LECTURE09
lecture的衍生词
lecture的衍生词
一、lecturer（名词）
1. 解析。
- 由“lecture”（讲座、讲课）加上后缀 -er构成。
在英语中，后缀 -er通常可以加在动词后面，表示“做某事的人”。
在这里，“lecturer”的意思就是“讲课者；演讲者；（大学或学院中的）讲师”。
例如：The lecturer gave an interesting talk on modern English literature.（这位讲师就现代英国文学作了一个有趣的演讲。
）
二、lectureship（名词）
1. 解析。
- 由“lecture”加上 -ship后缀构成。
-ship这个后缀可以表示“职位；资格；身份”等意义。
“lectureship”表示“讲师的职位（或资格）”。
例如：He applied for the lectureship in the university.（他申请了这所大学的讲师职位。
）
三、lecturette（名词）
1. 解析。
- 这是在“lecture”的基础上加上 -ette后缀。
-ette后缀有“小”的意思，在这里“lecturette”表示“简短的演讲；小型讲座”。
例如：The conference included several lecturettes on different topics.（会议包括几个关于不同主题的小型讲座。
）。
Lecture 09 第一、第二、第三宇宙速度
青岛科技大学
大学物理讲义
求卫星飞出太阳系的第三宇宙速度 解 第三宇宙速度 v3 ，是抛体脱离太阳引力所需的 最小发射速度 .
h
v
设
地球质量 mE , 抛体质量 m , 地球半径 RE , 太阳质量 mS , 抛体与太阳相距 RS .
青岛科技大学 大学物理讲义
取抛体和地球为一系统，抛体首先要脱离 地球引力的束缚, 其相对于地球的速率为 v' . 取地球为参考系,由机械能守恒得
mmE m G 2 RE h ( RE h)
解得 v1
v
2
h
``````
v
2GmE GmE RE RE h
大学物理讲义
青岛科技大学
v1
2GmE GmE RE RE h
h
``````
v
RE GmE ) g 2 v1 gRE (2 RE h RE
1 mE m 2 E mv2 (G ) 2 RE Ek Ep 0
青岛科技大学 大学物理讲义
v
h
``````
1 2 mE m E mv2 (G )0 2 RE
v2
计算得
v
h
``````
2GmE RE
2 gRE
E 0
第二宇宙速度
v2 11.2km/s
2
第三宇宙速度
青岛科技大学 大学物理讲义
抛体的轨迹与能量的关系
E 0 v
h
E 0 椭 圆(包括圆) v1 7.9km/s
E 0
E0
E 0 抛物线 v2 11.2km/s
E 0 双曲线 v3 16.4km s
厦门大学软件工程专业课程表
Assembly 汇编语 TX15 Language 言 Programming
本课程是计算机类专业的基础 课程。介绍 8086/8088 CPU 的主要 学 指令集，学会从汇编语言的角度考 科 计算机导 虑问题的解决方法，掌握微机系统 通 论、计算机 的编程结构。强调对底层编程能力 修 硬件基础 和意识的培养，从具体应用出发， 课 对 CPU 所提供的指令集的典型特 程 点、寻址方式等内容进行理论上的
件体系结构的整体分析和处理能 力，为开发大型软件打下基础 在熟悉 Java 程序开发的基础 上，学生应该进入更高一个层次， 使用 Java2 平台企业版（J2EE）来 创建企业级应用程序。本课程应以 丰富的实例提供当前这一技术的最 新动向， 对于每一位在 J2EE 平台上 开发和部署应用程序的开发人员， 提供入门以及进一步提高的基础。 本课程将使用许多构思精巧的实例 程序（包括源码）来描述 J2EE 关键 面 向 对 象 的概念。此外，也有助于学生快速 程 序 设 计 掌握 J2EE 平台的多项技术，包括: (Java) Enterprise JavaBean, Java Servlet, JavaServer Pages, Java Message Service (JMS，Java 消息 服务), Java Naming and Directory Interface (JNDI， Java 名字与目录 接 口 ), XML, J2EE Connector Architecture（J2EE 连接器架构） , JavaMail, JDBC。 为学生以后创建 自己的大型企业级应用系统时，通 过本课程获得进一步的技术指导、 知识和经验。 本课程也称为“用户界面设计 （User’s interface design）， ” 主要介绍软件界面的基本概念、要 其 它 专 业 求、基本特性和分析方法；介绍人 课程 机界面的快速原型及其开发工具、 可视化界面、 人机接口技术等内容， 同时使学生了解当前最新的界面设 计技术 本课程的任务是培养学生计算 机软件开发与应用的综合能力。介 绍软件的基本概念和软件工程的目 标，通过对传统的面向过程的软件 开发方法和面向对象的软件开发方 法的介绍，使学生掌握开发高质量 软件的方法；通过对软件开发过程 和过程管理技术的学习，使学生了 解如何进行软件度量和管理，怎样 进行质量保证活动，从而能够有效
量子力学09
−α T α 1− α T = λ ξ α ⊗ ( ρ1 1 ) + ( 1 − λ ) ξ 2 ⊗ ( ρ2 ) .
