Lecture2_Solution
江苏省徐州市王杰中学高一英语课件：Unit3 Words(牛津译林版必修3)
paint
v.①在某物上涂颜料或油漆 paint a house blue ②画画描绘 The girl is good at painting landscapes.
Practice:
I want to have the table painted yellow ____________(漆成黄色) paint the door in different colors Why not ___________________________ （把门漆成不同的颜色)
discover
vt.发现，发觉
Who discovered the new continent?
feed
1.给人或动物食物，喂养，饲养 ①feed sb./sth.on sth. feed sth.to sb./sth. What do you feed your dog on? She fed some more milk to the baby. ②以 ---为主食 Cows feed on grass . ③提供原料
3.I can allow you to play football on one condition ________,you must finish your homework first.
uncover vt.揭开，揭露
He uncovered the scandal.他揭露那件丑闻。 Police have uncovered a plot to steal the money of the bank. ______________________________ 警察已揭露一起偷钱阴谋..
lecture2
0
u x
e p, u u
L x, u
性质六的证明
e p, u 在价格 p 上为凹函数
固定效用为 u ，取价格 p, p, p ，p p 1 p ，设 在价格为 p时最优解为 x ，支出函数为
求支出函数。
对偶问题
效用最大化： max u x x 支出最小化： min px x
s.t. u x u x* x h p , u
s.t. px y x* x p , y
当 y e p, u 、 u v p, y 时，效用最大化问题的解和支出最小化问题的解相同，即：
本讲涉及到的数学知识之一: 包络定理 三、包络定理的图形描述
M (a) g ( x(a), a)
a
a
第一节
间接效用函数
一、间接效用函数的定义
二、间接效用函数的性质 三、间接效用函数的应用
一、间接效用函数的定义
①
x2
收入变化
x2 x2
x1
x1
x1
②
价格变化
x2
x2
x1
x1
ＰＳ：关于吉芬商品
① ②
③
吉芬商品是由英国人Robert Giffen发现的，地 点在爱尔兰，时间19世纪中叶。 吉芬商品的存在一般得具备两个条件：一是很 少相近的替代品；二是其开支占收入的比例很 大。 一般来说，劣质商品对财富水平比较低的家庭 来说很可能是吉芬品。即土豆价格下降时，家 庭实际上更富有了，就愿意购买其它更为合意 的商品，从而减少对土豆的消费。
Lecture2-Problem & Solution1
Directions: For this part, you are allowed 30 minutes to write a short essay entitled Digital Age. You should write at least 150 words following the outline given below. 1. 如今数字化产品得到越来越广泛的使用，举例说明 2. 数字化产品的使用对人类工作、学习和生活产生的 影响
My Summer Vacation
Sometimes in the afternoon I, in company with my cousins, took a rod and went to the river to fish. It made the household happy when they saw us returning with a basketful of large fresh fish. When night came, I would sit at the door and tell the most marvelous and interesting stories to my cousins, who listened to the tales with open eyes and mouths. I wish I could enjoy such calm, pure and beautiful rural life forever.
总之,数码产品有它好的方面和不好的方面。我们应该充分利 用他们,尽量避免可能的副作用。
情景式作文:
2005.12六级真题
• Directions: You should write at least 150 words following the situation given below: Situation: Students have different ways to spend their summer vacations, and you decide to take your summer vacation with your family in the countryside. Your feeling is quite different with the life in the city, so you think about it deeply. • 作文的中心思想是：作者暑期在乡下度假的情 形及感受。因此，我们可以断定该作文题要求 考生写一篇记叙文. • 例文见教材P26--30 14
lecture2
Validation – An Introduction
• The process of acquiring the conviction that a model actually works is commonly called validation. • When people are persuaded or convinced that a model is useful in some basic context, they will speak of it as a valid model. However, the validity is often restricted to a certain context. Hence, it is important to know the limitations of the model. Some people may never accept a model.
