武汉理工大学光纤通信考试

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光纤通信期末试题及答案

光纤通信期末试题及答案一、选择题1. 下列哪个不是光纤通信的优点？A. 带宽大B. 传输距离长C. 抗干扰能力弱D. 体积小答案：C2. 光纤通信系统主要包括哪些组成部分？A. 发光器、光纤、接收器B. 发射机、中继机、收发器C. 发送端、传输介质、接收端D. 中继器、光纤放大器、光开关答案：A3. 光纤通信中，衰耗是指光信号在光纤中逐渐减弱的现象。

下列哪个不是光纤衰耗的类型？A. 吸收衰耗B. 反射衰耗C. 散射衰耗D. 倍增衰耗答案：D4. 光纤通信中使用的主要光源有哪些？A. 红外激光器、氦氖激光器B. LED、半导体激光器C. 氦氖激光器、LEDD. 钬铝激光器、碘钨灯答案：B5. 光纤光谱宽度对光纤通信有何影响？A. 光纤传输距离增加B. 光纤衰耗增加C. 光纤通信速率下降D. 光纤通信质量提高答案：D二、填空题1. 光纤通信的传输介质是_________。

答案：光纤2. 光纤通信的主要优点之一是其_________。

答案：带宽大3. 光纤通信中，光纤信号经过_________转换为电信号。

答案：接收器4. 光纤通信中，信号传输时受到的减弱现象称为_________。

答案：衰耗5. 光纤通信中使用的光源有_________和_________。

答案：LED和半导体激光器三、简答题1. 请简述光纤通信的工作原理。

答案：光纤通信是利用光的全反射原理，通过光纤传输光信号的一种通信方式。

发光器会将电信号转换为光信号，通过光纤传输到接收器。

光纤的光在一定角度范围内从光纤内壁全反射，使光信号能够在光纤中一直传输。

接收器将接收到的光信号转换为电信号，实现光纤通信。

2. 光纤通信相比于其他通信方式有何优点？答案：光纤通信相比于其他通信方式具有以下优点：- 带宽大：光纤通信传输带宽能力远大于传统的铜缆通信，可以支持高速宽带传输。

- 传输距离长：光纤传输损耗较低，信号可以在光纤中传输较长距离而不损失质量。

通信工程《光纤通信》考试题(含答案)

1、1966年7月，英籍华人（高锟）博士从理论上分析证明了用光纤作为传输介质以实现光通信的可能性。

2、光在光纤中传输是利用光的（折射）原理。

3、数值孔径越大，光纤接收光线的能力就越( 强 )，光纤与光源之间的耦合效率就越( 高 )。

4、目前光纤通信所用光波的波长有三个，它们是：（ 0.85μm 、1.31μm 、1.55μm ）。

5、光纤通信系统中最常用的光检测器有：（PIN光电二极管；雪崩光电二极管）。

6、要使物质能对光进行放大，必须使物质中的( 受激辐射)强于( 受激吸收 )，即高能级上的粒子数多于低能级上的粒子数。

物质的这一种反常态的粒子数分布，称为粒子数的反转分布。

7、在多模光纤中，纤芯的半径越( 大 )，可传输的导波模数量就越多。

8、光缆由缆芯、( 加强元件(或加强芯) )和外护层组成。

9、（波导色散）是指由光纤的光谱宽度和光纤的几何结构所引起的色散。

10、按光纤传导模数量光纤可分为多模光纤和( 单模光纤)。

11、PDH的缺陷之一：在复用信号的帧结构中，由于( 开销比特 )的数量很少，不能提供足够的运行、管理和维护功能，因而不能满足现代通信网对监控和网管的要求。

12、光接收机的主要指标有光接收机的动态范围和（灵敏度）。

13、激光器能产生激光振荡的最低限度称为激光器的（阈值条件）。

14、光纤的（色散）是引起光纤带宽变窄的主要原因，而光纤带宽变窄则会限制光纤的传输容量。

15、误码性能是光纤数字通信系统质量的重要指标之一，产生误码的主要原因是传输系统的脉冲抖动和（噪声）。

二、选择题：（每小题2分，共20分。

1-7:单选题，8-10：多选题）1、光纤通信是以（ A ）为载体，光纤为传输媒体的通信方式。

A、光波B、电信号C、微波D、卫星2、要使光纤导光必须使（ B ）A、纤芯折射率小于包层折射率B、纤芯折射率大于包层折射率C、纤芯折射率是渐变的D、纤芯折射率是均匀的3、（ D ）是把光信号变为电信号的器件A、激光器B、发光二极管C、光源D、光检测器4、CCITT于（C）年接受了SONET概念，并重新命名为SDH。

《光纤通信原理》测试题

光纤通信原综合测试题（一）一、填空题：(30分，每题1分)1、光纤通信系统基本组成部分有、和。

光纤的三个工作波长分别为、和。

G、652光纤的波长为，G、655光纤的波长为。

2、光纤的主要损耗是和，其色散主要有、、。

3、目前常用的光源有LD和LED，其中LD是物质发生过程而发光，而LED是因物质发生过程而发光，在数字光纤通信系统中对光源采用调制，而收端采用方法进行检测。

4、影响接收灵敏度的主要是光电检测器的和噪声，以及放大电路的噪声。

5、光无源器件主要有、、和。

6、SDH的复用过程主要步骤是、和。

其主要设备有、、和。

二、名词解释（2*2=4分）1、 SES2、消光比三、判断题(每题2分，共计10分)1、渐变型多模光纤具有自聚焦特性，故其没有模式色散。

（）2、接收机能接收的光功率越高，其灵敏度越高。

（）3、一个STM—1中能装入63个2Mb/s的信号，故其频带利用率比PDH高。

（）4、LED的P-I特性呈线性，故无论用模拟信号和数字信号对其进行调制均不需要加偏置电流。

（）5、SDH采用同步字节间插复用方式，故能实现一次性上下低速信号。

（）四、简答题（每题6分，共计18分）1、 1、什麽是自愈网？画图表示二纤单向复用段倒换环的工作原理。

2、 2、温度对LD有什麽样的影响？采用什麽方法可以抑制温度对它的影响？3、OTDR是什麽？在光纤通信中有何用途？五、计算题（38分）1、（12分）已知阶跃型多模光纤的包层折射率为1、48，相对折射率差为1%，试求：（1）① 纤芯折射率为多少？（2）② 数值孔径为多少？（3）③ 在5km长的光纤上，脉冲展宽为多少？（4）④ 若要求光纤的模式色散为30ns/km，相对折射率差为１％，包层折射率应为多少？2、 2、（10分）四次群系统采用4B1H码码型，（1）① 试求其光接口速率为多少？（2）② 其传输容量为多少个话路？（3）③ 其监控信息的传输速率为多少？3、 3、（8分）系统发端平均发送光功率为10μW，经损耗系数为0、5dB/km的光纤传输20km后刚好能被接收，（1）① 试求其接灵敏度。

