Gio trnh L thuyt tn hiu v truyn tinLi ni uNgay nay, cc lnh vc khoa hc my tnh v truyn thng thm nhp ln nhau v gn kt dn n lm thay i rt nhiu lnh vc cng ngh v sn xut. Chnh iu ny lm cho rt nhiu lnh vc khoa hc cng ngh c nhng iu kin c s pht trin mnh m. Trong hon cnh , vic nghin cu tm hiu v l thuyt tn hiu v truyn tin ngy cng tr nn quan trng v cn c t trong mt tnh hnh mi.Vi yu cu cn c mt gio trnh cho sinh vin ngnh in t - Vin thng, gio trnh L thuyt tn hiu v truyn tin c bin son. Trong qu trnh bin son khng trnh khi thiu st mong c gi gp gio trnh ngy cng hon thin hn.ThS. on Hu Chc B mn K thut in t 3Gio trnh L thuyt tn hiu v truyn tinChng 1 Cc khi nim c bn ca l thuyt tn hiu v truyn tin1.1. V tr, vai tr v lch s pht trin 1.1.1. V tr, vai tr ca l thuyt thng tinTrong cuc sng con ngi lun c nhu cu trao i, giao tip vi nhau, ci m mi ngi trao i vi nhau gi l thng tin. Nh s pht trin ca khoa hc cng ngh, con ngi cng c th nhn thng tin t i, bo, mng internet, truyn hnh ... v.v. Hay n gin nh cc bn sinh vin nhn thng tin t ging vin v phc tp hn na l s lin lc, thng tin gia cc mng my tnh vi nhau. iu c ngha l thng tin l ci g c truyn t i tng ny ti i tng khc chuyn, thng bo mt iu g . Thng tin s ch c ngha khi iu g m bn nhn cha bit.Nh nhng v d trn trnh by, thng tin c th c cha trong nhiu dng nh hnh nh, m thanh, vn bn. Nhng dng ny l v bc vt cht ca thng tin. V bc c th hiu l phn xc, thng tin l phn hn.Mt trong nhng phng tin din t thng tin l ngn ng, thng tin ch c th c truyn t, hiu nu c hai bn truyn v nhn hiu c ng ngha ca nhau.Thng tin c th c truyn hoc lu tr. Mi trng thc hin vic c gi l mi trng cha tin hay knh tin.ThS. on Hu Chc B mn K thut in t 4Gio trnh L thuyt tn hiu v truyn tinCc i tng sng lun c nhu cu tm hiu v th gii xung quang thch nghi tn ti v pht trin. Thng tin tr thnh mt nhu cu c bn, mt iu kin cn cho s tn ti v pht trin. Ngy nay, khi khoa hc pht trin mnh m th thng tin ngy cng tr nn quan trng i vi mi con ngi. Mi hnh ng ca con ngi u xut pht t nhng suy ngh ca ngi . Mi suy ngh li chu s nh hng ca nhng thng tin m ngi c c, do vy hnh ng ca con ngi chu s nh hng ca thng tin.ngvkhacnhkhoahclthuyt tnhiuvtruyntin nghin cu nhm to ra mt iu kin tt cho vic x l phn tch tn hiu v truyn tin nhanh chng, an ton v lu tr hiu qu. Mt cch tng qut l thuyt tn hiu v truyn tin nghin cu cc vn v x l thng tin.Mt s lnh vc nghin cu ca mn hc:- Phn tch, tng hp tn hiu- M ho chng nhiu- Nn d liu- Mt m hoL thuyt thng tin c nhiu ng dng trong cuc sng v khoa hc k thut. S bng n v thng tin ang lm thay i din mo cuc sng ca con ngi, to ra s pht trin mnh m ca cc phng thc truyn thng, truyn tin v lu tr thng tin. Cng chnh nh thng tin m con ngi c cung cp nhng c s l thuyt v ci nhn trit hc su sc hn v cc vn con ngi gp phi hm nay v trong tng lai. L thuyt thng tin c p dng trong hu ht cc lnh vc cng ngh nh truyn thng, nn, bo mt, c bit c p dng vo trong lnh vc gio dc nhiu kha cnh nh o to, nghin cu v pht trin cng ngh.Vy l thuyt tn hiu v thng tin c lch s hnh thnh pht trin nh th no?ThS. on Hu Chc B mn K thut in t 5Gio trnh L thuyt tn hiu v truyn tin1.1.2. Lch s hnh thnh v pht trinNgi t vin gch u tin xy dng l thuyt thng tin l Hartley R.V.L. Nm 1928, ng a ra s o 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 calthuyt thngtintrongbi boVkhnngthngquaca khng trung v dy dn trong h thng lin lc in.Nm 1935, D.V Ageev a ra cng trnh L thuyt tch tuyn tnh, trong ng pht biu nhng nguyn tc c bn v l thuyt tch cc tn hiu. Nm 1946, V.A Kachenhicov thng bo cng trnh L thuyt th chng nhiu nh du mt bc pht trin rt quan trng ca l thuyt thng tin. Trong hai nm 1948 1949, Shanon C.E cng b mt lot cc cngtrnh v i,a s pht trin cal thuyt thng tin ln mt bc tin mi cha tng c. Trong cc cng trnh ny, nh vic a vo khi nim lng thng tin v tnh n cu trc thng k ca tin, ng chng minh mt lot nh l v kh nng thng qua ca knh truyn tin khi c nhiu v cc nh l m ho. 