lecture14
英国文学史及作品选读教案 Lecture 14
Lecture 14The Modern Period (II)ⅠTeaching ContentModernism; the Stream of Consciousness; James JoyceⅡTime Allotment2 periodsⅢTeaching Objectives and Requirements1 Help the students know about modernism.2 Help the students know the characteristics of the novels of the Stream ofConsciousness.3 Help the students know clearly about James Joyce.ⅣKey Points and Difficult Points in Teaching1Modernism2 The Stream of Consciousness3 James JoyceⅤTeaching Methods and MeansLecture; Discussion; Multi-mediaⅥTeaching Process1 Modernism●Modernism is a general term applied retrospectively to the wide range ofexperimental and avant-garde trends in literature of the early 20th century, including Symbolism, Futurism, Expressionism, Imagism, Vorticism, Dada, and Surrealism, along with the innovations of the unaffiliated writers.●Modernism takes the irrational philosophy and the theory of psycho-analysis as itstheoretical base. It is a reaction against realism. It rejects rationalism which is the theoretical base of realism; it excludes from its major concern the external, objective, material world, which is the only creative source of realism; by advocating a free experimentation on new forms and new techniques in literary creation, it casts away almost all the traditional elements in literature such as story, plot, character, chronological narration, etc., which are essential to realism. As a result, the works created by the modernist writers can often be labeled as anti-novel, anti-poetry or anti-drama.(See Chang Yaoxin, 380-381)2 The Stream of consciousness●The tradition of the stream of consciousness can be traced to various sources. Theearliest attempt to concentrate the subject matter of fiction on the innerconsciousness of the c haracter can be found in Laurence Sterne’s Tristram Shandy.● Stream of Consciousness is a literary technique which was pioneered by DorthyRichardson, Virginia Woolf, and James Joyce. Stream of consciousness is characterized by a flow of thoughts and images, which may not always appear to have a coherent structure or cohesion. The plot line may weave in and out of time and place, carrying the reader through the life span of a character or further alonga timeline to incorporate the lives (and thoughts) of characters from other timeperiods. It was represented by James Joyce (Ulysses), Virginia Woolf (To the Light house, The Waves) and William Faulkner (The Sound and Fury).●Writers, who create stream-of-consciousness works of literature, focus on theemotional and psychological processes that are taking place in the minds of one or more characters. Important character traits are revealed through an exploration of what is going on in the mind.3 James Joyce (1882-1941)3.1 Life and achievements● James Joyce is an Irish novelist, noted for his experimental use of language in suchworks as Ulysses(1922) and Finnegans Wake(1939). During his career Joyce suffered from rejections from publishers, suppression by censors, attacks by critics, and misunderstanding by readers. From 1902 Joyce led a nomadic life, which perhaps reflected in his interest in the character of Odysseus. Although he spent long times in Paris, Trieste, Rome, and Zürich, with only occasional brief visit to Ireland, his native country remained basic to all his writings.“But when the restraining influence of the school was at a distance I began to hunger again for wild sensations, for the escape which those chronicles of disorder alone seemed to offer me. The mimic warfare of the evening became at last as wearisome to me as the routine of school in the morning because I wanted real adventures to happen to myself. But real adventures, I reflected, do not happen to people who remain at home: they must be sought abroad.”(From Dubliners)●Joyce’s life’s work was to write only and always about his hometown Dublin. Hetreated it in such a way that in fact he was writing about all of human experience.His works include Dubliners, The Portrait of an Artist as a Young Man, Ulysses, and Finnegans Wake.3.2 Comments on James Joyce●He was serious with both his thematic and technical concerns.◆In Thematic terms, he never forgot to promote spiritual freedom of his nativecountry with his writings. Irish nationalism features prominently in his work.The theme of Irish discontentment and struggle for freedom always appear asa clear narrative thread. Furthermore, he loves the common people. He findsthe great value in love and passion. He loves life itself and believes that love makes life worth living and makes man invincible. Finally, there is his preoccupation with art and its mission in life. All of his major works related to art and artist and their bearing on life.