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 Example1: Mt HTTT viba có công sut phát Pt = 8!" thông tin t#$ng c% l&'() *+m" t,n s-'.) /0h" su& ha$ ($ cáp 2 3,u phát'4.tx) v5 3,u thu'4.#x) l,n l67t l5: /(" 9( H; s- t<ng ch c>a anten phát'0tx) v5 thu'0#x) l,n l67t l5: /(i" 9*(i Má& thu'ch6a ?@ su& ha$ ($ cáp) có h; s- tAp Bm'CD9) "F( T-c 3bt l5'Gb) 89Mbps" v5 ph6ng pháp 3iIu chJ KPL+ Chi;t 3tAp Bm anten T a = 9*+" ? = 1"/8e9/ NO+ H; th-ng 2 nhi;t 3tiu chuQn T = 9R+ a Tnh công sut bSc xA 3ng h6Ung t6ng 36ng: EVGP b Tnh P$ut c Tnh tW s- sóng mang t#n tAp Bm 'XOC) L$luti$n: a Xông sut bSc xA 3ng h6Ung t6ng 36ng: EVGP: EVGPY(Z = PtY(!Z 4.txY(Z [ 0txY(iZ = 1l$g8 / [ / = \F(! b P$ut: P$utY(Z = EVGPY(Z 4.sY(Z [ 0#xY(iZ 4.#xY(Z  T #$ng 3ó 4.s l5 t]n ha$ t#$ng ?hông gian t % ($: 4.sY(bZ = R9"\\ [ 9l$g(Y+mZ [ 9l$g.Y0hZ =^ 4.sY(bZ = 1/F(b =^ P$utY(bZ = \F 1/F [ 9* 9 = F(! c TW s- sóng mang t#n tAp Bm 'XOC): 'XOC)Y(bZ = P$utY(bZ 1l$g'?T) H; s- tAp Bm má& thu: CD = 4.#x [ 'CD9 1)O'1O4.#x) = 1_"9 [ '1_"F 1)O 1_"9 = 1"*8 [ "*1 = R"1  T #$ng 3ó T l5 nhi;t 3tAp Bm t6ng 36ng: T = T a [ T'CD 1) = 9 *[ 9R'R"1 1) = 9**\+ <ng thông = GbO9 = 891_F O 9 = \11_F H =^ 'XOC)Y(bZ = F 1l$g'1"/8e9/ 9**\ \1eF) = F [ 118"\ = *1"\ (

BT HTVT

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BT HTVT

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Example1:Mt HTTT viba c cng sut pht Pt = 80w, thng tin trong c ly(d) 50Km, tn s(f) 3Ghz, suy hao do cp u pht(Lftx) v u thu(Lfrx) ln lt l: 3dB, 2dB. H s tng ch ca anten pht(Gtx) v thu(Grx) ln lt l: 30dBi, 25dBi. My thu(cha k suy hao do cp) c h s tp m(NF2) 7,6dB. Tc bt l(Rb) 82Mbps, v phng php iu ch QPSK. Nhit tp m anten Ta = 205K, k = 1,38e-23 J/K. H thng nhit tiu chun T0 = 290K.a. Tnh cng sut bc x ng hng tng ng: EIRP.b. Tnh Pout.c. Tnh t s sng mang trn tp m (C/N).

Solution:a. Cng sut bc x ng hng tng ng: EIRP:EIRP[dB] = Pt[dBw] - Lftx[dB] + Gtx[dBi] = 10log80 - 3 + 30 = 46dBw.b. Pout:Pout[dB] = EIRP[dB] - Lfs[dB] + Grx[dBi] - Lfrx[dB].Trong Lfs l tn hao trong khng gian t do: Lfs[db] = 92,44 + 20logd[Km] + 20logf[Ghz]=> Lfs[db] = 136db.=> Pout[db] = 46 - 136 + 25 - 2 = -67dBw.c. T s sng mang trn tp m (C/N):(C/N)[db] = Pout[db] - 10log(kTB).H s tp m my thu:NF = Lfrx + (NF2 - 1)/(1/Lfrx) = 10^0,2 + (10^0,76 - 1)/ 10^-0,2 = 1,58 + 7,51 = 9,1Trong T l nhit tp m tng ng: T = Ta + T0(NF - 1) = 205+ 290(9,1 - 1) = 2554KBng thng B = Rb/2 = 82.10^6 / 2 = 41.10^6 Hz=> (C/N)[db] = -67 - 10log(1,38e-23. 2554 . 41e6) = -67 + 118,4 = 51,4 dB.

