1.Make a choice （共十题 每题1分）10p(1).Which of the following dispersion (色散)dose not exist in single-mode optical fiber? （ D ）A .material （材料）dispersion B.waveguide （波导） dispersionC.polarization-mode （偏振）dispersionD.intermodal （联合） dispersion(2).The unit （单位）of the fiber attenuation coefficient （衰减系数）is (C)A. dBB. dBmC. dB/kmD. dBm/km(3).the bands of Optical fiber communication is (B)A.0.01um-0.39umB. 0.8um-2.0umC.0.39um-0.79umD.100um-1000um(4).If the optical input power of a Optical amplifier is 10mW ，and the optical output power is 100mW as well ,then its output gain level is (A)A.10dBB.20dBC.30dBD.40dB(5)In order to make sure of the system BER （比特误差率） conditions , if the minimum optical input power of the receiver is 1 uW, the sensitivity of the receiver must be (A)A.－30dBmB.－40dBmC.－20dBmD.－43dBm(6)The principal （主要）light sources used for fiber optical communications applications are ：（D ）A.OA 、LDB.PIN 、APDC.PD 、LEDD.LD 、LED(7)laser （激光） action is the result of three key process ，which one of the following is not be included ？（D ）A.photon absorption （光子吸收）B.spontaneous emission （自发发射）C.stimulated emission （受激发射）pound radiation （复合发射）(8) A single mode fiber usually has a core diameter （芯径）of A.A. 10mB. 62.5nmC. 125nmD. 50mm (9)To make sure that the APD photo-detector works properly, a sufficiently D is applied across the p-n junction.A. high forward-bias voltage （高的前置偏压）B. low forward-bias voltageC. low reverse-bias voltage （低的反相偏压）D. high reverse-bias voltage(10) When DFA fiber amplifier uses as light Repeaters （中继器）, its main effect is B.A. amplifying and regenerating the signalB. regenerating the signalC. amplifying the signalD. reducing the signal noise(11) In graded-index optical fiber, the numerical aperture （数值孔径） NA can be expressed as C. A. 21n n - B. ∆2a C. ∆2n 1 D. 21n n a -(12) In practical SMFs, the core diameter is just below the cutoff （截止） of the first higher-order （高阶）mode; that is, for V slightly A.A. <2.4B. > 2.4C. =3D. =3.5(13) It is well known that the total dispersion in the single-mode regime is composed of two components: C.A. mode-partition noise （电流分配噪声）, inter- symbol InterferenceB. frequency chirp, modal dispersionC. material dispersion, waveguide dispersionD. modal dispersion, waveguide dispersion(14)At present, erbium doped （涂饵的）fiber amplifier’s maximum small signal gain （小信号增益）is around A.A. 40 dBB. 30 dBC. 20 dBD. 10 dB(15)Which of the following doesn’t belong to passive optical components （无源光学组件）BA. Directional coupler（定向耦合器）B. Semiconductor laser（半导体激光）C. Optical fiber connector（光纤连接器）D. Optical attenuator（光学衰减器）(16) The A mode has no cut off（截止）and ceases（停止）to exist only when the core diameter （芯径）is zero.A. HE11B. TE01C. TM01D. EH11(17)When the phase difference（相位差）is an integral multiple（整数倍）of A, the two modeswill beat and the input polarization（偏振）state will be reproduced.A. 2πB. πC. 1800D. π/2(18)which one of the following model can transmit in the single-mode optical fiber ?(A)A.HE11B. TE01C. TM01D. EH112. Write the full name of the following acronym（共十题每题1分）10p(1)DCF: dispersion compensating fiber（色散补偿光纤）(2)CNR: carrier to noise ratio（载噪比）(3)SRS: Stimulated Raman Scattering（受激的拉曼色散）(4)SOA: Semiconductor optical amplifier（半导体激光放大器）(5)NA: numerical aperture（数值孔径）(6)PON: passive optical network（无源光网络）(7)SLM: single longitudinal mode（单纵向模式）(8)NEP: noise-equivalent power(噪声等效功率)(9)DSF: dispersion shift fiber（色散转移光纤）(10)SONET: synchronous optical network（同步光网络）(11)A TM: asynchronous transfer mode（异步传输模式）(12)ISDN: integrated services digital network（综合业务数字网）(13)WDM: wavelength-division multiplexing(波分多路复用)(14)SDH: Synchronous digital hierarchy(同步数据系列)(15)TLLM: transmission-line laser model(传输线激光模式)(16)ONSL: optical network simulation layer（光网络模拟层）(17)OVPO: outside vapor-phase oxidation（外气相沉积法）(18)V AD: vapor-phase axial deposition(汽相轴向沉积)(19)CDMA: code-division multiple access(码分多址)(20)FDM: frequency-division multiplexing（频分复用）(21)DSP: digital signal processing（数字信号处理）(22)MCVD: modified chemical vapor deposition（修正的化学汽相沉积）(23)EDFA: erbium-doped fiber amplifier(惨饵光纤放大器)(24)FDDI: fiber distributed data interface(光纤分布式数据接口)(25)SIOF: step index optical fiber(阶跃指数光纤)(26)GIOF: graded index optical fiber(渐变性光纤)(27)SQW: single quantum-well（单量子井）(28)ARQ: automatic reapt request(自动重发请求)(29)FEC: forward error correction（前向纠错）3. Filling blanks(共20空每空1分) 20p(1) According the mode which propagate（传播）in the fiber, the fiber can be divided into (single mode) fiber) and (multimode) fiber.