光纖通信(第5版)課后習(xí)題答案要點(diǎn)

1、CHAPTER 1 nBEK OPTIC COMMUNICATIONS SYSTEMS J-l dB= 10 logjQfPPi) Loss Fractional Fg航 (dB) （pypi） 0 I -1 Q.8 -3 0.5 -6 0,25 10 OJ -20 001 30 0.001 40 0.0001 -50 0.00001 29 32 1-2 dB = 10 logiQ (Pp/PJ dB/10 = log|C(P3fl*1) dB/10 P2/Pl，10 P2 = P1Jt 10 dB/IU = (1.001 x 10 JBdO P 2 mW P嚴(yán) Pi 10 dB/10 -3d

2、l/10 2 x 10 x 10=0159 mW P2 = Pl0dB/l= lOxlO9 P嚴(yán)二叭nW -956-9-3 lOx 10 x 10 = 1() x 10 = 10 W = 1 mW 1-5 From *he lext, wc find that RG-19/U weighs 1110 kg/ktir 1 mile of cable x 1110 kg/km x 1.609 km/milc x 2 2 Lbs/kg = 3929 lb 乩 1-6 From the text, we find lhaL RG-19/U has an attenuauon of 22,6 dB/kn

3、i at 1D0 MHz Using RG- 19/U, lhe allowed lass is: Loss = 10 log|Q n1 10 logio = -40 dB 卩 210 2 Maximum coaxial cable length 二 022.6 = 1.8 km Using a fiber with los備 the maximum length of fiber is： Length = 40/5 = 8 km J-7445 x IO6 bps x 1 message/64.000 bps = 69$ messages I -8 Wilh manually operated

4、 blinker lighu, I would guess about 2 or 3 bps. J -9 Conducting Cable 900 (pairs/cabJc) x 24 (messages/pair) = 21,600 meKsages Fiber 144 (hbera/cible) x 672 (messages /fiber) = 96t768 messages 96768 (fiber tahlCy21,600 (copper cable) = 4,48 About 45 copper cables arc needed to carry lhe same aiiwunt

5、 infonnaiion as the single fiber cable. At the DS-4 rate* each fiber carries 4032 messages. The coinpardiive message rale a are Ltien: 144 x 4032/21600 = 26,88 or about a factor of 27. 1 10 2 二 126.67 U1T1P A吶號町=3,8484s bll Frequency (Hz) Wavelength (m) g X = c/f = 3 x 10 /f Region of EM Spectrum 10

6、 3 x 107 Power C 3x10* n 1aW / v =2x10 m/s n L5 lhe disLancc traveled in one second (and, ihu the fibci length) is: = v/ = 2 x 10s x 1 = 2 x 10h m Solve for D=2r, with the result: Dlr U77w CHAPTER 3 LIGHTWAVE FUNDAMENTALS 3-1 A(VL) = (RL)2 (VL)i = * M 2, AX = -2 - 幾 For M positive (Z 1.3 pm), the lo

7、nger wavelength travels faster. A(t/L) 1.3 pm), the shorter wavelength travels faster. A(t/L) 0 32 X = O,85 pg AX = 30 nm, M =90 ps/nni x kin d(x/L)二-M 從二-30 k 90 三 2700 ps/km For AX - 2 nm, A(t/L = - 2 x 90 = -180 ps/kni 331.55 gm, M = - 20ps/nm x km, A(t/L) - -M A. For AX 30 tini. A(t/L = - M AA.

8、- -20) x 30 = 600 ps/km For AZ = 2 run, ( VL) = - (- 20) x 2 = 40 ps/km 3-4 fx L = 0.5/A(i/L)f OplicaL-RZ x L = 0.3 5/A(i/L) f 3血 x L = 0.350譏)、ElectricalRmkz x L 二 0.7/A(t/L) X = O.5J5 pm:(】)M = 30 nm, A(t/L) = 2.7 ns/km (2) AX = 2 nm, A(xL) = 0J8 ns/km X = 1.55 pm: (I) A?. = 30 run, A(t/L) = 0.6 n