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 c s ton hc cht ch ca l thuyt thng tin. Cng hin ch yu trong lnh vc ny thuc v cc nh bc hc li lc nh: N.Wiener, A. Feinstain,C.E Shanon, A.N. Kanmgorov,A.JA Khintrin.Lthuyt thngtinngdng:(lthuyt truyntin)Chuyn nghin cu cc bi ton thc t quan trng do k thut lin lc t ra c lin quan n vn chng nhiu v nng cao tin cy ca vic ThS. on Hu Chc B mn K thut in t 6Gio trnh L thuyt tn hiu v truyn tintruyntin. CcbchcC.EShanon, S.ORiCe, D. Midleton, W. Peterson, A.AKhakevich, V. Kachenhicovcnhngcngtrnh qu bu trong lnh vc ny.1.2. Tin tc v cc khi nim c bn1.2.1. Cc nh ngha1.2.1.1. Thng tinThng tin l mt khi nim tru tng, phi vt cht v rt kh nh ngha. C nhiu cch nh ngha v thng tin. Di y l mt s nh ngha:Thng tin l s cm hiu ca con ngi v th gii xung quanh thng qua vic tip xc vi n.Thng tin l mt h thng nhng tin bo v mnh lnh gip loi trskhngchcchncani nhntin. Ni mt cchngngn, thng tin l ci m loi tr s khng chc chn.nhngha utin cha nurbn chtca thngtin, nh ngha th hai ni r hn v bn cht ca thng tin v c dng nh lng thng tin trong k thut.Ngoi rac ngi cnnh ngha thngtinl nhng tnhcht xc nh ca vt cht m con ngi (hoc h thng k thut) nhn c t th gii vt cht bn ngoi hoc t nhng qu trnh xy ra trong bn thn n.Vi nh ngha ny, mi ngnh khoa hc l khm ph ra cc cu trc thng qua vic thu thp, ch bin, x l thng tin. y thng tin l mt danh t ch khng phi l ng t ch mt 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 nh nng lng, khi lng). Thng tin khng c to ra m ch c s dng bi h th cm.ThS. on Hu Chc B mn K thut in t 7Gio trnh L thuyt tn hiu v truyn tinThng tin tn ti mt cch khch quan, khng ph thuc vo h th cm. Trong ngha khi qut nht, thng tin l s a dng. S a dng y c th hiu theo nhiu ngha khc nhau: Tnh ngu nhin, trnh t chc,1.2.1.2. TinTin l dng vt cht c th biu din hoc th hin thng tin. C hai dng l tin ri rc v tin lin tc.V d cc bc nh, bn nhc, bi ni, bng s liu, ...v.v l cc tin.1.2.1.3. Tn hiuThng tin l mt hin tng vt l, n thng tn ti v c truyn i di dng vt cht no .Nhng dng vt cht mang thng tin c gi l tn hiu.Trong k thut c th hiu, tn hiu l cc i lng vt l bin thin, phn nh tin cn truyn.Cn ch rng khng phi bn thn qu trnh vt l l tn hiu m s bin i cc tham s ring ca qu trnh vt l mi l tn hiu. Ccctrngvt lcthldngin, inp, nhsng, m thanh, ...v.v.1.2.2. S khi h thng thng tinNgay nay, vi s pht trin mnh m ca cng ngh in t vin thng,c rt nhiu cc h thngthng tin khc nhau c hnh thnh v pht trin. Khi vic phn loi cc h thng thng tin c th c da trn nhiu c s khc nhau. V d nh da trn c s v nng lng mang tin c th chia cc h thng truyn tin thnh cc loi nh:- H thng thng tin v tuyn dng sng in tThS. on Hu Chc B mn K thut in t 8Gio trnh L thuyt tn hiu v truyn tin- H thng thng tin quang hu tuyn dng nh sng-Hthngthngtindngsngm, sium(nnglngc hc)...Hay da vo cc biu hin bn ngoi m ta chia thnh cc h thng nh:- H thng truyn s liu- H thng truyn hnh- H thng in thoi c nh cng cng- H thng thng tin di