光纤通信考试试题

光纤通信考试试题一、选择题（每题 2 分，共 40 分）1、光纤通信中使用的光源一般是（）A 发光二极管B 半导体激光器C 白炽灯D 日光灯2、以下哪种光纤的传输损耗最小（）A 多模光纤B 单模光纤C 塑料光纤D 石英光纤3、光纤通信系统中，光接收机的主要作用是（）A 将光信号转换为电信号B 将电信号转换为光信号C 对光信号进行放大D 对电信号进行放大4、下面哪个不是光纤通信的优点（）A 传输容量大B 抗干扰能力强C 成本低D 保密性好5、光在光纤中传输时，其模式数量取决于（）A 光纤的长度B 光纤的芯径C 光的波长D 光纤的材料6、以下哪个不是光纤通信系统的基本组成部分（）A 光源B 光放大器C 光检测器D 光耦合器7、单模光纤的纤芯直径一般为（）A 50μmB 625μmC 8μmD 100μm8、光纤通信中常用的波长窗口是（）A 850nmB 1310nmC 1550nmD 以上都是9、光信号在光纤中传输时产生的损耗主要有（）A 吸收损耗B 散射损耗C 弯曲损耗D 以上都是10、以下哪种不是光纤的分类方式（）A 按照传输模式分类B 按照折射率分布分类C 按照工作波长分类D 按照制造材料分类11、光纤通信系统中，能够延长传输距离的器件是（）A 光调制器B 光隔离器C 光中继器D 光环行器12、对于光的偏振态，在单模光纤中传输时通常是（）A 线偏振B 椭圆偏振C 随机偏振D 圆偏振13、下面哪个参数不是衡量光检测器性能的指标（）A 响应度B 量子效率C 带宽D 发光功率14、光纤的数值孔径反映了光纤的（）A 集光能力B 传输能力C 抗干扰能力D 损耗特性15、以下哪种光纤适用于短距离通信（）A 多模阶跃型光纤B 多模渐变型光纤C 单模光纤D 塑料光纤16、光放大器中，掺铒光纤放大器（EDFA）的工作波长范围是（）A 850nm 附近B 1310nm 附近C 1550nm 附近D 1625nm 附近17、在光纤通信系统中，用于实现不同类型光纤连接的器件是（）A 光纤连接器B 光分路器C 光耦合器D 光开关18、以下哪种不是常见的光纤通信系统（）A 强度调制直接检测（IM/DD）系统B 相干光通信系统C 光孤子通信系统D 无线光通信系统19、光在光纤中传输时，引起色散的主要原因是（）A 材料色散B 波导色散C 模式色散D 以上都是20、光纤通信系统的设计中，需要考虑的主要因素有（）A 传输距离B 传输速率C 误码率D 以上都是二、填空题（每空 2 分，共 20 分）1、光纤通信是以为信息载体，以为传输媒介的通信方式。

光纤通信技术考试试题

光纤通信技术考试试题一、选择题（每题 3 分，共 30 分）1、以下哪个不是光纤通信的优点？（）A 通信容量大B 传输损耗小C 抗干扰能力强D 成本低2、光纤的主要成分是（）A 铜B 铝C 石英D 塑料3、光在光纤中传输时，其模式数量与（）有关。

A 光纤芯径B 光纤长度C 光的波长D 光纤折射率4、以下哪种光纤适用于长距离通信？（）A 多模光纤B 单模光纤C 塑料光纤D 渐变型光纤5、光纤通信系统中，用于将光信号转换为电信号的器件是（）A 光源B 光检测器C 光放大器D 光耦合器6、以下哪种光源在光纤通信中应用最广泛？（）A 发光二极管B 半导体激光器C 固体激光器D 气体激光器7、光放大器的作用是（）A 产生光信号B 增强光信号C 转换光信号D 分配光信号8、以下哪种光放大器的增益带宽最宽？（）A 半导体光放大器B 掺铒光纤放大器C 拉曼光纤放大器D 布里渊光纤放大器9、光纤通信系统的工作波长通常在（）附近。

A 850nmB 1310nmC 1550nmD 1800nm10、以下哪种复用技术可以在一根光纤中同时传输多个波长的光信号？（）A 时分复用B 波分复用C 频分复用D 码分复用二、填空题（每题 3 分，共 30 分）1、光纤通信是以______为载体，以______为传输媒介的通信方式。

2、光纤的结构包括______、＿_____和______。

3、按照折射率分布情况，光纤可分为______光纤和______光纤。

4、光在光纤中传播会产生______、＿_____和______等损耗。

5、常见的光检测器有______和______。

6、光纤通信系统的基本组成包括______、＿_____和______。

7、光源的调制方式有______调制和______调制。

8、波分复用系统的关键器件有______和______。

9、光纤通信中的色散包括______色散、＿_____色散和______色散。

光纤通信考试题库——附答案

一．单项选择题（每题1分,共20分）1、在激光器中，光的放大是通过（Ａ）Ａ．粒子数反转分布的激活物质来实现的Ｂ．光学谐振腔来实现的Ｃ．泵浦光源来实现的Ｄ．外加直流来实现的2、下列哪一项不是要求光接收机有动态接收范围的原因?( B )A．光纤的损耗可能发生变化B．光源的输出功率可能发生变化C．系统可能传输多种业务D．光接收机可能工作在不同系统中3、光纤通信中光需要从光纤的主传输信道中取出一部分作为测试用时，需用（Ｂ）Ａ．光衰减器Ｂ．光耦合器Ｃ．光隔离器Ｄ．光纤连接器4、使用连接器进行光纤连接时，如果接头不连续时将会造成（D）Ａ．光功率无法传输Ｂ．光功率的菲涅耳反射Ｃ．光功率的散射损耗Ｄ．光功率的一部分散射损耗或以反射形式返回发送端5、在系统光发射机的调制器前附加一个扰码器的作用是（A）A．保证传输的透明度 B．控制长串“1”和“0”的出现C．进行在线无码监测 D．解决基线漂移6、下列关于交叉连接设备与交换机的说法正确的是（A）A．两者都能提供动态的通道连接 B．两者输入输出都是单个用户话路C．两者通道连接变动时间相同 D．两者改变连接都由网管系统配置7、目前EDFA采用的泵浦波长是( C )A．0.85μm和1.3μm B．0.85μm和1.48μmC．0.98μm和1.48μm D．0.98μm和1.55μm8、下列不是WDM的主要优点是( D )A．充分利用光纤的巨大资源 B．同时传输多种不同类型的信号C．高度的组网灵活性，可靠性 D．采用数字同步技术不必进行玛型调整9、下列要实现OTDM解决的关键技术中不包括（ D ）A．全光解复用技术 B．光时钟提取技术C．超短波脉冲光源 D．匹配技术10、掺饵光纤的激光特性（A）Ａ．由掺铒元素决定Ｂ．有石英光纤决定Ｃ．由泵浦光源决定Ｄ．由入射光的工作波长决定11、下述有关光接收机灵敏度的表述不正确的是( A)A．光接收机灵敏度描述了光接收机的最高误码率B．光接收机灵敏度描述了最低接收平均光功率C．光接收机灵敏度描述了每个光脉冲中最低接收光子能量D．光接收机灵敏度描述了每个光脉冲中最低接收平均光子数12、光发射机的消光比，一般要求小于或等于（B）A．5％ B．10％ C．15％ D．20％13、在误码性能参数中，严重误码秒（SES）的误码率门限值为（D）A．10-6B．10-5 C．10-4 D．10-314、日前采用的LD的结构种类属于（D）A．F-P腔激光器（法布里—珀罗谐振腔）B．单异质结半导体激光器C．同质结半导体激光器D．双异质结半导体激光器15、二次群PCM端机输出端口的接口码速率和码型分别为（ B）A．2.048 Mb/s，HDB3 B．8.448 Mb/s，HDB3C．2.048 Mb/s，CMI D．8.448 Mb/s，CMI 16、数字光接收机的灵敏度Pr=100微瓦，则为（ A ）dBm。

