光纖通信必考填空題、計算題及答案知識點

1、光纖通信必考填空題、計算題及其答案 知識點一、填空題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.光纖通信鏈路的主要成分。的關鍵部分是一個發(fā)射機包括一個光源及其相關的 驅(qū)動電路，一個電纜提供機械和環(huán)境保護于光纖內(nèi)部包含，和一個接收器包括 一個光電探測器加放大和信號復原電路。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.因為它傳播沿纖維是在光通信系統(tǒng)的設計中的重要考慮因素的光信號的衰減， 在一個光纖中的基本衰減機制是吸收，散射，以及光學能量的輻射損失。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.模間色散或模延遲只出現(xiàn)在多模光纖。這個信號扭曲機構是在單一頻率具有群 速度的不同值的每個模式的結果。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.在一般情況下，使用LEg多模光纖，因為通常t是只有到多模光纖，從一個 LED的非相干光輸出功率可被耦合以足夠的量是有用的。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.在光傳輸線的輸出端，必須有用于解釋包含在光學信號中的信息的接收裝置。該接收器的第一個元素是光檢測器。它檢測在發(fā)光功率落下后，它與該光功率 的變化轉換成相應的不同的電流。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.光接收器由一個光檢測器，放大器，和信號處理電路。該接收器具有第一轉換 的光能新興從光纖的端部轉換成電信號，然后放大該信號到一個足夠大的電平 的任務。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.波分復用的一個特點是，該離散波長形成正交載波集合可被分離，路由和交換 而不相互干擾的。Chromatic dispersion causes pulse broadening , which leads bit-error rates.色散導致脈沖展寬，從而導致誤碼率。The function

9、of an OADM is to insert or extract one or more selected wavelengths at a designated point in an optical network.一個OADMJ功能是插入或在光網(wǎng)絡中提取在一個指定點的一個或多個選擇的 波長。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.雖然這樣的光放大器的結構類似于激光的，它不具有光學反饋機制。因而放大 器可以放大輸入的信號的水平，但它本身不能產(chǎn)生一個相干光輸出。The pump light is usually injected from the same direction as the signal flow, this is known as codirectional pumping泵浦光通常從相同的方向的信號流注入，這被稱為同向

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纖傳輸系統(tǒng)，網(wǎng)絡流量的動態(tài)特性，尤其是在一個城 域網(wǎng)絡，可以建立在輸入光功率電平到EDF從，快速波動。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通常被稱為自愈環(huán)，因為沿著某個路徑流動的流量能夠自動 切換到下面的鏈路段的故障或降解的替代或備用路徑。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長路由網(wǎng)絡和光分組交換網(wǎng)絡之間In single-mode fibers, chromatic and polarization-mode dispersions are important factors that can limit the transmission distance or data rate在單模光纖，套色和偏振模分散體是可以限制的傳輸距離或數(shù)據(jù)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數(shù)據(jù)速率10Gb / s的很重要，因為它的統(tǒng)計行為最終可以 限制可達到的最高數(shù)據(jù)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.當鏈路正在安裝和測試，感興趣的操作參數(shù)包括誤碼率，定時抖動和信噪比由 眼圖所指示的。二、計算題一 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.考慮的多模石英光纖具有纖芯折射率和包層的折射率。發(fā)現(xiàn)(a)該臨界角，(b)該數(shù)值孔徑，及(c)的接受角。解：(a)由式sin c n2可以求得臨界角n11(b)由式NA nsin A (n2 n2)2 r J可 可以計算出數(shù)值孔徑(c)由式nsin 0,max nsin A n1sin c (n； n2)2可以求得出空氣中的接受角為二、 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.考慮多模階躍光纖帶62.5-纖芯直徑以及1.5 %的芯-包層折射率差。如果芯 的折射率是1.480,估計為850nm的波長支撐在纖維模式的總數(shù)。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%何計算出V值，再利用M 1V2可以計算出總模數(shù)。假設我們有一個多模階躍折射率光纖，其具有 25的芯半徑，1.480芯指數(shù)和折 射率差。什么是在波長86

19、0, 1310,和1550模式在光纖的數(shù)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.假設我們有三個多模階躍折射率光纖

20、每一個都具有為 1.48和折射率差的芯指 數(shù)。假設這三個纖維具有50, 62.5和100。纖芯直徑是什么，在這些纖維在波 長of1550nm模式的數(shù)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.的雙異質(zhì)結的InGaAsP的LED發(fā)光波長1310nm的峰值波長具有30和100納秒 的輻射和非輻射內(nèi)部qudiative重組倍。驅(qū)動電流40毫安。發(fā)現(xiàn)(a)該本體 復合時間；(b)該內(nèi)部量子效率；和(c)的內(nèi)部電源電平。解：(a)由1 - 1得整體復合壽命二r nrr nr1(b) 由 int -得內(nèi)重子效率r nr r(c)由pint int hv int，得內(nèi)部發(fā)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?考慮一個1公里長的多模階躍折射率光纖，其中=1.480和=0,01 ,以便 =1.465。什么是在這種光纖單位長度的模延遲？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?考慮下面兩個多模光纖(a) 一種階躍光纖帶芯指數(shù)=1.458和芯-包層折射率 差=0.01;(b) 一種拋物線輪廓漸變折射率光纖與相同的值和。比較均方根脈沖每公里拓寬這

24、兩個光纖？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?一個給定的步長指數(shù)光纖具有1.4

