc sut tin nghim v l xc sut hu nghim th ta li tr li trng hp bin i n gin ni trn v c lng tin tng h gia v : .

Nu ta xem l xc sut tin nghim v l xc sut hu nghim, ta s xc nh c lng tin tng h gia v vi iu kin bit ;

Nu ta xem l xc sut tin nghim v l xc sut hu nghim th lng tin tng h gia v cp l:

Mt cch trc gic c th nhn thy lng tin v cha trong cp phi bng lng tin v cha trong cng vi lng tin tng h v cha trong khi bit . iu ny c th c xc minh mt cch d dng nh sau:

Nu thay th t cc tp th s c cc biu thc mi nhng ngha hon ton khng thay i.

V d: Tnh lng tin tng h gia tin v cc k hiu ln lt nhn c, trong v d m ha nh phn nu trong v d trc

Tng lng tin v c bit ln lt khi nhn : log16

4.2.3 Tnh cht ca lng tin(1) Lng tin ring bao gi cng ln hn lng tin v n cha trong bt k k hiu no c lin h thng k vi n.

Do vy khi v c lp thng k th lng tin tng h bng 0. Lng tin tng h cc i khi v bng lng tin ring.

=( Error! Objects cannot be created from editing field codes.iu ni trn cho thy lng tin tng h m t s rng buc gia xi v yj, nu s rng buc y cng cht ch th lng tin v cha trong cng ln, hay lng tin v cha trong cng tng ln. T cng c th gii thch ngha ca lng tin nh l lng tin tng h cc i gia v .

(2) Lng tin ring l mt i lng lun dng (v nn ). Nhng lng tin tng h c th dng, c th m do ph thuc lng tin c iu kin.

(3) Lng tin ca mt cp (xiyj) bng tng lng tin ring ca tng tin tr i lng tin tng h gia chng.

Khi v c lp thng k

i vi trng hp ngun phc tp U=XYZ:

Gii thch lng tin ring c iu kin , cng tng t nh gii thch lng tin ring. Lng tin ring c iu kin chnh l lng tin tng h vi cng mt iu kin xc nh mt cch n tr gia cc tin vi nhau.

Lng tin tng h c th phn thnh tng ca nhng lng tin tng h khc:

4.2.4 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 ch nh gi c 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 tham gia. Trong thc t iu ta quan tm l gi tr tin tc ca mt tp hp ch khng phi gi tr tin tc mt phn t no trong tp hp.

nh gi hon chnh gi tr tin tc ca mt tin 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 mt k hiu bt k ca ngun cho.

(3.6)

V d: Mt ngun tin c hai k hiu l , vi xc sut v . Nh vy khi nhn mt tin ta bit gn nh chc chn l l , tin ny khng cn bt ng nn gi tr tin tc rt nh. Th nhng xt lng tin ring ca :

(bit/k hiu)

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

(Bit/k hiu)

Nh vy lng tin trung bnh rt nh phn nh ng thc t gi tr ca ngun tin.

Ta cng c khi nim lng tin tng h trung bnh:

(3.7)

Lng tin c iu kin trung bnh:

(3.8)

Ta c quan h gia cc lng tin trung bnh:

i vi trng hp ngun phc tp

(3.9)

4.3 Entropi ca ngun ri rc

4.3.1 Khi nim entropi

Khi ta nhn c mt tin ta s nhn c mt lng tin trung bnh, ng thi bt ng v tin cng c gii thot, cho nn bt ng v lng tin v ngha vt l tri ngc nhau, nhng v s o th ging nhau v c xc nh theo cng thc sau:

bt ng trung bnh ca mt tin thuc ngun (entropi ca ngun) c xc nh theo cng thc sau:

(3.11)

4.3.2 Tnh cht ca entropi

(1) Entropi l mt i lng khng m : H(X) ( 0

(2) H(X) = 0 khi ngun c mt k hiu bt k c xc sut xut hin bng 1 v xc sut xut hin tt c cc k hiu cn li bng khng.Ngha l ngun c mt tin lun c xc nh, nh vy gi tr thng tin ca ngun bng khng.

(3) Entropi cc i khi xc sut xut hin ca cc k hiu bng nhau

Chng minh:

Cc gi tr lm cc i hm

Vi iu kin cng chnh l cc gi tr lm cc i hm

(= c gi l h s Lagrang. Cc gi tr p(x) lm cc i hm ( tha mn iu kin:

vi mi gi tr p(x).

hay =0 vi mi gi tr .

Tc l cc gi tr bng nhau vi tt c cc tin ca ngun, v khi gi tr cc i ca s l nu ly n v l bt v ngun c m tin.

Nu ngun c m k hiu ng xc sut th xc sut xut hin mt k hiu l 1/m khi :

4.3.3 Entropi ng thi v Entropi c iu kin

4.3.3.1 Entropi ng thi

Entropi ng thi l bt ng trung bnh ca mt cp bt k trong tp tch . Theo nh ngha v entropi c:

(3.12)

4.3.3.2 Entropi c iu kin

Khi cn nh gi s rng buc thng k gia cc cp ta dng khi nim entropi c iu kin hoc . l bt nh trung bnh ca mt k hiu bt k khi bit bt k mt k hiu . Xut pht t cc xc sut c iu kin v cng nh theo nh ngha v entropi ta c biu thc nh ngha sau:

(3.13)

So snh vi cc biu thc nh ngha cho cc entropi, ta c quan h sau:

(3.14)

Trng hp ngun phc tp

i vi m ha hay truyn tin phc tp hn ta m rng khi nim entropi cho nhng tp tch m s cc tp hp thnh nhiu hn hai, chng hn trng hp ca tp tch ta c nh ngha v entropi ng thi v c iu kin m rng nh sau:

(3.15)

4.3.4 Entropi ngun Markov

Ngun Markov gi vai tr quan trng trong lnh vc truyn thng. N c c trng bi quan h trong l k hiu ca ngun X xut hin thi im n. iu ny c ngha l xc sut to ra mt k hiu no ti thi im n ch ph thuc vo k hiu to ra thi im th n-1 v khng ph thuc vo cc k hiu to ra cc thi im n-2, n-3,

Ti thi im n, ngun c th trng thi j vi xc sut no khi khi thi im n-1 ngun trng thi i.

Xc sut gi l xc sut chuyn i t trng thi i sang trng thi j, trong (m l s tin thuc ngun).