9、s/kin (2) AX = 2 nm, A(t/L = 0.04 ns/km i X AA SdBfopt) (MHz) f 3dB( elect) R磁(Mb/s) rNRZ (Mb （呻） (tun) 100m I km 10km 100m 1km l Okra 100m Ikni ICkm o 0.85 - 30 150 15 18.5 1300 0 13 2600 260 26 0.85 2 277S0 2778 277.8 19444 944 194 388HU 388S 3K9 h55 30 8330 I 833 833 5R3Q 583 58.3 H66O 1166 117 U

10、S 2 125000 125000 I2JW i S7500 8750 Lj A75 175000 17500 1750 1 3-5 = ka - 2X n U 一 = 7,66 x 10c rad / m 為.82汎附 kg|ass kon (7.66 x IO6) (L5) = L15 x IO7 rad/m is =兀T = 0*82/1 B = 0.55 pn 32 32 口羅,侶：,Af咱嚴(yán)l， When AX = 20 nm, AT/ f = 20 / 820 - 0.024, or 2.4% Af=878x 10I2Hz When AA = 1 nm, Af/ f = l / 8

11、2050 0.0012. or 0.12% Af=439x IO11 Hz 10 logio 0.68 = -1.67 dB All “2 68% frtnsinission loss 二 1.67 dB 3-8 ti)= 1.48, n2 - L46 9 二 |pj . perpendiculai Rr -Ip 2 T parallel -n? cos Oi + ni Vhi njsin- Pp= / ； ns cos Oi +ni Vnj - nf siifOi 0n = tan1 (巾伽)=44.61* T 6C - sin (nj/np = 80.57* 0.0 01020304050

12、63 70 80 K and 90 Incident Angle,気(deg) 3-9 Prove tan 如=巾加 Set pp = 0 in Eq. (3-29). Solve for 比=9 29 4 0 2 n. u. 5 trrlT 22 n 63 32 Thus, Gj is as shown in the inangle, so (hat Um 9j = n/ni or tan = n2i 3-10 a - k(nf sin2 創(chuàng)胡嚴(yán) 9C sin-1 Sl= sin1 !也=80.57t nlI.4X k = 21 如=7.66 x 1 妙 1.82 x 1滬 耳（de卽 a(

13、m) z / X 0 0.5 1 2 4 0 1 1 1 1 S2 9.8 x 1C5 1 .67 ,45 JO .04 84 t.4x IO6 1 .56 31 .10 ,01 $6 1,7 x IO6 1 .50 .25 a 004 88 | L8x 106 1 .48 .23 .05 *003 90 i L9x IO6 1 47 .22 3 M2 3-11Lf= -10 log exp (-0.693 (的也胡旬 Let Lf = 1 dB, solve for 恨血: 呱血=058 3-12 PT-P14p2 = Po1+P02+P|COS(WIrf + ll)+ ?22 COS ml

14、 + * 2) = 4*2 cos (閃祁* i +cos / ?) where 二 樸 亠 如 and we used the idenlily cos A 十 cos B = 2 cos (0 $2-如=0 P r = 4 + 2 cos (他 (1 ) P三 P? = 2 + cos (at + m(+) P2 = 2 + cos (mmt 十啊+ n/2) Pt 4 十 cos + 0 + ccs (o)mt + | + k/2) P =2 + cos(H4i) P? u 2 十 cos + i + n) Pt =4 6- (4)P-* - 4 + 2 cos hi 十 cos (林)

15、 2 cos (M / 2) Pae P P 0 2 4 tt/4 1.85 3.7 n/2 1.414 2.8 3k/2 0.77 1.58 K 0 0 3-13 n1 = 1.48 n2= L465 X= 1.3x IO6 (a) From Eq. 3-29 、_ ” L465 2 cos 85 + L48 Y - 1”48, sin?丘 L4652 cos 85 + 1.48 if 1,4653 - L42 sin2 85 087+j 0.2456 =().309 凸以？ = 1 沖力 0.187+j 0.24560.309ei如 (b) The alienuaiion factor i

16、s given by Eq. (3-35): 2?r 1.3 x IO6 YL482sin2 85- L4652 如0.1659 = 0.802 radian /pm 13 x IO*6 The decay is given by: 嚴(yán)工10 =- 3 ps/km AX = 2 nru Thu% M = 1,5 ps/nm x km Assume = 1300 run 15= 03$ 4 b” 入 一 1300 = X4 - 130(/ 【3乜3 扎A X4- BOO4-63 a3 = 0 SoNe by ileraiion A 13004-63X5 1320 | 3.00 x l0J2 -

