73337118 Thong Tin Vo Tuyen

7/28/2019 73337118 Thong Tin Vo Tuyen

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My pht

My thu

My pht

My thu

Circiulator

Dy feeder

Anten Anten

Tn hiuvo

Tn hiuvo

Tn hiu ra Tn hiu ra Hnh 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 TUYN

I. Qu trnh pht trin ca thng tin v tuyn

1. Khi nim v thng tin v tuyn

Thng tin v tuyn s dng khong khng gian lm mi trng truyn

Phng php thng tin l: pha pht bc x cc tn hiu thng tin bng sng inkhng gian t do, pha thu nhn sng in t bn pht gi ti trong khng giatch ly tn hiu gc.

Hnh 1.1 m t cc thnh phn c bn ca mt ng thng tin v tuychiu.

2. Qu trnh pht trin ca thng tin v tuyn

V lch s ca thng tin v tuyn , vo u th k 20 Marconi thnh cng vic lin lc v tuyn qua i Ty Dong, Kenelly v Heaviside pht hin mt l tng in ly hin din tng pha trn ca kh quyn c th dng lm vt psng in t. Nhng yu t m ra mt k nguyn thng tin v tyun cao tquy m.

Chin tranh th gii ln th hai l mt bc ngot trong thng tin v tuthng tin tm nhn thng - lnh vc thng tin s dng bng tn s cc cao (VHc nghin cu lin tc sau chin tranh th gii - tr thnh hin thc nh s

trin cc linh kin in t dng cho VHF v UHF, ch yu l pht trin nRaa.Vi s gia tng khng ngng ca lu lng truyn thng, tn s ca th

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v tuyn vn ti cc bng tn siu cao (SHF) v cc cao (EHF). Vo nhng1960, phng php chuyn tip qua v tinh c thc hin v phngchuyn tip bng tn x qua tng i lu ca kh quyn xut hin. Do nhtnh u vit ca mnh, chng hn nh dung lng ln, phm vi thu rng, hikinh t cao. Thng tin v tuyn c s dng rt rng ri trong pht thanh thnh qung b, v tuyn o hng, hng khng, qun s, quan st kh tng, lisng ngn nghip d, thng tin v tinh - v tr v.v... Tuy nhin, can nhiu vi lnhthng tin khc l iu khng trnh khi, bi v thng tin v tuyn s dng c phn khng gian lm mi trng truyn dn.

i ph vi vn ny, mt lot cc cuc Hi ngh v tuyn Qucc t chc t nm 1906. Tn s v tuyn hin nay c n nh theo "Q

thng tin v tuyn (RR) ti Hi ngh ITU-T Geneva nm 1959. Sau ln lHi ngh v phn b li di tn s sng ngn s dng vo nm 1967, Hi b sung quy ch tn s v tuyn cho thng tin v tr vo nm 1971, v Hi ng phn b li tn s v tuyn ca thng tin di ng hng hi cho mc ch kinh vo nm 1974. Ti Hi ngh ca ITU-T nm 1979, di tn s v tuyn phn c m rng ti 9kHz - 400 Ghz v xem xt li v b sung cho Quy ch ttin v tuyn in (RR). gim bt can nhiu ca thng tin v tuyn, ITU-T tinghin cu nhng vn sau y b sung vo s sp xp chnh xc khongia 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 t

1. Khi nim

- Trng in t l mt dng vt cht c bit c c trng v nng lng tng tc vi cc mi trng . Trng in t lan truyn trong cc mi t(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 (d

hoc m) ca cc in t t do. V nguyn l, bt k h thng in t no nng to ra in trng hoc t trng bin thin u c bc x sng in nhin trong thc t, s bc x ch xy ra trong nhng iu kin nht nh.

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- 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 g

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/f

2. Phn loi sng in t v ng dng

Tn sPhn

loi bng

tn

C ch truynsng v tuyn Lnh vc s dng

3KHz~30 KHz VLF B mt Thng tin o hng30KHz~300KHz LF B mt Thng tin o hng300KHz~3MHz MF B mt , Sng

triPht thanh AM, Hng khng,o hng.

3MHz~30MHz HF B mt , Sngtri

Pht thanh sng ngn, ohng.

30MHz~300MHz VHF Sng triSng khng gian

Pht thanh FM, truyn hnh,thng tin ng, thng tin vtuyn c nh (viba)

300MHz~3GHz UHF Sng tri

Sng khng gian

Sng thng

Truyn hnh, thng tin ng,thng tin v tuyn c nh(viba), Raar, thng tin vtinh.

3GHz~30GHz SHF,Viba

Sng khng gian

Sng thng

Thng tin v tinh, thng tinv tuyn c nh, Raar, vtuyn thin vn.

30GHz~300GH EHF,Milimeter

Sng khng gian

Sng thng

Thng tin v tinh, thng tinv tuyn c nh, Raar, vtuyn thin vn.

Trong thng tin v tuyn, c ch truyn sng v tuyn v vic s dng th

truyn thng ph thuc vo tn s v tuyn s dng. Bng di y trnh b

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tn s v tuyn c phn loi theo tiu chun quc t hin hnh v theo c phng thc s dng sng v tuyn.

3 .Cc phng thc truyn lan ca sng in t

Sng v tuyn khng truyn lan theo ng thng khi chng trong khgian c nh hng ca cc yu t trn mt t v tng i lu. Tuy nhin, khst sng truyn trong khng gian ta cng c th coi l mi trng l tng in t truyn theo ng thng. Trong thc t ngi ta phn thnh cc phntruyn lan ch yu sau: Sng t, sng tri, sng tn x qua tng i lu vthng.

a. Sng t

Sng t l sng in t chu nh hng ca cc yu t kh hu, a hnhtrn b mt t khi truyn t an ten pht n anten thu. Sng t gm 2 loi: Smt b sng khng gian.

- Sng b mt.

Sng b mt lsng in t truyn dctheo b mt t. Sng b

mt ch yu l ccsng in t tn sthp. S lan truync thc hin nh nhiu x sng in t (Hin tng tia sng un cong vvt cn vi suy hao rt t). Do nhiu x t l nghch vi bc sng cho nns dng cng cao th suy hao ca sng t cng ln, kh nng lan truyn sncng yu. Hnh 1.2 M t m hnh truyn sng t.

Hin tng nhiu x c mi tng quan cht ch vi dn in v hin mi ca t trong ng lan truyn. Hng s in mi ca nc bin nca t nn c ly truyn sng trn mt bin di hn so vi mt t. Do , snt tn s thp c s dng rng ri trong thng tin v tuyn o hng. Bngcc thp c s dng ch yu cho truyn thanh AM v thng tin hng hi vtin o hng.

- Sng khng gian

Sng khng gianl sng truyn gia 2

5

Tia sng un cong theomt t

Anten pht

Anten thu

Hnh 1.2: Sng truyn dc theo mt t

Sng trc tip

Sng phn x t

Sng phn x i luAnten pht

Anten thu

Hnh 1.3: Sng truyn gia 2 anten t cao trong khng gian


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anten t cao trong khng gian nh m t hnh 1.3. Sng in t n im tth :

- Trc tip t anten pht: Sng trc tip.

- Phn x ti mt t trc khi ti im thu: Sng phn x t.- Phn x trong tng i lu trc khi ti im thu: Sng phn tng i

Sng trc tip l sng in t pht x trc tip t anten pht n anten thkhng b phn x u c. Trong iu kin truyn lan bnh thng sng trc 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 phti mt t hoc ti cc vt th xung quanh. Do , di ng i ca tia snx t di hn so vi sng trc tip. Nh vy, ti im thu, sng phn x bin v pha khc vi sng trc tip. Nu chnh lch v ng truysng trc tip v sng phn x t khc nhau mt khong bng s l ln bth 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 nhchng s trit tiu nhau.

sng phn x tng i lu l cc sng in t bc x ti anten pht b u

khi truyn trong tng i lu - do h s khc x ca khng kh thay i theo trc khi n anten thu. Cng ging nh trng hp sng phn x t, nu lch v ng truyn gia sng trc tip v sng phn x tng i lu khc nhkhong bng s l ln bc sng th ti im thu pha ca hai sng ny khc nhkhong 180o v nu khi sng trc tip v sng phn x tng i lu c bi bng nhau th chng s trit tiu nhau.

Sng khng gian c s dng cho cc tn hiu ln hn VHF. S thay s khc x theo cao ca kh quyn gy nh hng n sng khng gianquyn tiu chun l mt kh quyn l tng c mt t l bin i h s khc cao mt cch u n, bi v n c mt h s thay i c nh ca p suquyn theo cao, nhit v m.

V c s bin i h s khc x mt cch lin tc, cho nn ng i thc sng khng gian l khc vi ng trc tip (thng). b li s khc nhau nly thng tin cc i thc t c tnh ton theo ng trc tip da trn quy knh hiu qu ca tri t KR (K=4/3 trong kh quyn tiu chun) nh m t

1.4.

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Hnh 1.4 ng i ca sng khng gian : a) iu kin thc t, b) iu king ng ca bn knh tri t c tnh bng R (K=4/3)

b. Sng tri

Sng tri l sng in t b thay i hnh trnh ca mnh ti tng in quay tr v tri t .

- Tng in ly

Tng in ly hnh thnh ti cao 100Km - 400Km l do kt qu ca viho trng thi ca tng i lu bng cc tia cc tm v tia X do mt tri bc x

in ly c phn chia thnh mt vi lp c gi tr mt in t cc i. Mc phn chia thnh cc lp D, E, Ftheo cao ca n. Lp F li c phnchia thnh lp F1, F2. Hnh 1.5 trnh bymt tnh theo cao ca cc lp ionin hnh.

- Truyn sng trong lp Ion

C th xem lp ion ca tng inly ging nh mt tm in mi khng lm h s khc x ca n bin i lintc, v s bin i ca mt ion theo bc sng l khng ng k trong bng tn s cao (bc sng ngn hn). Hntrnh by ng i ca sng v tuyn trong tng in ly. C th gii thch hin phn x ti tng in ly nh sau: Ta chia tng in ly thnh cc lp rt mng

trong lp chit sut c th coi l u. Nh vy khi truyn t lp Ion ny saIon khc th tia sng in t b khc x. Qua nhiu lp Ion nh vy tia sng i

7

600

400

200

0

Mt (cm3)

Ngym

Lp D

Lp E

Lp F LpF1

LpF2

10 102 104 106

Hnh 1.5 Mt tnh theo cao ca cc lp ion

R KR

(a) (b)

ng thc t

ng trc tip

h1

h1

h2

h2


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b un cong v coi nh phnx quay v tri t. Nu tmt my thu im tia sngin t tr v th ta s thuc thng tin pht i t my pht.

V c tnh ca tngin ly thay i theo caonn chng c nh hng khcnhau n qu trnh truynsng in t. cc sng in t

c tn s cao qu s i xuynthng qua tng in ly (Hnh1.6b). Sng in t c tn sthp qu th kh nng chuynti nng lng yu, s khngn c tng in ly. tngin ly ch phn x sng int trong bng sng ngn vsng trung. Bng thc nghimngi ta xc nh crng: sng ngn phn x tilp F1 v F2, sng trung phn x ti lp E, Lp D khng gy ra phn x sng ngc li lp D hp th nng lng sng in t trong bng sng trung. Tuy lp D ch xut hin vo ban ngy, v nh vy ban ngy khng th dng sng truthng tin qua tng in ly c.

Bng thc nghim ngi ta xc nh crng, kh nng phn x sng in t ti tng in ly ph thuc 3 yu t chnh l: Mt cc ht mangin (N) trong tng in ly, tn s v gc ti ca tiasng in t.