Xc sut ngun trng thi j ti thi im n l:

Biu din di dng ma trn ta c:

Mi quan h trn c th vit:

Nu ngun ang trng thi i th s c mt bt nh v trng thi ca ngun thi im sau, l trng thi j, trng thi ny l mt trong cc trng thi c th ca ngun. Gi tr trung bnh ca bt nh ny c xc nh bi entropi

Nu tnh ti tt c cc trng thi ca ngun, entropi ca ngun l gi tr trung bnh ca entropi ngun X mi trng thi:

4.4 Mi quan h gia lng tin tng h trung bnh v Entropi

Vy:

Suy ra

Nu X,Y c lp thng k th :

V ta cng chng minh c

V d:

Cho s truyn tin:

Hnh 3.3

Bit:

+ Tnh Entropi u vo ca knh:

+ Tnh Entropi u ra:

+ Tnh H(X,Y)

p dng cng thc:

+ Tnh

4.5 Tc lp tin ngun ri rc v thng lng knh ri rc

4.5.1 Tc lp tin

Khi nim

Ngoi thng s c bn ca ngun l Entropi ta thy s hnh thnh thng tin nhanh hay chm a vo knh li tu thuc vo bn cht vt l ca ngun nh qun tnh, phn bit

Cho nn s k hiu lp c trong mt n v thi gian rt khc nhau. V d: con ngi v kt cu ca c quan pht m hn ch nn mt giy ch pht m c t 5-7 m tit trong li ni thng thng, trong khi my in bo c th to ra t 50-70 k hiu trong mt giy.

Nh vy thng s th hai ca ngun l tc thit lp tin R (Lng thng tin ngun lp c trong mt n v thi gian), Tc thit lp tin ti u vo knh bng tch ca entropi vi s k hiu n0 lp c trong mt n v thi gian, trong trng hp dng loga c s hai th n v ca R l bit/sec.

(3.16)Nh vy mun nng cao th hoc l tng n0 hoc l tng . Tng n0 ph thuc vo thit b phn cng. Tng ta c th thay i cu trc thng k ca ngun nh vy s n gin khng tn km.

Ta bit nu xc sut xut hin cc k hiu bng nhau th cc i, do vy ta dng php m ho thc hin vic ny m ho ngun tin ban u thnh ngun tin m ho sao cho xc sut cc k hiu tng ng nhau.

V d:

Cho ngun tin vi xc sut tng ng l:

Nu c mt tin gm cc k hiu: . c cc i phi c xc sut cc k hiu bng nhau bng 1/8

Mun vy ta m ho ngun X trn thnh ngun Ynh sau:

Ta c dy cc k hiu ca tin:

Ngun ny c

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

Ta c dy cc k hiu ca tin:

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

Vy bng php m ho ngun thnh ngun ta c th nng Entropi ca ngun l = 7/4 thnh Entropi m vn m bo lng tin trong cc bn tin c bo ton v c cng gi tr l 14 (bit)

d ca 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:

(3.17)

Ngoi ra cng c th dng d tng i ca ngun nh gi:

(3.18)

4.5.2 Thng lng knh

Thng lng knh l lng tin cc i knh cho i qua trong mt n v thi gian m khng gy ra sai nhm.

Vy:

n v ca thng lng knh l bit/giy. Nh vy

Nhim v ca m ho thng k l bng cch m ho thay i Entropi ca ngun thay i tc lp tin sao cho xp x vi , gi l phi hp gia ngun vi knh v phng din tc truyn tin. Khi truyn tin trong knh c nhiu th nhim v ca m ha l li dng iu kin xy dng m chng nhiu ng thi tng tc lp tin.

4.5.2.1 Thng lng knh ri rc khng c nhiu

Khi knh ri rc khng c nhiu ton b tin tc c thit lp u c th truyn qua knh m khng b sai. Vy u thu ta nhn c lng tin bng vi u vo hay ta c:

l tc lp tin u vo, lng tin ny nhn c nguyn vn u ra. Nu knh c R < C th ta c th m ho tng R sao cho:

C R < ( vi ( nh tu

Ta khng th m ho cho c, l gii hn ca vic m ho. Trong trng hp m ho sao cho c gi l phng php m ho ti u.

Sau khi m ho ta c . Gia v ban u ta c chnh lch gi l d tng i ca ngun.

(3.19)Vy php m ho ti u cng c th coi l phng php lm gim d ca ngun ban u

4.5.2.2 Thng lng knh ri rc c nhiu

Thng thng tc lp tin b hn nhiu so vi thng lng knh, nhim v ca m ha thng k l thay i tc lp tin ca ngun bng cch thay i entropi, tc lp tin tim cn vi thng lng, gi l phi hp vi ngun v knh v phng din tc truyn tin.

Trong trng hp tin nhn c sau khng ph thuc nhng tin nhn c trc, ni cch khc chng c lp thng k vi nhau th chnh xc ca tin truyn i trong knh ch cn b nh hng ca nhiu l gim i, khi tc lp tin ti u ra ca knh c nh ngha nh sau:

(3.20)

v mt ln l lng tin b nhiu ph hy trong mt n v thi gian, vy mun nng cao tc lp tin th nht thit phi thay i thng s ca ngun. Lc lng tin ti a m knh cho i qua khng xy ra sai nhm s l tc lp tin cc i trong knh c nhiu:

(3.21)

d tng icn c th c xc nh theo cng thc sau: ; Hiu qu s dng knh:

5 5.1 Khi nim v iu kin thit lp mTrong cc h thng truyn tin ri rc hoc truyn cc tn hiu lin tc nhng c ri rc ha, bn tin thng phi thng qua mt s php bin i: i thnh s (thng l s nh phn), m ha.. pha pht. pha thu bn tin (gi l t m) phi thng qua nhng php bin i ngc li l gii m (m thm), lin tc ha..

S m ha thng tin cho php ta k hiu ha thng tin hay s dng cc k hiu quy c biu din bn tin dng ph hp cho ni s dng. Chnh nh m ha, ta c th nhnh thy hay hin th c thng tin thng tin c bn cht l cc khi nim (thng tin l s hiu bit ca con ngi). i vi mt h thng truyn tin, vic m ha cho php tng tnh hu hiu v tin cy ca h thng truyn tin, ngha l tng tc truyn tin v tng 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 lp tin tip cn vi kh nng truyn ca knh. Trong trng hp truyn tin trong knh c nhiu, iu cn quan tm nhiu l chnh xc ca s truyn tin, hay cc tin truyn i t b sai nhm. y chnh l nhim v th hai ca m ha.

Trong chng chng ny, trc tin ta cp ti cc khi nim v nh v m: Th no l m hiu? Cc thng s c bn ca m hiu l g? cc iu kin v yu cu i vi m hiu l g?

5.1.1 M hiu v cc thng s c bn

nh ngha: M hiu l mt ngun tin vi mt s thng k c xy dng nhm 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 du hiu ring hay bng ch 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 ho vi cc t m do cc du m to thnh

Cho ngun S (A,P). Khi php m ha l song nh f: A -> M. Trong A: l tp ngun, M l tp cc t m.

- S cc k hiu khc nhau trong bng ch ca m gi l c s m (m) mi k hiu c mt s tr no tu theo cu trc ca b m(v d m nh phn m = 2 v mi k hiu c hai tr l 0 v 1 l loi m c dng rng ri nht..).

