Thong Tin Vo Tuyen

\1My phtMy thuMy phtMy thuCirciulatorDy feederAnten AntenTn hiu voTn hiu voTn hiu ra Tn hiu raHnh 1.1 S tng qut v cc thnh phn c bn ca mt ng thng tin v tuyn hai chiu.CHNG ITNG QUAN V THNG TIN V TUYNI. Qu trnh pht trin ca thng tin v tuyn1. Khi nim v thng tin v tuynThng tin v tuyn s dng khong khng gian lm mi trng truyn dn. Phng php thng tin l: pha pht bc x cc tn hiu thng tin bng sng in t ra khng gian t do, pha thu nhn sng in tbn pht gi ti trong khng gian v tch ly tn hiu gc. Hnh 1.1 m t cc thnh phn c bn ca mt ng thng tin v tuyn hai chiu.2. Qu trnh pht trin ca thng tin v tuynV lch s ca thng tin v tuyn , vo u th k 20 Marconi thnh cng trong vic lin lc v tuyn qua i Ty Dong, Kenelly v Heaviside pht hin mt yu t l tng in ly hin din tng pha trn ca kh quyn c th dng lm vt phn x sng in t. Nhng yu t m ra mt k nguyn thng tin v tyun cao tn i quy m.Chin tranh th gii ln th hai l mt bc ngot trong thng tin v tuyn . thng tin tm nhn thng - lnh vc thng tin s dng bng tn s cc cao (VHF) c nghin cu lin tc sau chin tranh th gii - tr thnh hin thc nh s pht trin cc linh kin in t dng cho VHF v UHF, ch yu l pht trin ngnh Raa.Vi s gia tng khng ngng ca lu lng truyn thng, tn s ca thng tin 2v tuyn vn ti cc bng tn siu cao (SHF) v cc cao (EHF). Vo nhng nm 1960, phngphpchuyntipquavtinhcthchinvphngphp chuyn tip bng tn x qua tng i lu ca kh quyn xut hin. Do nhng c tnh u vit ca mnh, chng hn nh dung lng ln, phm vi thu rng, hiu qu kinh t cao. Thng tin v tuyn c s dng rt rng ri trong pht thanh truyn hnh qung b, v tuyn o hng, hng khng, qun s, quan st kh tng, lin lc sng ngn nghip d, thng tin v tinh - v tr v.v... Tuy nhin, can nhiu vi lnh vc thng tin khc l iu khng trnh khi, bi v thng tin v tuyn s dng chung phn khng gian lm mi trng truyn dn.i ph vi vn ny, mt lot cc cuc Hi ngh v tuyn Quc t c t chc t nm 1906. Tn s v tuyn hin nay c n nh theo "Quy ch thng tin v tuyn (RR) ti Hi ngh ITU-T Geneva nm 1959. Sau ln lt l Hi ngh v phn b li di tn s sng ngn s dng vo nm 1967, Hi ngh v b sung quy ch tn s v tuyn cho thng tin v tr vo nm 1971, v Hi ngh v phn b li tn s v tuyn ca thng tin di ng hng hi cho mc ch kinh doanh vo nm 1974. Ti Hi ngh ca ITU-T nm 1979, di tn s v tuyn phn b c m rng ti 9kHz - 400 Ghz v xem xt li v b sung cho Quy ch thng tin v tuyn in (RR). gim bt can nhiu ca thng tin v tuyn, ITU-T tip tc nghin cu nhng vn sau y b sung vo s sp xp chnh xc khong cch gia cc sng mang trong Quy ch thng tin v tuyn:- Dng cch che chn thch hp trong khi la chn trm.- Ci thin hng tnh ca anten.- Nhn dng bng sng phn cc cho.- Tng cng ghp knh.- Chp nhn s dng phng php iu ch chng li can nhiu.II.Sng in t1. Khi nim- Trng in t l mt dng vt cht c bit c c trng v nng lng, vn ng tng tc vi cc mi trng . Trng in t lan truyn trong cc mi trng (nh chn khng, khng gian t do, ng dn sng...) gi l sng in t.- Nguyn nhn pht sinh sng in t: l do s chuyn ng c gia tc (dng hoc m) ca cc in t t do. V nguyn l, bt k h thng in t no c kh nng to ra in trng hoc t trng bin thin u c bc x sng in t. Tuy nhin trong thc t, s bc x ch xy ra trong nhng iu kin nht nh.3- Cc i lng c trng:+ Cng in trng E+ Cng t trng H. c im ca 2 vc t E v H l lun vung gc vi nhau v vung gc vi phng truyn sng.+ Vn tc truyn sng v xp x vn tc nh sng (c = 300.000 km/s)+ Tn s f v bc sng := c/f2. Phn loi sng in t v ng dngTn sPhn loi bng tnC ch truyn sng v tuynLnh vc s dng3KHz~30 KHz VLF B mt Thng tin o hng 30KHz~300KHz LF B mt Thng tin o hng 300KHz~3MHz MF B mt , Sng tri Pht thanh AM, Hng khng, o hng.3MHz~30MHz HF B mt , Sng triPht thanh sng ngn, o hng.30MHz~300MHz VHF Sng triSng khng gianPht thanhFM, truynhnh, thngtinng, thngtinv tuyn c nh (viba)300MHz~3GHz UHF Sng triSng khng gianSng thngTruynhnh, thngtinng, thng tin v tuyn c nh (viba), Raar, thngtinv tinh.3GHz~30GHz SHF, VibaSng khng gianSng thngThngtinvtinh, thngtin vtuyncnh, Raar, v tuyn thin vn.30GHz~300GH EHF, MilimeterSng khng gianSng thngThngtinvtinh, thngtin vtuyncnh, Raar, v tuyn thin vn.Trong thng tin v tuyn, c ch truyn sng v tuyn v vic s dng thit b truyn thng ph thuc vo tn s v tuyn s dng. Bng di y trnh by bng 4tn s v tuyn c phn loi theo tiu chun quc t hin hnh v theo c ch v phng thc s dng sng v tuyn. 3 .Cc phng thc truyn lan ca sng in tSngv tuyn khng truyn lan theo ng thng khi chng trong khng gian c nh hng ca cc yu t trn mt t v tng i lu. Tuy nhin, khi kho st sng truyn trong khng gian ta cng c th coi l mi trng l tng sng in t truyn theo ng thng. Trong thc t ngi ta phn thnh cc phng thc truyn lan ch yu sau: Sng t, sng tri, sng tn x qua tng i lu v sng thng.a. Sng tSng t l sng in t chu nh hng ca cc yu t kh hu, a hnh, .v.v. trn b mt t khi truyn t an ten pht n anten thu. Sng t gm 2 loi: Sng b mt b sng khng gian.- Sng b mt. Sng b mt l sng in t truyndc theo b mt t. Sngb mt ch yu l cc sng in t tn s thp. S lan truyn c thc hin nh nhiu x sng in t (Hin tng tia sng un cong vt qua vt cn vi suy hao rt t). Do nhiu x t l nghch vi bc sng cho nn tn s s dng cng cao th suy hao ca sng t cng ln, kh nng lan truyn sng t cng yu. Hnh 1.2 M t m hnh truyn sng t.Hintng nhiu x c mi tng quan cht ch vi dn in v hng s in mi ca t trong ng lan truyn. Hng s in mi ca nc bin nh hn ca t nn c ly truyn sng trn mt bin di hn so vi mt t. Do , sng in t tn s thp c s dng rng ri trong thng tin v tuyn o hng. Bng tn s cc thp c s dng ch yu cho truyn thanh AM v thng tin hng hi v thng tin o hng.-Sng khng gianSng khng gian lsngtruyngia2 5Tia sng un cong theo mttAnten phtAnten thuHnh 1.2: Sng truyn dc theo mt tSng trc tipSng phn x tSng phn x i luAnten phtAnten thuHnh 1.3: Sngtruyngia 2 anten t cao trong khng giananten t cao trong khng gian nh m t hnh 1.3. Sng in t n im thu c th: - Trc tip t anten pht: Sng trc tip.- Phnx ti mt t trc khi ti im thu: Sng phn x t.- Phnx trong tng i lu trc khi ti im thu: Sng phn tng i lu. Sng trc tip l sng in t pht x trc tip t anten pht n anten thu m khng b phn x u c. Trong iu kin truyn lan bnh thng sng trc tip c bin ln hn tt c cc sng khc cng ti in thu.Sng phn x t l sng truyn t anten pht n anten thu sau khi phn x ti mt t hoc ti cc vt th xung quanh. Do , di ng i ca tia sng phn x t di hn so vi sng trc tip. Nh vy, ti im thu, sng phn x t s c bin v pha khc vi sng trc tip. Nu chnh lch v ng truyn gia sng trc tip v sng phn x t khc nhau mt khong bng s l ln bc sng th ti im thu pha ca hai sng ny khc nhau mt khong 180o, c ngha l ngc pha nhau. Nu khi sng trc tip v sng phn x t c bin bng nhau th chng s trit tiu nhau. sng phn x tng i lu l cc sng in t bc x ti anten pht b un cong khi truyn trong tng i lu - do h s khc x ca khng kh thay i theo cao - trc khi n anten thu. Cng ging nh trng hp sng phn x t, nu chnh lch v ng truyn gia sng trc tip v sng phn x tng i lu khc nhau mt khong bng s l ln bc sng th ti im thu pha ca hai sng ny khc nhau mt khong 180ov nu khi sng trc tip v sng phn x tng i lu c bin bng nhau th chng s trit tiu nhau. Sng khng gian c s dng cho cc tn hiu ln hn VHF. S thay i h skhcxtheocaocakh quyngynhhngnsngkhnggian. Kh quyn tiu chun l mt kh quyn l tng c mt t l bin i h s khc x theo cao mt cch u n, bi v n c mt h s thay i c nh ca p sut kh quyn theo cao, nhit v m. V c s bin i h s khc x mt cch lin tc, cho nn ng i thc t ca sng khng gian l khc vi ng trc tip (thng). b li s khc nhau ny, c ly thng tin cc i thc t c tnh ton theo ng trc tip da trn quy nh bn knh hiu qu ca tri t KR (K=4/3 trong kh quyn tiu chun) nh m t Hnh 1.4.6Hnh 1.4 ng i ca sng khng gian : a) iu kin thc t, b) iu kin t-ng ng ca bn knh tri t c tnh bng R (K=4/3)b. Sng triSng tri l sng in t b thay i hnh trnh ca mnh ti tng in ly v quay tr v tri t . - Tng in lyTng in ly hnh thnh ti cao 100Km - 400Km l do kt qu ca vic ion ho trng thi ca tng i lu bng cc tia cc tm v tia X do mt tri bc x. Tng in ly c phn chia thnh mt vi lp c gi tr mt in t cc i. Mi tng cphnchiathnhcclpD, E, F theo cao ca n. Lp F li c phn chia thnh lp F1, F2. Hnh 1.5 trnh by mt tnh theo cao ca cc lp ion in hnh.-Truyn sng trong lp IonC th xemlp ion ca tng in ly ging nh mt tm in mi khng l mhskhcxcanbini lin tc, vsbin icamt iontheo bc sng l khng ng k trong bng tn s cao (bc sng ngn hn). Hnh 1.6a trnh by ng i ca sng v tuyn trong tng in ly. C th gii thch hin tng phn x ti tng in ly nh sau: Ta chia tng in ly thnh cc lp rt mng sao cho trong lp chit sut c th coi l u. Nh vy khi truyn t lp Ion ny sang lp Ion khc th tia sng in t b khc x. Qua nhiu lp Ion nh vy tia sng in t s 76004002000Mt (cm3)Ngy mLp DLp ELp FLp F1Lp F210 102104106Hnh 1.5 Mt tnh theo cao ca cc lp ionRKR(a)(b)ng thc tng trc tiph1h1h2h2b un cong v coi nh phn xquayvtri t. Nut mt my thu im tia sng inttrv th ta s thu c thng tin pht i t my pht.Vc tnh ca tng inlythayi theocao nn chng c nh hng khc nhau n qu trnh truyn sng in t. cc sng in t c tn s cao qu s i xuyn thng qua tngin ly(Hnh 1.6b). Sng in t c tn s thp qu th kh nng chuyn ti nng lng yu, s khng nc tnginly. tng in ly ch phn x sng in ttrongbngsngngnv sng trung. Bng thc nghim ngi ta xc nh c rng: sngngnphnxti lp F1 v F2, sng trung phn x ti lp E, Lp D khng gy ra phn x sng in t, ngc li lp D hp th nng lng sng in t trong bng sng trung. Tuynhin lp D ch xut hin vo ban ngy, v nh vy ban ngy khng th dng sng trung thng tin qua tng in ly c.Bng thc nghim ngi ta xc nh c rng, kh nng phn x sng in t ti tng in ly ph thuc 3 yu t chnh l: Mt cc ht mang in (N) trong tng in ly, tn sv gc ti ca tia sng in t.Gistiasngintti tnginlyvi gc ti0(Hnh 1.7), sng in t ny phn x c ti tng in ly th0 phi tho mn iu kin sau:8n1n0n2n3n4nnTia sngt tri t pht ti tng in lyTia sng khc x v i ra tng in ly tr v tri tTng in lyTng Bnh lu(a)Mt tKhong nhyTng in lySng v tuyn xuyn qua tng in lySng v tuyn khc x ti hnHnh 1.6.- C ch phn x ca tng in lya. Khc x trong tng in lyb. iu kin ti hn c phn x(b)Tng in ly0Gc tiGc tTip tuyn tri tTia tiTia phn xMt tHnh 1.7208 , 80 1 sinfN 2028 , 801fNSin 2028 , 801 1fNCos 09 CosNfKhi N=Nmax th tn s ny c gi l tn s ln nht fmax0 ng vi 0. Tn s ln nht s t c khi N=Nmax v0= 0max. Khi0= 0max th tia ti chnh l tip tuyn catrit.tuynhintartkhcthxcnh chnh xc c gc0, bi vy ngi ta tm cch tnh tn s f thng qua gc , v v0quan h vi nhau nh sau: Khi0= 0max th =0, khi 0=0 th =90o.Vic tnh tn s f thng qua gc c m t hnh 1.8. Trong a l bn knh tri t, h l cao im phn x, 0: gc ti. : gc t.Tac: ahSinahNf2)21 ( . 8 , 80) (2maxmax+ + Khi0= 0max th =0 nn Sin20=0. Suy ra: hh a Nf2) 2 ( . 