Thit k tuyn vi ba s

Thit k tuyn vi ba s

CHNG 3 THIT K TUYN VI BA S3.1 Gii thiu chngChng ny gii thiu v : Mc tiu v yu cu Quy nh chung cho thit k tuyn vi ba s Cc bc thit k tuyn vi ba s+ Xem xt dung lng, tc bit v yu cu v cht lng ca ng truyn.+ Chn bng tn v tuyn cao tn.+ Sp xp cc knh v tuyn cao tn.+ Quyt nh cc tiu chun thc hin.+ Chn v tr v tnh ton ng truyn.+ Xy dng cu hnh h thng.+ Sp xp bo tr.+ Xem xt cc tiu chun k thut.+ Lp t, o th, vn hnh khai thc.3.2 Mc tiu v yu cu Khi thit k tuyn truyn dn vi ba s th chng ta phi m bo cc ch tiu k thut ca tuyn t ra, p ng cc yu cu phc v, v m bo tnh kinh t. Vi bt k h thng k thut no u tun th theo quy lut tng tc gia chi ph u t v hiu qu ca sn xut c th hin qua cht lng ca sn phm. H thng vin thng cng vy. Nu t s BER m thp th cht lng dch v s tng v nh vy th chi ph u vo s cao. Vy mc ch kinh t u tin l thit k tuyn c cht lng cao m ch ph hp l nht. Do vy, ngi thit k phi tnh ton chnh xc cc thng s k thut theo tiu chun quy nh, tnh ton n mc ch s dng ca h thng v c tnh hnh ti chnh ca n v thi cng, t la chn thit b cho ph hp, nhm trnh lng ph v t hiu sut cao nht.3.3 Mt s quy nh chung cho thit k tuyn vi ba s Vic thit k tuyn vi ba s c tin hnh trn c s: + D n bo co kh thi c cc cp c thm quyn ph duyt. + H s kho st, thuyt minh chnh xc v ni dung xy lp, cc s liu tiu chun cn t c. + Cc vn bn th tc hnh chnh ca c quan trong v ngoi ngnh lin quan n a im,mt bng xy dng trm. + Cc tiu chun,quy trnh,quy phm xy dng ca nh nc v ca ngnh. + Cc nh mc v d ton c lin quan p dng trong thit k. + H s ti liu thu thp c trong qu trnh kho st v o c. Vic thit k cn phi m bo ng tiu chun, quy trnh, quy phm ca nh nc ban hnh, nh : + ng k tn s lm vic ca thit b vi cc tn s v tuyn in,Quc gia. + An ton v phng chng thin tai,bo lt. + An ton khi c ging st,m bo cht lng ca cc h thng chng st,tip a cho thit b v thp anten theo quy nh ca ngnh3.4 Cc bc thit k tuyn vi ba s Nhn chung vic thit k tnh ton mt ng truyn vi ba im ni im c th c trnh by theo cc bc sau: +Bc 1: Xem xt dung lng,tc bt v yu cu v cht lng ca ng truyn. +Bc 2: Chn bng tn v tuyn cao tn. +Bc 3 : Sp xp cc knh v tuyn cao tn. +Bc 4: Quyt nh cc tiu chun thc hin. +Bc 5: Chn v tr v tnh ton ng truyn. +Bc 6: xy dng cu hnh h thng. +Bc 7: Sp xp bo tr. +Bc 8: Xem xt cc tiu chun k thut. +Bc 9: Lp t,o th v vn hnh khai thc.3.5 Xem xt dung lng, tc bit v yu cu v cht lng ca ng truyn Trong vic thit k mt h thng thng tin vi ba s im ni im, vic xem xt k v dung lng,tc bt v yu cu v cht lng ca ng truyn cn l rt quan trng v rt cn thit. N l c s cho cc quyt nh cc bc tip theo. Khi xem xt dung lng cn ch n vic d on kh nng pht trin trong tng lai. Vic d on ny da vo cc c s nh: + Ngun s liu d bo ng tin cy v vic pht trin dn s. + c im vng min( thnh ph, nng thn,vng nng