武汉理工大学光纤通信复习题(答案版)

一、画图Drawing1.Please draw out the basic setup for an automatic-repeat-request(ARQ) error-correctionscheme2.Please draw out the fundamental concept of a coherent lightwave system相干光系统的基本原理图3.Please draw out the general heterodyne receiver configurations.(a)Synchronous detection uses a carrier-recovery circuit(b)Asynchronous detection uses a one-bit delay line一般的外差接收机结构(a)使用载波恢复电路的同步检测(b)使用1比特延迟线的异步检测4.Please draw out the basic constituents of a generic RF--over--fiber link光载射频系统的基本框图5.Please draw out the configurations of an EDFA:(a)codirectionalpumping,(b)counterdirectional pumping(c)dual pumpingEDFA 三种结构(a)前向泵浦(b)后向泵浦(c)双向泵浦6.Please draw out a simple 4*4 optical crossconnect architecture using optical space switchesand wavelength converters使用光空分交换和波长交换的4*4光交叉连接结构7.Please draw out the components of a typical WDM link典型WDM链路的构成8.Please draw out the components of the intensity modulated digital optical receiver强度调制数字光接收机的方框图9.Please draw out the generic configuration of a large SONET or SDH network consisting oflinear chains and various types of interconnected rings10. Please draw out the schematic diagram of NNI position in the SDH network二、计算题1、2利用NA 、归一化频率求总模数 3、4求耦合损耗 5求信噪比 6求入射光功率 7求SOA 的泵浦功率和零信号增益 8 EDFA 的功率 9/10 求其模式色散τ∆及传输容量BL 11、12带宽距离积 13、14系统设计1. Suppose we have a multimode step--index optical fiber that has a core radius of 25um ，a coreindex of 1.48，and an index difference △ = 0.01. What are the number of modes in the fiber at wavelengths 860，1310，and1550m μ ? Solution ：(a)First, from and at an operating wavelength of 860nm the value of V is =38.2Using the total number of modes at 860nm is(b)Similarly,at 1310nm we have V = 25.1 and M = 315.()2221222212V n n a M =-⎪⎭⎫ ⎝⎛=λπ∆≈-==2sin 1212221A n n n n NA ）（θ01.0286.048.1252221⨯⨯⨯=∆≈mm n aV μμπλπ()2221222212V n n a M =-⎪⎭⎫ ⎝⎛=λπ72922=≈V M 72922=≈V M(c)Finally at 1550nm we have V = 21.2 and M = 224.2. Suppose we have three multimode step--index optical fibers each of which has a core indexof 1.48，and an index difference △ = 0.01. Assume the three fibers have core diameters of 50,62.5 and 100m μ.What are the number of modes in these fibers at wavelength of 1550m μ ? Solution ：(a)First, from and at 50mμ diameter the value of V isUsing the total number of modes at 50m μ core diameter fiber is(b)Similarly,at 62.5m μ we have V = 26.5 and M = 351.(c)Finally at 100m μ we have V = 42.4 and M = 898.3. A GaAs optical source with a refractive index of 3.6 is coupled to a silica fiber that has arefractive index of 1.48. What is the power loss between the source and the fiber? Solution ：If the fiber end and the source are in close physical contact,then,from , the Fresnel reflection at the interface isThis value of R corresponds to a reflection of 17.4 percent of the emitted optical power backinto the source.Given that()emittedcoupled P R P -=1the power loss L in decibels is found from:This number can be reduced by having an index-matching material between the source andthe fiber end.4. An InGaAsP optical source that has a refractive index of 3.540 is closely coupled to astep-index fiber that has a core refractive index of 1.480.Assume that the source size is smaller than the fiber core and that the small gap between the source and the fiber is filled211⎪⎪⎭⎫⎝⎛+-=n n n n R 174.048.160.348.160.32211=⎪⎭⎫⎝⎛+-=⎪⎪⎭⎫ ⎝⎛+-=n n n n R ()()dBR P P L emitted coupled 83.0826.0log 101log 10log 10=-=--=⎪⎪⎭⎫⎝⎛-=∆≈-==2sin 1212221A n n n n NA ）（θ()2221222212V n n a M =-⎪⎭⎫⎝⎛=λπ2.2101.0255.148.1252221=⨯⨯⨯=∆≈mm n aV μμπλπ()2221222212V n n a M =-⎪⎭⎫ ⎝⎛=λπ22422=≈V Mwith 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? Solution ：(a)Here we need to consider the reflectivity at two interfaces.First, using we have that the reflectivity sg R at the source-to-gel interface isSimilarly,usingwe have that the reflectivity gfR at the gel-to-fiber interface isThe total reflectivity then is()()0064.0040.0159.0=⨯=⨯=gf sg R R R .The power loss in decibels is ()()dB R L 0028.0994.0log 101log 10=-=--= (b)If no index-matching gel is used, and if we assume there is no gap between the source and the fiber, then fromwe have that the reflectivity isIn this case the power loss in decibels is()()dB R L 799.0832.0log 101log 10=-=--=5. Consider a Si APD operating at 300o K and with a load resistor R L = 1000Ω.For thisAPD assume the responsivityW A65.0=ℜ and let x = 0.3 (a)If dark current isneglected and 100nW of optical power falls on the photodetector , what is the optimum avalanche gain? (b)What is the SNR if B e = 100MHz? (c)How does the SNR of this APD compare with the corresponding SNR of a Si pin photodiode? Assume the leakage current is negligible. Solution ：(a)Neglecting dark current and withPI p ℜ= , we have()()()()()()421010065.010001060.13.03001038.1443.2191923=⎥⎦⎤⎢⎣⎡⨯⨯⨯=⎪⎪⎭⎫ ⎝⎛ℜ=---P xqR T k M L B opt211⎪⎪⎭⎫ ⎝⎛+-=n n n n R 159.0520.1540.3520.1540.32=⎪⎭⎫⎝⎛+-=sg R 211⎪⎪⎭⎫⎝⎛+-=n n n n R 040.0520.1480.1520.1480.12=⎪⎭⎫ ⎝⎛+-=gf R 211⎪⎪⎭⎫⎝⎛+-=n n n n R 168.0480.1540.3480.1540.32=⎪⎭⎫⎝⎛+-=R(b)Neglecting dark current and with ()()3.042==xM M F , we have()()()()[]()()()()()()()6591010010003001038.14421010065.0106.12421010065.0426233.2919293.22=⨯⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⨯+⨯⨯⨯=⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛+ℜℜ=----e L B B R T k PM q PM SNRor in decibels , SNR = 10log659 = 28.2dB(c)For a pin photodiode with M = 1, the above equation yields SNR(pin) = 2.3 = 3.5dB. Thus, compared to a pin photodiode, the APD improves the SNR by 24.7dB.6. Consider an analog optical fiber system operating at 1550nm, which has an effective receivernoise bandwidth of 5MHz. Assuming that the received signal is quantum noise limited, what is the incident optical power necessary to have a signal-to-noise ratio of 50dB at the receiver?Assume the responsivity is 0.9A/W and that m = 0.5. Solution ：First we note that a 50dB SNR means that S/N = 105 . Then, solving ere p qB P m qB I m N S 4422ℜ=≈for P r yields()()()()()()mWnW m qB N S P er32619521042.1142090.05.0105106.141014--⨯==⨯⨯⨯=ℜ=or in dBm()dBm P dBm P r r 5.281042.1log 10log 103-=⨯==-7. Consider the following parameters for a 1300nm InGaAsP SOA:System Parameter Value w Active area width 3um d Active area thickness 0.3um L Amplifier length 500um Γ Confinement factor 0.3 τr Time constant 1nsa Gain coefficient 2×10-20m 2 n th Threshold density 1.0×10-24m -3 (a)What is the pumping rate for the SOA? (b)What is the zero-signal gain? Solution ：(a)If a 100mA bias current is applied to the device, then, from ()()qd t J t R p =, the pumpingrate is()()()()()s m electrons m m m C A qdwL qd J R p 333191039.150033.0106.11.01⨯=⨯===-μμμ (b)From ⎪⎪⎭⎫⎝⎛-Γ=r th r n qd J a g ττ0 , the zero-signal gain is ()()11324133322004.2323400.1100.11039.11102.03.0------==⎪⎪⎭⎫ ⎝⎛⨯-⨯⨯⨯=cm m ns m s m ns m g8. Consider an EDFA being pumped at 980nm with a 30mW pump power. If the gain at 1550nmis 20dB, what are the maximum input and output powers? Solution ：From1,,-⎪⎭⎫ ⎝⎛≤G P P inp s p ins λλ , the maximum input power is()()WmW P ins μ1901100301550980,=-≤Fromin p spin s out s P P P ,,,λλ+≤ , the maximum output power is()()()dBm mW mW W P P P in p spin s out s 8.121.193063.0190max max ,,,==+=+=μλλ9. 一根突变型多模光纤的长度km L 1=，纤芯的折射率5.11=n ，相对折射率差01.0=∆，求其模式色散τ∆及传输容量BL 。