◆With regarded to his formal features, Joyce is noted for his frankrepresentation of reality. He embraces realism against romanticism. H e is outspoken on important social issues, and insists on portraying all the aspects of man—the good as well as the evil side. For him, characterization matters more than does plot. He is also well renowned for his adroit use of the stream-of-consciousness technique and his contribution to its subsequent popularity as an effective stylistic medium. In addition, Joycean language has always been a topic of immense interest to people. It is poetic, accurate, forceful, connotative, rhythmic, musical, picturesque, aptly polyglottic, and humorous beyond description.●By pushing both thematic and formal boundaries infinitely further back, he hasbecome “the writes’ writer” in literary history. His influence over the writers of his own and later generation can never be overstated.3.3 Discussion of “Araby”●In “Araby,” the allure of new love and distant places mingles with the familiarityof everyday drudgery, with frustrating consequences.◆Mangan’s sister embodies this mingling, since she is part of the familiarsurroundings of the narrator’s street as well as the exotic promise of the bazaar.She is a “brown figure” who both reflects the brown façades of the buildings that line the street and evokes the skin color of romanticized images of Arabia that flood the narrator’s head.◆Like the bazaar that offers experiences that differ from everyday Dublin,Mangan’s sister intoxicates the narrator with new feelings of joy and elation.His love for her, however, must compete with the dullness of schoolwork, his uncle’s lateness, and the Dublin trains. Though he promises Mangan’s sister that he will go to Araby and purchase a gift for her, these mundane realities undermine his plans and ultimately thwart his desires. The narrator arrives at the bazaar only to encounter flowered teacups and English accents, not the freedom of the enchanting East. As the bazaar closes down, he realizes that Mangan’s sister will fail his expectations as well, and that his desire for her is actually only a vain wish for change.●The narrator’s change of heart concludes the story on a moment o f epiphany, butnot a positive one. Instead of reaffirming his love or realizing that he does not need gifts to express his feelings for Mangan’s sister, the narrator simply gives up.He seems to interpret his arrival at the bazaar as it fades into darkness as a sign that his relationship with Mangan’s sister will also remain just a wishful idea and that his infatuation was as misguided as his fantasies about the bazaar. What might have been a story of happy, youthful love becomes a tragic story of defeat. Much like the disturbing, unfulfilling adventure in “An Encounter,” the narrator’s failure at the bazaar suggests that fulfillment and contentedness remain foreign to Dubliners, even in the most unusual events of the city like an annual bazaar.● The tedi ous events that delay the narrator’s trip indicate that no room exists forlove in the daily lives of Dubliners, and the absence of love renders the characters in the story almost anonymous. Time does not adhere to the narrator’s visions ofhis relationship. The story presents this frustration as universal: the narrator is nameless; the girl is always “Mangan’s sister” as though she is any girl next door, and the story closes with the narrator imagining himself as a creature. In “Araby,”Joyce suggests that all people experience frustrated desire for love and new experiences.ⅦReflection Questions and Assignments1. Self study Virginia Woolf.2. What is the function of Big Ben in Mrs.Dalloway?3. How does the bazaar experience become an epiphany for the boy in “Araby”?ⅧMajor References1 Abrams, M. H. ed. The Norton Anthology of English Literature, (6th edition),Norton: 1993.2 Baugh, Albert C. A Literary History of England. 1967.3 Drabble, Margaret.The Oxford Companion to English Literature. OxfordUniversity Press and Foreign language and Research Press, 1998.4 陈嘉.《英国文学史》. 北京：商务印书馆，1986.5 陈嘉.《英国文学作品选读》. 北京：商务印书馆，1982.6 侯维瑞. 《英国文学通史》. 上海：上海外语教育出版社，1999.7 刘炳善. 《英国文学简史》. 郑州：河南人民出版社，1993.8刘洊波. 《英美文学史及作品选读》（英国部分），北京：高等教育出版社，2001.9 罗经国. 《新编英国文学选读》. 北京：北京大学出版社，1997.10 孙汉云. 《英国文学教程》. 南京：河海大学出版社，2005.11 王佩兰等. 《英国文学史及作品选读》. 长春：东北师范大学，2006.12 王松年. 《英国文学作品选读》. 上海：上海交通大学出版社，2002.13 吴伟仁. 《英国文学史及选读》（第二册）. 北京：外语教学与研究出版社，1990.14 杨岂深，孙铢.《英国文学选读》. 上海：上海译文出版社，1981.15 张伯香.《英国文学教程》. 武汉：武汉大学出版社，2005.16 张伯香.《英美文学选读》. 北京：外语教学与研究出版社，1998.17 张定铨. 《新编简明英国文学史》. 上海：上海外语教育出版社，2002.。
Lecture14
