chapter12集成运算器习题答案
计算机组成课后习题答案
闪存可以分为NOR Flash和NAND Flash两种类型。
工作原理
闪存的主要功能是可擦写、非易失性存储数据。NOR Flash读取速度快，适用于代码存储；NAND Flash容量大 、成本低，适用于大量数据的存储。闪存的擦除和编程操作需要特定的电压和时序。
04
输入输出设备
键盘的工作原理与分类
THANKS
感谢观看
成字符或汉字。
05
数据传输方式
并行传输与串行传
并行传输
指数据以多个通道同时传输，适用于近距离、高速传输，如 CPU与内存之间的数据交换。
串行传输
指数据一位一位地顺序传输，适用于远距离、低速传输，如 网络通信。
数据传输速率
数据传输速率
指单位时间内传输的数据量，通常以比特率（bit per second）表示。
影响数据传输速率的因素
包括信号带宽、信噪比、误码率等。
数据传输过程中的同步与异步问题
同步传输
指发送方和接收方的时钟信号保持同 步，适用于高速数据传输。
异步传输
指发送方和接收方的时钟信号不需同 步，适用于低速数据传输。
06
操作系统基础知识
操作系统的功能与分类
要点一
功能
操作系统是计算机系统的核心软件，负责管理计算机硬件 和软件资源，提供用户与计算机交互的界面。
硬盘的分类与工作原理
分类
硬盘可以分为机械硬盘（HDD）和固态硬盘（SSD）。
工作原理
硬盘的主要功能是永久存储数据。HDD采用磁记录技术，将数据以磁道的形式存 储在旋转的盘片上；SSD则采用闪存技术，通过电子方式存储数据。HDD的读取 速度较慢，但价格较低；SSD的读取速度较快，但价格较高。
电子电路第十二章习题及参考答案
习题十二12-1 写出题图12-1所示逻辑电路输出F 的逻辑表达式，并说明其逻辑功能。
解：由电路可直接写出输出的表达式为：301201101001301201101001D A A D A A D A A D A A D A A D A A D A A D A A F +++==∙∙∙由逻辑表达式可以看出： 当A 1A 0=00 F =D 0 A 1A 0=01 F =D 1A 1A 0=10 F =D 2 A 1A 0=11 F =D 3这个电路的逻辑功能是，给定地址A 1A 0以后，将该地址对应的数据传输到输出端F 。
12-2 组合逻辑电路如题图12-2所示。
（1）写出函数F 的表达式；（2）将函数F 化为最简“与或”式，并用“与非”门实现电路； （3）若改用“或非”门实现，试写出相应的表达式。
解：（1）逻辑表达式为：C A D B D C B A F += （2）化简逻辑式CA DB D BC A C AD B D C A D B C A D C B BC A C A D B A C A D B D C B A C A D B D C B A F +=+++++=++++++=++++=+=∙)1()1())(()(这是最简“与或”表达式，用“与非”门实现电路见题解图12-2-1，其表达式为： C A D B F ∙=（3）若用“或非”门实现电路见题解图12-2-2，其表达式为： C A D B C A D B C A D B C A D B F +++=+++=++=+=))((由图可见，对于同一逻辑函数采用不同的门电路实现，所使用的门电路的个数不同，组合电路的速度也有差异，因此，在设计组合逻辑电路时，应根据具体不同情况，选用不同的门电路可使电路的复杂程度不同。
A A3210题图12-1 习题12-1电路图12-3 组合逻辑电路如题图12-3分析的结果，列成真值表的形式。
解：对于图12-3电路可以写出逻辑函数表达式为：ABCC AB C B ABC C A C B ABC C A F +=++==∙∙ =（AB ）⊙C真值表如右图所示，由真值表可以看出，该电路是实现AB 与C 的“同或”，及当AB 与C 的值相同时，电路输出为“1”，否则输出为“0”。
Chapter 12 课后答案
新编语言学教程Chapter 12答案Applied Linguistics1. Define the following terms briefly.(1)applied linguistics: the study of language and linguistics in relation to practicalissues, e.g. speech therapy, language teaching, testing, and translation.More often than not nowadays, it is used in the narrow sense, and refers tolanguage teaching in particular.(2)grammar-translation method: a method of foreign or second languageteaching which makes use of translation and grammar study as the mainteaching and learning activities.(3) audiolingual method: the teaching of a second language through imitation,repetition, and reinforcement. It emphasizes the teaching of speaking andlistening before reading and writing and the use of mother tongue in theclassroom is not allowed.(4)communicative language teaching: an approach to foreign or second languageteaching which emphasizes that the goal of language learning is toachieve communicative competence.(5)testing: the use of tests, or the study of the theory and practice of their use,development, evaluation, etc.(6)achievement test: a test which measures how much of a language someone haslearned with reference to a particular course of study or program of instruction.(7)validity: (in testing) the degree to which a test measures what it is supposedto measure, or can be used successfully for the purposes for which it is intended.A number of different statistical procedures can be applied to a testto estimate its validity. Such procedures generally seek to determine what thetest measures, and how well it does so.(8)reliability: (in testing) a measure of the degree to which a test gives consistentresults; a test is said to be reliable if it gives the same results when it isgiven on different occasions or when it is taken by different people.(9)proficiency test: a test which measures how much of a language someone haslearned without considering the syllabus, duration and manner of learning.(10) subjective test: a test which is scored according to the personal judgment ofthe marker, such as essay writing or translation.(11) objective test: a test that can be marked without the use of the examiner’spersonal judgment.(12) language aptitude test: a test which measures a person’s aptitude for secondor foreign language learning and it can be used to identify those learners who are most likely to succeed.(13) diagnostic test: a test which is designed to show what skills or knowledge alearner knows and doesn’t know. For example, a diagnostic pronunciationtest may be used to measure the learner’s pronunciation of English sounds.It would show which sounds a student is and is not able to pronounce. Diagnostictests may be used to find out how much a learner knows or to measure how successful an instruction program has been.(14) backwash effect: Tests strongly affect what actually occurs in the classroomand the effect of tests on classroom L2 teaching and learning is known as thebackwash effect.2. The advantages of grammar-translation method:(1)As the grammars described in this method contain very detailed descriptionsof the correct construction of phrases and sentences of a language, accuracyis stressed and improved.(2)Students’ ability to read and write is encouraged and improved since themethod focuses on the written work.(3)This method is less demanding than some other approaches for a teacherwhose oral proficiency may not be adequate.(4)This method is popular with people who would like to study English independently,especially the adult learners who want to learn grammar rules anduse them to approach new materials by themselves.The disadvantages of grammar-translation method:(1)It emphasizes language at the sentence level regardless of context, so the organizationof language above the sentence level is not so carefully studied.(2)As the focus is on written work, oral fluency and spontaneity is not so welldeveloped and common everyday language is not taught enough.(3)The basic techniques in this method are rote learning of the rules and vocabulary,and grammar rules are taught deductively as general statements tobe applied in particular exercises in translation, so the learners may find itboring to learn.(4)With the emphasis on grammar, students typically know a lot about the languagebut are unable to actually use it. As a result, their use of the new languageoften tends to be literal or unnatural.3.Changes required would include:(1)Change in teacher’s role. The teacher can no longer be the source of knowledgeand trut h about the language. The teacher’s role has more to do withinitiating activities.(2)Change in learner’s role. The learner can no longer be passive. The learnermust actively participate in the activities.(3)Change of materials. These should, as far as possible, preserve the features ofauthentic instances of language use.(4)Change of techniques. These should emphasize the tasks (not drills) to beperformed and identify the skills being practiced.(5)Change in attitude. If the above are to be achieved then we are involved inchanging our attitudes towards teaching and learning in general.4.Achievement tests are based on a particular language syllabus, or part of a syllabus,or chapters in a textbook that learners are known to have studied and theyaim to know how well learners know what they have been taught. For example,the Chinese MET test, which is based on the Middle School English Syllabus andtaken by students leaving Senior Middle School, and Mid-Term tests, designed forUniversity English Majors based on just a few chapters from a textbook.5.The validity of a test relates to what the test claims to measure and how well itdoes so. If we know that a test is valid, then we know what we can confidently sayabout a person who passes or fails it. The two most important aspects of validityare content validity and construct validity. If a test has content validity it meansthat the test questions cover a fair sample of the language structures and skillsthat the test claims to be measuring. If a test has construct validity, it shows that itmeasures only what it claims to measure and nothing else.6.A test is said to be reliable if it gives the same results when it is given on differentoccasions or when it is taken by different people. There are two aspects to reliability:test reliability and scorer reliability. Test reliability refers to how consistent scoresare on a test. If, for example, there are two versions of a particular test and the sameperson takes them on consecutive days and his scores are almost the same on eachversion, then such a test has test reliability. A test has scorer reliability if there is ahigh level of agreement between different people marking the same test paper.。
