Lecture2_Solution

托福听力lecture怎么练最好托福听力考试中有分为两个部分，conversation对话部分和lecture讲座部分。

日常备考中，如何练习托福听力lecture?下面就是店铺给大家整理的托福听力lecture的练习方法，希望对你有用!托福听力lecture的练习方法第一项：进行概括托福听力怎么练，我们首先推荐就是进行概括，概括就是在听过听力材料后，备考的同学可以对文章大意进行总结和概括。

考生不需要听懂托福材料的每一个细节，但要抓住说话人的思路和框架。

据谷天田研究表明：这个训练对于托福听力至关重要。

练习概括的方法也很简单，可以用托福听力中的lecture为材料，一边听一边记笔记，录音结束后，根据大脑记忆和笔记，对全文重点内容用英文进行复述，这样反复练习，直到可以将一篇lecture文章完全复述出来为止。

第二项：听写托福听力怎么练，大家比较常见的一种练习方法就是听写，这是大家比较熟悉的练习方法，也是相对来说对所有考生都行之有效的一种方法。

它就是将一篇托福听力录音中的原文全部听录下来，一句一句丝毫不落，它是提高听力最有效的方法。

对于准备托福考试的考生而言，最好的听写材料是老托福的lecture。

推荐part 3 的93篇文章，对于基础不同的同学，可以选择其他的适合自己水平的材料进行练习，再用老托福听力，同时辅助一些长篇的段落听写练习。

第三项：精听练习听力，对于任何一种英语考试来说，托福听力怎么练都少不了精听这一环节。

到底该怎么精听呢?其步骤如下：一边听一边看听力材料原文，划出自己不认识或似曾相识但反应不过来的词;录音结束，开始查生词，写在原文的旁边，只要写出这个单词在这篇文章中的意思即可;然后将这篇材料当阅读文章快速精读一遍，彻底看懂;再一边看原文，一边放录音，嘴上要跟读，反复几遍，一直听到可以不看原文彻底听懂为止。

如大家比较常用的SSS(Scientific American’s Sixty-second Science)，非常短，但信息量很大，是很好的练习材料。

英语六级听力真题长对话英语六级听力真题（长对话）（通用8篇）随着时间的推移，一年一度的六级考试马上就要到来了。

听力一直是六级考试的难点。

下面是yjbys网店铺提供给大家关于英语六级听力真题(长对话)，供大家参考。

英语六级听力真题长对话篇1Conversation OneM: So how long have you been a Market Research Consultant?W: Well, I started straight after finishing university.M: Did you study market research?W: Yeah, and it really helped me to get into the industry, but I have to say that it's more important to get experience in different types of market research to find out exactly what you're interested in.M: So what are you interested in?W: Well, at the moment, I specialize in quantitative advertising research, which means that I do two types of projects. Trackers, which are ongoing projects that look at trends or customer satisfaction over a long period of time. The only problem with trackers is that it takes up a lot of your time. But you do build up a good relationship with the client. I also do a couple of ad-hoc jobs which are much shorter projects.M: What exactly do you mean by ad-hoc jobs?W: It's basically when companies need quick answers to their questions about their consumers' habits. They just ask for one questionnaire to be sent out for example, so the time you spend on an ad-hoc project tends to be fairly short.M: Which do you prefer, trackers or ad-hoc?W: I like doing both and in fact I need to do both at the sametime to keep me from going crazy. I need the variety.M: Can you just explain what process you go through with a new client?W: Well, together we decide on the methodology and the objectives of the research. I then design a questionnaire. Once the interviewers have been briefed, I send the client a schedule and then they get back to me with deadlines. Once the final charts and tables are ready, I have to check them and organize a presentation.M: Hmm, one last question, what do you like and dislike about your job?W: As I said, variety is important and as for what I don't like, it has to be the checking of charts and tables.Questions 1 to 4 are based on the conversation you have just heard.Q1: What position does the woman hold in the company?Q2: What does the woman specialize in at the moment?Q3: What does the woman say about trackers?Q4: What does the woman dislike about her job?Conversation TwoW: Hello, I'm here with Frederick. Now Fred, you went to university in Canada?M: Yeah, that's right.W: OK, and you have very strong views about universities in Canada. Could you please explain?M: Well, we don't have private universities in Canada. They’re all public. All the universities are owned by the government, so there is the Ministry of Education in charge of creating the curriculum for the universities and so there is not much room for flexibility. Since it's a government operatedinstitution, things don't move very fast. If you want something to be done, then their staff do not have so much incentive to help you because he's a worker for the government. So I don't think it's very efficient. However, there are certain advantages of public universities, such as the fees being free. You don't have to pay for your education. But the system isn't efficient, and it does not work that well.W: Yeah, I can see your point, but in the United States we have many private universities, and I think they are large bureaucracies also. Maybe people don't act that much differently, because it’s the same thing working for a private university. They get paid for their job. I don’t know if they're that much more motivated to help people. Also, we have a problem in the United States that usually only wealthy kids go to the best schools and it's kind of a problem actually.M: I agree with you. I think it's a problem because you're not giving equal access to education to everybody. It’s no t easy, but having only public universities also might not be the best solution. Perhaps we can learn from Japan where they have a system of private and public universities. Now, in Japan, public universities are considered to be the best.W: Right. It's the exact opposite in the United States.M: So, as you see, it's very hard to say which one is better.W: Right, a good point.Questions 5 to 8 are based on the conversation you have just heard.Q5: What does the woman want Frederick to talk about?Q6: What does the man say about the curriculum in Canadian universities?Q7: On what point do the speakers agree?Q8: What point does the man make at the end of the conversation?英语六级听力真题长对话篇2Lecture 1The negative impacts of natural disasters can be seen everywhere. In just the past few weeks, the world has witnessed the destructive powers of earthquakes in Indonesia, typhoons in the Philippines, and the destructive sea waves that struck Samoa and neighboring islands.A study by the Center for Research on the Epidemiology of Disasters finds that, between 1980 and 2007, nearly 8,400 natural disasters killed more than two-million people. These catastrophic events caused more than $1.5 trillion in economic losses.U.N. weather expert Geoffrey Love says that is the bad news. "Over the last 50 years, economic losses have increased by a factor of 50. That sounds pretty terrible, but the loss of life has decreased by a factor of 10 simply because we are getting better at warning people. We are making a difference. Extreme events, however, will continue to occur. But, the message is that they may not be disasters."Love, who is director of Weather and Disaster Risk Reduction at the World Meteorological Organization, says most of the deaths and economic losses were caused by weather, climate, or water-related extremes. These include droughts, floods, windstorms, strong tropical winds and wildfires.He says extreme events will continue. But, he says extreme events become disasters only when people fail to prepare for them."Many of the remedies are well-known. From a planning perspective, it is pretty simple. Build better buildings. Don’tbuild where the hazards will destroy them. From an early-warning perspective, make sure the warnings go right down to the community level. Build community action plans. ”The World Meteorological Organization points to Cuba and Bangladesh as examples of countries that have successfully reduced the loss of life caused by natural disasters by taking preventive action.It says tropical cyclones formerly claimed dozens, if not hundreds of lives, each year, in Cuba. But, the development of an early-warning system has reversed that trend. In 2008, Cuba was hit by five successive hurricanes, but only seven people were killed.Bangladesh also has achieved substantial results. Major storm surges in 1970 and 1991 caused the deaths of about 440,000 people. Through careful preparation, the death toll from a super tropical storm in November 2007 was less than 3,500.Q16. What is the talk mainly about?Q17. How can we stop extreme events from turning into events?Q18. What does the example of Cuba serve to show?Lecture 2As U.S. banks recovered with the help of American government and the American taxpayers, president Obama held meetings with top bank execut ives, telling them it’s time to return the favor. “The way I see it are banks now having a greater obligation to the goal of a wide recovery,” he said. But the president may be giving the financial sector too much credit. “It was in a free fall, and it was a very scary period.” Economist Martin Neil Baily said. After the failure of Lehman Brothers, many of the world’s largest banks feared the worst as the collapse ofthe housing bubble exposed in investments in risky loans.Although he says the worst is just over, Bailey says the banking crisis is not. More than 130 US banks failed in 2009. He predicts high failure rates for smaller, regional banks in 2010 as commercial real estate loans come due."So there may actually be a worsening of credit availability to small and medium sized businesses in the next year or so."Analysts say the biggest problem is high unemployment, which weakens demand and makes banks reluctant to lend. But US Bankcorp chief Richard Davis sees the situation differently."We're probably more optimistic than the experts might be.With that in mind, we're putting everything we can, lending is the coal to our engine, so we want to make more loans. We have to find a way to qualify more people and not put ourselves at risk."While some economists predict continued recovery in the future, Baily says the only certainty is that banks are unlikely to make the same mistakes - twice. "You know, forecasting's become a very hazardous business so I don't want to commit myself too much. I don't think we know exactly what's going to happen but it's certainly possible that we could get very slow growth over the next year or two.”If the economy starts to shrink again, Baily says it would make a strong case for a second stimulus -- something the Obama administration hopes will not be necessary.Q19. What dose president Obama hope the banks will do?Q20. What is Martin Neil Baily’s prediction about the financial situation in the future?Q21. What does U.S. Bankcorp chief Richard Davis say about its future operation?Q22. What does Martin Neil Baily think of a second stimulus to the economy?英语六级听力真题长对话篇3Section ADirections: In this section, you will hear two long conversations. At the end of eachconversation, you will hear four questions. Both the conversation and the questions will bespoken only once. After you hear a question, you must choose the best answer. from the fourchoices marked A), B),C) and D). Then mark the corresponding letter on Answer Sheet 1 with asingle line through the centre.注意：此部分试题请在答题卡1上作答。