4.5e−p深度非弹性散射4.5.1从弹性散射到非弹性散射考虑e−和有结构p的弹性散射。
当入射电子的能量增加时，四动量转移的平方q2的平均值也随之增加。
在高q2情形下，τ=−q24M2≫1，弹性散射的Rosenbluth公式变为：(dσdΩ)elastic=α24E21sin4θ2E3E1(q22M2G2M sin2θ2)由于在高q2情形时，质子的磁形状因子以q−4迅速减小，因此：(dσdΩ)elastic∝q−6即弹性散射截面以q−6迅速减少，如图4.15中点划线所示。
这说明在高q2时发生弹性散射的可能性很小。
图4.15:弹性和非弹性散射的截面随q2的变化。
点划线为弹性散射的Rosenbluth公式预言，数据点为实验测量值，取自M.Breidenbach et al.,Phys.Rev.Lett23(1969)935。
图中W是非弹性散射时末态强子系统的不变质量，我们用M X表示。
p→图4.17所示是E1=4.879GeV射角θ=10◦处放置探测器，测量散射电子的能量E3，从而得到微分散射截面。
系统运动学完全由E3和θ决定。
例如，在给定θ和E3下，末态强子系统的不变质量是：M2X=10.06−2.03E3z Elastic Scattering¾Proton remains intact¾W=Mz Inelastic Scattering¾Produce “excitedstates”or proton.e.g.'(1232)¾W=M'z Deep Inelastic Scattering¾Proton breaks upresulting in amany particlefinal state¾DIS = large W图4.17:弹性和非弹性散射的截面随E3和W的变化。
2许多文献中末态强子系统的不变质量M用W表示，例如图4.15中所示。
机器学习讲义14
RBF Network: distance similarity-to-centers as feature transform
Radial Basis Function Network
RBF Network Hypothesis
Fun Time
Which of the following is not a radial basis function? 1 φ(x, µ) = exp(−γ x − µ 2) 2 φ(x, µ) = − xT x − 2xT µ + µT µ 3 φ(x, µ) = x = µ 4 φ(x, µ) = xT x + µT µ
Machine Learning Techniques (機器學習技法)
Radial Basis Function Network
Roadmap
1 Embedding Numerous Features: Kernel Models 2 Combining Predictive Features: Aggregation Models 3 Distilling Implicit Features: Extraction Models
Interpolation by Full RBF Network
full RBF Network for squared error regression:
N
h(x) = XOuXtpXu Xt
βmRBF(x, xm)
m=1
• just linear regression on RBF-transformed data
Radial Basis Function Network
RBF Network Learning
Lecture14-平摊分析
• • • • • • • • • 对不同的栈操作赋予以下的平摊代价： PUSH — 2 POP — 0 MULTIPOP — 0. 栈操作的实际代价为： PUSH — 1 POP — 1 MULTIPOP — min(k, s). 对于初始为空的栈，任何的栈操作序列，当一个元素压入栈时，我们需要支 付2元，其中一元用于支付该push操作的实际代价，另一元作为“存款”保 存在该元素上。因此在执行pop或multipop操作时，可以用元素上保存的“存 款”来支付操作的实际代价。因此，对任意的包含n次push、pop、multipop 操作的序列，总的平摊代价就是其总的实际代价的一个上界。在此例中，总 的平摊代价为O(2n)，故总的实际代价为O(n)。
12
势能方法工作过程
• 从直觉上看，若第i个操作的势差 Φ(Di) Φ(Di-1) 是正的，则平摊代价Ci尖表示对第i 个操作多收了费，同时数据结构的势也增 加了。若势差是负值，则平摊代价表示对 第i个操作的不足收费，通过减少势来支付 该操作的实际代价。
13
栈操作的“势能”分析
• 对于栈的三个操作PUSH, POP, and MULTIPOP. 定义势能函数Φ为栈中数据对 象的个数。对空栈 D0 有 Φ(D0) = 0。由于 栈中对象个数不会为负，因此栈 Di 具有非 负势能，即Φ(Di)≥ 0 =Φ(D0). • 因此以 Φ 表示的n个操作的平摊代价的总和 就是实际代价的一个上界。
15
二进制计数器增1
• 定义第 ith次 INCREMENT 操作后的“势能”为 bi , 即第 ith 次操作后 计数器中1的个数。 • 计算1次 INCREMENT 操作的平摊代价如下。设第 ith 次 INCREMENT 操作后复位了 ti 个二进制位。则实际代价最多为 ti + 1, （复位数加置位数）。若 bi = 0，则第 ith 次操作复位了l k 位。所以 bi-1 = ti = k。若 bi > 0, 则 bi = bi-1 - ti + 1。在则两种情况下都有， bi ≤ bi-1 - ti + 1, 势能差位： • Φ(Di) - Φ(Di-1)≤ (bi-1 - ti + 1) - bi-1 =1 - ti. • 因此一次INCREMENT操作的平摊代价为： • ci + Φ(Di) - Φ(Di-1) • ≤ (ti + 1) + (1 - ti) =2. • 若计数器从0开始，则有 Φ(D0) = 0. 因为对任意i,有 Φ(Di) ≥ 0 。所以 n 次INCREMENT 操作的总平摊代价是总实际代价的一个上界。因此 the worst-case cost of n INCREMENT operations is O(n).
Lecture14_Lattice_Structures
Digital Signal ProcessingLecture 14Lattice StructuresTesheng Hsiao,Associate ProfessorThe lattice structure is a modular structure consisting of cascaded stages.Digital filters implemented by the lattice structure can be transformed into direct from and vice versa.In this lecture,we are going to investigate the implementation of the lattice form and the conversion between it and the direct form.1Recursive Lattice StructureFig.(1)is a single lattice stage.It is a two-input,two-output system consisting of two multipliers,two adders,and one delayelement.Figure 1:A single lattice stageThe constant κn is referred to as reflection coefficient .The input and output relation can be expressed in the z-domainU n (z )=U n +1(z )−κn z −1V n (z )(1)V n +1(z )=κn U n (z )+z −1V n (z )(2)Rearrange the equation and we have[U n +1(z )V n +1(z )]=[1κn z −1κn z −1][U n (z )V n (z )](3)The lattice structure of an N th order LTI digital filter is a cascade of N stages in the way shown in Fig.(2).Given the lattice structure of Fig.(2),we are going to answer the following questions:(1)how the input signal x propagates to the output signal y ,(2)what is the system function implemented by the lattice structure,and (3)how a system function can be converted to a lattice structure.Figure2:Recursive lattice structure•Lattice FilteringThe output of the lattice structure can be calculated in a recursive way.Assume that the system is at initial rest;hence v n[−1]=0,n=0,1,···,N−1.From Fig.(2),for each time step k≥0,we haveinitial conditions:v n[−1]=0,n=0,1,···,N−1for k=0,1,2,···u N[k]=x[k]for n=N−1to0u n[k]=u n+1[k]−κn v n[k−1]v n+1[k]=κn u n[k]+v n[k−1]endv0[k]=u0[k]y[k]=N∑n=0λn v n[k]end•The system function of the lattice structureLetP n(z)=U n(z)U0(z),Q n(z)=V n(z)V0(z),n=0,1,2,···,NHence,Eq.