The identity vec(1 )∗ ( X ⊗ Y T ) vec(1 ) = Tr( XY ) then gives the desired result. Now, toward the proof of Theorem 9.4, we require the following lemma. Lemma 9.5. Let P1 , P2 , Q1 , Q2 , R1 , R2 ∈ Pd (X ) be positive definite operators that satisfy these conditions: 1. [ P1 , P2 ] = [Q1 , Q2 ] = [ R1 , R2 ] = 0,
′ S(ρ ξ ) = − log(e) f ρ ,ξ (0) = − log( e) lim
α ↓0
Tr ξ α ρ1−α − 1 , α
where the second equality follows by substituting f ρ,ξ (0) = 1 into the definition of the derivative. Now consider the following lemma that concerns the relationship among the functions f ρ,ξ for various choices of ρ and ξ . Lemma 9.3. Let ξ 1 , ξ 2 , ρ1 , ρ2 ∈ Pd (X ) be positive definite operators. Then for every choice of α, λ ∈ [0, 1] we have
Lecture 09
Development ofA Trans-national Approach Course: Eurocode 3Module 4 : Member designLecture 9 : Local Buckling and Section ClassificationSummary:Pre-requisites:Notes for Tutors:Objectives:References:Contents:1. IntroductionStructural sections, be they rolled or welded, may be considered as an assembly of individual plate elements, some of which are internal (e.g. the webs of open beams or the flanges of boxes) and others are outstand (e.g. the flanges of open sections and the legs of angles) - see figure 1. As the plate elements in structural sections are relatively thin compared with their width, when loaded in compression (as a result of axial loads applied to the whole section and/or from bending) they may buckle locally. The disposition of any plate element within the cross section to buckle may limit the axial load carrying capacity, or the bending resistance of the section, by preventing the attainment of yield. Avoidance of premature failure arising from the effects of local buckling may be achieved by limiting the width-to-thickness ratio for individual elements within the cross section. This is the basis of the section classification approach.Flange(a) Rolled I-section (b) Hollow sectionFlange (c) Welded box sectionFigure 1 Internal or outstand elements2. ClassificationEC3 defines four classes of cross section. The class into which a particular cross section falls depends upon the slenderness of each element (defined by a width-to-thickness ratio) and the compressive stress distribution i.e. uniform or linear. The classes are defined in terms of performance requirements for resistance of bending moments:5.3.2 (1)Class 1cross-sections are those which can form a plastic hinge with the required rotationalcapacity for plastic analysis.Class 2cross-sections are those which, although able to develop a plastic moment, havelimited rotational capacity and are therefore unsuitable for structures designed by plastic analysis.Class 3 cross-sections are those in which the calculated stress in the extreme compression fibrecan reach yield but local buckling prevents the development of the plastic moment resistance.Class 4cross-sections are those in which local buckling limits the moment resistance (or compression resistance for axially loaded members). Explicit allowance for the effects of localbuckling is necessary.Table 1 summarises the classes in terms of behaviour, moment capacity and rotational capacity.Table 1 Cross-section classifications in terms of moment resistance androtation capacity3. Behaviour of plate elements in compressionA thin flat rectangular plate subjected to compressive forces along its short edges has an elastic critical buckling stress (σcr ) given by:()222112⎪⎭⎫ ⎝⎛-=b t E k crνπσσ(1)Wherek σ is the plate buckling parameter which accounts for edge support conditions, stress distribution and aspect ratio of the plate - see figure 2a. ν= Poisson’s coefficient 3.2.5 (1) E = Young’s modulus3.2.5 (1) The elastic critical buckling stress (σcr ) is thus inversely proportional to (b/t)2and analogous to the slenderness ratio (L/i) for column buckling.Open structural sections comprise a number of plates which are free along one longitudinal edge (see figure 2b) and tend to be very long compared with their width. The buckled shape for such a plate is illustrated in figure 2c. The relationship between aspect ratio and buckling parameter for a long thin outstand element of this type is shown in figure 2d, from which it is clear that the buckling parameter tends towards a limiting value of 0.425 as the plate aspect ratio increases.For a section to be classified as class 3 or better the elastic critical buckling stress (σcr ) must exceed the yield stress f y . From equation (1) (substituting ν = 0.3 and rearranging) this will be so if3.2.5 (1)()0,5y E/f k 0,92<b/t σ(2)This expression is general as the effect of stress gradient, boundary