nghip.). + Tc pht trin ca cc hot ng kinh t. + Tc ci thin iu kin sng trong tng lai. + Kh nng m rng thm trong tng lai. Tuy nhin, cc nc ang pht trin( nh thc trng nc ta) thng kh d on chnh xc dung lng cn thit trong khong thi gian di. Do khng nn lp t cc h thng c dung lng qu ln cho cc yu cu cho tng lai. S kinh t hn khi chn cc thit b c dung lng nh giai on u tin v nu dung lng ny khng p ng c sau khi s dng vi nm, h thng c th thay th bi mt h thng khc c dung lng ln hn cn h thng c c dng tuyn cn dung lng nh hn. Nn i khi xy dng mt h thng va phi v d dng thay th khi c k thut mi trong tng lai th kinh t hn.3.6 Chn bng tn v tuyn cao tn i vi cc ng dng ca k thut vi ba,bng tn hot ng ca n nm trong khong 1GHz 15GHz. Trong cc tn s v tuyn c cp pht cho cc dch v xc nh c quy nh bi CCIR. V vic la chn bng tn c nh hng nhiu n c tnh chung ca thit b. Chng hn,khi lm vic tn s thp th thit b lm vic n nh, khuch i ln hn,tn hao dy dn, tn hao phi v tn hao ng truyn u nh.Dung lng cng ng vai tr quan trng trong vic chn bng tn hot ng cho h thng. Nhn chung tn s v tuyn cao tn cng cao th dung lng h thng cng ln, h thng c th truyn tn hiu vi tc bt cao. Do , khi la chn bng tn lm vic ta cn cn i mt s yu t quan trng sao cho h thng t c hiu qu cao v lm vic n nh.3.7 Sp xp cc knh v tuyn cao tn S sp xp cc knh v tuyn cao tn l mt phn rt quan trng trong vic thit k h thng. Bng 3.1 trnh by s sp xp cc knh v tuyn cao tn theo khuyn ngh ca CCIR.

Bng 3.1 Khuyn ngh ca CCIR v s sp xp cc knh v tuyn cao tn i vi h thng vi ba s im ni im, do khng c cu hnh trm lp nn s sp xp knh v tuyn cao tn tr nn n gin hn rt nhiu. Khi ta cn quan tm n mt s im nh sau: cc tn s vi ba s khc c th s dng trong cc vng lin quan,cc trm vi ba s c th gy giao thoa n h thng, vic thit k mt h thng vi ba s mi khng gy nhiu cho mt h thng vi ba s ang c v khng b cc h thng ny gy nhiu.3.8 Quyt nh tiu chun thc hin Cc tiu chun k thut c th phn loi nh sau: + Tiu chun hnh chnh. +Mc tiu thit k thit b. +Mc tiu thit k cc h thng. +Vn hnh hay cc mc tiu bo dng. V trong lut v tuyn c quy nh v nhng tiu chun cho tn s cao tn v mi quc gia s dng cho tn s, bng tn v tuyn ring bit nn cn phi tham kho tiu chun ny. Ngoi ra,cn xt n nhng yu t quyt nh gi thnh h thng vi ba l cht lng truyn dn v tin cy ca h thng. Do , cn chn tun th theo cc tiu chun quy nh trong vic tnh ton thit k ng truyn ni chung v ng trung k ni ring.3.9 Chn v tr t trm v tnh ton ng truyn3.9.1 Kho st v tr t trm Trong vic chn v tr phi quan tm n cht lng truyn dn, tin cy v tnh kinh t. tin hnh cc tnh ton ng truyn mt cch hp l, cn phi thc hin cc vic sau y: + Xc nh tuyn,xc nh v tr ca cc trm trn bn . + V cc bn v mt ct nghing ca tuyn.3.9.1.1 Xc nh tuyn,xc nh v tr ca trm trn bn xc nh v tr t trm ta cn c: Bn t nhin cho bit cao so vi mt nc bin