光纤通信第五版_第八章讲义

光纤通信第5版课后习题答案要点《光纤通信第5版课后习题答案详细解析》。

在学习光纤通信的过程中，课后习题可是帮助咱们巩固知识的好帮手呢。

不过有时候，有些习题可能会让咱们绞尽脑汁也想不出答案。

别担心，下面我就来给大家分享一下光纤通信第5版课后习题的答案要点，还会讲讲为啥是这样的答案哦，让咱们一起把这些知识点搞清楚！第一章光纤通信概述。

习题1：简述光纤通信的优点。

答案：光纤通信有好多优点呢。

比如说传输损耗低，就像长途运输，损耗小就意味着能传得更远，不用频繁地“加油”（补充信号）。

举个例子哈，传统的电缆传输可能传个几十公里信号就弱得不行了，但是光纤能传几百公里都还很清晰。

再比如说带宽大，能同时传输好多好多信息，就像高速公路有很多车道一样，能容纳大量的车辆（数据）通行。

还有抗干扰能力强，它不受电磁干扰，就像在嘈杂的环境中，它能安静地传输自己的信号，不会被其他干扰影响。

原因：光纤的材料和传输原理决定了它的这些优点。

光纤是用玻璃或者塑料等材料制成的，光在里面传播损耗比较小。

而且它是通过光信号传输信息的，和电磁信号不“打架”，所以抗干扰能力强。

带宽大是因为光的频率很高，能携带大量的信息。

习题2：光纤通信系统由哪些部分组成？答案：光纤通信系统主要由光发送机、光纤、光接收机这几个部分组成。

光发送机就像是一个“快递打包员”，把要传输的信息打包成光信号，然后通过光纤这个“快递通道”送出去。

光纤就像一条专门的道路，让光信号在里面快速传输。

光接收机呢，就像是“快递收件员”，把光信号接收下来，再把里面的信息取出来。

原因：这几个部分各司其职，缺一不可。

光发送机负责把电信号转换成光信号，因为光纤只能传输光信号呀。

光纤提供了传输的路径，保证光信号能准确地到达目的地。

光接收机则要把光信号再变回电信号，这样我们才能使用这些信息。

第二章光纤和光缆。

习题1：光纤的结构是怎样的？答案：光纤一般由纤芯、包层和涂覆层组成。

纤芯就像是光纤的“心脏”，光信号主要在纤芯里面传播。

does not change the average power or the modulation frequencies,but it does lower the signal variation.The transmitted information is contained in this variation,so its attenua-We may think of this result as broadening the signal peak (lowering its amplitude) and filling in the valley (raising its level).Excessive broadening will cause Distortion caused by material (or waveguide) dispersion can be reduced by usingby using more coherent emitters.A laser diode has the advantage over an LED in this respect.In principle,dispersive distortion could be reduced by filtering the optic beam at the transmitter or receiver,allowing only a very narrow band of wavelengths to reach the photodetector.This technique hasA wave incident on a plane boundary between two dielectrics (refrac-) is partially transmitted and partially reflected.(3.30)Although somewhat formidable in appearance,these equations are easily evalu-ated when the two indices of refraction,the incident angle,and the polarization are (3.29) and (3.30) cannot be understated,because they predict the phenomenon by which dielectric fibers guide light.The reflectance is found by squaring the magnitudes of the reflection coeffi-Results are shown in Fig.3.22for an air-to-glass interface and for a glass-to-air interface.The general characteristics shown on the figures appear when there are reflections between any two dielectrics.Some interesting,and features can be noted:The reflectance does not vary a great deal for incident angles near zero.For thethe reflectance value calculated for normal incidence,4%,is a good approximation for angles as large as 20°.meaning full transmission,for certain incident angles andindicating total reflection,for a range of incident angles.-21n 22-n 12sin 2 u i2+21n 22-n 12 sin 2 u i 2The evanescent electric field decays exponentially according to the expression where the attenuation factor and is the free-space propagation factor.the attenuation coefficient discussed in the first section of this chapter.The attenuation coefficient is attributed to actual power losses,critical angle,decay.The decay rate merely indicates how far the field extends into the second medi-um before returning to the incident region.er and the fields decay faster.Rays incident at angles greater than,waves that decay slowly and penetrate deeply into the second medium,dent far above the critical angle produce waves that disappear after only a short pene-tration into the second medium.The reflection coefficient,tity,having a magnitude and an angle when is unity under the condition of total reflection.the reflected wave relative to the incident wave.SUMMARY AND DISCUSSIONThis chapter concentrated on developing fundamental ideas about light waves that apply directly to fiber optics.and polarization —should now be clear.was studied extensively because of its impact on the information-handling capacity of fibers.Other causes of pulse distortion will be considered in Chapter 5.The dependence of information rate on the spectral width of the optic source indicated the importance of this light-emitter property.longitudinal mode structure appearing in the output spectrum of a laser diode.shall see in Chapter 4,resonance also explains the mode structure in a dielectric wave-guide.Reflections at dielectric boundaries play a major role in fiber optics.nal reflection makes it possible for dielectrics to form waveguides for light rays.sin u i =n 2/k 0。