【数据库】ch6-theory-2

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Theory_2_solutions

Copyright notice© 2006-2008 byWired Communications Group (Fachgebiet Leitungsgebundene Übertragungstechnik) Institute for Communications Engineering (LNT)Technische Universität München (TUM)D-80290 MünchenGermanyPhone: +49 89 23472ContentsTheoretical Background (4)1. Mach-Zehnder Modulators (4)1.1. Physical Background and Mathematical Modeling (4)1.2. Pulse shaping and Modulation (8)1.2.1. NRZ (8)1.2.2. RZ (9)1.2.3. CSRZ (13)1.2.4. Duobinary (13)Bibliography (14)Theoretical BackgroundAs we have seen in the first session of this simulation lab, it is difficult to generate intensity modulation by direct modulation of a semiconductor laser for high-speed communications (e.g. > 5 Gbit/s) due to the limited bandwidth of direct modulation. This problem can be solved if an additional optical component is introduced. The laser is then operated with a constant injection current, simply generating a constant light carrier. This carrier is modulated in a new device, the so-called external modulator . The most frequently used external modulator is applying a Mach-Zehnder interferometer structure, and is therefore called Mach-Zehnder Modulator (MZM). This device is investigated in the following.1. Mach-Zehnder Modulators1.1. Physical Background and Mathematical ModelingIn a MZM the incoming non-modulated carrier according to ˆ()c j t in E t E e ω=⋅ is first split up into two optical paths (see Fig. 1).Figure 1: General structure of a Mach-Zehnder Modulator (dual-drive configuration)The incoming light power is split up in the left Y-branch according to the power splittingratios 21κ and 211κ−, i.e. 11ˆˆ(0)in Ez E κ==⋅, ˆˆ(0)E z E ==. The waveguide is made of a special material showing a strong electro-optic effect, typically Lithium-Niobate (LiNbO 3). Due to this electro-optical effect the refractive index of LiNbO 3changes with an externally applied electrical voltage u(t) according to0()().EO n t n u t ς=+⋅ (2.1)Here n 0 is the refractive index without external field, EO ς models the strength of the electro-optical effect. As a consequence, the phase velocity of a propagating electrical wave is modified as /n c c n =, and thus the wave number k changes according to ()/n k c n ω=⋅, where c is the speed of light in vacuum. Therefore the phase shift of the waves propagating in branch 1 and 2 of the waveguide can be tuned by the applied voltages u 1(t) and u 2(t), respectively, resulting in111ˆˆ()(0)j E z L E z e ϕ−∆===⋅ and 2ˆˆ()(0)j E z L E z e ϕ−∆===⋅, (2.2)Session 22-5with ()11n k u L ϕ∆=⋅and ()22n k u L ϕ∆=⋅, respectively. The attenuation of the waveguidesis neglected for the sake of simplicity. 1ˆEand ˆE are then combined at the output of the device according to212ˆˆˆ()().out E E z L E z L κ=⋅=+= (2.3)22κ and 221κ− are the power splitting ratios of the Y-branch on the right-hand side inFig. 1. This results finally in the output field ()out E t according to1212ˆ()c j t j j out E t E e e e ωϕϕκκ−∆−∆ =⋅⋅+⋅. (2.4)Under the condition of symmetrical splitting, i.e. 221212κκ==, we get121ˆ().2c j t j j in E t E e e e ωϕϕ−∆−∆ =⋅+⋅ (2.5)It can be seen from (2.5) that, depending on the relative phases ∆ϕ1 and ∆ϕ2, neither amplitude nor phase of the outcoming field are modified. This becomes more obvious after a simple modification of (2.5), resulting in121212122221221ˆ()2ˆ cos .2c c j j j j t out j j t in E t E e e e e E e e ϕϕϕϕϕϕωϕϕωϕϕ∆−∆∆−∆∆+∆−−∆+∆− =⋅+⋅⋅= ∆−∆ =⋅⋅⋅(2.6)There is a linear relation between the phase shift difference ∆ϕ1 - ∆ϕ2 and the voltage difference u 1-u 2. Also, it is important to notice that these quantities are usually time-dependent. This leads us to()()()()()1212.u t u t t t u t V V πππϕϕπ−∆−∆=⋅=⋅ (2.7)The parameter V π denotes the voltage difference u(t) = u 1(t)-u 2(t) that generates a relative phase shift of π between upper and lower branch of the MZM. The phase term in (2.6) is now re-written as()()()()121222,u t u t t t j jV eeππϕϕ− ∆+∆−Θ⋅ −= where ()()()()()()()()12121212t t u t u t t t u t u t ϕϕϕϕ∆+∆+Θ==∆−∆−. (2.8)Substituting (2.7) and (2.8) into (2.6) finally results in()()()()12212ˆ()cos .2c u t u t j V j t out in u t u t E t E ee V ππωππ−−Θ⋅⋅− =⋅⋅⋅⋅(2.9)Lab on Simulation of Optical Communication Systems2-6 The parameter Θ is called the asymmetry factor of the dual-drive MZM. To make Θ a time-independent constant, u 1(t) and u 2(t) can no longer be selected independently, but are constrained by()()12()()()1,()1,22u t u t u t u t =⋅Θ+=⋅Θ− (2.10)with the definition: 1 1.−≤Θ≤The way the selection of the asymmetry factor Θ acts on u 1 and u 2 is depicted in Figure 2.Figure 2: Effect of the asymmetry factor ΘAs we can see from (2.9), if Θ = 0 is selected, i.e. u 1(t) = u(t)/2, u 2(t) = -u(t)/2, the modulator generates pure amplitude (or intensity) modulation without any time-dependent modification of the signal phase. If asymmetry is introduced by selecting Θ ≠ 0, the intensity modulation is not modified; however, an additional phase modulation is generated. This parasitical, time-dependent phase modulation modifies the instantaneous frequency of the light wave according to()().2c du t t V dtππωωΘ=−⋅⋅ (2.11)As we know from Session 1 of this course, this frequency chirp, i.e. the deviation ∆f(t) = ∆ω(t) / 2π of the instantaneous signal from the carrier frequency, has a negative influence on the propagating signal due to the chromatic dispersion of the fiber. Therefore, strong efforts are made generally to reduce this chirp to negligible values. As we could see above, selecting Θ = 0, i.e. u 2(t) = -u 1(t), results in chirp-free intensity modulation. This operation of the dual-drive MZM is called push-pull mode .Another solution is a re-design of the Mach-Zehnder modulator: If the poles of the applied voltage are changed according to Figure 3, the condition u 2(t) = -u 1(t) is met automatically. In this single-drive configuration the voltage difference u(t) is directly impressed on the modulator. In this session, we will only be using dual-drive MZMs.Session 22-7Figure 3: Single-drive Mach-Zehnder ModulatorIf the dual-drive MZM is operated in chirp-free mode, the electrical output field is simplygiven as()ˆˆ()cos ().2c c j t j t out inout u t E t E e E t e V ωωππ =⋅⋅⋅=⋅(2.12)From (2.12) it can be seen that the amplitude ˆ()E t depends on the input voltage u(t) according to a cosine function. The power P(t) of the modulated carrier is therefore calculated as222()ˆˆ()()()cos .2out in u t P t E t E t V ππ=⋅⋅ ∼(2.13)The curves relating ˆoutE and P with u are depicted in Figures 4 and 5. Figure 4: ˆoutE (normalized) as a function of input voltage u 1-1-1 1 2 3 41X X X 23Lab on Simulation of Optical Communication Systems2-8Figure 5: P (normalized) as a function of input voltage uThe operation point of the MZM is defined by applying a specific bias voltage. The points 1 (trough ), 2 (quadrature ) and 3 (crest ) marked in Figure 4 are used to generate specific optical modulation schemes. Of course any other adequate operation point could be selected due to the periodicity of the curves.1.2. Pulse shaping and ModulationDespite their simple working principle, Mach-Zehnder modulators are extremely versatile and can be used to create many different pulse shapes and modulation formats at very high bit rates. The curves depicted in Figures 4 and 5 are the key to creating the various transmitter configurations.In this lab session, we will concentrate on the pulse shapes Non-Return-to-Zero (NRZ ), Return-to-Zero (RZ ) with various duty cycles and Carrier-Suppressed Return-to-Zero (CSRZ ). As a modulation format, we will consider On-Off-Keying (OOK ). However, if you have some time left at the end of the session, you can try and use Mach-Zehnder Modulators to set up a DPSK transmitter.1 Lastly, we are going to experiment with a duobinary transmitter. Although duobinary is commonly referred to as a modulation scheme, strictly speaking, it is achieved by a combination of a logical precoder (so-called line coding ) with Amplitude Shift Keying (ASK).21.2.1. NRZA single MZM can be used to produce an NRZ OOK signal. The MZM is biased in the quadrature point, i.e. /2Bias u V π=−. The driving or modulation voltage mod u has a range of V π and produces NRZ pulses corresponding to the binary data stream.3 Table 1 summarizes these parameters. In Figure 6, the modulation voltage is depicted along the time axis (which is drawn shifted by Bias u ) for the bit sequence …010011….1There is a multitude of other pulse shapes (e.g. Chirped RZ – CRZ pulses) and modulation schemes (e.g. multilevel ASK, DPSK, DQPSK, D8-PSK, as well as combinations of ASK and PSK, etc.), whosetransmission properties are in the center of research attention. Some of these advanced transmission systems are already deployed in undersea optical communication links. The transmitter structures covered in this lab session, however, are still by far the most popular in current systems. 