光纤通信必考填空题、计算题及答案知识点

1、光纤通信必考填空题、计算题及其答案 知识点一、填空题1The main constituents of an optical fiber communications link . The key sections are a transmitter consisting of a light source and its associated drive circuitry, a cable offering mechanical and environmental protection to the optical fibers contained inside, and a receiv

2、er consisting of a photodector plus amplification and signal-restoring circuitry.光纤通信链路的主要成分。的关键部分是一个发射机包括一个光源及其相关的 驱动电路，一个电缆提供机械和环境保护于光纤内部包含，和一个接收器包括 一个光电探测器加放大和信号复原电路。Attenuation of a light signal as it propagates along a fiber is an important consideration in the design of an optical communicatio

3、n system, the basic attenuation mechanisms in a fiber are absorption, scattering, and radiative losses of the optical energy.因为它传播沿纤维是在光通信系统的设计中的重要考虑因素的光信号的衰减， 在一个光纤中的基本衰减机制是吸收，散射，以及光学能量的辐射损失。Intermodal dispersion or modal delay appears only in multimode fibers. This signal-distorting mechanism is a

4、 result of each mode having a different value of the group velocity at a single frequency.模间色散或模延迟只出现在多模光纤。这个信号扭曲机构是在单一频率具有群 速度的不同值的每个模式的结果。In general, LEDs are used with multimode fibers, since normally t is only into a multimode fiber that the incoherent optical output power from an LED can be cou

5、pled in sufficient quantities to be useful.在一般情况下，使用LEg多模光纤，因为通常t是只有到多模光纤，从一个 LED的非相干光输出功率可被耦合以足够的量是有用的。At the output end of an optical transmission line, there must be a receiving device that interprets the information contained in the optical signal. The first element of this receiver is a photode

6、tecor. It senses the luminescent power falling upon it and converts the variation of this optical power into a corresponding varying electric current.在光传输线的输出端，必须有用于解释包含在光学信号中的信息的接收装置。该接收器的第一个元素是光检测器。它检测在发光功率落下后，它与该光功率 的变化转换成相应的不同的电流。An optical receiver consists of a photodetector , an amplifier, an

7、d signal-processing circuitry. The receiver has task of first converting the optical energy emerging from the end of a fiber into an electric signal, and then amplifying this signal to a large enough level.光接收器由一个光检测器，放大器，和信号处理电路。该接收器具有第一转换 的光能新兴从光纤的端部转换成电信号，然后放大该信号到一个足够大的电平 的任务。A characteristic of

8、WDM is that the discrete wavelengths form an orthogonal set of carriers that can be separated, routed ,and switched without interfering with each other.波分复用的一个特点是，该离散波长形成正交载波集合可被分离，路由和交换 而不相互干扰的。Chromatic dispersion causes pulse broadening , which leads bit-error rates.色散导致脉冲展宽，从而导致误码率。The function

9、of an OADM is to insert or extract one or more selected wavelengths at a designated point in an optical network.一个OADMJ功能是插入或在光网络中提取在一个指定点的一个或多个选择的 波长。Although the structure of such an optical amplifier is similar to that of a laser, it does not have the optical feedback mechanism. thus an amplifier

10、 can boost incoming signal levels, but it cannot generate a coherent optical output by itself.虽然这样的光放大器的结构类似于激光的，它不具有光学反馈机制。因而放大 器可以放大输入的信号的水平，但它本身不能产生一个相干光输出。The pump light is usually injected from the same direction as the signal flow, this is known as codirectional pumping泵浦光通常从相同的方向的信号流注入，这被称为同向

11、泵送In any fiber transmission system using optical amplifiers, the dynamic characteristics of network traffic, particularly in a metro network, can create large, rapid fluctuations in the input optical power level to an EDFA.在使用光放大器的任何光纤传输系统，网络流量的动态特性，尤其是在一个城 域网络，可以建立在输入光功率电平到EDF从，快速波动。The SONET and S

12、DH rings are commonly called self-healing rings because the traffic flowing along a certain path can automatically beswitched to an alternate or standby path following failure or degradation of the link segment.在SONE下口 SDK通常被称为自愈环，因为沿着某个路径流动的流量能够自动 切换到下面的链路段的故障或降解的替代或备用路径。The performance characteris

13、tics of OBS lie between those of a wavelength-routed network and an optical packet-switched networkOBS勺性能特征在于那些波长路由网络和光分组交换网络之间In single-mode fibers, chromatic and polarization-mode dispersions are important factors that can limit the transmission distance or data rate在单模光纤，套色和偏振模分散体是可以限制的传输距离或数据速率的

14、重要因 素Polarization-mode dispersion is important for data rates above 10Gb/s, since its statistical behavior ultimately can limit the highest achievable data rate.偏振模色散是以上的数据速率10Gb / s的很重要，因为它的统计行为最终可以 限制可达到的最高数据速率。When a link is being installed and tested, the operational parameters of interest inclu