8 , 80) (maxmax+ Phng thc truyn sng qua tng in ly c ng dng trong truyn thanh. Bngcchchosngintphnx nhiu ln gia tng in ly v mt t nh m t hnh 1.9 ta c th truyn c sng in t i rt xa. Khi truyn nh vy s c nhng vng khng nhn c sng in t, vng c gi 9Tri tTng in lyVng tiVng tiKhong nhyHnh 1.9Tng in ly0Gc tiGc tTip tuyn tri tTia tiTia phn xMt tHnh 1.8haOaTri tTng in lyHnh 1.10l vng ti. khc phc hin tng ny ngi ta pht ln tng in ly chm sng in t phn k nh m t hnh 1.10.c. Sng tn x qua tng i luVic lan truyn ca sng v tuyn nh hiu ng tn x ca vng kh quyn khng ngnht trongtngi lucgi ltruynsngtnxquatngi lu. Phng php ny cho php thng tin lin lc c ly xa cc bng tn VHF, UHF, v SHF. Hnh1.11mtbncht cavic truyn sng tn x qua tn i lu. Trong tng i lu, cao 10n 12 Km, thngcvngkhngkh khng ng nht. Trong , chit sut gia cc lp khng kh thay i rt ln, nn khi i vo vng ny sng in t b tn x ra nhiuhngkhcnhau. Mt stiab tn x n mc gn nh phn x v quay tr v tri t.Hin tng ny c gii thch gii thch nh sau: ta chia vng khng kh khng ng nht thnh cc lp rt mng sao cho trong lp chit sut ca khng kh c th coi l u. Nh vy khi truyn t lp khng kh ny sang lp khng kh khc th tia sng in t b khc x b khc x. Quanhiulpkh quynnh vy tia sng in t s b un cong v quayv trit. Nu t mt mythuimtia sng in t tr v th ta s thu c thng tin pht i t my pht.Phng php ny i hi cng sut pht ln v my thu c nhy cao.Mt khc tnh n nh ca h thng thng tin ny khng cao do vng kh khng ng nht thng xuyn thay i theo ngy - m, theo ma . . .10n1n0n2n3n4nnTia sngt tri t pht tivng kh khng ng nhtTia sng khc x v dn dn v quaytr v tri tVng kh khng ng nhtHnh 1.12: Khc x trong vng kh quyn khng ng nhtMt tim pht im thuVng khng kh khng ng nhtSng in t b tn xHnh 1.11: Truyn sng tn x trong tng i lud. Truyn sng thngL phng thc truyn sng trong khng gian t do. Phng thc ny ch c trong h thng thng tin v tinh, khi sngin t i ra ngoi tng kh quyn ca tri t.III.Tng quan v h thng thng tin vi ba1. Khi nim vthngtin vi baThng tin vi ba l thng tin v tuyn gii sng cc ngn v thc hin thng tin nhiu knh.Hin nay cc h thng thng tin vi ba c ng dng rt rng ri truyn tn hiu trong mng thng tin. K thut thng tin vi ba cng c pht trin mch m ngy cng ph hp vi nhu cu thng tin trn mng. Ngy nay hu ht cc thit b vi ba c s dng l thit b vi ba s dung lng cao. Ngoi cc h thng vi ba trn mt t cn c cc h thng vi ba v tinh vi c li thng tin ln ti hng chc ngn Km. Cc tn hiu truyn trn ng thng tin vi ba bao gm cc tn hiu thoi, truyn hnh, s liu, . . .2. Cu trc gi nh ca tuyn thngtin vi baHnh 1.13 m t s khi gi nh ca mt tuyn vi ba trn mt t (cn gi l h thng v tuyn chuyn tip c nh).Trmu cui cn c mt my pht, mt my thu, cc thit b ghp knh v mt anten. Nu h thng vi ba s ny c s dng truyn tn hiu s c s (lung s c tc 2 Mb/s) v lung s c ly t tng i hoc t thit b ghp knh cp cao hn th trm u cui khng cn thit b ghp knh. Cc trm u cui thng c t cc thnh ph ln. Trm u cui c nhim v:- hng pht: nhn thng tin t cc trung tm truyn hnh, truyn s liu, tng i hoc cc thu bao sau tin hng ghp knh to thnh mt lung s tc 11TxRxRxTxGhp knhTxRxRxTxGhp knhTxRxGhp knhTxRxTrm u cuiTrmlpTrm r xenTrm u cuiHnh 1.13: S khi gi nh ca h thng v tuyn chuyn tip c nh cao v a vo my pht iu ch vo sng mang siu cao tn pht sang pha i phng.- hng thu: thu nhn thng tin l tn hiu sng mang siu cao tn t pha i phng sau tin hnh gii iu ch to li lung s nguyn thu v a ti my phn knh x l thnh cc tn hiu ring bit a v cc trung tm truyn truyn hnh, truyn s liu, tng i v thu bao. Trm trung gian cn c hai my pht, hai my thu v hai anten thng tin vi hai pha khc nhau.Cc trm trung gian thng c t nhng t c s giao lu thng tin, kinh t v.v. Trm trung gian c nhim v thu nhn thng tin l tn hiu sng mang siu cao tn t trm pha trc sau tin hnh gii iu ch to li lung s nguyn thu v a ti my pht tip sang trm pha sau.Trm r xen (Cn gi l trm trung gian chnh) cn c hai my pht, hai my thu, cc my ghp tch knh v hai anten thng tin vi hai pha khc nhau.Cc trm r xen thng c t cc th x v cc ni c khu cng nghip ln.Trm trung gian c nhim v thu nhn thng tin l tn hiu sng mang siu cao tn t trm pha trc sau tin hnh gii iu ch to li lung s nguyn thu v a ti my ghp xen knh ly ra v/hoc xen vo mt s knh sau a ti my pht tip sang trm pha sau.3. c im v phn loia) c imThng tin vi ba lm vic di sng cc ngn nn c cc c im sau: - C di tn rng ph hp vi truyn thng tin nhiu knh (Thng tin bng rng).- Thng tin n nh.- Vi mt c ly thng tin cho trc th c th p dng cng ngh antenvtngtnscngtc gim nh cng sut my pht.rhntaxt biuthc tnh sut in ng u vo ca my thu nh sau: ( )

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mmVkm RG PEthP. 173 1210 30 100 300 103104105E(v)F(MHz)Nhiu kh quyn v nhiu cng nghipNhiu v trTp m niHnh 1.14:Nhiu v tp m ni ph thuc tn s.Trong : Eth: Sut in ng ca tn hiu ti u vo my thu, P: Cng sut ca my pht, R: Khong cch thng tin gia hai trm, GP: H s tng X (H s khuch i) ca anten, Shd GP24, Shd: Din tch ming m ca anten, : Tn s cng tc.- Trnh c can nhiu t bn ngoi nh: Nhiu kh quyn, nhiu v tr, nhiu cng nghip.Hnh1.14 biu din mc nhiu v tp m ni ph thuc tn s. t th ta thy: tn s thp hn 30 MHz th nhiu v tr, nhiu kh quyn, nhiu cng nghip ln.T 30 n 300 MHzch cn nhiu v tr, t 300 MHz tr ln khng cn nhiu v tr na nhng tp m ni ca thit b s tng ln. Do cc phn t cao tn trong ccthit b thu pht phi c cu trc c bit c th lm vic n nh tn s cao.- Kh nng nhiu x ca sng in t tn s cao rt yu nn ch thc hin thng tin trong tm nhn thng.b) Phn loiC nhiu phng php phn loi ng thng tin vi ba. Sau y ta xt mt s phng php c bn:* Phn loi theo Phng thc ghp knh+ ng thng tin vi ba ghp knh theo tn s+ ng thng tin vi ba ghp knh theo thi gian* Phn loi theo dung lng ca ng thng tin vi ba+ ng thng tin vi ba dung lng nh+ ng thng tin vi ba dung lng trung bnh+ ng thng tin vi ba dung lng ln* Phn loi theo tn hiu truyn trn ng thng tin vi ba+ ng thng tin vi ba tng t+ ng thng tin vi ba s* Phn loi theo phng thc truyn sng trn ng thng tin vi ba+ ng thng tin vi ba tm nhn thng+ ng thng tin vi ba tn x qua tng i lu+ ng thng tin vi qua v tinh nhn to134. Phn b tn s trong thng tin vi bang thng tin vi ba tm nhn thng thng c nhiu trm chuyn tip nhau, v vy nu ta cho tt c cc my pht trn ng thng tin hot ng cng mt tn s th s xy ra hin tng can nhiu, tn hiu ca trm ny s gy nh hng cho trmkialmgimcpcht lngthngtin. khcphchintngny, trn ng thng tin vi ba ngi ta b tr cho cc my pht lm vic cc tn s khc nhau thch hp m bo cht lng cho ng thng tin. Vic quy hoch tn s cho mt h thng c gi l phn b tn s. Hin nay, h thng thng tin vi ba s dng mt s phng php phn b tn s nh sau:a) H thng s dng 2 tn sHnh 1.15 m t phng php phn b 2 tn s. t hnh v ny ta thy: Ton h thngchs dng 2 tn s, myphtv my thu ca cctrmktipnhautrn hng ithay nhau lm vic ti cc tn s f1 v f2, my pht v my thu trn hng v ca cc trm k tip nhau thay nhau lm vic ti cc tn s f2 v f1.Phngphp ny tng i n gin, di tn ca anten khng cn ln lm. Tuy nhin, h thng ny c hin tngthu vt trm ca my thu, c ngha l my thu ca mt trm thu c tn hiu pht ra t nhiu my pht ca cc trm pha trc. Mt khc, c th xy ra hin tng thu theo hng ngc li do sng in t phn x tr li khi gp cc vt chn. Ngoi ra, do h s phng v ca anten khng cao lm nn my thu cn thu c tn hiu pht ra t my pht ca trm pha sau.Khi s dng phng php phn b 2 tn s th yu cu anten phi c phng v t nht l 70 dB.b) H thng s dng 4 tn s14My pht f1My thuf2My pht f2My thuf1My thu f1My phtf2My pht f1My thuf2My thu f2My phtf1f1f1f1f2f2f2Trm u cui Trm trung gian Trm trung gianHnh 1.15: Phng php phn b 2 tn s.My pht f1My thuf3My pht f3My thuf1My thu f4My phtf2My pht f4My thuf2My thu f3My phtf1f1f4f1f3f2f3Trm u cui Trm trung gian Trm trung gianHnh 1.16: Phng php phn b 4 tn s.Hnh 1.16 m t phng php phn b 4 tn s. t hnh v ny ta thy: Ton h thng s dng 4 tn s, my pht v my thu ca cc trm k tip nhau trn hng i nhau thaylm vic ti cc tn s f1 v f3, my pht v my thu ca cc trm k tip nhau trn hng vthay nhau lm vic ti cc tn s f2 v f4.Phng php khc phc c hin tng thu theo hng ngc li, cho php anten c phng v khng cao. Tuy nhin, phng php ny vn cn c hin tng thu vt trm. khc phc tnh trng ny trong c hai phng php phn b tn s ngi ta b tr cc trm trn tuyn sao cho khng nm trn cng mt ng thng.c) Phng php chia i gii tnNu mt s knh v tuyn c s dng trong cng mt tuyn truyn dn th bng tn xc nh c chia lm 2 bng tn con: Bng tn cao (Nhm tn s cao : Uper Band) v bng tn thp (Nhm tn s thp : Lower Band) nh m t hnh 1.18. Mi mt knh v tuyn s dng phng php phn b hai tn s.15TxRxRxTxGhp knhTxRxRxTxGhp knhTxRxTrm u cuiTrmlpTrm r xenTxRxRxTxTrmlpHnh 1.17: B tr cc trm trn tuynkhng nm trn cng mt ng thng.f0Bng tn thp Bng tn caof1 f1 f2 f2 fn fn X Y X: Khong cch knh ln cn Y: Khong cch thu phtHnh 1.18: Mt v d tiu biu v phng php chia i gii tnCHNG II.ANTEN, FEEDER (FI)V CC PHN T SIU CAO TNI. Cc loi phi (feeder)1. Nhim v v yu cua. Nhim vFeeder l phn t trung gian gia my pht hoc my thu v anten. Nhim v ca dy feeder l dn nng lng cao tn t u ra ca my pht ti anten v t u ra ca anten ti u vo my thu. b. Yu cu16My pht f1My thuf1My pht f2My thuf2My pht fnMy thufnMy pht f1My thuf1My pht f2My thuf2My pht fnMy thufnMy pht f1My thuf1My pht f2My thuf2My pht fnMy thufnMy pht f1My thuf1My pht f2My thuf2My pht fnMy thufnMy pht f1My thuf1My pht f2My thuf2My pht fnMy thufnHnh 1.19: T chc nhiu knh v tuyn trn cng mt tuynp dng phng php chia i gii tn- Khng c bc x v thu sng in t, hay ni cch khc l khng c hiu nganten. V nuFeeder chiungantenth ngoi vicnhhngntnh phng hng ca anten n cn l ngun gy nhiu cho my thu.- C tn hao nh khi chuyn ti nng lng tn hiu siu cao tn, hay ni cch khc l hiu sut ca feerder phi cao.- Chu c dng v p cao tn c bin ln.- Phi hp c tr khng vo/ra ca anten, tr khng ra ca my pht v tr khng vo ca my thu.- C bn c hc cao (Kt cu chc chn) khng b bin dng trong cc iu kin thi tit.- Cc thng s ca Feeder phi n nh.- Phi m bo tnh kinh t. 2. Phn loi fiC hai loi feeder l feeder cp ng trc v feeder ng dn sng. Trong mi loi li c phn thnh cc nhm khc nhau tu theo suy hao, kch thc . . .3. Cc tham s ca fi- Suy hao trn mi mt chiu di.- Tr khng c tnh WC4. Cc loi fi thng dnga ) Fid cp ng trca1. Cu to: Feeder cp ng trc c cu to nh m t hnh 2.1. 174 3 2 1Hnh 2.1 : Cu to ca Feeder co ng trca. Hnh di dng phi cnh, b. Hnh di dng mt ct4 2 1 3Dd- Dy dn trong 1: cn gi l l li c ch to bng vt liu dn in tt, thng l ng, nhm hoc kim loi do, c th rng hoc c. - Lp cch in cao tn 2: c ch to bng Poliethylen hoc nha do c nhim v cch in gia dy dn trong v dy dn ngoi.- Dy dn ngoi 3: c ch to bng ng thau pha trong c m vng hoc bc.- Lp v bo v 4:c ch to bng Poliethylen hoc cao su.a2. Nguyn l truyn sngKhi a nng lng cao tn vo cp ng trc th gia 2 dy dn trong v ngoi s pht sinh sng in t chy dc theo dy dn, do s thc hin c vic truyn nng lng (cn gi l truyn sng) t my pht ti anten v t anten ti my thu. Tr khng c tnh cafeeder cp ng trc c tnh theo theo cng thc sau: ) ( lg138 dDWC. TrongDlngknhmptrongcadydnngoi, dl ng knh ca dy dn trong, l hng s in mi ca cht cch in gia dy dn trong v dy dn ngoi. Thng thng =1 nn ) ( lg 138 dDWC .Ccloi feedercpngtrctiuchuntrongcngnghipthngctr khng c tnh nm trong khong 40 n 150 . Trong , loi feeder vi tr khng c tnh bng 50 c bn v in cao nht (Chu c dng v p cao tn ln nht)loi ny c D/d=2,7. Feeder vi tr khng c tnh bng 50 thng c s dng trong thng tin vi ba.Loi cp ng trc vi tr khng c tnh bng 75c hpthnnglngnhnhtnnthng c s dng truyn tnhiu bng gc (tn hiu truyn hnh, tn hiu s .v.v.) loi ny c D/d=3,6.a3. u nhc im -Cuto tngi dngin, tr khngc tnh bin ng trong phm vi ln nn c th chtoccloi congtrcgiaotipvi nhiuloi thit b ctrkhngvorakhc nhau, bng tn cng tc tng i rng.-Khi tn s cao qu 2 GHz th suy hao trn cp ng trc rt ln, do khng th s dng 18badd2d1Hnh 2.2 Cu to ca ng dn sng c v theo mt ct.Mt trong c trng nhnMt trong c trng nhnMt trong c trng nhnng kim loi rngng kim loi rngng kim loi rngcp ng trc dn tn hiu c tn s ln hn 2GHz, khi cn s dng Feeder ng dn sng.b) Fi ng dn sngb1. Cu tong dn sng l ng kim loi rng c dn in cao, c thit din hnh ch nht, hnh trn hoc hnh elip, mt trong ca ng c lm nhn c th phn x tt sng in t. Hnh 2.2 m t cu to ca ng dn sng. Ta c th tng tng ng dn sng l mt ng dy song hnh, trn gn cc on dy ngn mch u cui c chiu di /4. Khi t cc on dy st vo nhau th s to thnh ng dn sng. Xem hnh 2.3.V on dy c chiu di /4 ngn mch u cui nn tr khng vo ch ni vi ng dy bng v cng nn khngnhhngnchcng tc ca ca ng y v do khi cc on y c chiu di /4 t st vonhaustothnhmt hp kimloi, hpnykhngnhhng n qu trnh truyn sng trn ng dy. b2. Nguyn l truyn sng,iu kin sng in t truyn c trong ng dn sng.Nu kch thch vo ng dn sng mt nng lng siu cao tn th scsnginttruynlantrong ng dn sng. Sng in t truyn lan trong ng dn sng bng cch phn x nhiu ln trn thnh ng nh m t hnh 2.4.Sngin t truyn trong ng dn sng gm 2 loi: Sng in ngang TEmn (Ccthnhphnintrngnm ngang) v sng t ngang TMmn (Cc thnh phn t trng nm ngang), m-nlccchsccloi sng khc nhau. Trong mi loi sngli cchialmnhiuloi 19Hnh 2.4: Nguyn l truyn sng trong ng dn sngPhn t kch thch a sng in t vo trong ng dn sngng dn sngmt u bt kn /4 /4 /4 /4(a) (b)Hnh 2.3: Cu to tng ng ca ng dn sng.a. Cc on dy /4 ngn mch c ni ri nhau trn mt ng dy song hnhb. Cc on dy /4 ngn mch c ni st nhau trn mt ng dy song hnh to thnh ng dn sngkhc nhau gi l cc Mode. Cc mode ny c xc nh bi ch s m-n (m, n = 0,1,2,3 .v.v. v khng ng thi bng 0). Cc loi sng in t truyn trong ng dn sng cn ph thuc vo tn s cng tc v kch thc ca ng dn sng. Mi ng dn sng c kch thc khc nhau th c mt bc sng ti hn (th) khc nhau. Do cu to ca ng dn sng nn bc sng ti hn c gi tr bng 2 ln dichiu rng ca ng. iu c ngha l cc sng in t c bc sng ln hn hai ln kch thc chiu rng ca ng dn sng th khng th truyn c trong ng . Bc sng ti hn ca ng dn sng c xc nh theo biu thc :2 22