(2) The most common misalignment（非线性）occurring in practice, which also causes the greatest power loss, is (axial displacement轴向移位).(3) The electromagnetic energy（电磁能）of a guided mode is carried partly in the (core) and partly in the (cladding包层).(4)The basic attenuation mechanisms（衰减机制）in a fiber are (absorption吸收), (scattering散射) and (radiative losses辐射损耗) of the optical waveguide.(5) The two main optical amplifier types can be classified as (SOAs半导体激光放大器) and (DFAs).(6) Networks are traditionally divided into: (LANs局域网), (MANs城域网),(W ANs广域网).(7)The principal light sources used for fiber optic communications applications are (ILDs注入型激光二极管) and (LEDs发光二级管) .(8) The dominant noise generated in an optical amplifier is (ASE).(9) The two most common samples of these spontaneous fluctuations（波动）are (shot noise) and (thermal noise热噪声).(10)According to the refractive index（折射率）of the core, the fiber can be divided into (step-index) fiber and (graded-index) fiber .(11)The total dispersion in single-mode fibers consists mainly of (material) and (waveguide) dispersion.(12) The most meaningful criterion（标准）for measuring the performance of a digital communication system is the (average error probability平均概率误差). In an analog system the fidelity criterion is usually specified in terms of a (peak signal-to-noise ratio).(13) The simplest transmission link is a (point-to-point line) that has a transmitter on one end an a receiver on the other.(14)The commonly used materials for fiber lasers are (erbium) and (neodymium).(15)Absorption is related to the fiber material, whereas scattering is associated both with the (fiber material) and with (structural imperfections) in the optical waveguide.(16)The two basic LED configurations being used for fiber optics are (surface emitters) and (edge emitters).(17) The basic schemes for improving the reliability are (ARQ自动重发请求) and (FEC前向纠错).4. Give a brief description of following terms and questions（共5题每题3分）15p (1) Briefly describe the key system features of WDM.波分复用1) capacity upgrade（容量升级）2) transparency （透明度）3) wavelength routing（波长选路）4) wavelength switching（波长转换）5) the connectors used to join individual fiber cables to each other and to the source anddetector(2) Briefly describe there major goals of SDH.同步数据系列1) Avoid the problems of PDH2) Achieve higher bit rates3) Better means for operation, administration and Maintenance(3) List at least three advantages of SOA半导体激光放大器.1) Small size, and easy to be integrated with semiconductor circuits.2) Fabrication（制造）is simple and with low power consumption（功耗）, long life-span and low cost.3) Gain response is very quick and well suited for switching and signal processing in optical networks application.4) Can amplify optical signal and process signal in the same time such as switch, so can be used in wavelength converting and optical switch.(4) List more than three disadvantages of SOA.1) The coupling loss with optical fiber is too large2) Sensitive to polarization3) Noise figure is high(~8 dB)4) crosstalk串音干扰5) Easy to be affected by temperature, low stability(5) Stimulated Emissions受激发射If a photon（光子）of energy hv12 impinges on（撞击）the system while the electron is still in its excited state, the electron is immediately stimulated to drop to the ground state and give off a photon of energy hv12.(6)Dynamic rangeSystem dynamic range is the maximum optical power range to which any detector must be able to respond.(7) What conditions should be met to achieve a high signal-to-noise ratio?1) The photodetector（光电探测器）must have a high quantum efficiency（量子效率）to generate a large signal power.2) The photodetector and amplifier noises should be kept as low as possible.