2Line coding (for which AMI codes are another example) is a technique used to equip the signal with certain desired properties in the time or frequency domain (spectral shaping ). It must be distinguished from channel coding, where redundancy is added to the signal in order to combat transmission errors at the receiver. 3Bias u can be regarded as the constant offset (bias) of u(t). Hence mod ()()Bias u t u u t =+.1-1 1 2 3 4Session 22-9Table 1: Voltage Parameters of a MZM producing OOK NRZ modulation1.2.2. RZCompared to the simpler NRZ pulses, Return-to-Zero (RZ) pulses have several advantages:• In long sequences of ones , RZ pulses still allow the receiver to recover the signal clock.• For the same pulse energy, the peak power is higher in RZ than in NRZ pulses, resulting in a higher SNR at the sampling instant.• Depending on the system configuration, RZ pulses can yield a better systemperformance than NRZ pulses. It is, however, difficult to make general statements about the performance gain of RZ over NRZ, as this gain depends on many system parameters such as signal power, filter bandwidths, fiber nonlinearities, dispersion1-1uV π-1 1 2 3 4Lab on Simulation of Optical Communication Systems2-10map and data rate. As a rough rule, it can be said that RZ plays out its advantages with increasing data rates and signal powers (i.e. higher fiber nonlinearities) [5]. These advantages, however, come at the cost of an increased transmitter complexity. To generate a signal with RZ pulses, a setup employing two MZMs is used (Figure 7). The first MZM, called the pulse carver (PC), is generating a train of identical pulses, which serve as an input to the second MZM, the data or gating modulator (DM). The purpose of the DM, which is configured as an NRZ OOK modulator as introduced in the previous section, is simply to block or let pass single pulses depending on the data to be transmitted.Figure 7: Setup of RZ transmitterThe pulse carver is driven by a cosine function. Depending on the bias voltage and the modulation voltage’s range and frequency, different types of RZ pulses can be generated. RZ pulses are characterized by their duty cycle , which is the pulse’s full width at half intensity relative to the bit slot duration. RZ33 denotes an RZ pulse with 33% duty cycle; RZ50 has 50% duty cycle. In contrast to these two, RZ67 pulses have an alternating field sign. They are covered in the next section.RZ50 Pulse GenerationTo produce RZ pulses with 50% duty cycle, the PC is driven with the parameters given in Table 2.Table 2: Voltage Parameters of a MZM producing RZ 50 pulsesQuestionWhat are A and B in ()mod cos u A B t =⋅⋅ if we want to produce RZ50 pulses for a 10Gbit/ssignal? Sketch the modulation voltage versus time in Figure 8.mod ()(),bias u t u u t =+ where /2bias u V π= andmod ()/2cos 210 GHz B A u t V t ππ=⋅⋅⋅Figure 8: Generation of RZ50 pulses (E (solid) and P (dashed) normalized)RZ33 Pulse GenerationRZ33 pulses are even narrower in the time domain than RZ50 pulses and accordingly have a broader spectrum. They can be generated using the parameters given in Table 3.Table 3: Voltage Parameters of a MZM producing RZ 33 pulsesBias u 0 (crest)mod u function Cosinemod u frequency Half the bit frequency (e.g. 5 GHz for a 10 Gbit/s signal) mod u range[];V V ππ−1.2.3. CSRZAt V π, the power P is minimum and the electrical field changes its sign. When the bias voltage is set to this point, a pulse carver driven by a cosine function will produce pulses that have alternating field signs. Therefore, the generated signal is DC-free (in the electric field domain). In the frequency spectrum, this property corresponds to an absence of discrete spectral components at the carrier frequency. For that reason, these pulses are called Carrier-Suppressed RZ (CSRZ). The parameters used to create CSRZ pulses are given in Table 4. The duty cycle of CSRZ pulses is 67%, so they are also called RZ67.Table 4: Voltage Parameters of a MZM producing CSRZ pulsesBias u V π (trough)mod u function Cosinemod u frequency Half the bit frequency (e.g. 5 GHz for a 10 Gbit/s signal) mod u range[];V V ππ−CSRZ pulses have the advantage of being quite robust against ISI caused by dispersion. To understand why, imagine two neighboring pulses spreading in time such that they overlap partly. Adjacent CSRZ pulses have contrary signs, so that ideally, they completely cancel. In the lab, this can be observed from a larger eye opening compared to other RZ pulses.1.2.4. DuobinaryDuobinary modulation is, strictly speaking, not a modulation format, but rather 3-level Amplitude Shift Keying (ASK) with a logical precoder. The values transmitted are 0P = (i.e. u V π=) for a logical zero, and 1normalized P = (i.e. 0u = or 2u V π=) for a logical one. The data modulator is therefore driven by a three-level (0,,2u V V ππ=) electrical NRZ signal. The transformation of the binary data sequence to a ternary sequence is achieved by a logical precoder, whose coding rule is as follows: • A logical zero is always coded as 0.• A logical one is coded as +1 or -1. If an odd number of zeros is transmitted, thesign of the following one is changed.The block diagram of a duobinary coder is depicted in Figure 9. Figure 9: Block diagram of duobinary precoderFor the system in Figure 9 to make sense, two conventions are important:The output of the XOR element is bipolar, i.e. a 1 is output as +1, a 0 as -1.The output of the total system will have an absolute amplitude of 2, i.e. to obtainthe duobinary code described above, we would have to divide by 2.The block diagram in Figure 9, however, illustrates the way duobinary signals are generated in OptiSystem . The first dotted box corresponds to the OptiSystem element Binary NOT , the second to the element Duobinary Precoder , and the third box corresponds to the Duobinary Pulse Generator .The example given in Table 5 will help clarify the principle.Table 5: Example of duobinary line codingA 0 1 0 1 1 1 1 0 0 1 1 0 0 0 1 1 0 1B 1 0 1 0 0 0 0 1 1 0 0 1 1 1 0 0 1 0C 1 1 -1 -1 -1 -1 -1 1 -1 -1 -1 1 -1 1 1 1 -1 -1 D0 +2 0 -2 -2 -2 -2 0 0 -2 -2 0 0 0 +2 +2 0 -2The main advantage of duobinary modulation is its dispersion tolerance. In particular, isolated zeros (such as in …101…) cause the sign of the ones to change. Because of the opposed field signs, the neighboring pulses ideally cancel when leaking into the zero bit slot.Duobinary coding is usually used with NRZ pulses. Therefore, only one MZM is required at the transmitter. Because of the broad NRZ pulses used, duobinary signals have a narrow spectrum suitable for WDM systems with a small channel separation, which is another advantage.QuestionWhat bias voltage Bias u do you choose for duobinary modulation? Which values does mod u take (compare with Figure 6)?Bias u V π= (trough){}mod ,0,u V V ππ∈−Bibliography[1] N. Hanik; Optical Communication Systems, Lecture Notes at Technische UniversitätMünchen (TUM), 2008[2] G.P. Agrawal; Fiber-Optic Communication Systems, Third Edition, John Wiley &Sons, New York, 2002[3] E. Voges, K. Petermann (eds.); Optische Kommunikationstechnik, Springer, Berlin,2002[4] I. Kaminov, T. Li (eds.), Optical Fiber Telecommunications IV, Academic Press, SanDiego, 2002[5] P.J. Winzer, R.-J. Essiambre, Advanced Modulation Formats, Proceedings of theIEEE, Vol. 94, No. 5, pp. 952-985, May 2006。