25、800.25 %的芯-包層折射率差的芯折射率, 纖芯半徑等于4.5,和。什么是截止波長該纖維？2ao 32a .=解：由式 c(n2 n2)2 np/2,當V=2.405時，光纖的截止波長為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處輸出功率為p( ) 239 W在100M處的輸出光功率為 p( ) 76 W 由此可見這種器件的輸出光功率隨著調(diào)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一個給定的GaAlAs激光器和光的300and100-一腔的寬度。在正常工作溫度 下，該增益因子和損耗系數(shù)。假設反射率是鄰為每個端面。發(fā)現(xiàn)(a)該閾值電流

28、密度和(b)的閾值電流為這個設備_11_.一0解：(a)由 gtht ln()end 和 gth Jth 可得2LR1R2,1 1-1、Jth = Lln( R)(b)閾值電流路由下式給出Ith Jth腔體橫截面積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?考慮一個雙異質(zhì)邊發(fā)射法布里-珀羅的AlGaAs激光器，發(fā)射在900nm的。假 設激光芯片300長和激光材料的折射率是4.3。(一)有多少個半波長跨越法 布里-珀羅反射鏡表面之間的區(qū)域？(b)本人的激射模之間的間距是什么？解：(a)從式m 2Lnv可得法布里-珀羅腔兩鏡面間的半波數(shù)目m2n c2(b)從式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?為3.6的折射率的神化錢光學源耦合到一個石英光纖具有外徑1.480的折射率。什么是源和纖維之間的功率損耗？2解：如果光纖端面和光源在屋里上緊密銜接，則由 R，在光源和光纖頭端的分ni n界面上菲涅耳反射可用下式來表示2R 比 0.174這相當于有17.4

31、%的光功率被反射回光源，與這一R值相應的耦合n1 n功率由下式給定Pcoupled (1 R) Pemitted用分貝表示的功率損耗 L為Pc1adL101g(Rouped)10lg(1 R) 0.83dBPemitted這個值有可能因為在光源和光纖端面之間存在折射率匹配物質(zhì)而減小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、緊密耦合到一個階躍光纖具有1.480的芯折射率。假設源尺寸比纖維核心更小，并且在源和纖維之間的小間隙被填 充有具有1.520的折射率的凝膠。(一)什么是用分貝的功率損耗從源到光纖 多少？ ( b)是什么的功率損耗，如果不使用凝膠？2解：(a)這里需要考慮兩個界面反射率。首先由 R可以得出凝膠對光源的n1 n反射率Rsg 0.159同樣，可得光纖對凝膠的反射率 Rgf 0.040于是，總的反射率為 R Rsg Rgf 0.040 0.0064。功率損耗的分貝值則為L 10lg(1 R) 0.0028dB(b)如果不用凝膠作為折射率匹配材料，切假設光源與光纖之間沒有間隙，則由2R 叱可得此時反射率

34、為 R=0.168 n1 n此時的功率損耗分貝數(shù)為 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?一個給定的硅雪崩光電二極管具有 65%的900nm的波長的量子效率，假設光功 率的0.5

35、產(chǎn)生的10倍增光電流。什么是乘法 M解：初級光電流為IpPn Pin ,倍增因子M 顯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光電二極管具有在1300nm的波長以下參數(shù)：，且表面的漏電流 是可以忽略不計。入射光功率為 30nW(-35dBnj),并接U攵帶寬為20MHz找到 接收機的各種噪聲條件。解：首先計算初級光電流Ip,Pin光電二極管的均方量子噪聲電流為(i：hoj 2qI pBesiiop e均方暗電流為:i；B； 2qIDBe接收機的均方熱噪聲電流處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.考慮一個模擬光纖系統(tǒng)在1550nm的工作，其中有5 MHz的有效接收機噪聲帶 寬。假定r

38、eceiverd信號量子噪聲的限制，什么是入射光功率必須具有在接收 器中的信號與噪聲為50dB比率？假設響應是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?雙錐錐形光纖耦合器具有。輸出功率的輸入光功率電平在其它三個端口和。什 么是耦合比，過量損耗，插入損耗和回程損耗對這種聯(lián)接器？P2解：耦合比=(2 ) 100%P P2附加損耗=10lg(p)P插入損耗=10lg()Pj回波衰減=10lg(鳥 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?假定其驅(qū)動電路的LED在一起具有15ns的上升時間。取40納米的典型LED的 光譜寬度，我們有21ns在6公里鏈路的材料分散。假定接收器具有 25 MHz帶 寬，然后從接收的上升時間降解為 14ns。如果我們選才該纖維具有400-MHz帶 寬距離產(chǎn)品和其中q= 0.7 ,則模態(tài)色散引起的纖維的上升時間為 3.9納秒。什 么是鏈接的上升時間？解：鏈路的展寬時間為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?假設我們有一個3分貝耦合器，所以輸入功率的一半被耦合到所述第二光纖。什么是輸出功率和？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?考慮市售的單模耦合器從一個級聯(lián)的3-dB的熔合光纖耦合器，其中5%的功率損失在每個元素進行。什么是這個耦合器的過剩和分裂的損失？解：由1響武與可以得到附加損耗為mg。95*- 1.1dB八工1分路損耗=10lg( -)101g(32) =15dB最后的總損耗為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?考慮具有波導光柵復用器=10毫米。=5 , =1.45,和一個中央設計波長 二1550nm的。什么是信道間隔對于 m =1?解：對于m=1 ,由L m可以得到波導長度差nc由式 上必&則得到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?考慮對于一個130納米的InGaAsP的SOAZ下參數(shù)：有源區(qū)寬度是3,有源區(qū) 厚度d為0.3 ,放大器長度L為500,什么是排氣速度為 SOAJ 1解：如果米用100mA的偏置電流，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