ng,...v.v.Chng ta cng c th chia h thng thng tin thnh hai loi h thng:- H thng truyn tin ri rc- H thng truyn tin lin tcTuy nhin, mt cch tng qut h thng thng tin c th c biu din bi s khi sau:1.2.2.1. Ngun tinNgun tin l ni sn sinh ra hay cha cc tin cn truyn i. Khi mt ng truyn tin c thit lp truyn tin t ngun tin n ni 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. Do c th nh ngha:Ngun tin l tp hp cc tin m h thng truyn tin dng lp cc bn tin khc nhau truyn tin.ThS. on Hu Chc B mn K thut in t 9Ngun tin Knh tin Nhn tinNhiuHnh 1.1. S khi h thng thng tinGio trnh L thuyt tn hiu v truyn tinNu 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 nhau no l khc nhau.V d: P(x/ta) P(x/ba)Thng tin trc khi truyn i c th c x l m ho nn, chng nhiu v bo mt.1.2.2.2. Knh tinKnh tin l ni hnh thnh, truyn hay lu tr tn hiu mang tin ng thi xy ra cc tp nhiu ph hu tin tc.Trong l thuyt tn hiu v truyn tin, knh tin l mt khi nim tru tng i biu cho hn hp tn hiu v tp nhiu.truyntin, tathngsdngmt mi trngno truyn. Mi trng truyn tin thng rt a dng. Mi trng khng kh, ta c th truyn tin di dng m thanh, ting ni hay bng la (nh sng). Mi trng tng in ly thng l ni xy ra s truyn tin giaccvtinhnhntovcctrmradamt t. Haycth truyn tin bng nh sng qua cc mi trng truyn l si dn quang trong tn hiu mang tin c truyn di dng nh sng,...v.v.Cho d truyn tin di bt k mi trng no cng u b nh hng bi nhiu. Nhiu rt phong ph v a dng, ph thuc vo bn cht ca mi trng truyn tin. V d khi truyn tin bng m thanh th nhng ting n xung quanh chnh l nhiu hay khi truyn bng sng intquanhngni cinttrngmnhth cngsb nh hng. Nhiu c nhiu loi nh nhiu cng tnh hay nhiu nhn.ThS. on Hu Chc B mn K thut in t 10Gio trnh L thuyt tn hiu v truyn tin1.2.2.3. Nhn tinNi nhn tin l ni tip nhn thng tin t knh truyn v khi phc li thng tin ban u nh ngun tin pht i. Tin n c ni nhn tin thng khng thu c nh tin ban u truyn i v chu s tc ng ca nhiu. V th, ni nhn phi thc hin vic pht hin sai v sa sai. Hn th na, nu ngun tin c thc hin m ho nn hay bo mt th ni nhn tin cng phi thc hin vic gii nn hay gii m bo mt nhn li tin.Ni nhn tin thng c ba chc nng c bn:-Lugitin, v dnhbnhmytnh, bngghi m, ghi hnh,...- Biu th tin lm cho cc gic quan ca con ngi hay cc b cmbincmthcxltin, vdnhbngghim, hnh nh,...- X l tin a tin v dng d s dng. Chc nng ny c th thc hin bi con ngi hay bng cc thit b my mc.1.2.3. H thng thng tin sCc h thng thng tin tng t c tip tc pht trin t th k trc v t c nhiu thnh tu. Tuy nhin, h thng thng tin tng t c nhng nhc im c hu khng th khc phc. Cc h thng ny thng rt cng knh, khng hiu qu v chi ph rt cao. V d vn v nhiu trong h thng thng tin tng t lun lm au u cc nh khoa hc.Cc h thng ri rc (s) c nhiu u im v khc phc c nhng nhc im ca h thng lin tc. Ngy nay, cc h thng ri rc ngy cng c pht trin mnh m v thu c nhng thnh tu vt c ngoi s mong i.ThS. on Hu Chc B mn K thut in t 11Gio trnh L thuyt tn hiu v truyn tin thc hin c cc h thng ri rc trc ht phi ri rc ho tn hiu mang tin. C hai loi ri rc ho: Ri rc ho theo trc thi gian hay cn gi l ly mu v ri rc ho theo bin hay cn gi l lng t ho.Ly mu tn hiu l t mt hm ban u ta ly ra nhng mu nhng thi im nht nh. iu quan trng l lm th no c th thay th tng ng cc mu ly c vi tn hiu gc. iu ny c gii quyt bi nh l ly mu ca Shannon. Shannon chnh l cha ca l thuyt tn hiu v truyn tin.nh l ly mu Shannon c pht biu nh sau:Mt hm S(t) c ph hu hn, khng c thnh phn tn s ln hn max(= 2 fmax) c th thay th bng