db06

数据库系统及应用
6.14
金培权（jpq@）
二、函数依赖
什么是函数依赖基本概念平凡和不平凡的函数依赖
函数依赖集的闭包
属性集的闭包
最小函数依赖集
数据库系统及应用
6.15
金培权（jpq@）
1、什么是函数依赖？
函数依赖是指一个关系模式中一个属性集
F={A→BC, B→E, CD→EF}
AD→F对于函数依赖集F是否成立？ A→BC（已知） A→C （分解律） AD→CD（增广律） CD→EF（已知） AD→EF（传递律） AD→F（分解律）
数据库系统及应用
6.23
金培权（jpq@）
（2）码的形式化定义
设关系模式R（U），F是R的一个FD集，X
T3
数据库系统及应用
A3
C6
6.11
N6
金培权（jpq@）
4、问题（4）：删除异常
如果教师T3现在不带课了，则需将T3的元
组删去，但同时也把他的姓名和地址信息删掉了
Tname T1 T1 T1 T2 T2 T3
数据库系统及应用
Addr A1 A1 A1 A2 A2 A3
001
002 003
张三
李四王五
20
21 22
数据库
数据库系统及应用
6.4
金培权（jpq@）
问题的提出
如何把现实世界表达成数据库模式？针对一个具体应用，应该如何构造一个适
合于它的数据库模式？
这是数据库的逻辑设计问题
关系数据库的模式设计理论是数据库逻辑
设计的理论基础
和另一个属性集间的多对一关系
例如选课关系SC(S#, C#, Score)

Ch6-鸽巢原理与Ramsey定理

使 mk1 = mk 2 = L = mk n+1 = i, 1 ≤ i ≤ n
15
n 2 + 1个实数 a 1 , a 2 , L , a n 2 + 1 组成的序列例3: 任意中, 必有一个长度为 n + 1 的单调递增或单调递减的子序列.
下面证明对应这些下标 k1 , k 2 , L , k n +1 的实数序列必满足
= (b1 + b2 + L + bi + bi +1 + L + b j ) − (b1 + b2 + L + bi ) = bi +1 + bi + 2 + L + b j
说明从第 i+1天到第j天这连续j-i天中，该棋手刚好下了21 盘棋。
9
例7: 设 a1 , a2 , a3 是3个任意整数, b1 , b2 , b3 是 a1 , a2 , a3 的任一排列, 则 a1 − b1 , a2 − b2 , a3 − b3 至少有一个是偶数.
j +1
与 mk = mk 矛盾，从而可知 ak , ak ,L, ak 长度为 n + 1 的递减子序列。
j j +1
1
2
n +1
就是所存在的
16
例4: 网络服务器一家公司有15台网络服务器和10个因特网端口，任意时刻都不会有超过10台的服务器同时需要端口。每5分钟，这些服务器的某个子集请求端口。若希望以这样的方式把每一个服务器与某些端口相连接：使用尽可能少的连接，总是确保一台网络服务器有一个存取端口（一个端口一次至多被一个服务器使用），需要多少次连接？例5：调制解调器在15天的周期内，某调制解调器连接使用了300个小时，则在连续3天的某个周期内，该调制解调器至少使用了 60个小时。