17、2.856 x 1012-0.144 x 1013 = 0.03 x 1012 1315 2.99 x 1012-2.856 x IO12 - 0J43 x IO12 = 0,009 x 1012 ?i-Xo = 1315 - 1300= 15 nm Plot M vs” a from 1275 to 1325 nm to traphicuUy solve ibis problem. Exira for【his problems At: 1310nni ”310= Q93 pnm x km At: 1315 nm M二11315= L4 ps/nm k kni 4 I1315珅 3-21 T

18、= 20 pst soliton pulse willh (a) The maximum rate is; R =丄【=0.05 x 10 12 = 50 Gbps T 20 x 10 42 (b) The sysiem losses will limit the length of fiber that can be used. 3-22(My4)X - “” The slope is ihc value of dM/dX evaluated at 人” dM/dk - (Mo/4)l -為 1-3加) Evaluating at X = yields (dM/dX) = Mo as was

19、 to be shown. 3-23 Pwt/P,n = 10vL, In dB, the loss is; Loss = 10 log P9Ui/Pm = 10 log 0Ult,= 10 (yLUO) log 10 Loss = yL、y in dB/km and L in km. Compare this with the altemHtivc loss calculation dB m 10 log e2 Jl =-8.685a Loss (dB)=10 lag 嚴(yán)t = |0 Og 心見=X(KQ 2L) log e Loss = 23yL(0.434) - yL Thus, the

20、 two results are identical. 3-24 Poo/P,廠 1嚴(yán)” y = -0.5 dB/km L(km) Plt/Plh Efficiency (%) 1 0.89 89 10 0316 31.61 100 12 0.001 3-25 L = 02 dB/lnri. X = 1550 nm Use Beer s 1-aw P叔P刖W(wǎng) y = -0.2 dB/km L(km) P Out/Pth EiTiciency (%) I 0.955 95.5 10 0.631 63.1 100 0,01 J 3-26 X = 1.55 pm, = 2 nm M = -3 pi/

21、(nm.kin) A(t/L) = -MAX. = 3x2=6 ps/km This is much less than calculated in problem 3-2, as the dispersion M is much less at the longer wavelength. 3-27 For RZ coding: Rx L 二 035 A(r/I) 0.35 6x0*12 = 583xl0*mxft/ /fx = 5 is 0 dB/km Maximum allowed length： L - 24/50 = 0 48 km = 4B0 m This is a much sh

22、orter path Length than we calculated for the glas fiber m the precedint； problem CHAPTER 6 LIGHT SOURCES Solve using LapUce tiansfcanx Vin(s) Vo 也=sRCH V , =fl1 0U1 S(S4 1/RC) Is s + l/RCJ The 63% rise time is xr = RC. The voltage reaches 10% of its final value when 0J = 1 -e U05 tT The vohage reach

23、es 90% of its final value when t 二 230 ir The 10 to 90% nse time is then tr = (2.30405)ir = 2.19RC 丄 oul jwc 6-2 學(xué)訕=cos P 1530 io 1550 iim Al- 1550- 11530 = 20 nm. Find the corresponding bandwidth Af M = f2.兒=C/X2 5=35 (鳥曲-1.55 ：胡 Af-2.53 x IO12 Hz The channel spacing i$ ) 0 GHz = 101( H/ = Afc The

24、number of channels is: N - = 2.53 x W12/1010 = 253 chamw】軋 6-J 5 EDFA. Safiiraiion power - 20 mW This i$ the maximuni outpui power possible. Gain-G = i x525 dB Ps = 20 mW = 13 dBm - G =】325 -12 dBm - 10 log Pin (mW) P|a(mW) = 10d 2 = 0.06 mW 6 16 T 【TH 30 80 40 1 85 50 | 92 60 100 70 109 BO 1 120 Th

25、is is the increase in threshold current per uml of temperature nse. 6- 7 Emission wavclengih increase is 0.5 nm.C. Initial wavelength； Xj = 1510 nm Final wavelength： = 1310 nm + 10 C k 0r5 run/C =1310 + 5 二 1315 nm Using the analytic approximation for the dispersion M from Chapter 3: Al 1310 nm, M -