Chapter 2 – Solutions2.1 Dielectric slab waveguidea Consider the rays 1 and 2 in Figure 2.3. Derive the waveguide condition.b Consider the two rays 1 and 2 in Figure 2.4. Show that the phase difference when they meet at C at a distance y above the guide center isΦm = k12(a-y)cosθm - φmc Using the waveguide condition show thatΦm =Φm(y)=mπ-ya(mπ+φm)Solutiona) Ray 1 experiences a phase shift k1 AB – 2 φ (at B)Ray 2 experiences a phase shift k1 A΄B΄(at B΄)The phase difference between the two rays is k1(AB - A΄B) – 2 φFor a wave guide mode this phase difference must be m2πFrom the geometry:AB cos θ = 2aA΄B΄ = A΄C + CB΄A΄C = AC cos(π - 2θ)CB΄= CB cos(π - 2θ)Therefore A΄B΄ = A΄C + CB΄ = (AC + CB) cos(π - 2θ) = AB cos(π - 2θ)This means that the phase shift between the two beamsk1(AB - A΄B΄) – 2 φ = k1[AB - AB cos(π - 2θ)]– 2 φ = k1 AB [1 + cos(2θ)]= k1AB( 2 cos2 θ) – 2 φor the phase shift is equal to k1[2a/cosθ] 2 cos2 θ – 2 φ = k1 (4a cos θ) – 2 φFor a mode this phase shift must be m2πOr k14a cos θ – 2 φ = m2πk12a cos θ –φ = mπ2π n12a/λ cos θm–φm= mπ which is the text’s (page 52 equation 3) waveguide condition.b) c) From the geometry we have the following:(a - y)/AC = cosθand A'C/AC = cos(π - 2θ)The phase difference between the rays meeting at C isΦ = kAC - φ - kA'C = k1AC - k1AC cos(π - 2θ) - φ= k1AC[1 - cos(π - 2θ)] - φ = k1AC[1 + cos(2θ)] - φ = k1[(a - y)/cosθ][ 1 + 2cos2θ- 1]- φ= k1 [(a - y)/cosθ][2cos2θ]- φ= 2k1(a - y)cosθ- φGiven,2π(2a)n1λ⎡⎣⎤⎦cosθm-φm=mπ∴cosθm =λ(mπ+φm)2πn1(2a)=mπ+φmk1(2a)Then Φm =2k1(a-y)cosθm-φm=2k1(a-y)mπ+φmk1(2a)-φm∴Φm =(1-ya)(mπ+φm)-φm=mπ-ya(mπ+φm)Φm =Φm(y)=mπ-ya(mπ+φm)Additional problem:In class we traced the propagation of a 1 V/m wave between two mirrors that each have reflectivity r. We kept track of the phase shift between the two mirrors (separated by a distance L) and found that the total electrical field E (phasor) in the cavity was given by 1/[1-r 2 exp(-2jkL)]. If r is real and R=r 2, show that the total intensity in cavity can be written as (constant)/[(1-R)2 + 4R sin 2kL] (see equation 3 on zpage 30 of the text). It will help you to remember that intensity is proportional to |E|2 and |E|2 = E ⨯E * where E * is the complex conjugate of E.22222222422222222)2cos(21)2sin()2cos(1)]2(cos )2([sin )2cos(21)2sin()2cos(1)2(sin )]2cos(1[)2sin()2cos(1)2(sin )]2cos(1[)2sin()2cos(1)2sin()2cos(11)]2sin()2[cos(11)2exp(11R kL R kL jR kL R kL kL R kL R kL jR kL R kL R kL R kL jR kL R kL r kL r kL jr kL r kL jr kL r kL j kL r jkL r E +---=++---=+---=+---=+-=--=--=2*)2cos(21)2sin()2cos(1RkL R kL jR kL R E +-+-=)(sin 4)1(1)(sin 4211)](sin 21[211)2cos(211])2cos(21[)2cos(21])2cos(21[)2(sin )2(cos )2cos(21])2cos(21[)2(sin )]2cos(1[])2cos(21[)]2sin()2cos(1)][2sin()2cos(1[22222222222222222222222*2kL R R kL R R R kL R R R kL R R kL R R kL R R kL R kL R kL R kL R R kL R kL R kL R R kL R kL jR kL R kL jR kL R E E E I +-=+-+=--+=+-=+-+-=+-++-=+-+-=+---+-=⨯==2.2 TE field pattern in slab waveguide Consider two parallel rays 1 and 2interfering in the guide as in Figure 2.4. Given the phase difference (as in Question 2.1)Φm =Φm (y )=m π-ya(m π+φm ) between the waves at C , distance y above the guide center, find the electric field pattern E (y ) in the guide. Plot the field pattern for the first three modes taking a planar dielectric guide with a core thickness 20 μm, n 1 = 1.455 n 2 = 1.440, light wavelength of 1.3 μm. SolutionThe two waves interfering at C are out phase by Φ,E (y )=A cos(ωt )+A cos[ωt +Φm (y )]where A is an arbitrary amplitude. Thus, [])(cos )](cos[22121y y t A E m m ΦΦ+=ω orE =2A cos 12Φm (y )[]{}cos(ωt +' Φ ) = E o cos(ωt + Φ') in which cos(ωt + Φ') is the time dependent part that represents the wave phenomenon,and the curly brackets contain the effective amplitude. Thus, the amplitude E o isE o =2A cos m π2-y2a(m π+φm )⎡ ⎣ ⎤ ⎦To plot E o as a function of y, we need to find φm for m = 0, 1 and 2 , the first threemodes. From Example 2.1.1 in the textbook, the waveguide condition is(2a )k 1cos θm -m π=φmwe can now substitute for φm which has different forms for TE and TM waves to find,TE wavesta n ak 1cos θm -m π2⎛ ⎝ ⎫ ⎭ =sin 2θm -n 2n 1⎛ ⎝ ⎫ ⎭ ⎪ 2⎡ ⎣ ⎢ ⎤ ⎦⎥ 1/2cos θm=f TE (θm )TM waves tan ak1cosθm-mπ2⎛ ⎝ ⎫⎭ =sin2θm-n2n1⎛⎝⎫⎭⎪2⎡⎣⎢⎤⎦⎥1/2n2n1⎛⎝⎫⎭⎪2cosθm=fTM(θm)The above two equations can be solved graphically as in Example 2.1.1 to find θm for each choice of m. Alternatively one can use a computer program for finding the roots of a function. The above equations are functions of θm only for each m. Using a = 10 μm, λ = 1.3 μm, n1 = 1.455 n2 = 1.440, the results are:TE Modes m = 0 m = 1 m = 2θm (degrees) 88.84 87.67 86.51φm (degrees) 163.75 147.02 129.69TM Modes m = 0 m = 1 m = 2θm(degrees) 88.84 87.67 86.51φm (degrees) 164.08 147.66 130.60There is no significant difference between the TE and TM modes (the reason is that n1 and n2 are very close).We can set A = 1 and plot E o vs. y usingE o =2cosmπ2-y2a(mπ+φm)⎡⎣⎤⎦with the φm and m values in the table above.y = Distance f rom the guide axis (μm)E oFundamental modeSecond mode (m = 2)Third mode (m = 3)CladdingCladdingElectric f ield distribution in the core region of a planar dielectric waveguide. The f ield is harmonic inside the core but decays exponentially in the cladding.Figure 2Q22.3 TE and TM Modes in dielectric slab waveguide Consider a planar dielectric guide with a core thickness 20 μm, n 1 = 1.455 n 2 = 1.440, light wavelength of 1.3 μm. Given the waveguide condition, Eq. (3) in §2.1, and the expressions for phase changes φ and φ' in TIR for the TE and TM modes respectively,tan 12φm ()=sin 2θm -n 2n 1⎛ ⎝ ⎫ ⎭ ⎪ 2⎡ ⎣ ⎢ ⎤ ⎦⎥ 1/2cos θmand tan 12' φ m ()=sin 2θm -n 2n 1⎛ ⎝ ⎫ ⎭ ⎪ 2⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1/2n 2n 1⎛ ⎝ ⎫⎭⎪ 2cos θmusing a graphical solution find the angle θ for the fundamental TE and TM modes and compare their propagation constants along the guide. SolutionFrom Example 2.1.1 in the textbook, the waveguide condition is(2a )k 1cos θm -m π=φmwe can now substitute for φm which has different forms for TE and