bài giảng lttt hv bcvt

8/6/2019 Bi ging LTTT HV BCVT

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HC VIN CNG NGH BU CHNH VIN THNG- - - - - - - - - - - - - -

BI GING

L THUYT THNG TIN

Bin son : PGS.Ts. NGUYN BNH

Lu hnh ni b

H NI - 2006


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LI NI U

Gio trnh L thuyt thng tin l mt gio trnh c sdng cho sinh vin chuyn ngnh

in t Vin thng v Cng ngh thng tin ca Hc vin Cng ngh Bu chnh Vin thng.y cng l mt ti liu tham kho hu ch cho cc sinh vin chuyn ngnh in - in t.

Gio trnh ny nhm chun b tt kin thc cscho sinh vin hc tp v nm vng ccmn k thut chuyn ngnh, m bo cho sinh vin c thnh gi cc ch tiu cht lng cbnca mt h thng truyn tin mt cch c cn c khoa hc.

Gio trnh gm 6 chng, ngoi chng I c tnh cht gii thiu chung, cc chng cn lic chia thnh 4 phn chnh:

Phn I: L thuyt tn hiu ngu nhin v nhiu (Chng 2)

Phn II: L thuyt thng tin v m ha (Chng 3 v Chng 4)Phn III: L thuyt thu ti u (Chng 5)

Phn IV: Mt m (Chng 6)

Phn I: (Chng II). Nhm cung cp cc cng c ton hc cn thit cho cc chng sau.

Phn II: Gm hai chng vi cc ni dungch yu sau:

- Chng III: Cung cp nhng khi nim cbn ca l thuyt thng tin Shannon trong h

truyn tin ri rc v mrng cho cc h truyn tin lin tc.- Chng IV: Trnh by hai hng kin thit cho hai nh l m ha ca Shannon. V

khun kh c hn ca gio trnh, cc hng ny (m ngun v m knh) chc trnh by mc cc hiu bit cbn. c th tm hiu su hn nhng kt qu mi v cc ng dng c thsinh vin cn phi xem thm trong cc ti liu tham kho.

Phn III: (Chng V)Trnh by vn xy dng cc h thng thu ti u m bo tc truyn tin v chnh xc t c cc gi tr gii hn. Theo truyn thng bao trm ln ton bgio trnh l vic trnh by hai bi ton phn tch v tng hp. Cc v d trong gio trnh cchn lc k nhm gip cho sinh vin hiu c cc khi nim mt cch su sc hn. Cc hnh v,

bng biu nhm m t mt cch trc quan nht cc khi nim v hot ng ca s khi chcnng ca cc thit b c th

Phn VI: (Chng VI)Trnh by csl thuyt cc h mt bao gm cc h mt kha bmt v cc h mt kha cng khai. Do khun kh c hn ca gio trnh, mt s vn quan trngcn cha c cp ti (nh trao i v phn phi kha, xc thc, m bo tnh ton vn )

Sau mi chng u c cc cu hi v bi tp nhm gip cho sinh vin cng cc cc knng tnh ton cn thit v hiu su sc hn cc khi nim v cc thut ton quan trng.

Phn ph lc cung cp mt s kin thc b xung cn thit i vi mt s khi nim quantrng v mt s s liu cn thit gip cho sinh vin lm c cc bi tp c ra cc chng.


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Gio trnh c vit da trn cs cng mn hc L thuyt thng tin do B Gio dcv o to v c c kt sau nhiu nm ging dy v nghin cu ca tc gi. Rt mong csng gp ca bn c.

Cc ng gp kin xin gi v

KHOA K THUT IN T1 - HC VIN CNG NGH BU CHNH VIN THNGKM 10. NG NGUYN TRI - TH X H NG

Email: [email protected]

Hoc [email protected]

Cui cng ti xin chn thnh cm n GS. Hunh Hu Tu cho ti nhiu kin qu butrong cc trao i hc thut c lin quan ti mt s ni dung quan trng trong gio trnh ny.

NGI BIN SON


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Chng 1: Nhng vn chung v nhng khi nim cbn

3

CHNG I: NHNG VN CHUNG V NHNG KHINIM C BN

1.1. V TR, VAI TR V SLC LCH SPHT TRIN CA L THUYTTHNG TIN

1.1.1. V tr, vai tr ca L thuyt thng tin

Do s pht trin mnh m ca k thut tnh ton v cc h tng, mt ngnh khoa hcmi ra i v pht trin nhanh chng, l: L thuyt thng tin. L mt ngnh khoa hc nhngn khng ngng pht trin v thm nhp vo nhiu ngnh khoa hc khc nh: Ton; trit; ho;

Xibecnetic; l thuyt h thng; l thuyt v k thut thng tin lin lc v t c nhiu ktqu. Tuy vy n cng cn nhiu vn cn c gii quyt hoc gii quyt hon chnh hn.

Gio trnh L thuyt thng tin ny (cn c gi l Csl thuyt truyn tin) ch lmt b phn ca l thuyt thng tin chung N l phn p dng ca L thuyt thng tin vo kthut thng tin lin lc.

Trong cc quan h ca L thuyt thng tin chung vi cc ngnh khoa hc khc nhau, ta phic bit kn mi quan h ca n vi ngnh Xibecnetic.

Mi quan h gia cc hot ng khoa hc ca con ngi v cc qung tnh ca vt chtc m t trn hnh (1.1).

- Nng lng hc: L mt ngnh khoa hc chuyn nghin cu cc vn lin quan ti cckhi nim thuc v nng lng. Mc ch ca nng lng hc l lm gim s nng nhc ca lao

ng chn tay v nng cao hiu sut lao ng chn tay. Nhim v trung tm ca n l to, truyn,th, bin i, tch lu v x l nng lng.

Qung tnh ca vt cht

Khi lng

Cng ngh hc

Thng tinNng lng

Nng lng hc

iu khin hc(Xibecnetic)

Cc lnh vc hot ng khoa hc cacon ngi

Hnh 1.1. Quan h gia hot ng khoa hc v qung tnh ca vt cht


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4

- Xibecnetic: Bao gm cc ngnh khoa hc chuyn nghin cu cc vn c lin quan nkhi nim thng tin v tn hiu. Mc ch ca Xibecnetic l lm gim s nng nhc ca tr c vnng cao hiu sut lao ng tr c. Ngoi nhng vn c xt trong Xibecnetic nhi tng,mc ch, ti u ho vic iu khin, lin h ngc. Vic nghin cu cc qu trnh thng tin (nh

chn, truyn, x l, lu tr v hin th thng tin) cng l mt vn trung tm ca Xibecnetic.Chnh v vy, l thuyt v k thut thng tin chim vai tr rt quan trng trong Xibecnetic.

- Cng ngh hc: gm cc ngnh khoa hc to, bin i v x l cc vt liu mi. Cngngh hc phc vc lc cho Xibecnetic v nng lng hc. Khng c cng ngh hc hin ith khng th c cc ngnh khoa hc k thut hin i.

1.1.2. Slc lch spht trin

Ngi t vin gch u tin xy dng l thuyt thng tin l Hartley R.V.L. Nm 1928,ng a ra so lng thng tin l mt khi nim trung tm ca l thuyt thng tin. Da vo

khi nim ny, ta c th so snh nh lng cc h truyn tin vi nhau.Nm 1933, V.A Kachenhicov chng minh mt lot nhng lun im quan trng ca l

thuyt thng tin trong bi bo V kh nng thng qua ca khng trung v dy dn trong h thnglin lc in.

Nm 1935, D.V Ageev a ra cng trnh L thuyt tch tuyn tnh, trong ng phtbiu nhng nguyn tc cbn v l thuyt tch cc tn hiu.

Nm 1946, V.A Kachenhicov thng bo cng trnh L thuyt th chng nhiu nh dumt bc pht trin rt quan trng ca l thuyt thng tin.

Trong hai nm 1948 1949, Shanon C.E cng b mt lot cc cng trnh vi, a spht trin ca l thuyt thng tin ln mt bc tin mi cha tng c. Trong cc cng trnh ny,nhvic a vo khi nim lng thng tin v tnh n cu trc thng k ca tin, ng chngminh mt lot nh l v kh nng thng qua ca knh truyn tin khi c nhiu v cc nh l mho. Nhng cng trnh ny l nn tng vng chc ca l thuyt thng tin.

Ngy nay, l thuyt thng tin pht trin theo hai hng ch yu sau:

L thuyt thng tin ton hc: Xy dng nhng lun im thun tu ton hc v nhng cston hc cht ch ca l thuyt thng tin. Cng hin ch yu trong lnh vc ny thuc v ccnh bc hc li lc nh: N.Wiener, A. Feinstain, C.E Shanon, A.N. Kanmgorov, A.JA Khintrin.

L thuyt thng tin ng dng: (l thuyt truyn tin)

Chuyn nghin cu cc bi ton thc t quan trng do k thut lin lc t ra c lin quann vn chng nhiu v nng cao tin cy ca vic truyn tin. Cc bc hc C.E Shanon, S.ORiCe, D. Midleton, W. Peterson, A.A Khakevich, V. Kachenhicov c nhng cng trnh qu

bu trong lnh vc ny.


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5

1.2. NHNG KHI NIM CBN - SHTRUYN TIN V NHIM VCA N

1.2.1. Cc nh ngha cbn

1.2.1.1. Thng tin

nh ngha: Thng tin l nhng tnh cht xc nh ca vt cht m con ngi (hoc hthng k thut) nhn c t th gii vt cht bn ngoi hoc t nhng qu trnh xy ra trong bnthn n.

Vi nh ngha ny, mi ngnh khoa hc l khm ph ra cc cu trc thng qua vic thuthp, ch bin, x l thng tin. y thng tin l mt danh t ch khng phi l ng t chmt hnh vi tc ng gia hai i tng (ngi, my) lin lc vi nhau.

Theo quan im trit hc, thng tin l mt qung tnh ca th gii vt cht (tng t nhnng lng, khi lng). Thng tin khng c to ra m chc s dng bi h th cm.

Thng tin tn ti mt cch khch quan, khng ph thuc vo h th cm. Trong ngha khi qutnht, thng tin l sa dng. Sa dng y c th hiu theo nhiu ngha khc nhau: Tnh ngunhin, trnh t chc,

1.2.1.2. Tin

Tin l dng vt cht c th biu din hoc th hin thng tin. C hai dng: tin ri rc vtin lin tc.

V d: Tm nh, bn nhc, bng s liu, bi ni, l cc tin.

1.2.1.3. Tn hiuTn hiu l cc i lng vt l bin thin, phn nh tin cn truyn.

Ch : Khng phi bn thn qu trnh vt l l tn hiu, m s bin i cc tham s ringca qu trnh vt l mi l tn hiu.

Cc c trng vt l c th l dng in, in p, nh sng, m thanh, trng in t

1.2.2. S khi ca h thng truyn tin s (Hnh 1.2)


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M bomt

Mknh

Dnknh

Tp

Gii mmt

Gii mknh

Chiaknh p

Dng bitH thng ng b

( Synchronization )

T cc ngun khc

Ti cc b nhn tin khc

nh khundng

nh khundng

u vo s

u ra s

iu ch

Gii iuch

Khi cbn

Khi tu chnHnh 1.2. S khi h thng truy

m1 S1(t)

Nhntin

m1

Ngun

tin Mngun

Gii mngun


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1.2.2.1. Ngun tin

Ni sn ra tin:

- Nu tp tin l hu hn th ngun sinh ra n c gi l ngun ri rc.

- Nu tp tin l v hn th ngun sinh ra n c gi l ngun lin tc.

Ngun tin c hai tnh cht: Tnh thng k v tnh hm .

Vi ngun ri rc, tnh thng k biu hin ch xc sut xut hin cc tin l khc nhau.

Tnh hm biu hin ch xc sut xut hin ca mt tin no sau mt dy tin khc nhauno l khc nhau.

V d: P(y/ta) P(y/ba)

1.2.2.2. My pht

L thit b bin i tp tin thnh tp tn hiu tng ng. Php bin i ny phi l n trhai chiu (th bn thu mi c th sao li c ng tin gi i). Trong trng hp tng qut, my

pht gm hai khi chnh.

- Thit b m ho: Lm ng mi tin vi mt t hp cc k hiu chn nhm tng mt ,tng kh nng chng nhiu, tng tc truyn tin.

- Khi iu ch: L thit b bin tp tin ( hoc khng m ho) thnh cc tn hiu bc xvo khng gian di dng sng in t cao tn. V nguyn tc, bt k mt my pht no cng ckhi ny.

1.2.2.3. ng truyn tin

L mi trng vt l, trong tn hiu truyn i t my pht sang my thu. Trn ngtruyn c nhng tc ng lm mt nng lng, lm mt thng tin ca tn hiu.

1.2.2.4. My thu

L thit b lp li (sao li) thng tin t tn hiu nhn c. My thu thc hin php bin ingc li vi php bin i my pht: Bin tp tn hiu thu c thnh tp tin tng ng.

My thu gm hai khi:

- Gii iu ch: Bin i tn hiu nhn c thnh tin m ho.

- Gii m: Bin i cc tin m ho thnh cc tin tng ng ban u (cc tin ca ngungi i).

1.2.2.5. Nhn tin

C ba chc nng:

- Ghi gi tin (v d b nhca my tnh, bng ghi m, ghi hnh,)

- Biu th tin: Lm cho cc gic quan ca con ngi hoc cc b cm bin ca my th cm

c x l tin (v d bng m thanh, ch s, hnh nh,)


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- X l tin: Bin i tin a n v dng d s dng. Chc nng ny c th thc hinbng con ngi hoc bng my.