conditions and aspect ratio are all encompassed within the buckling parameter k σ. Table 2 gives values for high aspectFigure 2 Behaviour of plate elements in compressionTrahair andBradford5.2.1.4 (7)ESDEP Lecture 7.25.3.2 (3)ct fd t wElement Class 1 Class 2 Class 3 Flange c / t f = 10 εc / t f = 11 εc / t f = 15 ε Web subject to bendingd / t w = 72 εd / t w = 83 εd / t w = 124 εWeb subject to compression d / t w = 33 εd / t w = 38 εd / t w = 42 εTable 3 Maximum slenderness ratios for the elements of a rolledsection in compression and bendingTables 4-7 are extracts from EC3 giving the limiting proportions for compression elements of class 1 to 3. When any of the compression elements within a section fail to satisfy the limit for class 3 the whole section is classified as class 4 (commonly referred to as slender), and local buckling should be taken into account in the design using an effective cross section.5.3.2 (8) 4. Effective width approach to design of Class 4 sectionsCross-sections with class 4 elements may be replaced by an effective cross-section, taken as the gross section minus holes where the buckles may occur, and then designed in a similar manner to class 3 sections using elastic cross-sectional resistance limited by yielding in the extreme fibres. Effective widths of compression elements may be calculated by use of a reduction factor ρ which is dependent on the normalised plate slenderness λp (which is in turn dependent on the stress distribution and element boundaries through application of the buckling parameter k σ) as follows:5.3.5 (3) ()()ρλ=-⎛⎝ ⎫⎭⎪⎪p p0222,(6)The reduction factor ρmay then be applied to the outstand or internal element as shown in Tables 8 and 9. Figure 4 shows examples of effective cross-sections for members in compression or bending. Notice that the centroidal axis of the effective cross-section may shift relative to that for the gross cross-section. For a member in bending this will be taken into account when calculating the section properties of the effective section. For a member subject to an axial force, the shift of the centroidal axis will give rise to a moment which should be accounted for in member design.Summative test: Partially derive EC3 Table 5.3.1 Class 2 plate slenderness ratios for rolled sections in compression and bending ∙ Figure 3 uses λp < 0.6 as a normalised plate slenderness for class 2 rolled sections∙ Substituting the appropriate values of kσinto equation (5), use λp to determine limiting b/t ratios for a flange∙ Revise in terms of d/t w for a web in compression∙ Classify the section in exercise xxxx (this refers to one of the overall design exercises)6. Concluding summary∙Structural sections may be considered as an assembly of individual plate elements.∙Plate elements may be internal (e.g. the webs of open beams or the flanges of boxes) and others are outstand (e.g. the flanges of open sections and the legs of angles).∙When loaded in compression these plates may buckle locally.∙Local buckling within the cross-section may limit the load carrying capacity of the section by preventing the attainment of yield strength.∙Premature failure (arising from the effects of local buckling) may be avoided by limiting the width to thickness ratio - or slenderness - of individual elements within the cross section.∙This is the basis of the section classification approach.∙EC3 defines four classes of cross-section. The class into which a particular cross-section falls depends upon the slenderness of each element and the compressive stress distribution.*For a cross section in compression with no bending the classification 1,2,3 are irrelevant and hence the limit is the same in each case.Table 5 Maximum width-to-thickness ratios for compression elementsTable 6 Maximum width-to-thickness ratios for compression elementsTable 8 Effective widths of outstand compression elementsTable 9 Effective widths of compression elementsFigure 4 Effective cross-sections for class 4 in compression and bending。
Lecture-09-Constraint_Satisfaction
Instructor:
TA:
Sofus A. Macskassy, macskass@
Harris Chiu (chichiu@), Wed 2:45-4:45pm, PHE 328 Penny Pan (beipan@), Fri 10am-noon, PHE 328
CS561 - Lecture 9 - Macskassy - Fall 2010
7
Constraint graph
Binary CSP: each constraint relates at most two variables Constraint graph: nodes are variables, arcs show constraints
Simple example of a formal representation language
Allows useful general-purpose algorithms with more power than standard search algorithms
3
CS561 - Lecture 9 - Macskassy - Fall 2010
General-purpose CSP algorithms use the graph structure to speed up search. E.g., Tasmania is an independent subproblem!