ca vng c tuyn i qua. S phn b dn c ca vng ang kho st.Trong bc tm trm ny ta gi thit : tuyn ta thit k c hai trm u cui v ntrm lp, khng c trm xen k(trm xen k xem nh trm lp). Xc nh v tr t trm u cui Cn c vo phn b dn c xc nh trn bn a hnh v tr cc trm u cui,xen k. Xc nh nhng i ni, m t, ta nh cao tng trong khu vc tuyn. Chn trong cc v tr va xc nh trn, mt v tr thch hp t thp anten. Tm trm u cuiV tr va chn phi m bo hai tiu ch sau: C cao ng k( c th khng phi l cao nht Gn trung tm giao dch BCVT thun tin cho vic ko Feeder. Xc nh v tr trm lpTrm lp cn xc nh tha mn hai yu cu sau: C tng di ng truyn t trm u cui A qua trm lp v n trm u cui B l nh nht. C suy hao do nh hng ca a hnh l nh nht. Vic xc nh v tr trm lp c tin hnh nh sau: V ng thng ni hai trm u cui A v B. Tm trn ng thng hoc ln cn ng thng cc v tr c cao ng k c th t trm. Tm trm lpV tr trm lp phi lu : Tm nhn thng. Nu hai trm u cui kh gn nhau th khng cn trm lp.3.9.1.2 To nn cc bn v mt ct nghing ca tuyn T cc yu cu thc t ca mt tuyn vi ba gm: v tr trm,khong cch trm, dung lng truyn dn, a hnh tuyn s i qua. Ta tin hnh nh du hai u cui ca trm trn bn ca S o c xc nh chnh xc kinh ,v ca mi trm. Cc thng s ta ny c s dng iu chnh cc anten mi trm trong giai on lp t thit b. K hiu trn bn : trm A l trm th nht v trm B l trm th hai. Sau v mt mt ct nghing ca ng truyn. Hnh 3.2 th hin mt ct ng truyn gia 2 trm A v B. Hnh 3.1 Mt ct ng truyn gia 2 trm A v B Trong : + v: cao ca v tr t trm A v B so vi mc nc bin. + v : cao ca ct anten ti trm A v trm B. + B l cao tia v tuyn tng ng vi im kho st M. Trn mt ct nghing,xt im M l mt im nm khong gia trm A v trm B tng ng vi a hnh c vt chn hnh nm cao nht, E l li ca mt t ti M v c xc nh theo phng trnh sau: E = 1000 (3.1) l bn knh qu t = 6370[Km]. Thay gi tr = 6370[Km] vo (3.1) ta c biu thc: E= [m] (3.2) k l h s bn knh ca qu t. [Km]: ln lt l khong cch t trm A v trm B n im ang xt li ca mt t. cong ca cc ng chia trn b mt cho php v ng cong chnh xc ca ng truyn nh l mt ng thng da vo khi nim h s k. Gi tr ca h s k thay i theo thi gian v a im. Ni chung k thay i theo v , vng pha nam c gi tr k ln hn vng pha bc. Vo ma h k c gi tr ln hn ma ng. Khi chn v tr phi tnh n ma dao ng ca k so vi gi tr bnh thng v k cng ln th tnh n nh ca h thng cng km. Do trong qu trnh thit k,gi thit k khng i th vic tnh ton s tt hn. Thng chn k= iu kin bnh thng. Nh vy trn mt ct nghing ny th hin c b mt ca a hnh. Ngoi ra n cng c th biu din c c cao ca cy ci cc vt chn trn ng truyn ni hai trm A v B chng hn nh cc g,i,cc nh cao tng i vi khong c truyn dn di, cong ca mt t ln th cn phi tnh ton n nng ca v tr trm. nng c v dc cc ng thng ng nn khng i dc theo ng bn knh xut pht t tm qu t. Sau khi v mt ct tuyn, ta tnh min Fresnel th nht. l min c dng hnh elip t anten pht