- S cc k hiu trong mt t m gi l di t m n. Nu cc t m trong b m c di bng nhau gi l m ng u, di t m khng bng nhau gi l m khng u. M khng u ta c khi nim di trung bnh ca t m tnh nh sau:

(4.1)

: Xc sut xut hin ca tin xi c m ho thnh t m th i

: di t m ng vi tin xi

: Tng s t m ca b m tng ng vi tng s cc tin

S t m chnh l tng s cc t m c trong mt b m sau khi m ho mt ngun no . Vi b m u theo l thuyt ta c

V d: Mt b m nh phn mi t m c di 5 bit ta c

N = 25 = 32 t

Nu ta dng ht cc t m hay N = mn gi l m y

Nu ta dng s t m N < mn gi l m vi

Nh vy trong thc t s dng ta thy c th c hai loi m. Vi m u cc t m hay b sai nhm gia t ny vi t khc nn dng b m ny ta phi nghin cu cch pht hin sai v sa sai. Vi m khng u ta phi chn di cc t m sao cho di trung bnh ca t m l ngn nht gi l m thng k ti u.

Gi tr ring hay cn gi l tr ca mi k hiu m. Mi m hiu c m k hiu m khc nhau, nu c s ca n l m. Mi k hiu m l mt du hiu ring, v chng c gn mt gi tr xc nh theo mt o xc nh, cc gi tr ny c gi l cc gi tr ring ca cc k hiu m v k hiu l a, trong trng hp s, mi k hiu m c gn mt gi tr ring nm trong khong t 0 ti m-1.

Ch s v tr ca k hiu m trong t m: ta gi mt v tr m l mt ch trong t m t mt k hiu m vo . Mt t m c n k hiu m s c n v tr m. Mt v tr m nh phn cn c gi l v tr nh phn hay mt bit. Mt v tr m thp phn cn c gi l mt v tr thp phn hay mt digit. Ch s v tr l s hiu ca v tr m trong t m theo mt cch nh s c th. Hin nay ta thng nh s v tr m bn phi nht ca t m l v tr 0 v sau c dch sang tri mt v tr th ch s v tr tng thm mt.

Trng s v tr (i: i l ch s ca v tr cn xc nh trng s ca n. y l mt h s nhn lm thay i gi tr ca k hiu m khi n nm cc v tr khc nhau. Trng s v tr ny ph thuc vo mi m hiu c th. Trong trng hp s, trng s v tr l ly tha bc ch s v tr ca c s ca m.

Trng s ca t m b: Trng s ca t m l tng cc gi tr ca cc k hiu m c trong t m.

(4.2)

trong ak l gi tr ring ca k hiu m v tr k. Trong trng hp m hiu l h m th trng s ca t m l: b =

Khong cch m D: Khong cch m l khong cch gia hai trng s ca hai t m ty vo vic nh ngha gi tr ring v trng s v tr ca cc k hiu m, ta s c nhng nh ngha khc nhau cho khong cch m.

5.1.2 iu kin thit lp b mTa ni trong phn trn l mi t m l mt t hp m dng m ha mt tin hay mt khi tin ca ngun. Vn l c iu kin no rng buc mt t hp m s c hay khng c dng lm mt t m. Nhng iu kin ny s c gi l iu kin thit lp m. Chng s th hin nh th no v kim tra chng nh th no l vn chng ta s phi xc nh trong chng ny. Trc ht ta s pht biu nhng iu kin thit lp m v sau s i tm th hin ca chng v cch kim tra chng.

iu kin chung

Cc tin truyn i c m ho thnh dy cc k hiu lin tip. Khi nhn tin ta phi gii m c thu c thng tin, mun vy cc k hiu phi c sp xp theo quy lut no tch ng thng tin ban u.

V d: ta c 4 tin a, b, c, d c m ho bng b m nh phn nh sau:

a=00

b=01

c=10

d=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 bng cch t gc tr i chia thnh tng nhm hai k hiu m tng ng. Nh vy b m trn cho php phn tch cc t m mt cch duy nht v c gi l m phn tch c.

Cng tin trn nu ta m bng b m khc:

a=0

b=01

c=101

d=1

Vn ngun trn m theo b m ny ta c: 00001101101. Khi nhn tin v gii m ta c th gii m nh sau: aaaaddbdb hoc aaabcdb. Nh vy tin nhn c s sai lc so vi ngun v vy b m ny khng dng c. Vy khi 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 t m c 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 m 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.

5.2 Cc phng php biu din m5.2.1 Biu din bng bng lit k (Bng i chiu m)y l cch trnh by b m n gin nht. Ngi ta dng bng lit k nhng tin ca ngun v m tng ng ca n. Bng lit k c u im l cho thy c th tc thi tin v t m nhng c nhc im l cng knh v khng cho thy tm quan trng khc nhau ca tng t m.

V d: Biu din m bng bng nh sau:

Tina1a2a3a4a5

T m000110010101011

5.2.2 Biu din bng to m

Mi t m c hai thng s c th xc nh duy nht m khng b nhm ln gia cc t m vi nhau l di n v trng s b, ngha l khng tn ti hai t m bng nhau ng thi c di n v trng s bMt ta m l mt biu din da trn hai thng s ca t m l di t m n v trng s b lp mt mt phng c hai ta , trn mi t m c biu din bng mt im. Trng s b ca t m l tng trng s cc k hiu trong t m.

Trng s b c tnh theo cng thc:

b =

(4.3)

k: L s th t ca k hiu th k trong t m

ak: L tr ca k hiu th k (v d m nh phn c hai tr l 0 v 1)

m: L c s m (m nh phn m = 2)

V d 1: Tnh trng s ca cc t m nh phn sau:

T m 1011 ta c b = 1.20 + 1. 21 + 0. 22 + 1. 23 = 11

T m 011 ta c b = 1. 20 + 1. 21 + 0. 22 = 3

V d 2: Cho cc t m sau:

a1 = 00 n1 = 2 b1 = 0

a2 = 10 n2 = 2 b2 = 2

a3 = 100n3 = 0 b3 = 4

a4 = 101n4 = 3 b4 = 5

nh l. Khng c hai t m m ha hai tin khc nhau ca cng mt b m tha mn ng thi ni = nj v bi = bj ( i ( j )

5.2.3 hnh m

Cc phng php hnh s dng mt hnh biu din mt m hiu. N cho php trnh by m mt cch gn hn cc bng m, ng thi cho thy r cc tnh cht quan trng ca m hiu mt cch trc quan hn. Cc phng php biu din m hiu bng hnh gm c cy m v hnh kt cu. Biu din bng cy m

Cy m l mt th hnh cy biu din m c cc nt v nhnh, cy c mt nt gc duy nht t mt nt c nhiu nht l m nhnh (m l c s m) mi nhnh l tr ca k hiu, mi nhnh kt thc mt nt cao hn. Nt cui l c trng cho mt t m hnh thnh t cc tr trn cc nhnh. Cc t m mc trn c tm qua trng cao hn cc t m mc di.