Gi s tia sng in t ti tng in ly vigc ti 0 (Hnh 1.7), sng in t ny phn xc ti tng in ly th0 phi tho mn iu kinsau:

8

n1n0

n2

n3n4

nn

Tia sng t tri t phtti tng in ly

Tia sng khc x v i ratng in ly tr v tri t

Tngin ly

TngBnh lu

(a)

Mt t

Khong nhy

Tng in ly

Sng v tuyn xuyn qua tng in ly

Sng v tuynkhc x ti hn

Hnh 1.6.- C ch phn x ca tng in lya. Khc x trong tng in lyb. iu kin ti hn c phn x

(b)

Tngin ly

0

Gcti

Gct

Tip tuyntri t

Tiati

Tia phn x

Mt t

Hnh 1.7


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20 8,801sin f N

2028,80

1 f

N Sin

202 8,8011

f N

Cos = 0

9

Cos N

f

Khi N=Nmax th tn s ny c gi l tns ln nht fmax0 ng vi 0 . Tn s ln nhts t c khi N=Nmax v0 = 0 max.

Khi 0 = 0 max th tia ti chnh l tip tuynca tri t. tuy nhin ta rt kh c th xc nhchnh xc c gc 0 , bi vy ngi ta tm cch

tnh tn s f thng qua gc, v v0

quan hvi nhau nh sau: Khi0 = 0 max th =0, khi 0 =0 th=90o.

Vic tnh tn s f thng qua gc c mt hnh 1.8. Trong a l bn knh tri t, h l cao im phn x,0 : gc ti.:gc t.

Ta c:

ah

Sin

ah

N f

2

)2

1(.8,80)(

2

max

max

+

+=

Khi 0 = 0 max th =0 nn Sin2 0 =0.

Suy ra:h

ha N f

2

)2(.8,80)( maxmax

+=

Phng thc truyn sng qua tng in ly c ng dng trong truyn Bng cch cho sng in t phn xnhiu ln gia tng in ly v mt tnh m t hnh 1.9 ta c th truync sng in t i rt xa. Khi truynnh vy s c nhng vng khng nhnc sng in t, vng c gi

9

Tri t

Tng in ly

Vng tiVng ti

Khong nhy

Hnh 1.9

Tngin ly

0

Gcti

Gct

Tip tuyntri t

Tiati Tia phn x

Mt t

Hnh 1.8

h

a

O

a

Tri t

Tng in ly

Hnh 1.10


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l vng ti. khc phc hin tng ny ngi ta pht ln tng in ly chmin t phn k nh m t hnh 1.10.

c. Sng tn x qua tng i lu

Vic lan truyn ca sng v tuyn nh hiu ng tn x ca vng kh quyn ng nht trong tng i lu c gi l truyn sng tn x qua tng Phng php ny cho php thng tin lin lc c ly xa cc bng tn VHF, UHSHF.

Hnh 1.11 m t bn cht ca victruyn sng tn x qua tn i lu.

Trong tng i lu, cao 10 n

12 Km, thng c vng khng khkhng ng nht. Trong , chit sutgia cc lp khng kh thay i rtln, nn khi i vo vng ny sng int b tn x ra nhiu hng khc nhau.Mt s tia b 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 kng nht thnh cc lp rtmng sao cho trong lp chit sut ca khng kh c thcoi l u. Nh vy khi truynt lp khng kh ny sang lpkhng kh khc th tia sngin t b khc x b khc x.

Qua nhiu lp kh quyn nhvy tia sng in t s b uncong v quay v tri t. Nut mt my thu im tiasng in t tr v th ta s thuc thng tin pht i t my pht.

Phng php ny i hi cng sut pht ln v my thu c nhy cao. Mttnh n nh ca h thng thng tin ny khng cao do vng kh khng ng thng xuyn thay i theo ngy - m, theo ma . . .

10

n1n0

n2n3n4

nn

Tia sng t tri t pht ti vng khkhng ng nht

Tia sng khc xv dn dn v quay

tr v tri t

Vng khkhng ng

nht

Hnh 1.12: Khc x trong vng kh quyn khng ng nh

Mt t

im pht im thu

Vngkhng kh

khng ngnht

Sngin t b

tn x

Hnh 1.11: Truyn sng tn x trong tng i lu


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d. Truyn sng thng

L phng thc truyn sng trong khng gian t do. Phng thc ny chtrong h thng thng tin v tinh, khi sng in t i ra ngoi tng kh quyn c

t.III. Tng quan v h thng thng tin vi ba

1. Khi nim v thng tin vi ba

Thng tin vi ba l thng tin v tuyn gii sng cc ngn v thc hin ttin nhiu knh.

Hin nay cc h thng thng tin vi ba c ng dng rt rng ri truyhiu trong mng thng tin. K thut thng tin vi ba cng c pht trin mch ngy cng ph hp vi nhu cu thng tin trn mng. Ngy nay hu ht cc thi ba c s dng l thit b vi ba s dung lng cao. Ngoi cc h thng vi bmt t cn c cc h thng vi ba v tinh vi c li thng tin ln ti hng chcKm. Cc tn hiu truyn trn ng thng tin vi ba bao gm cc tn hiu thoi, thnh, s liu, . . .

2. Cu trc gi nh ca tuyn thng tin vi ba

Hnh 1.13 m t s khi gi nh ca mt tuyn vi ba trn mt t (cl h thng v tuyn chuyn tip c nh).

Trm u cui cn c mt my pht, mt my thu, cc thit b ghp knmt anten. Nu h thng vi ba s ny c s dng truyn tn hiu s c ss c tc 2 Mb/s) v lung s c ly t tng i hoc t thit b ghp kcao hn th trm u cui khng cn thit b ghp knh. Cc trm u cui tc t cc thnh ph ln. Trm u cui c nhim v:

- hng pht: nhn thng tin t cc trung tm truyn hnh, truyn s

tng i hoc cc thu bao sau tin hng ghp knh to thnh mt lung

11

TxRx

Rx Tx

Ghpknh

TxRx

Rx Tx

Ghpknh

TxRx

Ghpknh

TxRx

Trmu cui

Trmlp Trm

r xenTrm

u cui

Hnh 1.13: S khi gi nh ca h thng v tuyn chuyn tip c nh


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cao v a vo my pht iu ch vo sng mang siu cao tn pht sangi phng.

- hng thu: thu nhn thng tin l tn hiu sng mang siu cao tn t

i phng sau tin hnh gii iu ch to li lung s nguyn thu vmy phn knh x l thnh cc tn hiu ring bit a v cc trung tm trtruyn hnh, truyn s liu, tng i v thu bao.

Trm trung gian cn c hai my pht, hai my thu v hai anten thng tinhai pha khc nhau.Cc trm trung gian thng c t nhng t c s giathng tin, kinh t v.v. Trm trung gian c nhim v thu nhn thng tin l tn sng mang siu cao tn t trm pha trc sau tin hnh gii iu ch 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 thu, cc my ghp tch knh v hai anten thng tin vi hai pha khc nhau.Ctrm r xen thng c t cc th x v cc ni c khu cng nghip lntrung gian c nhim v thu nhn thng tin l tn hiu sng mang siu cao tn t pha trc sau tin hnh gii iu ch to li lung s nguyn thu v my ghp xen knh ly ra v/hoc xen vo mt s knh sau a ti m pht tip sang trm pha sau.

3. c im v phn loia) c im

Thng 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 rng).

- Thng tin n nh.

- Vi mt c ly thng tin chotrc th c th p dng cng nghanten v tng tn s cng tc gim nh cng sut my pht.

r hn ta xt biu thctnh sut in ng u vo camy thu nh sau:

( )

=m

mV km R

G P Eth P .173

12

10 30 100 300 103 104 105

E(v)

F(MHz)

Nhiu kh quynv nhiu cng

nghip

Nhiu v tr

Tp m ni

Hnh 1.14:Nhiu v tp m ni ph thuc tn s.


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Trong :

Eth: Sut in ng ca tn hiu ti u vo my thu, P: Cng sut c pht, R: Khong cch thng tin gia hai trm, GP: H s tng X (H s khuc

ca anten, Shd G P 24 = , 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, cng nghip.

Hnh 1.14 biu din mc nhiu v tp m ni ph thuc tn s. t thy: tn s thp hn 30 MHz th nhiu v tr, nhiu kh quyn, nhiu cngln. T 30 n 300 MHz ch cn nhiu v tr, t 300 MHz tr ln khng cn v tr na nhng tp m ni ca thit b s tng ln. Do cc phn t cao tncc thit b thu pht phi c cu trc c bit c th lm vic n nh tn

- Kh nng nhiu x ca sng in t tn s cao rt yu nn ch thc hin thtrong tm nhn thng.

b) Phn loi

C nhiu phng php phn loi ng thng tin vi ba. Sau y ta xt m 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 to13


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4. Phn b tn s trong thng tin vi ba

ng thng tin vi ba tm nhn thng thng c nhiu trm chuyn tipv vy nu ta cho tt c cc my pht trn ng thng tin hot ng cng m

s th s xy ra hin tng can nhiu, tn hiu ca trm ny s gy nh htrm kia lm gim cp cht lng thng tin. khc phc hin tng nng thng tin vi ba ngi ta b tr cho cc my pht lm vic cc tn snhau thch hp m bo cht lng cho ng thng tin. Vic quy hoch cho mt h thng c gi l phn b tn s. Hin nay, h thng thng tin vdng mt s phng php phn b tn s nh sau:

a) H thng s dng 2 tn s

Hnh 1.15 m t phng php phn b 2 tn s. t hnh v ny ta thy: Tothng ch s dng 2 tn s, my pht v my thu ca cc trm k tip nhahng i thay nhau lm vic ti cc tn s f 1 v f 2, my pht v my thu trn hngv ca cc trm k tip nhau thay nhau lm vic ti cc tn s f 2 v f 1.

Phng php ny tng i n gin, di tn ca anten khng cn lnTuy nhin, h thng ny c hin tng thu vt trm ca my thu, c ngha lthu ca mt trm thu c tn hiu pht ra t nhiu my pht ca cc trm phaMt khc, c th xy ra hin tng thu theo hng ngc li do sng in t p

tr li khi gp cc vt chn. Ngoi ra, do h s phng v ca anten khng cann 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 pv t nht l 70 dB.

b) H thng s dng 4 tn s

14

My phtf1

My thu

f2

My phtf2

My thu

f1

My thuf1

My pht

f2

My phtf1

My thu

f2

My thuf2

My pht

f1

f1

f1

f1

f2

f2

f2

Trm u cui Trm trung gian Trm trung gian

Hnh 1.15: Phng php phn b 2 tn s.


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My phtf1

My thuf3

My phtf3

My thuf1

My thuf4

My phtf2

My phtf4

My thuf2

My thuf3

My phtf1

f1

f4

f1

f3

f2

f3

Trm u cui Trm trung gian Trm trung gian

Hnh 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: Tothng s dng 4 tn s, my pht v my thu ca cc trm k tip nhau trn hnhau thay lm vic ti cc tn s f 1 v f 3, my pht v my thu ca cc trm k tinhau trn hng v thay nhau lm vic ti cc tn s f 2 v f 4.

Phng php khc phc c hin tng thu theo hng ngc li, choanten c phng v khng cao. Tuy nhin, phng php ny vn cn c hin tthu vt trm.

khc phc tnh trng ny trong c hai phng php phn b tn s ng

b tr cc trm trn tuyn sao cho khng nm trn cng mt ng thng.c) Phng php chia i gii tn

Nu mt s knh v tuyn c s dng trong cng mt tuyn truyn d

bng tn xc nh c chia lm 2 bng tn con: Bng tn cao (Nhm tn sUper Band) v bng tn thp (Nhm tn s thp : Lower Band) nh m t 1.18. Mi mt knh v tuyn s dng phng php phn b hai tn s.

15

TxRx

Rx Tx

Ghpknh

TxRx

Rx Tx

Ghpknh

TxRx

Trmu cui

Trmlp

Trmr xen

TxRx

Rx Tx

Trmlp

Hnh 1.17: B tr cc trm trn tuyn khng nm trn cng mt ng thng.

f0Bng tn thp Bng tn cao

f1 f1f2 f2fn fn

X Y X: Khong cch knh ln cnY: Khong cch thu pht

Hnh 1.18: Mt v d tiu biu v phng php chia i gii tn


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CHNG II.