TM waves to find,TE waves ta n ak1cosθm-mπ2⎛ ⎝ ⎫⎭ =sin2θm-n2n1⎛⎝⎫⎭⎪2⎡⎣⎢⎤⎦⎥1/2cosθm=fTE(θm)TM waves tan ak1cosθm-mπ2⎛ ⎝ ⎫⎭ =sin2θm-n2n1⎛⎝⎫⎭⎪2⎡⎣⎢⎤⎦⎥1/2n2n1⎛⎝⎫⎭⎪2cosθm=fTM(θm)The above two equations can be solved graphically as in Example 2.1.1 to find θm for each choice of m. Alternatively one can use a computer program for finding the roots of a function. The above equations are functions of θm only for each m. Using a = 10 μm, λ = 1.3 μm, n1 = 1.455 n2 = 1.440, the results are:TE Modes m = 0θm (degrees) 88.8361βm = k1sinθm 7,030,883 m-1TM Modes m = 0θ'μ(degrees) 88.8340β'μ = k1sinθm7,030,878 m-1Note that ∆β = 5.24 m-1 and the β-difference is only 7.5⨯10-5 %.The following intuitive calculation shows how the small difference between the TE and TM waves can lead to dispersion that is time spread in the arrival times of the TE and TM optical signals.Suppose that ∆τ is the delay time between the TE and TM waves over a length L. Then,∆τL =1vTE-1vTM=βTEω-βTMω≈∆βω=(5.24 m-1)(1.45⨯1015 ra d/s)= 3.6⨯10-15 s m-1 = 0.0036 ps m-1.Over 1 km, the TE-TM wave dispersion is ~3.6 ps. One should be cautioned that we calculated dispersion using the phase velocity whereas we should have used the group velocity.2.5 Dielectric slab waveguide Consider a dielectric slab waveguide which has a thin GaAs layer of thickness 0.2 μm between two AlGaAs layers. The refractive index ofGaAs is 3.66 and that of the AlGaAs layers is 3.40. What is the cut-off wavelength beyond which only a single mode can propagate in the waveguide assuming that the refractive index does not vary greatly with the wavelength? If a radiation of wavelength 870 nm (corresponding to bandgap radiation) is propagating in the GaAs layer, what is the penetration of the evanescent wave into the AlGaAs layers? What is the mode field distance of this radiation?SolutionGiven n1 = 3.66 (AlGaAs), n2 = 3.4 (AlGaAs), 2a = 2⨯10-7 m or a = 0.1 μm, for only a single mode we needV=2πaλn12-n22()1/2<π2∴λ>2πa n12-n22()1/2π2=2π(0.1 μm)3.662-3.402()1/2π2= 0. 542 μm.The cut-off wavelength is 542 nm.When λ = 870 nm,V=2π(1 μm)3.662-3.402()1/2(0.870μm)= 0.979 < π/2Therefore, λ = 870 nm is a single mode operation.For a rectangular waveguide, the fundamental mode has a mode field distance2wo =MFD≈2aV+1V=(0.2 μm)0.979+10.979= 0.404 μm.The decay constant α of the evanescent wave is given by,α=Va=0.9790.1 μm=9.79 (μm)-1 or 9.79⨯106 m-1.The penetration depthδ = 1/α= 1/ [9.79 (μm)-1] = 0.102 μm.The penetration depth is half the core thickness.2.8 A multimode fiber Consider a multimode fiber with a core diameter of 100 μm, core refractive index of 1.475 and a cladding refractive index of 1.455 both at 850 nm. Consider operating this fiber at λ = 850 nm.a Calculate the V-number for the fiber and estimate the number of modes.b Calculate the wavelength beyond which the fiber becomes single mode.c Calculate the numerical aperture. dCalculate the maximum acceptance angle.e Calculate the modal dispersion ∆τ and hence the bit rate ⨯ distance product given that rms dispersion σ ≈ 0.29∆τ where ∆τ is the full spread. SolutionGiven n 1 = 1.475, n 2 = 1.455, 2a = 100⨯10-6 m or a = 50 μm and λ = 0.850 μm. The V -number is,V =2πaλn12-n 22()1/2=2π(50 μm)1.4752-1.4552()1/2(0.850 μm)= 89.47Number of modes M ,M =V 22=89.4722≈ 4002 The fiber becomes monomode when,V =2πaλn12-n 22()1/2<2.405or λ>2πa n 12-n 22()1/22.405=2π(50 μm)1.4752-1.4552()1/22.405= 31.6 μmFor wavelengths longer than 31.6 μm, the fiber is a single mode waveguide. The numerical aperture NA isN A =n 12-n 22()1/2=1.4752-1.4552()1/2= 0.242If αmax is the maximum acceptance angle, then,αmax=arcsin NA n o ⎛ ⎝ ⎫⎭⎪ = arcsin(0.242/1) = 14︒Modal dispersion is given by ∆τi n t ermo de L =n 1-n 2c =1.475-1.4553⨯108m s-1= 66.7 ps m -1 or 67.6 ns per kmGiven that σ ≈ 0.29∆τ, maximum bit-rate is BL =0.25Lσtota l≈0.25Lσinte rmode=0.25(0.29)(66.7 ns km -1)i.e.BL = 13 Mb s -1 km (only an estimate!)We neglected material dispersion at this wavelength which would further decreaseBL . Material dispersion and modal dispersion must be combined byσt otal 2=σinte rmode 2+σmat eria l 2For example, assuming an LED with a spectral rms deviation σλ of about 20 nm,and a D m ≈ -200 ps km -1 nm -1 (at about 850 nm)we would findσm = -(200 ps km -1 nm -1)(20 nm)(1 km) ≈ 4000 ps km -1 or 4 ns km -1,which is substantially smaller than the intermode dispersion and can be neglected.Additional problem:Derive Equations (7) and (8) on page 54. Start with equations the equations for E1 and E2 that appear on the bottom of page 53 and the top of page 54. The basic idea is to prove that the mode of the cavity has a transverse profile that depends on modenumber. To do this derivation you will need the trigonometric identity (given on the bottom of page 54), sin A + sin B = 2 sin[½ (A + B)] cos[½ (A – B)]. Solution:also from trigonometric identitycosA+cosB= 2cos[½ (A + B)] cos[½ (A – B)]. From text,)cos(),,(01m m z wt E t z y E φβ+-= )cos(),,(02z wt E t z y E m β-=This two waves interfere to give)8()21cos()(2)7()21cos()21cos(2)cos()cos(),,(),,(),,(00021−→−+-=−→−+-=-++-=+=m m m m m m m m m z wt y E z wt E z wt E z wt E t z y E t z y E t z y E φβφβφβφβHere )21cos()(0m m E y E φ=2.9 A single mode fiber Consider a fiber with a SiO 2-13.5%GeO 2 core of diameter of 8 μm and refractive index of 1.468 and a cladding refractive index of 1.464 bothrefractive indices at 1300 nm where the fiber is to be operated using a laser source with a half maximum width of 2 nm. a Calculate the V -number for the fiber. Is this a single mode fiber? b Calculate the wavelength below which the fiber becomes multimode. c Calculate the numerical aperture. dCalculate the maximum acceptance angle.e Obtain the material dispersion and waveguide dispersion and hence estimate the bit rate ⨯ distance product (B ⨯L ) of the fiber. Solutiona Given n 1 = 1.475, n 2 = 1.455, 2a = 8⨯10-6 m or a = 4 μm and λ =1.3 μm. The V -number is,V =2πaλn12-n22()1/2=2π(4 μm)1.4682-1.4642()1/2(1.3 μm)= 2.094b Since V < 2.405, this is a single mode fiber. The