cc mu ca n c ly ti cc thi im cch nhau mt khong t /maxhay ni cch khc tn s ly mu F 2fmax. Hnh 1.2 minh ho iu ny.Bin ca tn hiu thng l mt khong lin tc (Smin , Smax). Lng t ho l phn chia khong ny thnh mt s mc nht nh, chng hn l: S0 = Smin, S1 = ...,..., Sn = Smax v qui cc gi tr bin khng trng vi cc gi tr ny v cc gi tr gn vi n nht. C ngha l s c sai s khi thc hin lng t ho. Nh vy vic lng t ho s bin hm S(t) thnh mt hm S'(t) c dng bc thang. S sai khc ThS. on Hu Chc B mn K thut in t 12S(t)SmaxSmintHnh 1.2. Qu trnh ly mu tn hiuGio trnh L thuyt tn hiu v truyn tingia S(t) v S'(t) c gi l sai s lng t ho. Sai s lng t cng nh th S'(t) cng gn vi S(t).Hnh 1.3 minh ho qu trnh lng t ho.Khi thc hin vic ri rc ho tn hiu ta s c cc ngun tin ri rc. Trong nhiu trng hp chng ta thng ch nghin cu cc ngun ri rc. Mt bng ch ci A gm m k hiu l mt ngun tin ri rc, A = {a1, a2, ..., am} vi nhng xc sut hin p(ai) vi i = 1,..., m. Ngun tin ny khng din t mi quan h gia cc tin trc v tin sau nn c gi l ngun tin khng nh ri rc.C nhiu phng php bin i trong h thng thng tin s nh di y minh ho.ThS. on Hu Chc B mn K thut in t 13S(t)SnS1tHnh 1.3. Qu trnh lng t haGio trnh L thuyt tn hiu v truyn tinThS. on Hu Chc B mn K thut in t 14Hnh1.4.Ccphngphpbinithngtin s trong cc khi chc nng ca h thng.Gio trnh L thuyt tn hiu v truyn tin1.3. o thng tinocamt i lnglcchtaxcnhlncai lng . Mi o phi tho mn 3 tnh cht sau:- o phi cho php ta xc nh c ln ca i lng. i lng cng ln, gi tr o c cng phi ln.- o phi khng m.- 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. xc nh o thng tin, chng ta nhn thy rng thng tin cng c nhiu ngha khi n cng him gp, do ln ca n phi t l nghch vi xc sut xut hin ca tin, hay n l hm f(1/p(xi)) cho tin xi c xc sut xut hin p(xi). Mt tin khng cho chng ta lng tin no khi chng ta bit trc v n hay c xc sut bng 1. xc nh dng hm ny, ngi ta s dng tnh cht th ba. Gi thit rng c hai tin xi v xj l c lp thng k mi tin khng cha thng tin v tin cn li. Nu hai tin c xc sut hin l p(xi) v p(xj), lng tin ca mi tin l f(1/p(xi)) v f(1/p(xj)). Gi thit hai tin ny cng ng thi xut hin, ta c tin (xi, xj), lng tin chung cho tin ny phi bng tng lng tin ca tng tin ring bit. Khi hai tin xut hin ng thi, xc sut xut hin ng thi ca chng l p(xi, xj), v ta c:f(1/ p(xi, xj)) = f(1/ p(xi)) + f(1/ p(xj)) (1.1)V hai tin l c lp thng k nn:p(xi, xj) = p(xi) + p(xj) (1.2)Vy nn:f(1/ (p(xi).p(xj))) = f(1/ p(xi)) + f(1/ p(xj))(1.3)ThS. on Hu Chc B mn K thut in t 15Gio trnh L thuyt tn hiu v truyn tinNh vy, trong trng hp ny hm f phi c dng hm loga. Vy hm log(1/p(xi)) l dng hm c th chn lm o thng tin. Ta cnkimtra tnh khng mcahmny.Vta c 0 p(xi) 1nn 1/p(xi) 1 hay log(1/p(xi)) l khng m.Thm vo khi mt tin lun lun xut hin th lng tin nhn c bng khng, ta s kim tra iu kin ny. Khi p(xi) = 1, do vy log(1/p(xi)) = 0.Vyhmlog(1/p(xi))csdnglmothngtinhay lng o thng tin ca mt tin ca ngun tin. Lng o thng tin ca tin xi ca ngun tin no thng c k hiu l I(xi) :I(xi) = log(1/p(xi)) (1.4)Trong biu thc trn c s ca hm loga cha c ch ra. Tu vo c s ca hm loga ny ta s c cc n v o ln thng tin xc nh. Hin nay, thng dng cc n v o sau:Bit hay n v nh phn khi c s loga l 2;Nat hay n v t nhin khi c s loga l e;Hartley hay n v thp phn khi c s loga l 10.V d 1. Ngun A c m k hiu ng xc sut, mt tin do ngun A hnh thnh l mt dy n k hiu ai bt k (aiA). Chng ta s xc nh lng tin cha trong mt tin nh vy. Trc ht hy tm lng tin cha trong mt tin