(3)can be rewritten as[P n+1(z) Q n+1(z)]=[1κn z−1κn z−1][P n(z)Q n(z)],n=0,1,2,···,N−1(4) =[1κnκn1][P n(z)z−1Q n(z)],n=0,1,2,···,N−1(5)Note thatU0(z)=V0(z),P0(z)=Q0(z)=1,X(z)U0(z)=P N(z),Y(z)U0(z)=N∑n=0λn Q n(z)Therefore the system function H (z )isH (z )=Y (z )X (z )=∑N n =0λn Q n (z )P N (z )(6)If we expand Eq.(4),we obtain [P n (z )Q n (z )]=[1κn −1z −1κn −1z −1]···[1κ0z −1κ0z −1][11],n =0,1,···,N (7)It is clear from Eq.(7)that P n (z )and Q n (z )are polynomials of z −1of order n .From Eq.(6),P N (z )is the denominator of H (z )while ∑N n =0λn Q n (z )is the numerator ofH (z ).Note that the number of parameters in the lattice structure (κn ,n =0,···N −1and λn ,n =0,···,N )is the same as the number of coefficients of an N th order rational function.In summary,the system function H (z )can be determined by applying Eq.(4)recur-sively to find P n (z )and Q n (z ),n =1,2,···,N ,given Q 0(z )=P 0(z )=1.Then use Eq.(6)to determine H (z ).•Convert the direct form to the lattice structureLet P n (z )=p n 0+p n 1z −1+···+p n n z−n and Q n (z )=q n 0+q n 1z −1+···+q n n z −n ;From Eq.(7),we have [P 1(z )Q 1(z )]=[1κ0z −1κ0z −1][11]=[1+κ0z −1κ0+z −1][P 2(z )Q 2(z )]=[1κ1z −1κ1z −1][1+κ0z −1κ0+z −1]=[1+(κ0+κ0κ1)z −1+κ1z −2κ1+(κ1κ0+κ0)z −1+z −2]...=...Hence we conclude by induction that p n n =κn −1and q n n =1for n =0,1,2,···,N .Moreover we have the following lemma.Lemma 1Q n (z )=z −n P n (z −1),n =0,1,···,NProof:This lemma can be proved by induction.The n =0case is trivialFor n =1,P 1(z )=1+κ0z −1and Q 1(z )=κ0+z −1.Thus the equality holds.Suppose that the equality holds for n =k ,i.e.Q k (z )=z −k P k (z −1).Equivalently,z −k Q k (z −1)=P k (z )For n =k +1,from Eq.(4)we haveP k +1(z )=P k (z )+κk z −1Q k (z )Q k +1(z )=κk P k (z )+z −1Q k (z )Thereforez −(k +1)P k +1(z −1)=z −k −1P k (z −1)+κk z −k Q k (z −1)=z −1Q k (z )+κk P k (z )=Q k +1(z )Thus,by mathematical induction Q n (z )=z −n P n (z −1)for n =0,1,···,NQ.E.D.Assume that κn =1for all n .If we inverse Eq.(5),we have[P n (z )z −1Q n (z )]=11−κ2n[1−κn −κn 1][P n +1(z )Q n +1(z )],n =0,1,···,N −1Hence,P n(z )=P n +1(z )−κn Q n +1(z )1−κ2n ,n =N −1,N −2,···,0(8)Let H (z )=B (z )A (z )=∑N n =0b n z −n 1−∑N n =1a n z −n ,where A (z )andB (z )are polynomials of z −1.Since p n n =κn −1for all n ,thereflection coefficients κn ’s can be determined recur-sively by first setting P N (z )=A (z )and Q N (z )=z −N P N (z −1).Then κN −1=p N N is determined.Applying Eq.(8)and Lemma 1recursively to find P n (z ),κn ’s can be determined successively.To determine λn ,we observe that the coefficient of z −N in the numerator must beλN since B (z )=∑Nn =0λn Q n (z )and q N N =1.Therefore λN =b N .We can removeλN Q N (z )from B (z ),resulting in a (N −1)th order polynomial,and determine λN −1by taking advantage of the property q n n =1for all n .The whole process continuous until all λn ’s are determined.In summaryP N =A (z ),S N =B (z ),λN =b Nfor n =N −1to 0κn =p n +1n +1Q n +1(z )=z −(n +1)P n +1(z −1)P n (z )=P n +1(z )−κn Q n +1(z )1−κ2nS n (z )=S n +1(z )−λn +1Q n +1(z )λn =s n nend•Stability of the Lattice Structure Form Eq.(4),we haveP 1(z )=1+κ0z −111is stable if and only if |κ0|<1.Since the lattice structure is a cascade of N similar stages,the stability of the filter can be verified easily as follows.Lemma 2The lattice structure in Fig.(2)is stable if and only if |κn |<1for all n .2All-pole SystemsAn all-pole system has no nonzero zeros,i.e.the system function is H (z )=1A (z ).In thelattice structure in Fig.(2),if λ0=1and λn =0for n >0,thenH (z )=∑N n =0λn Q n (z )P N (z )=1P N (z )Hence the all-pole system has a simpler lattice structure shown in Fig.(3).Figure 3:The lattice Structure for an all-pole systemOne interesting feature of the lattice structure in Fig.(3)is that the system function from x to v N is an all-pass system.This can be seen as follows.H all (z )=V N (z )X (z )=Q N (z )P N (z )=z −N P N (z −1)P N (z )If z 0is a pole of H all (z ),then 1/z 0must be a zero of H all (z )and vice versa.Due to the symmetry of poles and zeros,H all (z )is indeed an all-pass system.3Nonrecursive lattice structureIf H (z )=B (z ),i.e an FIR filter,the lattice structure becomes nonrecursive.We will explore its properties in this section.We would like to maintain the symmetric structure in Fig.(2)or Fig.(3)because previous results (e.q.Lemma 1)can be applied directly by doing so.In other words,Eq.(1)and Eq.(2)must hold for each stage.If H (z )is FIR,then G (z )=H −1(z )is an all-pole system.If we implement the all-pole system G (z )in the lattice form of Fig.(3),we haveG (z )=1H (z )=1P N (z )=U 0(z )U N (z )By exchanging its input and output,we get the desired FIR system.Note that in this FIR lattice structure,signals flow from u 0to u N .Hence Eq.