ANTEN, FEEDER (FI) V CC PHN T SIU CAO TN

I. Cc loi phi (feeder)

1. Nhim v v yu cu

a. Nhim v

Feeder l phn t trung gian gia my pht hoc my thu v anten. Nhimca dy feeder l dn nng lng cao tn t u ra ca my pht ti anten v tra ca anten ti u vo my thu.

b. Yu cu

16

My phtf1

My thuf1

My phtf2

My thuf2

My phtfn

My thufn

My phtf1

My thuf1

My phtf2

My thuf2

My phtfn

My thufn

My phtf1

My thuf1

My phtf2

My thuf2

My phtfn

My thufn

My phtf1

My thuf1

My phtf2

My thuf2

My phtfn

My thufn

My phtf1

My thuf1

My phtf2

My thuf2

My phtfn

My thufn

Hnh 1.19: T chc nhiu knh v tuyn trn cng mt tuynp dng phng php chia i gii tn


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- Khng c bc x v thu sng in t, hay ni cch khc l khng c ng anten. V nu Feeder c hiu ng anten th ngoi vic nh hng n 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 nikhc 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 vkhng vo ca my thu.

- C bn c hc cao (Kt cu chc chn) khng b bin dng trong ckin thi tit.

- Cc thng s ca Feeder phi n nh.- Phi m bo tnh kinh t.

2. Phn loi fi

C hai loi feeder l feeder cp ng trc v feeder ng dn sng. Trong 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 WC

4. Cc loi fi thng dng

a) Fid cp ng trc

a1. Cu to: Feeder cp ng trc c cu to nh m t hnh 2.1.

17

4 3 2 1

Hnh 2.1 : Cu to ca Feeder co ng trca. Hnh di dng phi cnh, b. Hnh di dng mt ct

4 213

Dd


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- Dy dn trong 1: cn gi l l li c ch to bng vt liu dn ithng 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 d

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

bc.

- Lp v bo v 4: c ch to bng Poliethylen hoc cao su.

a2. Nguyn l truyn sng

Khi a nng lng cao tn vo cp ng trc th gia 2 dy dn tronngoi s pht sinh sng in t chy dc theo dy dn, do s thc hin truyn nng lng (cn gi l truyn sng) t my pht ti anten v t anten tthu. Tr khng c tnh ca feeder cp ng trc c tnh theo theo cng thc

)(lg138 =

d D

W C

. Trong D l ng knh mp trong ca dy dn ngoi, d

ng knh ca dy dn trong, l hng s in mi ca cht cch in gia dy d

trong v dy dn ngoi. Thng thng=1 nn )(lg138 =d

DW C .

Cc loi feeder cp ng trc tiu chun trong cng nghip thng ckhng c tnh nm trong khong 40 n 150. Trong , loi feeder vi tr khngc tnh bng 50 c bn v in cao nht (Chu c dng v p cao tnnht)loi ny c D/d=2,7. Feeder vi tr khng c tnh bng 50 thng c sdng trong thng tin vi ba. Loi cp ng trc vi tr khng c tnh bng 75 c hp th nng lng nh nht nn thng c s dng truyn tn higc (tn hiu truyn hnh, tn hiu s .v.v.) loi ny c D/d=3,6.

a3. u nhc im- Cu to tng i dn gin, tr khng c tnh bin ng trong phm vi lnth ch to cc loi co ng trc giao tip vi nhiu loi thit b c tr khnkhc nhau, bng tn cng tc tng i rng.

18


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- Khi tn s cao qu 2 GHz th suy hao trn cp ng trc rt ln, do khng thdng cp ng trc dn tn hiu c tn sln hn 2GHz, khi cn s dng Feeder ngdn sng.

b) Fi ng dn sng

b1. Cu to

ng 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 ngc 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 ccon dy ngn mch u cui c chiu di/4.Khi t cc on dy st vo nhau th s tothnh ng dn sng. Xem hnh 2.3.

V on dy c chiu di/4

ngn mch u cui nn tr khngvo ch ni vi ng dy bng vcng nn khng nh hng n ch cng tc ca ca ng y v do khi cc on y c chiu di/4t st vo nhau s to thnh mt hpkim loi, hp ny khng nh hngn qu trnh truyn sng trn ngdy.

b2. Nguyn l truyn sng,iu kin sng in t truyn ctrong ng dn sng.

Nu kch thch vo ng dn sng mt nng lng siu cao tn th s cin t truyn lan trong ng dn sng. Sng in t truyn lan trong ng dn bng cch phn x nhiu ln trn thnh ng nh m t hnh 2.4.

19

b

a

d

d2

d1

Hnh 2.2 Cu to ca ng dn sng cv theo mt ct.

Mt trong c trngnhn

Mt trong ctrng nhn

Mt trong ctrng nhn

ng kimloi rng

ng kimloi rng

ng kimloi rng

/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 trnmt ng dy song hnh to thnh ng dn sng


20/152

Sng in t truyn trong ng dn sng gm 2 loi: Sng in ngang T(Cc thnh phn in trng nmngang) v sng t ngang TMmn(Cc thnh phn t trng nmngang), m-n l cc ch s cc loisng khc nhau. Trong mi loisng li c chia lm nhiu loikhc nhau gi l cc Mode. Ccmode ny c xc nh bi ch sm-n (m, n = 0,1,2,3 .v.v. v khng ng thi bng 0). Cc loi sng in t trutrong ng dn sng cn ph thuc vo tn s cng tc v kch thc ca n

sng. Mi ng dn sng c kch thc khc nhau th c mt bc sng ti h th)khc nhau. Do cu to ca ng dn sng nn bc sng ti hn c gi tr bn di chiu rng ca ng. iu c ngha l cc sng in t c bc shn hai ln kch thc chiu rng ca ng dn sng th khng th truyn ng .

Bc sng ti hn ca ng dn sng c xc nh theo biu thc :

22

2

+

=

b

n

a

mth

Trong : a l kch thc chiu rng ca ng; b l kch thchiu 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:

a

aba

thTE 212

01

22210

==

+

=

b4. u nhc im

- C cu trc chc chn, truyn c cc sng in t tn s cao vi sutng i thp.

- Ch to kh, tn s cng thp th kch thc ng phi cng ln, yu ciu kin bo qun tng i nghim ngt (Trong lng ng phi lun kho r bm kh kh-sch hoc kh tr vo lng ng).

20

Hnh 2.4: Nguyn l truyn sng trong ng dn sng

Phn t kch thch a sng in t vo trong ng dn

sng

ng dn sngmt u bt kn


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c .

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 (

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 lcc 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 ln

dn sng v ni u kia vi ngun tn hiu siu cao tn nh m t hnh 2.621

V bo v

Dy dn ngoi

Dy dn trong(u gi)

Lp in mi

Hnh 2.5: Cu to ca lng ccin (gi kch thch)

Ngun tnhiu cao tn

intrng

Ttrng

Hnh 2.6: Nguyn l kch thch bng lng cc in

V bo v

Dy dn ngoi

Dy dn trong(u gi)

Lp in mi

Hnh 2.7: Cu to ca lng cct (Vng kch thch)

Ngun tnhiu cao tn

intrng

Ttrng

Hnh 2.8: Nguyn l kch thch bng lng cc t


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ca u gi (Dy dn trong) hp vi dy dn ngoi to thnh mt t in c grt ln, khi c ngun siu cao tn th gia 2 bn t ny s c ng sc in bin thin, in trng ny to ra t trng bin thin v chng tip tc chuyln 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. nhlng cc t c mt vng dy c bit ni tt gia dy dn trong v day dnVng 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 lngdn sng v ni u kia vi ngun tn hiu siu cao tn nh m t hnh 2.8. ngun siu cao tn th trong vng dy s c dng in, v nh vy xung quanh t trng bin thin, t trng ny to ra t in bin thin v chng tip tc 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 v

Anten l mt thnh phn khng th thiu c trong thng tin v tuynchung v thng tin vi ba ni ring. Anten c nhim v bc x (Pht) v thu sng t.b. Yu cu

+ C kh nng tp trung nng lng v mt hng (Hoc mt s hngyu cu.

+ Kh nng bc x v thu sng in t ca cc hng ph v hng ng

rt nh.+ C kt cu chc chn.+ C gii tn cng tc rng.+ m bo tnh kinh t

2. Phn loi antenC nhiu cch phn loi anten:

* Phn loi theo bc sng cng tc ta c: Anten sng di, Anten sng trung, Asng ngn, Anten sng cc ngn . . .

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X

Y

(a)

X

Y

(b)

Hng chnh(Bp chnh)

Hng ph(Bp ph)

X

Y

(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 h

M M

Y

X X

Y

R R

Anten chun Anten khost

Hnh 2.10: ngha ca h s nh hng D

* Phn loi theo hnh dng v cu to ta c: Anten chn t, anten parabol, antengng, 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 hnht nh.- Anten v hng: Cn gi l anten ng hng, nng lng sng in t

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 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 anmt im cch anten mt khong R trn phng kho st v mt nng lx ca mt anten chun (Anten v hng) ti mt im cng cch anten mt k

R v cng trn phng kho st khi cng sut bc x ca 2 anten l nh nhau

23


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Nu gi0 l mt nng lng bc x ca anten chun ti im M, l mt nng lng bc x ca anten kho st ti im M th ta c : D=/0 (ln).

Vi 20..4 R

P

= ta c 2..4

.

R

P D

= Trong P l cng sut bc x ca anten, R l

khong cc t anten n im kho st.

Nu gi P1 l cng sut thu ti im M ng vi anten kho st, P0 l cngsut thu ti im M ng vi anten chun th D=P1/P0 (Ln). Nu tnh theo dB th :

0

101 lg10lg10lg10)( P

P P P decibel D ==

c) Hiu sut ca anten ( ): L t s gia cng sut bc x (cng sut c ch) c

anten v cng sut a vo anten:

= P P bx

a . Trong : Pbx v cng sut bc x ca

anten, P cng sut a vo anten. Thng thng a = 0,750,95.

d) H s khuch i (G)

H s khuch i G ca anten l t s gia mt nng lng bc x canten ti mt im cch anten mt khong R trn phng php st v mt lng bc x ca mt anten chun (Anten v hng) ti mt im cng cchmt khong R v trn phng kho st khi cng sut a vo 2 anten l nh

Nu anten chun c hiu sut bng 100% th ta c G= .D

e) Gc na cng sut

Gc na cng sut l gc hp bi haihng bc x m theo hng cng sut phtgim i mt na so vi hng chnh.

C mt cch nh ngha khc: Gc nacng sut l gc hp bi hai hng bc x mtheo hng cng sut pht gim i 3 dB so vi hng chnh.

f) Tr khng vo/ra

Tr khng vo ca anten l tr khng m tn hiu gp phi khi i vo uca anten. Tr khng ra ca anten l tr khng m tn hiu gp phi khi i ra ca anten.

Khi ta ni anten vo u ra ca my pht hoc u vo ca my thu bngtrc tip hoc gin tip qua dy feeder th tng ng vi vic ta u vo u

24

2

3dB

Y

X

Hnh 2.11: Gc na cng sut 2


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Hnh 2.12 m t cu to ca anten chn t. Cu to ca anten chn t g

- 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 hng x D c chiu di lD 0,87 .

Cc chn t trn c t song song vi nhau trn gi kim loi, vungv cch in vi gi kim loi.

Trong chn t P t cch chn t A mt khong t 0,15 n 0,25 . Chnt D u tin t cch chn t A mt khong t 0,1 n 0,3. Cc chn t D tcch nhau mt khong t 0,25.

Cc chn t c ch to bng kim loi c t thm cao. trnhFuco, cc chn t c ch to di dng cc ng rng. Chn t bc x ngun cao tn hoc my thu.

a2. Nguyn l hot ng

* Nguyn l bc x sng in t

Khi chn t bc x c ni vi ngun tn hiu cao tn th n bc x snt v c hai hng trc v sau theo phng vung gc vi cc chn t. sng bc x v ph sau s c chn t phn x phn x ngc tr li v pha trchn chiu di chn t phn x v khong cch gi n vi chn t bc x thnn sng in t phn x v ng pha vi sng in t bc x ra pha trc. Ct hng x cm nhn nng lng sng in t bc x t chn t bc x v tngun bc x th cp bc x sng in t v pha trc. V c di ngn hcc chn t hng x c kh nng tp trung nng lng cao hn v mt phanhiu chn t hng x th kh nng nh hng cng cao, tuy nhin s lhng n kt cu ca anten.