计算机组成原理课后习题答案（一到九章）
计算机组成原理课后习题答案（⼀到九章）作业解答第⼀章作业解答1.1 基本的软件系统包括哪些内容？答：基本的软件系统包括系统软件与应⽤软件两⼤类。
系统软件是⼀组保证计算机系统⾼效、正确运⾏的基础软件，通常作为系统资源提供给⽤户使⽤。
包括：操作系统、语⾔处理程序、数据库管理系统、分布式软件系统、⽹络软件系统、各种服务程序等。
1.2 计算机硬件系统由哪些基本部件组成？它们的主要功能是什么？答：计算机的硬件系统通常由输⼊设备、输出设备、运算器、存储器和控制器等五⼤部件组成。
输⼊设备的主要功能是将程序和数据以机器所能识别和接受的信息形式输⼊到计算机内。
输出设备的主要功能是将计算机处理的结果以⼈们所能接受的信息形式或其它系统所要求的信息形式输出。
存储器的主要功能是存储信息，⽤于存放程序和数据。
运算器的主要功能是对数据进⾏加⼯处理，完成算术运算和逻辑运算。
控制器的主要功能是按事先安排好的解题步骤，控制计算机各个部件有条不紊地⾃动⼯作。
1.3 冯·诺依曼计算机的基本思想是什么？什么叫存储程序⽅式？答：冯·诺依曼计算机的基本思想包含三个⽅⾯：1) 计算机由输⼊设备、输出设备、运算器、存储器和控制器五⼤部件组成。
2) 采⽤⼆进制形式表⽰数据和指令。
3) 采⽤存储程序⽅式。
存储程序是指在⽤计算机解题之前，事先编制好程序，并连同所需的数据预先存⼊主存储器中。
在解题过程（运⾏程序）中，由控制器按照事先编好并存⼊存储器中的程序⾃动地、连续地从存储器中依次取出指令并执⾏，直到获得所要求的结果为⽌。
1.4 早期计算机组织结构有什么特点？现代计算机结构为什么以存储器为中⼼？答：早期计算机组织结构的特点是：以运算器为中⼼的，其它部件都通过运算器完成信息的传递。
随着微电⼦技术的进步，⼈们将运算器和控制器两个主要功能部件合⼆为⼀，集成到⼀个芯⽚⾥构成了微处理器。
同时随着半导体存储器代替磁芯存储器，存储容量成倍地扩⼤，加上需要计算机处理、加⼯的信息量与⽇俱增，以运算器为中⼼的结构已不能满⾜计算机发展的需求，甚⾄会影响计算机的性能。
(完整版)集成运算放大器习题参考答案
（对）7．“虚地”是指该点与“地”点相接后，具有“地”点的电位。
（错）8、集成运放不但能处理交流信号，也能处理直流信号。
（对）9、集成运放在开环状态下，输入与输出之间存在线性关系。
（错）10、各种比较器的输出只有两种状态。
（对）11、微分运算电路中的电容器接在电路的反相输入端。
（对）三、选择题：（每小题2分，共16分）1.集成运算放大器能处理（C）。
A、直流信号；B、交流信号；C、交流信号和直流信号。
2. 为使电路输入电阻高、输出电阻低，应引入（A）。
A、电压串联负反馈；B、电压并联负反馈；C、电流串联负反馈；D电流并联负反馈。
3. 在由运放组成的电路中，运放工作在非线性状态的电路是（D）。
A、反相放大器；B、差值放大器；C、有源滤波器；D、电压比较器。
4. 集成运放工作在线性放大区，由理想工作条件得出两个重要规律是（C ）。
A、U＋＝U－＝0，i＋＝i－；B、U＋＝U－＝0，i＋＝i－＝0；C、U＋＝U－，i＋＝i－＝0；D、U＋＝U－＝0，i＋≠i－。
5．分析集成运放的非线性应用电路时，不能使用的概念是（B）。
A、虚地；B、虚短；C、虚断。
6. 集成运放的线性应用存在（C）现象，非线性应用存在（B）现象。
A、虚地；B、虚断；C、虚断和虚短。
7. 理想运放的两个重要结论是（B）。
A、虚短与虚地；B、虚断与虚短；C、断路与短路。
8. 集成运放一般分为两个工作区，它们分别是（B）。
A、正反馈与负反馈；B、线性与非线性；C、虚断和虚短。
四、问答题：1. 集成运放一般由哪几部分组成？各部分的作用如何？答：集成运放一般输入级、输出级和中间级及偏置电路组成。
输入级一般采用差动放大电路，以使运放具有较高的输入电阻及很强的抑制零漂的能力，输入级也是决定运放性能好坏的关键环节；中间级为获得运放的高开环电压放大位数（103～107），一般采用多级共发射极直接耦合放大电路；输出级为了具有较低的输出电阻和较强的带负载能力，并能提供足够大的输出电压和输出电流，常采用互补对称的射极输出器组成；为了向上述三个环节提供合适而又稳定的偏置电流，一般由各种晶体管恒流源电路构成偏置电路满足此要求。
电子技术习题册答案
第1章 逻辑代数基础1. 用真值表证明下列等式。
(1) (A B)C=A (B C)⊕⊕⊕⊕ (2) C B A C B A A +=++ A B C B A ⊕C B ⊕C B A ⊕⊕)( A )(C B A ⊕⊕0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 0 0 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 0 0 1 1111A B C AC B +C BC B A A ++ C B A +0 0 0 1 1 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 1 0 0 0 0 1 0 0 0 1 1 1 1 1 0 1 0 0 1 1 1 1 1 0 0 0 1 1 1 1110 0112. 用代数法化简下列各式。
(1) A+ABC+ABC+CB+CB(CA B B C BC BC A +=++++=)()1(2) ABC+ABC+ABC+ABCAABB AC C AB C C B A =+=+++=)()(3．将下列各函数化为最小项之和的形式。
(1) Y=ABC+BC+AB7543)()(mmmmCBACBABCAABCBCACCBAAABCBCA+++=++++=++++=(2))(ABY DCBCABD+++＝DCABDCBDCABDCBCDBDADCBCADBBDADCBCABDBA=+=+++++=+++++=++++=)()()()(4．根据下列各逻辑式，画出逻辑图。
①Y=（A+B）C；②Y=AB+BC；③Y=（A+B）（A+C）；5．试对应输入波形画出下图中Y1 ~ Y4 的波形。
6.如果“与”门的两个输入端中，A为信号输入端，B为控制端。
设当控制端B=1和B=0两种状态时，输入信号端A的波形如图所示，试画出输出端Y的波形。
如果A和B分别是“与非”门、“或”门、“或非”门的两个输入端，则输出端Y的波形又如何？总结上述四种门电路的控制作用。
计算机科学导论第12章参考答案
CHAPTER 12Abstract Data TypesReview Questions1.An abstract data type is a data declaration packaged together with the operationsthat are meaningful for the data type with the implementation hidden from the user.3.A linear list is a list in which each element has a unique successor.5.Two common implementations of a general list are an array and a linked list.7.A push operation adds an element to the top of the stack while a pop operationremoves an element from the top of the stack. A push can put the stack in an over-flow condition while a pop can put the stack in an underflow condition.9.The enqueue operation adds an element to the rear of a queue. The dequeue opera-tion removes the element at the front of the queue. The enqueue operation couldput the queue in an overflow state while the dequeue could put the queue in anunderflow state.11.A depth first traversal processes all of the nodes in one subtree before processingall of the nodes in the other subtree. In a breadth first traversal, all the nodes at onelevel are processed before moving on to the next level.13.In a depth-first traversal, all of a vertex's descendents are processed before movingto an adjacent vertex. In a breadth-first traversal, all adjacent vertices of a vertexare processed before going to the next level.15.A network is a graph with weighted lines.Multiple-Choice Questions17.d19.c21.a23.a25.b27.d29.a34CHAPTER 12ABSTRACT DATA TYPES31.c33.c35.d37.d39.c41.b43.a45.d47.a49.b51.b53.aExercises55.(top) 6 5 (bottom)57.moveStackInput: Source stack (s1) and destination stack (s2)1. While s1 is not empty1.1 push ( s2, pop(s1) )End loopEnd59.catStackInput: Source stack (s2) and destination stack (s1)1. While s2 is not empty1.1 push ( s1, pop(s2) )End loopEnd61.compareStackInput: The two stacks to compare (s1 and s2)1. Allocate memory for two temporary stacks (Temp1 and Temp2)2. copyStack ( s1, Temp1 ) (see #58)3. copyStack ( s2, Temp2 ) (see #58)4. While Temp1 is not empty AND Temp2 is not empty4.1 TempValue1 = pop ( Temp1 )4.2 TempValue2 = pop ( Temp2 )4.3 If TempValue1 is not equal to TempValue2SECTION 54.3.1 Return falseEnd ifEnd loop5. If Temp1 is not empty OR Temp2 is not empty5.1 Return falseEnd if6. Return trueEnd63.emptyQueueInput: Queue to empty (q3)1. While q3 is not empty1.1 dequeue( q3 )End loopEnd65.copyQueueInput: Source queue (q2) and destination queue (q3)1. Allocate memory for a temporary queue (Temp)2. While q2 is not empty2.1 enqueue ( Temp, dequeue(q2) )End loop3. While Temp is not empty3.1 TempValue = dequeue(Temp)3.2 enqueue ( q2, TempValue )3.3 enqueue ( q3, TempValue )End loopEnd67.compareQueueInput: The two queues to compare (q1 and q2)1. Allocate memory for two temporary queues (Temp1 and Temp2)2. copyQueue ( q1, Temp1 ) (see #65)3. copyQueue ( q2, Temp2 ) (see #65)4. While Temp1 is not empty AND Temp2 is not empty4.1 TempValue1 = dequeue ( Temp1 )4.2 TempValue2 = dequeue ( Temp2 )4.3 If TempValue1 is not equal to TempValue24.3.1 Return falseEnd if6CHAPTER 12ABSTRACT DATA TYPESEnd loop5. If Temp1 is not empty OR Temp2 is not empty5.1 Return falseEnd if6. Return trueEnd69. See Figure 12.1Figure 12.1Exercise 6971.This tree cannot be drawn because it is not a valid binary tree. Node C must be theroot because it is listed last in the postorder traversal. From the inorder traversal,we see that nodes A, B, and D must be in the left subtree (because they are listed tothe left of the root) and that nodes E, F, and G are in the right subtree. Lookingback at the postorder list, however, we see that nodes G and F are listed first,which is not possible.73.Assuming that arcs are stored in sequence by their destination, the traversal is: A,G, F, H, D, E, C, B75.See Figure 12.2.Figure 12.2Exercise 75SECTION 7 77.See Figure 12.3.Figure 12.3Exercise 778CHAPTER 12ABSTRACT DATA TYPES。
电子电路第十二章习题及参考答案
习题十二12-1 写出题图12-1所示逻辑电路输出F 的逻辑表达式，并说明其逻辑功能。
解：由电路可直接写出输出的表达式为：301201101001301201101001D A A D A A D A A D A A D A A D A A D A A D A A F +++==•••由逻辑表达式可以看出： 当A 1A 0=00 F =D 0 A 1A 0=01 F =D 1A 1A 0=10 F =D 2 A 1A 0=11 F =D 3这个电路的逻辑功能是，给定地址A 1A 0以后，将该地址对应的数据传输到输出端F 。
12-2 组合逻辑电路如题图12-2所示。
（1）写出函数F 的表达式；（2）将函数F 化为最简“与或”式，并用“与非”门实现电路； （3）若改用“或非”门实现，试写出相应的表达式。
解：（1）逻辑表达式为：C A D B D C B A F += （2）化简逻辑式CA DB D BC A C AD B D C A D B C A D C B BC A C A D B A C A D B D C B A C A D B D C B A F +=+++++=++++++=++++=+=•)1()1())(()(这是最简“与或”表达式，用“与非”门实现电路见题解图12-2-1，其表达式为： C A D B F •=（3）若用“或非”门实现电路见题解图12-2-2，其表达式为：C AD B C A D B C A D B C A D B F +++=+++=++=+=))((由图可见，对于同一逻辑函数采用不同的门电路实现，所使用的门电路的个数不同，组合电路的速度也有差异，因此，在设计组合逻辑电路时，应根据具体不同情况，选用不同的门电路可使电路的复杂程度不同。
A A3210题图12-1 习题12-1电路图12-3 组合逻辑电路如题图12-3分析的结果，列成真值表的形式。
解：对于图12-3电路可以写出逻辑函数表达式为：ABCC AB C B ABC C A C B ABC C A F +=++==••=（AB ）⊙C真值表如右图所示，由真值表可以看出，该电路是实现AB 与C 的“同或”，及当AB 与C 的值相同时，电路输出为“1”，否则输出为“0”。
模拟集成电路Chapter 12 Switched capacitor circuits
Switched-capacitor circuits
Gao Fei
1
Continuous-time circuits
• R2 degrades the open-loop gain of the opamp • Capacitors limit the bandwidth to highpass behavior.