ai. Do ng xc sut nn mi tin aiu c xc sut l 1/m, do :I(ai) = logmLng tin cha trong mt dy x gm n k hiu bng n ln lng tin ca mt k hiu (v chng ng xc sut):I(x) = nlogmn v lng tin tu thuc cch ta chn c s ca log, l bit, nat, hay Hartley nu c s ln lt l 2, e hay 10. R rng khi m k hiu ca ngun c nhng xc sut khc nhau v khng c lp thng k vi ThS. on Hu Chc B mn K thut in t 16Gio trnh L thuyt tn hiu v truyn tinnhau th lng tin ring tng k hiu ph thuc vo xc sut xut hin p(ai) ca n:I(ai) = log 1/p(ai)V lng tin cha trong mt dy k hiu ca ngun khng nhng ph thuc vo xc sut xut hin tng k hiu m cn ph thuc vo xc sut c iu kin. Khi nim ny s c cp n mt cch chi tit cc chng sau.V d 2:Hy xc nh lng tin ring cha trong mt nh ca b nh bn dn.Gii:Mt nh nh bit c th cha cc tin l 0 hay 1. Ngun tin l ngun tin nh phn N=2.Ta t nh sau: tin a1 tng ng vi 0 v a2 tng ng vi 1.V ng xc sut nn P(a1)=P(a2). V ta c cng c:P(a1) + P(a2) = 1.Hay P(a1) = P(a2) =1/2Vy: I(a1) = I(a2) = log22 = 1 (bit)Vy mt nh c lng tin l 1 bit nu tnh theo c s 2.V d 3:Cho mt ngun tin c 8 tin vi phn b xc sut nh sau:Tin ai a1 a2 a3a4a5a6a7a8Xc sut P(ai)1/4 1/4 1/8 x 1/16 1/16 1/16 1/16Hy xc nh lng tin ring ca a4.Gii:Ta c P(a4) =1/8 v I(a4) = log 8 = 3bit.Lng tin cha trong mt dy k hiu ca ngun khng nhng ph thuc vo xc sut xut hin tng k hiu m cn ph thuc vo ThS. on Hu Chc B mn K thut in t 17Gio trnh L thuyt tn hiu v truyn tinxc sut c iu kin. Khi nim ny s c cp n mt cch chi tit cc chng sau.1.3.3. Lng tin trung bnh thng k - Entropy ca ngun tin1.3.3.1. Lng tin trung bnh thng k ca ngun tinLngtintrungbnhcamt nguntinAllngtintrung bnh cha trong mt k hiu bt k ca ngun tin. Ta thng k hiu l I(A) v c tnh bi: A ai iA ai iA a iii i ia I a p a p a pa pa p A I ) ( ) ( ) ( log ) () (1log ) ( ) ((1-5)V d 4:Ta s dng li iu kin ca v d 3. By gi hy tnh lng tin ring ca cc k hiu v trung bnh ca ngun tin. Gii:Theo biu thc 1-4 ta c bng gi tr lng tin ring ca cc tin nh sau:Lng tin ringI(a1) a2 a3A4a5a6a7a82 2 3 3 4 4 4 4Theo biu thc (1-5) ta c:I(A) = 2*2*1/4 +2*3*1/8 +4*4*1/16 = 2,75 bit.iu ny cho ta thy rng c th biu din cc tin trong ngun tin A bng chui c chiu di trung bnh l 2,75 bit thay v dng 4 bit.Bygitaxt mt sbini mt ngunXthnhngunY thng qua s truyn lan trong mt knh truyn. Qua bt k mt knh truyn no cng u c nhiu do vy s truyn lan X qua knh thnh ThS. on Hu Chc B mn K thut in t 18Gio trnh L thuyt tn hiu v truyn tinY khng phi l mt - mt. Ta s i tm kh nng xi no c kh nng ln nht chuyn thnh yj trong qu tnh truyn tin.T y ta c khi nim lng thng tin tng h, lng tin cn li ca xi sau khi nhn yj (lng tin iu kin). Lng tin cn li ca xi sau khi nhn c l yj xc nh nh xc sut hu nghim:) / ( log) / (1log ) / (j ij ij iy x py x py x I (1-6)Lng tin tng h l hiu lng tin ring v lng tin cn li ca xi sau khi nhn c yj (lng tin iu kin). Do vy ta c:) () / (log ) / ( ) ( ) , (ij ij i i j ix py x py x I x I y x I (1-7)Mt khc ta c:jj i j iy x p y p x p ) / ( ) ( ) (Do vy kt qu ta c: jj i jj ij i i j iy x p y py x py x I x I y x I) / ( ) () / (log ) / ( ) ( ) , ((1-8)Tnh cht ca lng tin:1. Lng tin ring bao gi cng ln hn lng tin v n cha trong bt k mt k hiu no c lin h thng k vi n. Ngha l:I(xi)>=I(xi,yj)2. Lng tin ring l mt i lng khng m: I(xi)>=0 nhng I(xi,yj) c th m c th dng.3. Nu X v Y c lp th lng tin ca cp xiyj l:) ( ) ( ) (j i j iy I x I y x I + Mt cch tng qut ta c:) , ( ) ( ) ( ) (j i j i j iy x I y I x I y x I + 1.3.3.2 