15、de bit-error rate, timing jitter, and signal-to-noise ratio as indicated by the eye pattern.当链路正在安装和测试，感兴趣的操作参数包括误码率，定时抖动和信噪比由 眼图所指示的。二、计算题一 Consider a multimode silica fiber that has core refractive indexn1 1.480 and a cladding index n2 1.478 . Find (a) the critical angle, (b) the numerical apertur

16、e, and (c) the acceptance angle.考虑的多模石英光纤具有纤芯折射率和包层的折射率。发现(a)该临界角，(b)该数值孔径，及(c)的接受角。解：(a)由式sin c n2可以求得临界角n11(b)由式NA nsin A (n2 n2)2 r J可 可以计算出数值孔径(c)由式nsin 0,max nsin A n1sin c (n； n2)2可以求得出空气中的接受角为二、 Consider a multimode step-index fiber with a 62.5- mcore diameter and a core-cladding index differ

17、ence of 1.5 percent. If the core refractive index is 1.480, estimate the the total number of modes supported in the fiber at a wavelength of 850nm.考虑多模阶跃光纤带62.5-纤芯直径以及1.5 %的芯-包层折射率差。如果芯 的折射率是1.480,估计为850nm的波长支撑在纤维模式的总数。a Solution : the normalized frequency is V n1 J2 , the total number of modesis MV

18、2三 suppose we have a multimode step-index optical fiber that has a core radius of25 m ,a core index of 1.480 and an index difference0.01.What are thenumber of modes in the fiber at wavelengths 860, 1310, and 1550nm.解：首先由V 2且%何计算出V值，再利用M 1V2可以计算出总模数。假设我们有一个多模阶跃折射率光纤，其具有 25的芯半径，1.480芯指数和折 射率差。什么是在波长86

19、0, 1310,和1550模式在光纤的数目。四 suppose we have three multimode step-index optical fibers each of which has a core index of 1.48 and index difference 0.01. Assume the three fibers have core diameters of 50, 62.5 and 100m. What are the number of modes in these fibers at a wavelength of1550nm.假设我们有三个多模阶跃折射率光纤

20、每一个都具有为 1.48和折射率差的芯指 数。假设这三个纤维具有50, 62.5和100。纤芯直径是什么，在这些纤维在波 长of1550nm模式的数量。五 A double-heterojunction InGaAsP LED emitting at a peak wavelength of 1310nm has radiative and nonradiative internal qudiative recombination times of 30 and 100ns, respectively. The drive current is 40mA. Find (a) the bulk

21、recombination time; (b) the internal quantum efficiency; and (c) the internal power level.的双异质结的InGaAsP的LED发光波长1310nm的峰值波长具有30和100纳秒 的辐射和非辐射内部qudiative重组倍。驱动电流40毫安。发现(a)该本体 复合时间；(b)该内部量子效率；和(c)的内部电源电平。解：(a)由1 - 1得整体复合寿命二r nrr nr1(b) 由 int -得内重子效率r nr r(c)由pint int hv int，得内部发光功率 qq六 Consider a 1-km

22、long multimode step-index fiber in whichn1 =1.480 and =0,01,so that n2=1.465. what is the modal delay per length in this fiber?考虑一个1公里长的多模阶跃折射率光纤，其中=1.480和=0,01 ,以便 =1.465。什么是在这种光纤单位长度的模延迟？Solution: from TLn12yieldsT 立二50ns/kmcn2Lcn2七 Consider the following two multimode fibers(a) a step-index fiber

23、 with a core index n1 =1.458 and a core-cladding index difference =0.01;(b) a parabolic-profile graded-index fiber with the same values ofn1 and. Compare to the rms pulsebroadening per kilometer for these two fibers?考虑下面两个多模光纤(a) 一种阶跃光纤带芯指数=1.458和芯-包层折射率 差=0.01;(b) 一种抛物线轮廓渐变折射率光纤与相同的值和。比较均方根脉冲每公里拓宽这

24、两个光纤？2解：(a)由式s14.0ns/ kmLnL(NA)有 s n2 3c 4、3nlc L 2 J3c(b)由式s有20.3c L2 n120 3c14.0 ps/ km八 a given step-index fiber has a core refractive index of 1.480, a core radius equal to 4.5 m , and a core-cladding index difference of 0.25 percent. What is the cutoff wavelength for this fiber?一个给定的步长指数光纤具有1.4

25、800.25 %的芯-包层折射率差的芯折射率, 纤芯半径等于4.5,和。什么是截止波长该纤维？2ao 32a .=解：由式 c(n2 n2)2 np/2,当V=2.405时，光纤的截止波长为2 a -c n| v21230nm。V 1九 a particular LED has a 5-ns injected carrier lifetime. When no modulation current is applied to the device , the optical output power is 0.250mW for a specified dc bias . assuming p