数据库原理及应用 chp2课后习题答案

第 2 章关系数据库
2.1 试述关系模型的三个组成部分。答：关系模型的三个组成部分为关系结构、关系操作和关系完整性约束。在关系模型中，无论是实体集，还是实体集之间的联系均由单一的关系表示。关系模式
可以形式化地表示为：R（U，D，Dom，F），其中 R 为关系名，U 为组成该关系的属性集合，D 为属性组 U 中属性所来自的域，Dom 为属性向域的映像的集合，F 为属性间数据的依赖关系集合。
2.5 假定关系 R 和 S 分别有 n 和 m 个元组，试说明下列运算结果中的最小和最大元组个数：
（1） R U S
（2）Ｒ
Ｓ
（3）σ F (R) × S ，其中 F 是条件表达式（4） Π L (R) − S ，其中 L 是属性集合
答：
（1） R U S 的结果，最大元组个数为 m+n，最小为 m（ R ⊇ S ）个或者 n（ R ⊆ S ）
属性）为候选码。当一个关系有多个候选码时，应选定其中的一个候选码为主码；而如果关系中只有一个候选码，这个惟一的候选码就是主码。
设 F 是基本关系 R 的一个或一组属性，但不是关系 R 的主码（或候选码）。如果 F 与基本关系 s 的主码 KS 相对应，则称厅是基本关系 R 的外码。
2）给定一组域 D1，D2，…， Dn．这些域中可以有相同的部分，则 D1，D2，…， Dn 的笛卡地积为：D1×D2×…×Dn﹦｛（dl，d2，…，dn）∣di∈Di，i=1，2， …，n｝。
说明：SC 自乘之后，同一个学号下两个课程号不同的元组若修改为：检索至少选修一门课的学生学号
πSNO(SC)
（7）检索全部学生都选修的课程的课程号和课程名；
π (C CNO,CNAME
(πSNO,CNO(SC)÷πSNO(S)))

Ch6 利率决定理论_货币银行学

I I (r )

古典利率理论：储蓄供给和投资需求决定了利率水平
iff : when r r * has S (r ) I (r ) r * is equilibrium interestrate
S S (r ) Md Ms Bd Bs S (r ) Bd I I (r ) I S I (r ) Bs Bd Bs I (r ) Bd
6.5 流动性偏好理论和利率
流动性偏好理论关注货币供给对利率的影响 6.5.1 流动性需求： • 货币需求：M/p=f(Y, i)
6.5.2 均衡利率：货币需求和供给相等时的利率
6.5 流动性偏好理论和利率
6.5 流动性偏好理论和利率
LP 理论是分析投资者行为和政府政策对经济和金融影响的一个工具，它描述了中央银行如何影响短期利率
• 名义利率是以实际支付的货币数量计算的利率水平。 • 实际利率是即使不存在通货膨胀和风险的情况下借入者被要求支付的利率水平。实际利率是纯粹的放弃一段时间货币使用权所需的回报，由于时间具有价值，长期的资金出让所得到的回报应高于短期资金出让所得到的回报。长期利率也应高于短期利率。
6.3 古典利率决定理论
6.1 利率在经济中的作用 1. 利率是决定储蓄向投资转化规模的关键因素 2. 利率影响了储蓄和信贷资源的分配 3. 影响货币需求和供给的主要因素 4. 一个重要的确定其他金融工具的价格的基准（benchmark） 5. 一个重要的货币政策工具或指标
6.2 利率的概念
• 利息是资金借入者向借出者支付的对后者一段时间内出借一定数量资金的补偿。补偿的比率就是利率
•
一些批评：
– 关于公众如何形成预期的模型不可信。 – 零交易成本假设不现实。 – 决策信息过于简单，难以认为基于这些信息的决策是理性的行为。

第二章粘附分子

(五) 纤维粘连蛋白二、免疫球蛋白超家族三、selectin家族(一) selectin分子的基本结构(二) selectin家族的组成(三) selectin分子识别的配体四、Cadherin家族(一） Cadherin分子的结构(二) Cadherin家族的组成和分布(三) Cadherin分子识别的配体五、其它未归类的粘附分子(一) selectin分子结合的配体(二) CD44第二节粘附分子表达的调节一、粘附分子构型改变影响细胞的粘附作用(一) LFA-1分子构型改变对其粘附作用的影响(二) 其它粘附分子构型的改变对粘附作用的影响二、细胞粘附分子表达数量改变对粘附作用的调节第三节粘附分子的功能一、炎症过程中白细胞与血管内皮细胞的粘附二、粘附分子与淋巴细胞的归巢(一) T细胞前体向胸腺的归巢(二) 淋巴细胞向外周淋巴器官的归巢(三) 淋巴细胞向非淋巴组织的归巢(四) 淋巴细胞归巢过程中激活粘附作用的分子三、粘附分子参与免疫细胞的识别作用四、粘附分子参与细胞发育、分化、附着及移动(一) 粘附分子参与细胞间的附着(二) 粘附分子参与细胞与基质的附着(三)粘附分子参与细胞的移动五、粘附分子与肿瘤(一) 粘附分子与肿瘤的浸润与转移(二) 粘附分子对杀伤细胞杀伤肿瘤细胞的影响(三) 粘附分子与肿瘤的诊断六、粘附分子与凝血(一) 粘附分子与动脉凝血(二) 粘附分子与静脉凝血七、粘附分子与细胞内信号传导(一) 粘附分子与细胞内酪氨酸磷酸化(二) 粘附分子与细胞膜磷酯酰肌醇代谢第四节可溶性粘附分子(一) 可溶性L-selectin(sL-selectin)(二) 可溶性P-selectin(sP-selectin)(三) 可溶性E-selectin(sE-selectin)(四) 可溶性ICAM-1(sICAM-1)(五) 可溶性VCAM-1(sVCAM-1)参考文献(三) Cadherin分子识别的配体五、其它未归类的粘附分子(一) selectin分子结合的配体(二) CD44一、粘合素超家族国内将integrin译为粘合素、整合素等，本书暂命名为粘合素。

Ch06 The Theory of Consumer Choice(经济学,英文版-复旦,周翼)

Preferred region

d a
Dominated region
c

b
e
Quantity of meals
6.4
Modelling consumer preferences (3)

U2
An indifference curve like U2U2 shows all the consumption bundles that yield the same utility to the consumer

C E
H BL1
U2
U1 BL0
Meals
6.14
The substitution effect
H
Films
U2 U1 D

The SUBSTITUTION
EFFECT is from C to D
C E
H BL1
along U2U2.
U2
U1 BL0
Meals
– –
It is always negative a price increase leads
Four key elements in consumer choice

Consumer’s income
Prices of goods Consumer preferences The assumption that consumers maximize utility6.1源自The budget line
6.12
Response to a price change
The response to a price change comprises two effects: The SUBSTITUTION EFFECT