fiber becomes multimode whenV =2πaλn12-n22()1/2>2.405or λ<2πa n 12-n 22()1/22.405=2π(4 μm)1.4682-1.4642()1/22.405=1.13 μmFor wavelengths shorter than 1.13 μm, the fiber is a multi-mode waveguide. c The numerical aperture NA isN A =n 12-n 22()1/2=1.4682-1.4642()1/2= 0.108d If αmax is the maximum acceptance angle, then,αmax=arcsin NA n o⎛ ⎝ ⎫⎭⎪ = arcsin(0.108/1) = 6.2︒so that the total acceptance angle is 12.4︒. e At λ =1.3 μm, from the figure, D m ≈ -7.5 ps km -1 nm -1, D w ≈ -5 ps km -1 nm -1. ∆τ1/2L =D m +D w ∆λ1/2 = |-7.5-5 ps km -1 nm -1|(2 nm) = 15 ps km -1 + 10 ps km -1= 0.025 ns km -1Obviously materials dispersion is 15 ps km -1 and waveguide dispersion is 10 ps km -1 The maximum bit-rate distance product is thenBL ≈0.59L ∆τ1/2=0.590.025 ns km-1 = 23.6 Gb s -1 km.Figure 2Q92.10 A single mode fiber design According to Question 1.3 (Ch.1), the Sellmeier dispersion equation provides n vs. λ for pure SiO 2 and SiO 2-13.5 mol.%GeO 2. Therefractive index increases linearly with addition of GeO 2 to SiO 2 from 0 to 13.5 mol.%. A single mode step index fiber for use at 1300 nm is required to have the following properties: NA = 0.1, core diameter of 9 μm and a core of SiO 2-13.5% GeO 2. What should be the cladding composition? Solution Given2a = 9⨯10-6 m or a = 4.5 μm. From Ch 1, Question 1.3, the Sellmeier equation is,n G G G 21221222222322321-=-+-+-λλλλλλλλλwhere G 1, G 2, G 3 and λ1, λ2 and λ3 are constants given below where λ1, λ2, λ3 are in μm.The fiber is to operate at λ =1.3 μm, thus, using the Sellmeier equation abovewith the constants in the table we findn 1 = 1.4682The V -number is,V =2πaλNA =2π(4.5 μm)(1.3 μm)(0.1)= 2.175Apply N A =n 12-n 22()1/2≈2n 12∆()1/2or0.1≈2(1.4682)2∆[]1/2to obtain ∆ = 0.002320Apply∆=n 1-n 2n 1, i.e. 0.00232 = (1.4682-n 2)/1.4682Thus, the required cladding refractive index isn 2 = 1.4648Pure silica has n = 1.4473, SiO 2-13.5 mol.%GeO 2 has n 1 = 1.4682, by linear interpolation the composition corresponding to n 2 = 1.4648 is 11.3 mol.% GeO 2. Note, the refractive index n (x ) of SiO 2-x mol.%GeO 2, assuming a linear relationship, can be written asn (x )=n (0)1-x 13.5⎛⎝ ⎫ ⎭ +n (13.5)x 13.5where n (0) = 1.4473; n (13.5) = n 1 = 1.4682. Substituting n (x ) = 1.4648 gives x = 11.3 .2.11 Material dispersion If N g 1 is the group refractive index of the core material of astep fiber, then the propagation time (group delay time) of the fundamental mode isτ=L v g =LN g 1cSince N g will depend on the wavelength, show that the material dispersioncoefficient D m is given approximately by22λλλτd n d c Ld d D m -≈= Using the Sellmeier equation in Question 1.3 in Chapter 1 evaluate materialdispersion at λ = 1.55 μm for pure silica (SiO 2) and SiO 2-13.5%GeO 2 glass. SolutionFrom Ch. 1 we know thatN g ≈n -λdn d λDifferentiate τ with respect to wavelength λ using the above relationship betweenN g and n .τ=L v g =LN g 1c∴d τd λ=L c dN g 1d λ≈L c dn d λ-λd 2n d λ2-dn d λ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ =-L c λd 2n d λ2Thus,22λλλτd n d c Ld d D m -≈=From Ch. 1 we know that the Sellmeier equation isn G G G 21221222222322321-=-+-+-λλλλλλλλλ (2)where G 1, G 2, G 3 and λ1, λ2 and λ3 are constants; given in Table 2Q11 (called Sellmeiercoefficients) that are determined by fitting this expression to the experimental data. We can substitute Eq. (2) in Eq. (1) to obtain D m and plot D m vs . λ as shown in Figure 2Q11. The substitution, differentiation and the plot were done on Mathview but almost any other math-software package can do the same. Thus,At λ = 1.55 μm, D m = 15 ps km -1 nm -1Table 2Q11The Sellmeier coefficients for SiO 2-13.5%GeO 2. The λ1, λ2, λ3 are in μm.Wavelength (μm)D m (ps km -1 nm -1)15 ps km -1 nm -11.55 μmFigure 2Q112.14A graded index fibera Consider an optimal graded index fiber with a core diameter of 30 μm and arefractive index of 1.474 at the center of the core and a cladding refractive index of 1.453. Suppose that the fiber is coupled to a laser diode emitter at 1300 nm and a spectral linewidth (FWHM) of 3 nm. Suppose that the material dispersion coefficient at this wavelength is about -5 ps km -1 nm -1. Calculate, the total dispersion and estimate the bit rate ⨯ distance product of the fiber. How does this compare with the performance of a multimode fiber with the same core radius, and n 1 and n 2? What would be the totaldispersion and maximum bit rate if an LED source of spectral width (FWHM) ∆λ1/2 ≈ 80 nm is used?b If σintermode (γ) is the rms dispersion in a graded index fiber with a profile index γ, and if γo is the optimal profile index, thenσintermode (γ)σintermode (γo )=2(γ-γo )∆(γ+2)and γo is given by Eq. (2) in § 2.7. Calculate the new dispersion and bit rate ⨯ distance product if γ is 10% greater than the optimal value γo . SolutionThe normalized refractive index difference ∆ = (n 1 - n 2)/n 1 = (1.474 - 1.453)/1.474 = 0.01425 Modal dispersion for 1 km of graded index fiber is σinte rmode ≈Ln 1203c ∆2=(1000)(1.474)203(3⨯108)(0.01425)2= 2.9⨯10-11 s or 0.029 nsThe material dispersion (FWHM) is ∆τm (1/2)=LD m ∆λ1/2=(1000 m)(-5 ps ns -1 km -1)(3 nm)= 0.015 nsAssuming a Gaussian output light pulse shape, rms material dispersion is,σm = 0.425∆τ1/2 = (0.425)(0.015 ns) = 0.00638 nsTotal dispersion isσtotal =σintermode 2+σm2=0.0292+0.006382= 0.0295ns. so that B = 0.25/σtotal = 8.5 GbIf this were a multimode step-index fiber with the same n1 and n2, then the rms dispersion would roughly be∆τL ≈(n1-n2)c=1.474-1.453(3⨯108m s-1)= 70 ps m-1 or 70 ns per kmMaximum bit-rate isBL≈0.25Lσinte rmode≈0.25L(0.28)∆τ=0.25(0.28)(70ns km-1)i.e.BL = 12.8 Mb s-1 km (only an estimate!)The corresponding B for 1 km would be around 13 Mb s-1With LED excitation, again assuming a Gaussian output light pulse shape, rms material dispersion isσm =(0.425)∆τm(1/2)=(0.425)LDm∆λ1/2=(0.425)(1000m)(-5ps ns-1km-1)(80nm)= 0.17 nsTotal dispersion isσtotal =σintermode2+σm2=0.0292+0.172= 0.172 nsso that B = 0.25/σtotal = 1.45 GbThe effect of material dispersion now dominates intermode dispersion.。