光纤数字通信设备检验工试卷B

2014年湖北理工学院光纤、数字通信设备检验工培训笔试试卷（B）************************班级：********************************************************学号：********************************************************姓名：************************************************************************************************ ******** ******** ******** ******** 班级：******** ******** ******** ******** ******** ******** ******** 学号：******** ******** ******** ******** ******** ******** ******** 姓名：******** ******** ******** ******** ******** ******** ******** ******** A、吸收损耗B、散射损耗C、弯曲损耗D、固有损耗35、计算机网络类型按覆盖的地理范围分（）。

A、局域网B、城域网C、电线网D、广域网36、属于光纤光学特性的是（）。

A、折射率、数值孔径、截止波长B、纤芯、包层直径、偏心度C、损耗、带宽D、色散、损耗37、光缆的种类按照结构分为()。

A 松套层绞光缆B 中心束管光缆C 骨架式光缆D 架空光缆38、传输媒质不同，通信系统可分为（）。

A、有线通信系统B、无线通信系统C、电力线通信D、光纤通信39、下列（）属于单模光纤。

A、G.651B、G.652C、G.654D、G.65540、单模光纤的色散是由（）引起的。

A、波导色散B、材料色散C、结构色散D、模式色散41、PDH指的是( )。

武汉理工大学_通信原理期末考试

1
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四．已知 sAM (t) (1 K cos t K cos 2t) cos 0t ，且 0 。
1．画出此已调波的频谱示意图。
（3 分）
2．画出包络解调器的原理框图。
（3 分）
3．试求使此 AM 信号无包络失真的 K 值。
（4 分）
五．若消息代码序列为 110000010100001，
似为 fc 。包络服从瑞利分布，相位服从均匀分布。
2．若采用双边带调制，则每路信号带宽为 W＝2×1＝2MHz，6 路信号复用，至少需要带宽 12MHz。若采用单边带调制，则每路信号带宽为 W＝1MHz，至少需要带宽 6MHz。
3．均匀量化：在量化区内，大、小信号的量化间隔相同，因而小信号时量化信噪比太小；非均匀量化：量化级大小随信号大小而变，信号幅度小时量化级小，量化误差也小，信号幅度大时量化级大，量化误差也大，因此增大了小信号的量化信噪比。
1．试求此时编码器输出码组，并计算量化误差；（段内码采用自然二进制码）；（8 分） 2．写出对应于该 7 位码（不包括极性码）的均匀量化 11 位码。（2 分）
九．某 (n,k) 线性分组码的生成矩阵为
100101 G 010110
001011
1. 求监督矩阵 H，确定 n,k 。（4 分） 2. 写出监督位的关系式及该 (n,k) 的所有码字。（6 分）
2
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武汉理工大学教务处
试题标准答案及评分标准用纸
课程名称—通信原理 A———— （ A 卷、闭卷）
一．是非题（正确打√，错误打×，每题 2 分，共 10 分） 1. ×；2．×；3．√；4．×；5．×