n anten thu biu din mt mi trng vy quanh tia truyn thng nh c m t hnh 3.2. ng bin ca min Fresnel th nht to nn qu tch sao cho bt k tn hiu no i n anten thu qua ng ny s di hn so vi ng trc tip mt na bc sng(/2) ca tn s sng mang. Min Fresnel th nht cha hu ht cng sut n my thu. Nu tn ti mt vt cn ra ca min Fresnel th nht th sng phn x s lm suy gim sng trc tip,ma suy gim ty thuc vo bin ca sng phn x. Hnh 3.2 Mt ct ng truyn v min Fresnel th nht Do vic tnh ton i vi min Fresnel th nht i hi c tnh chnh xc vic thng tin gia hai trm khng b nh hng ng k bi sng phn x ny. Bn knh ca min Fresnel th nht ) c xc nh theo biu thc sau: [m] (3.3) [Km]: ln lt l khong cch t trm A v trm B n im bn knh min Fresnel c tnh ton. d [Km] l khong cch gia hai trm, d =. f l tn s sng mang [GHz]. Trong thc t, thng gp ng truyn i qua nhng a hnh khc nhau c th chn min Fresnel th nht gy nn tn hao ng truyn. cc loi a hnh ny c th c vt chn hnh nm trn ng truyn v cc loi chng ngi khc. Hnh 3.2 ch ra m hnh ca vt chn trn ng truyn dn, trong l bn knh min Fresnel th nht, F l khong h thc l khong cch gia tia trc tip v mt vt chn hnh nm ti im tnh ton min Fresnel th nht. Theo cc ch tiu thit k v khong h ng truyn c khuyn ngh th cao ti thiu ca anten m bo sao cho tn hiu khng b nhiu x bi vt chn nm trong min Fresnel th nht l F = 0,577. Ngha l ng trc tip gia my thu v my pht cn mt khong h trn mt t hoc trn mt vt chn bt k t nht l vo khong 60% bn knh min Fresnel th nht t c cc iu kin truyn lan trong khng gian t do.3.9.1.3 Xc nh cao ca anten tnh cao ca anten th trc tin phi xc nh c cao ca tia v tuyn truyn gia hai trm. Trn c s ta tnh ton cao ti thiu ca thp anten thu c tn hiu. Vic tnh ton cao ca tia v tuyn cng phi dng n s mt ct nghing ng truyn ni hai trm trong c xt n cao ca vt chn (O), cao ca cy ci(T) gia tuyn v bn knh ca min Fresnel th nht(). Biu thc xc nh cao ca tia v tuyn nh sau: B=E(k) + (O + T) + C.= + (O + T) + 17,32C [m] (3.4) d, ,f c nh ngha nh trong biu thc (3.3). C l h s h, C =1. Thng thng th cao ca tia B c tnh ton ti im c mt vt chn cao nht nm gia tuyn. Hnh 3.3 Xc nh cao tia B lm h mt vt chn Cc cao ca cy ci v vt chn gia tuyn c xc nh t bc kho st ng truyn.hnh 3.2 biu din mt ct ng truyn ca tuyn cng vi cc vt chn gia tuyn v c xt n min Fresnel th nht. Sau khi c c cao tuyn, ta tnh cao ca anten lm h mt vt chn nm gia tuyn( tc khng gy nhiu n ng truyn v tuyn). bc kho st nh v trm,ta xc nh c cao ca hai v tr t trm so vi mt nc bin tng ng l v . Hai thng s ny kt hp vi cao B ca tia nh tnh ton trn s tnh c cao ca ct anten cn li khi bit trc cao ca mt ct anten. [m] (3.5) [m] Trong : [m] l cao ca mt trong hai anten cn c tnh. [Km] l khong cch t mi trm n v tr tnh ton cao ca tia B. Nh vy khi bit c cao ca mt anten th c th tnh c cao ca anten kia sao