26、arasitic capacitances are negligible, what are the optical outputs at modulation frequencies of (a) 10MHz and (b)100Mhz?21 2解：(a)由p( )Po 1 ( i) 可得在10M处输出功率为p( ) 239 W在100M处的输出光功率为 p( ) 76 W 由此可见这种器件的输出光功率随着调制速率的增加而增加 十. A given GaAlAs laser has an optical cavity of 300 mand a 100- m width. At a norm

27、al operating temperature, the gain factor 21 10 3 A cm3 and the loss coefficient _ 10cm 1. Assume the reflectivity is o for each end face. Find (a) the threshold current density and (b) the threshold current for this device一个给定的GaAlAs激光器和光的300and100-一腔的宽度。在正常工作温度 下，该增益因子和损耗系数。假设反射率是邻为每个端面。发现(a)该阈值电流

28、密度和(b)的阈值电流为这个设备_11_.一0解：(a)由 gtht ln()end 和 gth Jth 可得2LR1R2,1 1-1、Jth = Lln( R)(b)阈值电流路由下式给出Ith Jth腔体横截面积consider a double-heterostructure edge-emitting Fabry-Perot AlGaAs laser, which emits at 900nm. Suppose that the laser chip is 300 m long and the refractive index of the laser material is 4.3.

29、(a) How many half-wavelengths span the region between the Fabry-Perot mirror surfaces? (b)What I the spacing between the lasing modes?考虑一个双异质边发射法布里-珀罗的AlGaAs激光器，发射在900nm的。假 设激光芯片300长和激光材料的折射率是4.3。(一)有多少个半波长跨越法 布里-珀罗反射镜表面之间的区域？(b)本人的激射模之间的间距是什么？解：(a)从式m 2Lnv可得法布里-珀罗腔两镜面间的半波数目m2n c2(b)从式1 可得卜2 2Ln A G

30、aAs optical source with a refractive index of 3.6 is coupled to a silica fiber that has a refractive index od 1.480. what is the power loss between the source and the fiber?为3.6的折射率的神化钱光学源耦合到一个石英光纤具有外径1.480的折射率。什么是源和纤维之间的功率损耗？2解：如果光纤端面和光源在屋里上紧密衔接，则由 R，在光源和光纤头端的分ni n界面上菲涅耳反射可用下式来表示2R 比 0.174这相当于有17.4

31、%的光功率被反射回光源，与这一R值相应的耦合n1 n功率由下式给定Pcoupled (1 R) Pemitted用分贝表示的功率损耗 L为Pc1adL101g(Rouped)10lg(1 R) 0.83dBPemitted这个值有可能因为在光源和光纤端面之间存在折射率匹配物质而减小An InGaAsP optical source that has a refractive index of 3.540 is closely coupled to a step-index fiber that has a core refractive index of 1.480. assume that

32、the source size is smaller than the fiber core and that the small gap between the source and the fiber is filled with a gel that has a refractive index of 1.520. (a ) what is the power loss in decibels from the source into the fiber?(b)what is the power loss if no gel is used?具有的3.540的折射率的InGaAsP光源被

33、紧密耦合到一个阶跃光纤具有1.480的芯折射率。假设源尺寸比纤维核心更小，并且在源和纤维之间的小间隙被填 充有具有1.520的折射率的凝胶。(一)什么是用分贝的功率损耗从源到光纤 多少？ ( b)是什么的功率损耗，如果不使用凝胶？2解：(a)这里需要考虑两个界面反射率。首先由 R可以得出凝胶对光源的n1 n反射率Rsg 0.159同样，可得光纤对凝胶的反射率 Rgf 0.040于是，总的反射率为 R Rsg Rgf 0.040 0.0064。功率损耗的分贝值则为L 10lg(1 R) 0.0028dB(b)如果不用凝胶作为折射率匹配材料，切假设光源与光纤之间没有间隙，则由2R 叱可得此时反射率

34、为 R=0.168 n1 n此时的功率损耗分贝数为 L101g(1 R) 0.799dB十四 6.7 a given silicon avalanche photodiode has a quantum efficiency of 65 percent an a wavelength of 900nm, Suppose 0.5W of optical power produces a multiplied photocurrent of 10 A. What is the multiplication M?一个给定的硅雪崩光电二极管具有 65%的900nm的波长的量子效率，假设光功 率的0.5

35、产生的10倍增光电流。什么是乘法 M解：初级光电流为IpPn Pin ,倍增因子M 显hcI p十五 An InGaAs pin photodiode has the following parameters at a wavelength of 1300nm: ID 4nA, 0.90, rl 1000 , and the surface leakage current is negligible . The incident optical power is 30nW(- 35dBm), and the receiver bandwidth is 20MHz . find the vari

