BI TP TRUYN THNG QUANG

Ging vin:Sinh Vin: MSSV:Trn Minh c 0920026Nguyn Tn PhtL Hng PhcV Tn Ti

Chng 2:2-1. Mt trng in c th hin nh sau:E = (100 ex + 100 ey + 100 ez)(1)Hy biu din min in trng 100 MHz.Gii:Vi f = 100 MHz, suy ra:= 2 = 2.108 (rad/s)Thay vo (1), Ta c:E = [100ej30t ex + 20e-j30tey +40ej210tez] = 100 ex + 100 ey + 100 ez2-2. Mt sng c dng y = 8 cos 2 (2t 0.8z), trong y tnh bng micrometers v hng s truyn l m-1. (a) Tm bin .(b) Tm bc sng.(c) Tm tn s gc.(d) Tnh dch chuyn khi thi gian t = 0 v z = 4 Gii:Theo dng chung th:y = (amplitude) cos(t - kz) = A cos [2 (t z/)]Do :(a) Bin = 8 (b) Bc sng = 1/0.8 = 1.25 (c) Tn s gc: = 2 =2 x 2 = 4 (d) Ti t = 0 v z = 4 , Ta c:y = 8 cos [2(-0.8 ) (4)] = 8 cos [2(-3.2)]=2.4722-3. Tnh nng lng ca nh sng c bc sng 820nm, 1320nm v 1550nm? Tnh gi tr ca hng s truyn k v bc sng ca n?Gii:Nng lng ca nh sng c xc nh theo cng thc sau:E = Trong c tnh bng m.Cng thc tnh h s truyn sng: k = Trong c tnh bng m.(a) Vi 0.82, E = 1.240/0.82 = 1.512 eV.Vi 1.32, E = 1.240/1.32 = 0.939 eV.Vi 1.55, E = 1.240/1.55 = 0.800 eV.(b) Vi 0.82 k = 2/ = 7.662m-1.Vi 1.32 k = 2/ = 4.760m-1Vi 1.55 k = 2/ = 4.054m-12.4 Xt 2 sng nh sng v di chuyn trn cng 1 hng, nu 2 sng c chung tn s nhng khc bin v pha , chng ta c th biu din nh sau:= cos (t - )= cos (t - )Theo nh lut chng chp, gi sng X l tng hp ca 2 sng v . Chng minh X c th c vit:X = Acos( t ) Vi = + + 2 [ cos()cos() + sin()sin() ]tan() = Gii:Ta c:= cos (t - ) = [ cos(t).cos() + sin(t).sin() ]= cos (t - ) = [ cos(t).cos() + sin(t).sin() ]Cng v , ta c:X= + = [ cos(t).cos() + sin(t).sin() ] + [ cos(t).cos() + sin(t).sin() ] = [ cos() + cos() ] .cos(t) + [ sin() + sin() ] .sin(t)Do , , , l cc hng s nn ta t:cos() + cos() = Acos() (1)sin() + sin() = Asin() (2) + v theo v, ta c:[ () + () + 2.cos().cos() ] + [ () + () + 2.sin().sin() ]= + + 2 [ cos()cos() + sin()sin() ] = + + 2 cos( - ) = (pcm) v theo v, ta c: = tan() (pcm)Vy, ta c th vit:X= + = Acos( t )2.8

: nh sng truyn trong khng kh va phi 1 b mt thy tinh vi gc = , l gc hp bi tia sng ti v b mt thy tinh. Khi va vo, 1 phn tia sng b phn x, 1 phn tia sng b khc x. Nu tia phn x v khc x hp thnh 1 gc th chit sut thy tinh l bao nhiu? Tm gc ti hn.Gii:a) Theo nh lut Snell: . sin(a1) = . sin(a2)=> 1 . sin (90-33) = . sin(90 (90-33)) (do khi phn x ton phn th gc ti bng gc phn x).=> = 1.54b) iu kin v cng thc tnh gc ti hn: > v = arcsin = arcsin () = Bi 2.9: Mt ngun sng di nc cch mt nc 12 cm (n= 1,33). Tnh bn knh vng sng trn mt nc?

Gc ti gii hn: =arcsin ()= 48.75 Bn knh vng sng l: r=tan(). D= tan(48.75). 12= 13.3 cmBi 2.11: Tnh khu s c n1=1.48, n2=1.46. Tnh gc vo ti a cho php nu truyn t khng kh n=1 khu s NA= = 0.242Gc cho php: = arcsin()= arcsin()= 14Bi 2.18: Bit khu s NA= 0.2 s mode truyn M=1000 modes v bc sng =850 nmTnh ng knh li, s modes bc sng 1320 nm v 1550 nmTa c: M=Suy ra: a= = 30.25 uma. D= 2a= 60.5 umb. M== 414c. M== 300

2.20 Cho: a = 25 m, n1 = 1.48, .a. Bc sng = 1320nm, tnh tn s chun ha, s mode.b. Nu , th c bao nhiu mode v tnh power flow cladding?Gii:a. NA = = 0.21. = 25. = 312.b. NA = 0.36. = 42.84

i vi Step-index fiber: = 4.4%.