26、 0.93 ps/nm x kni AH315 run, M = 1.4 ps/nm x km A(t/L) - MA；. A(t/L) = 0.93 (1.5)= 1.395 ps/km f3 x L = 0.5/1.395 = 0358 x 1015 Hz x m - 358 GHz x km at IJlOiun A(i/L) = 1.4 (1.5)w 2.1 ps/km f3 dB x L - 0.5/2=O,23S x 】() Hz x rn = 238 GHz x km at 1315 nm The decrease in optical bandwidth is from 358

27、 to 238 GHz x km. A difference of 120 GHz x krm 6-lfi 恥 k A(7f= a3 Af7c At X - 1 AZ_ - 05 x 10 ntVC (106) Af=0J5 x 10l2 = 150GHz/C 6-19 From Eq (6*10), the grating spacing is: A = X/2 Xo = 1300 nm, n - 3.51 X = Xq/d = 13/3*51 = 037 pm A = 037/2 = 0J85pm (1st order) The second order diffraciion has a

28、 grating spacing of A = 2(0J 85) = 037 gm (2nd order) 6-20 入24/W EMITTED WAVELENGTH Hl-ebB1UJAAA Q.60,81 L2 1.41.61.82 BANDGAP ENERGY (eV) 6-21 P = g W, 1.24/X - L24/13 = 0.953 n - P/iWs = 2/50(0.953) - 0.042 The conversion etticiency is 4%. 6-22= 035/tr 3-dB Bandwidth 6-23 From Fig. 6*26, the mode

29、spacing is about AX - I ntn, k= 1.3 pm = AE/f Af= (AVk)f= 3叩fk = c Af = 3xlOK(10y(l .3x IO6)2 = 0.1775 x 10,2= 177.5 GHz 6-24 C-bandi 1530 to 1565 nm The C-band wavelength range is: A 9.39 THz L-IUnd; 7,09 THz The Raman amplifier has abandwidlh of6 THh Thus: C-band requires 1 Raman amplifier S-band

30、requires 2 Raman amplifiers L-band requires 2 Raman amplifiers Or. we could calculate the bandw idth of the combined bandi. Ji is： Q 439 + 939 + 7.09 = 20.S7 THz The required number of amplifiers is then: N 社 20.87/6 3,47 Or a total of 4 amplifiers. The 4 amplifiers can be centered uni(bonly aver th

31、e 1460-(625 nni range. To cover the 1625 - 460 - 65 nm range, the 4 amplifiers need be about 40 run apart. They can be centered ar U80t 1520, 1560, and 1600 nm. 6 3) Fiber I Pj Receiver Pc P2 Pj Amplifiers: Aif Aj, and Aj Fiber I is 100 km, fiber 2 is 150 km. P, = 0.5 mW 10 log 0.5 - -3.01 dBm +25 =

32、 21.99 dBm Pj = Pz-100(0.25) = 22 - 25 - -3 dBm P4 = Pz = 22 dBm Gain (Aj) = P斗(dBm) - Pj (dBm) = 22 - (-3) = 25 dB P5 = P4 - 150(0.25) = 22 37J-1M dBm P$ = J dBm Gain (Aj) = Pt (dBm) - Ps (dBm) = -3 -(-15.5) = 1X5 dB 6*32 Device Wavelength (jxni) LED 0. * Lt執(zhí)f + 1 q = Lp 0 Lt = Shp + L“p. A 3-dB co

33、upler is defined by Lhp 35 3 dB For at) ideal coupler, Ljtp - 3 dB implies that L L = 2 Lthp t 10.92 dB 3r L * 3 Lthp + Ltap =】1*3R dB 4: I. = 4 L/pHP + LtaP 】*84 dB 5: L 4 IzfHP + Lx?=】1 席4 dB The best coupler depends on the application The minimum loss between any two adjacent terminals excluding

34、I and 5 is 20.92 dB. Lf = (35 dB/km) (0+l km/scction)3 3 5 dB/section Lq h 0-8 dBT Lg = 0h2 dB LtaP = LTAP +lE = 3 + b5 = 4,5 dB Loss from lerminal 1 to terminal: 2： L 2 Lihp + Ltap + 2Lf + 3+ Lj； = 19.2 dB 3: L = 3 Lthp + LtaP * ?Lc + 5L + 2Lp = 27.6 dB 4: L = 4 Lthp + LTAP + 2L匚 * 7LS + 3Lb - 36 d