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+ ,_

bnamth Trong : a l kch thc chiu rng ca ng; b l kch thc chiu cao ca ng;m, n l ch s ca cc loi sng truyn trong ng dn sng.V d vi sng in ngang TE10 th bc sng ti hn: aab athTE2120 122 210

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b4. u nhc im- C cu trc chc chn, truyn c cc sng in t tn s cao vi suy hao tng i thp.- Ch to kh, tn s cng thp th kch thc ng phi cng ln, yu cu v iu kin bo qun tng i nghim ngt (Trong lng ng phi lun kho ro, phi bm kh kh-sch hoc kh tr vo lng ng).20c .Cch kch thch cho ng dn sng (Cch a sng in t vo ng dn sng)C 2 cch kch thch cho ng dn sng l kch thch bng lng cc in (Kch thch bng gi) v kch thch bng lng cc t (kch thch bng vng).c1. Kch thch bng Gi (lng cc in).* Cu to:Gi kch thch l mt on dy ng trc nh m t hnh 2.5. nh ca lng cc in ng vai tr nh mt anten bc x sng in t. * Nguyn l hat ng kch thch c ta phi a u gi ca lng cc in vo trong lng ng dn sng v ni u kia vi ngun tn hiu siu cao tn nh m t hnh 2.6. nh 21V bo vDy dn ngoiDy dn trong(u gi)Lp in miHnh 2.5: Cu to ca lng cc in (gi kch thch)Ngun tn hiu cao tnin trngTtrngHnh 2.6: Nguyn l kch thch bng lng cc inV bo vDy dn ngoiDy dn trong(u gi)Lp in miHnh 2.7: Cu to ca lng cc t (Vng kch thch)Ngun tn hiu cao tnin trngTtrngHnh 2.8: Nguyn l kch thch bng lng cc tca u gi (Dy dn trong) hp vi dy dn ngoi to thnh mt t in c gc m rt ln, khi c ngun siu cao tn th gia 2 bn t ny s c ng sc in trng bin thin, in trng ny to ra t trng bin thin v chng tip tc chuyn ho ln nhau to thnh sng in t.c1. Kch thch bng vng (lng cc t).* Cu to:Vng kch thch l mt on dy ng trc nh m t hnh 2.7. nh ca lng cct c mt vng dy c bit ni tt gia dy dn trong v day dn ngoi. Vng dy ny ng vai tr nh mt anten bc x sng in t. * Nguyn l hat ng kch thch c ta phi a vng dy ca lng cc t vo trong lng ng dn sng v ni u kia vi ngun tn hiu siu cao tn nh m t hnh 2.8. Khi c ngun siu cao tn th trong vng dy s c dng in, v nh vy xung quanh n c t trng bin thin, t trng ny to ra t in bin thin v chng tip tc chuyn ho ln nhau to thnh sng in t.II. Cc loi an ten dng trong thng tin vi ba1. Nhim v v yu cu ca antena. Nhim vAnten l mt thnh phn khng th thiu c trong thng tin v tuyn ni chung v thng tin vi ba ni ring. Anten c nhim v bc x (Pht) v thu sng in t.b. Yu cu+ C kh nng tp trung nng lng v mt hng (Hoc mt s hng) theo yu cu.+ Kh nng bc x v thu sng in t ca cc hng ph v hng ngc li rt nh.+ C kt cu chc chn.+ C gii tn cng tc rng.+ m bo tnh kinh t2. Phn loi antenC nhiu cch phn loi anten:* Phn loi theo bc sng cng tc ta c: Anten sng di, Anten sng trung, Anten sng ngn,Anten sng cc ngn . . .22XY(a)XY(b)Hng chnh(Bp chnh)Hng ph(Bp ph)XY(c)Hng chnh(Bp chnh)Hng ph(Bp ph)Hnh2.9: a. Anten bc x v hng, b. Anten bc x n hng, c. Anten bc x 4 hngMMYX XYR RAnten chunAnten kho stHnh 2.10: ngha ca h s nh hng D* Phn loi theo hnh dng v cu to ta c: Anten chn t, anten parabol, anten mt gng, anten hai gng, anten tim vng . . .* Theo kh nng bc x sng in t Ta c: - Anten bc x n hng: nng lng sng in t ch bc x theo mt hng nht nh.- Anten v hng: Cn gi l anten ng hng, nng lng sng in t bc x theo theo mi hng.-. . .3. Cc tham s ca antena) c tnh bc x (Tnh phng hng)L c tnh biu din kh nng bc x ca anten theo cc hng khc nhau. c tnh bc x c biu th bng th tnh phng hng.Hnh 2.9 m t th tnh phng hng ca mt s anten khc nhau.b) H s nh hng (D)H s nh hng D l t s gia mt nng lng bc x ca mt anten ti mt im cch anten mt khong R trn phng kho st v mt nng lng bc x ca mt anten chun (Anten v hng) ti mt im cng cch anten mt khong R v cng trn phng kho st khi cng sut bc x ca 2 anten l nh nhau.23Nu gi 0l mt nng lng bc x caanten chun ti im M,l mt nng lng bc x caanten kho st ti im M th ta c : D= /0 (ln).Vi 20. . 4 RP ta c 2. . 4.RP D Trong Pl cng sut bc x ca anten, R l khong cc t anten n im kho st.Nu gi P1l cng sut thu ti im M ng vi anten kho st, P0l cng sutthu ti im M ng vi anten chun th D=P1/P0 (Ln). Nu tnh theo dB th :010 1lg 10 lg 10 lg 10 ) (PPP P decibel D c) Hiu sut ca anten ( ):L t s gia cng sut bc x (cng sut c ch) ca anten v cng sut a vo anten: PPbxa. Trong : Pbx v cng sut bc x ca anten, P cng sut a vo anten. Thng thng a = 0,75 0,95.d) H s khuch i (G)H s khuch i G ca anten l t s gia mt nng lng bc x ca mt anten ti mt im cch anten mt khong R trn phng php st v mt nng lng bc x ca mt anten chun (Anten v hng) ti mt im cng cch anten mt khong R v trn phng kho st khi cng sut a vo 2 anten l nh nhau.Nu anten chun c hiu sut bng 100% th ta c G= .De) Gc na cng sutGc na cng sut l gc hp bi hai hng bc x m theo hng cng sut pht gim i mt na so vi hng chnh.Cmt cchnhngha khc: Gc na cngsut lgchpbi hai hngbcxm theo hng cng sut pht gim i 3 dB so vi hng chnh.f) Tr khng vo/raTr khng vo ca anten l tr khng m tn hiu gp phi khi i vo u vo ca anten. Tr khng ra ca anten l tr khng m tn hiu gp phi khi i ra u ra ca anten.Khi ta ni anten vo u ra ca my pht hoc u vo ca my thu bng cch trc tip hoc gin tip qua dy feeder th tng ng vi vic ta u vo u u ra 2423dBYXHnh 2.11: Gc na cng sut 2ca my pht, u vo ca my thu hoc dy feeder mt in tr c gi tr ng bng tr khng vo hoc tr khng ra ca anten.Trong vng tn hiu cao tn, khi tn hiu truyn t mch ny sang mch khc th mt trong nhng yu cu quan trng cn c p ng l phi c s phi hp tr khng gia u vo v u ra gia cc mch, nu khng s gy ra s phn x sng in t ti im mt phi hp tr khng gy ra tn hao nng lng v nhiu tc ng xu khc. Do vy, tr khng vo - ra ca anten l mt tham s rt quan trng. Nu c s phi hp tr khng tt th hng pht tn hiu t my pht hoc dy feeders c chuyn vo anten mt cch tt nht. Tng t, hng thu tn hiu t anten s c chuyn vo my thu mt cch tt nht. Ngc li, nu mt phi hp tr khng th s c sng in t phn x ti cc im u ni. Do mt trong nhng yu cu i vi tr khng ra ca anten l phi phi hp c vi tr khng ra ca my pht, tr khng vo caanten v tr khng ca ng dy feeder.Tr khng ca anten l m s phc do anten c c tnh ging mt khung cng hng.g) Bng tn cng tcBng tn cng tc hay cn gi l gii thng ca anten l tham s c trng cho phm vi tn s hot ng ca anten.V tr khng ca anten l mt gi tr thay i theo tn s cng tc nn khi tn s ca tn hiu cao tn trn anten thay i n mt gi tr no th s xy ra hin tng mt phi hp tr khng. Do vic phi hp tr khng cho anten l rt phc tp, thng thng ngi ta phi hp tr khng cho anten vo tn s trung tm ca bng tn cng tc.h) H s phng v D0L t s gia mt cng sut bc x hng chnh (hng thun)v mt nng lng bc x hng ngc li (Hng ngc): ngthD04. Cc loi anten thng dnga) Anten chn ta1. Cu to251PADDD DDDHnh 2.12: cu to ca anten chn tlPlAlDHnh 2.12 m t cu to ca anten chn t. Cu to ca anten chn t gm c:- Gi 1 c gia c bng kim loi, c su trc chc chn.- Chn t phn x P c chiu di lP 0,97 .- Chn t bc x A c chiu di lA 0,95 .- Cc chn t hngx D c chiu di lD 0,87 . Cc chn t trn c t song song vi nhau trn gi kim loi, vung gc v cch in vi gi kim loi. Trong chn t P t cch chn t A mt khong t 0,15 n 0,25 . Chn t D u tin t cch chn t A mt khong t 0,1 n 0,3 . Cc chn t D t cch nhau mt khong t 0,25 .Cc chn t c ch to bng kim loi c t thm cao. trnh dng Fuco, cc chn t c ch to di dng cc ng rng. Chn t bc x c ni ti ngun cao tn hoc my thu.a2. Nguyn l hot ng* Nguyn l bc x sng in tKhi chn t bc x c ni vi ngun tn hiu cao tn th n bc x sng in t v c hai hng trc v sau theo phng vung gc vi cc chn t. sng in t bc x v ph sau s c chn t phn x phn x ngc tr li v pha trc. V chn chiu di chn t phn x v khong cch gi n vi chn t bc x thch hp nn sng in t phn x v ng pha vi sng in t bc x ra pha trc. Cc chn t hng x cm nhn nng lng sng in t bc x t chn t bc x v tr thnh ngun bc x th cp bc x sng in t v pha trc. V c di ngn hn nn cc chn t hng x c kh nng tp trung nng lng cao hn v mt pha. Cng nhiu chn thngxthkhnng nhhng cngcao,tuy nhinslm nh hng n kt cu ca anten.* Nguyn l thu sng in tAnten chn t thu sng in t theo nguyn l cm ng in t. Khi c sng in t t pha i phng pht n th cc chn t s cm nhn nng lng to thnh dng in cao tn a ti my thu.Trong , cc chn t hng x cm nhn nng lng sng in t v tr thnh ngun bc x sng in t th cp bc x v pha chn t bc x tp trung nng lng vo chn t ny lm tng kh nng thu ca anten.26a3. U nhc im- Bng tn cng tc khng ln do anten ny c c tnh gn ging mt khung cng hng.- H s nh hng khng cao. c h s nh hng cao th yu cu anten phi c rt nhiu chn t hng x lm cho kch thc ca anten cng knh, kh gia c chc chn.- Khi hot ng tn s cao, chng dng fuco,th yu cu cc chn t phi ch to rt mnh nn rt d gy. Mt khc khi hot ng tn s cao th cc chn t phi rt ngn nn nh hng n kh nng bc x ca antenb) Anten ng dn sngb1. Cu toAntenng dn sng thc cht l mt on ng dn sng c mt u bt kn, u kia ng c c mt phn t kch thch c t vo trong lng ng nh m t hnh 2.13.b2. Nguyn l hot ngKhi ni phn t kch thch vi ngun tnh hiu cao tn th nng lng sng in t s c kch thch vo ng dn snglc ny sng in t s c truyn v pha ming ng v i ra ngoi khng gian. Ming m ca ng dn sng ng vai tr nhmt antenbc x sngint theohngvnggcvi mingm ca ng.b3. u nhc im- Dintchmingmhpnn bc x yu.- Tnh phng hng v h s khuch i khng cao.27u bit knPhn t kch thchHng bc x ca sng in tHnh 2.14: nguyn l hot ng ca anten ng dn sngHnh 2.13: Cu to ca anten ng dn snga. Cu to c v phi cnh, b. Cu to c v theo mt ctu bit knMing antenMing antenu bit knPhn t kch thch- Tr khng ca anten v mi trng bn ngoi khc nhau nhiu nn nng lng sng in t b phn x ti ming ng dn sng gy ra tn hao v nhiu bp sng ph.b4. ng dng- Anten ng dn sng ch c dng ln cc b chiu x t trong anten parabol.c) Anten loaAnten loa l mt ci tin ca anten ng dn sng phi hp tr khng gia ngdnsngvmtrngtruynsng bn ngoi. bng cch m rng ming ng dn sng (thnh hnh ging mt chic loa) lm cho tr khng ca ng dn sng bin i mt cchttkhi chuynttrong lngngdnsngrangoi khnggian, iu ny c ngha l lm tng phi hp tr khnggiangdnsngvkhnggian nn nng lng sng phn x tr li ng s gim,nnglngbcxrangoikhng gian s tng. Mt khc din tch mt loa s lnhnnhiulmtngkh nng bc x ca anten.Hnh 2.15 m t cu to ca anten loa. Khi m rng ming ng dn sng trn nh hnh 2.15a th ta c anten loa nn, khi m rng ming ng dn sng theo chiu ng ta c anten loa E nh hnh 2.15b, khi m rng ming ng dn sng theo chiu ngangtacantenloaHnhhnh2.15c, khi mrngmingngdnsngtheocchiu ngang v chiu ng ta c anten loa thp nh hnh 2.15d.Hnh 2.16 m t cc tham s cn thit ca anten loa. Trong : O l tm loa, OO trc chnh ca loa, L: di ca loa, h: rng ming loa (ng kch ming loa), 2 o: Gc m ca loa.28(a)(b)(c)(d)Hnh 2.15: Mt s anten loaL2oOOHnh 2.16: Cc tham s cn thit ca anten loahKhi kch thch sng in t vo anten loa th n s bc x sng in t ra pha trc ming loa. Sng in t ny ta h nh xut pht t mt ngun bc x im t ti tm loa. Nh vy, sng in t pha trc ming loa khng ng pha trn mt phng vung gc vi trc chnh ca loa. c mt anten loa c cht lng tt, ngi ta phi chn gc m v chiu di ming loa thch hp.Ngi ta tnh c h s khuch i ca anten loa nh sau: lg 20 . . . 