On the stability of the linear functional equation

常微分方程的英文Ordinary Differential EquationsIntroductionOrdinary Differential Equations (ODEs) are mathematical equations that involve derivatives of unknown functions with respect to a single independent variable. They find application in various scientific disciplines, including physics, engineering, economics, and biology. In this article, we will explore the basics of ODEs and their importance in understanding dynamic systems.ODEs and Their TypesAn ordinary differential equation is typically represented in the form:dy/dx = f(x, y)where y represents the unknown function, x is the independent variable, and f(x, y) is a given function. Depending on the nature of f(x, y), ODEs can be classified into different types.1. Linear ODEs:Linear ODEs have the form:a_n(x) * d^n(y)/dx^n + a_(n-1)(x) * d^(n-1)(y)/dx^(n-1) + ... + a_1(x) * dy/dx + a_0(x) * y = g(x)where a_i(x) and g(x) are known functions. These equations can be solved analytically using various techniques, such as integrating factors and characteristic equations.2. Nonlinear ODEs:Nonlinear ODEs do not satisfy the linearity condition. They are generally more challenging to solve analytically and often require the use of numerical methods. Examples of nonlinear ODEs include the famous Lotka-Volterra equations used to model predator-prey interactions in ecology.3. First-order ODEs:First-order ODEs involve only the first derivative of the unknown function. They can be either linear or nonlinear. Many physical phenomena, such as exponential decay or growth, can be described by first-order ODEs.4. Second-order ODEs:Second-order ODEs involve the second derivative of the unknown function. They often arise in mechanical systems, such as oscillators or pendulums. Solving second-order ODEs requires two initial conditions.Applications of ODEsODEs have wide-ranging applications in different scientific and engineering fields. Here are a few notable examples:1. Physics:ODEs are used to describe the motion of particles, fluid flow, and the behavior of physical systems. For instance, Newton's second law of motion can be formulated as a second-order ODE.2. Engineering:ODEs are crucial in engineering disciplines, such as electrical circuits, control systems, and mechanical vibrations. They allow engineers to model and analyze complex systems and predict their behavior.3. Biology:ODEs play a crucial role in the study of biological dynamics, such as population growth, biochemical reactions, and neural networks. They help understand the behavior and interaction of different components in biological systems.4. Economics:ODEs are utilized in economic models to study issues like market equilibrium, economic growth, and resource allocation. They provide valuable insights into the dynamics of economic systems.Numerical Methods for Solving ODEsAnalytical solutions to ODEs are not always possible or practical. In such cases, numerical methods come to the rescue. Some popular numerical techniques for solving ODEs include:1. Euler's method:Euler's method is a simple numerical algorithm that approximates the solution of an ODE by using forward differencing. Although it may not provide highly accurate results, it gives a reasonable approximation when the step size is sufficiently small.2. Runge-Kutta methods:Runge-Kutta methods are higher-order numerical schemes for solving ODEs. They give more accurate results by taking into account multiple intermediate steps. The most commonly used method is the fourth-order Runge-Kutta (RK4) algorithm.ConclusionOrdinary Differential Equations are a fundamental tool for modeling and analyzing dynamic systems in various scientific and engineering disciplines. They allow us to understand the behavior and predict the evolution of complex systems based on mathematical principles. With the help of analytical and numerical techniques, we can solve and interpret different types of ODEs, contributing to advancements in science and technology.。

高二英语数学建模方法单选题20题1.In the process of mathematical modeling, the factor that determines the outcome is called_____.A.independent variableB.dependent variableC.control variableD.extraneous variable答案：B。

本题考查数学建模中的基本术语。