1.2.2.6. Knh truyn tin

L tp hp cc thit b k thut phc v cho vic truyn tin t ngun n ni nhn tin.

1.2.2.7. Nhiu

L mi yu t ngu nhin c nh hng xu n vic thu tin. Nhng yu t ny tc ngxu n tin truyn i t bn pht n bn thu. cho gn, ta gp cc yu t tc ng vo mt trn hnh 1.2.

Hnh 1.2 l s khi tng qut nht ca mt h truyn tin s. N c th l: h thng vtuyn in thoi, v tuyn in bo, raa, v tuyn truyn hnh, h thng thng tin truyn s liu,v tuyn iu khin t xa.

1.2.2.8. Cc phng php bin i thng tin s trong cc khi chc nng ca h thng


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nh dng/ M ngun

M ho k t

Ly muLng t hoiu ch m xung(PCM)

- PCM vi phn

- iu ch Delta (DM)- DM c tc bin ilin tc (CVSD)- M ho don tuyntnh (LPC)- Cc phng php nn:M Huffman, m s hc,thut ton Ziv_Lempel

iu ch

Kt hp

- PSK: Manip pha- FSK: Manip tn s- ASK: Manip bin - Hn hp- OQPSK: Manip phatng i 4 mc- MSK

Khng kt hp

- PSK vi phn- FSK- ASK- Hn hp

M knh

Dng sng

Tn hiu M_trTn hiu trc giaoTn hiu song trcgiao

Cc dy c cu trc

- M khi- M lin tc

Dn knh/ a truy cp

- Phn chia tn s:FDM/ FDMA

- Phn chia thi gian:TDM/ TDMA

- Phn chia m:CDM/ CDMA

- Phn chia khng gian:

SDMA- Phn chia cc tnh:

PDMA- OFDM

Tri ph

Dy trc tip (DS)Nhy tn (FH)Nhy thi gian (TH)Cc phng php hnhp

ng b

- ng b sng mang- ng b du

- ng b khung- ng b mng

- Hon v- Thay th- X l bit- Cc phng php hn hp

- Thut ton RSA- Thut ton logarit ri rc- Thut ton McElice- Thut ton Merkle-Hellman- Thut ton s dng ngcong Elliptic

M bo mt

M ho theo khiM ho dng s liu Mt m cinMt m kho cng khai


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1.2.3. Nhng ch tiu cht lng cbn ca mt h truyn tin

1.2.3.1. Tnh hu hiu

Th hin trn cc mt sau:- Tc truyn tin cao.

- Truyn c ng thi nhiu tin khc nhau.

- Chi ph cho mt bit thng tin thp.

1.2.3.2. tin cy

m bo chnh xc ca vic thu nhn tin cao, xc sut thu sai (BER) thp.

Hai ch tiu trn mu thun nhau. Gii quyt mu thun trn l nhim v ca l thuyt thng

tin.1.2.3.3. An ton

- B mt:

+ Khng th khai thc thng tin tri php.

+ Ch c ngi nhn hp l mi hiu c thng tin.

- Xc thc: Gn trch nhim ca bn gi bn nhn vi bn tin (ch k s).

- Ton vn:

+ Thng tin khng b bp mo (ct xn, xuyn tc, sa i).

+ Thng tin c nhn phi nguyn vn c v ni dung v hnh thc.

- Kh dng: Mi ti nguyn v dch v ca h thng phi c cung cp y cho ngidng hp php.

1.2.3.4. m bo cht lng dch v (QoS)

y l mt ch tiu rt quan trng c bit l i vi cc dch v thi gian thc, nhy cmvi tr (truyn ting ni, hnh nh, .)


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Chng 2: Tn hiu v nhiu

11

CHNG II: TN HIU V NHIU

2.1. TN HIU XC NH V CC C TRNG VT L CA CHNG

Tn hiu xc nh thng c xem l mt hm xc nh ca bin thi gian t (s(t)). Hmny c thc m t bng mt biu thc gii tch hoc c m t bng th. Mt trong cc

c trng vt l quan trng ca tn hiu l hm mt ph bin phc S( )

. Vi tn hiu s(t)kh tch tuyt i, ta c cp bin i Fourier sau:

j t

j t

S( ) s(t)e dt (2.1)

1s(t) S( )e d (2.2)

2

=

=

Sau y l mt sc trng vt l quen thuc ca tn hiu:

- Thi hn ca tn hiu (T): Thi hn ca tn hiu l khong thi gian tn ti ca tn hiu,trong khong ny gi tr ca tn hiu khng ng nht bng 0.

- B rng ph ca tn hiu (F): y l min xc nh bi tn s khc khng cao nht ca tnhiu.

- Nng lng ca tn hiu (E): Nng lng ca tn hiu c th tnh theo min thi gian haymin tn s.

22 1E s (t)dt S( ) d [J] (2.3)

2

= =

(nh l Parseval)

- Cng sut ca tn hiu (P):

EP [W]T

=

2.2. TN HIU V NHIU L CC QU TRNH NGU NHIN

2.2.1. Bn cht ngu nhin ca tn hiu v nhiu

Nh xt trn, chng ta coi tn hiu l biu hin vt l ca tin (trong thng tin v tuyn:dng vt l cui cng ca tin l sng in t). Qu trnh vt l mang tin din ra theo thi gian, do v mt ton hc th khi c thc, cch biu din trc tip nht cho tn hiu l vit biu thc

ca n theo thi gian hay v th thi gian ca n.


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Trong l thuyt cin, d tn hiu tun hon hoc khng tun hon nhng ta u coi l bit trc v biu din n bng mt hm tin nh ca thi gian. l quan nim xc nh v tnhiu (tn hiu tin nh). Tuy vy, quan nim ny khng ph hp vi thc t. Tht vy, tn hiutin nh khng th dng vo vic truyn tin tc c. Vi cch coi tn hiu l biu hin vt l ca

tin, nu chng ta hon ton bit trc n th v mt thng tin, vic nhn tn hiu khng c ngha g. Nhng nu ta hon ton khng bit g v tn hiu truyn i, th ta khng th thc hinnhn tin c. Bi v khi khng c ci g lm cn c phn bit tn hiu vi nhng ci khng

phi n, c bit l vi cc nhiu. Nh vy, quan nim hp l nht l phi kn cc c tnhthng k ca tn hiu, tc l phi coi tn hiu l mt qu trnh ngu nhin. Chng ta s gi cc tnhiu xt theo quan im thng k ny l cc tn hiu ngu nhin.

2.2.2. nh ngha v phn loi nhiu

Trong qu trnh truyn tin, tn hiu lun lun b nhiu yu t ngu nhin tc ng vo, lmmt mt mt phn hoc thm ch c th mt ton b thng tin cha trong n. Nhng yu t ngunhin rt a dng, chng c th l nhng thay i ngu nhin ca cc hng s vt l ca mitrng truyn qua hoc nhng loi trng in t cm ng trong cng nghi p, y hcvvTrong v tuyn in, ngi ta gi tt c nhng yu t ngu nhin y l cc can nhiu (hay nhiu).Tm li, ta c th coi nhiu l tt c nhng tn hiu v ch (tt nhin l i vi h truyn tin ta xt)c nh hng xu n vic thu tin. Ngun nhiu c thngoi hoc trong h. Nu nhiu xc nhth vic chng n khng c kh khn g v mt nguyn tc. V d nh ngi ta c nhng bin

php chng n do dng xoay chiu gy ra trong cc my khuch i m tn, ngi ta cng bitr nhng cch chng s nhiu ln nhau gia cc in i v tuyn in cng lm vic m chngc ph tn hiu trm nhauvv Cc loi nhiu ny khng ng ngi.

Ch :

Cn phn bit nhiu vi s mo gy ra bi c tnh tn s v c tnh thi gian ca cc thitb, knh truyn (mo tuyn tnh v mo phi tuyn). V mt nguyn tc, ta c th khc phcc chng bng cch hiu chnh.

Nhiu ng lo ngi nht vn l cc nhiu ngu nhin. Cho n nay, vic chng cc nhiungu nhin vn gp nhng kh khn ln c v mt l lun ln v mt thc hin k thut. Do ,trong gio trnh ny ta ch cp n mt dng no (sau ny s thy y thng xt nht lnhiu cng, chun) ca nhiu ngu nhin.

Vic chia thnh cc loi (dng) nhiu khc nhau c th lm theo cc du hiu sau:1. Theo b rng ph ca nhiu: c nhiu gii rng (ph rng nh ph ca nh sng trng gi

l tp m trng), nhiu gii hp (gi l tp m mu).

2. Theo quy lut bin thin thi gian ca nhiu: c nhiu ri rc v nhiu lin tc.

3. Theo phng thc m nhiu tc ng ln tn hiu: c nhiu cng v nhiu nhn.

4. Theo cch bc x ca nhiu: c nhiu thng v nhiu tch cc.

Nhiu thng l cc tia phn x t cc mc tiu gi hoc ta vt trvi ta xt khicc tia sng ca n p vo chng. Nhiu tch cc (chng) do mt ngun bc x nng lng

(cc i hoc cc h thng ln cn) hoc my pht nhiu ca i phng cha vo i hoc hthng ang xt.


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5. Theo ngun gc pht sinh: c nhiu cng nghip, nhiu kh quyn, nhiu v trvv

Trong gio trnh ny khi ni v nhiu, ta ch ni theo phng thc tc ng ca nhiu lntn hiu, tc l ch ni n nhiu nhn hoc nhiu cng.

V mt ton hc, tc ng ca nhiu cng ln tn hiu c biu din bi h thc sau:u(t) = s(t) + n(t) (2.4)

s(t) l tn hiu gi i

u(t) l tn hiu thu c

n(t) l nhiu cng

Cn nhiu nhn c biu din bi:

u(t) (t).s(t)= (2.5)

(t): nhiu nhn, l mt qu trnh ngu nhin. Hin tng gy nn bi nhiu nhn gi lsuy lc (fading).

Tng qut, khi tn hiu chu tc ng ng thi ca c nhiu cng v nhiu nhn th:

u(t) (t).s(t) n(t)= + (2.6)

y, ta coi h s truyn ca knh bng n v v b qua thi gian gi chm tn hiuca knh truyn. Nu kn thi gian gi chm ca knh truyn th (2.6) c dng:

u(t) (t).s(t ) n(t)= + (2.7)

2.3. CC C TRNG THNG K CA TN HIU NGU NHIN V NHIU

2.3.1. Cc c trng thng k

Theo quan im thng k, tn hiu v nhiu c coi l cc qu trnh ngu nhin. c trngcho cc qu trnh ngu nhin chnh l cc quy lut thng k (cc hm phn b v mt phn b)v cc c trng thng k (k vng, phng sai, hm t tng quan, hm tng quan). Cc quylut thng k v cc c trng thng k c nghin cu trong l thuyt hm ngu nhin, vvy y ta s khng nhc li.

Trong lp cc qu trnh ngu nhin, c bit quan trng l cc qu trnh ngu nhin sau:

- Qu trnh ngu nhin dng (theo ngha hp v theo ngha rng) v qu trnh ngu nhinchun dng.

- Qu trnh ngu nhin ergodic

Ta minh ho chng theo lc sau:


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Hnh 2.1

Trong nhng c trng thng k ca cc qu trnh ngu nhin, hm t tng quan v hmtng quan l nhng c trng quan trng nht. Theo nh ngha, hm t tng quan s bng:

{ }

[ ] [ ]

x 1 2 1 x 1 2 x 2

1 x 1 2 x 2 2 1 2 1 2 1 2

R (t , t ) M X(t ) m (t ) . X(t ) m (t )

x(t ) m (t ) . x(t ) m (t ) .W (x ,x , t , t )dx dx

=

= (2.8)

x 1 2R (t , t ) c trng cho s ph thuc thng k gia hai gi trhai thi im thuc cng

mt th hin ca qu trnh ngu nhin.

( )2 1 2 1 2W x , x , t , t l hm mt phn b xc sut hai chiu ca hai gi tr ca qu trnh

ngu nhin hai thi im 1t v 2t .

Khi t1 = t 2 th (2.8) trthnh:

[ ]{ }2x 1 2 x xR (t , t ) M X(t) m (t) D (t)= = (2.9)Nh vy, phng sai l trng hp ring ca hm t tng quan khi hai thi im xt trng

nhau.

i khi tin tnh ton v so snh, ngi ta dng hm t tng quan chun ho c nhngha bi cng thc:

x 1 2 x 1 2x 1 2

x 1 1 x 2 2 x 1 x 2

x 1 2

x 1 x 2

R (t , t ) R (t , t )(t , t )

R (t , t ).R (t , t ) D (t ).D (t )

R (t , t )

(t ). (t )

= =

=

(2.10)

D dng thy rng: x 1 2(t , t ) 1 .

QTNN QTNN

dng dng QTNNrng h p chun

QTNN chun dngQTNN

QTNN ergodic


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15

2.3.2. Khong tng quan

Khong tng quan cng l mt c trng kh quan trng. Ta thy rng hai gi tr ca mt

qu trnh ngu nhin (t) ch tng quan vi nhau khi khong cch gia hai thi im xt l

hu hn. Khi , th coi nh hai gi try khng tng quan vi nhau na. Tuy vy, trongthc t, i vi hu ht cc qu trnh ngu nhin ch cn ln th s tng quan gia hai gitr ca qu trnh mt. Do , i vi tnh ton thc t ngi ta nh ngha khong (thi gian)tng quan nh sau:

nh ngha 1:

Khong tng quan K l khong

thi gian trong ( ) khng nh hn

0,05. (hnh v 2.2). Nh vy, > K th

xem nh ht tng quan.