数据仓库与数据挖掘教程（第2版）课后习题答案第七章

数据仓库与数据挖掘教程（第2版）课后习题答案第七章第七章作业1.信息论的基本原理是什么？一个传递信息的系统是由发送端(信源)和接收端(信宿)以及连接两者的通道(信道)组成的。

信息论把通信过程看做是在随机干扰的环境中传递信息的过程。

在这个通信模型中，信息源和干扰(噪声)都被理解为某种随机过程或随机序列。

在进行实际的通信之前，收信者(信宿)不可能确切了解信源究竟会发出什么样的具体信息，也不可能判断信源会处于什么样的状态。

这种情形就称为信宿对于信源状态具有不确定性，而且这种不确定性是存在于通信之前的，因而又叫做先验不确定性。

在通信后，信宿收到了信源发来的信息，这种先验不确定性才会被消除或者被减少。

如果干扰很小，不会对传递的信息产生任何可察觉的影响，信源发出的信息能够被信宿全部收到，在这种情况下，信宿的先验不确定性就会被完全消除。

但是，在一般情况下，干扰总会对信源发出的信息造成某种破坏，使信宿收到的信息不完全。

因此，先验不确定性不能全部被消除, 只能部分地消除。

换句话说，通信结束之后，信宿仍具有一定程度的不确定性。

这就是后验不确定性。

2.学习信道模型是什么？学习信道模型是信息模型应用于机器学习和数据挖掘的具体化。

学习信道模型的信源是实体的类别，采用简单“是”、“非”两类，令实体类别U 的值域为{u1，u2}，U 取u1表示取“是”类中任一例子，取u2表示取“非”类中任一例子。

信宿是实体的特征（属性）取值。

实体中某个特征属性V ，他的值域为{v1，v2……vq}。

3.为什么机器学习和数据挖掘的分类问题可以利用信息论原理？信息论原理是数据挖掘的理论基础之一。

一般用于分类问题，即从大量数据中获取分类知识。

具体来说，就是在已知各实例的类别的数据中，找出确定类别的关键的条件属性。

求关键属性的方法，即先计算各条件属性的信息量，再从中选出信息量最大的属性，信息量的计算是利用信息论原理中的公式。

4自信息：单个消息ui 发出前的不确定性（随机性）称为自信息。

数据仓库与数据挖掘教程(第2版)课后习题答案第二章

第二章作业1.画出数据仓库的结构图，说明各部分内容。

P18当前基本数据是最近时期的业务数据，是数据仓库用户最感兴趣的部分数据量大。

随着时间的推移，有数据仓库的时间控制机制转为历史数据，轻度综合数据是从当前基本数据中提取出来的，最高一层是高度综合数据层，这一层的数据十分精炼，是一种准决策数据。

2.说明数据仓库结构图中包含轻度综合层与高度综合数据层的作用。

这些数据为什么不是临时计算出来的。

P18-19数据仓库除了存储按主题组织起来的当前详细数据外，还需要存储综合数据，这是为了适应决策需求而增加的。

在数据库中需要得到综合数据时，采用数据立方体的方法对详细数据进行综合。

在数据仓库中并不采取临时计算的方式得到综合数据，而在用户提出需要综合数据之前，就预先将可能的综合数据利用数据立方体计算好，存入综合数据层中，这种综合数据层在用户查询时，能迅速提供给用户。

3.说明数据集市与数据仓库的区别和联系。

P20联系：数据集市是一种更小，更集中的数据仓库，为公司提供了一条分析商业数据的廉价途径。

数据集市是指具有特定应用的数据仓库，主要针对某个具有战略意义的应用或者具体部门级的应用，支持用户利用已有的数据获得重要的竞争优势或者找到进入新市场的具体解决方案。

区别：（1）数据仓库是基于整个企业的数据模型建立的，它面向企业范围内的主题。

而数据集市是按照某一特定部门的数据模型建立的。

（2）部门的主题与企业的主题之间可能存在关联，也可能不存在关联。

（3）数据集市的数据组织一般采用星型模型。

4.1、规模是小的2、特定的应用3、面向部门4、由业务部门定义，设计和开发5、由业务部门管理和维护6、快速实现7、购买较便宜8、投资快速回收9、更详细的、预先存在的数据仓库的摘要子集10、可升级到完整的数据仓库5.独立型数据集市直接从操作型环境获取数据，从属型数据集市从企业级数据仓库获取数据，带有从属型数据集市的体系结构。

6.原因：仓库管理：安全和特权管理；跟踪数据的更新；数据质量检查；管理和更新元数据；审计和报告数据仓库的使用和状态；删除数据；复制、分割和分发数据；备份和恢复；存储管理。