(3)should be used to compute the signal propagation from stage to stage.The corresponding lattice structure is shown in Fig.(4).Notice that the structure is nonrecursive.The system function implemented by the nonrecursive lattice structure can be constructed in the same way as the recursive lattice structure:P 0(z )=Q 0(z )=1for n =1to N [P n (z )Q n (z )]=[1κn −1z −1κn −1z −1][P n −1(z )Q n −1(z )]endH(z)=P N(z)Figure4:Nonrecursive Lattice StructureTo convert from a system function to the nonrecursive lattice structure,the algorithm is similar to that of the recursive version:P N=B(z)=1+b1z−1+···+b N z−Nfor n=N to1κn−1=p nnQ n(z)=z−n P n(z−1)P n−1(z)=P n(z)−κn−1Q n(z)1−κ2n−1endNote that from Lemma1,p n0=q nn=1for all n.Therefore the coefficient of the constantterm in H(z),i.e.b0,must be1.If b0=1,an intuitive approach is to divide B(z)by b0. However,if b N=b0,as in the case of the linear phasefilter,this will result inκN−1=1, and again we will run into trouble in computing the reflection coefficients.A preferable way is to implement the FIR system H′(z)=1+(B(z)−b0)in a lattice structure and subtract 1−b0from its output.The idea is shown in Fig.(5)Figure5:Nonrecursive lattice structure for b0=1If we apply Lemma2to the nonrecursive lattice structure,we observe that B(z)is a minimum phase system if and only if|κn|<1for all n.If B(z)=P N(z)is a minimum phase system,then the system function from x=u0to v N,i.e.Q N(z),becomes a maximum phase system according to Lemma1,i.e.all its zeros are outside the unit circle.Afinal remark of this lecture:According to Lemma2,each stage of the stable(or minimum phase)lattice structure is an attenuator,i.e.it does not amplify the signals.Thisproperty gives the lattice structure great computational stability and this is the primary reason that the lattice structure is implemented.However,the price for this property is the complex computation of the signalflow.。
lecture 14 description of place
Rely, chiefly, on precise nouns and vivid verbs, which provide the
In objective description, you record details without making any personal comment.
In subjective description, on the other hand, you are
free to interpret details for the reader/audience.
CONTENTS
Few descriptions outside of science writing, however,
are completely objective.
Methods of organization
Using the senses: describing how the place looks,
CONTENTS content
General remarks
Places may be described for their own sake, as in essays on visits to famous
scenic places, but also for the purpose of revealing the personality and
is like. In a nutshell, description is a word picture.
Lecture 14 禁忌与委婉语
Euphemism
2．W.C./ toilet --bathroom, washroom, restroom, cloakroom, powder room, retiring room, public convenience, the necessary house, house of amusement, old soldier’s home, chamber of commerce,
种族歧视
种族歧视是英语禁忌语的一个组成部分。 黑人 nigger, niggra, Negro, Blacks (Afro-Americans) 美籍意大利人 Dagos 美籍犹太人 Kikes 美籍波兰人 Polacks 美籍华人 Chinks 美籍日本人 Japs
Sexist language
Euphemism
1． Pregnant --in the family way, be expecting, in a bad shape, in a delicate condition, have a white swelling, wear the apron high, fruitful, hopeful, big with child, to have swallowed a water-melon seed
For example:
Human excretion Sex and certain parts of the body are taboo in both cultures.
In English
cursing words --- “four letter words” swear words --
清华大学高等量子力学-Lecture-14
2)自旋角动量 S 的性质： 无经典对应。 与空间运动无关，是粒子内部自由度。微观粒子内部自由度还有宇称、色、味等等。 相对论效应。自洽处理在相对论量子力学中，Dirac 方程。 对易关系： ⎡⎣Sˆi , Sˆ j ⎤⎦ = i εijk Sˆk , Sˆ × Sˆ = i Sˆ 。
2
本征值： Sx , S y , Sz = ± 2 ， S 2x = S 2 y = S 2z = 4 ，
2
2
则一般态：
ψ
=
⎛ ⎜ ⎜
Cos
α 2
−
e
i
E+t
⎞ ⎟ ⎟
。
⎜ ⎜⎝
α Sin
2
e− i E−t
⎟ ⎟⎠
自旋 Sˆi 的平均值： Si = ψ Sˆi ψ ， 代入矩阵形式，有
⎧ ⎪⎪
E−
=
−e B, 2µ
c1
=
0,
c2
=1
和⎨ ⎪ ⎪⎩
ϕ−
=
⎛ ⎜ ⎝
0 1
⎞ ⎟ ⎠
−
e
i
E−t
。
4
Sz
=
⎛ ⎜ ⎝
σx
=
⎛0
⎜ ⎝
1
1⎞
0
⎟ ⎠
,
σy
=
⎛0
⎜ ⎝
i
−i ⎞
0
⎟, ⎠
σz
=
⎛ ⎜ ⎝
1 0
0⎞ −1⎠⎟
。
3）自旋态
引入自旋后，粒子的 3 个自由度 r → 4 个自由度 r , Sz ， Hilbert 空间是坐标（连续）空间与
自旋空间（分离， D = 2 ）的直积。
高等有机化学--lecture 14
Here are some estimates of the strain based on "homodesmotic reactions" (agrees with experiments for carbon skeletons): theoretical strain energy (kcal/mol) cyclopropane (C3H6) 26.8 25.7 34.9 15.2
E. Kwan
Lecture 14: Small Ring Conformational Analysis
Chem 106
Geminal Hyperconjugations The effect of the D-C-A bond angle is illustrated by the relatively small hyperconjugations in propane vs. cyclopropane: Propane (CC to *CC = 0.52 kcal/mol)
D A D A
vicinal interaction
D
A
D
A
gemin normally talk about geminal interactions? It turns out that their strength depends a lot on the D-C-A angle. In a small ring, the angle is small, the overlap is good, and therefore the interaction is strong.