Nu cho biu thc gii tch ca ( )

th K c tnh nh sau:

K1

( ) d2

= (2.11)

ngha hnh hc:

K l na cnh y ca hnh ch nht c chiu cao bng n v K, c din tch bng din

tch ca min gii hn bi trc honh v ng biu din ( ) .

Trong thc t, ta thng gp nhng qu trnh ngu nhin ergodic. V d: tp m ca ccmy thu v tuyn in, i vi cc qu trnh ngu nhin ergodic, ta c th xc nh cc ctrng thng k ca chng bng thc nghim mt cch d dng.

Ta bit rng, nu X(t) ergodic v vi T ln th ta c th vit:

[ ] [ ]{ }

[ ] [ ]

x x x

T

x x0

R ( ) M X(t) m . X(t ) m

1x(t) m . x(t ) m dt

T

=

+ (2.12)

Trung bnh thng k = trung bnh theo thi gian

1 ()

0,05

0 k t

Hnh 2.2


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16

2.4. CC C TRNG VT L CA TN HIU NGU NHIN V NHIU. BINI WIENER KHINCHIN

2.4.1. Nhng khi nim xy dng l thuyt ph ca qu trnh ngu nhin - mt ph

cng sutMc trc ta mi cha ra mt sc trng thng k ca cc qu trnh ngu nhin (tn

hiu, nhiu) m cha a ra cc c trng vt l ca chng. V mt l thuyt cng nh thc t,cc c trng vt l ca tn hiu ngu nhin (qu trnh ngu nhin) ng mt vai tr rt quan trngnhng chng sau khi ni n csl thuyt chng nhiu cng nh xt cc bin php thc tv cc thit b chng nhiu ta khng th khng dng n nhng c trng vt l ca tn hiu ngunhin v nhiu. Khi xt cc loi tn hiu xc nh trong gio trnh L thuyt mch, chng ta lm quen vi cc c trng vt l ca chng nh: nng lng, cng sut, thi hn ca tn hiu,

ph bin phc, mt ph, b rng ph, Cs hnh thnh cc c trng vt l ny l

chui v tch phn Fourier.i vi cc tn hiu ngu nhin v nhiu, ta khng th dng trc tip cc bin i Fourier

xy dng cc c trng vt l ca chng c v nhng l do sau:

- Tp cc th hin { }ix (t) , i 1, 2,...,= ca qu trnh ngu nhin X(t) cho trn khong Tthng l mt tp v hn (thm ch n cng khng phi l mt tp m c).

- Nu tn hiu ngu nhin l dng cht th tp v hn cc th hin theo thi gian ca nthng s khng kh tch tuyt i. Tc l:

T 2

TT 2

lim x(t) dt

=

trnh khi nhng kh khn trn, ta lm nh sau:

Ly hm Tx (t) trng vi mt th hin ca qu trnh ngu nhin trung tm X(t) (QTNN trung

tm l QTNN c k vng khng) trong onT T

,2 2

v n bng khng ngoi on :

T

x(t) t T 2x (t) 0 t T 2

= >

(2.13)

T (2.13), ta thy Tx (t) tho mn iu kin kh tch tuyt i nn c th dng bin i

Fourier cho n c. Ta bit rng ph bin phc ( )TS ca Tx (t) c xc nh bitch phn thun Fourier sau:

( ) ( )T 2

j tT T

T 2

S x t e dt

= (2.14)

Theo nh l Parseval, ta c biu thc tnh nng lng ca Tx (t) nh sau:


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17

22

TT T1

E x (t)dt S ( ) d2

= = (2.15)

Cng sut ca th hin Tx (t) s bng:2

T2T

TT

S ( )E 1 1

P S ( ) d dT 2 T 2 T

= = = (2.16)

Ta thy v tri ca (2.16) l cng sut ca th hin Tx (t) trong khong thi gian tn ti hu

hn T, cn v phi l mt tng lin tc ca cc i lng

2

TS ( ) T d

. R rng l m

bo s bnh ng v th nguyn gia hai v ca (2.16) th lng

2

TS ( )

dT

phi biu th cng

sut trong gii tn v cng b d . Nh vy,

2

TS ( )

T

s biu th cng sut ca th hin Tx (t)

trong mt n v tn s [W/Hz] tc l mt ph cng sut ca th hin Tx (t) . n y ta t:

2

T

T

S ( )

G ( )T

= (2.17)

v gi TG ( ) l mt ph cng sut ca th hin Tx (t) trong khong T hu hn.

TG ( ) c trng cho s phn b cng sut ca mt th hin Tx (t) trn thang tn s. Khi cho

T ta s tm c mt ph cng sut ca mt th hin duy nht Tx (t) ca qu trnhngu nhin:

2

T

x TT T

S ( )

G ( ) lim G ( ) limT

= = (2.18)

xG ( ) cng c ngha tng t nh TG ( ) .

T (2.18) ta thy rng xc nh mt ph cng sut ca c qu trnh ngu nhin (tc l

tp cc th hin ngu nhin) th phi ly trung bnh thng k i lng xG ( ) , tc l:


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18

{ }

2

T

xT

S ( )

G( ) M G ( ) M limT

= = (2.19)

(2.19) l cng thc xc nh mt ph cng sut ca cc qu trnh ngu nhin.

2.4.2. Cp bin i Wiener Khinchin

thy c mi quan h gia cc c trng thng k (ni ring l hm t tng quan) v ccc trng vt l (ni ring l mt ph cng sut) ta vit li v thc hin bin i (2.19) nh sau:

{ }

1 2

1 2

2 2

T T

T T

*

T TT

T 2 T 2j t j t

T 1 1 T 2 2T

T 2 T 2

T / 2 T / 2j (t t )

T 1 T 2 1 2

T T / 2 T / 2

S ( ) M S ( )

G( ) M lim lim

T T1

lim M S ( )S ( ) do (2.14)T

1lim M x (t )e dt . x (t )e dt

T

1lim M x (t ).x (t ) e dt dt

T

= = =

=

= =

=

Nhng theo nh ngha (2.8), ta thy ngay { }T 1 T 2M x (t ).x (t ) l hm t tng quan ca

qu trnh ngu nhin trung tm (c xm 0= ) nn ta c th vit:

{ }T 1 T 2 T 1 2M x (t ).x (t ) R (t ,t )=

Nu 2 1t t = + th i vi nhng qu trnh dng, ta c:

{ }T 1 T 2 TM x (t ).x (t ) R ( )=

Ta c th vit li biu thc cho ( )G :

2

2

2

2

T t T / 22j

T 2T T T / 2t2

T t T / 22j

T 2T T

T T / 2t2

1G( ) lim R ( )e d dt

T

1lim R ( )e d . lim dt

T

+

+

=

=


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jG( ) R( )e d

= (2.20)

Tt nhin y phi gi s tch phn v phi ca (2.20) tn ti. iu ny lun lun ng

nu hm t tng quan R( ) kh tch tuyt i, tc l:

R( )d

<

(2.20) l mt ph cng sut ca qu trnh ngu nhin dng. N biu din mt cch trungbnh (thng k) s phn b cng sut ca qu trnh ngu nhin theo tn s ca cc thnh phn daong iu ho nguyn t (tc l nhng thnh phn dao ng iu ho v cng b).

Nh vy, t (2.20) ta c th kt lun rng ph cng sut G( ) ca qu trnh ngu nhindng l bin i thun Fourier ca hm t tng quan R( ) . Hin nhin rng khi tn ti bini thun Fourier th cng tn ti bin i ngc Fourier sau:

j1R( ) G( )e d2

= (2.21)

Cp cng thc (2.20) v (2.21) gi l cp bin i Wiener Khinchin, l s mrng cpbin i Fourier sang cc tn hiu ngu nhin dng (t nht l theo ngha rng).

R rng tnh ngha (2.17) ca mt ph cng sut, ta thy hm G( ) l hm chnca i s . Do sau khi dng cng thc Euler ( je cos jsin = ) bin i(2.20) v (2.21), ta c:

0

0

G( ) 2 R( )cos d

1R( ) G( )cos d

=

=

(2.22)

Ch 1: T mt ph cng sut ca tn hiu ngu nhin, khng th sao li bt c mt th

hin no (l hm ca thi gian t) ca n, v G( ) khng cha nhng thng tin (nhng hiu bit)v pha ca cc thnh phn ph ring l. i vi tn hiu xc nh th t mt ph hon ton cth sao li chnh tn hiu nh tch phn ngc Fourier. l ch khc nhau v bn cht gia

bin i Fourier v bin i Wiener Khinchin.

Ch 2: Nu phi xt ng thi hai qu trnh ngu nhin th ngi ta cng a ra khinim mt ph cho. Mt ph cho v hm tng quan cho ca hai qu trnh ngu nhin clin h dng cng tho mn cp bin i Wiener Khinchi.

2.4.3. B rng ph cng sut


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20

Mt c trng vt l quan trng khc cacc tn hiu ngu nhin l b rng ph cng sut,n c nh ngha bi cng thc sau:

0

0

G( )d

G( )

=

(2.23)

Trong :

G( ) l mt ph cng sut ca tn hiungu nhin.

G( 0 ) l gi tr cc i ca G( ).

l b rng ph cng sut (cn gi lb rng ph) ca qu trnh ngu nhin.

ngha hnh hc:

B rng ph chnh l y ca hnh ch nht c chiu cao bng G( 0 ) v c din tchbng din tch ca min gii hn bi trc v ng cong biu din G( ). (Hnh 2.4).

ngha vt l:

B rng phc trng cho s tp trung cng sut (hoc nng lng) ca tn hiu ngu nhinquanh mt tn s trung tm, ngoi ra n cng c trng cho c s bng phng ca phquanh

tn s trung tm 0 .

2.4.4. Mrng cp bin i Wiener Khinchin cho trng hp R( )khng kh tch

tuyt i

Nu qu trnh ngu nhin X(t) cha cc thnh phn dao ng iu ho dng:

K K K K X (t) A cos( t )=

trong KA v K ni chung c th l cc i lng ngu nhin, th hm tng quan trung bnh:

K

2* KX K

AR ( ) cos

2 = khng tho mn iu kin kh tch tuyt i.

Nu s dng biu din sau ca hm delta:

ixye dx cos(xy)dx (y)

= =

v biu din ph nng lng ca KX (t) di dng:

0,05

0

G()

Hnh 2.3


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21

[ ]2

* KK K K

AG ( ) ( ) ( )

4 = + +

th nh l Wiener Khinchin sng ci vi nhng qu trnh ngu nhin c nhng

thnh phn tn s ri rc, k c thnh phn mt chiu tn s K = 0.

2.5. TRUYN CC TN HIU NGU NHIN QUA CC MCH V TUYN INTUYN TNH

i vi cc tn hiu xc nh, trong gio trnh L thuyt mch, ta xt bi ton phn

tch sau: Cho mt mch tuyn tnh c cu trc bit (bit hm truyn t K( )

hoc bit phnng xung g(t)). Ta phi xt tc ng u vo theo hng ng u ra v ngc li. i vi cc tnhiu ngu nhin nu s th hin l m c v hu hn th ta c th xt hng ng ra i vitng tc ng u vo nh bi ton trn. Nhng khi s th hin ca tn hiu ngu nhin l v hn

th ta khng th p dng c nhng kt qu ca bi ton phn tch i vi cc tn hiu xc nh.Sau y ta s xt bi ton ny.

2.5.1. Bi ton ti thiu

2.5.1.1. Bi ton:

Cho mt mch tuyn tnh (c tham s khng i v bit K( )

ca n. Bit mt ph

cng sut vG ( ) ca qu trnh ngu nhin tc ng u vo. Ta phi tm mt ph cng sut

raG ( ) v hm t tng quan raR ( ) ca qu trnh ngu nhin u ra.

2.5.1.2. Gii bi ton:

gio trnh L thuyt mch ta bit hm ph bin phc ca tn hiu u ra mchv tuyn in tuyn tnh bng:

ra vS ( ) K( ).S ( )

= (2.24)

K()GV() Gra())

Ph bin SK()

AK/2 AK/2

- K 0 K

Ph nng lng GK()

( + K) ( - K)

- K 0 K


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Trong : K( )

l hm truyn ca mch bit.

vS ( )

l ph bin phc ca tn hiu vo

Ch : i vi cc qu trnh ngu nhin ta khng bit c vS ( )

. Khng th tnh c

vS ( )

, mt khc ta bit theo (2.19):

22

v Tra T

vT T

2

ra T

ra2 2T

S ( )S ( )1

G ( ) M lim M limT T

K( )

S ( )1 1

M lim .G ( )T

K( ) K( )

= =

= =

Hay:

2

ra vG ( ) K( ) .G ( )

= (2.25)

Ngi ta chng minh c rng hng ng ra ca h thng tuyn tnh c tham s khngi l mt qu trnh ngu nhin khng dng ngay c khi tc ng u vo l mt qu trnh ngunhin dng.

Tuy vy, trong trng hp h thng tuyn tnh thng c suy gim th nhng thi im

t >> t 0 = 0 (thi im t tc ng vo) th qu trnh ngu nhin u ra sc coi l dng.

Khi hm t tng quan v mt ph cng sut ca qu trnh ngu nhin u ra slin h vi nhau theo cp bin i Wiener Khinchin. Ta c:

jra ra

1R ( ) G ( )e d

2

= (2.26)

Nhn xt:

T (2.25) ta thy mt ph cng sut ca hng ng ra c quyt nh bi bnh phngmun hm truyn ca mch khi cho ph cng sut ca tc ng vo, n khng ph thuc gvo c tnh pha tn ca mch.