Entropy ca ngunNh trn trnh by, lngtin trung bnhllng tintrung bnh cha trong mt k hiu bt k ca mt ngun tin cho. Khi ta ThS. on Hu Chc B mn K thut in t 19Gio trnh L thuyt tn hiu v truyn tinnhn c tin ng thi nhn c mt lng tin trung bnh ngha l bt ng v tin cng c gii thot. V vy bt ng v lng tin c ngha vt l tri ngc nhau nhng v s o li ging nhau. y ta ch xt ngun ri rc. bt ng ca tin xi trong ngun X c xc nh bi:) ( log) (1log ) (iiix px px H (1-9)Khi bt ng trung bnh ca mt ngun tin tnh bi: X xi iX xi iX x iii i ix H x p x p x px px p X H ) ( ) ( ) ( log ) () (1log ) ( ) ((1-10)T cc biu thc trn ta thy v s o H(X)=I(X).H(X) c gi l Entropy ca ngun, l mt thng s thng k c bn ca ngun tin.Tnh cht ca H(X) ( cng l ca I(X))1. H(X)>=0. Nu p(xi)=1 th ta c H(X)=02. H(X)lnnht nuxcsut xut hincacck hiuca ngun bng nhau. Lc bt nh ca mt tin bt k trong ngun l ln nht. Tc l:H(X)max logN. Vi ngun c N tin.Ta c th chng minh iu ny nh sau:Nu c N tin cc xc sut xut hin bng nhau th H(X) = logN.Xt:0 11) 11(1log log log log ) (1 1 1 1 1 NiNi iiNiNi ii iNii iN N ppN pp N p p p N X H l iu phi chng minh.V d 5:Cho X={0,1}, p(0) = p, p(1) = 1-p.Tnh H(X) v v th ph thuc vo p.Gii:ThS. on Hu Chc B mn K thut in t 2010 0,5 1pH(X)Hnh 1.5. S ph thuc ca H(X) vo pGio trnh L thuyt tn hiu v truyn tinH(X) = -plogp - (1-p)log(1-p). Ta thy p=1/2 th H(X) =1 t gi tr ln nht.3. Nu hai ngun X={x1, ..., xn} v Y={y1, ..., ym} l c lp th vi z={x,y} ta c:H(Z) = H(X) +H(Y)1.4. Tc lp tin v d ca ngunThng sthngk quantrngnhtca nguntin l Entropy. Thng s th hai chnh l tc lp tin. Tc lp tin ph thuc vo tnh cht vt l ca ngun tin.Tc lp tin R c tnh bi biu thc sau:R = n0H(X) (1-11)Trong n0l s k hiu lp c trong mt n v thi gian. n v ca R l bit/s nu H(X) tnh theo bit. d ca ngun c nh ngha l s chnh lch gia H(X) v H(X)max.Rs = H(X)max - H(X) (1-12) d tng i ca ngun c nh ngha nh sau:max maxmax) () (1) () ( ) (X HX HX HX H X Hrs (1-13)1.5. Tn hiu, biu din v phn loiTn hiu l s bin i ca mt hay nhiu thng s ca mt qu trnh vt l no theo qui lut ca tin tc. Nh vy truyn tin ta s dng cc dng vt cht no truyn. Cn ch rng chnh s bin i ca tham s ca qu trnh vt l mi l tn hiu.Trong phm vi hp ca mch in, tn hiu l hiu th hoc dng in. Tnhiu cthc trkhng i,v dhiuth camtpin, accu; c th c tr s thay i theo thi gian, v d dng in c trng cho m thanh, hnh nh. . . . ThS. on Hu Chc B mn K thut in t 21Gio trnh L thuyt tn hiu v truyn tin1.5.1. Cch biu din hm tn hiu. Di y ta trnh by mt s tn hiu thng gp.- Kiu lit k: hay cn gi l dng bng, cc gi tr ca tn hiu c lit k trong mt bng gi tr.T 01 4 ...S(t) 0 3 7 ...- Dng th: dng th c loi ta cc v ta cc (dng vc t):Khi tn hiu l thc ta c dng th l trc s. Nu tn hiu l tn hiu phc ta c mt phng phc nh hnh 1.7 minh ha.Biu din phc ca tn hiu thng c dng: jAe c (1-14)Trong A l bin v l gc pha.1.5.2. Phn loi tn hiuNu phn loi tn hiu theo dng ton hcth ta c hai loi tn hiu:- Tn hiu lin tc- Tn hiu gin onChng ta cng c th phn tn hiu thnh hai loi l tn hiu tun hon v khng tun hon.Theo dng vt l th ta c loi tn hiu ngu nhin v tin nh.ThS. on Hu Chc B mn K thut in t 22S(t)tHnh 1.6. a.H cc. B. H ta ccS(t1)S(t2)S(t0)a bImReHnh 1.7. H ta cc.0baGio trnh L thuyt tn hiu v truyn tinTrong thc t k thut in t vin thng ta cn chia cc tn hiu thnh cc loi tn hiu: tn hiu lng t, ri rc, s v tng t.1.6. Mt s dng tn hiu c bnDi y gii thiu mt s dng tn hiu thng gp khi phn tch h thng truyn tin.a. Tn hiu dng hm e mTn hiu