24
Unity-gain sampler/buffer (cont’d)
• Effect of charged injected by S2. • It introduced an offset, which can be removed by differential operation.
25
Unity-gain sampler/buffer (cont’d)
4
General SC circuit
• Key components: Opamp, switch and capacitor
5பைடு நூலகம்
MOSFET as switches
• MOSFET could be a gate-voltage controlled switch.
6
Charging and discharging
20
Charge injection cancellation (cont’d)
21
Sampling capacitor
• The bottom-plate suffers substantial junction capacitance.
22
Unity-gain sampler/buffer
《金融学（第二版）》讲义大纲及课后习题答案详解十二章
《⾦融学（第⼆版）》讲义⼤纲及课后习题答案详解⼗⼆章CHAPTER 12CHOOSING AN INVESTMENT PORTFOLIOObjectivesTo understand the process of personal investing in theory and in practice.To build a quantitative model of the tradeoff between risk and reward.Outline12.1 The Process of Personal Portfolio Selection12.2 The Trade-off between Expected Return and Risk12.3 Efficient Diversification with Many Risky AssetsSummaryThere is no single portfolio selection strategy that is best for all people.Stage in the life cycle is an imp ortant determinant of the optimal composition of a person’s optimal portfolio of assets and liabilities.Time horizons are important in portfolio selection. We distinguish among three time horizons: the planning horizon, the decision horizon, and the trading horizon.In making portfolio selection decisions, people can in general achieve a higher expected rate of return only by exposing themselves to greater risk.One can sometimes reduce risk without lowering expected return by diversifying more completely either withina given asset class or across asset classes.The power of diversification to reduce the riskiness of an investor’s portfolio depends on the correlations among the assets that make up the portfolio. In practice, the vast majority of assets are positively correlated with each other because they are all affected by common economic factors. Consequently, one’s ability to reduce risk through diversification among risky assets without lowering expected return is limited.Although in principle people have thousands of assets to choose from, in practice they make their choices from a menu of a few final products offered by financial intermediaries such as bank accounts, stock and bond mutual funds, and real estate. In designing and producing the menu of assets to offer to their customers theseintermediaries make use of the latest advances in financial technology.Solutions to Problems at End of Chapter1. Suppose that your 58-year-old father works for the Ruffy Stuffed Toy Company and has contributed regularly to his company-matched savings plan for the past 15 years. Ruffy contributes $0.50 for every $1.00 your father puts into the savings plan, up to the first 6% of his salary. Participants in the savings plan can allocate their contributions among four different investment choices: a fixed-income bond fund, a “blend” option that invests in large companies, small companies, and the fixed-income bond fund, a growth-income mutual fund whose investments do not include other toy companies, and a fund whose sole investment is stock in the Ruffy Stuffed Toy Company. Over Thanksgiving vacation, Dad realizes that you have been majoring in finance and decides to reap some early returns on that tuition money he’s been investing in your education. He shows you the most recent quarterly statement for his savings plan, and you see that 98% of its current value is in the fourth investment option, that of the Ruffy Company stock..a.Assume that your Dad is a typical risk-averse person who is considering retirement in five years. Whenyou ask him why he has made the allocation in this way, he responds that the company stock has continually performed quite well, except for a few declines that were caused by problems in a division that the company has long since sold off. Inaddition, he says, many of his friends at work have done the same. What advice would you give your dad about adjustments to his plan allocations? Why?b.If you consider the fact that your dad works for Ruffy in addition to his 98% allocation to the Ruffy stockfund, does this make his situation more risky, less risky, or does it make no difference? Why? SOLUTION:a.Dad has exposed himself to risk by concentrating almost all of his plan money in the Ruffy Stock fund. This is analogous to taking 100% of the money a family has put aside for investment and investing it in a single stock.First, Dad needs to be shown that just because the company stock has continually performed quite well is no guarantee that it will do so indefinitely. The company may have sold off the divisions which produced price declines in the past, but future problems are unpredictable, and so is the movement of the stock price. “Past performance is no guarantee of future results” is the lesson.Second, Dad needs to hear about diversification. He needs to be counseled that he can reduce his risk by allocating his money among several of the options available to him. Indeed, he can reduce his risk considerably merely by moving all of his money into the “blend” fund because it is diversifi ed by design: it has a fixed-income component, a large companies component, and a small companies component. Diversification isachieved not only via the three differing objectives of these components, but also via the numerous stocks that comprise each of the three components.Finally, Dad’s age and his retirement plans need to be considered. People nearing retirement age typically begin to shift the value of their portfolios into safer investments. “Safer” normally connotes less variability, so that the risk of a large decline in the value of a portfolio is reduced. This decline could come at any time, and it would be very unfortunate if it were to happen the day before Dad retires. In this example, the safest option would be the fixed-income bond fund because of its diversified composition and interest-bearing design, but there is still risk exposure to inflation and the level of interest rates. Note that the tax-deferred nature of the savings plan encourages allocation to something that produces interest or dividends. As it stands now, Dad is very exposed to a large decline in the value of his savings plan because it is dependent on the value of one stock.Individual equities over time have proven to produce the most variable of returns, so Dad should definitely move some, probably at least half, of his money out of the Ruffy stock fund. In fact, a good recommendation given his retirement horizon of five years would be to re-align the portfolio so that it has 50% in the fixed- income fund and the remaining 50% split between the Ruffy stock fund (since Dad insists) and the “blend” fund.Or, maybe 40% fixed-income, 25% Ruffy, 15% growth-income fund, and 20% “blend” fund. This latterallocation has the advantage of introducing another income-producing component that can be shielded by the tax-deferred status of the plan.b.The fact that Dad is employed by the Ruffy Company makes his situation more risky. Let’s say that the companyhits a period of slowed business activities. If the stock price declines, so will th e value of Dad’s savings plan. If the company encounters enough trouble, it may consider layoffs. Dad’s job may be in jeopardy. At the same time that his savings plan may be declining in value, Dad may also need to look for a job or go onunemployment. Thus, Dad is exposed on two fronts to the same risk. He has invested both his human capital and his wealth almost exclusively in one company.2. Refer to Table 12.1.a.Perform the calculations to verify that the expected returns of each of the portfolios (F, G, H, J, S) in thetable (column 4) are correct.b.Do the same for the standard deviations in column 5 of the table.c.Assume that you have $1million to invest. Allocate the money as indicated in the table for each of the fiveportfolios and calculate the expected dollar return of each of the portfolios.d.Which of the portfolios would someone who is extremely risk tolerant be most likely to select? SOLUTION:d.An extremely risk tolerant person would select portfolio S, which has the largest standard deviation but also thelargest expected return.3. A mutual fund company offers a safe money market fund whose current rate is4.50% (.045). The same company also offers an equity fund with an aggressive growth objective which historically has exhibited an expected return of 20% (.20) and a standard deviation of .25.a.Derive the equation for the risk-reward trade-off line.b.How much extra expected return would be available to an investor for each unit of extra risk that shebears?c.What allocation should be placed in the money market fund if an investor desires an expected return of15% (.15)?SOLUTION:a.E[r] = .045 + .62b.0.62c.32.3% [.15 = w*(.045) + (1-w)*(.020) ]4. If the risk-reward trade-off line for a riskless asset and a risky asset results in a negative slope, what does that imply about the risky asset vis-a-vis the riskless asset?SOLUTION:A trade-off line wit h a negative slope indicates that the investor is “rewarded” with less expected return for taking on additional risk via allocation to the risky asset.5. Suppose that you have the opportunity to buy stock in AT&T and Microsoft.a.stocks is 0? .5? 