Example2:Cho my thu gm 3 tng. Tng 1 l b lc thng di c bng thng 70Mhz ti tn s trung tm 2,4GHz, suy hao ca b lc l 2db. Tng 2 l tng khuch i c h s khuch i 10db v h s tp m 2db. Tng 3 c h s khuch i 10db v h s tp m 2db. Gi thit h thng nhit phng.a. Tnh h s tp m tng ca my thu.b. Tnh t s tn hiu - tp m u ra nu cng sut tn hiu u vo Pin = -80dBm. C th thay i th t cc tng cho h s tp m tt hn c khng?

Solution:Tng 1: NF1 = 2db = 10^0,2 = 1,58 ln => G1 = 1/NF1 = 0,63 ln.Tng 2: G2 = 10db = 10 ln; NF2 = 2db = 1,58 ln.Tng 3: G3 = 10db = 10 ln; NF3 = 2db = 1,58 ln.a. H s tp m tng ca my thu:NF = NF1 + (NF2 - 1)/G1 + (NF3 - 1)/G1.G2 = 1,58 + 0,58/0,63 + 0,58/6,3 = 2,6 = 4,15 dB.b. Tnh t s tn hiu - tp m (SNRin).C Pin[dbm] = k.T.B.NF.SNRout = -174 + 10logB + NF[db] + SNRout[db] = -174 + 10log70.10^6 + 4,15 + SNRout[db] = -80dBm=> SNRout[db] = 11,4 db = 13,8 ln.M NF = SNRin/SNRout. => SNRin = NF. SNRout = 2,6 . 13,8 = 35,88 = 15,5 dB.Nhn vo biu thc tnh tp m h thng ta thy: NF1 = NF2 = NF3; nhng G1 < G2 = G3; NF nh nht khi chuyn tng 1 thnh tng 3, sau tnh ton li cc thng s.

Example3:H thng thng tin viba s hot ng tn s 2Ghz c cng sut pht 80w, tng ch ca anten pht 25dBi. C ly thng tin gia trm pht v trm thu l 50Km. Anten trm thu c tng ch 20dbi v nhit tp m anten 100k. Anten c ni vi my thu c h s tp m 2db. Gi thit tc bit 34Mbit/s v dng phng thc iu ch QPSK v cc b lc l l tng.a. Tnh cng sut bc x ng hng tng ng( EIRP)b. Tnh cng sut tn hiu nhn c ti u vo my thu.c. Tnh t s sng mang - tp m.

Solution:

a. Cng sut bc x ng hng tng ng: EIRP:EIRP[dB] = Pt[dBw] + Gtx[dBi] = 10log80 + 25 = 44dB.b. Tnh cng sut tn hiu nhn c ti u vo my thu.

Prx[dB] = EIRP[dB] - Lfs[dB] + Grx[dBi]Trong Lfs l tn hao trong khng gian t do: Lfs[db] = 92,44 + 20logd[Km] + 20logf[Ghz]=> Lfs[db] = 132,44db.=> Prx[db] = 44 - 132,44 + 25 = -63,44dBw.c. Tng t example1.

Example4:Cho si quang trn c chit sut li n1= 1,53, n2= 1,5. Bm nh sng c bc sng ld= 1,3um, bn knh li 50um. Cho phn b chit sut trong li c dng nhy bc(g >>>1).a. Tm s mode sng c th lan truyn trong si quang.b. Tm bn knh si quang si quang l n mode.

Solution:a. S mode sng trong si quang:N = (g. V^2)/2.(g+2) = V^2/2Vi V l tham s kch thc chun ha: V= 2.pi.a.NA/ldtrong : a l bn knh si quang.NA l Khu s: NA = sqrt(n1^2 - n2^2) = sqrt(1,53^2 - 1,5^2) = 0,3.

=> V= 72,46=> S mode sng trong si quang:N = 72,46^2/2 = 2625 mode.b. Tm bn knh si quang si quang l n mode. si quang l n mode => N=1 => V=sqrt(2)=> a = 1,94um.

Example5:Mt my pht viba c u ra 0,1w ti 2Ghz. My pht ny c s dng trong thng tin viba c cc anten pht v thu l cc parabol, mi anten c ng knh 1,219m.a. Tnh h s khuch i ca anten.

Solution: n(pi.D.f/c)^2D: L bn knh anten.