35、B 5: L = 4 Lthp + Sp + 2LC + 8LS + 4LF = 397 dB 9-8 . = 40 log(l.0) + + 21 +2Lf Le = LLf=O Loss from! (o Rh Rg, Rj, or R#: Loss from Tj lo R: 9-10 L w Lgf+Le 十 2【 = Star loss is： Li = 6.99 dB (from problem 9-8) Le = 2 dBt ! = (U dB, Ls = 0.2 dB Lp = (35 dB/km) (QJ kitbsection) =3.5 dB/scction ?卜 =(3

36、5 dB/km) (0.02 krrvseclioi) = 0.7 dBfsection L = Lst +Le + 2氐 + 2Ls * 2Lf 二 17.99 dB L = Lt +Le + 2L 4 2L + Lp= 15.2 dB JO bg (1/N) + 3 + 2(0 8) = -10 log (1/N) + 4.6 No. of Loss Terminals (dB) 3 9.37 6 12.38 9 】44 14 16.06 18 17.15 20 17,61 O 2 4 5 _ 3 Analog to Digital Ootfivaflers X DEMULTIPLEXER

37、 DigiUl to Analog Oonven2 t) x cos Received current: iR = pPt p = 0.5 iR = 0,5 |2.5 + 0*835 1 + 0.5 coswmlt cos + 5PECTRUM: * 1.25 mA 0.835 1 + 0.5 cos(02 U cos (mA) 417 mA.417rtiA ,104 mA 八.ICMmA .104 mA 八.104 nA lPlQOO血 1MHzlxl-lOOO 2MHz 1(/+ 1000HzIxlCTlOOO The problem may be redone to include th

38、e second harmonic term. In this case P 耳 p + 討 ig + 電 ig * 站恥 is = i -10 is the signal current. 10J2100111000 I 0 bRZ RZ MANCHESTER BIPOLAR n , ni t_n_r n. . nnn, . . USl TJlKLmUWTLTTL 10-13 Let: Level 1:1 = 0 , q = ij =0. Level 2; I = 73 mA J | IB mAt = 0 Level 3: J = 146 mA, a 】*b mA,丘 n 1,6 mA I

39、= cl + VDC-5V Level 2: Load Jine equations： VDC = vd + klR + VCE1 甘DC =vd + iR + VCE2 . VDCJlvd At vCEl - vctl = 0, lcJ = g xc2 5 - L4 Set Li 立 icj 80 mA, R = Qg = 45 口 Al i創(chuàng)=ic2 -0. vCEl = V2 VDC * vd = 5 - L4 = 3 6 V From the transistor characteristics curve* if iQ L6 mA, then ic = 75 mA* Al satur

40、ation Vdc 切八CE (勉)5 - 1,4 * 0.3* icl=R= 45 =73 mA Thust at level two we have: ij 1.6 mA, ig 二 0 icl - 93 mA, Q = 0 I J + i2 73 mA Al level three we have, i L6mA, i2 .6 mA I = icl + ic2 = L46 mA 10-14 From Eq, (10-9) P = Pdc*a1 is + a2 The signal cunent is similar to the given by Eq. (10-20t but with

41、 two subcarriers and (wo modulation frequencies. S =心(1 + nij cos oin| t) cos osc1 t + 1S2 (1 + m2 cos scjt2 Expanding lhe 乜 lerm only: a2 is (1 十 m cos aml0 2 cos2 usct + l|(l +mcos omt)2 cos2 (Ogot + 2疋(1 + m cos wmlt) (1 + m cos cnt) cos(uscil cos Ci)sc3t Multiply out all sinusoidal terms to comp

42、lete【he solution. The detected current is proportional (o the optical power. The frequencies present in the detected current art; Frequency 0 oSCi wml C2 士 am2 2t0SCI 加SC2 唄 am2 2wSCl 士血 I 2m$Ci 士 2uml 加隨2 士 wni2 2SC2 土 2如誡 SCJSCI G)m + 0Jm2 SC3 + 肋 SCI + (Aml WSC2 十SCI +* 佛m2 SC2 + mSCi + ffiml + m