V d: C cy m nh phn c 5 t m sau:Mc 0

0 1

Mc 1

0 1 0

Mc 2

00 01 0 1

Mc 3

Mc 3

100 0 1

Mc 4

1010 1011

Hnh 4.1

Trong v d trn cho thy khi nhn vo cy m ta bit cy m c phi l cy m ng u hay khng ng u, loi m y hay vi. Cy m trn thuc loi m khng ng u.

hnh kt cuPhng php ny ta dng mt th c hng gm cc nt v cc nhnh, mi nhnh l mt cung c hng. Mi t m l mt vng khp kn theo chiu ca cung i t gc. Phng php ny biu din t m gn nh v trc quan

V d: Biu din b m nh phn bng th:

Gc 1

1or 0 0

0

0

1or 0

1

Hnh 4.2

S ny ta c 5 t m sau: 00, 01, 100, 1010, 1011

hnh kt cu khng nhng dng m t bn thn t m m cn dng xt cch vn hnh thit b m ha v gii m nh l mt hnh trng thi ca thit b.

5.2.4 Phng php hm cu trc m

Phng php ny ni ln mt c tnh quan trng ca m l s phn b cc t m c di khc nhau, k hiu bng : S t m c di l . T hm cu trc c th phn bit c m u hoc khng u. Cng t hm cu trc ta hon ton c th xc nh c b m c tha mn iu kin phn tch c hay khng.

5.3 M c tnh phn tch c, m c tnh prefixTrong mc ny ta xt cc tiu chun c s dng nh gi mt m hiu c tha mn iu kin thit lp m hay khng. Ta bit rng iu kin chung thit lp m l m phi phn tch c cho nn tiu chun thit lp m chnh l tiu chun m phn tch c hay chnh l nhng tiu chun cho php tch ng tng t m t chui m nhn c.

Lu gia t m v tin c m ha c quan h 1-1 th vic gii m pha thu s bao gm vic tch ng t m v chuyn ngc t m thnh tin tng ng.

Vic chuyn t m thnh tin c thc hin nh mt s gii m xc nh.

Vic tch ng cc t m l mt thut ton kim tra tnh ng ca mt s tiu chun c gi l iu kin phn tch ca m hiu. Vic kim tra ny s bt u t k hiu m u tin ca chui cho n khi c th ct c mt t m th n s ct t m v li coi k hiu tip sau lm k hiu u tin ca chui kim tra tip.

Mt trong nhng cch tip cn c bn nht l trong bng m, hy chn 1 t m trong vi phn u ca xu m sau xa phn u ca xu m v ghp k hiu tng ng vo xu gc, qu trnh s dng khi xu m b xa ht.

Thut ton gii m c th m phng nh sau

Procedure Giai_Ma;

Inputst:string;{Xau da ma hoa}

x:array[1..N] of char;{Bang ki hieu}

b:array[1..N] of string;{Bang ma tuong ung}

Output xaugoc:string;{xau goc ban dau}

BEGIN

xaugoc:=;

while length(st)>0 do

for i:=1 to N do

if b[i]=copy(st,1,lenght(b[i])) then

begin

xaugoc:=xaugoc+x[i];

delete(st,1,lenght(b[i]));

end;

END;5.3.1 iu kin m phn tch c

Ta thy khi nhn c mt dy k hiu m c th phn tch t m mt cch duy nht v ng n b m phi tho mn iu kin cn v l: Bt k dy t m no ca b m cng khng c trng vi mt dy t m khc.

V d 1:

Cho b m c cc t m sau: 00 01 11 100 1010 1011

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 c mi phn tch c mt t m. chm gii m c th l hu hn nhng cng 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 m ngn 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 ly phn 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 thc hin ging nh trn cho n khi c mt ct trng th dng

iu kin cn v m c th phn tch l trong ct j ( 2 khng c t hp no trng vi mt t m trong ct 1

r hn v thut ton ta quan st cc v d sau:V d 2:

123

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 cn v l bt k t m no cng khng c trng vi phn u ca t m khc trong cng b m.V d 3:

12345

10010

10011100

01-01

011-0011

Trong cc ct t 2,3,.. ca bng th ny khng c t hp m no trng vi cc t m trong ct 1, nhng c th in cc ct j n v hn m khng gp ct trng. B m ny phn tch c nhng v chm gii m l v hn nn trong trng hp ny c th coi b m l khng phn tch c. chm gii m c tnh theo cng thc sau: . Trong l s hiu ct rng; tng ng l di t m ngn nht v di t m di nht.

5.3.2 M c tnh prefix

Trong cc loi m c tnh phn tch c, loi m mang nhiu c im c ch cho vic khai thc s dng v c nghin cu nhiu l m c tnh prefix.

Prefix ca mt t hp m l b phn ca t hp m sau khi b i mt hay nhiu k hiu t bn phi

V d 1: Cho t hp m: 01100 1110

C cc prefix sau:

01100111

0110011

011001

01100

0110

011

01

0

M c tnh prefix c nh ngha nh sau: Mt b m c gi l m c tnh prefix nu bt k t m no cng khng phi l phn u ca bt k mt t m khc trong b m.

Khi biu din bng cy m, ta nhn thy b m c tnh prefix khi cc t m ch l nt l. M y l b m c tnh prefix.

V d 2: Cy m sau biu din mt b m prefix

0 1

0 1 1 00 0 1

010 011

Hnh 4.3

5.3.3 Bt ng thc Kraft

a ra c iu kin tng qut v tnh phn tch c ca t m, ta c nhn xt sau i vi m c tnh prefix vi c s m l :

V vy: hay c th vit tng ng nh sau:

. Bt ng thc ny c gi l Bt ng thc Kraft. Trong : S t m tng ng vi s tin ca ngun, : di t m m ha tin .

Ngc li mt dy s nguyn tha mn Bt ng thc Kraft th s tn ti b m c tnh prefix nh th tc to m prefix, iu ny c th m rng cho tt c cc loi m phn tch c c hoc khng c tnh prefix.

Th tc to m prefix

Bc 1: Sp xp cc t m theo th t tng dn ca t m: ; xy dng cy y , mi nt c nhnh, cao l .

Bc 2: mc chn mt nt bt k, v gn m l t m v xa cc nt k sau n.

Bc 3: Lp li Bc 2 i vi mc ta c cc t m .5.4 M thng k ti u

5.4.1 Gii hn di trung bnh ca t m

Gii hn di

Ta phi xc nh di trung bnh ca t m t c tiu chun ti u nh phn tch trn.

Gi s c ngun tin vi cc xc sut tng ng lng tin trung bnh l .

Ta chn b m c c s sao cho cc xc sut xp x bng nhau, khi lng tin ca mt k hiu m l.

- Nu m nh phn (bit/k.h)

Nu t m c di ni th lng tin cha trong t m s l:

Nu cc t m c di khng bng nhau th lng tin s bng . Trong l di trung bnh ca t m.