托福听力lecture的结构1. Lecture有结构托福听力的lecture基本上都是总分总的结构，即「引入话题——展开讨论——教授总结」，下面分点论述：1）引入话题其实托福听力lecture里面讲课的教授就跟大家碰到过的老师一样，有的喜欢直入主题，有的喜欢拉七杂八。

总的说来，lecture话题的引入方式有三种：•开门见山：啥也不扯，一言不合就开车，比如TPO1 Lecture2：Ok, let’s get started. Great. Today I want to talk about a way in which we are able to determine how old a piece of land, or some other geologic feature is –dating techniques.•课程回顾：回顾上堂课的内容，然后再切入本堂课的内容；一般来说本次是上次的细化或者与上次的对比。

比如TPO5 Lecture2：Last week, we covered some arguments against going back to the Moon. But there are compelling reasons in favor of another Moon landing too, um…not the least of which istrying to pinpoint the moon’s age.••啰哩吧嗦：有的教授实在是特别啰嗦，比如TPO2 Lecture2：Hi, everyone. Good to see you all today.【你也好啊】Actually, Iexpected the population to be a lot lower today. It typically runs between 50 and 60 percent on the day the researchpaper is due.【来的人多不好么，说明你受欢迎啊】Um, I was hoping to have your exams back today【啊，要放榜啦？】, but, uh, the situation was that I went away for the weekend, and I was supposed to get in yesterday at five, and Iexpected to fully complete all the exams by midnight or so, which is the time that I usually go to bed, but my flight was delayed, and I ended up not getting in until one o’clock in the morning【你飞机晚点关我什么事啊？】. Anyway, I’ll do my best to have them finished by the next time we meet【开始上课吧，please】. OK. In the last class, we started talking about...【终于开始了…】不论是用哪一种引入方式，话题总是要出来的。