1? -1? What do you notice about the change in the allocations between AT&T andMicrosoft as their correlation moves from -1 to 0? to .5? to +1? Why might this be?b.What is the variance of each of the minimum-variance portfolios in part a?c.What is the optimal combination of these two securities in a portfolio for each value of the correlation,assuming the existence of a money market fund that currently pays 4.5% (.045)? Do you notice any relation between these weights and the weights for the minimum variance portfolios?d.What is the variance of each of the optimal portfolios?e.What is the expected return of each of the optimal portfolios?f.Derive the risk-reward trade-off line for the optimal portfolio when the correlation is .5. How much extraexpected return can you anticipate if you take on an extra unit of risk?SOLUTION:a.Minimum risk portfolios if correlation is:-1: 62.5% AT&T, 37.5% Microsoft0: 73.5% AT&T, 26.5% Microsoft.5: 92.1% AT&T, 7.9% Microsoft1: 250% AT&T, short sell 150% MicrosoftAs the correlation moves from -1 to +1, the allocation to AT&T increases. When two stocks have negativec orrelation, standard deviation can be reduced dramatically by mixing them in a portfolio. It is to the investors’benefit to weight more heavily the stock with the higher expected return since this will produce a high portfolio expected return while the standard deviation of the portfolio is decreased. This is why the highest allocation to Microsoft is observed for a correlation of -1, and the allocation to Microsoft decreases as the correlationbecomes positive and moves to +1. With correlation of +1, the returns of the two stocks will move closely together, so you want to weight most heavily the stock with the lower individual standard deviation.b. Variances of each of the minimum variance portfolios:62.5% AT&T, 37.5% Microsoft Var = 073.5% AT&T, 26.5% Microsoft Var = .016592.1% AT&T, 7.9% Microsoft Var = .0222250% AT&T, short 150% Microsoft Var = 0c. Optimal portfolios if correlation is:-1: 62.5% AT&T, 37.5% Microsoft0: 48.1% AT&T, 51.9% Microsoft.5: 11.4% AT&T, 88.6% Microsoft1: 250% AT&T, short 150% Microsoftd. Variances of the optimal portfolios:62.5% AT&T, 37.5% Microsoft Var = 048.1% AT&T, 51.9% Microsoft Var = .022011.4% AT&T, 88.6% Microsoft Var = .0531250% AT&T, short 150% Microsoft Var = 0e. Expected returns of the optimal portfolios:62.5% AT&T, 37.5% Microsoft E[r] = 14.13%48.1% AT&T, 51.9% Microsoft E[r] = 15.71%11.4% AT&T, 88.6% Microsoft E[r] = 19.75%250% AT&T, short 150% Microsoft E[r] = -6.5%f.Risk-reward trade-off line for optimal portfolio with correlation = .5:E[r] = .045 + .66/doc/31dbf23b580216fc700afd59.html ing the optimal portfolio of AT&T and Microsoft stock when the correlation of their price movements is 0.5, along with the results in part f of question 12-5, determine:a.the expected return and standard deviation of a portfolio which invests 100% in a money market fundreturning a current rate of 4.5%. Where is this point on the risk-reward trade-off line?b.the expected return and standard deviation of a portfolio which invests 90% in the money market fundand 10% in the portfolio of AT&T and Microsoft stock.c.the expected return and standard deviation of a portfolio which invests 25% in the money market fundand 75% in the portfolio of AT&T and Microsoft stock.d.the expected return and standard deviation of a portfolio which invests 0% in the money market fundand 100% in the portfolio of AT&T and Microsoft stock. What point is this?SOLUTION:a.E[r] = 4.5%, standard deviation = 0. This point is the intercept of the y (expected return) axis by the risk-rewardtrade-off line.b.E[r] = 6.03%, standard deviation = .0231c.E[r] = 15.9%, standard deviation = .173d.E[r] = 19.75%, standard deviation = .2306. This point is the tangency between the risk-reward line from 12-5part f and the risky asset risk-reward curve (frontier) for AT&T and Microsoft.7. Again using the optimal portfolio of AT&T and Microsoft stock when the correlation of their price movements is 0.5, take $ 10,000 and determine the allocations among the riskless asset, AT&T stock, and Microsoft stock for:a. a portfolio which invests 75% in a money market fund and 25% in the portfolio of AT&T and Microsoftstock. What is this portfolio’s expected return?b. a portfolio which invests 25% in a money market fund and 75% in the portfolio of AT&T and Microsoftstock. What is this portfolio’s expect ed return?c. a portfolio which invests nothing in a money market fund and 100% in the portfolio of AT&T andMicrosoft stock. What is this portfolio’s expected return?SOLUTION:a.$7,500 in the money-market fund, $285 in AT&T (11.4% of $2500), $2215 in Microsoft. E[r] = 8.31%, $831.b.$2,500 in the money-market fund, $855 in AT&T (11.4% of $7500), $6645 in Microsoft. E[r] = 15.94%, $1,594.c.$1140 in AT&T, $8860 in Microsoft. E[r] = 19.75%, $1,975.8. What strategy is implied by moving further out to the right on a risk-reward trade-off line beyond the tangency point between the line and the risky asset risk-reward curve? What type of an investor would be most likely to embark on this strategy? Why?SOLUTION:This strategy calls for borrowing additional funds and investing them in the optimal portfolio of AT&T and Microsoft stock. A risk-tolerant, aggressive investor would embark on this strategy. This person would be assuming the risk of the stock portfolio with no risk-free component; the money at risk is not onl y from this person’s own wealth but also represents a sum that isowed to some creditor (such as a margin account extended by the investor’s broker).9. Determine the correlation between price movements of stock A and B using the forecasts of their rate of return and the assessments of the possible states of the world in the following table. The standard deviations for stock A and stock B are0.065 and 0.1392, respectively. Before doing the calculation, form an expectation of whether that correlation will be closer to1 or -1 by merely inspecting the numbers.SOLUTION:Expectation: correlation will be closer to +1.E[r A] = .05*(-.02) + .15*(-.01) + .60*(.15) + .20*(.15) = .1175, or, 11.75%E[r B] = .05*(-.20) + .15*(-.10) + .60*(.15) + .20*(.30) = .1250, or, 12.50%Covariance = .05*(-.02-.1175)*(-.20-.125) + .15*(-.01-.1175)*(-.10-.125) +.60*(.15-.1175)*(.15-.125) + .20*(.15-.1175)*(.30-.125) =.008163Correlation = .008163/(.065)*(.1392) = .90210.Analyze the “expert’s” answers to the following questions:a.Question:I have approx. 1/3 of my investments in stocks, and the rest in a money market. What do you suggestas a somewhat “safer” place to invest another 1/3? I like to keep 1/3 accessible for emergencies.Expert’s answer:Well, you could try 1 or 2 year Treasury bonds. You’d get a little bit more yie ld with no risk.b.Question:Where would you invest if you were to start today?Expert’s answer:That depends on your age and short-term goals. If you are very young – say under 40 –and don’tneed the money you’re investing for a home or college tuition or such, you would put it in a stockfund. Even if the market tanks, you have time to recoup. And, so far, nothing has beaten stocks overa period of 10 years or more. But if you are going to need money fairly soon, for a home or for yourretirement, you need to play it safer.SOLUTION:a.You are not getting a little bit more yield with no risk. The real value of the bond payoff is subject to inflationrisk. In addition, if you ever need to sell the Treasury bonds before expiration, you are subject to the fluctuation of selling price caused by interest risk.b.The expert is right in pointing out that your investment decision depends on your age and short-term goals. In addition, the investment decision also depends on other characteristics of the investor, such as the special character of the labor income (whether it is highly correlated with the stock market or not), and risk tolerance.Also, the fact that over any period of 10 years or more the stock beats everything else cannot be used to predict the future.。
有机化学chapter 12--芳环取代反应习题答案
一、芳环的亲电取代反应及机制
Activated Electrophiles (活化的亲电试剂)
—Cation formation by reaction of an elcerophile with an aromatic ring is accompanied by the loss of aromatic stabilization. 正离子的形成伴随着芳香性的丧失。 — Therefore, the electrophile that can effect aromatic subsitution of benzene must be more reactive than those addition to alkenes. 因此， 与苯环反应的亲电试剂需要更高的反应活性
一、芳环的亲电取代反应及机制
(4) Friedel-Crafts Alkylation (傅氏烷基化)
10
10
(4) Friedel-Crafts Alkylation (傅氏烷基化) 常用催化剂：路易斯酸、质子酸; 烷基化试剂：卤代烷、 烯、醇 、环氧化合物 · · · · · ·
+ CH2 O CH2 AlCl3 H2O CH2CH2OH
NH2 H3O+ NO2
SO3H
SO3H
一、芳环的亲电取代反应及机制
(4) Friedel-Crafts Alkylation (傅氏烷基化)
AlCl3催化量 0.2--0.3 eq.