4 lg 10 V S G(dBi). Trong : V l hiu sut ca anten loa so vi loa tt nht, S l din tch ming loa, l bc sng cng tc.So vi anten ng dn sng th anten loa c h s khuch i v h s phng v cao hn, kh nng phi hp tr khng tt hn. Tuy nhin vic ch to mt anten loa c kch thc ln khng m bo tnh kinh t v tnh k thut, nn anten loa thng c ch to c kch thc nh dng lm cc b chiu x t trong cc anten parabol, parabol 2 gng .v.v. Antenloa thngch s khuch i nh hn 25 dB.d) Anten Parabol 1 gngLi dngtnhcht phnx ca gng parabol, ngi ta ch to ra anten parabol.d1. Cu toHnh 2.17 m t cu to ca anten parabol:- Gng phn x 1: cn gi l a phn x. Gng c chto bng kim loi, b mt nhn c th phn x tt v t hp th nng lng sng in t. Gng c 2 loi:+ Gng kn: Gng c mt phn x hon ton kn.+ Gng h: Gng phn x l mt li kim loi hoc cc thanh kim loi c un cng v xp thnh hnh parabol.Mt s tham s ca gng:+ f: Tiu c ca gng.+ h: su ca gng291. Gng phn x2. Chiu x t t ti tiu im F ca gng3. ng dn sng4. Connectorsiu cao tn3. Tm gng lhfHnh 2.17 : Cu to ca gng parabolMing gng+ l: Khu ca gng ( rng ming gng)+ o: Gc m ca gng- ngphng: ngphnggmmt on ng dn sng 2 v mt b chiu xt3. onngdnsngcmt ucuvi dyFeeder thng qua connecter siu cao tn 4 cn u kia c u vi chiu x t. Chiu x t t ti tiu im F ca gng c 2 nhim v chnh l bc x nng lng siu cao tn thnh chm sng in t phn k pht ti gng v chuyn chmsnginthi ttgng thnhnnglngsiucaotntrong ng dn sng.- Ngoi ta cn c cc c cu gi nh v gng phn x v c nh chiu x t ti tiu im F ca gng.Hnh dng y ca mt anten parabol c m t hnh 2.18:d2. Nguyn l hot ng* Nguyn l pht sng in tHnh 2.19 m t nguyn l pht sng in t. Nng lng siu cao tn c atmypht quadyfeeder a vo u vo ng phng. Nng lng siu cao tn ny c chuyn thnh sng in t truyn trong ng dn sng v ti chiu x t, c chiu x t bin i thnh chm sng in t phn k truynti gngphnx. Chm sng ny ta h nh xut pht t mt ngun im t ti tin im cagng. Cctiasngint ti gng s c phn x to thnh chm tia song song truyn v pha trc gng. Chm sng in t ny cng song song vi trc chnh ca gng v ng pha trn mt phng vung gc vi trc chnh ca gng.30 Gng phn x Chiu x tng dn sngConnector siu cao tn Gi chiu x tGi Gng phn xGi Gng phn xHnh 2.18: Hnh dng y ca mt anten parabolTn hiu siu cao tn t ng dy FeederSng in t c bc x ra pha trcHnh 2.19: Nguyn l pht sng in tAnten phtAnten thu Hnh 2.21: Hot ng ca mt h thng anten Parabol trong h thng vi baTn hiu siu cao tn ti ng dy FeederSng in t pha pht pht tiHnh 2.20: Nguyn l pht sng in tSng in t do anten Parabol bc x ch truyn theo mt hng nht nh nn loi anten ny cn c gi l anten bc x n hng.Ngi ta tnh c h s khuch i ca gng nh sau: ). 4 . .lg( 10 ) (2 SdB GaTrong : a: Hiu sut ca anten, S: Din tch ca mt gng, : Bc sng cng tc.* Nguyn l thu sng in t.Hnh 2.20 m t nguyn l pht sng in t. Nu c chm sng in t t i phng pht ti v song song vi trc chnh ca gng th khi ti mt phn x ca gng chm sng ny c phn x to thnh mt chm sng in t hi t v tin in F, do chiu x t t ti tiu im F nn chm sng in t ny s i vo chiu x t, qua on ng dn sng v c chuyn thnh nng lng cao tn i vo dy feeder ti my thu.Hot ng ca mt h thng anten Parabol trong h thng vi ba c m t hnh 2.21.d3. u,nhc im* u im : - C h s nh hng cao - Gii thng rng v c kh nng cng tc tn s cao do hot ng da trn nguyn l phn x.- C kh nng bc x cng sut ln.31* Nhc im:- cn gi ln, c bit i vi loi c a phn x kn.- Chiu x t v ng phng t trc anten nn c th chn bt mt phn nng lng sng in t, to thnh vng ti trc anten.-Nu l a h th h s phng ca anten khng cao do mt phn sng in t i c ra pha sau anten v ngc li sng in t t pha sau anten cng c th i ti chiu x t .e) Anten Parabol hai gnge1. Cu toHnh 2.22 m t cu to ca anten Parabol 2 gng. Anten Parabol 2 gng gm c:-Mt gngParabol (Gngcu lm) c ch to bng vt liu phn x tt v t hp th nng lng sng in t. Gng c 2 loi:+Gngkn: Gngcmt phn x hon ton kn.+ Gng h: Gng phn x l mt li kimloi hoc cc thanhkimloi c un cng v xo thnh hnh parabol.- Mt gng Hypecbol (Gng cu li)cchtobngvtliuphnx rt tt v t hp th nng lng sng in t.Mt s tham s ca 2 gng:+ f1: Tiu c ca gng chnh.+ f2: Tiu c ca gng ph.+ h: su ca gng chnh32F1, F2f2f1hlGng chnhGng PhChiu x tHnh 2.22: Cu to ca anten parabol 2 gng1. Gng Parabol(Gng chnh)3. Chiu x t4. ngdn sng5. Connector siu cao tn2. Gng Hypecbol(Gng ph)6. Gi gng ph1. Gng Parabol(Gng chnh)3. Chiu x txa t4. ngdn sng5. Connector siu cao tn2. Gng Hypecbol(Gng ph)7. Gi gng chnh6. Gi gng chnhHnh 2.23: Cu to y ca mt anten parabol 2 gngHnh 2.24: Nguyn l pht sng in t ca anten parabol 2 gngNng lng siu cao tn t my pht a tiSng in t c bc x ra pha trc+ l: ng knh ca gng ( rng ming gng chnh)+ F1, F2 : Tiu im ca gng chnh v gng ph c t trng nhau.-Chiu x t t ti tiu im F1, F2 ca 2 gng v hng vo gng ph v cch tm ca gng ph mt khong d=f2/2.Chiu x t c 2 nhim v chnh l bc x nng lng siu cao tn thnh chm sng in t phn k pht ti gng ph v chuyn chm sng in t hi t t gngph thnh nng lng siu cao tn trong ng dn sng.- Ngoi ta cn c cc c cu gi nh v cc gng chnh v gng ph. Hnh dng y ca mt anten parabol c m t hnh 2.23.33e2. Nguyn l hot ng* Nguyn l pht sng in tNng lng siu cao tn c a t my pht qua dy feeder vo u vo ng phng. Nng lng siu cao tn ny c chuyn thnh sng in t truyn trong ng dn sng v ti chiu x t, c chiu x t bin i thnh chm sng in t phn k s cp truyn ti gng Hypecbol c gng ny phn x to thnh chm sng in t phn k th cp truyn ti gng Parabol. Chm sng phn k th cp ny ta h nh xut pht t mt ngun im t ti tin im ca gng Hypecbol. Do tiu im ca 2 gng trung nhau nn chm sng phn k th cp ny cng ta h nh xut pht t mt ngun im im t ti tin im ca gng Parabol. Cc tia sng in t ti gng Parabol s c phn x to thnh chm tia song song truyn v pha trc gng. Chm sng in t ny cng song song vi trc chnh ca gng v ng pha trn mt phng vung gc vi trc chnh ca gng.Sng in t do anten Parabol bc x ch truyn theo mt hng nht nh nn loi anten ny cng c gi l anten bc x n hng.Ngi ta tnhc h s khuch i ca gng nh sau: ). 4 . .lg( 10 ) (2 SdB GaTrong: a: Hiu sut ca anten, S: Dintch ca mt gng, : Bc sng cng tc.* Nguyn l thu sng in t.Nu c chm sng in t t i phng pht ti v song song vi trc chnh ca gng th khi ti mt phn x ca gng chm sng ny c phn x to thnh mt chm sng in t hi t v tin in F1, do tiu im F1 v F2 ca 2 gng trng nhau nn chm sng in t hi t ny cng ta h nh hi t tiu im F2 ca gng 2. ti gng 3 chng s c phn x v to thnh chm sng in hi t th cp. Do chn im t chiu 34Hnh 2.25: Nguyn l thu sng in t ca anten parabol 2 gngSng in t t i phng pht tiNng lng siu cao tn c a n my thux t thch hp nn chm sng in hi t thc cp i vo chiu x t, qua on ng dn sng v c chuyn thnh nng lng cao tn i vo dy feeder ti my thu.Hot ng ca mt h thng anten Parabol 2 gng trong h thng vi ba c m t hnh 2.26.e3.U nhc im* u im : - C h s nh hng cao - Gii thng rng v c kh nng cng tc tn s cao do hot ng da trn nguyn l phn x.- C kh nng bc x cng sut ln.- C kh nng gia c phn t bc x chc chn do c th gim nh ti thiu kch thc ca phn t bc x.* Nhc im:- cn gi ln, c bit i vi loi c a phn x kn.- Chiu x v gng ph t trc anten nn c th chn bt mt phn nng lng sng in t, to thnh vng ti trc anten.III.van ferit v xiculator :1.FeritKhc vi cc vt liu thng thng, Ferit c nhng tnh cht t kh c bit lm cho n c th ng dng rng ri trong thng tin vi ba. Ferit l hn hp cc -xt st, km, mangan, c ban, nhm hoc kn. Ngi ta to ra Ferit bng cch a hn hp cc -xt ni trn vo p trong nhng khun mu nht nh vi mt lc rt ln sau nung nng n nhit cao hn 2000oC. Qu trnh ny lm chohn hp cc -xt thay i c tnh nh chng tr nn cch in (in tr ni rt cao),h s t 35Anten phtAnten thuHnh 2.26: Hot ng ca mt h thng anten Parabol 2 gng trong h thng vi bathm tng v c kh nng gii phng t rt nhanh.2.Van feritVanFerit cncgi lb cchly,ctcdngchotnhiui theo mt chiu v hp th nng lng tn hiu theo chiu ngc li.a) Cu toHnh2.27mtcutoca mt van ferit. Gm c: Mt on ng dnsng1, tronglngngt mt thanh Ferit 2 v mt nam chm vnh cu 3 c 2 c p st vo 2 cnh ca ng dn sng sao cho thanh ferit nng chnh gia 2 cc ca nam chm. Nam chm c nhim v nh hng t tnh cho thanh Ferit, ngi ta cn gi nam chm ny l nam chm nh thin t. Theo hnh v, t trng ca nam chm hng t di ln trn.b) Nguyn l hot ngKhi c tc dng ca nam chm vnh cu th thanh Ferit s b nhim t v to ra mt t trng c chiu nht nh trong lng ng dn sng. Khi c sng in t tc ng vo th thanh Ferit tip tc nhim t v s bin thin t tnh trn thanh Ferit tun theo quy lut bin thin ca sng in t. tuy nhin chiu ca t tnh do thanh ferit to ras khng thay i.Do nu t trng do sng in t to ra trong lng ng dn sng trng vi t trng do thanh Ferit to ra th sng in t d dng i qua ngc li nu nu t trng do sng in t to ra trong lng ng dn sng ngc vi t trng do thanh Ferit to ra th sng in t s b trit tiu.V ch truyn sng c mt chiu nn cu trc trn c gi l Van ferit. Nu gi 2 ng vo ra ca van l A v B th nu sng in t truyn c t ng A n ng B th khng th truyn c t ng B n ng A. Mun thay i chiu truyn sng (truyn c t ng B n ng A) ta phi thay i cc tnh ca nam chm vnh cu.c) ng dngLm b cch ly y racabkhuchi cng sut cao trong mt pht vi ba trnh hin tng phn x sng in t khi ng raca my pht b mt phi hp tr khng.3. Xiculator36Thanh Ferit 2ng dn sng 1Nam chm vnh cu 3Mi tn ch chiu t trng ca Nam chmHnh 2.27: Cu to c bn ca phn t cch lySNMing FeritCng 1Cng 2Cng 3C cu bng kimloiHnh 2.28: Cu to ca Xicuilatora) Cu toXiculator cn c gi l b phn hng siu cao tn, l mt c cu c 3 hoc 4cng vo/ra nng lng. Hnh 2.28 m t mt Xiculator c 3 li vo ra nng lng. Nng lng i vo ngth nht s i ra ngth hai, nng lng i vo ngth hai s i ra ngth ba v nng lng i vo ngth ba s i ra ngth nht.Cu to ca Xiculator gm mt kt cu bng kim loi c 3 cng vo/ra, gia t mt li Ferit, ngoi ra cn c mt nam chm vnh cu lm nhim v nh thin t cho li Ferit (Khng v trong hnh).b) Nguyn l hot ngKhi c nh thin t, li Ferit s b nhim t v to ra mt t tnh c chiu nht nh. Trong trng hp li ferit c dng hnh tr trn nh hnh 2.28 th t trngcchiuxoytrnxungquanhli ferit. Khi sngintcavo xicuilator th li ferit tip t b nhim t di tc dng ca t trng do sng in t to ra. Tuy nhin chiu ca t trng do li Ferit to ra khng thay i. Nu tnh hiu c a vo ng th nht th t ng 1 sang ng 2 ng sc t do li Ferit to ra cng chiu vi ng sc do sng in t to ra nn tn hiu d dng i qua, cn t 1 sang ng 3 ng sc t do li Ferit to ra ngc chiu vi ng sc t do sng in t to ra nn tn hiu khng th i qua. Tng t nh vy, tn hiu s vo ng 2 ra ng 3 v vo ng 3 ra ng 1.c) ngdng:Xiculator cngdnglmbphnhngsiucaotntrong DIPLEXER cho php my pht v my thu s dng chung mt anten.CHNG IIIX L TN HIU TRONG THNG TIN VI BA SI. Khi phc xung nhp1. Khi nim H thng vi ba s ni ring v cc h thng x l tn hiu s ni chung cn phi c xung nhp hot ng. cc h thng hot ng ng b th cc xung nhp pha pht v pha thu phi ng b vi nhau, c ngha l cc tn hiu xung nhp pha pht 37v pha thu phi cng tn s v ng pha. C nhiu phng php thc hin yu cu trn:Phng php 1:dng hai b dao ng to xung nhp c cng tn s, mt b to xung nhp pha pht, mt b to xung nhp cho pha thu. Tuy nhin, do cc b dao ng ny hot ng c lp vi nhau nn nhng ni khc nhau chng s chu tc ng bi cc yu t khc nhau ca mi trng, dn n tn s v pha ca chng kh ng b vi nhau, lm h thng hot ng khng chnh xc. Phng php 2:Dng mt ng truyn dn truyn tn hiu xung nhp t pha pht n pha thu. Cch ny tuy p ng c yu cu v xung nhp nhng li phi tn thm kinh ph xy dng thm ng truyn m kinh ph ny khng phi nh. Phng php ny thng c s dng khi dung lng ng truyn cho php tng i ln. Vic s dng mt phn ng truyn truyn tn hiu xung nhp khng nh hng ln n lu lng ca tuyn truyn dn.Phng php 3: Bn pht to ra mt xung nhp ch, sau gi cc thng s v xung nhp ln lung s. pha thu, khi thu c tn hiu s da vo cc tham s to li xung nhp. Qu trnh to li tn hiu xung nhp bn thu theo phng php trn c gi l khi phc xung nhp. Mch thc hin cng vic ny c gi l mch khi phc xung nhp. Hin nay hu ht cc thit b thng tin vi ba s dng phng php khi phc xung nhp ng b gia my pht v my thu.2. Khi phc xung nhp t tn hiu c m ng HDB3a. S khi Hnh3.1 m t s khi ca mch khi phc xung nhp. + BA1: bin p vo c nhim v phi hp tr khng gia thit b v ng dy ng thi ghp tn hiu HDB3 vo mch. + N1, N2, N3: cc b khuch i thut ton hot ng theo kiu so snh.