E. Kwan
Lecture 14: Small Ring Conformational Analysis Small Ring Conformational Analysis
(英文)细胞生物学lecture14-细胞的信号
Recommended Reading: MBOC 5e, pages 124 - 158
Page 1
What is signaling and why is it needed ?
• The plasma membraneforms a barrier between the inside of a cell and its extracellular environment. • Cells require a method to sense their extra-cellular (and intra-cellular !) environments. • Signaling allows a cell to receive information from its external environmentand to respond to the signal.
Overview of a generic signaling pathway.
Types of signals; Properties of a good signal. Conformational change – a common way to regulate protein activity. Non-covalent binding of atoms and molecules can regulate a protein’s activity. Covalent modification is a common method to regulate protein activity. Changes in molecular structure affect specificity of molecular interactions.
Lecture 14(12.09)
knowledge=behaviors teaching/testing
of structures
the
communicative language teaching approach: psycholinguistic-socialinguistic testing use of language knowledge=behaviors from cognition & affections teaching/testing of knowledge+ skills +competences
validity （效度） —— the degree to which a test measures what it is supposed to measure, or can be used successfully for the purposes for which it is intended. internal validity: content validity (controlled by syllabus), construct validity (controlled by the viewpoints of linguistics, pragmatics and language learning) external validity: a test for a certain purpose, communicative language test
3. major techniques and methods in applied linguistics (CA/EA/DA, speech therapy, corpus building, language planning, syllabus design, computer language , testing, etc.) application in difference fields concerning language uses, guided by theories (linguistics, psychology, sociology, pedagogy, etc)
Lecture 14 习语翻译
有许多习语在另一种语言里虽然没 有 意 象 相 同 的 对 等 用 法， 但 却 能 找到 意 象 不 同 而 意 义 相 同 或 相 近 的 习 语， 也就是说可以归化成译入语中固有 的习 语， 只 不 过 意 象 发 生 了 变 化， 而 原 文 的 意 义 基 本 未 变。
Translation of Idioms
习语翻译
英汉两种语言都富于生动形象的习语 ， 英 汉 习 语 因 文 化 的 巨 大 差 异 而呈 现 出 各 自 独 有 的 特 色， 但 由 于各民族的 文 化 与 思 维有着巨大的“共核”(common core) ，所以英汉习语有时也有惊人的相似 之 处。 相 似 的 习 语 固 然 可 以 让 译 者惊 喜， 而 独 具 文 化 特 色 的 习 语 往 往 又 使 译 者 望而生畏。
Make hay while the sun shines. 抓紧时机 strike while the iron is hot. 趁热打铁。 Man proposes, God disposes. 谋事在人，成事在天。 Great minds think alike. 英雄所见略同。 burn one„s boats. 破釜沉舟。
一、异化译法
异化译法(Foreignization)就是保留原 文的文化意象(异国情调），也就是所 谓的直译法，目的是要使译文保持原 文的语言风格，向译文读者介绍源语 文化，并丰富译入语及其文化。
1) armed to the teeth： 武装到牙齿 2) an eye for an eye, a tooth for a tooth： 以眼还眼，以牙还牙。 3) a wolf in sheep's clothing：披着羊皮的狼 4) Kill two birds with one stone：一石二鸟 5) All roads lead to Rome： 条条大路通罗马 6) born with a silver spoon in one„s mouth： 嘴里衔着银匙出生的 7) a storm in a teacup： 茶杯里的风暴；小事惹起的轩然大波。
lecture 14 语篇的翻译
• 这样,马乔里在17 岁那年爱恋上了一个学医 的穷学生以后,她姑母立即就筹谋让她到国 外去。在19 世纪末叶,这仍旧是拆散情人所 采用的典型办法。
• 1.3 替代指的是用较短的语言形式来替代上 下文中的某些词语，其目的是行文简练流 畅。 • You must remember ,my dear mother ,that I have never considered this matter as certain . I have had my doubts , I confess , but they are fainter than they were, and they may soon be entirely done away…
• Mr. Kirsch, having lost all his money by this time, followed his master out into the moonlight, where the illuminations were winking out. (from Vanity Fair) • 基希先生这时输得两手空空，便跟着主人走出屋 来；月光下，彩灯闪闪烁烁，渐渐熄灭。 • 译文用分号把人物活动和环境描写拆开来译，译 文顺畅明了。同时把“月光下”置于两个分句之 间，又加强了语篇的衔接，使人物活动和环境描 写有机地连结在一起。
• 直到十九世纪初叶，人们才看到不仅需要 而且必要创建新的大学。第一批新建筑之 一就是伦敦大学，它沿用了古代建筑的组 织结构，由一系列成分学院组成。 • 直到十九世纪初叶，人们才看到不仅需要 而且必要创建新的大学。而这第一批新成 立的大学当中就有伦敦大学，它沿用了牛 津大学和剑桥大学的组织建构，也是由一 系列的学嚏，要说那个小 孩，那就不是打喷嚏就是叫，不是叫就是 打喷嚏。那厨房里只有两个不打喷嚏的， 一个就是那个做饭的老妈子，一个是只大 猫。
Lecture_14
Global calibration methods: deterministic versus random search
Andrew Western, Tim Peterson, Department of Infrastructure Engineering, The University of Melbourne
Source: Duan, Sorooshian, Gupta (1992), Effective and efficient global optimization for conceptual rainfall-runoff models, WRR, 28(4) pp 1015-1031.