Note: Rearrangement may occur in alkylation
亲电试剂是烷基正离子，可能会发生碳正离子重排
一、芳环的亲电取代反应及机制
2. Reaction coordinate diagrams:
电路基础与集成电子技术-第12章习题解答
第12章 集成逻辑门电路习题解答【12-1】填空 解：1． 开关 ， 截止 ， ， b ， 加深 ，减轻 ， 增大 ， 加速 。
2． 灌 ， ， 拉 ，160μA 。
3．3V ， ， ， ， ， ， ， ， ， 15mA ， 10 。
4． 高电平 ， 。
5． 输出与电源 。
6． 极低 ， 增加 ， 很大 ， 高 。
7． 30MHz 。
【12-2】电路如题图12-2(a)～(f)所示，试写出其逻辑函数的表达式。
解：F 1=A ，F 2=1，F 3=B A +，F 4=AB ，F 5=1，F 6=B【12-3】TTL 三态门电路如题图12-3所示。
在下图示输入波形的情况下，画出其输出端的波形。
解：当1=C 时，AB F = 当0=C 时，B A B A F +==于是，逻辑表达式 C B A C AB F )(++= F 的波形见解图12-3所示。
A B C F解图12-3【12-4】题图12-4中各电路中凡是能实现非功能的要打号，否则打×号。
图(a)为TTL门电路，图(b)为CMOS 门电路。
解：&1&&&1A 5VAAA1001M "1"A Ω=1Ω(a) TTL 门A(b) CMOS 门 解图12-4【12-5】要实现题图12-5中各TTL 门电路输出端所示的逻辑关系各门电路的接法是否正确如不正确，请予更正。
解：见解图12-5。
CB A X AXB X A +=AB=CDAB F +=解图12-5【12-6】由CMOS 传输门和反相器构成的电路如题图12-6（a ）所示，试画出在图（b ）波形作用下的输出u O 的波形（u I1=10V u I2=5V ）解：输出波形见解图12-6。
ttu解图12-6【12-7】甲乙两人用指针式万用表测量一个由TTL 门组成的电路，发现某点的电位为。
对此甲认为是由于该点的负载过重，导致灌电流或拉电流太大所致；乙则认为应先观察一下该点的波形，才能做出判断。
《计算机》各章课后习题答案
习题11-1 单向选择题1-5 B D A B B 6-10 B B C D A1-2 填空题1. 1946，ENIAC2. 四，电子管、晶体管、中小规模集成电路、大规模和超大规模集成电路3. 计算机硬件系统计算机软件系统4. 运算器、控制器、存储器、输入设备和输出设备1-3 思考题1. 计算思维是运用计算机科学的基础概念进行问题求解、系统设计、以及人类行为理解等涵盖计算机科学之广度的一系列思维活动。
计算思维具有以下特性：1)计算思维是概念化的，而不是程序化的。
2)计算思维是每个人需掌握的根本技能，而不是刻板的重复性工作。
3)计算思维不是计算机的思维方式，而是人类解决问题的一种思维方式。
4)计算思维是数学和工程思维的互补与融合5)计算思维是思想，不是产品6)计算思维面向所有的人，所有的地方2. 可以模拟那些不容易观察到的现象，可以求解用理论和实验手段无法解决的重大科学技术问题。
计算方法突破了实验和理论科学方法的局限，并进一步提高了人们对自然和社会的洞察力，为科学研究与技术创新提供了新的重要手段和理论基础。
计算已和理论、实验一起，被公认为科学的第三大支柱3. 计算机（computer）俗称电脑，是一种用于高速计算的电子计算机器，可以进行数值计算，又可以进行逻辑计算，还具有存储记忆功能。
是能够按照程序运行，自动、高速处理海量数据的现代化智能电子设备。
其主要特点有：1)运算速度快2)运算精度高3)可靠性高4)具有逻辑判断功能，逻辑性强5)存储容量大6)自动化程度高4. 存储程序的工作原理是：在计算机中设置存储器，将程序和数据存放到存储器中，计算机按照程序指定的逻辑顺序依次取出存储器中的内容进行处理，直到得出结果。
习题22-1 选择题1-5 D C B C B 6-10 D B D A B 11-15 B A B B C2-2 填空题1.13075O = 163DH= 5693D2.10101001.1B= 251.4O = a9.8H3.1110111111B= 1677O= 959 D4.原码：11000011 反码：10111100 补码：101111015.F4E8H6.位图矢量图2-3 思考题1.所谓信息是人们用于表示具有一定意义的符号的集合。
集成电子技术习题册答案
第十二章存储器和可编程器件12--1 填空1、按构成材料的不同，存储器可分为磁芯和半导体存储器两种。
磁芯存储器利用 正负剩磁 来存储数据；而半导体存储器利用 器件的开关状态 来存储数据。
两者相比，前者一般容量较 大 ；而后者具有速度 快 的特点。
2、半导体存储器按功能分有 ROM 和 RAM 两种。
3、ROM 主要由 地址译码器 和 存储矩阵 两部分组成。
按照工作方式的不同进行分类，ROM 可分为 固定内容的ROM 、 PROM 和 EPROM 三种。
4、某EPROM 有8数据线，13位地址线，则其存储容量为 213×8 。
5、在系统可编程逻辑器件简称为 ISPPLD 器件，这种器件在系统工作时 可以 （可以、不可以）对器件的内容进行重构，它包括 ISPGAL 、 ISPGDS 、 ISPLSI 三种系列的产品。
6、对isp 器件进行编程时不需要（需要、不需要）专门的编程器，对GAL 器件进行编程时需要 （需要、不需要）专门的编程器。
7、对GAL 器件和ispLSI 器件进行编程时可以选用下列那几种输入方式。
a)原理图方式 b)ABEL-HDL 语言c)VHDL 语言 d)原理图与ABEL 语言混合输入方式 e)FM 输入方式GAL 器件 : a) b ）c) d) e) ispLSI 器件: a) b ）c) d)12--2 D 0A 0D 1m (3,6,9,12,15)D 210D 3m (0,5,9,13)==∑=⋅=∑⎧⎨⎪⎪⎪⎩⎪⎪⎪ 12--3地址译码器A1A0D3 D2 D1 D0B1B0m 0m 1512--4 1。
F Q Q Q Q Q Q Q F Q Q Q Q Q Q F Q Q 110212102210210210310=⋅+⋅+⋅=⋅⋅+⋅+⋅⋅=⋅⎧⎨⎪⎪⎩⎪⎪2、CP F1F2F312--5AA B B C C i-1i-1S i C i12—61、状态转换表000111100001111000011110Qe =0D aQ a Q b c Q Q d Q e =1100011101注：卡诺图中的空格为约束项。
计量经济学导论CH12习题答案
CHAPTER 12TEACHING NOTESMost of this chapter deals with serial correlation, but it also explicitly considers heteroskedasticity in time series regressions. The first section allows a review of what assumptions were needed to obtain both finite sample and asymptotic results. Just as with heteroskedasticity, serial correlation itself does not invalidate R-squared. In fact, if the data are stationary and weakly dependent, R-squared and adjusted R-squared consistently estimate the population R-squared (which is well-defined under stationarity).Equation (12.4) is useful for explaining why the usual OLS standard errors are not generally valid with AR(1) serial correlation. It also provides a good starting point for discussing serial correlation-robust standard errors in Section 12.5. The subsection on serial correlation with lagged dependent variables is included to debunk the myth that OLS is always inconsistent with lagged dependent variables and serial correlation. I do not teach it to undergraduates, but I do to master’s students.Section 12.2 is somewhat untraditional in that it begins with an asymptotic t test for AR(1) serial correlation (under strict exogeneity of the regressors). It may seem heretical not to give the Durbin-Watson statistic its usual prominence, but I do believe the DW test is less useful than the t test. With nonstrictly exogenous regressors I cover only the regression form of Durbin’s test, as the h statistic is asymptotically equivalent and not always computable.Section 12.3, on GLS and FGLS estimation, is fairly standard, although I try to show how comparing OLS estimates and FGLS estimates is not so straightforward. Unfortunately, at the beginning level (and even beyond), it is difficult to choose a course of action when they are very different.I do not usually cover Section 12.5 in a first-semester course, but, because some econometrics packages routinely compute fully robust standard errors, students can be pointed to Section 12.5 if they need to learn something about what the corrections do. I do cover Section 12.5 for a master’s level course in applied econometrics (after the first-semester course).I also do not cover Section 12.6 in class; again, this is more to serve as a reference for more advanced students, particularly those with interests in finance. One important point is that ARCH is heteroskedasticity and not serial correlation, something that is confusing in many texts. If a model contains no serial correlation, the usual heteroskedasticity-robust statistics are valid. I have a brief subsection on correcting for a known form of heteroskedasticity and AR(1) errors in models with strictly exogenous regressors.100101SOLUTIONS TO PROBLEMS12.1 We can reason this from equation (12.4) because the usual OLS standard error is anestimate of σthe AR(1) parameter, ρ, tends to be positive in time series regression models. Further, the independent variables tend to be positive correlated, so (x t - x )(x t +j - x ) – which is what generally appears in (12.4) when the {x t } do not have zero sample average – tends to be positive for most t and j . With multiple explanatory variables the formulas are more complicated but have similar features.If ρ < 0, or if the {x t } is negatively autocorrelated, the second term in the last line of (12.4)could be negative, in which case the true standard deviation of 1ˆβis actually less than σ12.2 This statement implies that we are still using OLS to estimate the βj . But we are not using OLS; we are using feasible GLS (without or with the equation for the first time period). In other words, neither the Cochrane-Orcutt nor the Prais-Winsten estimators are the OLS estimators (and they usually differ from each other).12.3 (i) Because U.S. presidential elections occur only every four years, it seems reasonable to think the unobserved shocks – that is, elements in u t – in one election have pretty much dissipated four years later. This would imply that {u t } is roughly serially uncorrelated.(ii) The t statistic for H 0: ρ = 0 is -.068/.240 ≈ -.28, which is very small. Further, theestimate ˆρ= -.068 is small in a practical sense, too. There is no reason to worry about serial correlation in this example.(iii) Because the test based on ˆt ρ is only justified asymptotically, we would generally be concerned about using the usual critical values with n = 20 in the original regression. But any kind of adjustment, either to obtain valid standard errors for OLS as in Section 12.5 or a feasible GLS procedure as in Section 12.3, relies on large sample sizes, too. (Remember, FGLS is not even unbiased, whereas OLS is under TS.1 through TS.3.) Most importantly, the estimate of ρ ispractically small, too. With ˆρso close to zero, FGLS or adjusting the standard errors would yield similar results to OLS with the usual standard errors.12.4 This is false, and a source of confusion in several textbooks. (ARCH is often discussed as a way in which the errors can be serially correlated.) As we discussed in Example 12.9, the errors in the equation return t = β0 + β1return t-1 + u t are serially uncorrelated, but there is strong evidence of ARCH; see equation (12.51).12.5 (i) There is substantial serial correlation in the errors of the equation, and the OLS standarderrors almost certainly underestimate the true standard deviation in ˆEZβ. This makes the usual confidence interval for βEZ and t statistics invalid.102(ii) We can use the method in Section 12.5 to obtain an approximately valid standard error.