hm e m thng c biu din di dng biu thc sau.tKe t v ) ((1-15)Trong K, l cchs. Dngtnhiu phthucvoccgitr trn c minh ho hnh 1.8.b. Hmnhy bc n vHm c biu din bi biu thc sau:ThS. on Hu Chc B mn K thut in t 23Hnh 1.8. Dng tn hiu e mGio trnh L thuyt tn hiu v truyn tin'< a t k h ia t k h i01) a t ( uy l hm thay i gi tr t 0 ln 1 ( hoc gi tr bt k) ti thi im t=a. Hnh 1.19 l minh ha mt s trng hp ca hm nhy bc n v.c. Hm Dirac (hay hm xung n v)Khi vi phn hm nhy bc n v ta c hm xung n v hay hm Dirac. Thng k hiu l hm ny l hm (t).dtdu) t ( (1-17)Ta thy rng hm ny l mt hm ton hc khng cht ch v ti thi imt>0th vi phnnybng0 nhngli khngxcnhti t=0. Mt cch tng qut hm ny c nh ngha bi cc iu kin sau:1. (t) = 0 ti t 02. (t) = ti t=03.1 dt ) t ( Dng hm (t) c minh ha hnh 1.10.ThS. on Hu Chc B mn K thut in t 24Hnh 1.9. Hm nhy bc n vHnh 1.10. Hm xung n vGio trnh L thuyt tn hiu v truyn tind. Cc hm lin quan n hm sinDi y l minh ha mt s dng hm lin quan ti hm sin nh hm tt dn, tch ca hai hm sin.1.7. Cc c trng c bn ca tn hiu1.7.1. Cc thng s v c trng ca tn hiua. di ca tn hiu di ca tn hiu l khong thi gian tn ti ca tn hiu t khi xut hin n khi kt thc.b. Khong bin thin ca tn hiu:= Smax - Sminc. Tr trung bnh ca tn hiu:+001tttbdt ) t ( s sd. Tr hiu dng+002tthddt ) t ( s se. Cng sut ca tn hiuThS. on Hu Chc B mn K thut in t 25Hnh 1.11. a. Hm tt dn. b. Tch hai hm sin.S(t)SmaxSmintHnh 1.12. Cc thng s ca tn hiuGio trnh L thuyt tn hiu v truyn tinCng sut tc thi ca tn hiu s(t) c tnh bi:2) t ( s ) t ( p Cng sut trung bnh c tn bi:+001ttdt ) t ( p Pf. Nng lng ca tn hiu dt ) t ( s P . E21.7.2. Cc thnh phn c trng ca tn hiuMt tn hiu bt k bao gm hai thnh phn chnh l thnh phn tn hiu mt chiu v thnh phn xoay chiu. Ch thnh phn bin i mi cha tin tc. Gi tr trung bnh ca tn hiu chnh l thnh phn mt chiu.Ngoi ra mt tn hiu s(t) c th tch ra lm hai tn hiu chn v l.ThS. on Hu Chc B mn K thut in t 26Gio trnh L thuyt tn hiu v truyn tinChng 2. C s l thuyt phn tch tn hiuChng ny s cung cp cho chng ta nhng cng c c bn phn tch tn hiu. l cc chui Fourier v php bin i Fourier.2.1. M uc th phn tch tn hiu thnh dng tngca cctn hiu thnh phn, cc tn hiu n v thnh phn phi trc giao vi nhau tng i mt. V vy khi dng mt cng c ton hc phn tch tn hiu th tn hiu n v phi c dng hm e m ( hoc tng ca cos v sin.).Hai tn hiu f(t) v g(t) c gi l trc giao vi nhau trn on [a,b] nu chng tha mn iu kin:' g ( t ) f ( t ) k h i 0g ( t ) f ( t ) k h i 0 Cd t ) t ( g ) t ( fba*(2-1)Trong g*(t) l lin hp phc ca g(t).V d 2.1:Cho hai tn hiut jmt jne ) t ( ge ) t ( f00Chng minh rng hai tn hiu trn trc giao vi nhau trn on [-T0/2; T0/2] vi w0 = 2 /T0.Gii:Xt tch phn theo biu thc (2-1):ThS. on Hu Chc B mn K thut in t 27Gio trnh L thuyt tn hiu v truyn tin' 1]1

g ( t ) f ( t ) m n k h i 0g ( t ) f ( t ) m n k h i00000 02222000 0TT) m n (T) m n ( s i n Td t e e d t ) t ( g ) t ( f/ T/ Tt j m t j nba* iu ny t c khi ta ch n tnh cht sinx/x1 khi x0. l iu phi chng minh.Da vo tnh cht trc giao ta c th phn tch tn hiu phc tp thnh tng cc thnh phn ca tn hiu n gin tha mn iu kin trc giao.2.2. Chui FourierMt tn hiu c hm s(t) bt k c th c vit di dng tng ca cc thnh phn cos, sin nh sau:] t f n sin b t f n cos a [ a ) t ( snn n + + 10 0 02 2 (2-2)Vi tn hiu tun hon s(t) th ta c 0 002 1 fT. xc nh cc h s an v bn ta s dng cc biu thc di y.