Topic 2: Elliptic Partial Differential EquationsLecture 2-4: Poisson’s Equation: Multigrid MethodsWednesday, February 3, 2010Contents1 Multigrid Methods2 Multigrid method for Poisson’s equation in 2-D3 Simple V −cycle algorithm4 Restricting the Residual to a Coarser Lattice 2 35 71 MULTIGRID METHODS5 Prolongation of the Correction to the Finer Lattice6 Cell-centered and Vertex-centered Grids and Coarsenings7 Boundary points8 Restriction and Prolongation Operators9 Improvements and More Complicated Multigrid Algorithms8 8 11 11 151 Multigrid MethodsThe multigrid method provides algorithms which can be used to accelerate the rate of convergence of iterative methods, such as Jacobi or Gauss-Seidel, for solving elliptic partial differential equations.Iterative methods start with an approximate guess for the solution to the differential equation. In each iteration, the difference between the approximate solution and the exact solution is made smaller.One can analyze this difference or error into components of different wavelengths, for example by using Fourier analysis. In general the error will have components of many different wavelengths: there will beshort wavelength error components and long wavelength error components.Algorithms like Jacobi or Gauss-Seidel are local because the new value for the solution at any lattice site depends only on the value of the previous iterate at neighboring points. Such local algorithms are generally more efficient in reducing short wavelength error components.The basic idea behind multigrid methods is to reduce long wavelength error components by updating blocks of grid points. This strategy is similar to that employed by cluster algorithms in Monte Carlo simulations of the Ising model close to the phase transtion temperature where long range correlations are important. In fact, multigrid algorithms can also be combined with Monte Carlo simulations.2 Multigrid method for Poisson’s equation in 2-DWith a small change in notation, Poisson’s equation in 2-D can be written:∂ 2u ∂x2 +∂ 2u∂y2= −f (x, y) ,where the unknown solution u(x, y) is determined by the given source term f (x, y) in a closed region. Let’s consider a square domain 0 ≤ x, y ≤ 1 with homogeneous Dirichlet boundary conditions u = 0 on the perimeter of the square. The equation is discretized on a grid with L + 2 lattice points, i.e., L interior points and 2 boundary points, in the x and y directions. At any interior point, the exact solution obeysu i,j = 14u i+1,j + u i−1,j + u i,j+1 + u i,j−1 + h2f i,j .The algorithm uses a succession of lattices or grids. The number of different grids is called the number of multigrid levels . The number of interior lattice points in the x and y directions is then taken to be 2 , so that L = 2 + 2, and the lattice spacing h = 1/(L − 1). L is chosen in this manner so that the downward multigrid iteration can construct a sequence of coarser lattices with2−1→ 2−2→ . . . → 20 = 1interior points in the x and y directions.Suppose that u(x, y) is the approximate solution at any stage in the calculation, and u exact(x, y) is the exact solution which we are trying to find. The multigrid algorithm uses the following definitions:· The correctionv = u exact− uis the function which must be added to the approximate solution to give the exact solution. · The residual or defect is defined asr = 2 u + f .Notice that the correction and the residual are related by the equation2 v = 2 u exact+ f − 2 u + f = −r .This equation has exactly the same form as Poisson’s equation with v playing the role of unknown function and r playing the role of known source function!3 SIMPLE V −CYCLE ALGORITHM3 Simple V −cycle algorithmThe simplest multigrid algorithm is based on a two-grid improvement scheme. Consider two grids:· a fine grid with L = 2 + 2 points in each direction, and· a coarse grid with L = 2−1 + 2 points.We need to be able to move from one grid to another, i.e., given any function on the lattice, we need to able to· restrict the function from fine → coarse, and· prolongate or interpolate the function from coarse → fine.Given these definitions, the multigrid V −cycle can be defined recursively as follows:· If = 0 there is only one interior point, so solve exactly foru1,1 = (u0,1 + u2,1 + u1,0 + u1,2 + h2f1,1)/4 .· Otherwise, calculate the current L = 2 + 2.3 SIMPLE V −CYCLE ALGORITHM· Perform a few pre-smoothing iterations using a local algorithm such as Gauss-Seidel. The idea is to damp or reduce the short wavelength errors in the solution.· Estimate the correction v = u exact− u as follows:– Compute the residualr i,j = 1h2 [u i+1,j + u i−1,j + u i,j+1 + u i,j−1− 4u i,j] + f i,j .–Restrict the residual r → R to the coarser grid.– Set the coarser grid correction V = 0 and improve it recursively.–Prolongate the correction V → v onto the finer grid.· Correct u → u + v.· Perform a few post-smoothing Gauss-Seidel interations and return this improved u. How does this recursive algorithm scale with L? The pre-smoothing and post-smoothing Jacobi or Gauss- Seidel iterations are the most time consuming parts of the calculation. Recall that a single Jacobi or Gauss-Seidel iteration scales like O(L2). The operations must be carried out on the sequence of grids with2 → 2−1→ 2−2→ . . . → 20 = 1interior lattice points in each direction. The total number of operations is of orderL2n=0122n≤ L211 .1 − 44 RESTRICTING THE RESIDUAL TO A COARSER LATTICEThus the multigrid V −cycle scales like O(L 2), i.e., linearly with the number of lattice points N!4Restricting the Residual to a Coarser LatticeThe coarser lattice with spacing H = 2h is constructed as shown. A simple algorithm for restricting the residual to the coarser lattice is to set its value to the average of the values on the four surrounding lattice points (cell-centered coarsening):6 CELL-CENTERED AND VERTEX-CENTERED GRIDS AND COARSENINGSR I,J = 14[r i,j + r i+1,j + r i,j+1 + r i+1,j+1] , i = 2I − 1 , j = 2J − 1 .5 Prolongation of the Correction to the Finer LatticeHaving restricted the residual to the coarser lattice with spacing H = 2h, we need to solve the equation2 V = −R(x, y) ,with the initial guess V (x, y) = 0. This is done by two-grid iterationV = twoGrid(H, V, R) .The output must now be i nterpolated or prolongated to the finer lattice. The simplest procedure is to copy the value of V I,J on the coarse