武汉理工大学光纤通信复习题全

复习题题库1.Make a choice（共十题每题1分）10p(1).Which of the following dispersion dose not exist in single-mode optical fiber? （D ）(2).The unit of the fiber attenuation coefficient is (C)(3).the bands of Optical fiber communication is (B)(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)(5)In order to make sure of the system BER conditions , if the minimum optical input power of the receiver is 1 uW, the sensitivity of the receiver must be(6)The principal light sources used for fiber optical communications applications are ：(7)laser action is the result of three key process，which one of the following is not be included？(8) A single mode fiber usually has a core diameter of(9)To make sure that the APD photo-detector works properly, a sufficiently is applied across the p-n junction.(10) When DFA fiber amplifier uses as light Repeaters, its main effect isA. amplifying and regenerating the signal(11) In graded-index optical fiber, the numerical aperture NA can be expressed as(12) In practical SMFs, the core diameter is just below the cutoff of the first higher-order mode; that is, for V slightly(13) It is well known that the total dispersion in the single-mode regime is composed of two components:(14) At present, erbium doped fiber amplifier’s maximum small signal gain is aroundA. 40 dB(15) Which of the following doesn’t belong to passive optical components(16) The mode has no cut off and ceases to exist only when the core diameter is zero.A. HE11(17)When the phase difference is an integral multiple of A, the two modes will beat and the inputpolarization state will be reproduced.(18)which one of the following mode can transmit in the single-mode optical fiber ?(A)A.HE11(19)Light attenuation in an optical fiber is caused by ( B )(20)Which of the following will reduce the attenuation of fiber-optic cable assembly?(21)Increasing a fiber’s length can(22)What is the least important characteristic of a fiber-optic light source?(23)Which of the following codes cannot be transmitted in fibers?(24)Dispersion-shifted fiber (DSF) is a type of single-mode fiber designed to have zero dispersion near ___nm.(25) In graded-index optical fiber, the numerical aperture NA can be expressed as2. Write the full name of the following acronym(1)OTDM: optical time-division multiplexing(2)DTM: Dynamic Synchronous Transfer Mode(3)DWDM: Dense wavelength division multiplexing(4)DFF: Dispersiion-flattened fiber(5)AWF: All wave fiber(6)EPON: Ethernet Passive Optical Network(7)ARQ: automatic reapt request(8)SDH: Synchronous digital hierarchy(9)SONET: synchronous optical network(10)TDMA: time-division multiple access(11)ISDN: integrated services digital network(12)FDM: frequency-division multiplexing(13)SIOF: step index optical fiber(14)ATM: asynchronous transfer mode(15)SOA: Semiconductor optical amplifier(16)APD: Avalanche Photo Diode(17)WDM: wavelength-division multiplexing(18)PCM: pulse-code modulation(19)NRZ: nonreturn-to-zero(20)DSF: dispersion shift fiber(21)TLLM: transmission-line laser model(22)ONSL: optical network simulation layer(23)OVPO: outside vapor-phase oxidation(24)V AD: vapor-phase axial deposition(25)MCVD: modified chemical vapor deposition(26)EDFA: erbium-doped fiber amplifier(27)FDDI: fiber distributed data interface(28)GIOF: graded index optical fiber(29)SQW: single quantum-well(30)FEC: forward error correction(1) The transmission distance of any fiber-optic communication system is inherently limited by ?(2) The role of the optical transmitter is to convert the ( ) into the corresponding (optical signal) and then launch it into the ( ) serving as a communication channel.(3) The LED structures being used for fiber optics can be classified as ( ) or ( ).(4)The two basic LED configurations being used for fiber optics are ( ) and( ).(5) The three main components of the optical transmitter are ( ), ( ), ( ).(6) The role of the optical receiver is to convert the ( ) back into ( ) and recover the data transmitted through the ( ).(7) The design of the front end of receiver requires a trade-off between ( ) and ( ).(8) The two fundamental noise mechanisms responsible for current fluctuations in all optical receivers are the ( ) and ( ).(9) From an architectural standpoint, fiber-optic communication systems can be classified intothree broad categories-point-to-point links, ( ), and ( ).(10) Optical amplifiers are often cascaded to overcome ( ) in a long-haul lightwave system.(11) A fiber Bragg grating acts as an optical filter because of the existence of a ( ), the frequency region in which most of the incident light is reflected back.(12) The electromagnetic energy of a guided mode is carried partly in the ( ) and partly in the ( ).(13)The basic attenuation mechanisms in a fiber are ( ), ( ) and ( ) of the optical waveguide.(14) The two main optical amplifier types can be classified as ( ) and ( ).(15) The two most common samples of these spontaneous fluctuations are ( ) and ( ).(16)According to the refractive index of the core, the fiber can be divided into ( ) fiber and ( ) fiber .(17)The total dispersion in single-mode fibers consists mainly of(18) The most meaningful criterion for measuring the performance of a digital communication system is the(19) The simplest transmission link is a.(20)Absorption is related to the fiber material, whereas scattering is associated both with the (fiber material) and with (structural imperfections) in the optical waveguide.4. Give a brief description of following terms and questions（共5题每题3分）15p (1) Please write three main factors that influence the sensitivity of the optical receiver.1) Noise: shot noise and thermal noise2) Extinction Ratio3) Fiber dispersion(2) Briefly describe there major goals of SDH.1) Avoid the problems of PDH2) Achieve higher bit rates3) Better means for operation, administration and Maintenance(3) List at least three advantages of optical fiber in fiber communication.1) Wide bandwidth2) Anti-interference3) Low loss4) Large capacity(4) List at least three advantages of SOA.1) Small size, and easy to be integrated with semiconductor circuits.2) Fabrication is simple and with low power consumption, 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.(5) 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) crosstalk5) Easy to be affected by temperature, low stability(6)BERBit-Error rate, defined as the probability of incorrect identification of a bit by the decision circuit of the receiver.(7) What conditions should be met to achieve a high signal-to-noise ratio?1) The photodetector must have a high quantum efficiency 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 of degradation of light sources1) internal damage2) ohmic contact degradation3) damage to the facets of laser diodes(9)List the three factors largely determining the frequency response of an LED1) the doping level in the active region2) the injected carrier lifetime Ti in the recombination region3) the parasitic capacitance of the LED.(10) Please write the three different mechanisms causing absorption briefly1) Absorption by atomic defects in the glass composition.2) Extrinsic absorption by impurity atoms in the glass material.3) Intrinsic absorption by the basic constituent atoms of the fiber material.(11) 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.(12) Dynamic RangeSystem dynamic range is the maximum optical power range to which any detector must be able to respond.(13) List at least three advantages of fiber amplifiers in fiber communication1) low insertion loss2) large bandwidth3) low noise4) low crosstalk5) high gain(14) List at least three factors of attenuation in fiber communication1)material absorption2)Rayleigh scattering3)mie scattering4)connection losses5. 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:(2)Please draw out the basic elements of the optical receiver.(5p)Solution:(3)Please draw out the basic elements of an analog link and the major noise contributions. Solution:光光光光光光光光光光光光光光光光光光光RIN光光光光光光光光光光光光光光GVD光光光光ASE 光光光光光光光光光光光光光光光光光光APD 光光光光光光光光光光RF 光光光(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 is given in 1m μ-.Find (a) the amplitude,(b) the wavelength,(c) the angular frequency, 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 have18cos[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 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 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 witha 1002()W cm sr ∙axial radiance at a 100mA drive current. How much optical power can becoupled 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 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 has the 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 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 optP 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 between the LED and the photodetector is 40 dB and the photodetector quantum efficiency is 0.65, what is the probability that fewer than 5 electronhole pairs 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 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?Solution: In terms of wavelength,at acentral wavelength of 1546nm a 500-GHz channel spacing isnm s s m nm f c 410500/103)1546(19822=⨯⨯=∆=∆-λλThe number of wavelength channels fitting into the 1536-to-1556 spectral band then is 54/)15361556(=-=nm nm N(10) The output saturation 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 sat 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 =≈-=选择题、填空题、问答题、计算题题库一 Make a choice1) In graded-index optical fiber, the numerical aperture NA can be expressed as C.A. 21n n -B. ∆2aC. ∆2n 1D. 21n n a -2) In practical SMFs, the core diameter is just below the cutoff of the first higher-order mode; that is, for V slightly A.A. <2.4B. > 2.4C. =3D. =3.53) When the phase difference is an integral multiple of _2π_, the two modes will beat and the input polarization state will be reproduced.A. 2πB. πC. 1800D. π/24) It is well known that the total dispersion in the single-mode regime is composed of two components: C.A. mode-partition noise, inter- symbol InterferenceB. frequency chirp , modal dispersionC. material dispersion , waveguide dispersionD. modal dispersion , waveguide dispersion5) Which of the following codes cannot be transmitted in fibers B. A. CMI B. HDB3 C. 5B6B