+ Cc in tr c tc dngphnptoin p chun a vo chn (-) ca cc b so snh N1 v N2.CcintrRcnhim v hn ch dng bo v ccmchsosnhN1v N2. 38BA1N1N2N3BA2BA2ORICTn hiu NRZXung nhp khi phcTn hiu HDB3Hnh 3.1 S khi ca mch khi phc xung nhp.+ IC: mt mch tch hp chuyn dng lm nhim v bin i tn hiu t HDB3 sang NRZ.+ Cng OR: c nhim v cng hai tn hiu ng ra ca N1 v N2 li vi nhau kch thch cho mch dao ng LC.+ L v C: to thnh mt khung cng hng LC c tn s dao ng ring.+ BA2: bin p ghp tn hiu dao ng t khung dao ng LC vo mch so snh N3. b.Nguyn l hot ngNguyn l hot ng ca mch c th m t nh sau: Tnhiu s vi m ng HDB3 c a vo trn hai u cun s cp ca bin p BA1. Tn hiu ny c ghp qua bin p v t vo cc chn khng o (+) ca cc b so snh N1 v N2 (cun th cp ca bin p c im gia ni t), c N1 v N2 tch thnh cc lung HDB3 (+) v HDB3 (-). Cc lung HDB3 (+) v HDB3 (-) sau N1 v N2 c a vo IC bin i t m ng HDB3 sang m my NRZ n cc, ng thi cc lung s sau N1 v N2 c cng vi nhau thng quacngORtothnhmttnhiucnhiuchuyni mckchthchcho khung dao ng LC. Khi c kch thch th khung LC s dao ng vi tn s ng bng tc ca lung s u vo. V d: Nu lung s u vo l 2,048Mb/s th khung dao ng LC phi c tn s cng hng ring l 2,048MHZ, nu lung s u vo l 34,368Mb/s th khung dao ng LC phi c tn s cng hng ring l 34,368Mb/s.Tn hiu dao ng hnh sin c ghp qua bin p BA2 a vo mch so snh N3, cN3sosnh vi inpchun0v (chno(-) caN3 ni t). Kt qusau N3 ta thu c dy 39Xung HDB3 voXung HDB3+Xung HDB3-Xung sau cng ORTn hiu sin trn LCXung CK c khi phcttttttHnh 2.2 Dng sng ca mch khi phc xung nhp.Uxung nhp c tn s ng bng tc ca lung s u vo. Xung nhp ny c s dng nhp cho cc mch pha thu. Hnh 2.2 m t dng sng qua cc phn t mch ca mch khi phc xung nhp t m ng HDB3. II. Ngu nhin ho v gii ngu nhin ho1. Cc khi nim v lung s a. Lung s khng ngu nhinLung s khng ngu nhin l lung s c nhiu chui bit 0 hoc/v nhiu chui bit 1 xut hin c tnh quy lut (hnh 3.2a). Lung s khng ngu nhin cn c gi l lung s c t chuyn i mc.b. Lung s ngu nhinLung s ngu nhin l lung s c cc bit 0 v bit 1 xut hin mt cch ngu nhin. Lung s ngu nhin cn c gi l lung s c nhiu chuyn i mc trong cc chuyn i mc chuyn i khng theo mt quy lut no c (hnh3.2b).c. Lung s gi ngu nhin400 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1t 1ttUHnh 3.2a: Lung s khng ngu nhin1 0 0 1 0 1 1 0 01 1 1 0 1 1 0 0 1 0 1 1 0 1 0 10 1 0 1 1 01 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1t 0ttUHnh3.2b : Lung s ngu nhinLung s gi ngu nhin l lung s m cc bit trong mt nhm bit xut hin ngu nhin, nhng nu xt quan h gia cc nhm bit th cc bit xut hin khng ngu nhin m c tnh quy lut. Lung s gi ngu nhin cn c gi l lung s c nhiu chuyn i mc trong cc chuyn i mc chuyn i theo mt quy lut nht nh (hnh 3.2c).2. Ngu nhin haa. Khi nimNgu nhin ho l bin lung s khng ngu nhin thnh lung s ngu nhin, l php x l s liu s c thc hin trong my pht Vi ba s, sau khi ghp khung v tuyn v trc khi iu ch.b. Mc ch + Tng chuyn i mc ca lung s pha thu d dng khi phc xung nhp: Cc lung s nguyn thu c ly ra t my ghp knh hoc tng i thng c cc chui bit 0 lin tip do knh thoi b trng, hoc cc chui bit 1 do ting thoi qu ln, nn s chuyn i mc trong lung s rt t v do pha thu kh khi phc xung nhp, nn cn phi ngu nhin ho tng s chuyn i mc ca lung s pha thu d khi phc xung nhp.+ Chng nhiu: Nu lung s c nhiu chuyn i mc th ph tn ca n s tri u trong ton bng ( ph ca tn hiu sau iu ch cng ng u trong ton bng ( t b nh hng ca nhiu, nn pha thu thu tt. Nu lung s c t chuyn i mc th ph ca n ch tp trung vng tn s thp, nng lng vng tn s cao rt thp ( sau khi iu ch nng lng khng ng u trong ton bng ( nhiu d tc ng vo vng nng lng thp lm cho pha thu d thu nhm tn hiu. c. Nguyn l 411 0 0 1 1 0 1 1 0 0 1 1 0 1tttU0 0 1 0 1 0 1 0 0 1 0 1 0 11 1 0 1 0 1 0 1 1 0 1 0 1 0Hnh3.2c : Lung s ngu nhin0 0 01 1 01 00 0 11 01 0001 1 0 011 1 00 0 1 0000 0 01 11 1 1 100 UTn hiucn ngu nhin ho ATn hiugi ngu nhin BTn hiu ngu nhin ho StttHnh 3.3b Biu thi gian ca mch ngu nhin ho- thc hin ngu nhin ho, trcht phi toc tnhiugingunhinng b vi tn hiu nguyn thu(c cngtcvcngphavi lung s nguyn thu).- Vic ngu nhin ho c thc hin bng cch: em tn hiu cn ngu nhin ho cng modul hai vi tn hiu gi ngu nhin thng qua mt cng EX-OR. S khi ca mch ngu nhin ho c trnh by hnh 3.3.- Biu thi gian m t hot ng ca mch ngu nhin ho nh sau:3. Gii ngu nhin haa. Khi nim- Gii ngu nhin ho l bin lung s c ngu nhin ho pha pht thnh lung s khng ngu nhin.- Gii ngu nhin ho l php x l s liu s c thc hin trong my thu Vi ba s, sau khi gii iu ch v trc khi phn khung v tuyn.- Gii ngu nhin ho l to li lung s nguyn thu t lung s c ngu nhin ho.b. Nguyn l - thc hin gii ngu nhin ho trc ht phi 42Tn hiugi ngu nhin BTn hiu gii ngu nhin ho ATn hiu ngu nhin hoS = A BHnh 3.4a S khi ca mch gii ngu nhin hoTn hiuGi ngu nhin BTn hiu NGU NHIN HOS = A BTn hiu cn ngu nhin ho AHnh 3.3a S khi ca mch ngu nhin ho000110 1000 1101 0 00110 0111 0001000000 11111 100Tn hiu ngu nhin ho STn hiugi ngu nhin BTn hiu gii ngu nhin ho AtttUHnh 3.4b Biu thi gian ca mch gii ngu nhin hoto c 1 tn hiu gi ngu nhin ging ht v ng b vi tn hiu gi ngu nhin bn pht.- Vic gii ngu nhin c thc hin bng cch a tn hiu gi ngu nhin bn pht pht ti cng EX-OR vi tn hiu gi ngu nhin thng qua 1 cng EX-OR.S khi ca mch gii ngu nhin c m t nh sau: ng ra ca cng EX - OR ta thu c lung s l S BV: S = A B ta c: S B = A B B = A (B B) = A 0 = A- Biu thi gian m t nguyn l hot ng ca mch gii ngu nhin ho c m t hnh 3.4b.4. Totnhiugingu nhinMt trong nhng phng php to tn hiu gi ngu nhinnginnht c m t hnh 3.5. Mch ghi dch gm n Flip-Flop mc nitipvinhau. Ng ra ca Flip-Flop cui cng v mt ng ra ca Flip-Flopthkc hi tip v u vo ca 43FF2FFkFFn-1FFnTn hiu gi ngu nhinCLKEXOR 1EXOR 21FF1HNH 3.5: Mch to tn hiu gi ngu nhins dng nFlip - FlopS1 S2 S3 S4 S5 S6 S7 S8 S9 S10Tn hiu raXung nhp voEX- OR1EX- OR2Tn hiu voHNH 3.7: Mch gii ngu nhin ha t ng bFlip - Flop th nht thng qua hai cng X-OR. Nu dng n Flip - Flop th to c mt chui (mt nhm) tn hiu ngu nhin c 2n - 1 bit. Mc ngu nhin ca tn hiu gi ngu nhin ph thuc vo v tr ly hi tip ca Flip-Flop th k.Trong hnh 3.5: cc Flip - Flop c k hiu t FF1 n FFn. Cc cng X-OR1 v X-OR2 to nn mt mch hi tip. X-OR2 c mt u vo c u vi mc cao trnh hin tng to ra chui bit khng lin tip khi mch khi ng. Xung CK c a ng b ti tt c cc Flip - Flop. Ngoi ra, trong mt s mch, cc Flip - Flop cn c thm cc u iu khin SET (thit lp) v RESET (xo).5. Ngu nhin ho v gii ngu nhin ho t ng b. Phng php ngu nhin ho v gii ngu nhin ho trn c gi l khng ng b. V trong mch to tn hiu gi ngu nhin c to ra c lp vi lung s cn ngu nhin ho v lung s cn gii ngu nhin. Phng php ny m bo lung s sau khi ngu nhin ho lun lun l lung s ngu nhin. Tuy nhin vic ng b gia mch to tn hiu gi ngu nhin v lung s cn ngu nhin ho rt kh khn, c bit l ng b gia mch to tn hiu gi ngu nhin bn pht v mch to tn hiu gi ngu nhin bn thu. khc phc nhc im , ngi ta dng phng php ngu nhin ho v gii ngu nhin ho t ng b. phng php ny, khi thc hin ngu nhin ho th chnh lung s ngu nhin ho c s dng to tn hiu gi ngu nhin. Sau khi to c tn hiu gi ngu nhin, tn hiu ny c cng tri du vi tn hiu cn ngu nhin ho c tn hiu ngu nhin ho. Khi thc hin gii ngu nhin ho th tn hiu cn gii ngu nhin ho c s dng to tn hiu gi ngu nhin, sau tn hiu gi ngu nhin ny c cng tri du vi tn hiu cn gii ngu nhin to ra tn hiu nguyn thu.Hnh 3.6 m t mch ngu nhin ho tngbsdng mch to tn hiu gi ngu nhin gm10 Flip - Flop.T mch in ta thy tn hiu gi ngu nhin c ly sau cng X-OR1 c biu thc ton hc l: B = S7 S10. Tn hiu ngu nhin ho c ly sau cng X-44S10 S9 S8 S7 S6 S5 S4 S3 S2 S1Tn hiu voTn hiu raXung nhp voEX- OR1EX- OR2HNH 3.6:Mch ngu nhin ha t ng bOR2 c biu thc ton hc l S = A B = A S7 S10. Tn hiu ny c a vo iu ch pht sang pha thu, ng thi c s dng to tn hiu gi ngu nhin cho mch.Hnh 3.7 m t mch gii ngu nhin ho t ng b s dng mch to tn hiu gi ngu nhin gm 10 Flip - Flop ging ht bn pht. Tn hiu cn gii ngu nhin ho S t bn pht pht ti c s dng to tn hiu gi ngu nhin. D dng nhn thy rng tn hiu gi ngu nhin c ly ra sau cng X-OR2 v c biu thc:B = S7 S10 Tn hiu ny c cng tri du vi tn hiu cn gii ngu nhin ho S thng qua cng X-OR1. Sau cng X-OR1 ta thu c tn hiu: S B = (A S7 S10) ( S7 S10) (v S = A S7 S10) = A (S7 S7) (S10 S10) = A 0 0 = A A chnh l lung s nguyn thu a vo ngu nhin ho bn pht.III. M ho vi sai v gi m vi sai1- Khi nimCc h thng thng tin vi ba s hin nay thng s dng phng php iu ch pha PSK. Phng thc iu ch ny thng tin ca lung s c truyn i cc trng thi pha ca sng mang v tuyn. pha thu, thc hin gii iu ch ng s liu cn phi c sng mang ging ht bn pht c v tn s v pha gi l sng mang chun. Khi c sng mang chun th vic gii iu ch c thc hin bng cch so snh pha ca sng mang chun vi pha ca sng mang iu ch bn pht pht ti. Bnh thng pha thu khng th khi phc c sng mang chun. c c sng mang chun th pha pht cn phi pht n pha thu mt sng mang khng iu ch (sng mang chun). Tuy nhin, vic pha pht pht ti pha thu sng mang chun s lm gim hiu sut truyn dn gia cc trm vi ba s do sng mang chun khng mang thng tin trong khi li chim mt phn nng lng tng i ln trong s nng lng c my pht pht i. Mt khc, khi sng mang khng iu ch c pht sang pha thu th pha ca sng mang ny c khi khng cn chun na do chu nh hng ca mt s yu t trn ng truyn. trnh vic phi pht i sng mang chun v tng tin cy ca cc tuyn thng tin th hin nay cc h thng vi ba s s dng mt php x l s liu gi l m 45ho vi sai v gii m vi sai. bn pht, trc khi a vo iu ch ngi ta tin hnh m ho vi sai, cn pha thu sau khi gii iu ch ngi ta tin hnh gii m vi sai. Nh vy, m ho vi sai l mt php x l s liu c thc hin bn pht nhm gip cho bn thu gii iu ch ng s liu m khng cn sng mang chun, cn gii m vi sai c thc hin bn thu ly li tn hiu nguyn thu.2- Nguyn l thc hin m ho v gii m vi sai.a. M ho vi saiKhi thc hin m ho vi sai th iu ch pha sau c gi l iu ch pha m ho vi sai. Vi iu ch kiu ny th thng tin truyn i khng phi cc gi tr pha tuyt i ca sng mang v tuyn m l cc bc nhy pha ca sng mang. c s thay i th phi iu ch vo sng mang lung s sao cho pha ca sng mang ti thi im iu ch bng pha ca sng mang thi im iu ch trc cng vi mt bc nhy pha c xc nh bi tn hiu iu ch.C