* Nguyn l thu sng in t

Anten chn t thu sng in t theo nguyn l cm ng in t. Khi c in t t pha i phng pht n th cc chn t s cm nhn nng lng thnh dng in cao tn a ti my thu. Trong , cc chn t hng x cmnng lng sng in t v tr thnh ngun bc x sng in t th cp bc pha chn t bc x tp trung nng lng vo chn t ny lm tng kh nca anten.

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a3. U nhc im

- Bng tn cng tc khng ln do anten ny c c tnh gn ging mt khcng hng.

- H s nh hng khng cao. c h s nh hng cao th yu cu phi c rt nhiu chn t hng x lm cho kch thc ca anten cng knh, c chc chn.

- Khi hot ng tn s cao, chng dng fuco, th yu cu cc chn tch to rt mnh nn rt d gy. Mt khc khi hot ng tn s cao th cc phi rt ngn nn nh hng n kh nng bc x ca anten

b) Anten ng dn sng

b1. Cu toAnten ng dn sng thc cht l mt on ng dn sng c mt u b

u kia ng c c mt phn t kch thch c t vo trong lng ng nh mhnh 2.13.

b2. Nguyn l hot ng

Khi ni phn t kch thch vi ngun tnh hiu cao tn th nng lng

in t s c kch thch vo ng dn sng lc ny sng in t s c tru pha ming ng v i ra ngoi khng gian. Ming m ca ng dn sng ng vnh mt anten bc x sng in ttheo hng vng gc vi ming m ca ng.

b3. u nhc im

- Din tch ming m hp nn

bc x yu.- Tnh phng hng v h s khuch i khng cao.

27

u bit kn

Phn t kchthch

Hng bcx ca sng

in t

Hnh 2.14: nguyn l hot ng ca anten ng dn sng

Hnh 2.13: Cu to ca anten ng dn sng a. Cu to c v phi cnh, b. Cu to c v theo mt ct

u bit kn

Ming antenMing anten

u bit kn

Phn t kchthch


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- Tr khng ca anten v mi trng bn ngoi khc nhau nhiu nn nng lsng in t b phn x ti ming ng dn sng gy ra tn hao v nhiu b ph.

b4. ng dng - Anten ng dn sng ch c dng ln cc b chiu x t trong anten para

c) Anten loa

Anten loa l mt ci tin ca antenng dn sng phi hp tr khng giang dn sng v m trng truyn sng bn ngoi. bng cch m rng ming ng

dn sng (thnh hnh ging mt chic loa)lm cho tr khng ca ng dn sng bini mt cch t t khi chuyn t tronglng ng dn sng ra ngoi khng gian,iu ny c ngha l lm tng phi hp tr khng gia ng dn sng v khng giannn nng lng sng phn x tr li ng sgim, nng lng bc x ra ngoi khnggian s tng. Mt khc din tch mt loa sln hn nhiu lm tng kh nng bc xca anten.

Hnh 2.15 m t cu to ca anten loa. Khi m rng ming ng dn snnh hnh 2.15a th ta c anten loa nn, khi m rng ming ng dn sng theong ta c anten loa E nh hnh 2.15b, khi m rng ming ng dn sng theo ngang ta c anten loa H nh hnh 2.15c, khim rng ming ng dn sng theo c chiungang v chiu ng ta c anten loa thp nhhnh 2.15d.

Hnh 2.16 m t cc tham s cn thitca anten loa. Trong : O l tm loa, OO trcchnh ca loa, L: di ca loa, h: rngming loa (ng kch ming loa), 2o: Gc

m ca loa.

28

(a)

(b)

(c)

(d)

Hnh 2.15: Mt s anten loa

L

2oO O

Hnh 2.16: Cc tham s cn thit ca anten loa

h


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Khi kch thch sng in t vo anten loa th n s bc x sng in t ratrc ming loa. Sng in t ny ta h nh xut pht t mt ngun bc x ti tm loa. Nh vy, sng in t pha trc ming loa khng ng pha tr phng vung gc vi trc chnh ca loa. c mt anten loa c cht lng tt,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: lg20...4lg10 = V S G (dBi). Trong : V l hiu sut ca anten loa so v

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 phcao hn, kh nng phi hp tr khng tt hn. Tuy nhin vic ch to mt an

c kch thc ln khng m bo tnh kinh t v tnh k thut, nn anten loa thc ch to c kch thc nh dng lm cc b chiu x t trongcc anten parabol, parabol 2 gng.v.v. Anten loa thng c h skhuch i nh hn 25 dB.

d) Anten Parabol 1 gng

Li dng tnh cht phn xca gng parabol, ngi ta chto ra anten parabol.

d1. Cu to

Hnh 2.17 m t cu to caanten parabol:

- Gng phn x 1: cn gi l a

phn x. Gng c ch to bng kim loi, b mt nhn c th phn x tth 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 cng v xp thnh hnh parabol.

Mt s tham s ca gng:

+ f: Tiu c ca gng.

+ h: su ca gng

29

1. Gng phn x

2. Chiu x t tti tiu im F

ca gng3. ng dnsng

4. Connector siu cao tn

3. Tmgng

l

h

f

Hnh 2.17 : Cu to ca gng parabol

Minggng


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+ l: Khu ca gng ( rng ming gng)

+ o: Gc m ca gng

- ng phng: ng phng gm mton ng dn sng 2 v mt b chiux t 3. on ng dn sng c mtu c u vi dy Feeder thngqua connecter siu cao tn 4 cn ukia c u vi chiu x t. Chiu xt t ti tiu im F ca gng c 2nhim v chnh l bc x nng lngsiu cao tn thnh chm sng in t phn k pht ti gng v chuynchm sng in t hi t t gngthnh nng lng siu cao tn trongng dn sng.

- Ngoi ta cn c cc c cu gi nh v gng phn x v c nh 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 t

Hnh 2.19 m t nguyn l pht sng in t. Nng lng siu cao tn a t my pht qua dy feeder a vo u vo ng phng. Nnglng siu cao tn ny cchuyn thnh sng in t truyntrong ng dn sng v ti chiu xt, c chiu x t bin ithnh chm sng in t phn ktruyn ti gng phn x. Chmsng ny ta h nh xut pht tmt ngun im t ti tin imca gng. Cc tia sng in tti gng s c phn x to thnh chm tia song song truyn v pha trc

Chm sng in t ny cng song song vi trc chnh ca gng v ng phamt phng vung gc vi trc chnh ca gng.

30

Gng phn x Chiu x t

ng dn sngConnector siu

cao tn

Gi chiu x t

Gi Gng phn x

Gi Gng phn x

Hnh 2.18: Hnh dng y ca mt anten parabol

Tn hiu siu caotn t ng dy

Feeder

Sng

in tc bc xra phatrc

Hnh 2.19: Nguyn l pht sng in t


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Anten pht

Antenthu

Hnh 2.21: Hot ng ca mt h thng anten Parabol trong h thng vi ba

Tn hiu siu cao

tn ti ng dyFeeder Sng

in t pha pht

pht ti

Hnh 2.20: Nguyn l pht sng in t

Sng in t do anten Parabol bc x ch truyn theo mt hng nht nloi anten ny cn c gi l anten bc x n hng.

Ngi ta tnh c h s khuch i ca gng nh

).4..

lg(10)( 2 S

dBG a=

Trong : a: Hiu sut ca anten, S: Din tch ca mt gng,: Bc sng cngtc.* Nguyn l thu sng in t.

Hnh 2.20 m t nguyn l pht sng in t. Nu c chm sng in t t phng pht ti v song song vi trc chnh ca gng th khi ti mt phn xgng chm sng ny c phn x to thnh mt chm sng in t hi t in F, do chiu x t t ti tiu im F nn chm sng in t ny s i vox t, qua on ng dn sng v c chuyn thnh nng lng cao tn i vfeeder ti my thu.

Hot ng ca mt h thng anten Parabol trong h thng vi ba c mhnh 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 d

nguyn l phn x.- C kh nng bc x cng sut ln.

31


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* 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 ph

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 i

i c ra pha sau anten v ngc li sng in t t pha sau anten cng c tti chiu x t .e) Anten Parabol hai gng e1. Cu to

Hnh 2.22 m t cu to ca anten Parabol 2 gng. Anten Parabol 2 ggm c:

- Mt gng Parabol (Gng culm) c ch to bng vt liu phn xtt v t hp th nng lng sng in t.Gng c 2 loi:

+ Gng kn: Gng c mt phnx hon ton kn.

+ Gng h: Gng phn x l mtli kim loi hoc cc thanh kim loic un cng v xo thnh hnh

parabol.- Mt gng Hypecbol (Gng cu

li) c ch to bng vt liu phn xrt tt v t hp th nng lng sng int.

Mt s tham s ca 2 gng:+ f1: Tiu c ca gng chnh.

+ f2: Tiu c ca gng ph.+ h: su ca gng chnh

32

F1,F2

f2

f1

h

l

Gngchnh

GngPhChiux t

Hnh 2.22: Cu to ca anten parabol 2 gng


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1. Gng Parabol(Gng chnh) 3. Chiu

x t

4. ngdn sng


2. Gng Hypecbol(Gng ph)

6. Gi gng ph

1. Gng Parabol(Gng chnh)

3. Chiux t

xa t

4. ng

dn sng


2. Gng Hypecbol(Gng ph)

7. Gi gng chnh

6. Gi gng chnh

Hnh 2.23: Cu to y ca mt anten parabol 2 gng

Hnh 2.24: Nguyn l pht sng in t caanten parabol 2 gng

Nng lngsiu cao tnt my pht

a ti

Sngin t

c bcx ra phatrc

+ 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

tm ca gng ph mt khong d=f 2/2. Chiu x t c 2 nhim v chnh l bc xnng lng siu cao tn thnh chm sng in t phn k pht ti gng pchuyn chm sng in t hi t t gng ph thnh nng lng siu cao tnng dn sng.- Ngoi ta cn c cc c cu gi nh v cc gng chnh v gng ph.dng y ca mt anten parabol c m t hnh 2.23.

33


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e2. Nguyn l hot ng * Nguyn l pht sng in t

Nng lng siu cao tn c a t my pht qua dy feeder vo u vo phng. Nng lng siu cao tn ny c chuyn thnh sng in t truynng dn sng v ti chiu x t, c chiu x t bin i thnh chm sng phn k s cp truyn ti gng Hypecbol c gng ny phn x to thnsng in t phn k th cp truyn ti gng Parabol. Chm sng phn k thny ta h nh xut pht t mt ngun im t ti tin im ca gng HyDo tiu im ca 2 gng trung nhau nn chm sng phn k th cp ny cnh nh xut pht t mt ngun im im t ti tin im ca gng Parabtia sng in t ti gng Parabol s c phn x to thnh chm tia songtruyn v pha trc gng. Chm sng in t ny cng song song vi trc

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 n

loi anten ny cng c gi lanten bc x n hng.

Ngi ta tnh c h skhuch i ca gng nh sau:

).4..

lg(10)( 2 S

dBG a=

Trong : a: Hiu sut ca anten,S: Din tch ca mt gng,:Bc sng cng tc.* Nguyn l thu sng in t.