独立变量（independent variable）是指在实验或研究中被研究者主动操纵的变量；因变量dependent variable）是指随着独立变量的变化而变化的变量，在数学建模中决定结果的因素通常是因变量；控制变量（control variable）是指在实验中保持不变的变量；无关变量（extraneous variable）是指与研究目的无关，但可能会影响研究结果的变量。

2.The statement “The value of y depends on the value of x” can be represented by a mathematical model where y is the_____.A.independent variableB.dependent variableC.control variableD.extraneous variable答案：B。

在“y 的值取决于x 的值”这句话中，y 是随着x 的变化而变化的变量，所以y 是因变量。

3.In a mathematical model, the variable that is held constant toobserve the effect on other variables is_____.A.independent variableB.dependent variableC.control variableD.extraneous variable答案：C。

International Journal of ControlVol.78,No.18,15December2005,1447–1458Stability and persistent disturbance attenuation properties for a class of networked control systems:switched system approachH.LIN*and P.J.ANTSAKLISDepartment of Electrical Engineering,University of Notre Dame,Notre Dame,IN46556,USA(Received15November2004;infinal form16August2005)In this paper,both the asymptotic stability and l1persistent disturbance attenuation issues areinvestigated for a class of networked control systems(NCSs)under bounded uncertain accessdelay and packet dropout effects.The basic idea is to formulate such NCSs as discrete-timeswitched systems with arbitrary switching.Then the NCSs’stability and performance prob-lems can be reduced to the corresponding problems of such switched systems.The asymptoticstability problem is consideredfirst,and a necessary and sufficient condition is derivedfor the NCSs’asymptotic stability based on robust stability techniques.Secondly,the NCSs’l1persistent disturbance attenuation properties are studied and an algorithm is introduced tocalculate the l1induced gain of the NCSs.The decidability issue of the algorithmis also discussed.A network controlled integrator system is used throughout the paper forillustration.1.IntroductionBy networked control systems(NCSs),we mean feed-back control systems where networks,typically digital band-limited serial communication channels,are used for the connections between spatially distributed system components like sensors and actuators to controllers. These channels may be shared by other feedback control loops.In traditional feedback control systems these connections are established by point-to-point cables. Compared with point-to-point cables,the introduction of serial communication networks has several advan-tages,such as high system testability and resource utili-zation,as well as low weight,space,power and wiring requirements(Zhang et al.2001,Ishii and Francis 2002).These advantages have made the networks connecting sensors/actuators to controllers increasingly popular in many applications including traffic control, satellite clusters,mobile robotics,etc.Recently,model-ling,analysis and control of networked control systems with limited communication capability has emerged as a topic of significant interest to the control community, see for example Wong and Brockett(1999),Brockett and Liberzon(2000),Elia and Mitter(2001),Zhang et al.(2001),Ishii and Francis(2002),and recent special issue edited by Antsaklis and Baillieul(2004).Time delay typically has negative effects on the NCSs’stability and performance.There are several situations where time delay may arise.First,transmission delay is caused by the limited bit rate of the communica-tion channels.Secondly,the channel in NCSs is usually shared by multiple sources of data,and the channel is usually multiplexed by a time-division method. Therefore,there are delays caused by a node waiting to send out a message through a busy channel,which is usually called accessing delay and serves as the main source of delays in NCSs.There are also some delays caused by processing and propagation which are usually negligible for NCSs.Another interesting problem in NCSs is the packet dropout phenomenon.Because of the uncertainties and noise in the communication chan-nel there may exist unavoidable errors or losses in the transmitted packet even when an error control coding and/or Automatic Repeat reQuest(ARQ)mechanisms are employed.If this happens,the corrupted packetis*Corresponding author.Email:hlin1@International Journal of ControlISSN0020–7179print/ISSN1366–5820onlineß2005Taylor&Francis/journalsDOI:10.1080/00207170500329182dropped