cho khng lm gin on tia truyn ca hai trm. Hnh 3.4 minh ha cch tnh cao ca anten ni trn trong 2 trng hp v . Hnh 3.4 Minh ha vic tnh cao ca mt anten khi bit cao anten cn kia Tuy nhin nh cp phn trc, m bo cho h thn hot ng khng chu nh hng ca cc yu t trong tng lai th cao ca anten phi s dng mt khong d phng,ph thuc vo ngi thit k. Khi cc cao ca anten thc t phi l do c cng vi mt lng cao d phng l hoc nh sau: [m] (3.6) [m] (3.7)3.10 Tnh ton qu cng sut ca ng truyn3.10.1 Cc nhn t nh hng n ng truyn Cc tham s c s dng trong tnh ton ng truyn nh: mc suy hao trong khng gian t do,suy hao do ma,do pha inh,cng sut pht, ngng thu,cc suy hao trong thit b c vai tr quan trng xem xt tuyn c th hot ng c hay khng v hot ng mc tn hiu no. + d tr pha inh phng Do tc ng ca pha inh phng mc tn hiu thu c ti u vo my thu c th thp hn so vi mc tn hiu thu khng b pha inh. Khi pha inh mnh c th lm tn hiu nhn c ti u vo my thu nh hn mc ngng v lm gin on thng tin. Trong trng hp ny vic tnh ton mt lng pha inh phng d tr l cn thit cho ng truyn v tuyn khng lm gin on thng tin. d tr pha inh phng Fm [dB] lin quan n cng sut tn hiu u vo my thu trong trng hp khng c pha inh [dBm] v cng sut ngng ca my thu RX trong trng hp b nh hng bi pha inh phng, theo biu thc: Fm = 10log() [dB] = [dBm] RX [dBm] [dB] (3.8) + Tn hao trong khng gian t do Tn hao trong khng gian t do() l tn hao ln nht cn phi c xem xt trc tin. y l s tn hao do sng v tuyn lan truyn t trm ny n trm kia trong mi trng khng gian c tnh theo biu thc sau: [dB] (3.9) Hoc [dB] (3.10) Vi f : tn s sng mang tnh bng [GHz]. d: di tuyn [Km]. + Suy hao do ma Suy hao do ma v pha inh l cc nh hng truyn lan ch yu cc tuyn v tuyn tm nhn thng trn mt t lm vic cc tn s trong di tn GHz, chng nh hng ng k n khong cch lp cng nh gi thnh ca mt h thng v tuyn chuyn tip. Suy hao do ma tng nhanh theo s tng ca tn s s dng, c bit vi cc tn s trn 35GHz thng suy hao nhiu v do m bo th khong cch lp phi nh hn 20Km, ngoi ra vic gim di ca ng truyn s gim cc nh hng ca pha inh nhiu tia. + Tn hao phi y l tn hao thit b(gm dy dn v ng dn sng) truyn dn sng gia anten v my pht hoc my thu. Khi tnh ton suy hao ny th cn c vo mc suy hao chun c cho trc bi nh cung cp thit b. Chng hn vi phi s dng ng dn sng trn Andrew WC 109 c mc tiu hao chun l 4,5 dB/100 m v cng vi 0,3 dB suy hao ca vng trn chuyn tip ng dn sng elip(EW 127) th tn hao phi my pht () v my thu () c tnh nh sau: [dB] (3.11) [dB] (3.12) Vi v l cao ca cc anten c tnh ton lng d phng. + Tn hao r nhnh Tn hao r nhnh xy ra ti b phn nhnh thu pht,tn hao ny cng c cho bi nh cung cp thit b. Mc tn hao ny thng khong (2 8) dB. + Tn hao hp th kh quyn Cc thnh phn trong kh quyn gy ra cc tn hao m mc ca n thay i theo iu kin thi tit,thay i theo ma,theo tn s s dng.Khi tnh ton mc