Nu c chm sng in t ti phng pht ti v song song vi trc chnh ca gng th khi ti mt phca gng chm sng ny c phn x to thnh mt chm sng in t h

tin in F1, do tiu im F1 v F2 ca 2 gng trng nhau nn chm sng ihi t ny cng ta h nh hi t tiu im F2 ca gng 2. ti gng 3 chc phn x v to thnh chm sng in hi t th cp. Do chn im

34

Hnh 2.25: Nguyn l thu sng in t caanten parabol 2 gng

Sngin tt i

phng pht ti

Nng lngsiu cao tnc a

n my thu


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x t thch hp nn chm sng in hi t thc cp i vo chiu x t, qua odn sng v c chuyn thnh nng lng cao tn i vo dy feeder ti my th

Hot ng ca mt h thng anten Parabol 2 gng trong h thng vi bam 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 d

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 thi

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 ph

lng sng in t, to thnh vng ti trc anten.III. van ferit v xiculator :1. Ferit

Khc vi cc vt liu thng thng, Ferit c nhng tnh cht t kh clm cho n c th ng dng rng ri trong thng tin vi ba. Ferit l hn hp cc st, km, mangan, c ban, nhm hoc kn. Ngi ta to ra Ferit bng cch hp cc -xt ni trn vo p trong nhng khun mu nht nh vi mt lc rsau nung nng n nhit cao hn 2000oC. Qu trnh ny lm cho hn hp cc-xt thay i c tnh nh chng tr nn cch in (in tr ni rt cao), h

35

Anten phtAnten thu

Hnh 2.26: Hot ng ca mt h thng anten Parabol 2 gng trong h thng vi ba


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thm tng v c kh nng gii phngt rt nhanh.2.Van ferit

Van Ferit cn c gi l bcch ly, c tc dng cho tn hiu itheo mt chiu v hp th nng lngtn hiu theo chiu ngc li.a) Cu to

Hnh 2.27 m t cu to camt van ferit. Gm c: Mt on ngdn sng 1, trong lng ng t mtthanh Ferit 2 v mt nam chm vnhcu 3 c 2 c p st vo 2 cnh ca ng dn sng sao cho thanh ferit nng chnh2 cc ca nam chm. Nam chm c nhim v nh hng t tnh cho thanh Fngi ta cn gi nam chm ny l nam chm nh thin t. Theo hnh v, t trca nam chm hng t di ln trn.b) Nguyn l hot ng

Khi c tc dng ca nam chm vnh cu th thanh Ferit s b nhim t v tmt t trng c chiu nht nh trong lng ng dn sng. Khi c sng in tng vo th thanh Ferit tip tc nhim t v s bin thin t tnh trn thanhtun theo quy lut bin thin ca sng in t. tuy nhin chiu ca t tnh doferit to ra s khng thay i. Do nu t trng do sng in t to ra tronng dn sng trng vi t trng do thanh Ferit to ra th sng in t d dng ngc li nu nu t trng do sng in t to ra trong lng ng dn sngvi 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 feritgi 2 ng vo ra ca van l A v B th nu sng in t truyn c t ng A B th khng th truyn c t ng B n ng A. Mun thay i chiu truy(truyn c t ng B n ng A) ta phi thay i cc tnh ca nam chm vnh

c) ng dng Lm b cch ly y

ra ca b khuch i cngsut cao trong mt pht vi ba trnh hin tng phn xsng in t khi ng ra camy pht b mt phi hp tr khng.

3. Xiculator

36

Thanh Ferit 2

ng dn sng 1

Nam chm vnh cu 3 Mi tn ch chiu t trng ca Nam chm

Hnh 2.27: Cu to c bn ca phn t cch ly

S

N

Ming FeritCng 1

Cng 2

Cng 3 C cu bngkim loi

Hnh 2.28: Cu to ca Xicuilator


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a) Cu toXiculator cn c gi l b phn hng siu cao tn, l mt c cu c 3

4 cng vo/ra nng lng. Hnh 2.28 m t mt Xiculator c 3 li vo ra nng l Nng lng i vo ng th nht s i ra ng th hai, nng lng i vo ng ths i ra ng th ba v nng lng i vo ng th ba s i ra ng th nht.

Cu to ca Xiculator gm mt kt cu bng kim loi c 3 cng vo/ra, t mt li Ferit, ngoi ra cn c mt nam chm vnh cu lm nhim v nh thicho li Ferit (Khng v trong hnh).b) Nguyn l hot ng

Khi c nh thin t, li Ferit s b nhim t v to ra mt t tnh c cnht nh. Trong trng hp li ferit c dng hnh tr trn nh hnh 2.28 ttrng c chiu xoy trn xung quanh li ferit. Khi sng in t c axicuilator th li ferit tip t b nhim t di tc dng ca t trng do sng to ra. Tuy nhin chiu ca t trng do li Ferit to ra khng thay i. Nu tnc a vo ng th nht th t ng 1 sang ng 2 ng sc t do li Ferit cng chiu vi ng sc do sng in t to ra nn tn hiu d dng i qua, csang ng 3 ng sc t do li Ferit to ra ngc chiu vi ng sc t doin t to ra nn tn hiu khng th i qua. Tng t nh vy, tn hiu s vo2 ra ng 3 v vo ng 3 ra ng 1.

c) ng dng: Xiculator c ng dng lm b phn hng siu cao tn troDIPLEXER cho php my pht v my thu s dng chung mt anten.

CHNG III

X L TN HIU TRONG THNG TIN VI BA S

I. Khi phc xung nhp

1. Khi nim

H thng vi ba s ni ring v cc h thng x l tn hiu s ni chung cn pxung nhp hot ng. cc h thng hot ng ng b th cc xung nhp pht v pha thu phi ng b vi nhau, c ngha l cc tn hiu xung nhp pha

37


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v pha thu phi cng tn s v ng pha. C nhiu phng php thc hicu trn:

Phng php 1: dng hai b dao ng to xung nhp c cng tn s, mt

to xung nhp pha pht, mt b to xung nhp cho pha thu. Tuy nhin, do cdao ng ny hot ng c lp vi nhau nn nhng ni khc nhau chng stc ng bi cc yu t khc nhau ca mi trng, dn n tn s v pha cakh ng b vi nhau, lm h thng hot ng khng chnh xc.

Phng php 2: Dng mt ng truyn dn truyn tn hiu xung nh pha pht n pha thu. Cch ny tuy p ng c yu cu v xung nhp nh phi tn thm kinh ph xy dng thm ng truyn m kinh ph ny khngnh. Phng php ny thng c s dng khi dung lng ng truyn chtng i ln. Vic s dng mt phn ng truyn truyn tn hiu xunkhng 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 sxung nhp ln lung s. pha thu, khi thu c tn hiu s da vo cc tham to li xung nhp. Qu trnh to li tn hiu xung nhp bn thu theo phntrn c gi l khi phc xung nhp. Mch thc hin cng vic ny c mch khi phc xung nhp. Hin nay hu ht cc thit b thng tin vi ba s

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 HDB3

a. 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 nng 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

dng phn p to inp chun a vo chn (-)ca cc b so snh N1 v N2.

Cc in tr R c nhim

v hn ch dng bo vcc mch so snh N1 v N2.

38

BA1

N1

N2

N3BA2BA2OR

ICTnhiu NRZ Xung nhp

khi phc

Tn hiuHDB3

Hnh 3.1 S khi ca mch khi phc xung nhp.


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+ IC: mt mch tch hp chuyn dng lm nhim v bin i tn hiu t Hsang NRZ.

+ Cng OR: c nhim v cng hai tn hiu ng ra ca N1 v N2 li vi

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 N3.

b. Nguyn l hot ng

Nguyn l hot ng ca mch c th m t nh sau:

Tn hiu s vi m ng HDB3 c a vo trn hai u cun s c bin p BA1. Tn hiu ny c ghp qua bin p v t vo cc chn khng ca cc b so snh N1 v N2 (cun th cp ca bin p c im gia ni t) N1 v N2 tch thnh cc lung HDB3 (+) v HDB3 (-). Cc lung HDB3 (+)HDB3 (-) sau N1 v N2 c a vo IC bin i t m ng HDB3 sanmy NRZ n cc, ng thi cc lung s sau N1 v N2 c cng vi nhau qua cng OR to thnh mt tn hiu c nhiu chuyn i mc kch thckhung dao ng LC. Khi c kch thch th khung LC s dao ng vi tn s

bng tc ca lung s u vo.V d: Nu lung s u vo l 2,048Mb/s th khung dao ng LC ph

tn s cng hng ring l 2,048MHZ, nu lung s u vo l 34,368Mbkhung dao ng LC phi c tn s cng hng ring l 34,368Mb/s.

Tn hiu daong hnh sin cghp qua bin p BA2a vo mch so snh N3, c N3 so snhvi in p chun 0v(chn o (-) ca N3

ni t). Kt qu sau N3 ta thu c dy

39

XungHDB3vo

XungHDB3+

XungHDB3-

Xungsau cngOR

Tn hiusin trnLC

Xung CK ckhi phc

t

t

t

t

t

t

Hnh 2.2 Dng sng ca mch khi phc xung nhp.

U


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xung nhp c tn s ng bng tc ca lung s u vo. Xung nhp ny dng nhp cho cc mch pha thu.

Hnh 2.2 m t dng sng qua cc phn t mch ca mch khi phc xun

t m ng HDB3. II. Ngu nhin ho v gii ngu nhin ho

1. Cc khi nim v lung s

a. Lung s khng ngu nhin

Lung s khng ngu nhin l lung s c nhiu chui bit 0 hoc/v n

chui bit 1 xut hin c tnh quy lut (hnh 3.2a). Lung s khng ngu nhic gi l lung s c t chuyn i mc.

b. Lung s ngu nhin

Lung s ngu nhin l lung s c cc bit 0 v bit 1 xut hin mt cchnhin. Lung s ngu nhin cn c gi l lung s c nhiu chuyn i m cc chuyn i mc chuyn i khng theo mt quy lut no c (hnh3.2b).

c. Lung s gi ngu nhin

40

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

t1

t

t

U

Hnh 3.2a: Lung s khng ngu nhin

1 0 0 1 0 1 1 0 0

1 1 1 0 1 1 0 0 1 0 1 1 0 1 0 1

0 1 0 1 1 0

1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1

t0

t

t

U

Hnh3.2b : Lung s ngu nhin


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Lung s gi ngu nhin l lung s m cc bit trong mt nhm bit xutngu nhin, nhng nu xt quan h gia cc nhm bit th cc bit xut hin k

ngu nhin m c tnh quy lut. Lung s gi ngu nhin cn c gi l lungnhiu chuyn i mc trong cc chuyn i mc chuyn i theo mt qunht nh (hnh 3.2c).

2. Ngu nhin ha

a. Khi nim

Ngu nhin ho l bin lung s khng ngu nhin thnh lung s ngu nl php x l s liu s c thc hin trong my pht Vi ba s, sau khi ghp kv tuyn v trc khi iu ch.b. Mc ch

+ Tng chuyn i mc ca lung s pha thu d dng khi phc xungCc lung s nguyn thu c ly ra t my ghp knh hoc tng i thng chui bit 0 lin tip do knh thoi b trng, hoc cc chui bit 1 do ting tholn, nn s chuyn i mc trong lung s rt t v do pha thu kh khi xung nhp, nn cn phi ngu nhin ho tng s chuyn i mc ca lung

pha thu d khi phc xung nhp.+ Chng nhiu: Nu lung s c nhiu chuyn i mc th ph tn ca n

tri u trong ton bng ( ph ca tn hiu sau iu ch cng ng u tron bng ( t b nh hng ca nhiu, nn pha thu thu tt. Nu lung s c t chuymc th ph ca n ch tp trung vng tn s thp, nng lng vng tn sthp ( sau khi iu ch nng lng khng ng u trong ton bng ( nhiu ng vo vng nng lng thp lm cho pha thu d thu nhm tn hiu.

c. Nguyn l

41

1 0 0 1 1 0 1 1 0 0 1 1 0 1

t

t

t

U

0 0 1 0 1 0 1 0 0 1 0 1 0 1

1 1 0 1 0 1 0 1 1 0 1 0 1 0

Hnh3.2c : Lung s ngu nhin


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0 0 0 1 1 0 1 0 0 0 1 1 0 1

0 0 0 1 1 0 0 1 1 1 0 0 0 1

0 0 0 0 0 0 1 1 1 1 1 1 0 0

U

Tn hiucn ngu nhin ho

A

Tn hiugi ngu nhin B

Tn hiu ngu nhin ho

S

t

t

t

Hnh 3.3b Biu thi gian ca mch ngu nhin ho

- thc hin ngu nhinho, trc ht phi to c tnhiu gi ngu nhin ng bvi tn hiu nguyn thu(ccng tc v cng pha vilung snguyn thu).

- Vic ngu nhin ho c thc hin bng cch: em tn hiu cn nguho cng modul hai vi tn hiu gi ngu nhin thng qua mt cng EX-OR. Skhi 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 ha

a. Khi nim

- Gii ngu nhin ho l bin lung s c ngu nhin ho phathnh lung s khng ngu nhin.