Cng sut ca qu trnh ngu nhin u ra (khi qu trnh ngu nhin vo l dng):


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23

22

ra ra ra v1 1

R (0) G ( )d P K( ) G ( )d2 2

= = = = (2.27)

Nu ph cng sut ca tc ng vo khng ph thuc tn s, tc l vG ( ) = 0N (qutrnh ngu nhin c tnh cht ny c gi l tp m trng) th:

2

ra 01

P N K( ) d2

= (2.28)

V mun hm truyn lun l mt hm chn nn:

2

ra 0

0

2P N K( ) d

2

=

(2.29)

Mt khc, nu gi 0G l ph cng sut thc t (phn ph cng sut tri t 0 ) th

0G = 2 0N v (2.29) c th vit li nh sau:

20

ra0

GP K( ) d

2

=

(2.30)

Hm t tng quan ca qu trnh ngu nhin u ra trong trng hp ny s bng:

2j

ra v

2j

0

2j0

1R ( ) G ( ) K( ) e d

2

1N K( ) e d

2

NK( ) e d

2

=

=

=

2

0ra

0

GR ( ) K( ) cos d

2

=

(2.31)

2.5.1.3. V d 1

Mt mch v tuyn in tuyn tnh c tham s khng i v c tnh truyn t dng chnht (hnh 2.4b) chu tc ng ca tp m trng dng. Tm hm t tng quan ca tp m ra.

GV()

2N0

0 1 0 2

| K() |


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24

Theo gi thit: v 0G ( ) 2 N = v0 1 2

1 2

KK( )

0 ( , )

< < =

Theo (2.31), ta c:

2

1

220 0 0

ra 0 2 1

20 0

0

N N K R ( ) K cos d = (sin sin )

sinN K 2. cos2

=

=

2ra ra 0

sin2R ( ) cos2

=

(2.32)

th ( )raR nh hnh 2.5.

(2.32) c th vit gn li nh sau:

ra 0ra 0R ( ) R ( )cos = (2.32a)

Trong :

20ra ra

sin 2R ( )

2

=

(2.32b)

(2.32b) gi l bao ca hm t tng quan ca hng ng.

1 20 2

+ = (2.32c)

gi l tn s trung bnh.


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25

Vy, bao ca hm t tng quan ca tp m ra l mt hm ca i s dngsinx

x. Cc

i ca hm t tng quan ca tp m ra t ti = 0 v bng 2ra , tc l bng cng sut trung

bnh ca tp m ra.

By gita s chuyn sang xt mt tham s vt l na nh gi mc truyn tp m quamch tuyn tnh.

2.5.1.4. Gii thng tp m

nh ngha:

Gii thng t p m camch tuyn tnh (hay b lctuyn tnh) c xc nh theo

biu thc sau:

2

0t 2

K( ) d

K( ) m

ax

=

(2.33)

2/

Rra()

2ra

0

0 ta

| K()|2

| K()|2max

Hnh 2.6.

Hnh 2.5.


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26

ngha hnh hc: t chnh l y ca hnh ch nht c din tch bng din tch ca

min gii hn bi ng cong

2

K( )

v na trc honh (0, ); cn chiu cao ca hnh ch

nht ny l

2

K( )

max.

ngha vt l:

t c trng cho kh nng lm suy gim tp m ca cc b lc tuyn tnh. Vi cng

0K( )

, b lc no c t cng hp th cng sut tp m u ra ca b lc y cng b.

2.5.2. Bi ton ti a

RG ( ) v RB ( ) cha c trng y cho qu trnh ngu nhin.

Ni dung: Tm hm mt xc sut ca tn hiu u ra mch v tuyn in tuyn tnh.

2.5.2.1. Mu

Tm mt xc sut n chiu ca tn hiu ngu nhin u ra mch tuyn tnh l bi ton rtkh, n khng gii c di dng tng qut. Di y ch xt hai trng hp n gin:

- Tm mt xc sut mt chiu ca tn hiu ra b lc tuyn tnh khi tc ng u vo l tnhiu ngu nhin chun (c v hn th hin). Trong trng hp ny ngi ta chng minh ctn hiu ra cng l mt tn hiu ngu nhin chun.

- t vo b lc tuyn tnh mt tn hiu ngu nhin khng chun. Nu t 12 F


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27

Cng sut trung bnh ca c hai thnh phn ca nhiu bng nhau v bng hng s:2 2 2c s = = . Khi n(t) dng, ngi ta coi l hai thnh phn ca nhiu khng tng quan.

Tc ng n(t) ln b tch sng tuyn tnh. Hy tm mt xc sut mt chiu ca in p ra

b tch sng bit rng b tch sng khng gy mo ng bao v khng gy thm mt lng dchpha no. Thc cht ca bi ton l phi tm (A) v1 1W W ( ) .

Trong gio trnh l thuyt xc sut, ta c cng thc tm mt xc sut mt chiu catng i lng ngu nhin theo mt xc sut ng thi ca chng, nn ta c:

2

0 0

(A) (A, )d ; ( ) (A, )dA1 2 1 2W W W W

= =

Do , vn y l phi tm (A, )2W .

V b tch sng khng gy mo ng bao v khng gy thm mt lng dch pha no nn

(A, )2W u ra cng chnh l (A, )2W u vo.

Tm (A, )2W : V u bi ch cho (c) v (s)1 1W W nn ta phi tm (A, )2W theo

(c,s)2W .

Theo gi thit c(t) v s(t) khng tng quan nn:

(c,s)2W = (c). (s)1 1W W (2.34)

( )2 2 2 2 2 2c 2 s 2

2 2 21 1 1 c s

W c,s e . e exp2 2 2 2

+ = =

( ) 22 2 21 1

W c,s exp A2 2

=

(2.35)

Ta thy xc sut mt im c to (c,s) trong h tocac ri vo mt yu t din

tch dcds s bng: (c,s)dcds 2P W dcds= . n (*) ta thy xc sut ny cng chnh l xcsut mt im c to(A, ) trong h to cc ri vo mt yu t din tch dAd . Ta c:


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28

(c,s) (A, )dcds 2 2P W dcds = W dAd= (2.36)

T:

(A, ) (c,s)2 2dcds

W WdAd

=

(**)

T H.2.7 ta thy vi dA, d nh ta c: dc ds = Ad . DA

T (**) ta c:

( ) ( )2

2 2 2 21 AW A, W c,s exp

2 2 = =

(2.37)

Do : ( ) ( )2 22

1 2 2 20 0

A AW A W A, d exp d

2 2

= =

( )2

1 2 2

A AW A exp

2

=

(2.38)

(2.38) gi l phn b Reyleigh (H.2.8).

ddA

S + dS

S

S

0 c c + dc c

A

Hnh 2.7.


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29

Vy nhiu gii hp m tr tc thi c phn b chun th phn b ca ng bao l phn b

khng i xng Reyleigh. Sdnh vy v gi tr tc thi c c gi tr m v gi tr dng nnphn b mt xc sut si xng qua trc tung (phn b Gausse). Cn xt ng bao tc lch xt bin (gi tr dng) nn mt phn b xc sut l ng cong khng i xng v chtn ti na dng trc honh.

( ) ( )2

1 2 2 20 0

1 A AW W A, dA exp dA

2 2

= =

( ) ( )1 1

0

1W W A dA

2

= (2.39)

Vy mt phn b xc sut pha u ca nhiu gii h p, chun l phn b u trongkhong (0,2 ). (H.2.9).

2.5.2.3. V d 3:

u vo b tch sng tuyn tnh t hn hp tn hiu v nhiu:

y(t) = x(t) + n(t)

Vi: 0x(t) U c t0os= l tn hiu xc nh.

[ ]nn(t) A (t)c t (t)0os= l nhiu gii hp, chun.

Tm mt phn b xc sut ng bao v pha ca in p u ra b tch sng tuyn tnh.

Ta c:

[ ]

0 0

0 0 y y

y(t) U c t c(t)c t s(t)sin t

U c(t) c t s(t)sin t A (t)c t (t)

0 0

0 0

os os

os os

= + +

= + + =

Trong : [ ]2 2

y 0A (t) U c(t) s (t)= + + l bao ca hn hp tn hiu v nhiu.

y (t)

c(t)0

s(t)arctang

U

=

+

l pha ca hn hp tn hiu v nhiu.

Lm tng t nh VD2, ta c:

W1(A/)

0,6

0,4

0,2

0 1 2 3 A/

Hnh 2.8.

W1()

1/2

0 2

Hnh 2.9.


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31

Nhn xt:

- a = 0 ch c nhiu y( )1W chnh l ( )1W xt VD2.

- a >> 1 ng cong y( )1W cng nhn, hp.

Gii thch:

Vi a cng ln th c th b qua nh hng xu ca nhiu. Do ng bao (bin tn

hiu) khng c gia s (khng thng ging) v cng khng c sai pha. Khi y nhn gi tr 0

trong khong (- , ) vi xc sut ln.

2.6. BIU DIN PHC CHO TH HIN CA TN HIU NGU NHIN TN HIUGII HP

2.6.1. Cp bin i Hilbert v tn hiu gii tch

2.6.1.1. Nhc li cch biu din mt dao ng iu ho di dng phc

Cho: x(t) = ( )0 0A c t A(t)c0os os (t) + = (2.42)

Trong :

0 : tn s trung tm; (t) : pha y ;

0 : pha u.

Trong L thuyt mch, ngi ta rt haydng cch biu din x(t) di dng phc sau:

j (t)x(t) x(t) jx(t) A(t)e

= + = (2.43)

Trong :

x(t) = Re [ x(t)

];

x(t)

= Im [ x(t)

] = 0A sin (t)

Im[x(t)]

x(t) M

A

(t)

0 x(t) Re[x(t)]

Hnh 2.11


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32

Ta c th biu din x(t)

di dng mt vecteur trn mt phng phc.

Khi A(t) = const th qu tch ca im M s l mt vng trn tm O, bn knh OM.

(t) d (t) dt = l tn s ca dao ng (H.2.11)2.6.1.2. Cp bin i Hilbert Tn hiu gii tch

a. Cp bin i Hilbert v tn hiu gii tch:

d dng biu din di dng phc nhng th hin phc tp ca cc qu trnh ngu nhin,

ngi ta dng cp bin i Hilbert. N cho php ta tm x(t)

khi bit x(t) v ngc li.

Hilbert chng t rng phn thc v phn o ca hm phc (2.43) lin h vi nhau bi ccbin i tch phn n tr hai chiu sau:

x( )x(t) Im (t) d

t

1[x ] =

= =

h [x(t)] (2.44)

x( )x(t) d Re (t)

t

1[x ]

= = =

h1 [x(t)] (2.45)

Cp cng thc trn c gi l cp bin i Hilbert. Trong (2.44) gi l bin i thun

Hilbert, cn (2.45) gi l bin i ngc Hilbert.Ch :

Cng ging nh tnh cht ca cc tch phn, bin i Hilbert l mt php bin i tuyntnh.

(Mt php bin i fc gi l tuyn tnh nu c:

f(x1 + x 2 ) = f(x1 ) + f(x 2 )

f(kx) = k f(x), k = const)

Cc hm x(t) v x(t) c gi l lin hip Hilbert i vi nhau. Tn hiu phc x(t) cphn thc v phn o tho mn cp bin i Hilbert gi l tn hiu gii tch (tng ng vi tnhiu thc x(t)).

b. Bin i Hilbert i vi tn hiu hnh sin:

Trong mc ny ta s chng tc t0os v t0sin tho mn cp bin i H. Tht vy:


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33

1 c 1 c (t ) tx(t) d d

t t

1 c (t ).c t sin (t ).sin tdt

c t c (t ) sin t sin (t )d d

t t

0 0 0

0 0 0 0

0 0 0 0

os os[ ]

os os ]

os os

= = =

+ = =

= +

Ch rng:c

dz 0osaz

z

= vsin

dzaz

z

=

0x(t) sin t =

Vy ( 0sin t ) l lin hp H ca ( 0c tos )

Tng t ( - 0c tos ) l lin hp phc H ca ( 0sin t )

c. Bin i H i vi cc hm tng qut hn:

- i vi cc hm tun hon x(t):

Trong L thuyt mch ta bit, chui Fourier ca hm tun hon (tho mn iu kin

Dirichlet) l:

K K 0K 0

x(t) (a c K t b sin K t)0os

=

= + (2.46)

V bin i H l bin i tuyn tnh nn bin i H ca tng bng tng cc bin i H cacc hm thnh phn, nn:

x(t)

= h [x(t)] K K 0K 0

(a sin K t b c K t)0 os

=

= (2.47)

(2.46) v (2.47) gi l chui lin hip H.

- x(t) khng tun hon:

Nu hm khng tun hon x(t) kh tch tuyt i th khai trin Fourier ca n l:

0

1x(t) a( )c t b( )sin t

2[ os ]d

= + (2.48)

Khi :


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34

x(t)

= h [x(t)] =1

2h

0

[a( )cos t + b( )sin t]d

=

{ }0

1 H[a( )cos t] + H[b( )sin t] d2

=

0

1[a( )sin t - b( )cos t]d

2

= (2.49)

(2.48) v (2.49) gi l cc tch phn lin hip H.

d. Cc yu tca tn hiu gii tch:

T (2.46) v (2.47) (hoc t (2.48) v (2.49)) ta xy dng c tn hiu gii tch ng vi tnhiu thc x(t) nh sau:

j (t)x(t) x(t) jx(t) A(t)e

= + =

x(t) = Re [ x(t)

] = A(t)cos (t) (a)

x(t)

= Im [ x(t)

] = A(t)sin (t) (b)

- ng bao ca tn hiu gii tch:

T (a) v (b) ta thy:

22A(t) x (t) x (t)

= + (2.50)

A(t) c trng cho s binthin (dng bin thin) ca bin ca tn hiu (H.2.12).