lattice to the 4 neighboring cell points on the finer lattice: v i,j = v i+1,j = v i,j+1 = v i+1,j+1 = V I,J , i = 2I − 1, j = 2J − 1 .6 Cell-centered and Vertex-centered Grids and CoarseningsIn the cell-centered prescription, the spatial domain is partitioned into discrete cells. Lattice points are defined at the center of each cell as shown in the figure:The coarsening operation is defined by doubling the size of a cell in each spatial dimension and placing a coarse lattice point at the center of the doubled cell.Note that the number of lattice points or cells in each dimension must be a power of 2 if the coarsening operation is to terminate with a single cell. In the figure, the finest lattice has 23 = 8 cells in each dimension, and 3 coarsening operations reduce the number of cells in each dimension23= 8 → 22= 4 → 21= 2 → 20 = 1 .Note also that with the cell-centered prescription, the spatial location of lattice sites changes with each coarsening: coarse lattice sites are spatially displaced from fine lattice sites.A vertex-centered prescription is defined by partitioning the spatial domain into discrete cells and locating the discrete lattice points at the vertices of the cel ls as shown in the figure:The coarsening operation is implemented simply by dropping every other lattice site in each spatial dimension. Note that the number of lattice points in each dimension must be one greater than a power of 2 if the coarsening operation is to reduce the number of cells to a single coarsest cell. In the example in the figure the finest lattice has 23 + 1 = 9 lattice sites in each dimension, and 2 coarsening operations reduce the number of vertices in each dimension23+ 1 = 9 → 22+ 1 = 5 → 21 + 1 = 3 .The vertex-centered prescription has the property that the spatial locations of the discretization points are not changed by the coarsening operation.8 RESTRICTION AND PROLONGATION OPERATORS7 Boundary pointsLet’s assume that the outermost perimeter points are taken to be the boundary points. The behavior of these boundary points is different in the two prescriptions:· Cell-centered Prescription: The boundary points move in space towards the center of the region at each coarsening. This implies that one has to be careful in defining the “boundary values” of the solution.· Vertex-centered Prescription: The boundary points do not move when the lattice is coarsened.This make i t easier in principle to define the boundary values.These two different behaviors of the boundary points make the vertex-centered prescription a little more convenient to use in multigrid applications. However, there is no reason why the cell-centered prescription should not work as well.8 Restriction and Prolongation OperatorsIn the multigrid method it is necessary to move functions from a fine grid to the next coarser grid (Restric- tion), and from a coarse grid to the next finer grid (Prolongation). Many prescriptions for restricting andprolongating functions have been studied. Let’s consider two of the simplest prescriptions appropriate for cell- and vertex-centered coarsening:· Cell-centered Coarsening: In this prescription, a coarse lattice point is naturally associated with 2d neighboring fine lattice points in d-dimensions.· Suppose that f (x) is a function on the fine lattice at spatial position x, and F (X ) is the corresponding function on the coarse lattice, then this diagram suggests a simple prescription for restriction and prolongation.–Restriction: Average the function values at the 4 neighboring fine lattice sites x i:F (X ) = 144i=1f (x i) .– Prolongation: Inject the value of the function at the coarse lattice site to the 4 neighboring fine lattice sites:f (x i) = F (X ) , i = 1 . . . 4· Vertex-centered Coarsening: Consider a coarse lattice point and the 9 neighboring fine lattice points shown in the figure:2 2 · In this prescription, a coarse lattice point can naturally associated (in 2-D) with · the corresponding fine lattice point, or· the four nearest neighbor fine lattice points, left, right, up, and down, or · with the four diagonally nearest fine lattice points, etc.· It is a little more complicated here to define transfer operators. The problem is that the fine lattice points are associated with more than one coarse lattice point, unlike the cell-centered case: – The single red fine lattice point in the center coincides with an unique coarse lattice point. – Each of the 4 black fine lattice points however is equidistant from two coarse lattic e points. – Each of the 4 red fine lattice points is equidistant from four coarse lattice points. · This sharing of lattice points suggests the following prescriptions:· Prolongation: use bilinear interpolation in which the value of F at a coarse grid point is copied to 9 neighboring fine -grid points with the following weights:1 1 14 2 1 1 1 1 4 1 1.4 24This matrix is called the stencil for the prolongation.8 8 9 IMPROVEMENTS AND MORE COMPLICATED MULTIGRID ALGORITHMS· Restriction: The restriction operator is taken to be the adjoint of the prolongation operator:1 1 116 1 1 16 8 1 4 1 8 161 1 16.This choice of restriction operator is called full weighting.9Improvements and More Complicated Multigrid AlgorithmsThe algorithm implemented above is the simplest multigrid scheme with a single V-cycle. Section 19.6 of Numerical Recipes discusses various ways of improving this algorithm:· One can repeat the two-grid iteration more than once. If it is repeated twice in each multigrid level one obtains a W-cycle type of algorithm.· The Full Multigrid Algorithm starts with the coarsest grid on which the equation can be solved exactly. It then proceeds to finer grids, performing one or more V -cycles at each level along the way. Numerical Recipes gives a program mglin(u,n,ncycle) which accepts the source function −f in the first argument and implements the full multigrid algorithm with = log 2(n − 1) levels, performing ncycle V-cycles at each level, and returning the solution in the array parameter u. Note that this program assumes that the number of lattice points in each dimension L is odd, which leads to vertex centered coarsening:REFERENCESREFERENCESReferences[Recipes-C19-5] W.H. Press, S.A. Teukolsky, W. Vetterling, and B.P. Flannery, “Numerical Recipes in C”,Chapter 19 §6: Multigrid Methods for Boundary Value Problems, /a/bookcpdf/c19-6.pdf.。