[See equation (12.43).] While we might use g = 2 in equation (12.42), with monthly data we might want to try a somewhat longer lag, maybe even up to g = 12.12.6 With the strong heteroskedasticity in the errors it is not too surprising that the robuststandard error for 1ˆβ differs from the OLS standard error by a substantial amount: the robust standard error is almost 82% larger. Naturally, this reduces the t statistic. The robust t statistic is .059/.069≈ .86, which is even less significant than before. Therefore, we conclude that, once heteroskedasticity is accounted for, there is very little evidence that return t-1 is useful for predicting return t .SOLUTIONS TO COMPUTER EXERCISES12.7 Regressing ˆt uon 1ˆt u -, using the 69 available observations, gives ˆρ≈ .292 and se(ˆρ) ≈ .118. The t statistic is about 2.47, and so there is significant evidence of positive AR(1) serial correlation in the errors (even though the variables have been differenced). This means we should view the standard errors reported in equation (11.27) with some suspicion.12.8 (i) After estimating the FDL model by OLS, we obtain the residuals and run the regressionˆt uon 1ˆt u -, using 272 observations. We get ˆρ≈ .503 and ˆt ρ≈ 9.60, which is very strong evidence of positive AR(1) correlation.(ii) When we estimate the model by iterated C-O, the LRP is estimated to be about 1.110.(iii) We use the same trick as in Problem 11.5, except now we estimate the equation by iterated C-O. In particular, writegprice t = α0 + θ0gwage t + δ1(gwage t -1 – gwage t ) + δ2(gwage t-2 – gwage t )+ + δ12(gwage t -12 – gwage t ) + u t ,Where θ0 is the LRP and {u t } is assumed to follow an AR(1) process. Estimating this equationby C-O gives 0ˆθ≈ 1.110 and se(0ˆθ)≈ .191. The t statistic for testing H 0: θ0 = 1 is (1.110 – 1)/.191≈ .58, which is not close to being significant at the 5% level. So the LRP is not statistically different from one.12.9 (i) The test for AR(1) serial correlation gives (with 35 observations) ˆρ≈ –.110, se(ˆρ)≈ .175. The t statistic is well below one in absolute value, so there is no evidence of serial correlation in the accelerator model. If we view the test of serial correlation as a test of dynamic misspecification, it reveals no dynamic misspecification in the accelerator model.(ii) It is worth emphasizing that, if there is little evidence of AR(1) serial correlation, there is no need to use feasible GLS (Cochrane-Orcutt or Prais-Winsten).10312.10 (i) After obtaining the residuals ˆt ufrom equation (11.16) and then estimating (12.48), we can compute the fitted values ˆth = 4.66 – 1.104 return t for each t . This is easily done in a single command using most software packages. It turns out that 12 of 689 fitted values are negative. Among other things, this means we cannot directly apply weighted least squares using the heteroskedasticity function in (12.48).(ii) When we add 21t return - to the equation we get2ˆi u= 3.26 - .789 return t-1 + .297 21t return - + residual t (0.44) (.196) (.036)n = 689, R 2 = .130.So the conditional variance is a quadratic in return t -1, in this case a U-shape that bottoms out at .789/[2(.297)] ≈ 1.33. Now, there are no fitted values less than zero.(iii) Given our finding in part (ii) we can use WLS with the ˆth obtained from the quadratic heteroskedasticity function. When we apply WLS to equation (12.47) we obtain 0ˆβ≈ .155 (se ≈ .078) and 1ˆβ≈ .039 (se ≈ .046). So the coefficient on return t-1, once weighted least squares has been used, is even less significant (t statistic ≈ .85) than when we used OLS.(iv) To obtain the WLS using an ARCH variance function we first estimate the equation in(12.51) and obtain the fitted values, ˆth . The WLS estimates are now 0ˆβ≈ .159 (se ≈ .076) and 1ˆβ≈ .024 (se ≈ .047). The coefficient and t statistic are even smaller. Therefore, once we account for heteroskedasticity via one of the WLS methods, there is virtually no evidence that E(return t |return t -1) depends linearly on return t -1.12.11 (i) Using the data only through 1992 givesdemwins= .441 - .473 partyWH + .479 incum + .059 partyWH ⋅gnews (.107) (.354) (.205) (.036)- .024 partyWH ⋅inf (.028)n = 20, R 2 = .437, 2R = .287.The largest t statistic is on incum , which is estimated to have a large effect on the probability of winning. But we must be careful here. incum is equal to 1 if a Democratic incumbent is running and –1 if a Republican incumbent is running. Similarly, partyWH is equal to 1 if a Democrat is currently in the White House and –1 if a Republican is currently in the White House. So, for an incumbent Democrat running, we must add the coefficients on partyWH and incum together, and this nets out to about zero.104The economic variables are less statistically significant than in equation (10.23). The gnews interaction has a t statistic of about 1.64, which is significant at the 10% level against a one-sided alternative. (Since the dependent variable is binary, this is a case where we must appeal to asymptotics. Unfortunately, we have only 20 observations.) The inflation variable has the expected sign but is not statistically significant.(ii) There are two fitted values less than zero, and two fitted values greater than one.(iii) Out of the 10 elections with demwins = 1, 8 of these are correctly predicted. Out of the 10 elections with demwins = 0, 7 are correctly predicted. So 15 out of 20 elections through 1992 are correctly predicted. (But, remember, we used data from these years to obtain the estimated equation.)(iv) The explanatory variables are partyWH = 1, incum = 1, gnews = 3, and inf = 3.019. Therefore, for 1996,demwins= .441 - .473 + .479 + .059(3) - .024(3.019) ≈ .552.Because this is above .5, we would have predicted that Clinton would win the 1996 election, as he did.(v) The regression of ˆt uon 1ˆt u - produces ˆρ ≈ -.164 with heteroskedasticity-robust standard error of about .195. (Because the LPM contains heteroskedasticity, testing for AR(1) serial correlation in an LPM generally requires a heteroskedasticity-robust test.) Therefore, there is little evidence of serial correlation in the errors. (And, if anything, it is negative.)(vi) The heteroskedasticity-robust standard errors are given in [⋅] below the usual standard errors:demwins= .441 - .473 partyWH + .479 incum + .059 partyWH ⋅gnews (.107) (.354) (.205) (.036)[.086] [.301] [.185] [.030]– .024 partyWH ⋅inf(.028) [.019]n = 20, R 2 = .437, 2R = .287.In fact, all heteroskedasticity-robust standard errors are less than the usual OLS standard errors, making each variable more significant. For example, the t statistic on partyWH ⋅gnews becomes about 1.97, which is notably above 1.64. But we must remember that the standard errors in the LPM have only asymptotic justification. With only 20 observations it is not clear we should prefer the heteroskedasticity-robust standard errors to the usual ones.10512.12 (i) The regression ˆt uon 1ˆt u - (with 35 observations) gives ˆρ≈ -.089 and se(ˆρ)≈ .178; there is no evidence of AR(1) serial correlation in this equation, even though it is a static model in the growth rates.