2.21C: Step-index fiber n mode, bc sng 820nm, n1= 1.48, n2=1.478, bc sng: 1320nm.a. Tnh bn knh core.b. Tnh NA.c. Tnh Gii:a. Ta c i vi n mode th tn s chun ha b hn 2.405.NA = 0.08 6.3 .b. NA = 0.08.c.

2.28 Tnh NA:a. plastic step-index fiber c n1= 1.60, n2=1.49.b. Silaca core step-index (n1 = 1.458) v silicone cladding step-index fiber(n2= 1.405).Gii:a. NA = = 0.583.b. NA = 0.39.Bi 2.30: Mt ng silicon c bn knh bn trong v bn ngoi ln lt l 3 v 4 mm bn trong c ph mt lp thy tinh. Tnh dy lp thy tinh nu lp li c ng knh 50 um v ng knh ngoi lp m l 125um

Ta c:

T :R== 2.77 mm dy l 3- 2.77= 0.23 mm

Chng 3:3.1 Kim chng li cng thc sau:(dB/km) = (km-1) (1)

V mi lin h gia vi n v dB/km v p c n v km-1.Gii:Ta c:P(z) = P(0) Suy ra:= (2)Thay (2) vo (1), ta c:(dB / km) = log = log = 10 log e = 4.343 (1 / km)3-2. Mt si quang c suy hao ln lt l 0.6 dB/km vi bc sng 1300 nm v 0.3 dB/km vi bc sng l 1550 nm. Gi s hai tn hiu quang cng truyn ng thi vo trong si quang vi mt cng sut quang l 150 vi bc sng 1330 nm v mt cng sut quang 100 W vi bc sng 1550 nm. Hy tnh mc cng sut W ca hai tn hiu vi khong cch:(a) 8 km(b) 20km.Gii: Chuyn i cng sut quang hai bc sng ra dBm:+ Vi bc sng 1330 nm: P(100) = 10 log(100 / 1.0 mW) = 10 log (0.10) = -10.0 dBm.+ Vi bc sng 1550 nm:P(150) = 10 log(150 / 1.0 mW) = 10 log (0.15) = -8.24 dBm.(a) Vi khong cch 8 km ta c mc nng lng nh sau:+ Vi bc sng 1330 nm: P1300 (8 km) = -8.2 dBm (0.6 dB / km) (8 km) = -13.0 dBm = 50+ Vi bc sng 1550 nm:P1550 (8 km) = -10.0 dBm (0.3 dB / km) (8 km) = -12.4 dBm = 57.5(b) Vi khong cch 20 km ta c mc nng lng nh sau:+ Vi bc sng 1330 nm: P1300 (20 km) = -8.2 dBm (0.6 dB / km) (20 km) = -20.2dBm = 9.55+ Vi bc sng 1550 nm:P1550 (20 km) = -10.0 dBm (0.3 dB / km) (20 km) = -16.0 dBm = 25.13.3. : 1 tn hiu quang vi bc sng c nh mt 55% cng sut khi truyn trn dy dn 3.5km. Hi suy hao ca dy dn?Gii:Cng thc tnh suy hao theo dB/km: (dB/km) = () (*)vi L l di truyn tn hiu quang.Theo bi, tn hiu quang mt 55% cng sut, tc l cn li 45%, vy = 0.45 (dB/km) = 0.45 = 0.99 (dB/km)3.4. : Mt ng dy di lin tc 12km c suy hao 1.5dB/km.a) Cng sut ti thiu cn a vo ng dy duy tr cng sut u ra l 0.3 W?b) Cng sut a vo l bao nhiu nu ng dy c suy hao l 2.5dB/km?Gii:a) Theo cng thc (*), ta c:1.5 = ()=> = = 18.93 (W)b) Theo cng thc (*), ta c:2.5 = ()=> = = 300 (W)

Bi 3.13:a. Led lm vic bc sng 850 nm vi rng ph l 45 nm. Tnh gin xung ns/km do tn sc vt liu? Tnh gin xungkhi rng ph l 2 nmc. Tm gin xung ti 1550 nm cho led vi rng ph l 75 nmGii:a. Ta c =80 ps = 80ps .2nm= 0.16 ns /kmb. Ta c =22 ps.nm.km-1 Dmat()= 22. 75= 1.65 ns/km