8
CS561 - Lecture 9 - Macskassy - Fall 2010
Continuous variables
lecture_09
Binomial equations
We introduce in this section a particular kind of polynomial equations, that have nice computational properties. A binomial system of polynomial equations is one where each equation has only two terms. We also assume that the system has only a finite number of solutions, i.e., the solution set is a finite set of points in Cn . We are interested in determining the exact number of solutions, and in efficient computational procedures for solving the system. Let’s start with an example. Consider the binomial system given by 8x2 y 3 − 1 = 0 2x3 y 2 − yx = 0. (1)
2
Newton polytopes
Many of the polynomial systems that appear in practice are far from being “generic,” but rather present a number of structural features that, when properly exploited, allow for much more efficient computational techniques. This is quite similar to the situation in numerical linear algebra, where there is a big difference in performance between algorithms that take into account the sparsity structure of a matrix and those that do not. For matrices, the standard notion of sparsity is relatively straightforward, and relates mostly to the number of nonzero coefficients. In computational algebra, however, there exists a much more refined notion of sparsity that refers not only to the number of zero coefficients of a polynomial, but also to the underlying combinatorial structure. This notion of sparsity for multivariate polynomials is usually presented in terms of the Newton polytope of a polynomial, defined below. � Definition 2. Consider a multivariate polynomial p(x1 , . . . , xn ) = α cα xα . The Newton polytope of p, denoted by New(f ), is defined as the convex hull of the set of exponents α, considered as vectors in Rn . Thus, the Newton polytope of a polynomial always has integer extreme points, given by a subset of the exponents of the polynomial. Example 3. Consider the polynomial p(x, y ) = 5 − xy − x2 y 2 + 3y 2 + x4 . Its Newton polytope New(f ), displayed in Figure 1, is the convex hull of the points (0, 0), (1, 1), (2, 2), (0, 2), (4, 0). Example 4. Consider the polynomial p(x, y ) = 1 − x2 + xy + 4y 4 . Its Newton polytope New (p) is the triangle in R2 with vertices {(0, 0), (2, 0), (0, 4)}. Newton polytopes are an essential tool when considering polynomial arithmetic because of the fol lowing fundamental identity: New(g · h) = New(g ) + New(h), where + denotes the Minkowski addition of polytopes. 92
lecture09
§1集合的概念和表示法
3、幂集和索引集 幂集和索引集 （1）幂集(power set)： 定义》 《定义》 以集合A的所有子集作为元素构成的集合称 为A的幂集 幂集,记作ρ(A)或2A ，即： 幂集 ρ(A)={x | x ⊆A} ∅ 例：若A1=∅，则ρ(A1)= {∅} 若A2={a}，则ρ(A2)= {∅，{a}} ∅ ∅ ， ， ， 若A3={1，2}，则ρ(A3)= {∅，{1}，{2}，{1，2}}
§1集合的概念和表示法