(ii) We regress gc t on gc t-1 and obtain the residuals ˆt u. Then, we regress 2ˆt u on gc t -1 and 21t gc -(using 35 observations), the F statistic (with 2 and 32 df ) is about 1.08. The p -value isabout .352, and so there is little evidence of heteroskedasticity in the AR(1) model for gc t . This means that we need not modify our test of the PIH by correcting somehow for heteroskedasticity.12.13 (i) The iterated Prais-Winsten estimates are given below. The estimate of ρ is, to three decimal places, .293, which is the same as the estimate used in the final iteration of Cochrane-Orcutt:log()chn imp = -37.08 + 2.94 log(chempi ) + 1.05 log(gas ) + 1.13 log(rtwex )(22.78) (.63) (.98) (.51)- .016 befile6 - .033 affile6 - .577 afdec6(.319) (.322) (.342)n = 131, R 2 = .202(ii) Not surprisingly, the C-O and P-W estimates are quite similar. To three decimal places,they use the same value of ˆρ(to four decimal places it is .2934 for C-O and .2932 for P-W). The only practical difference is that P-W uses the equation for t = 1. With n = 131, we hope this makes little difference.12.14 (i) This is the model that was estimated in part (vi) of Computer Exercise 10.17. Aftergetting the OLS residuals, ˆt u , we run the regression 1垐 on ,2,...,108.t t u u t -= (Included anintercept, but that is unimportant.) The coefficient on 1ˆt u- is ˆρ=.281 (se = .094). Thus, there is evidence of some positive serial correlation in the errors (t ≈ 2.99). I strong case can be made that all explanatory variables are strictly exogenous. Certainly there is no concern about the time trend, the seasonal dummy variables, or wkends , as these are determined by the calendar. It is seems safe to assume that unexplained changes in prcfat today do not cause future changes in the state-wide unemployment rate. Also, over this period, the policy changes were permanent once they occurred, so strict exogeneity seems reasonable for spdlaw and beltlaw . (Given legislative lags, it seems unlikely that the dates the policies went into effect had anything to do with recent, unexplained changes in prcfat .(ii) Remember, we are still estimating the βj by OLS, but we are computing different standard errors that have some robustness to serial correlation. Using Stata 7.0, I get垐.0671, se().0267spdlaw spdlaw ββ== and 垐.0295, se().0331beltlaw beltlaw ββ=-=. The t statistic forspdlaw has fallen to about 2.5, but it is still significant. Now, the t statistic on beltlaw is less than one in absolute value, so there is little evidence that beltlaw had an effect on prcfat .(iii) For brevity, I do not report the time trend and monthly dummies. The final estimate of ρρ=is ˆ.289:prcf at= 1.009 + … + .00062 wkends-.0132 unem(.102) (.00500) (.0055)+ .0641 spdlaw -.0248 beltlaw(.0268) (.0301)n = 108, R2 = .641There are no drastic changes. Both policy variable coefficients get closer to zero, and the standard errors are bigger than the incorrect OLS standard errors [and, coincidentally, pretty close to the Newey-West standard errors for OLS from part (ii)]. So the basic conclusion is the same: the increase in the speed limit appeared to increase prcfat, but the seat belt law, while it is estimated to decrease prcfat, does not have a statistically significant effect.12.15 (i) Here are the OLS regression results:avgprc=-.073 -.0040 t- .0101 mon- .0088 tues + .0376 wed + .0906 thurs log()(.115) (.0014) (.1294) (.1273) (.1257) (.1257)n = 97, R2 = .086The test for joint significance of the day-of-the-week dummies is F = .23, which gives p-value = .92. So there is no evidence that the average price of fish varies systematically within a week.(ii) The equation isavgprc=-.920 -.0012 t- .0182 mon- .0085 tues + .0500 wed + .1225 thurs log()(.190) (.0014) (.1141) (.1121) (.1117) (.1110)+ .0909 wave2 + .0474 wave3(.0218) (.0208)n = 97, R2 = .310Each of the wave variables is statistically significant, with wave2 being the most important. Rough seas (as measured by high waves) would reduce the supply of fish (shift the supply curve back), and this would result in a price increase. One might argue that bad weather reduces the demand for fish at a market, too, but that would reduce price. If there are demand effects captured by the wave variables, they are being swamped by the supply effects.106107 (iii) The time trend coefficient becomes much smaller and statistically insignificant. We can use the omitted variable bias table from Chapter 3, Table 3.2 (page 92) to determine what is probably going on. Without wave2 and wave3, the coefficient on t seems to have a downward bias. Since we know the coefficients on wave2 and wave3 are positive, this means the wave variables are negatively correlated with t . In other words, the seas were rougher, on average, at the beginning of the sample period. (You can confirm this by regressing wave2 on t and wave3 on t .)(iv) The time trend and daily dummies are clearly strictly exogenous, as they are just functions of time and the calendar. Further, the height of the waves is not influenced by past unexpected changes in log(avgprc ).(v) We simply regress the OLS residuals on one lag, getting ˆ垐.618,se().081,7.63.t ρρρ===Therefore, there is strong evidence of positive serial correlation.(vi) The Newey-West standard errors are 23垐se().0234 and se().0195.wave wave ββ== Given thesignificant amount of AR(1) serial correlation in part (v), it is somewhat surprising that these standard errors are not much larger compared with the usual, incorrect standard errors. In fact,the Newey-West standard error for 3ˆwave βis actually smaller than the OLS standard error.(vii) The Prais-Winsten estimates arelog()avgprc = -.658 - .0007 t + .0099 mon + .0025 tues + .0624 wed +.1174 thurs (.239) (.0029) (.0652) (.0744) (.0746) (.0621)+ .0497 wave2 + .0323 wave3(.0174) (.0174)n = 97, R 2 = .135The coefficient on wave2 drops by a nontrivial amount, but it still has a t statistic of almost 3. The coefficient on wave3 drops by a relatively smaller amount, but its t statistic (1.86) is borderline significant. The final estimate of ρ is about .687.。
现代移动通信 蔡跃明 第三版思考题与习题参考答案 chapter_12
第十二章思考题与习题1.第四代移动通信系统和第一、第二和第三代移动通信系统有何不同？答：第四代移动通信系统具有以下特征：（1）第四代移动通信系统将是实时、宽带以及无缝覆盖的全IP多媒体通信系统。
它不是现在概念中的第一、二、三代，而是广泛用于各种电信环境的无线系统的总和，包括蜂窝系统、WLAN、无线广播系统等，支持各种空中接口。
它能够完成各个系统，包括无线LAN、室外宽带接入系统、2G、3G等之间的平滑切换。
用户可以根据不同系统自主选择，也可以自适应选择。
（2）第四代移动通信系统应该是符合全IP发展趋势的多媒体通信系统，核心网将基于IPv6，可以与Internet互连互通，并且承载与控制全程分离。
（3）第四代移动通信应该是适合于分组突发业务的系统，在步行环境中、车速环境中的峰值传输速率应分别达到1 Gbps和100 Mbps，适用于大动态范围业务（8 kbps～100 Mbps），频谱效率远远大于3G系统。
2.简述无线资源管理的基本原理，并说明它包括哪些基本部分？答：无线资源管理（RRM，Radio Resource Management）就是对移动通信系统的空中资源的规划和调度，它是在用户动态需求、信道动态时变和用户位置动态变化的环境下，对能量资源（如信号功率、能量）、时间资源（如时隙、业务帧、导频符号等）、频率资源（如信号带宽、保护频段、调制模式等）和空间资源（如天线角度、天线位置等）4类资源进行高效利用，以实现通信系统性能优化的技术。
无线资源管理的核心问题是在保证服务质量（QoS）的前提下，提高频谱利用率，其基本出发点在网内业务量和时延分布不均匀、且信道的状态因信号衰落和干扰而变化的状况时，动态分配和调整可用的资源。
3.举例说明MIMO技术利用多径衰落来改善系统性能的原理。
答：在这里我们就以两根天线发送和两根天线接收所组成的MIMO系统为例来说明多径衰落是怎样来改善系统的性能的，如下图：需要发送的信号经过串并转换把信号分成两路（这里假设采用的是水平编码），这两路先通过串并转换在进行编码，然后把编码后的信号经天线发送出去，由于天线之间的距离足够远，两天线所发送的信号是独立的，接收天线也满足独立分集的条件。
克鲁格曼-国际经济学理论与政策 (第七版)Chapter 12 课后习题答案
Chapter 12 第三题和第五题练习提示3. (a)Credit Debit An American buys a share of German stock(Financial account, U.S. asset import) -The American pays with a check on his Swiss bank account(Financial account, U.S. asset import) +(b) If the German stock seller deposits the U.S. check in its German bank,Credit Debit An American buys a share of German stock(Financial account, U.S. asset import) -The American pays with a check on his American bank account(Financial account, U.S. asset export) +(c)Credit Debit The sale of dollars by the Korean government(Financial account, U.S. asset export) -The Korean citizens who buy the dollars use them to buy American goods(Current account, U.S. goods export) +Credit Debit The sale of dollars by the Korean government(Financial account, U.S. asset export) -The Korean citizens who buy the dollars use them to buy American assets(Financial account, U.S. asset export) +(d) Suppose the company issuing the traveler’s check uses a checking account in France to make payments,Credit Debit The company issuing the traveler’s check pays the French restaurateur for the meal (Current account, U.S. service import) - Sale of claim on the company issuing the traveler’s check(Financial account, U.S. assets export) +(e)Credit Debit The California winemaker contributes a case of cabernet sauvignon abroad(Current account, U.S. unilateral current transfers) - Receivable of the California winemaker(Current account, U.S. goods export) +Credit Debit Receivable of the California winemaker(Current account, U.S. goods export) -The California winemaker contributes a case of cabernet sauvignon abroad(Current account, U.S. unilateral current transfers) +(f)Credit Debit The U.S. owned factory in Britain makes local earning(Current account, U.S. income receipts) +The U.S. owned factory in Britain deposits its local earning in a British bank(Financial account, U.S. asset import) -Credit Debit The U.S. owned factory in Britain uses its local earning to reinvest(Current account, U.S. income receipts) -The U.S. owned factory in Britain makes the payment for reinvestment(Financial account, U.S. asset import) +5.