Andrew Western, Tim Peterson, Department of Infrastructure Engineering, The University of Melbourne
Reducing objective function value with increasing normalised parameter space distance denotes the second (local) optima .
Thick band at a higher objective function value also denotes the second (local) optima .
ENEN90031 Quantitative Environmental Modelling ENEN90031 Quantitative Environmental Modelling Model Evaluation
Between grid and random search methods
Lecture 14 Keywords
• (3) Designated (指定的) Choice
– Keywords usually come from the title and /or the abstract of a paper, because both the title and the abstract contain the most important words or phrases of the paper.
• (2) Limited Numbers
– The number of keywords for a paper is limited. Four to six keywords are the average. Usually, there should be at least two and at most ten.
• 3. How to Prepare Keywords
– – – – (1) Using Required Terms (2) Placing in the Right Location (3) Spacing the Keywords (4) Adopting Standard Abbreviations
Lecture14 —— Keywords
• Question
– What are the functions of keywords?
1. Functions of Keywords
• (1) Easy to be Retrieved
– As the term implies, keywords are the key and most important words and phrases used in a paper. Readers can find out the theme of the paper by looking at the keywords.
Lecture14_differential pair
Lecture 14
Richard Li, 2009
4
I o = I DC + io I DC = ao + vio2 [(1/2)a2+ (3/8) a4vio2+…] , io= [a1vio+ (3/4) a3vio3 +… ]cosωt + + [(1/2) a2vio2 + (4/8) a4vio4 +… ]cos2ωt + + [(1/4) a3vio3 +… ]cos3ωt + + [(1/8) a4vio4 +… ] cos4ωt +… DC offset = vio2 [(1/2)a2+ (3/8) a4vio2+…]
4. CMRR (Common Mode Rejection Ratio)
o Expression of CMRR o CMRR in a Single-ended StageCMRR in a Pseudo Differential Pair o Enhancement of CMRR
Appendixes
A bipolar differential pair
A MOSFET differential pair
Lecture 14
Figure 2
Richard Li, 2009pairs. Typical differential
2
o Superficial Difference Between the Single-ended stage and Differential Pair
Lecture 14 :
Differential pair
Lecture14_Multi-omic_data_integration
Background - 12
ChIP-Seq have much better resolution and lower background noise level than
ChIP-on-chip;
Park, Nature Review Genetics 2009
Background - 11
Start from the computational pipeline for analyzing ChIP-Seq data sets?
• Now it’s widely used to probe the genome-wide binding patterns of certain TF or histone mark in given cell type.
regulation.
Outline
1. Background of gene transcriptional regulation.
Omic data sets related with gene transcriptional regulation and associated computational tools.
Sharp (e.g. TF binding) Mixture (e.g. Polymerase binding) Broad (e.g. histone modification)
Park, Nature Review Genetics 2009
Background - 16
Basic ideas for peak detection:
Park, Nature Review Genetics 2009
Lecture 14 Intensity (amplitude) modulation response
Small-signal IM response
2 PN R IM ( ) H ( ) 2 IM (0) R 2 j PN 2 R 2 R 2 j
8
The normalized IM response is described by the modulation transfer function
1 dN i d dI dN vg gdN p vg N p dg N dt qV dN p 1 d v g g dt p where 1 dN p vg N p dg dN ' N
' Rsp
Np
N
vg aN p , NP
' Rsp 1 vg a p N p p N p ' Rsp
N
'
vg aN p , PP
Np
vg a p N p
Well above threshold,
NN
1
N
vg aN p , NP
2
modulation or noises, the dynamic changes in the carrier and photon densities away from their steady-state values are small. Take differentials of the rate equations:
6
dI (t ) I1e jt dN (t ) N1e jt dN p (t ) N p1e jt
The differential rate equation becomes
Lecture 14
Poem Appreciation: She Walks in Beauty
Introduction
Written in 1814, when Byron was twenty-six years old, and published in Hebrew Melodies in 1815, the poem of praise "She Walks in Beauty" was inspired by the poet's first sight of his young cousin Anne Wilmot. According to literary historians, Byron's cousin wore a black gown that was brightened with spangles.
He had been worldwide known when only in his mid-twenties, but he was said to have a passionate love with his half-sister Augusta Leigh, which brought him notorious social reputation. After that, he left for Italy and lived with his friend Shelly, another very famous English poet.
Lecture_14
- Triad as harmonia perfetta (perfect harmony)
- Describes how Willaert handles dissonance.
- Beginning in 1550, worked in Roman churches.
- Prolific: 104 masses.
Mille regretz de vous habandonner Et d’eslonger vostre fache amoureuse? Jay si grand dueil et paine douloureuse,
Qu’on me verra brief mes jours definer.
- Benefited from the publishing industry of Venice.
- Sticks closely to Josquin’s style.
- But full triads at cadences and seamless texture.
- Gioseffo Zarlino publishes Le Institutioni harmoniche (1558)
?
Rondeau A B a a b A B
Virelai
Ballade
B a a b ro Aaron publishes Thoscanello de la Musica in 1523.
- Describes two aspects of Josquin’s style: Imitation Homorhythmic declamation
Head motive—Initial passage or motive of a piece or movement, used especially for a motive or phrase that appears at the beginning of each movement of a cantus firmus mass.