3-17. Xt mt si quang step index vi ng knh li v v ln lt l 62.5 v 125 m. Ch s khc x ca li n1 = 1.48 v bin thin ch s khc x l = 1.5%. So snh phng thc tn sc ns/km ca si quang ny phng trnhTmod = Tmax Tmin = (1)

Vi cng thc sau: (2)Trong L l chiu di si quang v n2 l ch s khc x ca v.Gii:Ta c:

Suy ra: n2 = n1 (1 - ). Thay vo (2) , ta s vit li nh sau:

= Vi :NA = = n1

Ta c:1 - 1 - 1 - 1 - Vi bc sng 1300 nm th h s ny l :

1- = 1- 0.127 = 0.8733.27 So snh dn xung hiu dng(rms pulse broading) trn km.a. multimode step-index fiber, n1=1.49, .b. Graded-index fiber, n1=1.49, .c. Graded-index fiber, n1=1.49, .Gii: (*nh nhn thm 10^3 vo kt qu cui cng v trn km*)a. = 14.33ns/km.b. 14.3ps/kmc. Nh vy a > b > c.Chng 4:4.3 Ga1-xAlxAs mt c band gap energy 1.540eV v mt cn li c x= 0.015.Gii:1.540 = 1.424 + 1.266x + .266x2 , suy ra x= 0.09V suy ra, .Bi 4.5 : Da vo cng thc E=hc/ chng minh v sao ph cng sut ca led li rng hn i vi bc sng di hn tng ng vi Cng mt bin thin nng lng t l vi bnh phng bc sng= =1.4

4.6.: 1 ngun LED InGaAsP pht ra bc sng cc i 1310nm, c thi gian ti hp sinh ra pht x v khng pht x tng ng l 25 v 90ns. Dng in a vo l 35mA.a) Tm hiu sut lng t ni v mc cng sut ni.b) Nu chit sut ca vt liu to ra ngun sng l 3.5, tm cng sut pht ra t thit b .Gii:a) Ta c: thi gian ti hp sinh ra pht x l = n/=25 ns.thi gian ti hp khng sinh ra pht x l = n/=90 ns.Hiu sut lng t ni l: = = = = 0.7826Mc cng sut ni l:= . = 0.7826. = 0.026 Wb) Cng sut pht ra t LED l:P = = = 0.00037 W4-10. Hy tn hiu sut bn ngoi ca Ga1-xAlxAs ca laser diode (vi x = 0.03) v laser diode c cng sut quang so vi dng in l 0.5 mW/mA.

Gii:Ta c: (1)Do vt liu l Ga1-xAlxAs:Eg = 1.424 + 1.266x + 0.266x2 (2)Thay (2) vo (1) vi x = 0.03, ta c:

= = 1.462M: = 0.8065 (m) V: = 0. 5 mW/mANn: = 0.8065 (1.462) (0.5) = 0.590Chng 6:6.5 Xt mt tn hiu quang c iu ch dng sin P(t) vi tn s , h s iu ch m v cng sut trung bnh P0 cho bi:P(t) = P0 (1 + m cos t)2Tnh bnh phng trung bnh dng c sinh ra do thanh phn DC Ip v dng tn hiu ip.

= + = ( P0)2 + ( m R0 P0 )2Trong p ng R0 cho phng trnh:R = Gii: = = (trong T = 2 / )== Ta c: = 0V: = T ta c: 6.6 PIN photodiode InGaAs, bc sng 1550 nm, ID = 1 nA, = 0.95, RL = 500 ohm, incedent optical power = 500 nW(-33dBm), reciever BW = 150 MHz.Giia. cng thc 6-13...b. Cng thc 6-14.

c. Cng thc 6-17.

6.7. Mt b nhn photodiode thc l c =1nA, =1nA, hiu sut lng t =0.85, li M=100, ch s nhiu F=, in tr ti =, bng thng B=10kHz. Gi s sng l sin c bc sng 850nm, photodiode c h s iu ch l m=0.85, nhit ng phng T=300K. so snh s ng gp gia cc phn nhiu khc nhau n SNR chung, v cc phn nhiu sau di dng decibels nh l hm ca cng sut quang trung bnh nhn c , cho t -70 n 0 dBm, tc l t 0.1nW ti 1mW.Gii: Ta c:==2qBF(M)=2qBF(M)=2qB=Vi = = = 0.58 A/W v = , P(dBm)= 10log[P(mW)].a) == == 6,548.b) == == 3,798.c) == == 3,798.d) == == 7,338.

a)

b)

c)

d)Bng thng t l nghch vi tt c SNR.ch nh hng ti v t l nghch vi .ch nh hng ti v t l thun vi . li M nh hng ti tt c SNR, c th: T l nghch vi v . T l thun vi v .