A(t) c gi l ng baoca tn hiu (cn gi l bin

bin thin hay bin tc thi catn hiu).

- Pha tc thi ca tn hiugii tch:

K hiu pha tc thi: (t) bng:

x(t)(t) arctg

x(t)

= (2.51)

t

x(t)

A(t)

Hnh 2.12


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35

- Tn sgc tc thi ca tn hiu gii tch (t) :

d (t) x(t)(t) arctgdt x(t)

= = =

2 22

2

x(t) x(t)

x(t) x (t) x(t)x (t)x (t) x (t) x (t)

1x (t)

=+

+

(2.52)

- Tnh cht ca A(t):

+ A(t) x(t)

+ Khi x(t)

= 0 A(t) = x(t)

+ Xt:2

2

x(t).x (t) x(t).x (t)A( t )

x (t) x (t)

+ =

+

Khi x(t)

= 0 A(t) = x(t)

Vy khi x(t)

= 0 th nghing ca A(t) v x(t) l nh nhau.- Kt lun:

i vi cc tn hiu ngu nhin th cc yu t ca tn hiu l ngu nhin. Nhc

khi nim tn hiu gii tch nn ta mi nghin cu cc tnh cht thng k ca cc yu

t ca n c thun li, c bit l trong tnh ton.

2.6.2. Tn hiu gii rng v gii hp

2.6.2.1. Tn hiu gii rng

Ngi ta gi mt tn hiu l tn hiu gii rngnu b rng ph ca n tho mn bt ng thcsau:

01

(2.53)

Nhn chung tn hiu gii rng l tn hiu m

b rng ph ca n c th so snh c vi 0 .

Trong 2 1 = v2 1

0 2 + = gi l tn s trung tm (xem H.2.13).

0 1 0 2

G()

Hnh 2.13


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36

V d: Cc tn hiu iu tn, iu xung, iu ch m xung, manip tn s, manip pha, lcc tn hiu gii rng.

2.6.2.2. Tn hiu gii hp

Nu tn hiu c b rng ph tho mn:

01

(2.54)

Th n c gi l tn hiu gii hp. (H.2.14).

V d: tn hiu gii h p l cc tn hiu nh:tn hiu cao tn hnh sin, tn hiu cao tn iu bin,tn hiu n bin .

Nhn chung tn hiu gii hp l tn hiu m brng ph ca n kh nh hn so vi tn s 0 .

2.6.2.3. Biu din tn hiu gii hp

Nu mt tn hiu gii hp c biu thc gii tch sau:

0x(t) A(t)cos[ t (t)] = A(t)cos (t)= (2.55)

Trong : 0t l thnh phn thay i tuyn tnh ca pha chy (pha tc thi)

(t) l thnh phn thay i chm ca pha chy

A(t) l ng bao ca tn hiu

Th (2.55) c th khai trin nh sau:

0 0

0 0

x(t) A(t)cos t cos (t) A(t)sin t sin (t)

A(t)cos (t)cos t A(t) sin (t) sin t

= +

= +

= c(t). cos 0t + s(t). sin 0t (2.56)

c(t). cos 0t l tn hiu iu bin bin i chm

s(t). sin 0t l tn hiu iu bin bin i chm

Vy mt tn hiu gii hp hnh sin bao gicng c th biu din di dng tng ca hai tnhiu iu bin bin i chm, vi cc yu t xc nh nh sau:

0 1 0 2

G()

Hnh 2.14


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38

n n

2 2K K

c cK 1 K 1

1E x x

2F= =

= =

(2.59)

(2.59) cho ta tnh c nng lng ca chui

2.7.1.2. Biu din x(t) thnh vect x

trong khng gian n chiu

Khai trin Kachennhicov (2.58) l mt dng khai trin trc giao. Cc hm

cK

c

sin (t K t)(t)

(t K t)

=

l cc hm trc giao.

cc c

c c

i Ksin (t K t) sin (t i t). dt

(t K t) (t i t) 0 i K

= =

V vy ta c th coi mi hm l mt vecteurn v trn h trc to trc giao. Khi T hu

hn th Kmax = n cng s hu hn. Khi ta c th coi x(t) l mt vect x

trong khng gian n

chiu c cc thnh phn (hnh chiu) trn cc trc to tng ng l x(K t) , (K = 1,n ).

{ }

{ }1 2 n

x(t) x(t t),x(t 2 t),..., x(t n t)

x(t) x ,x ,...,x x

Theo nh ngha, di (hay chun) ca vecteur x

s l:

n2K

K 1

x x ( x , x )

=

= =

(2.60)

n (2.59), ta c:

c cx 2F E 2F T.P nP

= = = (2.61)

( cT

n 2F Tt

= =

)

Trong P l cng sut ca th hin tn hiu trong khong hu hn T. Nh vy, vi thihn quan st v b rng ph ca th hin cho trc th di ca vecteur biu din t l vi cn

bc hai cng sut trung bnh ca n. Nu cho trc cng sut trung bnh P th di ca vecteur

x

s t l vi n (tc l t l vi cn bc hai ca y tn hiu B = cn

F T

2

= )

Nhn xt:


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39

Nh vy, vi cng mt cng sut trung bnh tn hiu no c y cng ln (tc l tn hiucng phc tp) th di ca vecteur biu din n cng ln. Khi y ca tn hiu cng ln th di ca vecteur tn hiu cng ln vecteur tng ca tn hiu v nhiu gii hp cng t khcvecteur tn hiu ta s nhn ng c tn hiu vi xc sut cao. tnh chng nhiu ca tn

hiu cng cao th yu cu B cng phi ln.

Trong trng hp x(t) khng ri rc ho:

T2

x0

E x (t)dt= . Khi chun ca vecteur s

l:

T2

c x c0

x ( x , x ) 2F E x 2F x (t)dt

= = = (2.62)

Ngi ta cn gi khng gian m chun ca vecteur cho bi tch v hng (2.62) l khnggian Hilbert v k hiu l L

2. Khng gian L

2l s mrng trc tip ca khng gian Euclide hu

hn chiu ln s chiu v hn.

2.7.2. Mt xc sut ca vecteur ngu nhin - Khong cch gia hai vecteur tn hiu

2.7.2.1. Mt xc sut ca vecteur ngu nhin

a. Vecteur tn hiu:

tip tc nhng vn sau ny c thun tin, ta a vo khi nim vecteur tn hiu.

nh ngha:

Vecteur tn hiu 0x

l vecteur sau: 0x

xn

= (2.63)

Trong x

l vecteur biu din tn hiu x(t) trong khng gian n chiu.

Tnh cht:

+ 0x

c phng v chiu trng vi x

+ ln (modul): 0x

x Pn

= =

b. Xc sut phn b ca mt vecteur 0x

v min xc nh ca n

Trong khng gian tn hiu, tn hiu c biu din bi vecteur. Do xc sut tn ti tn

hiu mt min (ni ring: ti mt im) no y ca khng gian chnh l xc sut mtvecteur tn hiu ri vo min y (ni ring: im y) ca khng gian.


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40

Nu x(t) l xc nh th mt ca vecteur 0x

ch chim mt im trong khng gian n chiu.

Cn nu x(t) l ngu nhin c mt tp cc th hin { }ix (t) th mt vecteur 0x

ca n s chim

mt min no trong khng gian n chiu vi th tch: 1 2 nV x . x .... x= . Khi y, xc sut tn ti tn hiu ngu nhin trong min c th tch dV s l:

{ }

( )1 2 n 1 2 n 0

P t / h NN dV P dV

dP x ,x ,..., x dx dx ...dx (x )dVn n

{mt vecteur t/h } =

W W

=

= = =(2.64)

Sau y ta s xt min xc nh ca mt s dng tn hiu ngu nhin:

- Cc th hin ca tn hiu pht c cng y, cng cng sut:

Khi min cc nh ca vecteur tn hiu pht s l mt cu c bn knh bng chun ca

vecteur tn hiu pht 0x P

= v c tm gc to ca vecteury. (Sdnh vy v 0x

c chun khng i nhng phng v chiu ca n thay i ngu nhin).

- Tp m trng:

Ta bit rng cc th hin in (t) ca tp m trng n(t) c cng cng sut P n . Nh vy

min xc nh ca tp m trng l mt cu c bn knh bng nP , c tm l gc ca vecteur tp

m 0n

.

- Tng ca tn hiu x(t) v tp m n(t):

y(t) = x(t) + n(t)

0 0 0 0 yy x n y P

= + =

Nu x(t) v n(t) khng tng quan th:

y x nP P P= + (v y x nB (0) B (0) B (0)= + )

2

x n x n0 0y P P y P P

= + = +

2 2 2

0 0 0y x n

= + (*)


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41

T (*) ta thy 0x

0n

v 0y

l cnh huyn ca mt tam gic vung c hai cnh l 0x

v 0n

.

Nu x(t) xc nh th min xc nh ca mt 0y

s l ng trn y ca hnh nn c nh

gc ta , chiu cao bng 0x

v bn knh bng 0n

. (H.2.15a).

Nu x(t) ch l mt th hin no ca qu trnh ngu nhin X(t) c cc th hin cng cng

sut th lc min xc nh ca mt 0y

s l mt mt cu c bn knh bng x nP P+ v ctm gc to (H.2.15b).

2.7.2.2. Khong cch gia hai vecteur tn hiu

nh gi nh lng s khc nhau gia hai vecteur tn hiu, ta a ra khi nim khongcch gia hai vecteur tn hiu.

nh ngha:

Khong cch gia hai vecteur tn hiu 0u

v 0v

c xc nh theo biu thc sau:

n0

x0

y0

0Hnh 2.15a

0

Hnh 2.15b


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42

0 0 0 0

n2

0 0 K K K 1

1d(u ,v ) u v u v

n

1d(u ,v ) (u v )n

=

= =

=

Hay:n n n

2 2 20 0 K K K K 2 2

K 1 K 1 K 1

1 1 2d (u ,v ) u v u .v

n( n ) ( n )

= = =

= +

Ta c:

2 2n2K 0 0 0 0 02

K 1

2 2n2K 0 0 0 0 02

K 1

n

K K 0 0 0 0 0 0K 1

1 1u u u u . u c (u ,u )

n( n )

1 1v v v v . v c (v ,v )

n( n )

1u .v (u ,v ) u . v c (u ,v )

n

os

os

os

=

=

=

= = =

= = = = =

2 22

0 0 0 0 0 0 0 0

2 22

0 0 0 0 0 0

d (u ,v ) u v 2 u . v c (u ,v )

d (u ,v ) u v 2 u . v c

os

os

= +

= +

Trong l gc hp bi 0u

v 0v

trong khng gian n chiu.

0 0

0 0

u .vc

u . v

os

= (2.65)

20 0 u v u vd (u ,v ) P P 2 P P cos

= + (2.66)

Nu ta khng ri rc ho tn hiu th:

T

0 0 0 00

1d(u ,v ) u v dt

T2[u(t) - v(t)]

= =


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43

Hay

T T T2

0 00 0 0

1 1 2d (u ,v ) dt v dt v dt

T T T2 2u (t) (t) u(t). (t)= +

u v u v

u v u vP P 2R (t, t)P P 2R (0)

= + = +

Trong u vR (0) l hm tng quan cho ca tn hiu u(t) v v(t).

u v u v u vR (0) D (t).D (t) (0)=

20 0d (u ,v ) = u v u v u vP P 2 P .P (0)+ (2.67)

So snh (2.66) v (2.67) ta thy ngay ngha hnh hc ca hm tng quan cho chun ho:u v (0) ng vai tr cosin ch phng ca hai vecteur tn hiu.

u vc (0)os = (2.68)

Kt lun:

- Vi mt mc nhiu xc nh, xc sut thu ng cng cao khi cc th hin ca tn hiucng cch xa nhau.

- Khong cch gia hai mt ca hai vecteur tn hiu cng ln khi di hai vecteur cng

ln.

2.7.3. Khi nim v my thu ti u

2.7.3.1. My thu ti u

Mt cch tng qut, ta coi mt my thu c trng bi mt ton t thu (H.2.17). Yu cu

ca ton t thu l tc dng vo y(t) (l tn hiu vo) phi cho ra tn hiu pht x(t).

Nu ta pht i mt th hin no ca mt qu trnh ngu nhin X(t):

{ }iX(t) x (t) (i 1,m)= =

Ta coi nhng th hin ny c cng cng sut P x , c cng thi

hn T v c cng b rng ph F c .

Gi thit: trong qu trnh truyn t ni pht n ni thu ch ctp m trng Gausse n(t), cc tn hiu pht l ng xc sut

Vecteur tn hiu ta nhn c: 0y y n

=

y(t) x(t)

Hnh 2.16.


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45

BI TP

2.1. th gi tr trung bnh a(t) v gi tr trung bnh bnh phng ( )t ca cc qu trnh ngunhin X(t), Y(t) v Z(t) v trn hnh 1 di y. Hy ch ra trn th min cc gi tr c th c

ca cc qu trnh ngu nhin ny, bit rng bin gii ca cc min c xc nh bi cc gi trca ( )t .

Hnh 1.

2.2. Trn hnh 2 v hm ngu nhin dng ri rc X(t), gi l dy xung in bo. Dy xung c bin khng i bng n v, c rng ngu nhin.

Phn b xc sut cc gi tr (0 hoc 1) ca X(t) tun theo lut Poisson:

ax(t)

0t

0

ay(t)

t t

az(t)

0

t tt

( )y t

00

( )z t

0

x(t)

1

0

Hnh 2.


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46

( )( )

nt

nt

P t e t 0n!