+++0 000 000 0000000022221T ttnT ttnT tttdt f n sin ) t ( sTbtdt f n cos ) t ( sTadt ) t ( sTa(2-3)V d 2.2:Xt mt sng vung c dng sau:ThS. on Hu Chc B mn K thut in t 28Gio trnh L thuyt tn hiu v truyn tin) 2 f ( x f ( x )0 x - k h i 0x 0 k h i+ '< v ) x ( f1Xc nh cc h s ca chui Fourier khi phn tch tn hiu thnh tng cc tn hiu cos v sin.Gii:Theo nh ngha ta c:212100 dx aTng t ta c: 00022n. xdx n cos an) n cos (nxdx n sin bn 112200Ta thy rng khi n chn th cosn =1 v khi n l th cosn =0, vy nn:'n l k h in2nc h n k h i 0nbCui cng ta c hm f(x) c th vit li nh sau:ThS. on Hu Chc B mn K thut in t 29Hnh 2.1. Tn hiu xung vung.Gio trnh L thuyt tn hiu v truyn tin1]1

+ + + + ...x sin x sinx sin ) x ( f5533 221V d 2.3.Xc nh chui Fourier ca tn hiu cost c chnh lu nh hnh sau:Gii:Ta c: 2 1 220 tdt cos aTng t ta c:1]1

++ +1 2 11 21 2221 22n) (n) (ntdt cos t cos an nn V bn = 0.Vy:nt cosn) (n) () t ( snn n21 2 11 21 2 211+1]1

1]1

+++ Dng cng thc EULER, c th a dng s(t) trn v dng gn hn ( dng hm m phc ). Theo cng thc EULER ej2nfot = cos 2nfot + j sin 2nfot Mt cch tng qut ta c th vit li biu din s(t) dng chui Fourier nh sau: nt nf jne C ) t ( s02(2-4)Trong n l s nguyn. Cn c tnh bi biu thc sau:ThS. on Hu Chc B mn K thut in t 30Hnh 2.2. Tn hiu cost chnh luGio trnh L thuyt tn hiu v truyn tin+T ttt nf jndt e ) t ( sTC00021(2-5)Ngoi ra ta c cc biu thc quan h sau y:) b a ( C) b a ( Ca Cn n nn n n+ 2121210 0(2-6)2.3. Bin i Fourier. Ph ca tn hiu.2.3.1. Bin i FourierBiniFouriercatnhius(t)btkcnhnghabi biu thc sau:[ ] dt e ) t ( s ) f ( S ) t ( s Fft j 2(2-7)Trong f l tn s ca tn hiu. Mt cch tng qut hm S(f) l mt hm phc ca tn s v c th c vit li nh sau:S(f) = X(f) +jY(f) (2-8)Trong X(f) l phn thc v Y(f) l phn o. Hoc S(f) cng c th biu din theo dng modul v pha nh sau:) f ( je ) f ( S ) f ( S(2-9)Vi:)) f ( X) f ( Y( arctg ) f () f ( Y ) f ( X ) f ( S+ 2 2(2-10) phc hi s(t) t bin i Fourier ta dng bin i ngc. Ta c biu thc sau: df e ) f ( S ) t ( sft j 2(2-11)Vy biu thc (2-7) v (2-11) cho ta cp bin i Fourier.Bin i Fourier dng chuyn tn hiu t min thi gian sang min tn s v ngc li. Ta thng s dng iu ny thc hin ThS. on Hu Chc B mn K thut in t 31Gio trnh L thuyt tn hiu v truyn tinvic phn tch ph tn hiu. Sau y ta xt mt s ph ca cc tn hiu c bit.2.3.2. Ph ca mt s tn hiu c bita. Hm cngHm cng c minh ha trn hnh 2.3. Biu thc ton hc ca hm l:' tt -0 A) t ( sT nh ngha ca php bin i Fourier ta c:) x ( c sin Aff sinAf je eAdt e A dt e ) t ( s ) f ( Sf j f jft j ft j 222222 22 2 nh ca hm cng c cho hnh 2.4.ThS. on Hu Chc B mn K thut in t 32Hnh 2.3. Hm cngGio trnh L thuyt tn hiu v truyn tinb. Hm e mXt mt hm e m c biu thc nh sau:'0 t k h i 00 t k h ite) t ( sTheo nh ngha ta c ph ca tn hiu ny c xc nh bi:f jdt e e ) f ( Sft j t2 1102+ c. Hm Dirac xc nh bin i F ca hm ny ta p dng tnh cht sau y ca hm.) ( s dt ) t ( ) t ( s 0 (2-12)Do ta c bin i F ca hm xung n v l:10 2 2 f j ft je dt e ) t ( ) f ( S )] t ( [ FT ThS. on Hu Chc B mn K thut in t 33Hnh 2.4. Dng ph ca hm cngGio trnh L thuyt tn hiu v truyn tinVy Hm Dirac c bin i F l 1. Da vo iu ny ta tm bin i F ca mt hm s l hng s A bt k.d. Hm s(t)=ATheo biu thc bin i F ca hm Dirac th ta c AA (f). Khi A=1 th F[1]=2 ( )= (f).Mt tnh cht na ca hm Dirac c nhc ti trong phn ny gip ta tm bin i F ca tn hiu t f jAe ) t ( s02 .Nu dch mt khong thi gian no th ta c:) t ( s dk ) k ( ) t k ( s dt ) t t ( ) t ( s0 0 0 + Ta c cp bin i F sau: ) f f ( A Aet f j020 e. Hm nhy bcHm nhy bc n v c m t bi biu thc sau:'