- Gii ngu nhin ho l php x l s liu s c thc hin trong my t 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 nhin ho.

b. Nguyn l

- thc hin giingu nhin ho trc ht phi

42

Tn hiugi ngu nhin B

Tn hiu gii ngu nhin

ho A

Tn hiu ngu nhin ho

S = AB

Hnh 3.4a S khi ca mch gii ngu nhin ho

Tn hiuGi ngu nhin B

Tn hiu NGU NHIN

HOS = A B

Tn hiu cn ngunhin ho A

Hnh 3.3a S khi ca mch ngu nhin ho


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0 0 0 1 1 0 1 0 0 0 1 1 0 1

0 0 0 1 1 0 0 1 1 1 0 0 0 1

0 0 0 0 0 0 1 1 1 1 1 1 0 0

Tn hiu ngu nhin

ho S

Tn hiugi ngu nhin B

Tn hiu gii ngunhin ho A

t

t

t

U

Hnh 3.4b Biu thi gian ca mch gii ngu nhin ho

to c 1 tn hiu gi ngu nhin ging ht v ng b vi tn hiu gi ngu bn pht.

- Vic gii ngu nhin c thc hin bng cch a tn hiu gi ngu bn pht pht ti cng EX-OR vi tn hiu gi ngu nhin thng qua 1 cngOR.

S khi ca mch gii ngu nhin c m t nh sau:

ng ra ca cng EX - OR ta thu c lung s l SB

V: S = AB ta c:SB = ABB = A (BB) = A 0 = A

- Biu thi gian m t nguyn l hot ng ca mch gii ngu nhin hom t hnh 3.4b.

4. To tn hiu gi ngunhin

Mt trong nhng phng php to tn hiu gi ngunhin n gin nht cm t hnh 3.5. Mch ghidch gm n Flip-Flop mcni tip vi nhau. Ngra ca Flip-Flop cuicng v mt ng ra ca

Flip - Flop th k chi tip v u vo ca

43

F F

2

FFk

FFn-1

FFn

Tn hiu gingu nhin

CLK

EXOR 1EXOR 2 1

F F

1

HNH 3.5: Mch to tn hiu gi ngu nhin s dng nFlip - Flop

S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

Tn hiu ra

Xung nhp vo

EX- OR1 EX- OR2

Tnhiuvo

HNH 3.7: Mch gii ngu nhin ha t ng b


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OR2 c biu thc ton hc l S = AB = A S7 S10. Tn hiu ny c a voiu ch pht sang pha thu, ng thi c s dng to tn hiu gi ngcho mch.

Hnh 3.7 m t mch gii ngu nhin ho t ng b s dng mch to tgi 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 tn hiu gi ngu nhin. D dng nhn thy rng tn hiu gi ngu nhin 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 qua cng X-OR1. Sau cng X-OR1 ta thu c tn hiu:

SB = (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 sai

1- Khi nim

Cc h thng thng tin vi ba s hin nay thng s dng phng phpch pha PSK. Phng thc iu ch ny thng tin ca lung s c truyn trng thi pha ca sng mang v tuyn. pha thu, thc hin gii iu ch liu cn phi c sng mang ging ht bn pht c v tn s v pha gi l snchun. Khi c sng mang chun th vic gii iu ch c thc hin bng snh pha ca sng mang chun vi pha ca sng mang iu ch bn phtti. Bnh thng pha thu khng th khi phc c sng mang chun. c

sng mang chun th pha pht cn phi pht n pha thu mt sng mang khngch (sng mang chun).

Tuy nhin, vic pha pht pht ti pha thu sng mang chun s lm gim sut truyn dn gia cc trm vi ba s do sng mang chun khng mang thntrong khi li chim mt phn nng lng tng i ln trong s nng lnmy pht pht i. Mt khc, khi sng mang khng iu ch c pht sang phth pha ca sng mang ny c khi khng cn chun na do chu nh hng ca

s yu t trn ng truyn.

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/2 0 1 /2 0 1

1 1 1 1

3/2

1 0 3/2 1 0

IV. m pht hin v sa li

1. Khi nim

truyn dn s ngi ta thng o cht lng truyn dn bng t s bit lBER 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, nhnng truyn dn lun lun thay i nn khng th gim t l ny xung 0. Ngngi ta phi chp nhn mt s lng li nht nh v tm kh nng khi pthng tin b li hay t nht cng c th pht hin c cc li khng s

thng tin ny. iu ny c bit quan trng trong cc h thng truyn dn s li 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

l qu trnh x l tn hiu s c thc hin sau ngun tin s v trc iu chtrong nhng nhim v ca m ho knh l kim sot li, thc cht y l qux l tn hiu s m bo thng tin c tin cy cao hn. M ho knh kili c thc hin bng cch thm vo tn hiu s cc bit kim tra c th thhai nhim v pha thu l PHT HIN v SA LI. Tu theo tm quan trng

tng h thng m thut ton m ho knh n gin hoc phc tp. Tn hiu sauc m ho knh, tu tng mc , c th cho php pha thu pht hin li ho pht hin li va c th sa li. Nh vy tn hiu sau khi m ho knh, tu tn, 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 phpchnh li trc (FEC) v yu cu pht li t ng (ARQ).

+ phng php hiu chnh li trc FEC (Forward Error Correction) bho nhn s liu s v b sung vo mt s bit kim tra theo mt quy lut nh

V th to ra mt lung bit mi c tc cao hn. B gii m s s dng cc btra xc nh gi tr thc ca cc bit thu c v trong mt s trng hp chiu chnh nu pht hin bit s liu b li.

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+ phng php ARQ (Automatic Retransmission Request) ngi ta cthm vo tn hiu s cn truyn mt s bit kim tra v nh vy cng lm cho lusau khi lp m c tc cao hn lung s nguyn thu. Pha thu da vo cc bittra pht hin li v nu pht hin thy li th my thu s gi yu cu v myu cu pht li phn thng tin b mc li ni trn. H thng vi ba s hin nyu s dng phng php FEC v vy ti liu ch s cp n phng php n

Nh ni trn, m pht hin v sa li c thit lp bng cch thmthng tin cn truyn mt s ckim tra kt qu l lm cho tc bit tng lnli t c an ton chng li cao hn. Ta hy xt mt v d minh hony.

Gi s ta mun gi i mt bt 0 hay mt bt 1. cc bt ny c bta b sung thm 3 bt kim tra theo cch sau:

Thng tin B sung Gi i0 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). thu c bt c ci g khc 0 0 0 0 hay 1 1 1 1 th c ngha l xy ra li trn truyn. T l m y l 1: 4 (1 bit thng tin: 4 bit phi truyn i). Nh vy vihin 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 0

Quyt 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 v s khng c vn g trong vic quyt nh pha pht pht i 0 hay 1. Nu

trong s 4 bit b li ta c th sa n, chng hn pht i 0 0 0 0, thu c 0 0 0quyt nh l 0 c pht hay pht i 1 1 1 1, thu c 0 1 1 1, ta quyt nh c pht.... Nhng nu c 2 bit b mc li ta ch c th pht hin nhng kth sa li, chng hn pht i 0 0 0 0, thu c 0 1 1 0 th ta khng th quyt bit 1 hay bit 0 c pht. Cui cng nu xy ra 3 hay 4 li th ta khng thhin c v nh vy thng tin s b li 1 bit. C th ni rng vi cch lp mtrn 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

khi v m xon.+ m khi ngi ta chia thng tin thnh cc khi, sau b sung vo khi mt s bit kim tra nht nh v cc bit kim tra trong mt khi m ch

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V d 1: Lp m chn, l cho khi tin 0 0 1 0. Bit rng m c lp thetnh chn.

Gii: Tng s bit 1 trong khi tin trn l 1 (l) nh vy m bo

chn ta phi thit lp bit kim tra l 1. Sau khi lp m ta thu c khi m l 0 (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 thetnh l.

Gii: Tng s bit 1 trong khi tin trn l 1 (l). m bo c tnh l c phi thit lp bit kim tra bng 0. Sau khi lp m ta thu c khi m l 0 0 1 0cui cng l bit kim tra chn l).c. M chn l theo ct c1. Nguyn tc lp m

Thng tin c chia thnh cc nhm, mi nhm c n khi, mi khi gm mthng tin, sau thm vo mi nhm mt khi m bit kim tra chn l [nh vtin sau khi m ho c (n + 1) khi]. Vic thit lp cc bit kim tra chn l hin bng cch kim tra c tnh chn, l ca cc bit 1 c cng th t trong ctin ca cng mt nhm v a gi tr vo v tr tng ng trong khi bit kim tral.

T bng trn ta thy vic kim tra c thc hin cc bit c cng th tt c cc khi trong nhm (kim tra theo ct). Sau khi lp m ta thu c bnm ho l:

Cc m chn, l theo hng v theo ct va xt trn c mt s mt hnh:

+ 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(2, 4, 6, 8...).

khc phc mt phn cc nhc im trn, ngi ta s dng phnm ho kt hp c theo hng v theo ct.

d. M chn, l theo hng v theo ct

d1. Nguyn tc lp m

M chn, l theo hng v theo ct c thit lp thng qua hai bc: Bc 1: Lp m theo hng

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Thng tin c chia thnh cc khi, mi khi c m bit, sau thm vo mkhi 1 bit kim tra kim tra c tnh chn, l ca cc bit 1 trong khi tin.

Bc 2: Lp m theo ct

Thng tin lp m chn, l theo hng c chia thnh cc nhm, mi nc n khi, mi khi c (m+1) bit, sau thm vo mi nhm mt khi bit kim (m+1) bit kim tra c tnh chn, l ca cc bit 1 c cng th t trong cc khca cng mt nhm v kt qu c a vo khi bit kim tra theo th t tn

Chng IVIU CH V GII IU CH

I/- Cc khi nim c bn v iu ch v gii iu ch

1- Tn hiu iu ch

L cc tn hiu tin tc cn truyn (tn hiu s v tn hiu tng t bao

Tn hiu thoi, truyn hnh, s liu. . . ) c tn s thp.2- Tn hiu sng mang

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L cc tn hiu in cao tn c th ti (mang) c thng tin. Tn hiu ca y mang tnh cht tng i. Mt tn hiu in cao tn c th lm sng mamt tn hiu iu ch ny nhng li khng th lm sng mang cho mt tn hich khc v chnh n c khi li tr thnh tn hiu iu ch cho mt sng mang 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

- = 2f 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 mtnhiu 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 tns) thng c tn s thp nn rt kh trc tip bc x thnh sng in t txa, nu c th bc x c th nng lng bc x cng rt yu v i hi phnx(anten)c kch thc ln. d dng truyn thng tin i xa bng sng i

ngi ta phi tin hnh iu ch tn hiu thng tin vo sng mang cao tn, nggi tin tc cn truyn vo sng mang sau mi cho sng mang iu ch bthnh sng in t truyn i xa. V sng mang c tn s cao nn d bc xsng 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 chmt bng tn truyn dn(V d: tn hiu thoi c bng tn t 0,3 3,4KHz)nn khitruyn nhiu tn hiu trn mt ng truyn dn th chng s b ln vo nhau l

pha thu khng thu c tn hiu. truyn c nhiu tn hiu trn cng mttruyn th ngi ta phi iu ch cc tn hiu cn truyn vo cc sng mangnhau, mc ch l chuyn ph ca thng tin cn truyn ln cc vng khc nhau s

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mi truyn chung trn mt ng truyn dn. Nh s khc nhau v vng ph ctn hiu truyn i m pha thu d dng thu c tn hiu.

4- Gii iu ch

V pha pht tin hnh iu ch nn pha thu mun ly li tn nguyn thy th phi tin hnh tch tn hiu tin tc ra khi sng mang iuQu trnh tch tn hiu tin tc cn truyn ra khi sng mang iu ch bnc gi l gii iu ch.

II/- Phn loi iu ch

1- 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 sng mang thay i theo quy lut ca tn hiu iu ch. iu ch bin tc gi l AM(Amplitude Modulation). iu ch bin s c gi l ASK(Amplitude Shift Keying).