D. 8B1H6) Dispersion-shifted fiber (DSF) is a type of single-mode fiber designed to have zero dispersion near A nm. A. 1550 B. 850 C. 1310 D. 15107) To make sure that the APD photo detector works properly, a sufficiently D is applied across the p-n junction.A. high forward-bias voltageB. low forward-bias voltageC. low reverse-bias voltageD. high reverse-bias voltage8) A single mode fiber usually has a core diameter of A.A. 10mB. 62.5nmC. 125nmD. 50mm二Blank filling1) Each SDH frame comprising three areas: a section overhead area （SOH ）, a pointer area, a payload area including the data to be transmitted.2) List three applications of FBG(fiber Bragg grating): filter 、Optical Add Drop Multiplexer(OADM) and dispersion compensator.3) According to whether there is electric or magnetic field in the direction of propagation or not, transverse modes of light waves are classified into different types: TEM modes, TE modes, TM modes and hybrid modes.4) Transmission of information in an optical format is carried out not by frequency modulation of the carrier, but by varying the intensity of the optical power.5) Largely due to attenuation and dispersion, the optical signals undergo waveform distortion and decreased amplitude.6) Material dispersion occurs because the index of refraction varies as a function of the optical wavelength.7)ZDSF is a dispersion shifted single mode fiber that has the zero dispersion wavelength nearthe 1550 nm window, but outside the window actually used to transmit signals.8) A laser is constructed from three principal parts: a pump source, a gain medium, and anoptical resonator.9)An optical fiber comprises at least two layers, a core and a cladding.10)Optical transmitter consists of optical source, a modulator and a channel coupler.11)Fiber refractive index profiles classify fibers as step-index fibers and graded-index fibers.12)BER (The bit error rate) performance and jitter are two important indicators in a opticaldigital communication system.13)LASER is a mechanism for emitting light within the electromagnetic radiation region of thespectrum, via the process of stimulated emission.14)There are two kinds of SOA：Fabry-Perot Amplifier (FPA) and Traveling-Wave Amplifier(TWA) .15)The principal noises associated with photo detectors that have no internal gain are quantumnoise, dark-current noise generated in the bulk material of the photodiode, and surface leakage current noise.16)In a step-index fiber, the refractive index of the core is uniform and undergoes an abruptchange at the core-cladding boundary. In graded-index fibers, the refractive index of the core varies gradually as a function of radial distance from the fiber center.17)If the input pulse excites both polarization components, it becomes broader as the twocomponents disperse along the fiber because of their different group velocities. This phenomenon is called the PMD.18)The most common semiconductor photo detector is the pin photodiode.19)The main cause of intrinsic absorption in the infrared region is the characteristic vibrationfrequency of atomic bonds.20)In SDH transmission structures, a TU (Tributary Unit) includes a low level VC and a TUPTR.21)Extrinsic absorption is caused by impurities introduced into the fiber material.22)Intramodal dispersion is a result of the group velocity being a function of the wavelength.23)The optical amplifiers is divided into 3 groups: semiconductor optical amplifier (SOA),optical fiber amplifier (OPA) and Raman amplifier (FRA).24)There are two kinds of directional couplers: a prism-fibre and a fibre-fibre lapped coupler.25)EDFA has three pumping structures: ①the forward pumping structure; ②the backwardpumping structure; ③ the double pumping structure.26) A laser consists of a gain medium inside a highly reflective optical cavity, as well as a meansto supply energy to the gain medium.27)Attenuation in an optical fiber is caused by absorption losses, scattering losses, and bendinglosses.28)STM-1 frames provide a transmission bit rate of 155 Mbit/s.29)From the point of view of the wave theory, light wave could be described as anelectromagnetic wave.30)Intermodal dispersion is a result of each mode having a different value of the group velocityat a single frequency.31) A typical optical digital communication system usually comprises three parts: transmitter,optical fiber channel, and receiver.32)The pin Photo detector structure consists of p and n regions separated by a very lightly n-doped intrinsic (i) region. 33)Please list three steps of SDH Multiplexing: mapping ， aligning ，multiplexing. 34)There are three variations of WDM that are commonly used: Broad WDM, Coarse WDM, and Dense WDM. 35)The STM-1 frame is the basic transmission format for SDH. The frame lasts for 125 microseconds; therefore there are 8000 frames per second. 36)In SDH frame structure, the SOH is made up of a regenerator section overhead (RSOH) and a multiplexing section overhead (MSOH). 37) The sensitivity of a photo detector in an optical fiber communication system is describable in terms of the minimum detectable optical power. 38)Please list three applications of EDFA in optical fiber communication system: as preamplifier of receiver 、as power amplifier of transmitter and as the optical repeater. 39) An optical isolator (optical diode), is an optical component which allows the transmission oflight in only one direction.三Interpretation of terms and phrases1) AON (all-optical network) 2) DBR (distributed Bragg reflector)3）FDDI (fiber distributed data interface) 4）HFC (hybrid fiber-coaxial)5）ISDN (integrated services digital network) 6）RA (raman amplifier)7）OC (optical carrier) 8）OTDM (optical time-division multiplexing)9）PON (passive optical network) 10）SCM (subcarrier multiplexing)11）SDH (synchronous digital hierarchy) 12）SPM (self-phase modulation)13）STS (synchronous transpor signal) 14）TW (traveling wave)15）WGA (waveguide-grating router) 16）IMD (intermodulation disortion)17）AWG (arrayed-waveguide grating) 18）AOTF (acousto-optic tunable filter)四画图题1) Draw the element block of a Distributed forward Raman amplifier2) Draw an element diagram of a Double pump EDFAOutputInput EDF WDM IsolatorPump Laser Isolator Pump Laser WDMPump Laser outputfiber input3) Draw a block diagram of a typical optical digital communication system and briefly describe the functions of each part.An optical communication system consists of a transmitter, which encodes a message into an optical signal, a channel, which carries the signal to its destination, and a receiver, which reproduces the message from the received optical signal. The optical repeater is to extend the transmission distance of optical signal.4) Draw the element diagram of the application of optical amplifier.四简答题1) Dispersion: Any phenomenon in which the velocity of propagation of any electromagnetic wave is wavelength dependent.2) Stimulated EmissionsIf a photon of energy hv 12 impinges on 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 hv 12.3) There are 3 dispersion types in the optical fibers in general:1- Material Dispersion2- Waveguide Dispersion3- Polarization-Mode Dispersion4) Polarization mode dispersion (PMD) is due to slightly different velocity for each polarization mode because of the lack of perfectly symmetric & anisotropic of the fiber5) Laser is an optical oscillator. It comprises a resonant optical amplifier whose output is fed back into its input with matching phase. Any oscillator contains:1. An amplifier with a gain-saturated mechanism Optical transmitter Repeater Opticalreceiverfiber fiber2. A feedback system3. A frequency selection mechanism4. An output coupling scheme6) In thermal equilibrium the stimulated emission is essentially negligible, since the density of electrons in the excited state is very small, and optical emission is mainly because of the spontaneous emission. Stimulated emission will exceed absorption only if the population of the excited states is greater than that of the ground state. This condition is known as Population Inversion. Population inversion is achieved by various pumping techniques.7) Turn on DelayWhen the driving current suddenly jumps from low (I1 < Ith) to high (I2 > Ith) , (step input), there is a finite time before the laser will turn on8) The Quantum LimitFor an ideal photo-detector having unity quantum efficiency and producing no dark current, it is possible to find the minimum received optical power required for a specific BER performance in a digital system. This minimum received power level is known as the quantum limit.9) Gain flatness: The difference between the biggest gain and the smallest gain of the different frequency signal.10) The advantage of Raman amplifier: Simple fabricationLow noise, because amplifying action take place inside the ordinarily fiber.The wavelength can be selected in the low loss waveband.Very wide gain bandwidth.11) Micro bending Loss: microscopic bends of the fiber axis that can arise when the fibers are incorporated into cables. The power is dissipated through the micro bended fiber, because of the repetitive coupling of energy between guided modes & the leaky or radiation modes in the fiber.12) Gain saturation: when near saturation, the gain is nonlinear; saturation, the signal cannot be amplified.13) The disadvantage of Raman amplifier:Need large output power pump laser. As Raman Scattering, the energy is transferred from high frequency to low frequency. Cross talk will affect signal.14) The principal noises associated with photo detectors are:1- Quantum (Shot) noise: arises from statistical nature of the production and collection of photo-generated electrons upon optical illumination. It has been shown that the statistics follow a Poisson process.2- Dark current noise: is the current that continues to flow through the bias circuit in the absence of the light. This is the combination of bulk dark current, which is due to thermally generated e and h in the pn junction, and the surface dark current, due to surface defects, bias voltage and surface area.15) List the advantages of fiber-optic communications over other types of communication technologies.The advantage of optical fiber communication:1. Weight and Size2. Material cost (SiO2 is plentiful)3. Information Capacity。