th minh ha bng ton hc nh sau: t = (t-1) + vi = t - (t-1)Trong : . (t-1): trng thi pha c. t: trng thi pha mi. : Bc nhy pha mang thng tin v lung s thi im a vo iu ch sng mang (thi im t). r hn v m ho vi sai ta xt mchmhovi sai choiuchpha2 PSK. Mchnycmthnh3.8. Mch gm mt cng X-NOR v mt mch tr 1 bit. S liu cn m ho vi sai c avomt uvocacngX-NOR, tnhiura chi tip vngvoth hai ca cng X-NOR thng qua mch tr 1 bit. bt u thc hin m ho vi sai th phi thit lp mt bit chun bng 1 ng ra ca cng X-NOR v mch tr.Bng 3.9 nguyn l lp m ca mch m ho vi sai v pha ca sng mang c pht i khi sng mang c iu ch bi lung s sau khi m ho vi sai.46To tr 1 bitS liu cn m ho vi saiS liu m ho vi saiHNH 3.8: M ha vi saiS liu vo A 1 1 0 1 1 1 0 0 1 1 0S liu sau m ha vi saiB 1 1 1 0 0 0 0 1 0 0 0 1 bit chunPha pht i 0 0 0 0 0to t1t2t3t4t5t6t7t8t9t10t11BNG 3.9 T bng ny ta thy trc ht pha ca sng mang c quyt nh bi mt bit chun.Sau pha ca sng mang c quyt nh bi bit hin ti v bit ngay trc . Quan h gia bit a vo iu ch A vi bc nhy pha v quan h gia trng thi pha vi tn hiu s sau khi m ho vi sai B nh sau:ABt 001 0 1 0 Khi thc hin iu ch QPSK th trc ht lung s cn iu ch c chia hai to thnh hai lung A v B sau mi tin hnh m ho vi sai ri mi thc hin iu ch. Hnh 3.10m t s khi ca mch iu ch QPSK c m ho vi sai:Quan h gia cc bc nhy pha vi cp bit A, B (tn hiu cn pht i) v gia trng thi pha vi cp bit P, Q (s liu m ho vi sai) nh sau:P Qt A B0 0 0 0 0 00 1 /20 1 /21 11 11 03 /21 03 /2b. Gii m vi sai47Chia 2M ho vi saiiu ch QPSKS liu iu chTn hiu QPSKA PQ BHNH 3.10: S khi tng qut ca mch iu ch QPSK c m ho vi saiKhi pha pht iu ch pha c m ho vi sai th pha thu gii iu ch tn hiu, trc ht cn phi khi phc sng mang. Sng mang khi phc c tn s ng bng sng mang bn pht nhng khng cn phi c pha chun. Gi s sng mang khi phc c pha l o th sau khi gii iu ch ta thu c mt s liu tng ng vi: = t + o Trong : t l pha ca sng mang u pht, l pha ca sng mang tng ng u thu. khi phc c thng tin tng ng u pht th pha thu phi thc hin m ho vi sai sao cho: = t - (t-1)= (T + o) - ((t-1) + o)= T + o - (t-1) - o= T - (t-1) = minh ho cho vn ny, ta xt mt v d gii m vi sai cho tn hiu sau gii iu ch 2 PSK. phn trc ta xt m ho vi sai cho iu ch 2 PSK v pha ca sng mang sau iu ch. Vi sng mang c pha nh bng 5.1, nu sng mang khi phc c pha th lung s sau khi gii iu ch s l 1 1 1 0 0 0 0 1 0 0 0 1. Nu sng mang khi phc c pha th lung s sau khi gii iu ch s l 0 0 0 1 1 1 1 0 1 1 1 0. Ta s xt nguyn l gii m vi sai vi hai trng hp ny. Hnh 3.11 m t mch gii m vi sai sau gii iu ch 2 PSK: Mch in hnh 5.3 gm c mt cng X-NOR 2 ng vo. S liu cn gii m vi sai c a n c hai ng vo ca cng X-NOR. Tuy nhin, s liu ch a trc tip n mt ng vo, cn ng vo kia th s liu phi qua mch tr 1 bit. Bng 3.12a v 3.12b cho thy nguyn l m ho vi sai trong hai trng hp ni trn.S liu cn gii m vi sai B 1 1 1 0 0 0 0 1 0 0 0 1S liu sau mch tr 1 1 1 0 0 0 0 1 0 0 0 1S liu thu c A 1 1 0 1 1 1 0 0 1 1 0BNG 3.12a : Cc lung bit trong mch gii m vi sai khi sng mang khi phc c pha o< / 248S liu vo gii m vi saiS liu ra gii m vi saiTr 1 btHNH 3.11: Mch gii m vi sai sau gii iu ch2 PSKS liu cn gii m vi sai B0 0 0 1 11 1 0 1 1 1 0S liu sau mch tr 0 0 0 1 1 1 1 0 1 1 1 0S liu thu c A 1 1 0 1 1 1 0 0 1 1 0BNG 3.12b : Cc lung s trong mch gii m vi sai khi sng mang khi phc c pha o> / 2 T bng 5.2 v 5.3 ta thy rng bt lun sng mang khi phc c pha nh th no th sau khi giim vi sai ta vn thu c tn hiu nguyn thu a vo m ho vi sai bn pht.Quan h gia bc nhy pha vi bt s liu A (s liu sau gii m vi sai) v quan h gia trng thi pha t vi bt s liuB (s liu thu c sau gii iu ch) nh sau: t BA000 1 0 1 cc thit b s dng phng thc iu ch v gii iu ch QPSK c m ho vi sai th sau khi gii iu ch ta thu c cc lung s P v Q. ly li lung s nguyn thu ban u th cc lung s P v Q c gii m vi sai to thnh cc lung s A, B sau c ghp li vi nhau (bin i song song sang ni tip). Hnh 3.14 m t s khi tng qut ca mch gii iu ch QPSK c gii m vi sai.HNH 3.14 : S khi ca mch gii iu ch QPSK c gii m vi sai Quan h gia cc bc nhy pha vi cp bt A, B v quan h gia trng thi pha t vi cp bt P, Q nh sau: tP QA B49Gii iu ch QPSKGii m vi saiBin i song song sang ni tipTn hiu QPSKLung s nguyn thuP'Q'AB0 0 0 0 0 0 /20 1 /20 11 11 13 /21 03 /21 0IV. m pht hin v sa li1. Khi nim truyn dn s ngi ta thng o cht lng truyn dn bng t s bit li BER. BER cho bit bao nhiu bit trong tng s bit thu c b pht hin l mc li.S bit b mc liBER = S bit thu c Tt nhin, ngi ta mong mun t s trn phi cng nh cng tt, nhng do ng truyn dn lun lun thay i nn khng th gim t l ny xung 0. Ngha l ngi ta phi chp nhn mt s lng li nht nh v tm kh nng khi phc cc thng tin b li hay t nht cng c th pht hin c cc li khng s dng thng tin ny. iu ny c bit quan trng trong cc h thng truyn dn s liu v cc li bit s nh hng rt nghim trng n kt qu s liu pha thu. p ng c yu cu trn, ngi ta thc hin m ho knh. M ho knh l qu trnh x l tn hiu s c thc hin sau ngun tin s v trc iu ch. Mt trong nhng nhim v ca m ho knh l kim sot li, thc cht y l qu trnh x l tn hiu s m bo thng tin c tin cy cao hn. M ho knh kim sot li c thc hin bng cch thm vo tn hiu s cc bit kim tra c th thc hin hai nhim v pha thu l PHT HIN v SA LI. Tu theo tm quan trng ca tng h thng m thut ton m ho knh n gin hoc phc tp. Tn hiu sau khi c m ho knh, tu tng mc , c th cho php pha thu pht hin li hoc va pht hin li va c th sa li. Nh vy tn hiu sau khi m ho knh, tu tng mc , c th gi l m PHT HIN LI hay m PHT HIN v SA LI.M ho knh kim sot li c th thc hin bng hai phng php: hiu chnh li trc (FEC) v yu cu pht li t ng (ARQ). + phng php hiu chnh li trc FEC (Forward Error Correction) b m ho nhn s liu s v b sung vo mt s bit kim tra theo mt quy lut nht nh. V th to ra mt lung bit mi c tc cao hn. B gii m s s dng cc bit kim 50tra xc nh gi tr thc ca cc bit thu c v trong mt s trng hp c th hiu chnh nu pht hin bit s liu b li.+phngphpARQ(AutomaticRetransmissionRequest)ngitacng thm vo tn hiu s cn truyn mt s bit kim tra v nh vy cng lm cho lung s sau khi lp m c tc cao hn lung s nguyn thu. Pha thu da vo cc bit kim tra pht hin li v nu pht hin thy li th my thu s gi yu cu v my pht yu cu pht li phn thng tin b mc li ni trn. H thng vi ba s hin nay ch yu s dng phng php FEC v vy ti liu ch s cp n phng php ny. Nh ni trn, m pht hin v sa li c thit lp bng cch thm vo thng tin cn truyn mt s ckim tra kt qu l lm cho tc bit tng ln nhng li t c an ton chng li cao hn. Ta hy xt mt v d minh ho iu ny.Gi s ta mun gi i mt bt 0 hay mt bt 1. cc bt ny c bo v ta b sung thm 3 bt kim tra theo cch sau:Thng tinB sung Gi i 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1 i vi mi bit (0 hay 1) ch c mt khi m ng (0 0 0 0 hay 1 1 1 1). Nu thu c bt c ci g khc 0 0 0 0 hay 1 1 1 1 th c ngha l xy ra li trn ng truyn. T l m y l 1: 4 (1 bit thng tin: 4 bit phi truyn i). Nh vy vic pht hin v sa li xy ra nh th no ? Ta hy xt cc khi m di y lm v d.Pht i 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0Thu c 1 1 1 1 0 0 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0Quyt nh 1 0 0 x 1 1 Nu ng truyn khng c li th cc nhm bit s l 0 0 0 0 v / hoc 1 1 1 1 v s khng c vn g trong vic quyt nh pha pht pht i 0 hay 1. Nu 1 bit trong s 4 bit b li ta c th sa n, chng hn pht i 0 0 0 0, thu c 0 0 0 1, ta quyt nh l 0 c pht hay pht i 1 1 1 1, thu c 0 1 1 1, ta quyt nh l 1 c pht.... Nhng nu c 2 bit b mc li ta ch c th pht hin nhng khng th sa li, chng hn pht i 0 0 0 0, thu c 0 1 1 0 th ta khng th quyt nh l bit 1 hay bit 0 c pht. Cui cng nu xy ra 3 hay 4 li th ta khng th pht hin c v nh vy thng tin s b li 1 bit. C th ni rng vi cch lp m nh trn ta c th pht hin 2 li v sa 1 li. C th chia m pht hin v sa li theo phng php FEC thnh hai loi: m khi v m xon.51+ m khi ngi ta chia thng tin thnh cc khi, sau b sung vo mi khi mt s bit kim tra nht nh v cc bit kim tra trong mt khi m ch ph thuc vo cc bit thng tin trong khi. S ca b m ho khi c m t hnh 3.15a:+ m ho xon, cc khi m c to ra khng ch ph thuc vo cc khi bn tin hin thi m cn ph thuc vo cc bit ca khi bn tin trc n. Nguyn l ca m xon c m t nh hnh 3.15b : HNH 3.15a : M ho khiHNH 3.15b: M ho xon2. M chn la.Khi nim M chn - l l mt loi m khi, trong bit kim tra c thit lp da trn vic nh gi tnh chn hoc l ca cc bit 0 hay 1 ca cc bit tin trong khi tin. C th c mt hoc nhiu bit kim tra chn, l trong mt khi m.b. M chn l theo hngb1. Nguyn tc lp m- Thng tin cn truyn c chia thnh cc khi, mi khi gm m bit thng tin, sau thm vo mi khi mt bit kim tra kim tra tnh chn, l ca cc bit 1 trong khi tin (thng thng ngi ta kim tra tnh chn, l ca cc bit 1 trong khi tin m khng kim tra tnh chn, l ca cc bit 0).- Nu m c lp theo c tnh chn th bit kim tra phi c thit lp sao cho tng s bit 1 trong mi khi tin sau khi lp m phi l chn.- Nu m c lp theo c tnh l th bit kim tra phi c thit lp sao cho tng s bit 1 trong mi khi tin sau khi lp m phi l l.52M HA KHIThng tinThng tinK.traKhi tin Khi tin sau m haM HA XONKhi tin N-1Khi m NKhi tinKhi tin sau m haKhi tin Nb2 .V d minh hoV d 1: Lp m chn, l cho khi tin 0 0 1 0. Bit rng m c lp theo c tnh chn.Gii: Tng s bit 1 trong khi tin trn l 1 (l) nh vy m bo c tnh chn ta phi thit lp bit kim tra l 1. Sau khi lp m ta thu c khi m l 0 0 1 0 1 (bit cui cng l bit kim tra chn l).V d 2: Lp m chn l cho khi tin 0 0 1 0. Bit rng m c lp theo c tnh l.Gii: Tng s bit 1 trong khi tin trn l 1 (l). m bo c tnh l ca m ta phi thit lp bit kim tra bng 0. Sau khi lp m ta thu c khi m l 0 0 1 0 0 (bit cui cng l bit kim tra chn l).c. M chn l theo ctc1. Nguyn tc lp m Thng tin c chia thnh cc nhm, mi nhm c n khi, mi khi gm m bit thng tin, sau thm vo mi nhm mt khi m bit kim tra chn l [nh vy bn tin sau khi m ho c (n + 1) khi]. Vic thit lp cc bit kim tra chn l c thc hin bng cch kim tra c tnh chn, l ca cc bit 1 c cng th t trong cc khi tin ca cng mt nhm v a gi tr vo v tr tng ng trong khi bit kim tra chn l.T bng trn ta thy vic kim tra c thc hin cc bit c cng th t ca tt c cc khi trong nhm (kim tra theo ct). Sau khi lp m ta thu c bn tin m ho l: Cc m chn, l theo hng v theo ct va xt trn c mt s mt hn ch nh:+ Ch pht hin li, khng th sa li.+ Ch pht hin s li l (1, 3, 5, 7...), khng th pht hin c s li chn (2, 4, 6, 8...). khc phc mt phn cc nhc im trn, ngi ta s dng phng php m ho kt hp c theo hng v theo ct.d.M chn, l theo hng v theo ctd1. Nguyn tc lp m M chn, l theo hng v theo ct c thit lp thng qua hai bc:53Bc 1: Lp m theo hngThng tin c chia thnh cc khi, mi khi c m bit, sau thm vo mi khi 1 bit kim tra kim tra c tnh chn, l ca cc bit 1 trong khi tin.Bc 2: Lp m theo ctThng tin lp m chn, l theo hng c chia thnh cc nhm, mi nhm c n khi, mi khi c (m+1) bit, sau thm vo mi nhm mt khi bit kim tra c (m+1) bit kim tra c tnh chn, l ca cc bit 1 c cng th t trong cc khi tin ca cng mt nhm v kt qu c a vo khi bit kim tra theo th t tng ng.Chng IV IU CH V GII IU CH I/- Cc khi nim c bn v iu ch v gii iu ch1- Tn hiu iu chL cc tn hiu tin tc cn truyn (tn hiu s v tn hiu tng t bao gm: Tn hiu thoi, truyn hnh, s liu. . . ) c tn s thp.542- Tn hiu sng mangL cc tn hiu in cao tn c th ti (mang) c thng tin. Tn hiu cao tn y mang tnh cht tng i. Mt tn hiu in cao tn c th lm sng mang cho mt tn hiu iu ch ny nhng li khng th lm sng mang cho mt tn hiu iu ch khc v chnh n c khi li tr thnh tn hiu iu ch cho mt sng mang c tn s cao hn. Mt sng mang tiu biu c biu thc ton hc l:U(t) = A sin ( t + )Trong : - A l bin ca sng mang- = 2 f l tn s gc ca sng mang (f l tn s ca sng mang)- : pha ca sng mang- Cc tham s A, f v u c th mang thng tin.3- iu ch a) iu ch iu ch l a tn hiu iu ch tc ng vo sng mang lm cho mt hoc nhiu tham s ca sng mang thay i theo quy lut ca tn hiu iu ch.b) S cn thit ca iu ch - Trong thng tin v tuyn: do cc tn hiu tin tc (tn hiu tng t v tn hiu s) thng c tn s thp nn rt kh trc tip bc x thnh sng in t truyn i xa, nu c th bc x c th nng lng bc x cng rt yu v i hi phn t bc x (anten)c kch thc ln. d dng truyn thng tin i xa bng sng in t, ngi ta phi tin hnh iu ch tn hiu thng tin vo sng mang cao tn, ngha l gi tin tc cn truyn vo sng mang sau mi cho sng mang iu ch bc x thnh sng in t truyn i xa. V sng mang c tn s cao nn d bc x thnh sng in t v i hi phn t bc x c kch thc khng ln.