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KMEM 1168 Engineering MechanicsLecture 14: Kinematic of Rigid Bodies 14.1 Introduction14.2 Translation14.3 Rotation About a Fixed Axis: Velocity14.4 Rotation About a Fixed Axis: Acceleration14.5 Rotation About a Fixed Axis: Representative Slab14.6 Equations Defining the Rotation of a Rigid Body About a Fixed Axis14.7 Sample Problem 15.514.8 General Plane Motion14.9 Absolute and Relative Velocity in Plane Motion14.10 Sample Problem 15.314.1 Introduction•Kinematics of rigid bodies: study of relations between time and the positions, velocities, and accelerations of the particles forming a rigid body.•Classification of rigid body motions:•translation:•rectilinear translation•curvilinear translation•rotation about a fixed axis•general plane motion•motion about a fixed point•general motion14.2 Translation•Consider rigid body in translation:- direction of any straight line inside the body is constant, - all particles forming the body move in parallel lines.• For any two particles in the body,A B A B r r r r r r += •Differentiating with respect to time,AB AA B A B v v r r r r rr &r &r &r &r ==+= All particles have the same velocity.•Differentiating with respect to time again,AB AA B A B a a r r r r rr &&r &&r &&r &&r ==+= All particles have the same acceleration.14.3 Rotation about a Fixed Axis: Velocity• Consider rotation of rigid body about a fixed axis AA’•Velocity vector dt r d v rr =of the particle P is tangent to the path with magnitude dt ds v =,()()()φθθφθφθsin sin lim sin 0&r t r dt ds v r BP s t =∆∆==∆=∆=∆→∆•The same result is obtained fromlocity angular ve k k rdt r d v ===×==r &r rr r rr θωωω14.4 Rotation about a fixed axis: acceleration•Differentiating to determine the acceleration,()v r dt d dt r d r dt d r dt d dt v d a r r r rr r r r r v rr ×+×=×+×=×==ωωωωω •k k k celerationangular ac dtd r&&r &r rr θωααω===== l tangentia•Acceleration of P is combination of two vectors,componenton accelerati radial component on accelerati l tangentia =××=×××+×=r r r r a rr r rr r r r r r r ωωαωωα14.5 Rotation About a Fixed Axis: Representative Slab• Consider the motion of a representative slab in a plane perpendicular to the axis of rotation. •Velocity of any point P of the slab,ωωωr v rk r v =×=×=r r r r r form scalar form vector −−•Acceleration of any point P of the slab,r r k r r a rr r r r r r r r 2ωαωωα−×=××+×= form scalar formvector −−•Resolving the acceleration into tangential and normal components,22ωωααr a r a r a r k a n n t t =−==×=r r r r r14.6 Equations Defining the Rotation of a Rigid Body about a Fixed Axis• Motion of a rigid body rotating around a fixed axis is often specified by the type of angular acceleration. •Recall,θωωθωαωθθωd d dt d dt d d dt dt d =====22or•Uniform Rotation, α = 0:t ωθθ+=0•Uniformly Accelerated Rotation, α= constant:()02022210002θθαωωαωθθαωω−+=++=+=t t t14.7 Sample Problem 15.5When the power to an electric motor is turned on, the motor reaches its rated speed of 2400 rpm in 4 s, and when the power is turned off, the motor coasts to rest in 40 s. Assuming uniformly accelerated motion, determine the number of revolutions that the motor executes (a ) in reaching its rated speed, (b ) in coasting to rest.Solution:a) To reach the rated speed:b) To come to rest14.8 General Plane Motion•General plane motion is neither a pure translation nor a pure rotation.•General plane motion can be considered as the sum of a translation and rotation.•Displacement of particles A and B to A2and B2 can be divided into two parts:B′•translation to A2and1B′about A2 to B2•rotation of114.9 Absolute and Relative Velocity in Plane Motion•Any plane motion can be replaced by a translation of an arbitrary reference point A and a simultaneous rotation about A .A B A B v v v r r r += ωωr v r k v A B A B A B =×=r r rA B A B r k v v r r rr ×+=ω• Assuming that the velocity v A of end A is known, and we wish to determine the velocity v B of end B and the angular velocity w in terms of v A , l , and θ.•The direction of v B and v B/A are known. Complete the velocity diagram.θθtan tan A B ABv v v v == θωθωcos cos l v l vv v A A A B A ===• Selecting point B as the reference point and solving for the velocity v A of end A and the angular velocity w leads to an equivalent velocity triangle.• v A/B has the same magnitude but opposite sense of v B/A . The sense of the relative velocity is dependent on the choice of reference point.•Angular velocity w of the rod in its rotation about B is the same as its rotation about A. Angularvelocity is not dependent on the choice of reference point.14.10 Sample Problem 15.63In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE.You can solve the problem by using vector or scalar method. The common equation to solve is: ݒൌݒݒ/Vector method:Bar AB - rotation about A:ݒൌ߱ൈݎ/ൌെ4݇ൈെ0.25݆ൌെ1.00݅Bar ED – rotation about E:ݒൌ߱ா݇ൈݎ/ாൌ߱ா݇ൈሺെ0.075݅െ0.15݆ሻൌ0.15߱ா݅െ0.075߱ா݆Bar BD – Translation with B + Rotation about B:ݒ/ൌ߱݇ൈݎ/ൌ߱݇ൈ0.2݅ൌ0.2݆߱Now,ݒൌݒݒ/0.15߱ா݅െ0.075߱ா݆ൌെ1.00݅0.2݆߱By equating coefficients:i => 0.15߱ாൌെ1.00߱ாൌെ6.67ݎܽ݀/ݏ࣓ாൌ6.67ݎܽ݀/ݏclockwisei => െ0.075߱ாൌ0.2߱߱ൌ2.50ݎܽ݀/ݏ࣓ൌ2.50ݎܽ݀/ݏanticlockwise。