M ho ngun trn vi b m . m bo khng b mt mt thng tin th

V vy: (

Nh vy di trung bnh ca t m khng nh hn t s Entropi ca ngun v lng tin trung bnh cc i ca mt k hiu m. Vi m nh phn ta c: ( H(U). l gii hn di ca di trung bnh ca t m.

Du = xy ra khi : Nu m ha tin bng mt m nh phn c chiu di th lng tin cha trong t m s l bit. Nu ly di trung bnh t m th ta s c lng tin trung bnh cha trong t m. Thng thng khng phi l s nguyn nn iu kin trn ch l iu kin gii hn m da vo c th xy dng c cc thut ton xc nh m thng k ti u.

Gii hn trn

tha mn tiu chun ca m thng k ti u th di trung bnh phi c gii hn sau:

(4.4)M nh phn th:

Nh vy c th xy dng b m vi di trung bnh khng ln hn t s Entropi ca ngun vi lng tin trung bnh cc i cha trong mt k hiu cng 1 n v.

Tm li b m c di trung bnh tho mn cc iu kin trn gi l m thng k ti u. Mt b m nh vy phi tho mn nhng c im sau:

- Cc k hiu phi c cng xc sut

- S xut hin ca cc k hiu trong t m l c lp vi nhau (xc sut xut hin ca k hiu sau khng ph thuc vo s c mt ca cc k hiu ra trc)

5.4.2 Tiu chun m kinh t ti u

Nh cp phn trc v chiu di trung bnh ca t m, tiu chun ca m thng k ti u l t n chiu di trung bnh ca t m ti thiu. y l mt hng ln ca m ha (m nn d liu). Do cc tin ca ngun tin c xc sut xut hin khc nhau, nn vic dng cc t m c di nh m ha cho cc tin c xc sut xut hin cao s lm cho s k hiu cn thit m ha cho mt chui cc tin nh hn v tnh kinh t cao hn (tnh kinh t ca b m c o bng cng thc

)

(4.5)

Vy nguyn tc c bn ca m thng k ti u l da trn c s di t m , t l nghch vi xc sut xut hin . Ngha l cc t m di s dng m ha cho cc tin c xc sut xut hin nh v ngc li.

5.4.3 M thng k Fano Shanon

Fano v Shanon c lp nghin cu v xut phng php xy dng b m thng k ti u, bn cht ca hai phng php l tng ng nhau. thun tin trong vic trnh by, cc b m xy dng c t cc thut ton ny c c s m m=2

5.4.3.1 Phng php m theo Fano

Gi s c ngun tin vi cc xc sut tng ng

...

...

Nh ton hc Fano xut thut ton m ho nh sau:

Bc 1: Sp xp cc tin ui theo th t gim dn ca xc sut.

Bc 2: Chia cc tin lm hai nhm c xc sut xp x bng nhau. Nhm u ly tr 0, nhm sau ly tr 1.

Bc 3: Lp li bc 2 i vi cc nhm con cho ti khi tt c cc nhm ch cn li mt tin th kt thc thut ton.

r hn v thut ton, ta xt v d sau:

Cho ngun tin gm 7 tin:

Ln 1Ln 2Ln 3Ln 4Ln 5T m

0,340000

0,230101

0,191010

0,10110110

0,0711101110

0,061111011110

0,011111111111

di trung bnh ca t m:

= 0,01.5 + 0,06 .5 + 0,07.4 + 0,10. 3 + 0,19. 2 + 0,23. 2 + 0,34 .2 = 2,41

Tr s kinh t = 2,37 /2,41 = 0,98

Nhn xt:

1. Vic sp xp ngun theo xc sut gim dn nhm mc ch y cc tin c xc sut cao ln u bng cng vi vic chia i ngun dn ti cc lp trn s kt thc rt nhanh, v vy cc tin c xc sut cao s c di t m ngn dn ti di trung bnh ca b m l nh.

2. phc tp ca thut ton ph thuc vo vic s dng thut ton sp xp. Nu s dng thut ton sp xp quy th phc tp s l .

3. Xut pht t thut ton Fano, ta c th m rng cho vic to b m vi c s bt k bng cch trong bc 2 ta chia thnh m lp v ly cc tr t 0 n

4. Trong trng hp khi c nhiu tin vi xc sut bng nhau cng nh c nhiu phng n phn lp th b m thu c c th khng duy nht.

Thut ton trn c m phng bi chng trnh sau:Program Fano;

uses wincrt;

var st:string;

A:array[1..255] of char;

Ma:array[1..255] of string[20];

P:array[1..255] of real;;

n,i,j,k:integer;

Procedure Input;

Begin

write('cho xau can ma hoa:');readln(St);

end;

Procedure Output;

var k:integer;xauma:string;

Begin

xauma:='';

for i:=1 to length(st) do

for k:=1 to n do

if 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) then

Begin

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 do

begin

for k:=1 to length(St) do

if 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;

var i,j,tgp:integer;

TgA: char;

Begin

For i:=1 to n-1 do

For j:=i+1 to n do

If P[i]0 ,1

nh l ny cho thy rng, s kho trong m Affine trn Zm bng m((m), trong ((m) c cho theo cng thc trn. ( S cc php chn ca b l m v s cc php chn ca a l ((m) vi hm m ho l e(x) = ax + b). V d, khi m = 60, ((60) = 2 ( 2 ( 4 = 16 v s cc kho trong m Affine l 960.

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 d y (ax+b (mod 26) theo x. T tho lun trn thy rng, phng trnh ny c mt nghim duy nht trong Z26 . Tuy nhin ta vn cha bit mt phng php hu hiu tm nghim. iu cn thit y l c mt thut ton hu hiu lm vic . Rt may b l mt s kt qu tip sau v s hc modulo s cung cp mt thut ton gii m hu hiu cn tm.

nh ngha 1.3Gi 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 . Nu p l s nguyn t th mi phn t khc khng ca ZP u c nghch o. Mt vnh trong mi phn t u c nghch o c gi l mt trng.

Trong phn sau s m t mt thut ton hu hiu tnh cc nghch o ca Zm vi m tu . Tuy nhin, trong Z26 , ch bng phng php th v sai cng c th tm c cc nghch o ca cc phn t nguyn t cng nhau vi 26: 1-1 = 1, 3-1 = 9, 5-1 = 21, 7-1 = 15, 11-1 = 19, 17-1 =23, 25-1 = 25. (C th d dng kim chng li iu ny, v d: 7 ( 15 = 105 ( 1 mod 26, bi vy 7-1 = 15).

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

ax ( y-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)

p dng tnh kt hp ca php nhn modulo:

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

Kt qu l x ( a-1(y-b) (mod 26). y l mt cng thc tng minh cho x. Nh vy hm gii m l:

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

e. nh ngha dng ton hc:

Cho P = C = Z26 v gi s:

P = { (a,b) ( Z26 ( Z26 : UCLN(a,26) =1 }

Vi k = (a,b) ( K , ta nh ngha:

ek(x) = ax +b ( mod 26 )

dk(y) = a-1(y-b) ( mod 26 ) (vi x,y ( Z26)

V d Gi s K = (7,3). Nh nu trn, 7-1 mod 26 = 15.