Validation is a considerably weaker term than "proof"
The Scientific Method (in natural science)
Inductive (归纳) generalization Hypothesis “Real” system Testing and revision Verification Application “Real” conclusions Models are invented ------发明 ------发明 – Theories are discovered -----揭示 -----揭示 Theory
Deduction
Deduction: (演绎 演绎) 演绎
托福听力lecture怎么练最好
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。
chapter2 solution
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
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。
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1/62建工学院土木系Dept. of Civil EngineeringAutumn 2008Solution to Lecture 2Dr. Shao WeiyunDept. of Civil EngineeringEmail:shaowy@2/62建工学院土木系Dept. of Civil Engineering Autumn 20082.10The basic elements of a hydraulic press are shown in Fig. P2.10.The plunger has an area of 1 in.2, and a force, F 1, can be applied to the plunger through a lever mechanism having a mechanical advantage of 8 to 1. If the large piston has an area of 150 in.2, what load, F 2, can be raised by a force of 30 lb applied to the lever? Neglect the hydrostatic pressure variation.3/62建工学院土木系Dept. of Civil Engineering Autumn 20082.16A U-tube manometer is connected to a closed tank as shown in Fig. P2.16. The air pressure in the tank is 0.50 psi and the liquid in the tank is oil (γ= 54.0 lb/ft3). The pressure at point A is 2.00 psi. Determine: (a) the depth of oil, z , and (b) the differential reading, h , on the manometer.4/62建工学院土木系Dept. of Civil Engineering Autumn 20082.21•The inverted U-tube manometer of Fig. P2.21 contains oil (SG = 0.9) and water as shown. The pressuredifferential between pipes A and B , pA −pB , is −5 kPa. Determine the differential reading, h .5/62建工学院土木系Dept. of Civil Engineering Autumn 20082.22• A piston having a cross-sectional area of 0.07 m2 is located in a cylinder containing water as shown in Fig. P2.22. An open U-tube manometer is connected to the cylinder as shown. For h 1 = 60 mm and h= 100 mm, whatis the value of the applied force, P , acting on the piston? The weight of the piston is negligible. 6/62建工学院土木系Dept. of Civil Engineering Autumn 20082.36• A rectangular gate having a width of 5 ft is located in the sloping side of a tank as shown in Fig. P2.36. The gate is hinged along its top edge and is held in position by the force P.Friction at the hinge and the weight of the gate can be neglected . Determine the requiredvalue of P.7/62建工学院土木系Dept. of Civil Engineering Autumn 20082.46A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless hinge at point O , as shown in Fig. P2.46. The horizontal portion of the gate covers a 1-ft-diameter drain pipe which contains air at atmospheric pressure. Determine the minimum water depth, h , at which the gate will pivot to allow water to flow into the pipe.8/62建工学院土木系Dept. of Civil Engineering Autumn 20082.49•Water backs up behind a concrete dam as shown in Fig. P2.69. Leakage under the foundation gives a pressure distribution under the dam as indicated. If the water depth, h , is too great, the dam will topple over about its toe (point A ). For the dimensions given, determine the maximum water depth for the following widths of the dam: ℓ= 20, 30, 40, 50, and 60 ft. Base your analysis on a unit length of the dam.The specific weight of the concrete is 150 lb/ft3.9/62建工学院土木系Dept. of Civil EngineeringAutumn 200810/62建工学院土木系Dept. of Civil Engineering Autumn 20082.50• A 4-m-long curved gate is located in the side of a reservoir containing water as shown in Fig. P2.50. Determine the magnitude of the horizontal and vertical components of the force of the water on the gate. Will this force pass through point A ? Explain.11/62建工学院土木系Dept. of Civil Engineering Autumn 20082.51•The air pressure in the top of the 2-liter pop bottle shown in Video V2.4and Fig. P2.51 is 40 psi, and the pop depth is 10 in. The bottom of the bottle has an irregular shape with a diameter of 4.3 in. (a)If the bottle cap has a diameter of 1 in. what is the magnitudeof the axial force required to hold the cap in place? (b)Determine the force needed to secure the bottom 2 in. of the bottle to its cylindrical sides. For this calculation assume the effect of the weight of the pop is negligible. (c)By how much does the weight of the pop increase the pressure 2 in. above the bottom? Assume the pop has the same specific weight as that of water.12/62建工学院土木系Dept. of Civil Engineering Autumn 20082.56• A closed tank is filled with water and has a 4-ft-diameter hemispherical dome as shown in Fig. P2.56. A U-tube manometer is connected to the tank. Determine the vertical force of the water on the dome if the differential manometer reading is 7 ft and the air pressure at the upper end of the manometer is 12.6 psi.13/62建工学院土木系Dept. of Civil Engineering Autumn 20082.57•Three gates of negligible weight are used to hold back water in a channel of width b as shown in Fig. P2.57. The force of the gate against the block for gate (b) is R.Determine (in terms of R ) the force against the blocks for the other two gates. 14/62建工学院土木系Dept. of Civil EngineeringAutumn 200815/62建工学院土木系Dept. of Civil EngineeringAutumn 20082.59•The homogeneous wooden block A of Fig. P2.83 is 0.7 m by 0.7 m by 1.3 m and weighs 2.4 kN. The concrete block B (specific weight = 23.6 kN/m3) is suspended from A by means of the slender cablecausing A to float in the position indicated. Determine the volume ofB.16/62建工学院土木系Dept. of Civil Engineering Autumn 20082.61•An inverted test tube partially filled with air floats in a plastic water-filled soft drink bottle as shown in Video V2.5and Fig. P2.61. The amount of air in the tube has been adjusted so that it just floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause the test tube to sink to the bottom of the bottle. Explain this phenomenon.。