Complex Numbers Theory Sheet 2

Complex NumbersTheory Sheet 2Manipulation of Complex Numbers in Rectangular and Polar formLearning Outcomes• … have been outlined in Theory Sheet 1BackgroundTheory Sheet 1 showed how the real number system, , was insufficient when trying to solve problems containing the square roots of negative numbers. It was necessary to extend the number system to incorporate ‘imaginary’ numbers and hence to develop the more comprehensive ‘complex’ number system. Although strange at first sight, the imaginary number, j , is no more or less imaginary than the number -1. Why ask, “What do j sheep look like?”, when it is equally ridiculous to ask, “What do -1 sheep look like?” Counting objects, using either negative or imaginary numbers is an inappropriate visualisation. A good visualisation of complex numbers uses the Argand Diagram. This diagram is extremely useful in an engineering context, as we shall see, as it allows the use of complex numbers to solve a variety of engineering problems. However, first it is important to illustrate how complex numbers behave with respect to the basic operations of addition, subtraction, multiplication and division.Before that, let us remind ourselves of where complex numbers arise and their general format. (See also Tutorial Sheet 1 in this series)The solution of quadratic equations (of the form ax 2 + bx + c = 0) will have complex solutions if b 2 < 4ac . So for example, 24130x x ++= gives, using the formula for solving a quadratic with a = 1, b = 4 and c = 13,46232j x j −±=====−±The quadratic equation has two complex solutions, 23x j =−+ and 23x j =−−The answers here are numbers of the form a + bj (sometimes written a + jb ) with realnumbers a and b , and a + bj is a complex number.• a is called the real part of a + bj , and b is called the imaginary part of a + bj .• If z denotes a complex number, the real part of z is sometimes written as Re z , and the imaginary part as Im z .• If a = 0, the complex number is wholly, or purely, imaginary, e.g. 3j ()03j ≡+• If b = 0, it is wholly real, e.g. 220j ≡+ (so all real numbers are complex numbers with a zero imaginary part! i.e. the real number set is a subset of the complex number set)• Two complex numbers are equal if and only if they have the same real and imaginary part.• The complex number 0 is just 0 + j 0.Manipulation of Complex Numbers in Rectangular formPractically the first operations we performed with numbers in early schooldays were addition, subtraction, multiplication and division. As we were introduced to each new number system, we learned how to perform these operations. So again, with complex numbers, we need to be able to know how to add, subtract, multiply and divide.It is safe to assume that the ordinary rules of arithmetic apply, so that the following rules for addition, subtraction and multiplication apply to any two complex numbers of the form z 1 = a + bj and z 2 = c + dj :Addition: z 1 + z 2 = (a + bj ) + (c + dj ) = (a + c ) + (b + d )jSubtraction:z 1 - z 2 = (a + bj ) - (c + dj ) = (a - c ) + (b - d )j Multiplication: z 1. z 2 = (a + bj ).(c + dj ) = ac + adj + bcj +bdj 2but remember j 2 = -1, soz 1. z 2 = (a + bj )(c + dj ) = (ac − bd ) + (bc + ad )jNote:• In addition (or subtraction), add (or subtract) corresponding real parts and imaginary parts• Multiplication is carried out in exactly the same way that you would multiply, for example, (x + 2)(2x – 3)• don’t try to remember these formulae, work each one out ‘on the fly’.As you become more adept at such calculations, it becomes possible to write down the answers directly. The first three examples that follow are fairly straightforward – just beware of “minus minuses”!Examples of Complex Addition and Subtraction1. (2 + 3 j ) + (6 - j 5) = 8 – 2j (remember a + jb = a + bj )2.j3.()(5665j j −=−++There’s nothing wrong with leaving your answer in this square root format – although you can write it in decimals if you really want to.Examples of Complex Multiplication1. ()()2563j j +−=12 – 6j + 30j – 15j 2 12 +24j -15(-1) = 27 + 24j2. 3j (4 – 6j ) = 12j -18j 2 = 18 + 12 j(remember j 2 = -1)Before looking at complex division it is necessary to first consider …The Complex Conjugate of z , z(i.e. just change the sign of the imaginary part)Example : If z = - 2 – 3j then z =- 2 + 3j .Note that these were the two roots of the quadratic equation on page 1 of this tutorial sheet. In fact, when a quadratic equation has complex solutions these always occur as complex conjugate pairs.If z = a + bj , then the complex conjugate , z , is z a bj =−Obviously, complex conjugating twice returns us to the original number: z = z .On its own, the complex conjugate doesn’t appear useful, but it has one important property that allows us to perform complex division, that is, z z is always wholly real .Proof : let z be any complex number a + bj so that z a bj =−, so()()22222z z a bj a bj a abj abj b j a b =+−=−+−=+ - wholly realNote that a complex number multiplied by its own complex conjugate is easily found from(real part)2 + (imaginary part)2In general this will always look like:If (c + jd ) ≠ 0 + j 0 then a + bjc +dj can be calculated by writinga + bj c + dj as a + bj c -dj c + dj c - dj× = ac + bdc 2+ d 2 + j bc − ad c 2 + d 2A Bj ≡+ So when two complex numbers are divided, the result is a complex number. Again, don’t try to remember this formula, work each case out ‘on the fly’.Example of Complex Division3564j j −=−+3564381838181991996464361652262626j j j j j j j j −−−−+−+−+×====−+−+−−+Complex Numbers in Polar FormIn Tutorial Sheet 1 we saw that complex numbers can be visualised in Rectangular Form (or Cartesian Form) as points in a 2-dimensional plane (an Argand Diagram), unlike real numbers, which exist only along a 1-dimensional line.Rectangular form is not the only way to represent a complex number on an Argand Diagram. Instead of merely a point, a complex number can be represented by a directed line segment (arrow) from the origin of the Argand Diagram to the point itself. So 2 + 3j can also be represented thus:Complex division is always performed by multiplying both numerator and denominator of the complex division by the complex conjugate of the numerator .This will ensure a real denominator.So a complex number such as 2 + 3j exists at a point reached by moving +2 in the ‘real’ direction and +3 in the imaginary direction.angle (to the positive real axis),θ.One consequence of being able to express complex numbers in polar form is that they can be used to solve 2-dimensional vector and phasor problems.Conversion from Rectangular to Polar form and vice versaConsider the following Argand Diagram in which the complex number z = a + bj is illustrated with both its rectangular and polar‘coordinates’ shown.Dropping a vertical line from the end of the complexnumber to the real axis produces a right-angled triangle in which the non-hypotenuse sides are a and b and the hypotenuse is r . The angle measured from the positive real axis is denoted θ.Applying trigonometrical ratios to this triangle givescos and sin a br rθθ== or cos and sin a r b r θθ==So, the complex number written in Rectangular Form, z = a + bj , can be written in Polar Form as cos sin z r jr θθ=+ of course, with a common factor of r , this becomes()cos sin z r j θθ=+. This is often abbreviated to z r θ=∠ or ,z r θ=Needless to say, it is possible to determine a complex number’s Rectangular form from its polar form. Returning to the triangle above, Pythagoras’ Theorem givesr =1tan b a θ−=NOTE: When using this latter expression, it is important to determine the correct quadrant for θ.To this end it is always advisable to sketch the Argand Diagram to ensure the correct angle is found.In summary thenRectangular Form: z = a + bjPolar Form:()cos sin z r j θθ=+, or z r θ=∠ or ,z r θ=Rectangular to Polar Conversion: cos and sin a r b r θθ== Polar to Rectangular Conversion: 1tan b r a θ−==Note that this line segment can be thought of as a vector, or phasor, with a length r and a direction θ anticlockwise from the positive real axis. Since this is equivalent to getting to the point directly from origin, or ‘pole’, when a complex number is expressed in terms of r and θ it is said to be in Polar Form . (note in this case how it is more usualto write the j at the beginning of the imaginary term)NOTE: Many scientific calculators include rectangular to polar conversion, and vice versa, as standard. If your calculator has keys labelled “R P”, or “ (r, θ)”, or equivalent, read your calculator manual to see how these keys work. It will be well worth your while, since these return any angles in the correct quadrant – automatically!In Polar Form a complex number has a magnitude , r , and a direction ,θ. These are properties shared with vectors and phasors. This means that complex numbers can be used to solve vector, or phasor, problems and can be illustrated in the following.Addition of Complex Numbers on an Argand Diagram(… and its application to “Parallelogram of Forces” problems)Consider adding the two complex numbers z 1 = 1 + 3j and z 2 = 4 – j . The answer here is, of course, easily determined by adding the real parts and the imaginary parts separately.So z 1 + z 2 = (1 + 4) + (3 – 1)j = 5 + 2j .The Argand Diagram visualisation of this is:Now notice that, by joining the ends of the original complex numbers to the end of the resultant complex number, the following diagram is obtained:Some of you will recognise this as the “Parallelogram of Forces” diagram. Here, the forces (or vectors, or complex numbers) z 1 and z 2 combine to give a resultant 5 + 2j .In a vector/force/phasor, problem, the vector will not usually be specified in Rectangular Form. In the case above the complex number, or force/vector/phasor, z 1 will normally be expressed with a magnitude and a direction, i.e. in polar form.RealImaginary1 + 3j4 - j5 + 2jSo here z 1 = 1 + 3j = 3.1623∠71.565o (5 s.f.) and z 2 = 4 – j = 4.1231∠-14.036o (5 s.f.) The resultant is5 + 2j= 5.3852∠21.801o(5 s.f.)In words this says that there are two vectors, one with a magnitude 3.1623 making an angle of 71.565o with some datum axis (usually the positive real axis), the second with a magnitude 4.1231 and angle -14.036o which, when added, result in a vector with magnitude 5.3852 and angle 21.801o .A most important point to note here is that it is not possible to add complex numbers (or phasors or vectors) in polar form (3.1623 + 4.1231 is NOT equal to 5.3852 and 71.565o + (-14.036o ) is NOT equal to 21.801o ). Here, as in all such cases, the addition has to be done in rectangular form.Example A vector has a magnitude of 10 units and makes an angle of 35o to a datum axis. A second vector has a magnitude of 20 units and makes an angle of 120o to the same axis. Find their resultant.A problem like this is probably best tabulated. (The datum axis described is quite arbitrary, so the positive real axis is automatically chosen)Vector Polar Form Rectangular Formz 1 (Vector 1)10∠35o 8.1915 + 5.7358j z 2 (Vector 2)20∠120o -10 + 17.3205j z 1 + z 2 (Resultant)23.1271∠94.48o -1.8085 + 23.0563jSo the resultant vector has a magnitude of 23.1271 and makes an angle of 94.48o with the datum axis.Always sketch an Argand Diagram when solving such problems. A reasonable sketch should at least give an indication that the resultant ‘looks about right’! Try it now with this problem. You can, of course, check this result using the accompanying MathinSite ‘Complex Numbers’ applet available from/html/applets.htmlusing the ‘Polar Addition’ option from the drop-down menu in the top right of the applet window.Needless to say, this process can be extended to three, or indeed any number, of vectors.Subtraction of Complex Numbers on an Argand DiagramConsider again the two complex numbers z 1 = 1 + 3j and z 2 = 4 – j . To subtract z 1 from z 2, simply subtract the real parts and the imaginary parts separately.So z 2 – z 1 = (4 - 1) + (-1 - 3)j = 3 - 4j .The Argand Diagram visualisation of this is:Step 1: polar to rectangular conversionStep 2: add vectors in rectangular formStep 3: convert back to polar formFirst notice that subtracting z 1 is exactly the same as adding –z 1. In which case, the “Parallelogram of Forces” strategy can be used again on the Argand Diagram, but this time by adding vectors z 2 and MINUS z 1.Here -z 1 = -(1 + 3j )= 3.1623∠-108.435o and z 2= 4 – j = 4.1231∠-14.036o resultant isz 2 – z 1 = z 2 + (– z 1) = 3 - 4j= 5∠-53.130oNote that the required angle for -z 1 is -(180 - 71.565)o = -108.435oNote also that, as with addition, it is not possible to subtract complex numbers (phasors /vectors) in polar form (3.1623 + 4.1231 is NOT equal to 5 and -108.435o + (-14.036o ) is NOT equal to -53.130o ). Here again, any subtraction has to be done in rectangular form.Exercises (It is possible to corroborate certain answers using the accompanying “Complex Numbers” applet.)1. Convert the following complex numbers into Polar Form, ensuring that the angle obtained is in the correct quadrant.(a) 3 + 4j , (b) -3 + 5j , (c) -6 – 7j , (d) 5 – 12j2. Convert the following complex numbers into Rectangular Form.(a)0530∠, (b) 01530∠−, (c) 03230∠, (d) 03130∠−3. The answers you obtained in Q 2 (c) and (d) are the same. Explain why.4. If z 1 = 3 – 5j , z 2 = 5 + 6j and z 3 = -2 + 5j , find, in Rectangular form, the values ofa. z 1 + z 2b. z 1 + z 3c. z 1 - 4z 2 +5z 3d. 13z z ×e.132z z z ×5. Convert all your answers in Q. 4 into Polar Form.6. Given z 1 =0650∠, z 2 = 0350∠− and z 3 =05120∠, find, in Polar Form z 1 - 4z 2 +5z 3.7. Three forces have magnitudes 3N (newton), 6N and 15N each making an angle to a datum axis of 30o , 60o and -240o respectively. Find the resultant of these three forces and sketch this system of forces on an Argand Diagram.8. Show that the real and imaginary parts of z = 2312j j ++ − 86j j+− are 61185 and 107185−.。