and the receiver(controller or actuator)uses the packet that it received most recently.In addition, packet dropout may occur when one packet,say sampled values from the sensor,reaches the destination later than its successors.In this situation,the old packet is dropped and its successive packet is used instead. There is another important issue in NCSs,namely the quantization effect.With thefinite bit-rate constraints, quantization has to be taken into consideration in NCSs.Therefore,quantization and limited bit rate issues have attracted many researchers’attention with the aim to identify the minimum bit rate required to sta-bilize a NCS,see for example Delchamps(1990), Brockett and Liberzon(2000),Elia and Mitter(2001), Nair et al.(2004),Tatikonda and Mitter(2004).In this paper we will focus on packet exchange networks, in which the minimum unit of data transmission is the packet which typically is with the size of several hundred bits.Therefore,sending a single bit or several hundred of bits does not make a significant difference in the network resource usage.Hence,we will omit the quanti-zation effects here and focus our attention on the effects of network induced delay and packet dropout on NCSs’stability and performance.The effects of network induced delay on the NCSs’stability have been studied in the literature.In Branicky et al.(2000),the delay was assumed to be con-stant and then the NCSs’could be transformed into a time-invariant discrete-time system.Therefore,the NCSs’stability could be checked by the Schurness of certain augmented state matrix.Since most network protocols introduce delays that can vary from packet to packet,Zhang et al.(2001)extended the results to non-constant delay case.They employed Lyapunov methods,in particular a common quadratic Lyapunov function,to study bounds on the maximum delay allowed by the NCSs.However,the choice of a common quadratic Lyapunov function could make the conclu-sion for maximum allowed delay conservative in some cases.The packet dropouts have also been studied and there are two typical ways to model packet dropouts in the literature.Thefirst approach assumes that the packet dropouts follow certain probability distributions and describes NCSs with packet dropouts via stochastic models,such as Markovian jump linear systems.The second approach is deterministic,and spe-cifies the dropouts in the time average sense or in terms of bounds on maximum allowed consecutive dropouts. For example,Hassibi et al.(1999)modelled a class of NCSs with package dropouts as asynchronous dynami-cal systems,and derived a sufficient condition on packet dropouts in the time-average sense for the NCSs’stabi-lity based on common Lyapunov function approach. Note that most of the results obtained so far are for the NCSs’stability problem and the delay and packet dropouts are usually dealt with separately.Here,we will consider both network induced delay and packet dropouts in a unified switched system model.In addi-tion,the disturbance attenuation issues for NCSs are investigated as well as stability problems.In this paper,the asymptotic stability and l1persistent disturbance attenuation properties for a class of NCSs under bounded uncertain access delay and packet drop-out effects are investigated.The basic idea is to formulate such NCSs as discrete-time switched systems with arbitrary switching signals.Then the NCSs’stability and performance problems can be studied in the switched system framework.The strength of this approach comes from the solid theoretic results existing in the literature of switched systems.By a switched system,we mean a hybrid dynamical system consisting of afinite number of subsystems described by differential or difference equations and a logical rule that orchestrates switching between these subsystems.Properties of this type of model have been studied for the pastfifty years to consider engineering systems that contain relays and/or hysteresis.Recently,there has been increasing interest in the stability analysis and switching control design of switched systems(see,e.g.,Liberzon and Morse(1999), Decarlo et al.