(8) Please write the three basic categories（类别）of degradation of light sources1) internal damage2) ohmic contact degradation3) damage to the facets of laser diodes（激光二极管）(9)List the three factors largely determining the frequency response of an LED1) the doping level（掺杂度）in the active region2) the injected carrier lifetime Ti in the recombination region3) the parasitic capacitance(寄生电容) of the LED.(10)Write the three basic types of two-level binary line codes that can be used for optical fiber transmission links.1) non-return-to-zero (NRZ) format2) return-to-zero (RZ) format3) phase-encoded (PE) format(11) Please write the three different mechanisms causing absorption briefly1) Absorption by atomic defects （原子缺陷） in the glass composition.2) Extrinsic （非固有的） absorption by impurity atoms in the glass material.3) Intrinsic absorption by the basic constituent atoms of the fiber material.(12) The disadvantage of Raman amplifierNeed large output power pump laser. As Raman Scattering, the energy is transferred from high frequency to low frequency. Cross talk will affect signal.5. Figure （共1题 每题5分）5p（1）Please draw the basic step for an automatic-repeat-request (ARQ) error-correction scheme. Solution:（ARQ纠错机制）(2)Please draw out the basic elements of the optical receiver.(5p)（光接收器）Solution:(3)Please draw out the basic elements of an analog link and the major noise contributions. Solution:（模拟链路及噪声源）光发送机电模拟输入信号谐波失真互调失真RIN激光削波光纤信道模式失真损耗GVD 光放大器ASE 噪声光检测器散弹噪声热噪声放大器噪声APD 倍增噪声电模拟输出到RF 接收机(3) consider the encoder shown in Fig.1that changes NRZ data into a PSK ing this encoder,draw the NRZ and PSK waveforms for the data sequence 0001011101001101.clock/2PSK dataNRZ datafrequency Afrequency BFig.1Solution:6. Calculation Problems(共3-4题，统计40分) 40p(1) A wave is specified by 8cos 2(20.8)y t z π=-,where y is expressed in micrometers and the propagation constant （传播常数） is given in 1m μ-.Find (a) the amplitude,(b) the wavelength,(c) the angular frequency （角频率）, and (d) the displacement at time 0t = and 4z m μ=. Solution:The general form is:y = (amplitude) cos()cos[2(/)]t kz A vt z ωπλ-=-.Therefore(a) amplitude 8m μ=(b) wavelength: 11/0.8m λμ-= so that 1.25m λμ=(c) 22(2)4v ωπππ===(d) At 0t = and 4z m μ= we have 18cos[2(0.8)(4)]8cos[2( 3.2)] 2.472y m m πμμπ-=-=-=(2) A certain optical fiber has an attenuation of 0.6dB/km at 1300nm and 0.3dB/km at 1550nm.Suppose the following two optical signals are launched simultaneously into the fiber: an optical power of 150W μ at 1300nm and an optical power of 100W μ at 1550nm. What are the Solution:power levels in W μof these two signals at (a) 8km and (b) 20km?Since the attenuations are given in dB/km, first find the power levels in dBm for100W μ and 150W μ. These are, respectively,P(100W μ) = 10 log (100 W μ/1.0 mW) = 10 log (0.10) = - 10.0 dBmP(150W μ) = 10 log (150 W μ/1.0 mW) = 10 log (0.15) = - 8.24 dBm(a) At 8 km we have the following power levels:P 1300(8 km) = - 8.2 dBm – (0.6 dB/km)(8 km) = - 13.0 dBm = 50W μP 1550(8 km) = - 10.0 dBm – (0.3 dB/km)(8 km) = - 12.4 dBm = 57.5W μ(b) At 20 km we have the following power levels:P 1300(20 km) = - 8.2 dBm – (0.6 dB/km)(20 km) = - 20.2 dBm = 9.55W μP 1550(20 km) = - 10.0 dBm – (0.3 dB/km)(20 km) = - 16.0 dBm = 25.1W μ(3) A double-heterojunction （异质结） InGaAsP LED emitting at a peak wavelength of 1310nm has radiative and nonradiative recombination times of 25 and 90ns, respectively. The drive current is 35mA.(a) Find the internal quantum efficiency and the internal power level.(b) If the refractive index （折射率） of the light source material is n=3.5, find the power emitted from the device.Solution: (a) From Eq. int 11/r nr rτητττ==+, the internal quantum efficiency is int 10.783125/90η==+， and from Eq.int intint I hcI p hv q q ηηλ== the internal power level is int (35)(0.783)26(1310)hc mA p mW q nm ==(b) From Eq.int e int 2p (1)n t p P n n η==+, 21260.373.5(3.51)P mW mW ==+ (4) An LED with a circular emitting area of radius 20m μ has a lambertian emission pattern with a 1002()W cm sr ∙axial radiance at a 100mA drive current. How much optical power can be coupled into a step-index fiber having a 100m μ core diameter and NA=0.22? How much optical power