- Trong ghp knh theo tn s: V cc tn hiu thng tin cng loi u c chung mt bng tn truyn dn (V d: tn hiu thoi c bng tn t 0,3 3,4KHz) nn khi truyn nhiu tn hiu trn mt ng truyn dn th chng s b ln vo nhau lm cho pha thu khng thu c tn hiu. truyn c nhiu tn hiu trn cng mt ng truyn th ngi ta phi iu ch cc tn hiu cn truyn vo cc sng mang khc 55nhau, mc ch l chuyn ph ca thng tin cn truyn ln cc vng khc nhau sau mi truyn chung trn mt ng truyn dn. Nh s khc nhau v vng ph ca cc tn hiu truyn i m pha thu d dng thu c tn hiu.4- Gii iu ch V phapht tinhnhiuchnnphathumunlyli tnhiu nguyn thy th phi tin hnh tch tn hiu tin tc ra khi sng mang iu ch. Qu trnh tch tn hiu tin tc cn truyn ra khi sng mang iu ch bn pht c gi l gii iu ch.II/- Phn loi iu ch1- Phn loi theo tn hiu a vo iu ch - iu ch tng t: tn hiu iu ch l tn hiu tng t.- iu ch s: tn hiu iu ch l tn hiu s.2- Phn loi theo s thay i ca cc tham s sng mang- iu ch bin : l tc ng tn hiu iu ch vo sng mang lm cho bin ca sng mang thay i theo quy lut ca tn hiu iu ch. iu ch bin tng t cgi lAM (AmplitudeModulation). iuchbinscgi lASK (AmplitudeShift Keying).- iu ch tn s: l tc ng tn hiu iu ch vo sng mang lm cho tn s ca sng mang thay i theo quy lut ca tn hiu iu ch. iu ch tn s tng t cgi lFM (FrequencyModulation). iuchtnsscgi lFSK (Frequency Shift Keying).- iu ch pha: l tc ng tn hiu iu ch vo sng mang lm cho pha ca sng mang thay i theo quy lut ca tn hiu iu ch. iu ch pha tng t c gi lPM(PhaseModulation).iuchphascgi lPSK(PhaseShiftKeying).- iu ch QAM (Quature Amplitude Modulation): l phng php iu ch kt hp c iu ch bin ASK v iu ch pha PSK. Vi iu ch ny th khi tn hiu iu ch tc ng vo sng mang th c bin v pha ca sng mang u thay i theo quy lut ca tn hiu iu ch.III. iu ch v gii iu ch bin 1) iu ch bin (AM)56C LDR3R1R2Tn hiu AMT/h iu chSngmangHNH 4.1: iu ch AM bng mt diodeC nhiu phng php thc hin iu ch AM. Hnh 4.1 m t mt phng php tng i n gin thc hin iu ch AM. Mch gm mt mng trn tn hiu bng in tr (R1, R2, R3), mt diode nn v mt khung cng hng LC.Tn hiu iu ch c a n u vo R1, tn hiu sng mang c a n u vo R2. Mng in tr R1, R2 v R3 thc hin trn tuyn tnh hai tn hiu vi nhau theo nguyn tc cng s hc. Nu tn hiu iu ch l tn hiu hnh sin th tn hiu sau khi trn (ly trn in tr R3) c dng nh hnh 4.2 (c). Ta thy rng sng mang bin thin trn nn ca tn hiu iu ch, nhng y cha phi l tn hiu iu ch bin . y hai tn hiu mi c cng vi nhau, trong khi iu ch l nhn hai tn hiu vi nhau. Tn hiu sau khi cng c nn qua diode D. Sau khi nn ta thu c mt dy xung dng l tp hpca cc na chu k dng ca tn hiu tng, bin ca cc xung thay i theo quy lut ca tn hiu iu ch (hnh 4.2 d). Cc xung ny c a n kch thch cho mt mch cng hng song song LC. Khung cng hng LC ny c tn s cng hng ring fLC12 ng bng tn s ca sng mang. Khi c kch thch th khung LC s dao ng vi tn s ng bng tn s sng mang, cn bin th ph thuc vo bin ca tn hiu kch thch. Khi xung kch thch c bin ln th dao ng ly ra trn khung c bin ln, khi bin ca xung kch thch nh th dao ng ly ra trn khung c bin nh. Nh ni trn, bin ca cc xung kch thch thay i theo quy lut ca tn hiu iu ch nn dao ng ly ra trn khung LC cng c bin bin thin ng theo quy lut ca tn hiu iu ch. y chnh l tn hiu iu ch bin (hnh 4.2 e). Trong iu ch bin , nu gi tin tc cn truyn c tn s l Vt vi Vt = V . cos t (V l bin cc i ca tn hiu iu ch), sng mang cao tn l V = 57 t(e) tUUU tUU t(d)(a)(b)(c) tHnh 4.2: a) Tn hiu iu ch b) Sng mang c) Tn hiu trn in tr R3 d) Dng tn hiu qua diode e) Tn hiu trn khung cng hngVo . cos ot th sau khi iu ch ta thu c mt tn hiu mi: VAM (t) = (Vo + V. cos t) (Vo: bin ln nht ca sng mang). Biu thc trn c th vit:V VVVt tAM t ooo ( )( cos ) cos + 1 + V m t to( cos ) cos 1 (3.1)Trong : mVVol h s iu ch v cn c gi l su iu ch. H s iu ch m phi tha mn iu kin m 1. Num > 1 th mch c hin tng qu iu ch gy ra mo tn hiu sau iu ch. Trong thc t, m bo tn hiu khng b mo th m c chn vo khong 0,7 0,8.p dng bin i lng gic cho biu thc 4.1 ta c:V V tmVtmVtAM t o ooooo ( ).cos .cos( ) .cos( ) + + + 2 2 (3.2)58T biu thc 3.2 ta thy rng: tn hiu sau khi iu ch bin c 3 thnh phn chnh: sng mang c tn s gc o v bin Vo, hai thnh phn mang tin c tn s gc o tv bin mVo2 nh m t hnh 4.3b. Nu tn hiu tin tc c di tn t min n max th sau khi iu ch vo sng mang ta s thu c tn hiu c ph nh m t hnh 3.3d. Ngoi sng mang c bin Vo v tn s gc o cn c hai bin tn: bin tn trn c tn s t (o + min) n (o + max) v bin tn di c tn s t (o - max) n (o - min) i xng vi nhau qua sng mang.2.- Gii iu ch bin tng t Hnh 4.4 m t mt phng php gii iu ch bin n gin nhng c s dng kh rng ri trong k thut gii iu ch. Mch gm c mt bin p cng hng T1 ghp tn hiu AM vo mch, diode nn D, in tr ti R v t lc C3. in p trn cc phn t R v C chnh l tn hiu gii iu ch.Tn hiu AMc t vohai ucunscpca bin p cng hng T1. Cc t C1 v C2 iu chnh tn s cng hng ca bin p. Tn hiu AM cm ng qua cun th cp v t vo mch nn gm D v 59 min max 0 0 - 0 + 0 - min 0 + min 0 - max 0 + max 0UUUU(a)(b) (d)(c)HNH 4.3:a) Ph ca tn hiu iu ch n tnb) Ph ca tn hiu AM iu ch n tnc) Ph ca tn hiu iu ch a tnd) Ph ca tn hiu AM iu ch a tn Tn hiu gii iu chT1C2C1C3DTn hiu AMHNH4.4: Mch gii iu ch bin RR3. Diode D s dn cc na chu k dng ca tn hiu AM, cn na chu k m th D khng dn. Kt qu l trn ti ta thu c mtdy xungdng m bincanthayi theo quy lut ca tn hiu iu ch bnpht. khiphclitn hiu th phi u song song vi Rmt t C3.TC3phic chn sao cho c tr khng nh i vi tn s sng mang v c trkhnglni vi tnhiu tintc. Ktqulthnhphn sng mang s b ngn mch xung mass, cn thnh phn tintcscarau ra.Mt cch khc phn tch nguyn l hot ng ca mch l gi s khi diode dn t C3 c np rt nhanhngitr nhca xung. Khi diode tt th t C3 phng in qua R. Hng s thi gian = RC3 c chn sao cho ln hn rt nhiu so vi chu k ca sng mang. Kt qu l t phng in in p trn t gim khng nhiu, khi diode dn th t li c np in tr li ti gi tr nh. in p trn t gn ging vi tn hiu iu ch bn pht. in p ny chnh l tn hiu gii iu ch. 60Tn hiu gii iu chC2RD1D2Tn hiu AMC2T1HNH 4.6a: Mch gii iu ch bin kiu nn ton sngUUU t t t(a)(b)(c)HNH 4.5: Dng sng ca mch gii iu ch AM dng dioden n na sng. a) Tn hiu AMb) Dng qua diode. c) Tn hiu sau gii iu ch Do t phng np nn tn hiu khi phc c s c nhnggnsnggyramo tn hiu sau gii iu ch. Tuy nhin khi sng mang c tn s lnhnrt nhiulnsovi tn s ca tn hiu iu ch th nhng gn sng ny c th b qua. Hnh 4.5 m t dng sng ca cc tn hiu trong mch gii iu ch bin . tng hiu qu gii iu ch, ngi ta s dng mchgii iuchtngt theokiunntonsngnh m t hnh 4.6a.Hnh 4.6b m t mch in v dng sng ca cc tn hiutrong mch gii iu ch bin kiu nn ton sng.3. iu ch ASK iu ch ASK l iu ch binpdngchotnhiu iu ch l tn hiu s.Hnh 4.7 m t s khi ca mch iu ch ASK.+S(t): tnhiuiuch(tn hiu s)+ LO: mch dao ng to sng mang (Local Ocsilator)+ A.sin t: sng mang c to ra t mch dao ng LO+ M: B nhn+ BPF: b lc thng bng (BPF: Band Pass Filter)61U t(a)U t(b)U t(c)HNH 4.6b: Dng sng ca mch gii iu ch AM dng diodenn ton sng. a) Tn hiu AMb) Dng qua diode. c) Tn hiu sau gii iu chBPFLOS(t).Asin tS(t)MA(t).sin tHNH 4.7: iu ch ASKUtUtUt1 1 1 1 1 0 0 0 0(a)(b)(c)HNH 4.8: Dng sng ca iu ch ASKa) Tn hiu iu ch. b) Sng mang.c) Sng mang sau iu ch+ S(t).Asin t: sng mang c iu ch bin .Tn hiu iu ch S(t) l tn hiu s c a n u vo th nht ca b nhn M. Sng mang A.sin t to ra t b dao ng LO c a n u vo th hai ca b nhn. B nhn M s nhn tn hiu iu ch S(t) vi tn hiu sng mang A.sin t. u ra ca b nhn ta thu c thnh phn c bn l S(t).Asin t v mt s thnh phn khc (v d A.sin2 t, S(t).A sin3 t ...). Tn hiu ny c a qua b lc thng gii loi b cc thnh phn khng mong mun. Kt qu u ra ca b lc ta thu c tn hiu iu ch bin (ASK) l S(t).Asin t. Dng sng ra sau iu ch c m t hnh 4.8. iu ch ASK c u im l mch in iu ch v gii iu ch u rt n gin. Tuy nhin c nhc im rt ln l kh nng chng nhiu thp v nhiu d dng tc ng vo vng bin (vng mang tin ca sng mang iu ch ASK).4- Gii iu ch ASK Hnh 4.9 m t mt mch gii iu ch ASK tiu biu. + Bin p (T1) kt hp vi t C1 to thnh mt bin p cng hng ghp tn hiu vo mch gii iu ch. + D l diode nn cao tn+ R1 v C2 to thnh ti ca mch gii iu ch.+ N l b khuch i thut ton hot ng theo kiu so snh. + Uchun l in p ln hn 0V. Khi khng c sng mang a vo u vo ca bin p cng hng BA th trn cun th cpca bin p khng ctn hiu. Lcnyinpt vo chn(+) caNl0Vlmcho chn (+) m hn in p chn (-)caNnnngracaNc mc 0. 62DUchun+_Tn hiu ASKRC2Tn hiu s raT1HNH 4.9: Gii iu ch ASK1 0(a)tUchun(b)(c)0 1 1ttUUUHNH 4.10: Biu thi gian m t dng sngca mch gii iu ch ASKLCD+VccR1R2CoC2C1Tn hiu iu chTimch dao ngHNH 4.11: Mch iu ch tn s tng t FMKhi c sng mang a vo u vo ca bin p cng hng th tn hiu ny s c ghp qua cun th cp. Sau tn hiu c diode D nn theo kiu nn na chu k (trong mt s mch gii iu ch ASK, hiu qu hn ngi ta s dng nn c chu k) vi ti l in tr R. T C2 thc hin lc san bng v to ra trn in tr ti R mt in p tng i bng phng. in p ny c t vo chn (+) ca N lm cho in p chn (+) ca N dng hn in p chn (-) nn ng ra ca N c mc cao. Kt qu l ng ra ca mch so snh N ta thu c cc bit 0, 1 nh a vo iu ch bn pht. Biu thi gian m t nguyn l hot ng ca mch c m t hnh 4.10:IV. iu ch v gii iu ch tn s1. iu ch tn s tng t (FM)C nhiu phng php thc hin iu ch tn s FM. Hnh 4.11gii thiu mt phng php n gin thc hin iu ch tn s. Mch in y cha c v y m ch trch mt phn mch c bn lin quan n vic iu ch tn s. Trong mch in hnh 4.11: + CD l mt diode bin dung+ R1, R2: to thnh mt mch phn p nh im lm vic cho CD.+ Co: t dn tn hiu iu ch vo mch+ C2: mt t dn tn hiu c tr khng gn nh bng 0 i vi bng tn cng tc nhm m bo cho CD dng nh c mc song song vi khung LC1. Nh vy 3 phn t CD, L v C1to thnh mt khung cng hng vi tn s cng hng ring: fL C CD+121 ( )63Khi cha c tn hiu iu ch a vo mch th trn diode bin dung c mt in p tnh c xc nh bi mch phn p R1 v R2 do diode ny c mt gi tr in dung CD khngi. Lc ny mchdao ngs hotng vitn s khngi: fL C CD+121 ( ) Khi c tn hiu iu ch a vo mch, nu bin ca tn hiu iu ch tng th in p trn diode bin dung tng, lm cho tr s in dung CD ca diode gim, dn n tn s dao ng ca mch dao ng tng. Nu bin ca tn hiu iu ch gim th in p trn diode bin dung gim lm cho in dung CD tng dn n tn s cng tc ca mch dao ng gim. Nh vy, tn s dao ng ca mch dao ng b khng ch bi tn hiu iu ch u vo v bin thin ng theo quy lut ca tn hiu iu ch. ng ra ca mch dao ng ta thu c mt tn hiu c iu ch tn s. Hnh 4.12 m t dng sng ca mch iu ch tn s. Trong iu ch tn s, nu tn hiu iu ch l V(t) vi V(t) = V. cos t, tn hiu sng mang l V vi V = Vo . cosotth sau iu ch ta thu c mt tn hiu iu chVFM:V V t tFM o o + .cos( .sin ) Trong : l di tn cc i. Hsiuchmfcamchiuchtnsctnhnhsau: m kVf ., k: h s t l ph thuc vo c im ca tng mch iu ch. T biu thc trn ta c: = k.V. Nh vy, ta thy rng khiV = const th =const, nhng khi thay ith mf cng thay i.