托福听力tpo45 lecture1、2、3、4 原文+题目+答案+译文Lecture1 (2)原文 (2)题目 (4)答案 (6)译文 (6)Lecture2 (7)原文 (7)题目 (10)答案 (12)译文 (12)Lecture3 (15)原文 (15)题目 (17)答案 (19)译文 (19)Lecture4 (21)原文 (21)题目 (24)答案 (26)译文 (26)托福听力tpo45 lecture1、2、3、4 原文+题目+答案+译文Lecture1原文NARRATOR: Listen to part of a lecture in an art history class.MALE PROFESSOR: As I was saying, the Renaissance period—which started in the fourteen hundreds in Europe—the Renaissance was still a pretty religious period. And that's reflected in the artwork of that time.But artists were starting to experiment with a more secular point of view as well—a tendency to also use the natural world as the subject matter for their art.And there were different ways that these natural themes were explored. For instance, many artists would paint portraits, while others—although this was more common in northern Europe—would make landscapes the subject of their works.But today I'd like to consider an influential Italian Renaissance artist, Leon Battista Alberti, who took a slightly different approach.Leon Battista Alberti was a painter, sculptor, architect, musician, poet—very wide-ranging interests—like daVinci or Michelangelo, the sort of guy for whom the term “Renaissance man” was in fact created.Alberti believed that the most important approach for a painter was to capture a story or narrative. Now, as I've indicated, this narrative could be either religious or secular, depending on what the work of art was for. If the work was to be placed in a church then obviously it'd have a religious theme, whereas if it was for someone's home then it could deal with a different subject matter. The exact narrative didn't really matter, so long as it was one that captivated the audience—that held the viewer's attention.So what is actually needed to tell a story?Well, Alberti needed characters, right? Human figures.And he wanted to represent them as realistically as possible to capture the viewer's attention. One way he achieved this was to make use of what's known as the contrapposto pose.A contrapposto pose basically entails showing a slight twist in the body. The shoulders and hips are usually bent in different directions. In other words, if the left shoulder is bent so that it's slightly higher than the right shoulder, then the hips will be bent so that the left side will be slightly lower than the right side. Similarly, in sculptures, most of the weight seems to be on one foot, which also results in this slanted position—making it seem like the figure is about to walk or move. This adds to the realistic aspect of the figure.But there are actually a lot of things that could go wrong in the attempt to create such a pose. You could make a figure’s arms bigger than its legs, or the head too small for the body. Messing up the proportions can leave a figure looking cartoon-like and unnatural. But Alberti had a solution: He encouraged artists to visualize a figure's bones and structure. This would give the artist an idea of the proportions of the figure. From there, Alberti suggested the artist imagine attaching the tendons and muscles, then covering those with flesh and skin.Now, although this method may seem complicated, artists since antiquity have used anatomical observations to try to get the proportions of the human figure as accurate as possible—though obviously not to the degree that Alberti was recommending.Now, in addition to characters, the setting is extremely important, especially when attempting to tell a story realistically. Renaissance artists essentially needed to create a three-dimensional scene on a two-dimensional surface. They accomplished this by the use of perspective—a relatively new idea for artists at the time. In particular, the type of perspective that Alberti advocated was called linear one-point perspective. In fact, Alberti was one of the artists who developed the geometry behind linear one-point perspective.Linear perspective basically consists of drawing straight lines that extend from the forefront of the painting into the background—lines that seem to be parallel to each other, but which actually converge on a single point in the horizon, called the vanishing point. By drawing figures and objects smaller and smaller as the lines get closer together, the artist is able to create depth in a painting. This gives the illusion of a third dimension and makes the work of art more realistic.题目1.What is the lecture mainly about?A. Reasons for the transition from religious to secular themes in Renaissance artB. The disproportionate influence of Italian artists during the Renaissance periodC. Techniques used during the Renaissance to produce realistic works of artD. A comparison of themes in paintings and sculptures during the Renaissance2.What is the professor's opinion of Leon Battista Alberti as an artist?A. Alberti's interests were too diverse for him to succeed in any one field.B. Alberti was ineffective in imposing his own theories on other artists.C. Alberti was a much more skilled artist than da Vinci or Michelangelo.D. Alberti represents the Renaissance ideal of wide-ranging achievement.3.According to the professor, what did Alberti consider to be the most important aspect of a Renaissance painting?A. That it convey an appealing narrativeB. That its figures be posed symmetricallyC. That its theme not be religiousD. That its characters be positioned within a landscape4.Why did some artists begin to use the contrapposto pose?A. To create a cartoon-like effectB. To help viewers identify the main figure in a work of artC. To show the relative sizes of human figuresD. To make human figures appear more natural5.Why does the professor discuss tendons and muscles?A. To emphasize that Alberti's study of anatomy led to his interest in artB. To show the emphasis Alberti placed on using physically fit modelsC. To illustrate the difficulty of maintaining a contrapposto pose in real lifeD. To explain one of Alberti's methods for creating accurate proportions6.Why was the development of linear one-point perspective important to Renaissance artists?A. It helped painters to place figures more symmetrically within their paintings.B. It allowed painters to create an illusion of three dimensions.C. It enabled artists to paint large landscapes for the first time.D. It encouraged artists to take an interest in geometry.答案C D A D D B译文旁白：听一篇艺术史学科讲座。