(a) Since Pecunia had a current account deficit of $1b and a nonreserve financial account surplus of $500m in 2002, the balance of Pecunia’s official reserve transaction should be +$500m as follow:Pecunia international transactionCredit Debit Current account -$1b Financial accountThe balance of Pecunia’s official reserve transaction +$500mThe balance of nonreserve assets +$500mThe balance of payment of Pecunia = the negative value of the balance of Pecunia’s official reserve transaction= -$500m.Pecunia had a financial account surplus of $1b in 2002; it implies Pecunia’s net foreign assets decreased by $1b in 2002.(b) Pecunian central bank had to sell $500m, so Pecunian central bank’s foreign reserves decreased by $500m:Pecunia international transactionCredit Debit Current account -$1b Financial accountPecunian official reserve assets +$500mForeign official reserve assets 0 0The balance of nonreserve assets +$500m(c) There was no need for Pecunian central bank to sell dollar, and Pecunian central bank’s foreign reserves increased by $100m as shown below:Pecunia international transactionCredit Debit Current account -$1b Financial accountPecunian official reserve assets -$100m Foreign official reserve assets + $600mThe balance of nonreserve assets +$500m(d)Pecunia international transactionCredit Debit Current account -$1b Financial accountPecunian official reserve assets -$100m Foreign official reserve assets + $600mThe balance of nonreserve assets +$500mThe following is for your reference:3.(a) The purchase of the German stock is a debit in the U.S. financial account. There is acorresponding credit in the U.S. financial account when the American pays witha check on his Swiss bank account because his claims on Switzerland fall by theamount of the check. This is a case in which an American trades one foreign assetfor another.(b) Again, there is a U.S. financial account debit as a result of the purchase of a Germanstock by an American. T he corresponding credit in this case occurs when theGerman seller deposits the U.S. check in its German bank and that bank lends themoney to a German importer (in which case the credit will be in the U.S. currentaccount) o r to an individual or corporation that purchases a U.S. asset (in whichcase the credit will be in the U.S. financial account). Ultimately, there will be someaction taken by the bank which results in a credit in the U.S. balance of payments.(c) The foreign exchange intervention by the French government involves the sale of aU.S. asset, the dollars it holds in the United States, and thus represents a debititem in the U.S. financial account. The French citizens who buy the dollars mayuse them to buy American goods, which would be an American current accountcredit,or an American asset, which would be an American financial accountcredit.(d) Suppose the company issuing the traveler’s check uses a checking account inFrance to make payments. When this company pays the French restaurateur for themeal, its payment represents a debit in the U.S. current account.The company issuing the traveler’s check must sell assets (deplete its checking account in France) to make this payment. This reduction in the French assetsowned by that company represents a credit in the American financial account.(e) There is no credit or debit in either the financial or the current account sincethere has been no market transaction.(f) There is no recording in the U.S. Balance of Payments of this offshore transaction.5.(a) Since non-central bank financial inflows fell short of the current-account deficit by$500 million, the balance of payments of Pecunia (official settlements balance) was–$500 million. The country as a whole somehow had to finance its $1 billioncurrent-account deficit, so Pecunia’s net foreign assets fell by $1 billion.(b) By dipping into its foreign reserves, the central bank of Pecunia financed theportion of the country’s current-account deficit not covered by private financialinflows. Only if foreign central banks had acquired Pecunian assets could thePecunian central bank have avoided using$500 million in reserves to complete the financing of the current account. Thus,Pecunia’s central bank lost $500 million in reserves, which would appear as anofficial financial inflow (of the same magnitude) in the country’s balance ofpayments accounts.(c) If foreign official capital inflows to Pecunia were $600 million, the Central Banknow increased its foreign assets by $100 million. Put another way, the countryneeded only $1 billion to cover its current-account deficit, but $1.1 billion flowed into the country (500 million private and600 million from foreign central banks). The Pecunian central bank must, therefore, have used the extra $100 million in foreign borrowing to increase its reserves. The balance of payments is still –500 million, but this is now comprised of 600 million in foreign Central Banks purchasing Pecunia assets and 100 million of Pecunia’s Central Bank purchasing foreign assets, as opposed to Pecunia selling 500 million in assets. Purchases of Pecunian assets by foreign central banks enter their countries’balance of payments accounts as outflows, which are debit items. The rationale is that the transactions result in foreign payments to the Pecunians who sell the assets.(d) Along with non-central bank transactions, the accounts would show an increase inforeign official reserve assets held in Pecunia of $600 million (a financial account credit, or inflow) and an increase Pecunian official reserve assets held abroad of $100 million (a financial account debit, or outflow). Of course, total net financial inflows of $1 billion just cover the current-account deficit.。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
总目录 章目录 返回
上一页 下一页
12.2.6 求图12.04所示电路的uo与ui的运算关系式。
RF R R uo1 R/2 R1 R2
u _i
_
∞
_
∞
uo2 uo
_
总目录 章目录 返回
上一页 下一页
12.2.8 在图12.06中，已知RF= 2R1，uo=-2V，试求输出电压uo。
* 5. 理解电压比较器的工作原理和应用。
总目录 章目录 返回 上一页 下一页
名 称 反 相 比 例 1
R1
+ ui – R2
电路
RF
+ uo+ –
输入输出的关系
闭环电压放 大倍数
–
+
2 同 相 比 例 R1 u–
RF
+ u –
3
R
u+
– + +
RF – + +
+ uo –
uo (1
u dt
i
总目录 章目录 返回
上一页 下一页
名 称 微 分 8
电路
输入输出的关系
闭环电压 放大倍数
RF C1 – + R2 R1
+ ui –
uo RF C1
dui dt
+ ui + UR R2 – –
– + +
+
uo –
+
+ uo –
总目录 章目录 返回 上一页 下一页
名 称 电 压 比 较 器 9
)ui （
）
R1
)（
R3 R3 R2
）ui
R2
–
18 20
ui 5.4ui
总目录 章目录 返回
上一页 下一页
12.2.5 电路如图12.03所示，已知ui1＝1V，ui2＝2V，ui3 ＝2V，ui4＝2V，R1＝R2＝2kΩ，R3＝R4＝RF＝1kΩ，试计 算输出电压uo。 RF
ui1 ui2 ui3 ui4 R1 R2 R3 R4
RF + ui1 + ui2 R2 – – R1 – + + R3
总目录 章目录 返回
+ uo –
上一页 下一页
(1)根据 uo = Auo(u+– u–)得
1 3 1 0 (u u )
5
K 2 0 lg (V o / V i ) 5 Au o 1 0
5
u u 1 .3 1 0 V
4
( 2 ) I m ax
u Байду номын сангаас rid
1 .3 1 0 2 *10
_
∞
u _o
总目录 章目录 返回
上一页 下一页
12.2.5 电路如图12.03所示，已知ui1＝1V，ui2＝2V，ui3 ＝2V，ui4＝2V，R1＝R2＝2kΩ，R3＝R4＝RF＝1kΩ，试计 算输出电压uo。 RF
ui1 ui2 ui3 ui4 R1 R2 R3 R4
_
∞
u _o
上一页 下一页
名 称 减 法 运 算 6
电路
输入输出的关系
闭环电压 放大倍数
RF C1 – + R2
uo (1 RF ) R3 ui 2 RF R1
+ ui –
7
+ CF
uo –
+
R1 R2 R3
ui1
积 分
+ ui –
R1 R2
– +
+
+ uo –
uo
1 R1C F
+ uo –
总目录 章目录 返回
上一页 下一页
12.2.2 在图12.01的同相比例运算电路中，已知R1=2kΩ， RF=10kΩ，R2=2kΩ，R3=18kΩ，ui＝1V，求uo？
RF R1 R3 R3 R2
RF + + + ui – + uo – R3
uo (1 (1 RF R1 (1 5)
第12章 集成运算放大器
本章要 求 1. 了解集成运放的基本组成及主要参数的意义。
2.理解运算放大器的条件。
3. 理解运算放大器的电压传输特性 线性区分析的依据：虚短、虚断 饱和区：虚断 *4. 同相比例运算、反相比例运算、同相加法、 反相加法、减法、微分和积分运算电路的构成、 工作原理、闭环放大倍数和平衡电阻的推导及应 用。
电路
输入输出的关系
闭环电压 放大倍数
R1 + ui + UR R2 – –
– + +
+ uo –
10
R1
+ UR + ui R2 – –
– + +
+ uo –
总目录 章目录 返回
上一页 下一页
12.1.1 已知F007运算放大器的开环电压放大倍数Auo= 100 dB，差模 输入电阻rid= 100 MΩ ，最大输出电压Uopp= ±13V 。为了保证工作在 线性区，试求：(1)u+和u-的最大允许差值；(2)输入端电流的最大允 许值。
12.2.10 图12.08是利用两个运算放大器组成的具有较高输入电阻的差 分放大电路。试求出输出电压uo与ui1，ui2的运算关系式。
R1/K R1 KR2 R2 ui2
_ _
∞
ui1
_
∞
_
uo
总目录 章目录 返回
上一页 下一页
12.2.11在图12.2.5所示的差分运算电路中，R1＝R2＝4kΩ， RF＝R3＝20kΩ，ui1＝1.5V，ui2＝25V，试求输出电压uo。
RF R1
)ui
Au f 1
RF R1
R1
uo (1
+ uo –
RF R1
)ui （
R3 R3 R2
）
+ ui –
R2 R3
(1
RF R1
)（
R3 R3 R2
）ui
上一页 下一页
总目录 章目录 返回
名 称 同 相 加 法 4
电路
输入输出的关系
闭环电压 放大倍数
ui2 ui1
4
6
= 6 .5 1 0
11
A
总目录 章目录 返回
上一页 下一页
12.2.1 在图12.2.1的反相比例运算电路中，设R1=10 kΩ ，RF=500kΩ 。 试求闭环电压放大倍数和平衡电阻R2。若ui=10mV，则uo为多少？
RF + ui – R1 R2 – + +
平衡电阻R2 = R1 // RF =9.83KΩ
R12 R11
RF
R2
5
– +
uo (
RF R11
ui1
RF R12
ui 2 )
RF R1
+
+ uo –
反 相 加 法
ui1
ui2
– + +
R21 R22
+ uo –
uo (1
RF R1
)(
R22 R21 R22
ui1
R21 R21 R22
ui 2 )
总目录 章目录 返回
RF
_
∞
_
ui
uo1
R1 R2
_
∞
_o
u
总目录 章目录 返回
上一页 下一页
12.2.9 求图12.07所示电路中uo与各输入电压的运算关系 式。
10kΩ 10kΩ ui1 1kΩ 910Ω
_
∞
ui2 ui3
10kΩ 5kΩ 2kΩ
_
∞
1.1kΩ
uo _
总目录 章目录 返回
上一页 下一页