Hm m ho l ek(x) = 7x+3

Hm gii m tng ng l: dk(x) = 15(y-3) = 15y -19

y, tt c cc php ton u thc hin trn Z26. Ta s kim tra liu dk(ek(x)) = x vi mi x ( Z26 khng?. Dng cc tnh ton trn Z26 , ta c

dk(ek(x)) =dk(7x+3) =15(7x+3)-19= x +45 -19 =x

minh ho, ta hy m ho bn r "hot". Trc tin bin i cc ch h, o, t thnh cc thng du theo modulo 26. Ta c cc s tng ng l 7, 14 v 19.

M ho: 7 ( 7 +3 mod 26 = 52 mod 26 = 0

7 ( 14 + 3 mod 26 = 101 mod 26 =23

7 ( 19 +3 mod 26 = 136 mod 26 = 6

Bi vy 3 k hiu ca bn m l 0, 23 v 6 tng ng vi xu k t AXG.

6.3.2.5 M Vigenre

Trong c hai h MDV v MTT (mt khi kho c chn) mi k t s c nh x vo mt k t duy nht. V l do , cc h mt cn c gi h thay th n biu. Ta s trnh by ( trong hnh 1.2.4.1) mt h mt khng phi l b ch n, l h m Vigenre ni ting. Mt m ny ly tn ca Blaise de Vigenre sng vo th k XVI.S dng php tng ng A ( 0, B ( 1, . . . , Z ( 25 trn, ta c th gn cho mi kho K vi mt chui k t c di m c gi l t kho. Mt m Vigenre s m ho ng thi m k t: Mi phn t ca bn r tng ng vi m k t. nh ngha dng ton hc:

Cho m l mt s nguyn dng c nh no . nh ngha P=C=K=(Z26)m Vi kho K = (k1, k2, . . . ,km) ta xc nh :

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: thiscryptosystemisnotsecureTa 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:

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

Bi vy dy k t ca xu bn m s l

V P X Z G I A X I V W P U B T T M J P W I Z I T W Z T

gii m ta c th dng cng t kho nhng thay cho cng, ta tr cho n theo modulo 26.

Nhn xt : Ta thy rng cc t kho c th vi s di m trong mt m Vigenre l 26m, bi vy, thm ch vi cc gi tr m kh nh, phng php tm kim vt cn cng yu cu thi gian kh ln. V d, nu m = 5 th khng gian kho cng c kch thc ln hn 1,1 ( 107 . Lng kho ny ln ngn nga vic tm kho bng tay( ch khng phi dng my tnh).

Trong h mt Vigenre c t kho di m, mi k t c th c nh x vo trong m k t c th c (gi s rng t kho cha m k t phn bit). Mt h mt nh vy c gi l h mt thay th a biu (polyalphabetic). Ni chung, vic thm m h thay th a biu s kh khn hn so vic thm m h n biu.

6.3.2.6 Mt m Hill

Trong phn ny s m t mt h mt thay th a biu khc c gi l mt m Hill. Mt m ny do Lester S.Hill a 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 m k t mt 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, y1cng nh y2 u l mt t hp tuyn tnh ca x1v x2.

Chng hn, c th ly

y1 = 11x1+ 3x2

y2 = 8x1+ 7x2

Tt nhin c th vit gn hn theo k hiu ma trn nh sau:

(y1, y2) = (x1, x2)

Ni chung, c th ly mt ma trn K kch thc m ( m lm kho. Nu mt phn t hng i v ct j ca K l ki,j th c th vit K = (ki,j), vi x = (x1, x2, . . . ,xm) ( P v K (K , ta tnh y = ek(x) = (y1, y2, . . . ,ym) nh sau:

Ni mt cch khc y = xK

Chng ta ni rng bn m nhn c t bn r nh php bin i tuyn tnh. Ta s xt xem phi thc hin gii m nh th no, tc l lm th no tnh x t y. Ta lm quen vi i s tuyn tnh nn s thy rng phi dng ma trn nghch o K-1 gi m. Bn m c gii m bng cng thc yK-1 .

Sau y l mt s nh ngha v nhng khi nim cn thit ly t i s tuyn tnh. Nu A = (xi,j) l mt ma trn cp l ( m v B = (b1,k ) l mt ma trn cp m ( n th tch ma trn AB = (c1,k ) c nh ngha theo cng thc:

Vi 1 ( i ( l v 1 ( k ( l. Tc l cc phn t hng i v ct th k ca AB c to ra bng cch ly hng th i ca A v ct th k ca B, sau nhn tng ng cc phn t vi nhau v cng li. Cn rng AB l mt ma trn cp l ( n.

Theo nh ngha ny, php nhn ma trn l kt hp (tc (AB)C = A(BC)) nhng ni chung l khng giao hon ( khng phi lc no AB = BA, thm ch vi ma trn vung A v B).

Ma trn n v m ( m (k hiu l Im ) l ma trn cp m ( m c cc s 1 nm ng cho chnh v cc s 0 v tr cn li. Nh vy ma trn n v 2 ( 2 l:

y2 =

Im c gi l ma trn n v v AIm = A vi mi ma trn cp l ( m v Im B =B vi mi ma trn cp m ( n. Ma trn nghch o ca ma trn A cp m ( m ( nu tn ti) l ma trn A-1 sao cho AA-1 = A-1A = Im. Khng phi mi ma trn u c nghch o, nhng nu tn ti th n duy nht.

Vi cc nh ngha trn, c th d dng xy dng cng thc gii m nu: V y = xK, ta c th nhn c hai v ca ng thc vi K-1 v nhn c:

yK-1 = (xK)K-1 = x(KK-1) = xIm = x

C th thy rng, ma trn m ho trn c nghch o trong Z26

V

( Mi php ton s hc u c thc hin theo modulo 2)

Sau y l mt v d minh ho cho vic m ho v gii m trong h mt m Hill

V d

Gi s cn m ho bn r "July". Ta c hai phn t ca bn r m ho: (9,20) (ng vi Ju) v (11,24) (ng vi ly). Vi kha

Ta c

Ta tnh nh sau

Bi vy bn m ca July l DELW. gii m ta s tnh

V

Nh vy ta nhn c bn ng.