Chng 7:.30. :Xt 1 cng thc tnh SNR nh sau:= Da vo hnh 7-21, v SNR theo dB nh l 1 hm ca mc cng sut nhn c dB khi c dng ti =10nA, x=1, B=5MHz, m=0.8, = 0.5A/W, T=300K, =, bit = .Gii: = = =

Bi 7.7: Ta c in th cho mc 1 l V1 v in th ngng l V1/2.a. Nu =0.2 V1 cho p(y|0) v =0.24 V1 cho p(y|1) tnh xc sut li P0(vth) v P1(vth)b. Nu a=0.65 v b=0.35 tnh Pe.c. a=b=0.5 tnh Pe.Gii:a. Ta c:P0(vth)=dv=VP1(vth)=dv=Thay V=V1 v =0.2V1 ,v =0.24V1Ta c P0(vth)= 0.0065 v P1(vth)= 0.0185b. Pe=0.65(0.0185)+0.35(0.0065)= 0.0143c. Pe=0.5(0.0185)+0.5(0.0065)= 0.0143

7.8. LED, = 1300 nm, P = 25W, attenuation = 40dB, quntum eficiency = 0.65, t= 1ns, 5 cp electron-l trng.N = ..

Chng 11:Bi 11.1: a. Tc bm:=1.25x1027 (electrons/cm3)/sb. li tn hiu b ln nht:=7.5 cm-1c. Mt photon bo ha l:Nph.sat== 1.67x1015 photons/ cm3d. Mt photonNph==1.32x 10 10 photons/ cm311.7 (a) So snh PCE ln nht cho pumping 980 nm v 1475 nm trong EDFA cho tn hiu 1545 nm. Tri ngc vi kt qu o thc t cho PCE = 50.0% v 75.6% hoc 980 nm v 1475 nm pumping, tng ng.(b) S dng kt qu thc t trong cu (a), v cng sut tn hiu ng ra ln nht vi cng sut pump l 0 Pp,in 200 mW cho pump vi bc sng 980nm v 1475 nm.Gii:Ta c : PCE = Nh vy:PCE = = 63.4 % cho 980 nm pumping.PCE = = 95.5 % cho 1475 nm pumping.11.8 Khuych i cng sut EDFA Ps,out = 27 dBm, input level 2dB, bc sng = 1542nm.Gii:a. Tm li khuych i..b. minimum pump power...11.9. a) thy c s nh hng tng i ca nhiu loi nhiu ln b khuch i quang, hy tnh ton gi tr ca 5 loi nhiu vi li G=20dB v 30dB. Gi s bng thng quang bng vi bng thng truyn t pht ( rng ph 30nm), s dng cc thng s sau:

Trng i Hc Khoa Hc T Nhin TPHCMPage 1

=0.6=0.73 A/W=1W= 1550 nm= 3,77.HzB= Hz= 2=1000

b) thy c hiu qu ca vic s dng b lc quang bng hp ti u thu, cho = 1,25.Hz, hy tnh gi tr ca 5 loi nhiu vi li G=20dB v 30dB.Giia) = = + + + + Xt G=20dB, ta c:= = = 1,62.= 2qGB= 2.1,6..0,73.100.= 2,336.= 2q B= 2q(..G) B = 2.1,6..0,73.(.200). 3,77..=2,26.

= 4(G)()= 4(G)() = 4(0,73.100.)(0,73..200.)= 5,47.= (2 - B)B= (2 - B)B = .(2.3,77.-)= 2,65. = = 8,182. Xt G=30dB, ta c:= = = 1,62.= 2qGB= 2,336.= 2q B= 2q(..G) B= 2,26.= 4(G)()= 4(G)()= 5,47.= (2 - B)B= (2 - B)B= 2,65. = = 8,125.b) G=20dB; = 1,25.Hz= = = 1,62.= 2qGB= 2.1,6..0,73.100.= 2,336.= 2q B= 2q(..G) B = 2.1,6..0,73.(.200). 1,25..=7,49.= 4(G)()= 4(G)() = 4(0,73.100.)(0,73..200.)= 5,47.= (2 - B)B= (2 - B)B = .(2.1,25.-)= 8,74. = = 5,6. G=30dB; = 1,25.Hz= = = 1,62.= 2qGB= 2,336.= 2q B= 2q(..G) B= 7,49.= 4(G)()= 4(G)()= 5,47.= (2 - B)B= (2 - B)B= 8,74. = = 5,6.