03光纤通信试卷

ABC武汉理工大学教务处试题标准答案及评分标准用纸课程名称光纤通信原理B （ A 卷）I. True(T) or False(F). Please fill in the bracket with T or F. (10%, 1 mark each) 1F 2F 3T 4T 5F 6F 7T 8T 9F 10FII. Filling blanks (20%, 1 marks each )1 meridional rays2 Intrinsic3 Extrinsic4 Wavelength Selective Splitters 5.Wavelength Selective Couplers6 spontaneous emission7 stimulated emission8 incoherent9 continuous 10 diffusion 11 homojunction 12 sensitivity 13 response 14 noise 15 cost 16 L iability 17 speed 18 resistance 19 voltage 20. electric fieldIII. Explain the following terms (10% 2 marks each)1.A particular electromagnetic field configuration. For a given electromagnetic problem, many field distributions exist that satisfy the wave equation, Maxwell ’s equations, and the boundary condition. The particular distribution does not change with propagation2. When the atomic densities in the excited states is larger than ground states , the rate of stimulated emission is larger than spontaneous emission. This condition is referred to as population inversion3. the minimum amount of gain that can compensate the loss of stimulated emission caused by cavity4. A junction between the p-layer and n-layer there is a thin layer which has a bandgap smaller than the layers surrounding it.5. The minimum average received power required by the receiver to operate at BER of 10-9VI. Solving the following problems. (60%, 15 marks each)1. 132110006.51ln 21-⨯==m R R L a mir 5 marks sec 10258.1)(112int -⨯=+=αατmir g p v 5 marks111sec 10946.7/1-⨯==p th G τ 8010324.2⨯=+=N G G N Nthth 5 marks 2. m R R g L m μα200)13.3(1ln )302314.0(211ln )(21421int =+-⨯=-Γ=15marks3. η= 0.5. 7 marksi=RP opt = 6 x 1010 x hν= 6 x 1010 x ch/λ≈ 7.9 nA. 8marks4. For λ=1.3 µm, R= 0.84 A/W, 5 marksfor λ=1.55 µm, R= 1.0 A/W. 5 marksThe reason for the increase in the responsivity is that the photocurrent is proportional to the photon flux (photon arrivals per second), since each photogenerated electron-hole pair produces the same electrical pulse regardless of the energy of the photon that generated it. For a fixed power, a 1.3-µm light has less photon flux than 1 .55-µm light (since photonflux = P optical/hν). Note, however, that in this problem we held the quantum efficiencyconstant. This may or not be the case in general, depending on the type of materials used in the absorber.5 marksD。

《光纤通信》试卷题及答案(最全最经典)

《光纤通信》试卷题及答案（最全最经典）2020-2020年度教学质量综合评估测验试卷《光纤通信》试题注：1、开课学院：通信与信息工程学院。

命题组：通信工程教研组・张延锋2、考试时间：90分钟。

试卷满分：100分。

3、请考生用黑色或蓝色中性笔作答，考试前提前带好必要物件（含计算器）。

4、所有答案请写于相应答题纸的相应位置上，考试结束后请将试卷与答题纸一并上交。

总分一二三四五六七八九十1002018302012试题如下：一、选择题（每小题仅有一个选项是符合题意要求的，共10小题，每小题2分，共20分）1、表示光纤色散程度的物理量是A.时延B.相位差C.时延差D.速度差2、随着激光器使用时间的增长，其阈值电流会A.逐渐减少B.保持不变C.逐渐增大D.先逐渐增大,后逐渐减少3、当平面波的入射角变化时，在薄膜波导中可产生的三种不同的波型是A.TEM波、TE波和TMB.导波、TE波和TM波C.导波、衬底辐射模和敷层辐射模D.TEM波、导波和TM波4、平方律型折射指数分布光纤中总的模数量等于A.B.C.D.5、光接收机中将升余弦频谱脉冲信号恢复为“0”和“1”码信号的模块为A.均衡器B.判决器和时钟恢复电路C.放大器D.光电检测器6、在光纤通信系统中，EDFA以何种应用形式可以显著提高光接收机的灵敏度A.作前置放大器使用B.作后置放大器使用C.作功率放大器使用D.作光中继器使用7、EDFA中用于降低放大器噪声的器件是A.光耦合器B.波分复用器C.光滤波器D.光衰减器8、关于PIN和APD的偏置电压表述，正确的是A.均为正向偏置B.均为反向偏置C.前者正偏,后者反偏D.前者反偏,后者正偏9、下列哪项技术是提高每个信道上传输信息容量的一个有效的途径？A.光纤孤子(Soliton)通信B.DWDMC.OTDMD.OFDM10、光纤数字通信系统中不能传输HDB3码的原因是A.光源不能产生负信号光B.将出现长连“1”或长连“0”C.编码器太复杂D.码率冗余度太大二、填空题（本题共三部分，每部分6分，共18分）（一）、基本概念及基本理论填空（每空1分，共4小题6小空，共6分）11、以色散为基，对于单模光纤来说，主要是材料色散和，而对于多模光纤来说，占主要地位。