有限个元素集合： S3={1,2,3,4,5}= {x | x ∈I ∧ (1 ≤ x ≤ 5) } S4={F,T}= {x | x=T ∨ x=F} S5={1,4}= { x | (x²-5x+4=0) } 同一集合可以用多种不同的形式表示。 (c)集合也可作为某一集合的元素。 例：S={a,{1,2},p,{q}}
§1集合的概念和表示法
(d)文氏图 Venn Diagram) 文氏图( 文氏图 用一个大的矩形表示全集，在矩形内画一些圆或 其它的几何图形，来表示集合，有时也用一些点 来表示集合中的特定元素。 例如：集合V={a,e,i,o,u} ，用文氏图表示如下:
a V u
E
§1集合的概念和表示法
(3)三个特殊集合：空集、全集、集合族 空集、全集、 空集 定义》 《定义》如果一个集合包含了所要讨论的每一个集合， 全集(universal set)，用E表示，即： 则称该集合为全集 全集 E={x | p(x) ∨ ¬ p(x)} p(x)为任何谓词公式 空集或称零集 《定义》不拥有任何元素的集合称为空集 定义》 空集 (empty set)，用∅表示，即： ∅ ∅={x | p(x) ∧ ¬ p(x) }={ } ∅ ≠ {∅} 前者是空集，后者是以∅作为元素的集合。
Lecture_09
P13
四、搭接技术
2 方法
直接搭接：将欲连接的两个金属导体直接接触： (1) 直接搭接：将欲连接的两个金属导体直接接触： 间接搭接：利用各种搭接物使两者接触。 (2) 间接搭接：利用各种搭接物使两者接触。 直接搭接的性能优于间接搭接，但是在某些情况下则要求使用间接搭接。 直接搭接的性能优于间接搭接，但是在某些情况下则要求使用间接搭接。如果 要求设备可移动，或者能抗机械冲击，则应采用间接搭接。 要求设备可移动，或者能抗机械冲击，则应采用间接搭接。 理想的搭接方式是焊接，特别是熔焊，能保持最佳的导电性。 理想的搭接方式是焊接，特别是熔焊，能保持最佳的导电性。用螺钉或铆钉等 实现搭接能够保证连接处的可靠连接，其它部分可能存在缝隙或由于氧化造成不导 实现搭接能够保证连接处的可靠连接， 对于非永久性搭接，可以采用电磁密封衬垫进行搭接。 电。对于非永久性搭接，可以采用电磁密封衬垫进行搭接。当考虑机械性能要求 (如减振)而不能直接搭接时，需要用搭接条进行搭接。 如减振)而不能直接搭接时，需要用搭接条进行搭接。
P15
四、搭接技术
4 良好搭接的原则
(1)金属表面之间紧密接触。接触区应光滑、干净并去除非导电物质。非焊接时，两搭接 (1)金属表面之间紧密接触。接触区应光滑、干净并去除非导电物质。非焊接时， 金属表面之间紧密接触 体应加以紧固，搭接应能耐受机械扭曲、冲击和振动。软金属比硬金属接触更好。 体应加以紧固，搭接应能耐受机械扭曲、冲击和振动。软金属比硬金属接触更好。 (2)尽量采用同类金属进行搭接，如必须采用两种不同金属时，应考虑腐蚀问题。 (2)尽量采用同类金属进行搭接，如必须采用两种不同金属时，应考虑腐蚀问题。 尽量采用同类金属进行搭接 (3) 在采用焊料时应考虑焊料作为搭接材料的影响。 在采用焊料时应考虑焊料作为搭接材料的影响。 (4) (5) 在搭接前应使搭接面干燥，搭接后要防潮，必要时加保护层。 在搭接前应使搭接面干燥， 搭接后要防潮，必要时加保护层。 采用搭接带时应尽量短，以保证低的阻抗，不要使搭接带在电化学序列中低于被 采用搭接带时应尽量短，以保证低的阻抗，
Lecture_09_CP3_
Chapter
3
Ratio Control System
3.1 Introduction 3.2 Architectures 3.3 Computation of instrumental ratio 3.4 Realization 3.5 Implementation and tuning 3.6 Other issues Lab work：Ratio control system
2
F F K′ = • F F
3.3.2 Nonlinear case
Summary： (1) Feed ratio K and instrumental ratio K’ are two explicitly separate factors. (2) Instrumental ratio K’ relies on both K and measurement ranges of transmitters. (3) The measurement linearity / nonlinearity should be concerned.
F = k Δp
DDZ-Ⅲ instrumentation QDZ instrumentation
F I = × 16 + 4 2 F max
Instrumental ratio
2 2 2 1 2 1 max 2 2 max
2
F2 p = 2 × 80 + 20 F max
F1 max 2 =K ( ) F2 max
IF
F = × 16 + 4 Fmax
3.3.1 Linear case
F 2 ×16+4−4 F Fmax Fmax I2 −4 F max 2 K′ = = = 2 • 1 =K 1 I1 −4 F F F max F max 1 2 ×16+4−4 1 2 Fmax 1
职场英语写作--Chapter 9 道歉信与祝贺信
5
Tips for an apology letter
6
职场英语写作--Lecture 9: Apologies and Congratulations
Apology
Letters of Apology 1. Guidelines：
(1) 言辞恳切，诚心诚意，指出自己并非故意伤害或开罪对 方，切忌对自己的过失作过多辩解。
Apology
What are included in APOLOGETIC letters? The apologies
（The attitude counts.）
The mistake you made
（Why you write this letter of apology?）
The solution or remedial measures taken (The promises)
As per your request, I have reviewed the White & Sons Hardware account. According to our
records, the correct cost of three hundred Slam Bam hammers is $2,975.00.