suy hao ny ta da theo cc ch tiu c khuyn ngh cc nc Chu u,chng hn i vi h thng thit b v tuyn 18GHz, 23GHz v 38GHz th mc suy hao chun c cho trong khuyn ngh vo khong 0,04 dB/Km, 0,19dB/Km v 0,9 dB/Km khi tn hao cho c ng truyn c xc nh: = [dB] (3.13) Vi d l khong cch ca ng truyn tnh bng Km. Phng trnh cn bng cng sut trong tnh ton ng truyn: [dB] (3.14) Trong : l cng sut pht : Tn hao tng = tn hao trong khng gian t do + tn hao phi + tn hao r nhnh + tn hao hp th kh quyn G : Tng cc khuch i = khuch i ca anten A + khuch i ca anten B vi khuch i ca anten A v anten B: , (3.15) Trong : l hiu sut ca anten. : ng knh ca anten A,B. : cng sut ti u vo my thu. l tham s quan trng khi thit k ng truyn vi ba, tham s ny l mt ch tiu quyt nh xem tuyn c hot ng c hay khng khi em so snh n vi mc ngng thu RX ca my thu. iu kin cn tuyn hot ng c: >RX.3.11 Tnh ton cc tham s cht lng ca ng truyn V cht lng ng truyn c nh gi da trn t s BER, cc t s BER khc nhau s cho mt mc ngng tng ng v cng c d tr pha inh khc nhau. Cc t s BER thng c s dng trong vi ba s l: BER= v BER= tng ng vi hai mc ngng R v R. d tr pha inh ng vi R v R l F v F c tnh theo biu thc: F = R vi BER = (3.16) F = R vi BER= (3.17) Xc sut pha inh phng nhiu tia() l mt h s th hin kh nng xuthin pha inh nhiu tia c nh gi theo biu thc: (3.18) Trong ; B =1; C = 3,5 l cc tham s lin quan n iu kin truyn lan v kh hu v a hnh ca sng v tuyn v cc gi tr c s dng theo khuyn ngh ca CCIR. Xc sut t n ngng thu R, R. Gi l xc sut t ti cc gi tr ngng thu tng ng R v R c tnh nh sau: = (3.19a) = (3.19b) Vi F v F l d tr pha inh ng vi BER = , BER = c tnh ton trn. Khong thi gian pha inh v l cc gi tr c trng cho cc khong thi gian tn ti pha inh, cng tng ng vi F, F v tnh theo biu thc sau: [s] (3.20a) [s] (3.20b) Vi f[GHz]; = 56,6 , = 0,5; = - 0,5. Xc sut pha inh phng di hn 10s v 60s P(10) v P(60) l xc sut xut hin pha inh phng di hn 10s v 60s tng ng vi cc t s BER khc nhau v c xc nh theo biu thc: P(10) = 0,5[1 erf(] = 0,5 erfc() (3.21a) P(60) = 0,5[1 erf()] = 0,5 erfc() (3.21b) Vi Vi erfc(t) = 1 erf(t) l hm sai s b Trong erf(t) = l hm sai s. Xc sut BER vt Xc sut BER vt th hin s gin on thng tin nhng trong thi gian khng qu 10s. Xc sut (BER >) = (3.22) Xc sut mch tr nn khng th s dng c do pha inh phng trong khong thi gian ln hn 10s. (10) l xc sut mch s c BER > trong khong thi gian ln hn 10s tc l mch tr nn khng s dng c v c tnh theo biu thc: (3.23) Kh nng s dng tuyn Kh nng s dng tuyn c biu th bng phn trm v c xc nh theo nh sau: (3.24) Xc sut BER vt Xc sut BER vt th hin s gin on thng tin nhng trong thi gian khng qu 60s. Xc sut (BER > ) = (3.25) Xc sut mch tr nn khng th s dng c do pha inh phng trong khong thi gian ln hn 60s. l xc sut mch s c BER > trong khong thi gian ln hn 60s