= >

Trong l s cc bc nhy ca hm X(t) trong mt n v thi gian, cn ( )nP t lxc sut xy ra n bc nhy ca hm X(t) trong thi gian t.

Hy tm hm t tng quan, hm tng quan chun ho v thi gian tng quan ca qutrnh ngu nhin, bit rng P(1) = P(0) = 0,5.

2.3. Tm hm t tng quan ca qu trnh ngu nhin dng sau:

( ) ( )0X t A cos 2 f t= +

Trong A = const, 0f = const, l i lng ngu nhin c phn bu trong khong

( ), .2.4. Tm hm t tng quan v mt ph ca tn hiu in bo ngu nhin X(t) cho bi hnhdi y. Bit rng n nhn cc gi tr + a; - a vi xc sut nh nhau v bng 1/2. Cn xc sut trong khong c N bc nhy l:

( )( )

N

P N, e 0N!

= >

(theo phn b Poisson).

2.5. Hy chng t rng ng bao ca tn hiu gii tch c th biu din bng cng thc sau:

( ) ( ) ( )*a aA t S t .S t=

Trong : ( )*aS t l hm lin hp phc ca ( )aS t :

( ) ( ) ( )aS t x t jx t

= + l tn hiu gii tch.

2.6. Mt qu trnh ngu nhin dng c hm t tng quan:

a. ( )1

2xR .e

=

b. ( )2

2x 0R .e .cos

=

Hy tnh ton v v th mt ph ca cc qu trnh ngu nhin trn.


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Chng 3: Csl thuyt thng tin thng k

47

CHNG 3 - C S L THUYT THNG TIN THNG K

3.1. THNG TIN - LNG THNG TIN XC SUT V THNG TIN N VO THNG TIN

3.1.1. nh ngha nh tnh thng tin v lng thng tin

3.1.1.1. Thng tin

chng trc, ta hc khi nim v thng tin. y ta s xy dng nh ngha nhtnh ca thng tin theo quan im thng k. i ti nh ngha nh tnh ca thng tin, ta s xtv d sau:

Ta nhn c mt bc in (th) t nh n. Khi cha mbc in ra c th ta ch c th don hoc th ny hoc th khc v bc in, m khng dm chc ni dung ca n l g. Ni khci, khi cha mbc in ra c th ta khng th xc nh c ni dung ca n, tc l ta cha bitgia nh bo cho ta thng tin g. Nhng khi xem xong bc in th ni dung ca n i vi ta hon ton r rng, xc nh. Lc , ni dung ca bc in khng cn bp bnh na. Nh vy, ta nirng: ta nhn c mt tin v gia nh. Ni dung ca bc in c th c 3 c im sau:

- Ni dung ta tha bit. (VD: Cc em con c ngh h 3 thng). Khi bc inkhng cho ta mt hiu bit g mi v tnh hnh gia nh. Hay ni theo quan im thng tin, th bcin vi ni dung ta tha bit khng mang n cho ta mt thng tin g.

- Loi ni dung ta c thon th ny hoc th n (tc l loi ni dung c bp bnh noy). VD: Em An i hc. V em An hc lc trung bnh nn thi vo i hc c th, cth khng. in vi ni dung ta khng bit chc (ni dung cha mt bt nh no ) tht sc mang n cho ta mt thng tin nht nh.

- Loi ni dung m ta hon ton khng ng ti, cha h ngh ti. VD: Em An trng giinht trong t x s. Bc in nh vy, ng v mt thng tin m ni, a n cho ta mtthng tin rt ln.

Ch : y ta ni ti nhng ni dung cha h ngh ti phi hiu theo hon ton

khch quan ch khng phi do s khng y v t duy ca con ngi em li.T nhng v d trn, ta rt ra nhng kt lun sau v khi nim thng tin:

- iu g xc nh (khng nh c, on chc c, khng bp bnh,) th khng cthng tin v ngi ta ni rng lng thng tin cha trong iu y bng khng.

- iu g khng xc nh (bt nh) th iu c thng tin v lng thng tin cha trong nkhc khng. Nu ta cng khng th ngti iu th thng tin m iu mang li cho ta rt ln.

Tm li, ta thy khi nim thng tin gn lin vi s bt nh ca i tng ta cn xt. C sbt nh v mt i tng no th nhng thng bo vi tng s cho ta thng tin. Khikhng c s bt nh th s khng c thng tin vi tng . Nh vy, khi nim thng tin chl mt cch din t khc i ca khi nim s bt nh.


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Trc khi nhn tin (c thng bo) v mt i tng no y th vn cn s bt nh vi tng , tc l bt nh vi tng khc khng (c th ln hoc nh). Sau khi nhntin ( c hiu r hoc hiu mt phn) vi tng th bt nh ca n gim n mc thpnht, hoc hon ton mt. Nh vy, r rng Thng tin l bt nh b th tiu hay ni mt

cch khc Lm gim bt nh kt qu cho ta thng tin.3.1.1.2. Lng thng tin

Trong l lun trn, ta tng ni n lng thng tin v lng thng tin ln, lng thngtin nh m khng hnh ngha cc danh t. Di y ta s tr li vn .

trn ta cng ni: trc khi nhn tin th bt nh ln nht. Sau khi nhn tin (hiu rhoc hiu mt phn vi tng th bt nh gim n mc thp nht, c khi trit hon ton.

Nh vy, c mt s chnh lch gia bt nh trc khi nhn tin v bt nh sau khi nhn tin.S chnh lch l mc th tiu bt nh. ln, nh ca thng tin mang n ta ph thuc

trc tip vo mc chnh . Vy:Lng thng tin l mc b th tiu ca bt nh Lng thng tin = chnh ca

bt nh trc v sau khi nhn tin = bt nh trc khi nhn tin - bt nh sau khi nhntin ( bt nh tin nghim - bt nh hu nghim).

3.1.2. Quan h gia bt nh v xc sut

3.1.2.1. Xt v d sau

Ta phi chn mt phn t trong mt tp no . Php chn nh th (hoc chn hiu theongha rng: th, tm hiu, iu tra, trinh st, tnh bo,) bao gicng c bt nh.

- Nu tp ch c mt phn t th ta chng phi chn g c v nh vy khng c bt nhtrong php chn .

- Nu tp c hai phn t th ta phi chn. Nh vy, trong trng hp ny php chn c bt nh. Nu s phn t ca tp tng th bt nh s tng.

- Cc bc tip theo s cho bi bng sau:

S phn t ca tp bt nh ca php chn Xc sut chn mt phn t trong tp12

3...n...

00

0...

0

.

.

.

11/2

1/3...

1/n...

1/ 0 =

Ch : Bng ny a ra vi gi s vic chn cc phn t l ng xc sut.

Tng

Gim


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49

3.1.2.2. Kt lun

- Bng ny cho thy: bt nh gn lin vi bn cht ngu nhin ca php chn, ca bin c.

- bt nh (k hiu I) l hm ca s phn t thuc tp KI(x ) f (n)= (a)

- bt nh c lin quan vi xc sut chn phn t ca tp K KI(x ) E (x )[p ] = (b)

tm mi quan h gia bt nh I v xc sut chn mt phn t K Kx ( (x ))p trongtp, ta xut pht t cc tiu sau:

Theo suy nghthng thng, bt nh I phi tho mn:

+ KI(x ) 0

+ K K K(x ) 1 I(x ) E (x )p [p ] = E[1] = 0= = (3.1)

+ Tnh cng c:

Nu Kx v ix c lp, th:

K i K i K iE (x x ) (x ) (x ) (x ) (x )[p ] = E[p p ] = E[p ] + E[p ]

Nu Kx v ix ph thuc th:

K i K i K K i K E (x x ) (x ) (x x ) (x ) (x x )[p ] = E[p p ] = E[p ] + E[p ]

t K(x ) pp = v i K(x x ) qp ] = , th khi vi mi p, q (0 p 1, 0 q 1)< < , tac:

E[p] + E[q] = E(pq) (3.2)

T (3.2) ta c th tm c dng hm I(p). Ly vi phn 2 v ca (3.2) theo p, ta c:

E(p) = q E(pq)

Nhn c 2 v ca phng trnh ny vi p v k hiu p.q = , ta c:

pE(p) = E( ) (3.3)

(3.3) ng p, 0. Nhng iu ny ch c th c khi c hai v ca (3.3) bng mthng s k no :

pE(p) = E( ) = k = const

T chng ta c phng trnh vi phn pI(p) = const = k, ly tch phn phng trnh ny,ta tm c:

E(p) = k.lnp + C (3.4)

Kn iu kin ban u (3.1), chng ta c:

E(p) = k.lnp (3.5)

Nh vy, ta c: K KI(x ) k.ln x )[p( ]= (3.6)


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H s t l k trong (3.6) c th chn tu , n ch xc nh hn vo ca KI(x ) . V

Kln x )[p( ] 0 nn KI(x ) 0 th k < 0.

Nu ly k = -1 th K KK

1I(x ) ln x ) ln x )[p( ] = p(

= (3.7)

Khi , n vo bt nh s l n v t nhin, k hiu l nat.

Nu ly1

kln2

= th KK 2 Kln x )

I(x ) log p(x )ln 2

p(= = (3.8)

Khi n vo bt nh s l n v nh phn, k hiu l bit (1 nat = 1,433 bit)

Mt bit chnh l bt nh cha trong mt phn t (bin c ca tp xc sut chn (xuthin) bng 1/2. Ngi ta thng s dng n v [bit] do trong k thut tnh v k thut lin lc

thng dng cc m nh phn.

Ngoi ra, ngi ta cn c th s dng nhng n vo khc tu theo cch chn cs calogarit. V vy trong trng hp tng qut, ta c th vit:

K KI(x ) log x )p(= (3.9)

3.1.3. Xc nh lng thng tin

mc 1, ta c kt lun sau:

Lng thng tin = bt nh tin nghim - bt nh hu nghim. V bt nh s tr

thnh thng tin khi n b th tiu nn ta c th coi bt nh cng chnh l thng tin. Do :

Lng thng tin = thng tin tin nghim thng tin hu nghim (*)

Thng tin tin nghim (hay cn gi l lng thng tin ring) c xc nh theo (3.9). Cnthng tin hu nghim xc nh nh sau:

Gi Kx l tin gi i, y l tin thu c c cha nhng du hiu hiu bit v Kx (c

cha thng tin v Kx ). Khi xc sut r v Kx khi thu c y l K(x y )p . Nh

vy bt nh ca tin Kx khi r ybng:

(3.9)

K Kx y ) (x y )I( - logp= (3.10)

(3.10) c gi l thng tin hu nghim v Kx (thng tin ring v Kx sau khi c y ).

Thay (3.9) v (3.10) vo (*), ta c:


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51

K

L I( ) I( y )

L I( ) I( y )

1 1I( , y ) log logp( ) p( y )

K K K

K K K

hiu

KK K

ng thng tin v x x x

ng thng tin v x x x

xx x

=

=

=

KK

K

p(x y )I(x , y ) log

p(x ) = (3.11)

(3.11) gi l lng thng tin v Kx khi r tin y hay cn gi l lng thng tin cho v

Kx do y mang li.

Nu vic truyn tin khng b nhiu th Ky x . Tc l nu pht Kx th chc chn nhnc chnh n. Khi :

K K Kp(x y ) p(x x ) 1= =

T (3.11) ta c:

K K K K K

1I(x , y ) I(x ,x ) I(x ) log

p(x )= = = (**)

Nh vy khi khng c nhiu, lng thng tin nhn c ng bng bt nh ca s kin

Kx , tc l ng bng thng tin tin nghim ca Kx .

Vy lng thng tin tn hao trong knh s l:

I( ) I( , y ) I( y )K K Kx x x =

n vo ca thng tin (lng thng tin) cng chnh l n vo bt nh.

Nu c s ca logarit l 10 th n vo thng tin c gi l Hartley, hay n v thpphn.

Nu cs ca logarit l e = 2,718 th n vo thng tin c gi l nat, hay n vot nhin.

Nu cs ca logarit l 2 th n vo thng tin c gi l bit, hay n v nh phn.

1 Harley = 3,322 bit

1 nat = 1,443 bit


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52

3.2. ENTROPIE V CC TNH CHT CA ENTROPIE

3.2.1. Tnh cht thng k ca ngun ri rc v sra i ca khi nim entropie

Trong mc trc, ta mi ch xt n lng thng tin v mt bin c (hay mt tin) trong mt

tp cc bin c (hay tin) xung khc, ng xc sut.

Thc t tn ti ph bin loi tp cc bin c (hay ngun tin, tp tin) xung khc, khng ngxc sut. Tc l xc sut xut hin cc bin c khc nhau trong tp l khc nhau. Ta gi s khcnhau gia cc xc sut xut hin bin c ca tp (hay tin ca ngun ri rc) l tnh cht thng kca n.

V d 1: S xut hin cc con ch trong b ch Vit c xc sut khc nhau: p(e) = 0,02843;p(m) = 0,02395; p(k) = 0,02102, (Theo s liu trong n tt nghip Kho st cu trc thngk ch Vit ca on Cng Vinh HBK HN).

V d 2: Xc sut xut hin ca 26 ch ci trong ting Anh: (S liu theo Beker v Pipe)K t Xc sut K t Xc sut

ABCDEFG

HIJKLM

0,0820,0150,0280,0430,1270,0220,020

0,0610,0700,0020,0080,0400,024

NOPQRST

UVWXYZ

0,0670,0750,0190,0010,0600,0630,091

0,0280,0100,0230,0010,0200,001

Trong mt ngun tin nh th, ngoi thng tin ring ca mi tin (hay du) ca n, ngi tacn phi quan tm n thng tin trung bnh ca mi tin thuc ngun. Ngi ta cn gi thng tintrung bnh do mi du ca ngun mang li l entropie. Di y ta s xt knh ngha ventropie.