（For a late payment）
As for a whole, your apology should be brief and to the point. Your reputation as a trustworthy customer and credit risk may be in question, so rebuild your reputation by consistently making
简明法语教程Lecture翻译
L’hôtelparisi en巴黎的旅馆L’hôtelparisi en est-il différ ent de l’hôtelnew-yorkai s ou london ien? Peutêr e...巴黎的旅馆不同于纽约或伦敦的旅馆吗？也许可以这么说吧……On dit que l’hôtelparisi en est vieux.C’estvrai.Mais il y a beauco up d’hôtelmodern es.有人说巴黎的旅馆是古老的，的确如此，但巴黎也有很多现代旅馆。
Onditqu’ilestpasconfor table.C’estmoinsvrai : les hôtelparisi ens sont très confor table s.有人说巴黎的旅馆不舒适，这与实际情况不太相符，巴黎的旅馆其实是很舒适的。
Les prix? Les mêmesque ceux des autres capita les.至于巴黎旅馆的价格，同其他首都城市的旅馆价格是一样的。
Mais ils sont tout de même différ ents,par exempl e, des hôtelsanglai s.Le living-roomn’estjamais très grand,mais les chambr es sont grande s. L’hôtelfrança is ressem ble en cela à la maison frança ise. On y vit commechez soi, mais on ne fait pas facile mentla connai ssanc e de ses voisin s.但这些旅馆也不是完全相同的，就拿英国旅馆来举例说明吧。
lecture的名词
lecture的名词【释义】lecturen.讲座，讲课，演讲；训斥，告诫v.（尤指在大学里）开讲座，讲课；训斥，告诫复数lectures第三人称单数lectures现在分词lecturing过去式lectured过去分词lectured【短语】1Lecture Room百家讲坛;演讲室;教学室2lecture theatre大教室;阶梯教室;大讲堂;阅览室3lecture hall阶梯教室;大讲堂4The Last Lecture最后的演讲;最后一课;英文版5to attend a lecture听课;听讲6lecture notes讲稿;讲义;课堂笔记;教学笔记7attend a lecture参加演讲;听演讲;听讲座8Lecture Notes in Computer Science计算机科学讲义9He found my lecture interesting他觉得我讲课有趣;他觉得我讲课滑稽;他感觉我授课有趣儿;他感到我讲课有趣【例句】1I found her lecture very obscure.我觉得她的讲座非常费解。
2Can I borrow your lecture notes?我可以借你的讲稿看看吗？3She gave them a chastening lecture.她给他们做了一次令他们对自己的行为感到内疚的演讲。
4She wasn't taking notes on the lecture.她没记讲座笔记。
5She recast her lecture as a radio talk.她把讲稿修改成了一篇广播讲话。
6His lecture ranged over a number of topics.他的讲座涉及许多话题。
评价中心测评技术精选全文
第9讲 评价中心测评技术
总结：
• 能测出笔试和单一面试不能检测出的能力和素质 • 能观察个体之间的相互作用 • 能依据求职者的行为特征对其进行全面合理的评价 • 能够涉及到求职者的多种能力和个性特征 • 使求职者在无意中显露自己的各方面的特点 • 使求职者有公平的竞争机会 • 节省时间，并且便于比较 • 应用范围广
第9讲 评价中心测评技术
无领导小组讨论的试题形式(三)
C 多项选择问题 是指让考生在多种备选答案中选择其中有效的几种或者对 备选答案的重要性进行排序。 考察：考生分析问题的实质，抓住问题本质的能力。 举例：某个市场研究部门收集到了15条关于市场动态的信 息，只能向主管经理上报5条，请讨论出结果？ 回答：分析是否透彻有理（题目难出，但容易引发争论）
第9讲 评价中心测评技术 人员素质测评讲义·09级人力专业
第9讲 评价中心测评技术
几种典型测评技术及其效度
人员素质测评讲义·09级人力专业