(2000)and the references cited therein). The paper is organized as follows.The assumptions on the network link layer of the NCSs are described in x2,and the NCSs with bounded uncertain access delay and packet dropout effects are modelled as a class of discrete-time switched linear systems with arbitrary switching in x3.The stability for such NCSs is studied in x4,and a necessary and sufficient matrix norm condi-tion is derived for the NCSs’global asymptotic stability. This result also improves the sufficient only conditions found in the literature of asymptotic stability for arbi-trarily switching systems.The persistent disturbance attenuation properties for such NCSs are studied in x5,and a non-conservative bound of the l1induced gain for the NCS is calculated.The techniques are based on the recent progress on robust performance of switched systems(Lin and Antsaklis2003).A networked controlled integrator with disturbances is used through-out the paper for illustration.Finally,concluding remarks are presented.Notation:The letters E,P,S...denote sets,@P the boundary of set P,and int fPg its interior.A bounded polyhedral set P will be presented either by a set of linear inequalities P¼f x:F i x g i,i¼1,...,s g,and compactly by P¼f x:Fx g g,or by the dual represen-tation in terms of the convex hull of its vertex set vert fPg¼f x j g,denoted by Conv f x j g.For x2R n,the l1 and l1norms are defined as k x k1¼P ni¼1j x i j and k x k1¼max i j x i j respectively.l1denotes the space of bounded vector sequences h¼f hðkÞ2R n g equipped with1448H.Lin and P.J.Antsaklisthe norm k h k l 1¼sup i k h i ðk Þk 1<1,where k h i ðk Þk 1¼sup k !0j h i ðk Þj .2.The access delay and packet dropoutFor the network link layer we assume that the delays caused by processing and propagation are ignored,and we only consider the access delay which serves as the main source of delays in NCSs.Dependent on the data traffic,the communication bus is either busy or idle (available).If the link is available the communica-tion between sender and receiver is assumed to be instantaneous.Errors may occur during the communica-tion and destroy the packet and this is considered as a packet dropout.The model of the NCS used in this paper is shown in figure 1.For simplicity,but without loss of generality,we may combine all the time delay and packet dropout effects into the sensor to controller path and assume that the controller and the actuator communicate ideally.We assume that the plant can be modelled as a continuous-time linear time-invariant system described by_xðt Þ¼A c x ðt ÞþB c u ðt ÞþE c d ðt Þz ðt Þ¼C c x ðt Þ&,t 2R þ,ð1Þwhere R þstands for non-negative real numbers,x ðt Þ2R n is the state variable,u ðt Þ2R m is control input,and z ðt Þ2R p is the controlled output.The disturbanceinput d (t )is contained in D &R r .A c 2R n Ân ,B c 2R n Âm and E c 2R n Âr are constant matrices related to the system state,and C c 2R p Ân is the output matrix.For the above NCS,it is assumed that the plant output node (sensor)is time driven.In other words,after each clock cycle (sampling time T s ),the output node attempts to send a packet containing the most recent state (output)samples.If the communication bus is idle,then the packet will be transmitted to the controller.Otherwise,if the bus is busy,then the output node will wait for some time,say $<T s ,and try again.After several attempts or when newer sampled data become available,if the transmission still cannot be completed,then the packet is discarded,which is also considered as a packet dropout.On the other hand,the controller and actuator are event driven and work in a simpler way.The controller,as a receiver,has a receiving buffer which contains the most recently received data packet from the sensors (the overflow of the buffer may be dealt with as packet dropouts).The controller reads the buffer periodically at a higher frequency than the sampling frequency,say every T s =N for some integer N large enough.Whenever there are new data in the buffer the controller will calculate the new control signal and transmit it to the actuator.Upon the arrival of the new control signal,the actuator updates the output of the Zero-Order-Hold (ZOH)to the new value.Based on the above assumptions a typical time delay and packet dropout pattern is shown in figure 2.InthisDropoutτFigure 1.The networked control systems’model.Figure 2.The illustration of uncertain time