3.2.2. nh ngha entropie ca ngun ri rc

3.2.2.1. t vn

php o c chnh xc, trong vt l, khi o lng mt i lng, ta khng quan tmn tng tro c ca i lng m thng xt tr trung bnh ca chng. Khi ta ly cc tro c cng vi nhau ri chia cho s lng ca chng:

n

tb rr 1

i i n=

=


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53

y cng c iu tng t: ta khng quan tm n tng thng tin ring ca mi du mli ch n gi tr trung bnh ca cc thng tin . Ch khc ch mi mt thng tin ring ntng ng vi mt xc sut xut hin no , tc l ta c th xem cc thng tin ring l m ilng ngu nhin I. Do gi tr trung bnh ca cc thng tin ny (lng thng tin trung bnh hay

entropie) chnh l k vng ca i lng ngu nhin I. Ta i ti nh ngha sau:3.2.2.2. nh ngha

Entropie ca ngun tin ri rc l trung bnh thng k ca lng thng tin ring ca cc du

thuc ngun A, k hiu 1H (A) :

1H (A) M )i[I(a ]= (3.12)

Trong ia l cc du ca ngun A (Ta hiu du l cc con ch, hoc cc k hiu v.v

ca ngun). Cn ngun A l mt tp ri rc cc du ia vi cc xc sut xut hin ca chng. Taquy c vit A nh sau:

2 sa ... aAp( ) p( ) ... p( )

1i

1 2 s

a{a } =

a a a

=

(3.13)

Vi i0 p(a ) 1 vs

ii 1

p(a ) 1=

= (3.14)

A c cho bi (3.13) v (3.14) cn gi l trng tin (hay trng bin c). T (3.12) v(3.13), ta c:

1H (A) M ) )I )s

i i ii=1

[I(a ] = p(a (a=

1H (A) )log p )s

i ii=1

= p(a (a (3.15)

1H (A) cn gi l entropie mt chiu ca ngun ri rc:V d: 1H (Vit) = 4,5167 bit 1H (Nga) = 4,35 bit

1H (Anh) = 4,19 bit

3.2.3. Cc tnh cht ca entropie mt chiu ca ngun ri rc

3.2.3.1. Tnh cht 1

Khi kp(a ) 1= v rp(a ) 0= vi r k th:

1 1 minH (A) H (A ) 0= = (3.16)


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Chng minh:

Ta c: i0 p(a ) 1 i ilog p(a ) 0 log p(a ) 0

1 1 minH (A) 0 H (A ) 0 =

By gi ta ch cn phi chng t 1 minH (A ) 0= khi kp(a ) 1= v

rp(a ) 0= ( r k ).

Tht vy, rp(a ) 0= r rp(a )log p(a ) 0 ( r k)=

kp(a ) 1= k kp(a )log p(a ) 0 ( r k)=

s

1 i i

i 1s

k k i ii 1,i k

H (A) p(a )log p(a )

p(a )log p(a ) p(a )log p(a ) 0

=

=

=

= =

ngha:

Thc ra khng cn phi chng minh nh vy, m lp lun nh sau cng cho ta cng thc(3.16):

rp(a ) 0= cc ra khng xut hin

kp(a ) 1= cc ka chc chn xut hin

Khng c bt nh no v cc ia lng thng tin ring khng c lngthng tin trung bnh cng khng c.

3.2.3.2. Tnh cht 2

Mt ngun ri rc gm s du ng xc sut (vtho mn (3.14)) th entropie ca n t cc i v cci bng log s.

1 mH (A ) logsax = (3.17)

Chng minh:

Khi i jp(a ) p(a ), i, j (i, j 1,s)= =

Khi i1

p(a )s

= , tc l ngun gm cc du

xung khc v ng kh nng.

yx 1

lnx

0 1 x

- 1

Hnh 3.1.


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s

1i 1

1 1H (A') log logs

s s= = =

Xt hiu:

s

1 iii 1

s s

i iii 1 i 1

s

iii 1

s s

i iii 1 i 1

1H (A) logs p(a )log logs

p(a )

1p(a )log p(a )logs

p(a )

1p(a ) log logs

p(a )

1p(a )log p(a )log x

p(a )s

=

= =

=

= =

=

=

=

= =

Ta c: lnx x 1 x (xem hnh 3.1)

s

i ii 1

p(a ) l p(a )(x 1)ogx=

M:

s s s

i iii 1 i 1 i 1

1 1p(a ) 1 p(a ) 0p(a )s s= = =

= =

Vy: 1 1H (A) logs 0 H (A) logs

Tm li, ta thy 10 H (A) logs (entropie ca ngun ri rc)

Entropie l mt i lng gii ni.

K hiu m 0H(A) H (A)ax =

V d: 0H (Vi 36 5,1713bit2t) = log =

0H (Nga 32 5bit2) = log =

0H (Anh 27 4,75489bit2) = log =

3.2.4. Entropie ca ngun ri rc, nh phn

Ngun ri rc nh phn l ngun ch c hai du:

1

2

a "0" v ) p

a "1" v ) 1 p

1

2

i xc sut p(a

i xc sut p(a

=

=


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56

Ta c ngay:

2

1 i ii 1

H (A) p(a ) log p(a ) plogp (1 p) f (p)=

= = = (3.18)

th f(p) c biu din trn hnh 3.2.

Ta thy 1H (A) f (p)= ch ph thucvo c tnh thng k ca cc tin.

Nu n v dng l bit th

max 1H (A) 1=

Nhn xt:

- 1H (A)t max ti1

p 2= . S d nh

vy v tp ch c hai phn t, nn bt nh ca php chn s ln nht khi hai du c xc sutxut hin nh nhau.

- p = 0 1 minH (A) 0= . Khi 1 p = 1 l xc sut xut hin du 2a . Vy 2a l mtbin c chc chn. Php chn ny khng c bt nh lng thng tin trung bnh trit.

- p = 1 1 minH (A) 0= . Gii thch tng t.

3.2.5. Entropie ca trng skin ng thi

nh ngha 1:

C hai trng s kin A v B:

1 2 s

1 2 s

a a ... aA

p(a ) p(a ) ... p(a )

=

v

1 2 t

1 2 t

b b ... bB

p(b ) p(b ) ... p(b )

=

Cc ia v jb l cc s kin.

Ta xt mt s kin tch: k i jc a .b=

k i jp(c ) p(a .b )= . Ta xt trng C l giao ca hai trng A v B, nu:

1 1 1 2 1 t 2 j s t

1 1 1 2 1 t 2 j s t

a b a b ... a b ... a b ... a bC A.B

p(a b ) p(a b ) ... p(a b ) ... p(a b ) ... p(a b )

= =

Trng C c gi l trng s kin ng thi (trng giao, tch) ca hai trng s kin cbn A v B.

nh ngha 2:

H1(A)

1

H1(A)max

0 0,5 1 p

Hnh 3.2.


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58

trng tin kia

3.3.1.1. Mu

Trong phn trc, ta ni nu truyn tin c nhiu th tin pht ik

a v tin thu c b

l

khc nhau. V khi lng thng tin ring v ka do b mang li l:

kk

1I(a / b ) log

p(a / b )=

Vn : ta khng quan tm n lng thng tin ring v mt du ka c th no ca ngun

tin pht { ia } do b mang li m ch quan tm n lng thng tin ring trung bnh v mt du

no ca tp { ia } do b mang li. Ta thy rng kI(a / b ) l mt i lng ngu nhin. Do

tng t nhnh ngha ca entropie mt chiu, ta i ti nh ngha sau.

3.3.1.2.nh ngha

Entropie c iu kin v mt trng tin ny khi r mt tin ca trng tin kia c xc

nh bng k vng ca lng thng tin ring c iu kin v ka do mtb mang li:

s

i 1

H(A / b ) M / b ) / b ) I / b )i i i[I(a ] = p(a (a

=

=

s

i 1/ b ) log p / b )i ip(a (a

== (3.20)

ngha:

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

Tng t:

t

i i ij 1

H(B/ a ) / a )logp / a )j jp(b (b=

=

ngha:

iH(B / a ) l lng thng tin ring trung bnh cha trong mi tin u thu khi u pht

pht i mt tin ia .

3.3.2. Entropie c iu kin v trng tin ny khi r trng tin kia

Ta thy rng do nhiu ngu nhin nn bn thu khng phi ch thu c mt tin duy nht m

l c tp tin B = { jb } no , ( j 1, t)= . Vy jH(A / b ) cng l mt i lng ngu nhin, do


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ta phi xt n lng thng tin ring trung bnh v mi tin u pht khi u thu thu cmt du no .

Tng t nh trn, ta cng phi ly trung bnh thng k ca i lng ngu nhin ny.

nh ngha:Entropie c iu kin ca trng s kin A khi r trng s kin B c xc nh bi k

vng ca i lng jH(A / b ) .

t

j j jj 1

t s

j i j i jj 1 i 1

s t

j i j i ji 1 j 1

H(A / B) M H(A / b ) p(b )H(A / b )

p(b ) p(a / b )log p(a / b )

p(b )p(a / b )log p(a / b )

=

= =

= =

= =

=

=

s t

i j i ji 1 j 1

H(A / B) p(a b ) log p(a / b )= =

= (3.21)

ngha:

H(A/B) l lng thng tin tn hao trung bnh ca mi tin u pht khi u thu thuc mt du no .

Tng t:

s t

j i j ii 1 j 1

H(B/ A) p(b a )log p(b / a )= =

= (3.22)

ngha:

H(B/A) l lng thng tin ring trung bnh cha trong mi tin u thu khi u pht

pht i mt tin no .Ch :

Ta xt mt b ch A. c trng cho lng thng tin ring trung bnh cha trong mi conch khi kn xc sut xut hin cc c p ch (VD: trong ting Vit: p(a/b) 0, p(b/a) = 0,

p(t/a) 0, p(a/t) 0), ngi ta dng H(A/A) v k hiu l H 2 (A).

V d: H 2 (Vit) = 3,2223 bit

H 2 (Nga) = 3,52 bit

H 2 (Anh) = 3,32 bit


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Vic tnh H 3 , H 4 rt phc tp.

Khakevich tnh c n H5 . Shannon tnh c n H8 .

3.3.3. Hai trng thi cc oan ca knh truyn tin3.3.3.1. Knh bt (b nhiu tuyt i)

Trong trng hp ny, cc tin thu c hon ton khc cc tin pht i. Ni khc i v b

nhiu tuyt i nn trong mi tin jb B khng cha du hiu hiu bit no v cc tin pht i.

Nh vy, A v B l c lp nhau: i j ip(a / b ) p(a )= ; j i jp(b / a ) p(b )=

i j i jp(a b ) p(a )p(b ) =

Khi ta c:s

j i ii 1

t

i j jj 1

t s

j i ij 1 i 1

s t

i j ji 1 j 1

H(A / b ) p(a )log p(a ) H(A)

H(B/ a ) p(b )log p(b ) H(B)

H(A / B) p(b ) p(a )log p(a ) H(A)

H(B/ A) p(a ) p(b )log p(b ) H(B)

=

=

= =

= =

= =

= =

= =

= =

(3.23)

3.3.3.2. Knh khng nhiu

Khi : t = s. Vi i ii 1,s a b = =

i ip(a ) p(b ) = nn H(A) = H(B)

k k k k

i k i k

p(a / b ) p(b / a ) 1p(a / b ) p(b / a ) 0 vi i k

= == =

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

H(A / B) 0 H(B/ A) 0

= =

= =(3.24)

V khi khng nhiu, coi A v B ph thuc mnh nht, c ia th chc chn c ib , nn

bt nh v ia khi thu c ib l khng c bt nh trung bnh cng khng c.


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3.3.4. Cc tnh cht ca entropie c iu kin

3.3.4.1.Tnh cht 1

Nu A v B l hai trng bin c bt k (hai ngun tin bt k) th entropie ca trng bin

cng thi A.B bng:

H(A.B) = H(A) + H(B/A) = H(B) + H(A/B) (3.25)

Chng minh:

{ }

s t

i j i ji 1 j 1

s t

j i j j i j

i 1 j 1

H(A.B) p(a b ) logp(a b )

p(b )p(a / b )log p(b )p(a / b )

= =

= =

= =

= =

s t s t

j i j j j i j i ji 1 j 1 i 1 j 1

s t s t

i j j j i j i ji 1 j 1 i 1 j 1

H(A.B) p(b )p(a / b )logp(b ) p(b )p(a / b )log p(a / b )

p(a / b ) p(b )log p(b ) p(a b )log p(a / b )

H(B) H(A / B)

= = = =

= = = =

= =

=

= +

Trong :s

i ji 1

p(a / b ) 1=

= .

3.3.4.2.Tnh cht 2

Entropie c iu kin nm trong khong:

0 H(A / B) H(A) (3.26)

Chng minh:

+ H(A / B) 0 :

i j i j

i j

0 p(a / b ) 1 logp(a / b ) 0

logp(a / b ) 0 H(A / B) 0

N s nhn du bng khi A v B l ng nht (knh khng nhiu).