Cho ti lc ny ta ch ra rng c th thc hin php gii m nu K c mt nghch o. Trn thc t, php gii m l c th thc hin c, iu kin cn l K phi c nghch o. Bi vy, chng ta ch quan tm ti cc ma trn K kh nghich

Tnh kh nghch ca mt ma trn vung ph thuc vo gi tr nh thc ca n. trnh s tng qut ho khng cn thit, ta ch gii hn trong trng hp 2(2.

nh ngha 1.4nh thc ca ma trn A = (a,i j ) cp 2( 2 l gi tr det A = a1,1 a2,2 - a1,2 a2,1 Nhn xt

nh thc ca mt ma trn vung cp mm c th c tnh theo cc php ton hng s cp: hy xem mt gio trnh bt k v i s tuyn tnh.Hai tnh cht quan trng ca nh thc l det Im = 1 v quy tc nhn det(AB) = det A ( det B.Mt ma trn thc K l c nghch o khi v ch khi nh thc ca n khc 0. Tuy nhin, iu quan trng cn nh l ta ang lm vic trn Z26 . Kt qu tng ng l ma trn K c nghch o theo modulo 26 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 Ki j ma trn thu c t K bng cch loi b hng th i v ct th j. V nh ngha ma trn K* c phn t (i,j) ca n nhn gi tr(-1) det Kj i (K* c gi l ma trn b i s ca K). Khi c th chng t rng: K-1 = (det K)-1K*

Bi 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

Bi vy det K c nghch o trong Z26 .

Nhn xt: Cng thc i vi trn khng phi l mt cng thc tnh ton c hiu qu tr cc trng hp m nh ( chng hn m = 2, 3). Vi m ln, phng php thch hp tnh cc ma trn nghch o phi da vo cc php ton hng s cp.

Trong trng hp 2(2, ta c cng thc sau:

nh l 1.3

Gi s A = (ai j) l mt ma trn cp 2 ( 2 trn Z26 sao cho det A = a1,1a2,2 - a1,2 a2,1 c nghch o. Khi

Xt v d trc ta c:

V 1-1 mod 26 = 1 nn ma trn nghch o l

y chnh l ma trn c trn.

nh ngha dng ton hc:

Cho m l mt s nguyn dng c nh. Cho P = C = (Z26 )m v cho

K = { cc ma trn kh nghch cp m ( m trn Z26}

Vi mt kho k ( K ta xc nh:

ek(x) = x*k

dk(y) = y*k -1

Trong tt c cc php ton c thc hin trong Z26.6.3.2.7 Cc h m dng

Trong cc h mt nghin cu trn, ccb phn t lin tip ca bn r u c m ho bng cng mt kho K. Tc 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 s dng khc l mt m dng. tng c bn y 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 fi c 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 t u 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 th c thc hin bng cch tnh lin tip: z1, x1, z2 , x2 ...

Sau y l nh ngha di dng ton hc:

nh ngha

Mt m dng l mt b (P,C,K,L,F,E,D) tho mn dc 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 ( P i -1 (L

Vi mi z (L c mt quy tc m ez ( E v mt quy tc gii m tng ng dz (D . ez : P (C v dz : C (P 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 cng vi cc v d minh ho. 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. Khi ta coi K l mt "mn" m rng thnh dng kho z1z2 . . .

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 t u 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

Cc m dng thng c m t trong cc b ch nhi 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 ta coi "0" biu th gi tr "sai" v "1" biu th gi tr "ng" trong i s Boolean th php cng theo moulo 2 s ng vi php hoc c loi tr. Bi vy php m (v gii m ) d dng thc hin bng mch cng.

Ta xem xt mt phng php to mt dng kho (ng b ) khc. Gi s bt u vi (k1, . . , km) v zi = ki, 1 ( i ( m ( cng ging nh trc y), tuy nhin by gi ta to dng kho theo mt quan h quy tuyn tnh cp m:

trong c0, . . , cm-1 ( Z2 l cc hng s cho trc.

Nhn xt:

Php quy c ni l c bc m v mi s hng ph thuc vo m s hng ng trc. Php quy ny l tuyn tnh bi v Zi+m l mt hm tuyn tnh ca cc s hng ng trc. Ch ta c th ly c0= 1 m khng lm mt tnh tng qut. Trong trng hp ngc li php quy s l c bc m-1.

y kho K gm 2m gi tr k1, . . , km , c0, . . , cm-1. Nu (k1, . . , km)= (0, . . . , 0) th dng kho s cha ton cc s 0. D nhin phi trnh iu ny v khi bn m s ng nht vi bn r. Tuy nhin nu chn thch hp cc hng s c0, . . , cm-1 th mt vc t khi u bt k khc (k1, . . , km) s to nn mt dng kho c chu k 2m -1. Bi vy mt kho ngn s to nn mt dng kho c chu k rt ln. y l mt tnh cht rt ng lu tm v ta s thy phn sau, mt m Vigenre c th b thm nh tn dng yu t dng kho c chu k ngn.

Sau y l mt v d minh ho:

V d

Gi s m = 4 v dng kho c to bng quy tc:

zi+4 = zi + zi+1 mod 2

Nu dng kho bt u mt vc t bt k khc vi vc t (0,0,0,0) th ta thu c dng kho c chu k 15. V d bt u bng vc t (1,0,0,0), dng kho s l:

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1

Mt vc t khi u khc khng bt k khc s to mt hon v vng (cyclic) ca cng dng kho.

Mt hng ng quan tm khc ca phng php to dng kho hiu qu bng phn cng l s dng b ghi dch hi tip tuyn tnh (hay LFSR). Ta dng mt b ghi dch c m tng. Vc t (k1, . . , km) s c dng khi to ( t cc gi tr ban u) cho thanh ghi dch. mi n v thi gian, cc php ton sau s c thc hin ng thi.

k1 c tnh ra dng lm bit tip theo ca dng kho.

k2, . . , km s c dch mt tng v pha tri.

Gi tr mi ca s c tnh bng:

(y l hi tip tuyn tnh)

Ta thy rng thao tc tuyn tnh s c tin hnh bng cch ly tn hiu ra t mt s tng nht nh ca thanh ghi (c xc nh bi cc hng s cj c gi tr "1" ) v tnh tng theo modulo 2 ( l php hoc loi tr ). Hnh 1.8 cho m t ca LFSR dng to dng kho cho v d 1.7.

Thanh ghi dch hi tip tuyn tnh (LFSR)

Mt m kho t sinh

L do s dng thut ng "kho t sinh" l ch: bn r c dng lm kho ( ngoi "kho khi thu" ban u K).

V d :

Gi s kho l k = 8 v bn r l rendezvous. Trc tin ta bin i bn r thnh dy cc s nguyn:

17 4 13 3 4 25 21 14 20 18

Dng kho nh sau:

17 4 13 3 4 25 21 14 20

By gi ta cng cc phn t tng ng ri rt gn theo modulo 26:

21 17 16 7 3 20 9 8 12

Bn m dng k t l: ZVRQHDUJIM

By gi ta xem A gii m bn m ny nh th no. Trc tin A bin i xu k t thnh dy s:

21 17 16 7 3 20 9 8 12

Sau ta tnh:

x1 = d8(25) = 25 - 8 mod 26 = 17

v x2 = d17(21) = 21 - 17 mod 26 = 4

v c tip tc nh vy. Mi khi A nhn c mt k t ca bn r, A s dng n lm phn t tip theo ca dng kho.