64ttU(a)(b)(c)tHnh 4.12: Dng sng ca mch iu ch tn s.a) Tn hiu iu chb) Sng mang c) Tn hiu sau khi iu ch Khi iu ch n tn, ph ca tn hiu iu tn cha thnh phn o v nhiu thnh phn tn s bin (o tn ) vi n = 1, 2, 3... . Bin ca cc thnh phn tn s bin bin i khng ng u. Mt cch gn ng, rng ph ca tn hiu iu tn sau iu ch c th tnh nh sau: D m mFM f f + + 2 1 ( ) DFM: rng ca ph sau iu ch tn s. Khi tn hiu iu ch l mt bng tn th ph ca tn hiu iu tn DFM s l:D m mFM f f + + 2 1 ( )max(max: tn s ln nht trong tn hiu iu ch).+ Khi mf > 1 th ph ca tn hiu iu tn c th tnh gn ng: DFM 2.mf. max = 2+ Khi mf 1 th ph ca tn hiu iu tn c th tnh gn ng: DFM 2max2. Gii iu ch tn s tng t:Hnh 4.13 m t mch in gii iu ch tn s theo kiu tch sng t l c s dng rng ri trong k thut gii iu ch tn s v quan h pha ca cc in p trong mch.Mch gm c:+ Bin p cng hng T1 ghp tn hiu FM vo mch. + Cun th cp ca bin p T1 c im gia c u chung.+ Cun cm RFC to in p lch pha 90o.+ T C3 dn tn hiu FM vo cun cm RFC.+ D1, C7, R1 v D2, C5, R2: to thnh hai mch tch sng bin , trong : D1 ging D2, C7 = C5 v R1 = R2.+ Cc t C7, C5 v cc in tr R1, R2 to thnh mt mch cu.+ in p trn C7 v C5 l in p vo ca cu, in p gia C v D l in p ra ca cu. y chnh l im ly tn hiu ra ca mch gii iu ch.65V1V2V3V1-3V2-3V1 V1 V1-3V1-3V2 V2 V2-3V3V3(d) (c) (b)C1T1D2D1Tn hiu FMC7C5C6R1R2C2RFCTn hiu gii iu chV2V1V3(a)HNH 4.13: Gii iu ch tn s(a)Mch in gii iu ch tn s ,(b) Quan h pha ca cc in p khi tn s trong mch bng tn s sng mang, (c)Quan h pha ca cc in p khi tn s trong mch ln hn tn s sng mang, (d)Quan h pha ca cc in p khi tn s trong mch nh hn tn s sng mang+ C6: t c tr s ln, c nhim v hn ch bin , trnh hin tng iu bin k sinh xy ra trong qu trnh gii iu ch tn s.+ V1: in p na trn ca cun th cp+ V2: in p na di ca cun th cp+ V3: in p trn cun cm RFCV1-3 = V1 + V3 V2-3 = V2 + V3Khi tn hiu FM u vo l sng mang khng iu ch th in p V1-3cp vo D1 ng bng in p V2-3 cp vo D2 nn cc t C7v C5 c np vi in p bng nhau, du ca in p nh m t trn hnh 4.13 (b).+ V: C6 u song song vi C7 v C5 nn in p np cho C6 bng tng in p trn C7 v C5. V C6 c gi tr rt ln (thng dng t ha) nn khi c np y n lun duy tr mt in p n nh. + Do R1 = R2 v C1 = C2 nn U UR R1 2v U UC C1 2. Nh vy mch cu t trng thi cn bng. Lc ny, in p gia hai im C v D bng 0.Gi s rng khi c sng mang khng iu ch a vo, in p trn cc t C7 v C5 l U U VC C4 52 c ngha l U VC64 , khi in p trn cc in tr R1 v R2 l: U U VR R1 22 . 66Nu tn s ca tn hiu FM tng th quan h pha trong mch s thay i nh m t hnh 4.13c lm choV1-3 > V2-3 v nh vy in p trn t C7 (UC4) s ln hn in p trn t C5 (UC5).Gi s in p trn t C7 l 3V th in p trn t C5 l 1V, in p trn R1 v R2 vn bng nhau v bng 2V (v in p np cho C6 khng i). Lc ny cu khng cn bng, in p gia C v D s khc 0. D dng nhn thy rng in p UCDby gi l -1V. Khi tn s ca tn hiu FM gim th quan hpha trong mch s thay i nh m t hnh 4.13d lm cho V1-3 < V2-3 v nh vy in p np cho t C7 nh hn in p np cho C5. Nh phn tch trn, gi s in p trn C7 l 1V th in p trn C5 l 3V,UR1vn bng UR2v bng 2V. Lc ny cu khng cn bng nhng theo hng ngc li vi trng hp tn s FM tng. D dng nhn thy UCD by gi bng 1V. Nh vy, khi tn s ca tn hiu FM thay i th in p ra ca cu UCD cng thay i theo. y chnh l tn hiu tin tc gii iu ch. V UCD ph thuc vo t l in p trn C7 v C5 nn mch c gi l mch tch sng t l.3. iu ch tn s vi tn hiu s (iu ch FSK)Nh cp phn trc, khi tn hiu iu ch l tn hiu s th iu ch tn s c gi l FSK (Frequency Shift Keying: kha dch tn s). Hin nay c nhiu kiu iu ch FSK: 2 FSK, 4 FSK, 8 FSK ... . iu ch 2 FSK l iu ch tn s m sau khi iu ch sng mang ch c hai tn s. Tng t vi 4FSK th saukhi iuchsngmangc4tn s ... . n gin trc ht ta xt iu ch 2FSK n gin ta xt iu ch 2FSK. Skhi camchiuch 2FSK c m t hnh 4.14. + LO1 v LO0: hai b dao ng c lp: LO1 to ra sng mang c tn s f1, LOo to ra sng mang c tn s fo.+ Mo v M1: cc cng truyn dn c iu khin bi tn hiu iu ch. + : b cng c nhim v kt hp cc sng mang fo v f1 to thnh tn hiu 2 FSK.Nguyn l hot ng ca mch hnh 4.14 nh sau: Mch dao ng LO0 to ra tn hiu sng mang c tn s fo a n ng vo ca cng truyn dn Mo. Mch dao ng LO1 to ra sng mang c tn s f1 a n ng vo ca cng truyn dn M1. Tn hiu iu ch c a n iu khin cc cng truyn dn Mo v M1. Khi tn hiu iu ch l bit 0 th cng truyn dn M1khng c iu khin nn khng cho tn 67LO0LO1Tn hiuiu chM0M1Tn hiu iu chHNH 4.14: S khi mch iu ch 2FSKUUUUtttt(a)(b)(c)b)(d)b)HNH 4.15: Dng sng iu ch 2FSKa) Tn hiu iu ch, b) Sng mang f0 , c) Sng mang f1, d) Sng mang sau iu chhiu sng mang f1 i qua, cng truyn dn Mo c iu khin nn cho tn hiu sng mang fo i qua. Khi tn hiu iu ch l bit 1 th cng truyn dn Mo khng c iu khin nn khng cho sng mang fo i qua, cng truyn dn M1 c iu khin cho sng mang f1 i qua. Mch cng kt hp cc tn hiu ng ra ca cc cng truyn dn Mo v M1.Kt qu l ng ra ca mch cng ta thu c sng mang iu ch 2 FSK gm hai tn s fo v f1 ng vi cc bit s liu a vo iu ch. Hnh 4.15 m t dng sng ca iu ch 2 FSK.4. Gii iu ch FSKC hai phng php thc hin gii iu ch FSK: Phng php gii iu ch nht qun v phng php gii iu ch khng nht qun.* Phng php gii iu ch khng nht qun: Hnh 3.17 m t s khi ca mch gii iu ch 2 FSK khng nht qun. Mch gm hai b lc thng gii, hai mch tch sng v mt mch khuch i so snh. B lc thng gii BPFo c tn s trung tm l fo, b lc thng gii BPF1 c tn s trung tm l f1. Cc mch tch sng c nhim v bin tn hiu vo thnh in p ra ln hn 0. Mch so snh thc hin so snh in p ng ra ca hai mch tch sng. C th m t hot ng ca mch hnh 3.17 nh sau:68Tn hiu 2 FSK gm 2 tn s fo v f1 c a n ng vo ca cc mch lc thng gii BPFo v BPF1. Khi tn hiu 2 FSK c tn s l fo th mch lc thng gii BPF1 khng cho tn hiu qua, ng vo ca mch tch sng f1 khng c tn hiu do ng ra ca n c in p l 0V. Cn mchlc BPFo cho tn hiu c tn s fo i qua, c tn hiu sng mang fo a vo mch tch sng fo nn ng ra ca n c in p l +V. Lc ny chn (-) ca mch khuch i so snh c in p dng hn in p chn (+) ca n nn ng ra c in p l 0V. Khi tn hiu 2 FSK c tn s l f1 th mch lc BPFo khng cho tn hiu sng mang f1 i qua, mch tch sng fo khng c tn hiu vo a ra in p 0V t vo chn (-) ca mch khuch i so snh, cn mch lc thng gii BPF1 cho sng mang f1 i qua, mch tch sng f1 c tn hiu vo nn a ra in p +V. Lc ny in p chn (+) ca mch khuch i so snh dng hn in p chn (-) ca n nn ng ra c mc cao. Nh vy, ty theo tn hiu 2 FSK u vo m ng ra ca mch khuch i thut ton c mc 0 hoc 1. chnh l cc bit s liu c gii iu ch. Phng php gii iu ch FSK khng nht qun c u im l mch in n gin, khng cn phi khi phc sng mang. Tuy nhin do c tnh ca mch nn yu cu tn hiu FSK c bin tng i ln. khc phc nhc im ny, ngi ta s dng gii iu ch FSK nht qun.* Phng php gii iu ch FSK nht qun: S khi n gin ca mt mch gii iu ch 2 FSK nht qun c m t hnh 3.18. Mch gm c:+ 2 b dao ng LO1 v LOo khi phc li hai sng mang c tns f1v fo saochosng mang khiphcphingphavisng mang iu ch 2 FSK mt cch tng ng.+ 2 b nhn Mo v M1 nhn hai sng mang khi phc vi tn hiu 2 FSK.+ 2 mch lc thng thp loi b cc thnh phn tn s cao sau khi nhn v mt mch khuch i so snh.Nguyn l gii iu ch 2 FSK nht qun nh sau: Tn hiu 2 FSK c a n u vo th nht ca cc b nhn Mo v M1. u vo th hai ca Mo l tn hiu sng 69LO0M0M1LO1Lc thng thpLc thng thp_+S liu giiiu chTn hiu 2FSKHNH 4.18: S khi ca mch gii iu ch2 FSK nht qunmang fo c to ra t b dao ng ni LOo. u vo th hai ca b nhn M1l sng mang c tn s f1c to ra t b dao ng ni LO1. LO1v LO2u c khng ch bi tn hiu 2 FSK nhm m bo fo do LOo to ra ng pha vi fo trong tn hiu 2 FSK v f1 do LO1 to ra ng pha vi f1 trong tn hiu 2 FSK. Cc b nhn Mo v M1thc hin nhn sng mang 2 FSK vi cc sng mang khi phc fo v f1 tng ng. Nu sng mang 2 ng vo ca b nhn ng pha v cng tn s th ng ra ca b nhn s c in p l +V. Nu sng mang 2 ng vo ca b nhn khng ng pha v cng tn s th b nhn s to ra in p 0V. in p sau cc b nhn c a qua mch lc thng thp loi b cc thnh phn tn s cao to ra sau khi nhn. in p sau khi lc c a vo mch so snh. ng ra ca mch khuch i so snh ta thu c tn hiu s gii iu ch. T phn tch trn, ta thy rng nu tn hiu 2 FSK u vo c tn s l fo th ng ra ca Mo c in p l +V, ng ra ca M1 c in p bng 0V v tn hiu ra sau gii iu ch l 0. Nu tn hiu 2 FSK c tn s l f1 th ng ra ca Mo bng 0V, ng ra ca M1 l +V v tn hiu ra sau gii iu ch l 1.Mch gii iu ch FSK nht qun khc phc c nhc im ca mch iu ch FSK khng nht qun. Tuy nhin phng php ny c nhc im ln l mch rt phc tp, vic khi phc cc sngmangfov f1 cngrt khkhnnnt c ng dng trong cc h thng FSK thng thng.V.iuchvgii iuch pha V iu ch pha tng t t c s dng trong thc t nn trong phn ny ch xt iu ch v gii iu ch pha vi tn hiu s.1) iu ch - Gii iu ch pha hai trng thi (2 PSK)a- iu ch pha hai trng thi(2 PSK)70LBFTo sng mangMi mTn hiu iu chTn hiu iu ch Sng mang(a)0 10 1UtttS liu iu chSng mang cha iu chSng mang iu ch(b)Cos01(c)SinHNH 4.19: iu ch 2 PSKa) S khi ca mch iu ch 2 PSK. b) Dng sng ca iu ch 2 PSK. c) Biu pha ca iu ch pha 2 PSKiu ch pha hai trng thi l iu ch m sau khi iu ch sng mang iu ch c hai trng thi pha so vi pha ca sng mang cha iu ch.Hnh 3.19 m t s khi v dng sng ca mch iu ch pha hai trng thi.S liu nh phn cn iu ch trc ht c a vo mch chuyn m bin i t m NRZ n cc sang NRZ lung cc. Sau , s liu c a n b nhn M nhn vi sng mang 2. cos( ) A to + to ra t mch to dao ng sng mang. ngra ca bnhnta thuc mt tnhium thnhphnchnhl d(t) 2. cos( ) A t to + +v mt s thnh phn pht sinh khng mong mun. Cc tn hiu ny c a qua mch lc bng loi b cc thnh phn khng cn thit.Kt qu, ng ra ta thu c mt tn hiu iu ch 2 PSK: U t d t A to( ) ( ) . cos( ) + 2 . Trong : d(t) l s liu s cn iu ch,o: pha ban u ca sng mang, A 2 : bin ca sng mang. V b nhn M nhn tn hiu iu ch d(t) vi sng mang2. cos A t theo min pha nn c th vit li U(t) nh sau:U t A tt o( ) .cos( ) + + 2 Trong t l gc lch pha do s liu iu ch gy ra ti thi im iu ch (t = 0 hoc t = ). D dng nhn thy rng khi s liu iu ch l bit 0 thd(t) = -1 v sng mang sau iu ch l:U t A to( ) .cos( ) + + 2 0 .Khi s liu iu ch l bit 1 th d(t) = 1 v sng mang sau iu ch l:U t A to( ) .cos( ) + + 2 Nh vy sng mang sau iu ch pha c hai trng thi l ng pha vi sng mang cha iu ch (khi iu ch bit 0) v ngc pha vi sng mang cha iu ch (khi iu ch bit 1). Pha thu s da vo s chuyn pha ca sng mang thc hin gii iu ch.7190oVCO Lc vngLc thng thpLc thng thpPhn nhnhM1M2 cos( t +o) sin( t +o)M3Tn hiu gii iu chTn hiu 2PSKHNH 4.20: Mch gii iu ch 2 PSK bng vng COSTASb- Gii iu ch pha hai trng thi:Hnh 4.20 m t s khi ca mch gii iu ch 2 PSK bng vng kha pha COSTAS.Chc nng ca cc khi hnh 3.20 trn nh sau:+ B phn nhnh c nhim v chia tn hiu 2 PSK u vo thnh hai ng c bin bng nhau.+ M1 v M2 l cc b nhn thc hin gii iu ch.+ 90o: mch dch pha 90o bin sng mang A to2.cos( ) + thnh sng mang A to2.sin( ) +.+ VCO: b dao ng to sng mang c tn s v pha c iu khin bng in p.+ Lc thng thp: lc b cc thnh phn tn s cao pht sinh sau khi nhn tn hiu 2 PSK vi sng mang khi phc.+ M3: b nhn tm kim sai pha gia sng mang 2 PSK v sng mang khi phc. Kt qusai pha s th hin bng in p li.+ Lc vng: lc in p t ng ra ca b nhn a vo iu chnh tn s v pha ca b dao ng VCO.C th phn tch nguyn l hot ng ca mch da trn s khi hnh 3.20 nh sau:Sng mang 2 PSK c chia thnh hai ng a n cc b nhn M1 v M2. M1 thc hin nhn sng mang 2 PSK c biu thc: U t A to t( ) .cos( ) + + 2 vi sng mang khi phc c to ra t b dao ng VCO: cos( ) to+. B nhn M2 nhn sng mang 2 PSK: A to t2.cos( ) + + vi sng mang khi phc to ra t b dao ng VCO v qua dch pha 90o: sin( ) to+. ng ra ca M1 ta thu c tn hiu l:A to t2.cos( ) + +. cos( ) to+=t t o t t oA t A t A cos . ) 2 2 cos( . ] cos ) 2 2 2 [cos( 2 .21+ + + + + + ng ra ca b nhn M2 ta thu c tn hiu l:A to t2.cos( ) + +. cos( ) to+=t t o t t oA t A t A sin . ) 2 2 sin( . ] sin ) 2 2 [sin( 2 .21+ + + + + + Nh vy sau hai b nhn ta u thu c 2 thnh phn: 1 thnh phn tn s cao c tn s v gc pha l: ( ) 2 2 2 to t+ +, 1 thnh phn tn s thp c gc pha l t. 72Bin i ni tip sang song songChuyn mChuyn m90oVCOZ1Z2M1M2S2(t)S1(t)Tn hiu QPSKTn hiu iu chHnh 4.21 S khi ca iu ch pha 4 trng thi:Cc thnh phn ny u mang tin, tuy nhin thnh phn tn s cao l khng cn thit. Cc b lc thng thp sau M1 v M2 s lc b cc thnh phn tn s cao, thnh phn tn s thp thu c sau cc b lc u l s liu gii iu ch. Tuy nhin ch cn ly s liu u ra ca mt b lc thng thp l . Tn hiu sau 2 mch lc thng thp c nhn vi nhau ti b nhn M3 pht hin s sai pha ca sng mang 2 PSK vi sng mang to ra t b dao ng VCO. Nu c sai pha hoc sai tn th M3 s to ra mt in p li. in p ny c lc qua mch lc vng a v iu chnh tn s v pha ca b dao ng VCO. Phng php iu ch, gii iu ch pha 2 trng thi c u im l mch in tng i n gin, khong cch gia cc trng thi pha ln nn c kh nng chng nhiu cao. Tuy nhin, do mi trng thi pha ch th hin c mt bit thng tin nn hiu qu truyn tin khng cao. tng hiu qu truyn tin, ngi ta thc hin iu ch pha nhiu trng thi hn.2) iu ch pha 4 trng thi - Gii iu ch pha 4 trng thia- iu ch pha 4 trng thi (QPSK) iu ch pha 4 trng thi cn c gi l iu ch 4 PSK hay

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