ch6_信息检索1

2013-9-10
第一节信息存储的基本程序与方法
五、信息的存储

计算机信息检索系统的构成硬件设备：主机、检索终端、通信设备、
输入输出设备
软件设备：系统软件、应用软件、数据库
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第一节信息存储的基本程序与方法
五、信息的存储
数据库：至少由一种文档组成，并能
满足某一特定目的或某一特定数据处理系统需要的一种数据集合。
存储：原始文献 --- 文献标识.存储 --- 检索工具 19
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第一节信息存储的基本程序与方法
五、信息的存储

广义的信息检索系统就是信息的存贮和检索的系统狭义的信息检索系统就是信息检索工具

20
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第一节信息存储的基本程序与方法
五、信息的存储

分类：手工信息检索系统和计算机信息检索系统发展：手工信息检索系统

43
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第二节信息检索的类型程序与方法
二、信息检索的基本程序
2013-9-10
第一节信息存储的基本程序与方法
五、信息的存储

手工信息检索工具的著录对象单位出版物：目录：单位出版物以文献独自名称作为一个完整出版单位的题录：单位出版物中的单篇文献出版物文摘：单位出版物中的单篇文献一本书《现代信息检索》索引：单位出版物或单篇文献中的知识单元一种刊《四川农业大学学报》 24
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第一节信息存储的基本程序与方法
五、信息的存储