D nhin l m dng kho t sinh l khng an ton do ch c 26 kho.

6.3.3 H m DES

6.3.3.1 M t

tng ca ngi pht minh Des l to ra mt thut ton bin i d liu tht phc tp i phng khng th tm ra mi lin quan ca on tin m vi bn r cng nh khng th thit lp c mi quan h no gia on tin c m ha v kha.

Thut ton tin hnh theo 3 giai on:

1.Bn r cho trc x (64 bit), mt xu bt x0 s c xy dng bng cch hon v cc bt ca x theo php hon v c nh ban u IP. Khi d liu chia thnh 2 na (na tri v na phi). Ta vit: x0= IP(X) = L0R0, (L0 gm 32 bt u v R0 l 32 bt cui).

Ta vit: x0= IP(X) = L0R0, (L0 gm 32 bt u v R0 l 32 bt cui).

2. Tnh ton 16 ln lp theo mt hm xc nh. Ta s tnh LiRi, 1 ( i (16 theo quy tc sau:

Li = Ri-1Ri = Li-1 ( f(Ri-1,Ki)

( k hiu php hoc loi tr ca hai xu bt (cng theo modulo 2).

f l mt hm m ta s m t sau.

K1,K2, . . . ,K16 l cc xu bt di 48 c tnh nh hm ca kho K. ( trn thc t mi Ki l mt php chn hon v bt trong K). K1, . . ., K16 s to thnh bng kho

3. p dng php hon v ngc IP -1 cho xu bt R16L16, ta thu c bn m y. y = IP -1 (R16L16).

Hy ch th t o ca L16 v R16.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 v o: IP-1 (cho sn)

6.3.3.2 Tnh bng kha

Kha K l mt xu bt di 64, trong 56 bt l kho v 8 bt kim tra tnh chn l nhm pht hin sai. Cc bt 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 bt. Cc bt kim tra b b qua trong qu trnh tnh ton bng kho.

Cc bc tnh kha:

1. Vi mt kho K 64 bt cho trc, ta loi b cc bt kim tra tnh chn l bng cch p dng hon v cc bt ca K theo php hon v c nh PC-1. Sau chia kha thnh 2 phn C0: 28 bt u, D0: 28 bt cui.

PC-1(K) = C0D0

2. Vi i thay i t 1 n 16: Thc hin dch tri 1 hoc 2 bt ph thuc vo s vng lp.

S bt dch ca vng.

3. Cc bt ca kha c chn ra theo hon v nn PC-2 (hon v la chn): 56 bt -> 48 bt.

Ci = LSi(Ci-1)

Di = LSi(Di-1)

Tnh bng kho DES.6.3.3.3 Tnh hm F

Hm f c hai bin vo: bin th nht A ( chnh l thnh phn Ri-1) l xu bt di 32, bin th hai J (kha Ki) l mt xu bt di 48. u ra ca f l mt xu bt di 32.

Cc bc sau c thc hin:

1. A( na phi ca d liu Ri-1 c m rng thnh xu bt di 48 theo mt hm m rng c nh . (m rng c cng s t vi kha k).

E(A) gm 32 bt c m rng thnh 48 bt theo hon v m rng E. (nhm mc ch to ra d liu c cng kch c vi dng kha thc hin php ton XOR).

2. Tnh E(A) ( J v vit kt qu thnh mt chui 8 khi 6 bt = 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 ( B1 S1, B8 S8,).

Hp S l bng gm 4 hng v 16 dng. Vi khi bt c di 6, K hiu Bi = b1b2b3b4b5b6

Tnh Sj(Bj) nh sau:

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

Sj(Bj) = Sj(r,c); (r hng, c - ct) trong hp S). Phn t ny vit di dng nh phn l mt xu bt c di 4.

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

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

6.4 Mt s h m ho hin i

6.4.1 Khi nim chung

Chng ta xem xt 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 m c=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, ch tnh cht xc thc ca n l i hi m bo rng A qu thc l ngi c kho ring tng ng. 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 mt v s tin cy.

S m kho cng khai v cn bn chm hn so vi thut ton m ho i xng. V l do ny, m kho cng khai c s dng ph bin nht trn thc t cho vic truyn ti kho, sau c s dng m ho khi d liu bng thut ton i xng, v nhng ng dng khc bao gm s ton vn d liu v s thm nh quyn, m ho nhng mc d liu nh nh s Credit card, s PINs.

- M kha cng khai c tnh cht bt i xng, s dng 2 kha ring bit tng phn vi m ha qui c c tnh i xng l ch s dng 1 kha. Vic s dng 2 kha c tm quan trng su sc trong lnh vc cn tnh b mt, phn b kha v s chng thc.

- Mt kha cho m ha (kha cng khai) v mt kha khc (c quan h vi kha trn) cho gii m (Kha b mt, kha ring ch c ngi nhn mi bit m ny).

- Mt i tng B mun gi tin cho A th phi dng kha cng khai ca A m ha thng tin, gii m c bn tin B gi th A s dng kha ring ca mnh gii m.

- Gii thut m ha c c im bit gii thut m v kha m ha (kho cng khai) nhng vic tnh ra kha b mt hay kh nng gii m l khng kh thi.

- Cc bc cn thit trong qu trnh m ha cng khai:

+ Ngi nhn bn m t to ra mt cp kha dng cho m ha v gii m on tin m mnh s nhn.

+ Cng b rng ri kha m ha bng cch t kha vo mt thanh ghi hay mt file cng khai. y l kha cng khai, kha cn li c gi ring.

+ Nu A mun gi mt on tin ti B th A m ha on tin bng kha cng khai ca B.

+ Khi B nhn on tin m ha, gii m bng kha b mt ca mnh. Khng mt ngi no khc c th gii m on tin m ny bi v ch c mnh B bit kha b mt thi.

6.4.2 Mt s h m cng khai thng dng

6.4.2.1 H m RSA (R.Rivest, A.Shamir, L.Adleman)

Khi nim h mt m RSA c ra i nm 1976 bi cc tc gi R.Rivets, A.Shamir, v L.Adleman. H m ho ny da trn c s ca hai bi ton :

+ Bi ton Logarithm ri rc (Discrete logarith)

+ Bi ton phn tch thnh tha s

S m ho RSA l s m ho khi, on tin c m ho tng khi vi mi khi c gi tr < n, n l s nguyn v ln.

H m RSA l h m da vo bi ton logarithm ri rc v bi ton phn tch mt s nguyn thnh tch cc tha s nguyn t, l h m c s dng rng ri nht. N cung cp c s b mt v ch k in t, v tnh bo mt ca n l c s cho kh trong vn tm tha s nguyn.

Thut ton

To kho:

Mi thc th to mt kho cng khai v mt kho ring tng ng.

Thc th A cn lm cng vic sau:

1. To 2 s nguyn t ln p v q bt k c c xp x nhau.

2. Tnh n = p*q v (n)= (p-1)(q-1).

3. Chn 1 s nguyn e bt k, 1