tc l mch tr nn khng s dng c v c tnh theo biu thc: (3.26) Kh nng s dng tuyn Kh nng s dng tuyn c biu th bng phn trm v c xc nh theo nh sau: (60))% (3.27) Nh vy ton b cc tham s tnh cho ng truyn cng nh cc tham s nh gi cht lng ng truyn c s dng ngi thit k a ra cc quyt nh v kh nng lm vic ca ng truyn, tnh xem ng truyn c cng sut cung cp cho my thu v c hot ng tt hay khng. Ngoi ra cng da vo cc tham s ny c th hiu chnh li cng sut my pht, quyt nh dng cc bin php phn tp.3.12 Cc ch tiu k thut nh gi cht lng ng truyn - Ba ch tiu ch yu nh gi cht lng ng truyn: + khng s dng ng cho php ( i vi ng trc) vi L < 600Km (3.28) Trong L [Km]. + khng s dng c ca mng ni ht( mi mt u cui) = 0,0325% + khng s dng c ca hnh trnh ngc(mi mt u cui) = 0,0225% Mc ch ca tnh ton ch tiu cht lng l nhm xc nh xc sut vt cc ch tiu BER, bng cch s dng cc gi tr ca cc xc sut tm ra trong cc tnh ton ng truyn. - Cc mc tiu t l li bit BER c s dng sao cho BER khng c ln hn cc gi tr sau: + 1. trong hn ca thng bt k i vi thi gian hp thnh 1 pht, vi 280km < d < 2500km. + 1. trong hn 0,045% ca thng bt k i vi thi gian hp thnh 1 pht, vi d < 280km. + 1. trong hn ca thng bt k i vi thi gian hp thnh 1 giy, vi 280km < d < 2500km. + 1. trong hn 0,006% ca thng bt k i vi thi gian hp thnh 1 giy, vi d < 280km.3.13 nh gi cht lng ng truyn, lp t thit b a vo hot ng y l mt bc c tin hnh sau khi tnh ton c kh nng lm vic ca ng truyn v tnh xong cc tham s cn thit thit lp ng truyn, c ngha l trn tnh ton thit k th ng truyn hot ng. Tuy nhin vn cn nhiu vn tn ti s tc ng ln ng truyn v c th lm cho kh nng lm vic ca n khng nh mong mun ca ngi thit k. Ni chung vic nh gi cht lng ca ng truyn l da vo cc gi tr tnh c cc bc thit k trn. Cng vic cui cng l lp t thit b a vo vn hnh, tin hnh cn chnh anten thu c tn hiu t my pht. y cng l lc i chiu gia thc t v tnh ton l thuyt ph hp vi nhau hay khng bng cch th cc tn hiu hai bn thu v pht. Chng hn, tin hnh o cc thng s sau khi lp t nh(cng sut my pht, phn tch khung 2Mbit/s, t s bt li BER = v BER = )3.14 Kt lun chng Chng ny trnh by kh chi tit v l thuyt thit k tuyn vi ba s. Vic truyn bng ng truyn vi ba s, tuy cht lng khng tt v n nh bng ng truyn hu tuyn nh cp ng trc,quang. Nhng vic tnh ton thit k n gin hn v chi ph xy dng ca mt h thng l thp hn. Mt khc,vi nhng tuyn truyn c ly ngn th vic s dng ng truyn vi ba s tn dng c hiu qu m cht lng vn m bo. Thit k tuyn vi ba s tun theo cc quy nh vi cc bc tnh ton r rng, nhng mang tnh hin thc th phi bm st vo thc t. Theo bi thit k tng i hon chnh ca h thng vi ba s trn, vi vic la chn cc thit b cng nh tnh ton ng truyn l tng i chnh xc, do vy n c th a vo thc t. Mt khc vi thit k nh vy cng m bo cc ch tiu v kinh t, ph hp vi iu kin ca t nc hin c.

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