- iu ch tn s: l tc ng tn hiu iu ch vo sng mang lm cho tca sng mang thay i theo quy lut ca tn hiu iu ch. iu ch tn s tc gi l FM(Frequency Modulation). iu ch tn s s c gi l FSK(Frequency Shift Keying).

- iu ch pha: l tc ng tn hiu iu ch vo sng mang lm cho phsng mang thay i theo quy lut ca tn hiu iu ch. iu ch pha tng tgi l PM(Phase Modulation).iu ch pha s c gi l PSK (Phase Shift Keying).

- iu ch QAM(Quature Amplitude Modulation):l phng php iu ch kthp c iu ch bin ASK v iu ch pha PSK. Vi iu ch ny th khi tiu ch tc ng vo sng mang th c bin v pha ca sng mang u thatheo quy lut ca tn hiu iu ch.

III. iu ch v gii iu ch bin

1) iu ch bin (AM)

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C L

D

R3

R1

R2

Tn hiu AM

T/h iuch

Sng mang

HNH 4.1: iu ch AM bng mt diode

C nhiu phng php thc hin iu ch AM.Hnh 4.1m t mt phng php tng i n gin thc hin iu ch AM. Mch gm mt mng hiu bng in tr (R 1, R 2, R 3), mt diode nn v mt khung cng hng LC.

Tn hiu iu ch c a n u vo R 1, tn hiu sng mang c a nu vo R 2. Mng in tr R 1, R 2 v R 3 thc hin trn tuyn tnh hai tn hiu vi nhatheo nguyn tc cng s hc. Nu tn hiu iu ch l tn hiu hnh sin th tsau khi trn (ly trn in tr R 3) c dng nh hnh 4.2 (c). Ta thy rng sngmang bin thin trn nn ca tn hiu iu ch, nhng y cha phi l tn h

iu ch bin . y hai tn hiu mi c cng vi nhau, trong khi iunhn hai tn hiu vi nhau. Tn hiu sau khi cng c nn qua diode D. Sau khta thu c mt dy xung dng l tp hp ca cc na chu k dng ca ttng, 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

Khung cng hng LC ny c tn s cng hng ring f LC

= 12 ng bng tn s

ca sng mang. Khi c kch thch th khung LC s dao ng vi tn s ng

tn s sng mang, cn bin th ph thuc vo bin ca tn hiu kch thchxung kch thch c bin ln th dao ng ly ra trn khung c bin ln, khi 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 chiu iu ch nn dao ng ly ra trn khung LC cng c bin bin thintheo quy lut ca tn hiu iu ch. y chnh l tn hiu iu ch bin hnh4.2 e). Trong iu ch bin , nu gi tin tc cn truyn c tn s l Vt vi Vt =

V . cost (V l bin cc i ca tn hiu iu ch), sng mang cao tn l =

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t(e)

t

U

U

U

t

U

U

t(d)

(a)

(b)

(c)

t

Hnh 4.2: a) Tn hiu iu ch b) Sng mang c) Tn hiu trn in tr R3d) Dng tn hiu qua diodee) Tn hiu trn khung cng hng

Vo . cosot 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 V V

V t t

AM t o o o( )( cos ) cos= +1

= +V m t t o( cos ) cos1 (3.1)

Trong :mV

V o= l h s iu ch v cn c gi l su iu ch. H s

ch m phi tha mn iu kin m 1. Nu m > 1 th mch c hin tng qu ich gy ra mo tn hiu sau iu ch. Trong thc t, m bo tn hiu khmo 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 t mV

t mV

t AM t o oo

oo

o( ) .cos .cos( ) .cos( )= + + + 2 2 (3.2)

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T biu thc 3.2 ta thy rng: tn hiu sau khi iu ch bin c 3 thnhchnh: sng mang c tn s gco v bin Vo, hai thnh phn mang tin c tn s

gco v bin mV o

2nh m t hnh 4.3b. Nu tn hiu tin tc c di tn t

min nmax th sau khi iu ch vo sng mang ta s thu c tn hiu c ph

m t hnh 3.3d. Ngoi sng mang c bin Vo v tn s gco cn c hai bintn: 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 4.4m t mt phng php gii iu ch bin n gin nhng dng kh rng ri trong k thut gii iu ch. Mch gm c mt bin p cngT1 ghp tn hiu AM vo mch, diode nn D, in tr ti R v t lc C3. in ptrn cc phn t R v C chnh l tn hiu gii iu ch.

Tn hiu AM c tvo hai u cun s cp ca bin p cng hng T1. Cc tC1 v C2 iu chnh tn s cnghng ca bin p. Tn hiu

AM cm ng qua cun th cpv t vo mch nn gm D v

59

min

max

0

0 - 0 +

0 - min 0 +min

0 - max 0 +max

0

U

U U

U

(a)

(b) (d)

(c)

HNH 4.3: a) Ph ca tn hiu iu ch n tnb) Ph ca tn hiu AM iu ch n tn

c) Ph ca tn hiu iu ch a tnd) Ph ca tn hiu AM iu ch a tn

Tn hiu gii

iu ch

T1

C2C1C3

D

Tn hiuAM

HNH4.4: Mch gii iu ch bin

R


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L

CD

+Vcc

R1

R2

CoC2

C1

Tn hiuiu ch

Ti mchdao ng

HNH 4.11: Mch iu ch tn s tng t FM

chn (+) ca N l 0V lm cho chn (+) m hn in p chn (-) ca N nn nca N c mc 0.

Khi c sng mang a vo u vo ca bin p cng hng th tn hiu nc ghp qua cun th cp. Sau tn hiu c diode D nn theo kiu nn nk (trong mt s mch gii iu ch ASK, hiu qu hn ngi ta s dngchu k) vi ti l in tr R. T C2 thc hin lc san bng v to ra trn in tr tmt in p tng i bng phng. in p ny c t vo chn (+) ca N lin p chn (+) ca N dng hn in p chn (-) nn ng ra ca N c mc

Kt qu l ng ra ca mch so snh N ta thu c cc bit 0, 1 nh 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 t

Mch in y cha c v y m ch trch mt phn mch c bn lin vic iu ch tn s.

Trong mch in hnh 4.11:

+ CD l mt diode bin dung

+ R 1, R 2: 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 tcm bo cho CD dng nh c mc song song vi khung LC1.

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Nh vy 3 phn t CD, L v C1 to thnh mt khung cng hng vi tn s

cng hng ring: f L C C D=

+1

2 1 ( )

Khi cha c tn hiu iu ch a vo mch th trn diode bin dung c mp tnh c xc nh bi mch phn p R 1 v R 2 do diode ny c mt gi tr indung CD khng i. Lc ny mch dao ng s hot ng vi tn s khn

f L C C D

=+

1

2 1 ( )

Khi c tn hiu iu ch a vo mch, nu bin ca tn hiu iu chth in p trn diode bin dung tng, lm cho tr s in dung CD ca diode gdn n tn s dao ng ca mch dao ng tng. Nu bin ca tn hiu gim th in p trn diode bin dung gim lm cho in dung CD tng dn s cng tc ca mch dao ng gim.

Nh vy, tn s dao ng ca mch dao ng b khng ch bi tn hiu u vo v bin thin ng theo quy lut ca tn hiu iu ch. ng ra cdao 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. cost, tn hiusng mang l V vi V = Vo . cosot th sau iu ch ta thu c mt tn hiu

iu ch VFM:V V t t FM o o= +.cos( . sin )

Trong : l di tn cc i.

H s iu ch mf ca mch iu ch tn s c tnh nhm k

V f

= =.

, 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 khi V = const th = const, nhng khi thay i th mf cng thay i.

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Khi iu ch n tn, ph ca tn hiu iu tn cha thnh phno v nhiuthnh phn tn s bin (o n) vi n = 1, 2, 3... . Bin ca cc thnh phn tn bin bin i khng ng u.

Mt cch gn ng, rng ph ca tn hiu iu tn sau iu ch c tnh sau: D m m FM 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

D m m FM 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 2max

2. 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

dng rng ri trong k thut gii iu ch tn s v quan h pha ca cc in 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.

65

t

t

U

(a)

(b)

(c)

t

Hnh 4.12: Dng sng ca mch iu ch tn s.a) Tn hiu iu ch b) Sng mang c) Tn hiu sau khi iu ch


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V iu ch pha tng t tc s dng trong thc t nntrong phn ny ch xt iu chv gii iu ch pha vi tn hius.

1) iu ch - Gii iu ch phahai trng thi (2 PSK)

a- iu ch pha hai trng thi(2 PSK)

iu ch pha hai trng thi

l iu ch m sau khi iu chsng mang iu ch c haitrng thi pha so vi pha casng mang cha iu ch.

Hnh 3.19 m t s khiv dng sng ca mch iu ch pha hai trng thi.

S liu nh phn cn iuch trc ht c a vomch chuyn m bin i tm NRZ n cc sang NRZlung cc. Sau , s liu c a n b nhn M nhn vi sng

2. cos( ) A t o + to ra t mch to dao ng sng mang. ng ra ca b nhn tac mt tn hiu m thnh phn chnh l d(t)2. cos( ) A t t o + + v mt s thnh phn pht sinh khng mong mun. Cc tn hiu ny c a qua mch lc b

loi b cc thnh phn khng cn thit. Kt qu, ng ra ta thu c mt tn iu ch 2 PSK:

U t d t A t o( ) ( ) . cos( )= +2 . Trong : d(t) l s liu s cn iu ch,o: pha ban uca sng mang, A2 : 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 t t o( ) .cos( )= + +2

Trong t l gc lch pha do s liu iu ch gy ra ti thi im iu ch (t = 0hoct = ). D dng nhn thy rng khi s liu iu ch l bit 0 th d(t) = sng mang sau iu ch l:U t A t o( ) .cos( )= + +2 0 .

71

LBF

To sng mang

M

i mTn hiuiu ch

Tn hiu iu ch Sng mang

(a)

0 10 1

U

t

t

t

S liuiu ch

Sngmang

cha iuch

Sng mang iu

ch(b)

Cos

01

(c)

Sin

HNH 4.19: iu ch 2 PSK a) S khi ca mch iu ch 2 PSK. b) Dng sng ca

iu ch 2 PSK. c) Biu pha ca iu ch pha 2 PSK


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Khi s liu iu ch l bit 1 th d(t) = 1 v sng mang sau iu ch l:U t A t o( ) .cos( )= + +2

Nh vy sng mang sau iu ch pha c hai trng thi l ng pha vi

mang cha iu ch (khi iu ch bit 0) v ngc pha vi sng mang cha i(khi iu ch bit 1). Pha thu s da vo s chuyn pha ca sng mang thgii iu ch.

b- Gii iu ch pha hai trng thi:

Hnh 4.20 m t s khi ca mch gii iu ch 2 PSK bng vng khCOSTAS.

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 bin bng nhau.

+ M1 v M2 l cc b nhn thc hin gii iu ch.

+ 90o

: mch dch pha 90o

bin sng mang A t

o2.cos( ) +

thnh sng mang A t o2.sin( ) + .

+ VCO: b dao ng to sng mang c tn s v pha c iu khin bnp.

+ Lc thng thp: lc b cc thnh phn tn s cao pht sinh sau khi nhhiu 2 PSK vi sng mang khi phc.

+ M3: b nhn tm kim sai pha gia sng mang 2 PSK v sng mang k

phc. Kt qu sai pha s th hin bng in p li.

72

90o

VCO Lc vng

Lc thng thp

Lc thng thp

Phn nhnh

M1

M2

cos(t +o)

sin(t +o)

M3

Tn hiu gii iu ch

Tn hiu 2PSK

HNH 4.20: Mch gii iu ch 2 PSK bng vng COSTAS


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+ Lc vng: lc in p t ng ra ca b nhn a vo iu chnh tn pha ca b dao ng VCO.