+ H(A/B) H(A):

Xt hiu: H(A/B) H(A) = G


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62

s t s

j i j i j i ii 1 j 1 i 1

G p(b )p(a / b )logp(a / b ) p(a )log p(a ).1= = =

= +

Ch : ta thay

t

j ij 1

1 p(b / a )=

=

s t s t

i j i j i j ii 1 j 1 i 1 j 1

s ti j

i jii 1 j 1

s tii j

i ji 1 j 1

G p(a b )logp(a / b ) p(a b )log p(a )

p(a / b )p(a b )log

p(a )

p(a )p(a b )logp(a / b )

= = = =

= =

= =

= +

=

=

p dng log x x 1 :

s ti

i ji ji 1 j 1

s ti

j i ji ji 1 j 1

s t s t

i j j i ji 1 j 1 i 1 j 1

p(a )G p(a b ) 1

p(a / b )

p(a )G p(b )p(a / b ) 1

p(a / b )

G p(a ) p(b ) p(b )p(a /b )

G 1 . 1 1 0

= =

= =

= = = =

=

H(A/B) H(A).

H(A/B) = H(A) khi A v B l c lp (knh bt).

3.3.4.3. Tnh cht 3Entropie ca trng s kin ng thi khng ln hn tng entropie ca cc trng s kin

cbn.

H(A.B) H(A) + H(B) (3.27)

Chng minh:

(3.27) rt ra trc tip t (3.25) v (3.26).


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63

3.3.5. Lng thng tin cho trung bnh

phn trc, chng ta bit lng thng tin cho v mt tin ia pht i do mt tin jb

thu c mang li l:

i ji j

i

p(a / b )I(a ,b ) log

p(a )=

Thng thng, v bn pht pht i mt tp tin A = { ia } v bn thu nhn c mt tp tin B

= { jb }. Do ta khng quan tm n lng thng tin cho v mt tin c th ia pht do mt

tin jb c th thu c, m ta ch quan tm n lng thng tin cho trung bnh v mi tin ca tp

pht A do mi tin ca tp thu B mang li. i jI(a ,b ) l mt i lng ngu nhin, do ta phi ly

trung bnh thng k ca n.nh ngha:

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

i jI(A,B) M I(a ,b )

= (3.28)

Xc sut c thng tin i jI(a ,b ) l i jp(a b ) , do ta c:

s ti j

i jii 1 j 1

p(a / b )

I(A,B) p(a b ) log p(a )= == s t s t

i j i j i j ii 1 j 1 i 1 j 1

I(A,B) p(a b ) log p(a / b ) p(a b ) logp(a )

H(A / B) H(A)

= = = =

=

= +

Tm li: I(A,B) = H(A) H(A/B) (3.29a)

Tng t, ta c: I(A,B) = H(B) H(B/A) (3.29b)

Hay: I(A,B) = H(A) + H(B) H(A.B)

I(A,B) cn gi l lng thng tin trung bnh c truyn theo knh ri rc.

3.3.6. Tnh cht ca I(A,B)

3.3.6.1. Tnh cht 1

I(A,B) 0: (3.30)

Theo tnh cht 2 mc 3.3.4: H(A/B) H(A) H(A) H(A/B) 0.

I(A,B) = 0 khi knh bt.


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64

3.3.6.2.Tnh cht 2

I(A,B) H(A): (3.31)

Tht vy: H(A/B) 0 I(A,B) = H(A) H(A/B) H(A)

I(A,B) = H(A) khi knh khng c nhiu.

T (3.31) ta thy khi truyn tin trong knh c nhiu, thng tin s b tn hao mt phn.Lng thng tin tn hao trung bnh chnh l H(A/B).

3.3.6.3.Tnh cht 3

I(A,A) = H(A)

3.3.6.4. Tnh cht 4

I(A,B) = I(B,A)

3.3.7. M hnh ca knh truyn tin c nhiu

Da vo (3.29a), ta c m hnh knh truyn khi c nhiu nh sau:

Hnh 3.4. Lc Wenn m t mi quan h gia cc i lng.

A B

H(A)

A B

H(A/B)

A B

I(A,B)

A B

H(AB)

A B

H(B/A)

H(A) I(A,B) = I(B,A) H(B)

H(A/B) H(B/A)

Tn hao Lng tin tc b bng hi lin

Hnh 3.3.


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66

D 1= , trong :m

H(A)

H(A) ax = c gi l h s nn tin.

i vi ngun tin c s du: H(A) max = H 0 (A) = log s.

ngha:

tha c trng cho hiu sut, kh nng chng nhiu v mt ca tin. Nu D cng lnth hiu sut cng thp, mt cng thp nhng kh nng chng nhiu cng cao.

V d:

- i vi ting Vit: H1 (Vit) = 4.5167; H 0 (Vit) = 5,1713

1 87 = % 1D = 13 %

22 H (A) 3,2223l 5,1713ogs = = = 62% 2D =38%

- i vi ting Nga: 1 = 87% 1D = 13 %

3 = 60% 3D = 40 %

- i vi ting Anh: 1 = 84% 1D = 16 %

8 = 38% 8D = 62 %

3.4.4. Cc c trng ca knh ri rc v cc loi knh ri rc

Mt knh ri rc hon ton c c trng bi ba tham s sau:

- Trng du li vo v trng du li ra ca knh.

- Xc sut chuyn j ip(b / a )

- Tc truyn tin ca knh K

nh ngha 1:

Nu mt knh c j ip(b / a ) t th c gi l knh ng nht; j ip(b / a ) vo du

pht trc n th c gi l knh khng nh. Ngc li, j ip(b / a ) t th knh c gi l

khng ng nht; j ip(b / a ) vo du pht trc n th knh c gi l knh c nh

( i, j ).

nh ngha 2:

Nu mt knh c xc sut chuyn:

j i c , j 1, tp(b / a )p c i j

s

p onst vi i j, i=1,sonst

= == = =


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67

th knh sc gi l knh i xng.

V d:

xc sut sai bng nhau, xc sut ng bng nhau.

i vi knh i xng nh phn (Hnh v): sp p 1+ = .

3.4.5. Lng thng tin truyn qua knh trong mt n v thi gian

nh ngha:

KK

I(A,B)I '(A,B) I(A,B)

T

= = [bit/s] (3.35)

Trong : KK

1

T = , KT : thi gian trung bnh truyn mt du qua knh. K biu th

s du m knh truyn c (c truyn qua knh) trong mt n v thi gian. I(A,B) llng thng tin truyn qua knh trong mt n v thi gian.

Nu knh gin tin: K nT T>

Nu knh nn tin: K nT T<

Thng thng: K nT T=

3.4.6. Kh nng thng qua ca knh ri rc

nh gi nng lc ti tin ti a cu mt knh truyn, ngi ta a ra khi nim kh nngthng qua.

3.4.6.1. nh ngha

Kh nng thng qua ca knh ri rc l gi tr cc i ca lng thng tin truyn qua knhtrong mt n v thi gian, ly theo mi kh nng c th c ca ngun tin A. (Cc i ny st

c ng vi mt phn b ti u ca cc xc sut tin nghim p( ia ), ia A ).

KA A

C' m I '(A,B) m I(A,B)ax ax= = [bit/s] (3.36)

p(b1/a1) = pa1 b1

a2 b2p(b2/a2) = p

p(b2/a1)p(b1/a2)


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68

KA

C' .C v m I(A,B)i C = ax=

C c gi l kh nng thng qua ca knh i vi mi du.

C l mt tham s rt quan trng ca mt knh.

3.4.6.2. Tnh cht

- C 0, C = 0 khi v ch khi A v B c lp (knh bt).

- C Klogs , ng thc ch xy ra khi knh khng nhiu. (3.37)

Chng minh:

K K K

K K

K K

K

I(A,B) H(A)

I(A,B) H(A) ( 0)

m I(A,B)) m H(A))m I(A,B) m H(A)

C' logs

ax( ax(

ax ax

>

3.4.7. Tnh kh nng thng qua ca knh nh phn i xng khng nh, ng nht

3.4.7.1. t bi ton

Ta c mt knh nh phn nh hnh 3.4. Trong :Xc sut sai: 2 1 1 2 sp(b / a ) p(b / a ) p= =

Xc sut ng: 2 2 1 1p(b / a ) p(b / a ) p= =

p(a1) = p; p(a 2 ) = 1 - p;

Cc du a1 v a 2 c cng thi hn T. Vn : tnh C?

3.4.7.2. Gii bi ton

Ta c:

[ ]K K

1 1C' m I(A,B) m H(B) H(B/ A)

T TA Aax ax= =

Ta c ngay:

2 2

i j i j ii 1 j 1

H(B/ A) p(a )p(b / a )log p(b / a )= =

=

p(b1/a1) = p

a1 b1

a2 b2p(b2/a2) = p

p(b1/a2)= ps


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69

[ ]

[ ]

[ ]

[ ]

1 1 1 1 1 2 1 2 1

2 1 2 1 2 2 2 2 2

s s s s

s s s s

H(B/ A) p(a ) p(b / a )log p(b / a ) p(b / a ) log p(b / a )

p(a ) p(b / a )log p(b / a ) p(b / a )logp(b / a )

p (1 p )log(1 p ) p logp

(1 p) p log p (1 p )log(1 p )

= +

+

= +

+

[ ]s s s sH(B/ A) p logp (1 p )log(1 p )= +

Ta thy H(B/A) ch ph thuc vo sp , m khng ph thuc vo xc sut tin nghim cacc du thuc ngun tin A. Do :

[ ]K

K K

1C' m H(B) H(B/ A)

T

1 1m H(B) H(B/ A)T T

A

A

ax

ax

=

=

y H(B/A) khng i i vi mi trng thi (c tnh thng k) ca ngun A.

M:

m 2 2m H(B) H(B) log s log 2 1axAax = = = =

Vy: [ ]s s s sK

1C' 1 p log p (1 p )log(1 p )

T

= + +

( )s KC' f p ,T=

m sK

1C' p 0

Tax= = Knh khng nhiu.

s s s sm

C'1 p log p (1 p ) log(1 p )

C' ax= + +

(3.38) th (3.38) biu din trn hnh 3.5.

3.4.8. nh l m ho thhai ca Shannon

nh l:Nu kh nng pht H(A) ca ngun tin ri rc A b hn kh nng thng qua caknh: (H(A) < C) th tn ti mt php m ho v gii m sao cho vic truyn tin c xc sut gpli b tu (nu H(A) > C th khng tn ti php m ho v gii m nh vy) khi di t m ln.

Nhn xt:y l mt nh l tn ti v n khng ch cho ta cch thit lp mt m c th

no. L thuyt m knh trong chng 4 chnh l hng dn cn thit cho nh l ny.

C/ Cmax

1

0 0,5 1 ps

Hnh 3.5.


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70

3.4.9. Kh nng thng qua ca knh nh phn i xng c xo

3.4.9.1.t bi ton

Cho knh truyn, cc du1

a v2

a nh hnh v. Cc du1

a v2

a c cng thi hn T.Hy tnh kh nng thng qua C ca knh ny vi iu kin:

Xc sut xo: 3 ip(b / a ) q=

Xc sut thu ng:

1 1 2 2 sp(b / a ) p(b / a ) 1 p q= =

Xc sut thu sai:

2 1 1 2 sp(b / a ) p(b / a ) p= =

3.4.9.2. Gii bi ton

Tng t bi ton trn, ta c:

[ ]1

C' m H(B) H(B/ A)T A

ax=

Trong :

( ) ( )

( ) ( ) ( )

( ) ( )

2 3

i j i j i

i 1 j 1

s s s s

s s s s

s s s s

H(B/ A) p(a )p(b / a )log p(b / a )

p 1 p q log 1 p q p log p qlogq

1 p p log p 1 p q log 1 p q qlogq

1 p q log 1 p q p log p qlogq

= =

=

= + +

+ +

= + +

Ta thy H(B/A) vo tnh cht thng k ca ngun A. Do :

[ ]m H(B) H(B/ A) m H(B) H(B/ A)A Aax ax =

( ) ( ) ( )3

j jj 1

H B p b logp b=

=

Trong :

( ) ( ) ( ) ( ) ( )

( )3 1 3 1 2 3 2p b p a p b / a p a p b / a

pq 1 p q q

= +

= + =

khng ph thuc vo tnh cht thng k ca ngun A.

Nh vy, H(B) st max ng vi phn b ca cc xc sut p( ia ) m bo c:

1- ps - qa1 b1

psq

b3q

ps

a2 b21 ps - q


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71

( ) ( )1 21 q

p b p b2

= =

( ) ( )

( )1 qm H B qlogq 1 q log

2Aax

=

( ) ( ) ( ) ( ){ }s s s sC' F 1 q 1 log 1 q p logp 1 p q log 1 p q = + +

Trong 1

FT

=

3.5. ENTROPIE CA NGUN LIN TC. LNG THNG TIN CHO TRUNGBNH TRUYN QUA KNH LIN TC KHNG NH

3.5.1. Cc dng tn hiu lin tc

i vi cc tn hiu cao tn lin tc s(t) th gi tr ca n c th nhn mt cch lin tc cc

gi tr khc nhau trong mt khong xc nh min ms s ax , cn i s thi gian t li c th lintc hay ri rc (hnh 3.6)

V vy, ta s phn cc tn hiu lin tc ra 2 loi.

- Tn hiu lin tc vi thi gian ri rc (hnh 3.6a).

- Tn hiu lin tc vi thi gian lin tc (hnh 3.6b).

Cc tham sc trng ca tn hiu lin tc l:

- Cng sut ph trung bnh

- B rng ph

3.5.2. Cc c trng v tham s ca knh lin tc

Ta bit rng cc c trng ca knh ri rc l:

- Trng du li vo trc hay sau b m ho: A

s(t)smax

tsmin

t

s(t)smax

smin

t

a. b.Hnh 3.6.


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72

- Trng du li ra sau b gii iu ch hoc sau b gii m B.

- Xc sut chuyn ( )i jp a / b hoc ( ) ( )( )n ni ip /

i vi knh lin tc, cc c trng ca n l:- Trng du li vo (sau biu ch): {s(t)}

- Trng

bài giảng lttt hv bcvt

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