43、2 To minimize the crossialk, we can: (1) Reduce (he nonlinearity coefficient a?. (2) Spread (he subcamers so there is no overlap (crosstalk) In this case, set 10d51 R= 10 Gbpst 碑一 RZ code4 n T = 1/R - l/IO10 = 10“ pulse duration X = 1550 nm f=c/. = 3x 108/1.55x 10*6- L935 x 10i4Hz l/f=O,517 x 10 14

44、seconds per cycle. The number of oscillations is then; T Vf 10-16 Total fiber capacity is: R = 8 x 10 = 80 Gbps = 80 x 109 bps The requirement is 64,000 bits per second per voice channel. The total number of voice channels allowed is then: 80 x 10*64,000 = L25 x 10r, L25 million voice messages. 10-1

45、7 Increasing the data rate to 80 Gbps requires that the electronics work at this rale, greatly increasing electronics complexity and cost. 10-18 The available capacity is 80 Gbps, Assuming a compressed video rate of 5 Mbps per video channel, the allowed number of channels is: 80 x 109/(5 x 10b) - L6

46、 x 10416,000 video signals 10-19 We have an OTDM system with wn 40 Gb?s chaiujels Total system data rale and capacity: R= 10 x40 = 400 Gb/s Maximum allowable pulse width: Im 丄三- =2.5xl0_t2s R 400 x!09 T = 2.5 ps 10-20 We can combine OTDM and DWDM io increase the total capacity of a single fiber. For

47、 the described QTDIWDWDhl system, the total capacity would be： R =35 x 4 x 10ll = 14x J0H b/s- 14Tb/s 10-21 Types of possible modulators： External moduIators-Electrooptic and Electroabsorption- 10-22 When: V = 0. Wg = 0.9473 eV. Slate I. Transmission. V 工2一 We = 0.9458 eV. Sate 2. Absorption. 1 = 13

48、10 ran The photon energy oi this wavelength is: 比=0 = 094656 eV p 131 A photon is transmitted when iis energy is leas than th*: bandgap energy. That is: 仆化 This is true in state !. A photon is absorbed when its energr is greater than the bandgap energy. Ihui is; W W P s This is true in stale 2. 10-2

49、3 Elec (to absorption modulator L 150 pm = 0.015 cm a = 75 cm L Exljnction ratio: f/f=：30=101oge1Mo 30=10k)g 嚴(yán)譏 103 =嚴(yán)加 lnl03 = 6f90X=0.03Aa Aa = 23(X2 cml 十= 75 + 230,2 = 305.25 cm 10-24 ER - Extinction Ratio Transm isxioit State Powerc 2tJL ER =- = Absorrion State Power 2 Irtenion loss: 寸at iOlog

50、e - lOloge2 - lOloge CHAPTER 1! NOISE AND DETECTION AW 叱 T = 35C = 308 K RL = 50 Q Af=6x 10 HzR 50 k fl k- L38 x QT23 J/K 喬=- 2.04 Ji 045 A2 iNT(RMS)= -33 dB 11J PNS = 2eAfRLpP + Ip) Set pP - 1口 P 二*P = 05 =】071 * 11- 4 Given: p - 0.5 A/WT Ip = 2 x 10A, R_= 2000 Q, Af= 50 x 10bHz, T = 313 K.k = 1.38

51、 x 10-23 J/K. (a) PNS = 2eAf(pP + 1D) RL2 x IO9(0.5P + 2x IO*9) NT = kTAf = 864 x IO-1Set P ss P NT P = 54 nW (b) Ps = (pPopt2 Rl = (OJ k54x IO 6)2 2000 = 1 *46 pW S = 3!25 = 34.9 dB S/N = ES Want (S/NjjB = 30 dB, then S/N = 1000, L25 x IO-11 l A1 103(8.28x IO*12) 8.28 x IO-12 - 0 00,03 M = 3.2 11 -7(a) NHP = 氏勺十iRi - V2ElpAf +(4kTAfh3. X erf (X) 3.0 1 . 2.32 (10_5) = 0.999976791 3.5 1 - 7.71 (107)-0.99999922S 4.0 * 1.59 (lO g) = 0,999999841 4.5 0.999999999799 5.0 0.99999999999843 5.5 0.9999999999999252 6.0 0.99999999999999997S2 11-il Given BER= 10*9, RL= 50, T - 30() Kf R = 500 Mb