一、评价中心的概念
第9讲 评价中心测评技术
评价中心是一种包含多种评价方法和形式的测评系统。它通过创设一种 逼真的模拟管理系统或工作场景，将被试纳入到该环境中，使其完成该 系统环境下对应的各种工作，如主持会议、处理公文、进行决策、处理 各种日常事务和突发事件等。在这个过程中，主试采取多种测评技术和 方法，观察和分析被试在模拟的各种情境压力下的心理行为表现及工作 绩效，以测量和评价被试的管理能力和潜能等素质。
人员素质测评讲义·09级人力专业
第9讲 评价中心测评技术
AC的典型操作流程
• 前期准备（评价中心的开发） • 岗位分析：测评要素的内涵解释及评价标准（设置评分等
级及标准） • 选择具体可行的评价中心技术（二种或多种结合） • 根据评价中心的测评思想进行具体的方案设计 • 评委培训 • 场地布置 • 实施测评，评委观察评分 • 后期处理 • 数据统计处理 • 综合撰写测评报告
应用PDE讲解09StumLiouville问题
11
2.1 典则方程
特别关注二阶常微分方程的不同表现形式。
0
如果在有关区间上
0，可以换成
0 在方程两边分别乘以
合并，得到
exp
exp
0
从而得到方程
0 有导数项的系数本身与微分相联系的特性。
§2 解的零点与振荡性态 .................................................................... 11 2.1 典则方程..................................................................................... 12 2.2 Sturm 比较定理 ........................................................................ 13 2.3 Liouville 定理............................................................................. 19
§3 特征值与特征函数 ......................................................................... 23 3.1 特征函数正交性 ....................................................................... 23 3.2 特征值非负性 ........................................................................... 24 3.3 例子：热传导问题 ................................................................. 27
Lecture-09
3.一维束缚态问题若粒子只能在有限空间运动，即()lim ,0r r t ψ→∞→G ，称态(),r t ψG为束缚态。
1）若()()12,x x ψψ是属于同一能量E 的两个态，由能量本征方程()()()()11222220,20mE V x mE V x ψψψψ′′+−=′′+−===有 12210ψψψψ′′′′−=， 即()'122112210,, const ψψψψψψψψ′′−=′′−=对于束缚态，()0x ψ→∞→，有 ''12210ψψψψ−=。
在()0x ψ≠的区间，有1212ψψψψ′′=， 即 12ln 0ψψ′⎛⎞=⎜⎟⎝⎠，12lnln C ψψ=， ()()12x C x ψψ=说明：1）一维束缚态无简并；2）当()()V x V x −=，一维束缚态有确定宇称。
2）波函数的连续性定态方程：()()()()22mx E V x x ϕϕ′′=−−=， a) 若连续，则()V x ,,ϕϕϕ′′′连续。
b) 若不连续，则()V x ϕ′′不连续。
在不连续点附近的邻域积分：a ()()()()()22a a ma a E V x x εεϕεϕεϕ+−′′+−−=−−∫=dx若有限，则积分，V ∆0→ϕ′和ϕ均连续。
若积分为, V ∆→∞()()(), '' 'a a E V x x dx εεϕϕϕϕϕϕ+−∞⎧⎪−=⎨⎪⎩∫和均不连续有限, 不连续,连续0,和均连续ϕ3）δ势阱()()()V x x V x γδ=−=−，在尝试0,, ,x V V =→−∞∆→处 ∞ϕ连续的束缚定态解。
0x ≠的定态Schroedinger 方程 220mEϕϕ′′+== 考虑的情形，令 0E <222mEk =−=， 则 ()00kx kx kxkxAe A e x x Ce C e x ϕ−−′⎧+<=⎨′+>⎩ . 要求有限，故 (x ϕ→±∞)()0'0, 0kx kxAe x A C x C e x ϕ−⎧<===⎨′>⎩ ， 此时，为束缚态（这就是取()0x ϕ→±∞→0E <的原因，E>0时无束缚定态解）。
09戴维宁定理及最大功率定理
例1.