36、ous noise terms of the receiver.的InGaAs PIN光电二极管具有在1300nm的波长以下参数：，且表面的漏电流 是可以忽略不计。入射光功率为 30nW(-35dBnj),并接U攵带宽为20MHz找到 接收机的各种噪声条件。解：首先计算初级光电流Ip,Pin光电二极管的均方量子噪声电流为(i：hoj 2qI pBesiiop e均方暗电流为:i；B； 2qIDBe接收机的均方热噪声电流处BT BeRl十六 Consider an analog optical fiber system operating at 1550nm, which has an effe

37、ctive receiver noise bandwidth of 5 MHz . assuming that the receiverd signal is quantum noise limited, what is the incident optical power necessary to have a signal- to-noise ratio of 50 dB at the receiver? Assume the responsivity is 0.9A/W and that m=0.5.考虑一个模拟光纤系统在1550nm的工作，其中有5 MHz的有效接收机噪声带 宽。假定r

38、eceiverd信号量子噪声的限制，什么是入射光功率必须具有在接收 器中的信号与噪声为50dB比率？假设响应是0.9A/ W, H m= 0.5。解：注意到50dB的SNR即S/N=10 5,由式P (S / N2)4qBe可得入射光功率 m十七 A 2 2 biconical tapered fiber coupler has an input optical power level of P0 200 W .The output powers at the other three ports are P 90 w,P2 85 w and P3 6.3 w . What are the co

39、upling ratio, excess loss, insertion losses, and return loss for this coupler?双锥锥形光纤耦合器具有。输出功率的输入光功率电平在其它三个端口和。什 么是耦合比，过量损耗，插入损耗和回程损耗对这种联接器？P2解：耦合比=(2 ) 100%P P2附加损耗=10lg(p)P插入损耗=10lg()Pj回波衰减=10lg(鸟 Po十八 Assume that the LED together with its drive circuit has a rise time of 15ns. Taking a typical LE

40、D spectral width of 40nm, we have a materialdispersion of 21ns over the 6-km link. Assuming the receiver has a 25MHz bandwidth, then the rise-time degradation from the receiver is 14ns. If the fiber we select has a 400-MHz km bandwidth-distance product and with q= 0.7,then the modal-dispersion-induc

41、ed fiber rise time is 3.9 ns. What is the link rise time?假定其驱动电路的LED在一起具有15ns的上升时间。取40纳米的典型LED的 光谱宽度，我们有21ns在6公里链路的材料分散。假定接收器具有 25 MHz带 宽，然后从接收的上升时间降解为 14ns。如果我们选才该纤维具有400-MHz带 宽距离产品和其中q= 0.7 ,则模态色散引起的纤维的上升时间为 3.9纳秒。什 么是链接的上升时间？解：链路的展宽时间为tsys (ttx tmat Ood )1/2十九 assume we have a 3-dB coupler, so th

42、at half of the input power gets coupled to the second fiber. What are the output powerPut,1 and d?假设我们有一个3分贝耦合器，所以输入功率的一半被耦合到所述第二光纤。什么是输出功率和？二十 consider a commercially available 32 32single-mode coupler made from a cascade of 3-dB fused-fiber 2 2 couplers, where 5 percent of the power is lost in eac

43、h element. What are the excess and splitting losses for this coupler?考虑市售的单模耦合器从一个级联的3-dB的熔合光纤耦合器，其中5%的功率损失在每个元素进行。什么是这个耦合器的过剩和分裂的损失？解：由1响武与可以得到附加损耗为mg。95*- 1.1dB八工1分路损耗=10lg( -)101g(32) =15dB最后的总损耗为16.1dB二H consider an N N waveguide grating multiplexer having Lf=10mm.x d =5 m, nc =1.45,and a centra

44、l design wavelengthc =1550nm. What is thechannel spacing for m=1?考虑具有波导光栅复用器=10毫米。=5 , =1.45,和一个中央设计波长 二1550nm的。什么是信道间隔对于 m =1?解：对于m=1 ,由L m可以得到波导长度差nc由式 上必&则得到Lf m ng二十二 consider the following parameters for a 130-nm InGaAsP SOA :the active area widthw is 3 m, active area thickness d is 0.3m, ampli

45、fier length L is 500 m, what is the pumping rate for the SOA?考虑对于一个130纳米的InGaAsP的SOAZ下参数：有源区宽度是3,有源区 厚度d为0.3 ,放大器长度L为500,什么是排气速度为 SOAJ 1解：如果米用100mA的偏置电流，由 Rp 可以彳#到SOAqd qdwL二十三 consider a specific bus coupler that extracts 5 percent of the light to an optical power monitor. Furthermore, assume 2 percent of the optical power inserted into the coupler is lost internally .Wha