C th phn tch nguyn l hot ng ca mch da trn s khi

3.20 nh sau: Sng mang 2 PSK c chia thnh hai ng a n cc b nh1v M2. M1 thc hin nhn sng mang 2 PSK c biu thc:U t A t o t ( ) .cos( )= + +2 vi sng mang khi phc c to ra t b dao ng VCO:cos( ) t o+ . B nhn M2nhn sng mang 2 PSK: A t o t 2.cos( ) + + vi sng mang khi phc to ra t bdao ng VCO v qua dch pha 90o: sin( ) t o+ .

ng ra ca M1 ta thu c tn hiu l:A t o t 2.cos( ) + + . cos( ) t o+ =

t t ot t o At At A cos.)22cos(.]cos)222[cos(2.21 +++=+++=

ng ra ca b nhn M2 ta thu c tn hiu l:

A t o t 2.cos( ) + + . cos( ) t o+ =

t t ot t o At At A sin.)22sin(.]sin)22[sin(2.21 +++=+++=

Nh vy sau hai b nhn ta u thu c 2 thnh phn: 1 thnh phn tn sc tn s v gc pha l:( )2 2 2 t o t + + , 1 thnh phn tn s thp c gc pha lt. Cc

thnh phn ny u mang tin, tuy nhin thnh phn tn s cao l khng cn thi b lc thng thp sau M1 v M2 s lc b cc thnh phn tn s cao, thnh phn ts thp thu c sau cc b lc u l s liu gii iu ch. Tuy nhin ch s liu u ra ca mt b lc thng thp l . Tn hiu sau 2 mch lc thc nhn vi nhau ti b nhn M3 pht hin s sai pha ca sng mang 2 PSK vsng mang to ra t b dao ng VCO. Nu c sai pha hoc sai tn th M3 s to ramt in p li. in p ny c lc qua mch lc vng a v iu chnv pha ca b dao ng VCO.

Phng php iu ch, gii iu ch pha 2 trng thi c u im l mctng i n gin, khong cch gia cc trng thi pha ln nn c kh nng nhiu cao. Tuy nhin, do mi trng thi pha ch th hin c mt bit thng tihiu qu truyn tin khng cao. tng hiu qu truyn tin, ngi ta thc hich pha nhiu trng thi hn.

2) iu ch pha 4 trng thi - Gii iu ch pha 4 trng thi

a- iu ch pha 4 trng thi (QPSK)

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Bin ini tip

sang songsong

Chuynm

Chuynm

90o

VCO

Z1

Z2

M1

M2

S2(t)

S1(t) Tnhiu

QPSK

Tnhiuiuch

Hnh 4.21 S khi ca iu ch pha 4 trng thi:

iu ch pha 4 trng thi cn c gi l iu ch 4 PSK hay iu ch phng pha hay iu ch QPSK, l phng thc iu ch pha m sng mang sach c 4 trng thi pha, mi trng thi pha mang hai bit thng tin.

Hnh 4.21 m t s khi ca iu ch pha 4 trng thi:Mch hnh 3.21 gm c cc khi:

+ S/P: mch chuyn i lung s t ni tip sang song song c nhim vlung s iu ch thnh hai lung c tc bng nhau v bng 1/2 lung s v

+ Z1 v Z2: cc mch chuyn i m c nhim v bin cc lung s S1 v S2 tm NRZ n cc thnh m NRZ lng cc.

+ VCO: b to dao ng sng mang c tn s iu khin bng in p.

+ 90o: mch dch pha 90o bin sng mang A.cos (t + ) thnh A.sin (t +).

+ M1 v M2: hai b nhn thc hin nhn cc tn hiu s S1(t) v S2(t) vi ccsng mang A.cos(t + ) v A.sin (t +) tng ng.

+ : mch cng c nhim v cng cc tn hiu sau cc b nhn M1 v M2 c tn hiu QPSK.

Nguyn l hot ng ca mch iu ch pha nh sau: S liu iu ch mch bin i t ni tip sang song song c bin thnh hai lung s c tc nhau v bng 1/2 lung s vo. Cc lung s S1 v S2 sau c chuyn m t NRZn cc sang NRZ lng cc nh cc mch chuyn m Z1 v Z2. Sau khi chuyn mta thu c cc lung s tng ng l S1(t) v S2(t). B nhn M1 nhn tn hiu S1(t)vi sng mang A.cos(t + ) do b dao ng VCO to ra, b nhn M2 nhn tn

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VCO90

o

Lc thngthp

Lc thngthp

LPF

M1

M2

M3

M4

M5

Lung s S1

Lung s S2

Tnhiu

QPSK

Hnh 4.23 S khi ca mch gii iu ch QPSK s dng vng kho pha COSTAS

00 (45o)01 (135o)

11 (225o) 10 (315o)

Y

X

0 0 0 1 1 1 1 0

1 10 0 1 1 1

0 1 1 0 1

U

t

t

t

t

t

S liuiu ch

Lung s S1

Lung s S2

Sng mangcha iu ch

Sng mang iu ch

(a)

(b) HNH 4.22: Dng sng (a) v biu pha ca

iu ch QPSK (b)

hiu S2(t) vi sng mang A.sin(t + ) do b dao ng VCO to ra v qua dch pha 90o. Sau cc b nhn M1, M2 ta thu c cc tn hiu 2 PSK l S1(t).Acos(t + )v S2(t).Asin(t + ). Hai tn hiu ny c cng vi nhau qua mch cng v

ra ca mch cng ta thu c tn hiu 4 PSK (QPSK) c biu thc ton hc l:S1(t).Acos(t +) + S2(t).Asin(t + ) = A.cos (t + +t)

t: gc dch pha do iu ch pha sinh ra

Hnh 4.22m t dng sng v biu pha ca iu ch pha 4 trng thi:

b- Gii iu ch pha 4 trng thi (gii iu ch QPSK)

C nhiu phng php thc hin gii iu ch pha 4 trng thi, trong cs phng php nh: vng nhn pha, iu ch li, vng kha pha COSTAS. Tny cp n phng php gii iu ch QPSK bng vng COSTAS.

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+ x4: l mch nhn 4

+ LPF: Mch lc thng thp

+ BPF: mch lc thng gii

+ :4: mch chia 4.+ M1 v M2 l cc b nhn

+ N1 v n2 l cc b khuch i so snh

* Nguyn l hot ng

Trc ht tn hiu u vo X(t) = sin+

4)1(2 i

t C

(trong 4)1(2 i l pha

do tn hiu iu ch bn pht to ra, i = 1, 2, 3, 4 ty theo trng thi pha) vo mt mch nhn 4. Sau khi nhn ta thu c tn hiu ra Y(t) c cc thnh ph

Y(t) =8

3-

21 cos[ ])1(2 + it C + 8

1cos[ ])1(24 + it C

Mch lc thng gii s lc ly thnh phn8

1cos[ ])1(24 + it C c tn s

bng 4 ln tn s ca tn hiu 4PSK trong cc bc nhy pha ti2

c loi b.

Tn hiu Y(t) c a qua mt mch chia 4 v u ra ca mch ny ta c tsng mang khi phc. sng mang ny c chia ln 2 ng trong c ng c a qua 1 qua mng dch pha 90 ti 2 mng nhn. Tn hikhi nhn c a qua c mch lc thng thp ly li thng tin cn truyntin ny c sa dng bi cc mn khuch i so snh, sau khi sa dng tnc ghp li vi nhau thng qua mch bin i song song sang ni tip lung s nguyn thy.

VI. iu ch v gii iu ch QAM

1. Khi nim

iu ch QAM (Quature Amplitude Modulation: iu ch bin vung g phng php iu ch kt hp gia iu bin v iu pha. phng php nmang sau iu ch c bin v pha thay i theo quy lut ca tn hiu iu ch

2. iu ch QAM 16 trng thi (iu ch 16 QAM)

p ng yu cu thng tin tc cao, ngi ta phi tng s trng thi p

bin ca sng mang c c s trng thi pha nhiu v khong cch gi

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Chia 4

Bin

iA/D

BiniA/D

VCO

90 o

A1

A2M2

M1S(1)

S(2)

Tn hiu 16QAM

Tn hiuiu ch

HNH 4.23: S khi iu ch 16 QAM

trng thi pha ln. hiu r v vn ny, ta xt mch iu ch 16 QAM nt hnh 4.23.

S khi:

Mch iu ch 16 QAM gm c:

+ Mch bin i ni tip sang song song c nhim v chia lung s thnh 4 lung c tc bng nhau v bng 1/4 lung s vo.

+ Mch dao ng VCO: l b dao ng to sng mang c tn s iu k bng in p.

+ Mch dch pha 90o: bin i sng mang A.cos (t + ) thnh sng mang A.sin(t + ).

+ B bin i D/A c nhim v nhn hai lung s c hai mc bin bin i thnh mt tn hiu c 4 mc bin (tn hiu tng t) ng ra.

+ M1, M2: cc b nhn c nhim v nhn cc sng mang vi tn hiu sau cc bin i D/A.

+ B cng c nhim v cng cc tn hiu sau khi nhn M1 v M2 to thnhsng mang 16 QAM.

Biu pha ca sng mang sau iu ch 16 QAM c m t hnh 3.24:

79

0000 0001

0010 0011

1100

1110

0100

1101

0101

0111 0101

1000 1010

101110011111

Y

X

HNH 4.24: Biu pha ca sng mang 16


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Nguyn l hot ng:

S liu cn iu ch c a n mch chuyn i ni tip sang sonc chia thnh 4 lung D1, D2, D3, D4 c tc bng nhau v bng 1/4 lung s v

Cc lung D1, D2 c a n b bin i D/A 1 chuyn thnh tn hiu tnmc A1. Cc lung D3, D4 c a n b bin i D/A 2 chuyn thnh tn htng t 4 mc A2. Sau tn hiu tng t A1 c nhn vi sng mang A.cos(t +) do b dao ng VCO to ra ti M1, u ra ca M1 ta thu c tn hiu:

S1 = A1.Acos . (t + ). Tn hiu tng t A2 c nhn vi sng mang A.sin(t + )do b dao ng VCO to ra v qua dch pha 90o ti M2, u ra ca M2 ta thuc tn hiu: S2 = A2.Asin . (t + ). Cc tn hiu S1 v S2 c cng vi nhau

to thnh tn hiu 16 QAM.Hnh 4.25m t dng sng ca iu ch 16 QAM:

3. Gii iu ch 16 QAM

S khiS khi ca mch gii iu ch 16 QAM c m t hnh 3.26.

80

0 0 0 0 0 0 1 1 1 1 0 0

U

t

t

t

Tn hiuiu ch

Sngmang

cha iuch

Sngmang iu ch

Hnh 4.25: dng sng mch iu ch 16 qam


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Chc nng cc khi

+ M1, M2: cc b nhn c nhim v nhn tn hiu 16 QAM u vo visng mang khi phc.

+ VCO: mch dao ng to sng mang c tn s iu khin bng in p.

+ 900: mch dch pha 90o bin sng mang cos (t +o) thnh sin (t +o).

+ Mch lc thng thp c nhim v lc b cc thnh phn tn s cao sanhn.

+ A/D: b bin i tng t - s c nhim v bin i mt tn hiu tmc u vo thnh hai tn hiu s 2 mc bin u ra.

+ Mch khi phc xung nhp c nhim v ti to li xung nhp phc vcc mch x l tn hiu s pha thu. Xung nhp khi phc phi ng b v pht.

+ Mch P/S: l b bin i song song sang ni tip, c nhim v bin i 4s c tc bng nhau u vo thnh mt tn hiu u ra c tc ln htc ca mt lung s u vo v phi mang y ni dung ca 4 luu vo.

Nguyn l hot ng:

Tn hiu 16 QAM c a n cc b nhn M1 v M2. B nhn M1 nhn tnhiu 16 QAM vi sng mang khi phc cos(t + o). B nhn M2 nhn tn hiu 16QAM vi sng mang sin(t + o). Sau cc b nhn M1 v M2 ta u thu c haithnh phn mang tin: mt thnh phn c tn s thp, mt thnh phn c tn sCc b lc thng thp s loi b thnh phn tn s cao, u ra ca cc b lc cc tn hiu 4 mc. Cc tn hiu ny c quyt nh mc v bin i A

81

VCO

90 o

Lc thngthp

Lc thngthp

M1

M2

Tn hiu16QAM

A

73337118 Thong Tin Vo Tuyen

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