4 a 6
Rx 6
I 4
b
10V
+
–
计算Rx分别为1.2、 5.2时的I；
解
保留Rx支路，将其余一端口 网络化为戴维南等效电路：
广东海洋大学
信息学院
徐国保
Lecture_09 戴维南定理及最大功率定理
不能采用功率匹配条件。但是在测量、电子与信息工程中，
常常着眼于从微弱信号中获得最大功率，而不看重效率的
高低。
广东海洋大学
信息学院
徐国保
Lecture_09 戴维南定理及最大功率定理
例4－17 电路如图4-28(a)所示。 试求：(l) RL为何值时获得最大功率；
(2) RL获得的最大功率； (3) 10V电压源的功率传输效率。
a
A
Isc
b
a
Geq(Req) b
广东海洋大学
信息学院
徐国保
Lecture_09 戴维南定理及最大功率定理
a
a
N
iSC
N0
Gab=G0
b
b
a
N0为将N中所有独立
Isc
Geq
源置零后所得无源二 端网络。
b 诺顿等效电路可由戴维南等效电路经电源等效变换得到。
诺顿等效电路可采用与戴维南定理类似的方法证明。证明过
Rx I= Uoc /(Req + Rx) =0.333A Rx =5.2时，
b
I= Uoc /(Req + Rx) =0.2A
信息学院
徐国保
Lecture_09 戴维南定理及最大功率定理
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of bond market risk
• Bonds with longer maturities more sensitive
to fluctuations in interest rate
16
Bond Pricing
• Bond Pricing between Coupon Dates
• Accrued interest and quoted bond prices
• Call provisions on corporate bonds
• Callable bonds: May be repurchased by issuer
at specified call price during call period
• Convertible bonds
• Allow bondholder to exchange bond for
Asked Yield 0.151 0.358 0.769 1.234 1.847 2.598 2.941 3.263 3.697 3.718
3
Maturity 8/15/2012 8/15/2014 12/31/2015 8/15/2017 2/15/2020 8/15/2023 2/15/2027 2/15/2031 11/15/2039 5/15/2041
bond may be exchanged
• Market conversion value: the current value of the shares
for which the bonds may be exchanged
• Conversion premium: the excess of the bond price over
• Floating-rate bonds
• Coupon rates periodically reset according to
specified market rate
9
Bond Characteristics
•
10
Table: Principal and Interest Payments
• Price =
$40 60 ∑������������=1 (1.04)������������ $40 60 ∑������������=1 (1.05)������������
• If the interest rate is 10% (5% per six months) • Price =
specified number of common stock shares
5
Example: A Callable Bond
• Issue size: US$100 million • Issue date: September 15, 2015 • Maturity: September 15, 2025 (10 years) • Coupon: 8% • Call Provision: Callable in 5 years (a five-year
• Bond • Security that obligates issuer to make payments to holder over time • Face Value, Par Value • Payment to bondholder at maturity of bond • Coupon Rate • Bond’s annual interest payment per dollar of par value • Zero-Coupon Bond • Pays no coupons, sells at discount, provides only payment of par value at maturity
with par value of $1,000 paying 60 semiannual coupon payments of $40 each. Suppose that the interest rate is 8% annually, or r = 4% per sixmonth period.
7
Example: A Convertible Bond
• Suppose a convertible bond is issued at par value
of $1,000 and is convertible into 40 shares of a firm’s stock.
• Conversion ratio: the number of shares for which each
Coupon 1.750 4.250 2.125 4.750 8.500 6.250 6.625 5.375 4.375 4.375
Bid 101.570 111.547 105.789 120.219 152.063 137.406 145.547 130.266 111.766 111.719
Asked 101.594 111.594 105.820 120.266 152.094 137.438 145.594 130.297 111.813 111.750
Change -0.016 -0.094 -0.164 -0.234 -0.344 -0.688 -0.719 -0.953 -0.813 -0.938
Figure: Listing of Corporate Bonds
4
Bond Characteristics
• Corporaes at Different Interest Rates
• 8% coupon bond, coupons paid semiannually
15
Bond Pricing
• Prices fall as market interest rate rises • Interest rate fluctuations are primary source
• Invoice price = Flat price + Accrued interest
×
������������������������������������������������ ������������������������������������������������������������ ������������������������������������������������ ������������������������������������������������������������������������ ������������������������������������������������������������������������������������ ������������������������������������������������ ������������������������������������������������������������������������������������������������������������������������ ������������������������������������������������������������������������ ������������������������������������������������������������������������������������������������
+ +
$1,000 (1.04)60 $1,000 (1.05)60
= $1,000
= $810.71
13
Figure: Inverse Relationship between Bond Prices and Yields
Price of an 8% coupon bond with 30-year maturity making semiannual coupon payments
17
Example: Accrued Interest
• Suppose that the coupon rate is 8%. Then the
semiannual coupon payment is $40. Because 30 days have passed since the last coupon payment, the accrued interest on the bond is $40 × (30/182) = $6.59. If the quoted price of the bond is $990, then the invoice price will be $990 + $6.59 = $996.59.
period of call protection)
• Call price: 103% of par value
6
Exercise
• Suppose that Verizon issues two bonds with
identical coupon rates and maturity dates. One bond is callable, however, while the other is not. Which bond will sell at a higher price?