can be coupled from this source into a 50m μ core-diameter graded-index fiber having 12.0, 1.48n α== and 0.01∆=?Solution:The source radius （半径） is less than the fiber radius,so Eq. 222222,1()2LED step s o s o P r B NA r B n ππ==∆ holds: 22223222,()(210)(100/)(0.22)191LED step s o P r B NA cm W cm W ππμ-==⨯= From Eq. 222,122[1()]2s LED graded s o r P r B n απαα=∆-+ 232222,122(210)(100/)(1.48)(0.01)[1()]15925LED graded P cm W cm W πμ-=⨯-=(5)Suppose an avalanche photodiode hasthe following parameters ：1/231,1,0.85,,10L D L I nA I nA F M R η=====Ω, and 1B kHz=.Consider a sinusoidally(正弦） varying 850nm signal, which has a modulation index m=0.85 and an average power level 050P dBm =-, to fall on the detector at room temperature. At what value of M does the maximum signal-to-noise ratio occur?Solution: Using Eq.2222()()24/p P D L B Li M S N q I I M F M B qI B k TB R <>=+++ we have 22005/2001()22()24/D L B LR P m M S N qB R P I M qI B k TB R =+++ 162235/2191.215102.17610 1.65610M M ---⨯=⨯+⨯ The value of M for maximum S/N is found from Eq.224/()x L B L opt P D qI k T R M xq I I ++=+, with x = 0.5:Moptimum = 62.1.(6)An LED operating at 1300 nm injects 25W μ of optical power into a fiber. If the attenuation （衰减） between the LED and the photodetector （光电探测器） is 40 dB and the photodetector quantum efficiency （量子效率） is 0.65, what is the probability that fewer than 5 electronhole pairs （电子空穴对） will be generated at the detector in a 1-ns interval ?Solution: From ⎰==τηη0)(hvN hv E dt t P ,the average number of electron-hole pairs generated in a time t is 6.10)/103)(106256.6()103.1)(101)(1025(65.0/8346910=⨯⨯⨯⨯⨯===----s m Js m s W hc Pt h E N ληνη Then,from Eq.(7-2)%505.0120133822!5)6.10(!)(6.106.105=====---e e n e N n P N n(7) An engineer has the following components available:(a) GaAlAs laser diode operating at 850 nm and capable of coupling 1 mW (0 dBm) into a fiber. (b) Ten sections of cable each of which is 500 m long, as a 4-dB/km attenuation, and hasconnectors on both ends.(c) Connector loss of 2dB/connector.(d) A pin photodiode receiver.(e) An avalanche photodiode receiver.Using these components, the engineer wishes to construct a 5-km link operating at 20 Mb/s. If the sensitivities of the pin and APD receivers are -45 and -56 dBm, respectively, which receiver should be used if a 6-dB system operating margin is required?Solution:(a)Use margin 2system L l P P P f C R S T ++=-=α，to analyze the link power budget.(a) For the pin photodiode,with 11 jointsdB L km dB dB dBm dBm insystemm L l P P P f C R S T 6)/4()2(11)45(0arg )(11++=--=++=-=αWhich gives L=4.25km. the teansmission distance cannot be met with these components. (b)For the APDdB L km dB dB dBm dBm 6)/4()2(11)56(0++=--Which gives L=7.0km. the transmission distance can be met with these components.(8) Suppose we want to frequency-division multiplex 60 FM signals. If 30 of these signals have a per-channel modulation index i m =3 percent and the other 30 signals have i m =4 percent, find the optical modulation index （调制指数） of the laser.Solution:The total optical modulation index is%4.27])04(.30)03(.30[][2/1222/12=+==∑ii m m(9) An optical transmission system is constrained to have 500-GHz channel spacings. How many wavelength channels can be utilized in the 1536-to-1556-nm spectral band （频谱带）? Solution: In terms of wavelength,at acentral wavelength of 1546nm a 500-GHz channel spacing isnm s sm nm f c 410500/103)1546(19822=⨯⨯=∆=∆-λλ The number of wavelength channels fitting into the 1536-to-1556 spectral band then is 54/)15361556(=-=nm nm N(10) The output saturation （饱和） power sat out P , is defined as the amplifier output power for which the amplifier gain G is reduced by 3 dB (a factor of 2) from its unsaturated value 0G . Assuming 0G >>1, show that in terms of the amplifier saturation power sat amp P ,, the output saturation power issat amp sat out P G G P ,00,)1(2ln -=Solution: Let 2/0G G = and 0,/22/G P P P sat out out in ==.then Eq.(11-15) yields2ln 212,,00satout sat amp P P G G += Solving for s a t out P , and with 10>>G ,we havesat amp sat amp sat amp sat out P P P G G P ,..00,693.0)2(ln 22ln =≈-=。