光纤通信考试题及答案

光纤通信考试题及答案一、单项选择题（每题2分，共20分）1. 光纤通信中，光信号的传输介质是（）。

A. 铜线B. 光纤C. 无线电波D. 微波答案：B2. 光纤通信中，光信号的调制方式不包括（）。

A. 调幅B. 调频C. 调相D. 调时答案：D3. 下列哪个不是光纤通信的优点（）。

A. 传输距离远B. 抗干扰能力强C. 传输速率低D. 保密性好答案：C4. 光纤通信中，单模光纤的中心折射率（）。

A. 与包层折射率相同B. 与包层折射率不同C. 比包层折射率低D. 比包层折射率高答案：B5. 光纤通信中，下列哪种光纤损耗最小（）。

A. 多模光纤B. 单模光纤C. 塑料光纤D. 石英光纤答案：B6. 光纤通信中，光信号的传输速率与光纤的（）有关。

A. 长度B. 直径C. 折射率D. 材料答案：D7. 光纤通信中，下列哪种光纤最适合长距离传输（）。

A. 多模光纤B. 单模光纤C. 塑料光纤D. 石英光纤答案：B8. 光纤通信中，光信号的传输损耗主要来源于（）。

A. 光纤材料的吸收B. 光纤材料的散射C. 光纤材料的反射D. 光纤材料的折射答案：B9. 光纤通信中，下列哪种光纤最适合局域网（）。

B. 单模光纤C. 塑料光纤D. 石英光纤答案：A10. 光纤通信中，光信号的传输损耗与光纤的长度（）。

A. 成正比B. 成反比C. 无关D. 成对数关系答案：A二、多项选择题（每题3分，共15分）1. 光纤通信中，下列哪些因素会影响光纤的传输损耗（）。

A. 光纤材料的吸收B. 光纤材料的散射C. 光纤的弯曲D. 光纤的直径答案：ABC2. 光纤通信中，下列哪些是光纤的优点（）。

A. 抗电磁干扰B. 传输距离远C. 传输速率高D. 易于安装答案：ABC3. 光纤通信中，下列哪些是光纤的类型（）。

A. 多模光纤B. 单模光纤D. 石英光纤答案：ABCD4. 光纤通信中，下列哪些是光信号的调制方式（）。

A. 调幅B. 调频C. 调相D. 调时答案：ABC5. 光纤通信中，下列哪些因素会影响光纤的传输速率（）。

武汉理工大学通信原理期末考试试题及答案

武汉理⼯⼤学通信原理期末考试试题及答案武汉理⼯⼤学⼀. 填空题1.按照调制⽅式分类，可将通信系统分为通信系统按照信号特征分，包括_______和_______。

2.如果⼆进制独⽴等概信号的码元宽度为0.4ms，则码元速率为_______，信息速率_______；若改为四进制信号，码元宽度不变，码元速率为_______，信息速率_______。

3.线性调制包含四种⽅法：_______、_______、_______和_______。

4.时域均衡准则包括_______和_______。

5.数字带通系统三种基本键控⽅式为：_______、_______和_______。

6.在数字通信系统中，同步包括_______、_______、_______和⽹同步四种同步类型。

7.采⽤时域均衡的⽬的是_______。

8.对⼀个频带限制在（0，4）kHz的连续信号进⾏抽样时，应要求抽样间隔不⼤于_______。

⼆. 简答题（4⼩题，每⼩题5分，共20分）1.简述通信系统中采⽤调制的⽬的。

2.简述奈奎斯特第⼀准则及其物理意义。

3.简述⾹农公式，并简要说明信道容量和各个参数的关系。

4.已知信息代码写出相应的HDB3码。

三. 综合分析题1.设发送的⼆进制信息为10101，码元速率为1200B1)当载频为2400Hz时，试分别画出2ASK、2PSK及2DPSK信号的波形；2)2FSK的两个载波频率分别为1200Hz和2400Hz时，画出其波形；3)计算2ASK、2PSK、2DPSK和2FSK信号的带宽和频带利⽤率。

2.对抑制载波双边带信号进⾏相⼲解调，设接收信号功率为2mW，载波为100kHz，设调制信号m(t)的频带限制在4kHz，信道噪声双边功率谱密度P n(f)=2×10-6mW1)求该理想带通滤波器的传输特性H(ω);2)求解调器输⼊端的信噪⽐；3)求解调器输出端的信噪⽐；3.采⽤13折线A律编码，已知抽样脉冲值，计算编码器输出码组、量化误差、均匀量化11位码。

光纤通信试题资料

光纤通信试题资料一、单项选择题(在每小题列出的四个选项中只有一个选项是符合题目要求的，请将正确选项前的字母填在题干的括号内。

每小题1分，共10分。

)1.在薄膜波导的薄膜中形成导波的条件是平面波的入射角θ1必须满足(C)A.θ1>θC13C13著提高光接收机的灵敏度(A)A.作前置放大器使用B.作后置放大器使用C.作功率放大器使用D.作光中继器使用8.在下列选项中，光纤通信的监控系统不能监控的项目为(C)A.光中继器的接收光功率B.电源的电压C.光缆线路的色散特性D.光纤通信系统的误码性能9.光接收机中，雪崩光电二极管引入的噪声为(D)A.光电检测器的暗电流噪声、雪崩管倍增噪声、光接收机的电路噪声B.量子噪声、雪崩管倍增噪声、光接收机的电路噪声C.量子噪声、光电检测器的暗电流噪声、光接收机的电路噪声D.量子噪声、光电检测器的暗电流噪声、雪崩管倍增噪声10.表示光放大器放大能力的特性指标是(A)A.功率增益B.带宽C.输出饱和功率D.噪声系数B.θ1C12C12C.90°>θ1>θD.90°>θ1>θ2.薄膜波导中的单模传输条件为(B)A.λc(TE0)λc(TE0)C.λc(TE1)λc(TE1)3.光纤相对折射指数差的定义为(C)A.Δ=n1－n2B.Δ=n2－n1C.Δ=(n21－n22)/2n21D.Δ=(n21－n22)/n214.阶跃型光纤中，导模的传输条件为(C)A.V2.405C.V>VcD.V>05.STM －4每秒可传输的帧数是(D)A.1000B.2000C.4000D.80006.随着激光器温度的上升，其输出光功率会(A)A.减少B.增大C.保持不变D.先逐渐增大，后逐渐减少7.在光纤通信系统中，EDFA以何种应用形式可以显二、填空题(每小题1分，共20分)1.光通信2.传输频带宽，通信容量大3.将电信号变换成光信号4.薄膜5.横向谐振6.斜光线7.光纤的纤芯8.映射9.时分复用10.250MHz11.开销比特(或管理比特)12.270某N13.激活物质(或增益物质)14.耦合进光纤的功率(或入纤功率)15.5%～10%16.反向泵辅17.延长中继距离18.灵敏度19.0.98μm20.0.5A/W1.利用光波作为载波的通信方式称为光通信。