辅助索引字段
表达文献外表特征的字段

植物生理学ch6 植物体内有机物的运输

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2.运输的速度和溶质种类
1）速度 • 平均100cm/h • 不同植物、同一植物不同生育
期运输速度不同。
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2）溶质种类研究方法蚜虫吻针法
筛管分子
蚜虫吻针
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主要成分： 1）蔗糖100~250g/L，占干物质
90%，是糖类运输的主要形式 2）氨基酸和酰氨 3）磷酸、核苷酸和蛋白质 4）植物激素 5）糖醇、钾、磷等无机离子
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第一节有机物运输的途径、速度和溶质的种类
1.运输途径
1）韧皮部筛管
筛分子伴胞复合体伴胞韧皮薄壁细胞
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筛板孔薄壁细胞
伴胞
筛管分子筛板
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Transfer cell
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2）证明方法（1）环割法：在木本植物的树
干上，环割一圈，深度以到形成层为止，剥去圈内的树皮。
第六章
Chapter 6
Translocation in the Phloem
植物体内有机物的运输
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高等植物的各个器官是具有明确的分工的，不同功能的各器官之间必然需要物质、能量和信息的交流，植物体才能保持一个统一的整体，这些都依赖于有效的运输机构——维管束。
高等植物有机物的运输，是通过韧皮部完成的。
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第二节韧皮部装载
韧皮部装载：是指光合产物从叶肉细胞到筛分子- 伴胞复合体的整个过程。
韧皮部装载要经过3个步骤的：
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第一步：白天，磷酸三碳糖从叶绿体运到胞质溶液；晚上，叶绿体中贮存的淀粉可能以葡萄糖状态离开叶绿体。
后来在细胞质转变为蔗糖。

结构化学-Ch6-复习习题-杨媛

o Eeg Et2g 10 Dq 2Eeg 3Et2g 5Es 0
3.分裂能（△o 或10Dq)(P186)：
金属原子或离子的5个d轨道在球形场作用下分裂成2个高能级eg轨道和3个低能级t2g 轨道，高能的d轨道与低能的d轨道的能量之差即是分裂能。
Eeg=6Dq(或0.6Δ0) Et2g=-4Dq(或-0.4Δ0)
配位化合物（简称配合物）又称络合物。配合物是由中心金属原子或离子（M）与周围若干配体(L:分子或离子)所形成的化合物(MLn)。
配合物：单核配合物 MLn 多核配合物 MxLn
有关配合物的理论
➢晶体场理论 CFT (Crystal Field Theory)
静电作用模型，把M-L间作用看作类似离子晶体中正负离子的静电作用
• 以八面体场为例进行说明
➢ 假设中心原子M处于直角坐标系原点，6个配体处于坐标轴上。按M和L组成的分子轨道是还是轨道，可将M的价轨道分成两组：
：s、px、py、pz、dx2 y2、dz2 ：dxy、dxz、dyz
➢ 配体L按照能与中心原子生成或键分别组合成新
的群轨道，使之与M的原子轨道对称性匹配。
物的构型偏离理想构型，有长、短键之分。当遇到简并态时，配位化合物会发生变形，使一个轨道能级降低，消除简并。
1
2
dx2 y2
dz2
2
1
dx2 y2
dz2
➢大畸变：在高能的eg轨道上出现简并态，变形较大。
➢小畸变：在低能的t2g轨道上出现简并态，变形较小。

第六章配位化合物和簇合物的结构与性质
教学目标
对常见多原子分子所包含的化学键配位键，多电子键，金属多重键等，能作出正确的判断，从而可推测其几何结构与化学性质。

第六章宏观计量经济模型简介

“从一般到简单”： 20世纪60——70年代，在模型的估计计算很方便的情况下，模型中变量的多少对计算量几乎没有影响。
起点一般化的模型
依据已有的经济理论和经济行为规律假设，把对被解释变量有影响（不论其是否显著）的变量都作为解释变量。
相同的前提假设与研究对象不同的研究者会得到相同的最终模型
宏观计量经济模型是在西方国家首先发展起来的，20
世纪80年代初才在我国传播与发展。
宏观计量经济模型的设定和应用目前尚无成熟的理论
方法体系，在建模和应用实践中既有成功的经验，也有很多失败的教训。
CH6 宏观计量经济模型
6.1 宏观经济模型及其分类 6.2 宏观计量经济模型的设定理论
6.3 西方国家宏观计量经济模型简介
宏观经济模型
按建模范围分类
国家（地区）模型：把一个国家（地区）当作是较为独立和完整的宏观经济系统，描述其经济运行机制和运行规律。是一类被研究最多、应用最广泛的宏观计量经济模型。
国家（地区）间模型：主要用于研究国家（地区）之间的经济联系以及研究国际经济活动的发展对经济稳定性的影响。它以商品的进出口和资本的流入流出为纽带，把各国（地区）的宏观计量经济模型连接起来而形成。
专门模型：以重点描述宏观经济系统中的某一个组成部分为目标，例如能源宏观经济模型、外贸宏观经济模型、环境宏观经济模型，等。在模型技术上与一般宏观经济模型的最大区别在于对宏观经济系统的分解。例如能源宏观经济模型将能源部门作详细分解，而对经济的其他部分则相当总量化。
6.1 宏观经济模型及其分类
宏观计量经济模型是相当广泛的一类模型的总称，可按不同方法将其分类。
宏观经济模型

cha6 Statistical decision theory II

∫ ∫ ∞ −∞
p(x
|
A;
H1)
p(
A)dA
=
1 (2πσ2
)N
/2
1 2πσ2A
∞ −∞
exp
⎡ ⎢⎣
−
1 2
Q(
A)
⎤ ⎥⎦
dA
∑ Q( A)
=
1 σ2
N −1
(x[n] −
n=0
A)2
+
A2 σ2A
∑ =
1 σ2
A|x
⎛ ⎜⎜⎝
A−
Nx
σ2 A|x
σ2
⎞2 ⎟⎟⎠
−
N2x2 σ4
σ2 A|x
LG (x) =
p(x;θˆ1, H1) p(x;θˆ 0 , H0 )
>
γ
Example 6.4 DC level in WGN with Unknown Amplitude-GLRT
LG (x) =
p(x; Aˆ, H1) p(x; H0 )
>
γ
Aˆ = x
∑∑ ln
LG (x)
=
1 (2πσ2 )N / 2
(u[n] − u )2
N n=0
N n=0
T (x) | H0 is independent on σ2
PFA is independent on σ2(CFAR)
18
Instructor: Professor LuoPengfei
6.6 Equivalent Large Data Records Tests
2x
N −1 n=0
x[n] +
Nx 2
−
N −1 n=0

CH6 预测

Group of experts responds to questionnaire.
Panel Consensus
Delphi Method
l. Choose the experts to participate. There should be a variety of knowledgeable people in different areas. 2. Through a questionnaire (or E-mail), obtain forecasts (and any premises or qualifications for the forecasts) from all participants. 3. Summarize the results and redistribute them to the participants along with appropriate new questions. 4. Summarize again, refining forecasts and conditions, and again develop new questions. 5. Repeat Step 4 if necessary. Distribute the final results to all participants.
PUSH
PURCHASING PUSH
SALES
PUSH
C U S T O M E R
Inventory is ‘Pushed’ to the Distributor through whatever means necessary (Discounts, etc.)
Inventory is ‘Pushed’ to the Customer through whatever means necessary (Discounts, Golf, Hunting trips, etc.)