delay and packet dropout of networked control systems.Disturbance attenuation for network control systems1449figure the small bullet stands for the packet being transmitted successfully from the sensor to the control-ler’s receiving buffer,maybe with some delay,and being read by the controller,at some time t¼kT sþhðT s=NÞ(k and h are integers).The new control signal is sent to the actuator and the actuator holds this new value until the next update control signal comes.The symbol denotes the packet being dropped due to error,bus being busy,conflict or buffer overflow etc.3.A switched system model for NCSsIn this section,we will consider the sampled-data model of the plant.Because we do not assume time synchroni-zation between the sampler and the digital controller, the control signal is no longer of constant value within a sampling period.Therefore the control signal within a sampling period has to be divided into subintervals corresponding to the controller’s reading buffer period, T¼T s=N.Within each subinterval,the control signal is constant under the assumptions of the previous section. Hence the continuous-time plant may be discretized and approximated by the following sampled-data modelx½kþ1 ¼Ax½k þ½B BÁÁÁB|fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}Nu1½ku2½k...u N½k266664377775þEd½kð2Þwhere A¼e A c T s,B¼ÐT s=Ne A c B c d and E¼ÐT se A c E c d .The controlled output z½k is given byz½k ¼Cx½k ð3Þwhere C¼C c.Note that for a linear time-invariant plant and constant-periodic sampling,the matrices A, B,C and E are constant.3.1.Modelling uncertain access delayDuring each sampling period,there are several different cases that may arise.First,if there is no delay,namely ¼0, u1½k ¼u2½k ¼ÁÁÁ¼u N½k ¼u½k ,then the state transi-tion equation(2)for this case can be written asx½kþ1 ¼Ax½k þ½B BÁÁÁBu½ku½k...u½k2666666437777775þEd½k¼Ax½k þNÁBu½k þEd½k :Secondly,if the delay ¼hÂT,where T¼T s=N,and h¼1,2,...,d max(the value of d max is determined as the least integer greater than the positive scalar max=T, where max stands for the maximum access delay), then u1½k ¼u2½k ¼ÁÁÁ¼u h½k ¼u½kÀ1 ,u hþ1½k ¼u hþ2½k ¼ÁÁÁ¼u N½k ¼u½k ,and(2)can be written as x½kþ1 ¼Ax½k þ½B BÁÁÁBu½kÀ1...u½kÀ1u½k...u½k266666666666664377777777777775þEd½k¼Ax½k þhÁBu½kÀ1þðNÀhÞÁBu½k þEd½k :Note that h¼0implies ¼0,which corresponds to the previous‘‘no delay’’case.Finally,if a packet-dropout happens,which may be due to a corrupted packet or sending it out with delay greater than max,then the actuator will implement the previous control signal,i.e.u1½k ¼u2½k ¼ÁÁÁ¼u N½k ¼u½kÀ1 .Therefore,the state transition equation(2)for this case can be written asx½kþ1 ¼Ax½k þ½B BÁÁÁBu½kÀ1u½kÀ1...u½kÀ12666666437777775þEd½k¼Ax½k þNÁBu½kÀ1 þEd½k :In the following,we will model the uncertain multiple successive packet dropouts.3.2.Modelling packet dropoutHere,we assume that the maximum number of the consecutive dropped packets is bounded,by some inte-ger D max.In this subsection,we will analyse the bounded uncertain packet dropout pattern and model the NCSs as switched systems with arbitrary switching.Wefirst consider the simplified case when the packets are dropped periodically,with period T m.Note that T m is integer multiple of the sampling period T s,i.e., T m¼mT s.In case of m¼T m=T s!2,thefirst(mÀ1) packets are dropped.Then,for thesefirst(mÀ1)steps,1450H.Lin and P.J.Antsaklisthe previous control signal is used.ThereforexðkT mþT sÞ¼AxðkT mÞþNBuðkT mÀT sÞþEdðkT mÞxðkT mþ2T sÞ¼A2xðkT mÞþNÁðABþBÞuÂðkT mÀT sÞþAEdðkT mÞþEdðkT mþT sÞ...xðkT mþðmÀ1ÞT sÞ¼A mÀ1xðkT mÞþNÁX mÀ2i¼0A i BuðkT mÀT sÞþ½A mÀ2E,...,EÂdðkT mÞ...dðkT mþðmÀ2ÞT sÞ26643775:Note that the integer N¼T s=T,where T is the period of the controller reading its receiving buffers.During the period t2½kT mþðmÀ1ÞT s,ðkþ1ÞT mÞ,the new packet is transmitted successfully with some delay,say ¼hðT s=NÞ,where h¼0,1,2,...,d max.Thenxððkþ1ÞT mÞ¼AxðkT mþðmÀ1ÞT sÞþhBuðkT mÀT sÞþðNÀhÞBuðkT mþðmÀ1ÞT sÞþEdðkT mþðmÀ1ÞT sÞ¼A m xðkT mÞþNÁX mÀ1i¼1A iþh"#BuðkT mÀT sÞþðNÀhÞBuðkT mþðmÀ1ÞT sÞþ½A mÀ1E,...,EdðkT mÞ...dðkT mþðmÀ1ÞT sÞ2 66 43 77 5Note that xðkT mþmT sÞequals xððkþ1ÞT mÞ,and xðkT mþðmÀ1ÞT sÞ¼xððkþ1ÞT mÀT sÞ.Let us assume that dðkT mÞ¼dðkT mþ1Þ¼ÁÁÁ¼dðkT mþmÀ1Þ,and that the controller uses just the time-invariant linear feedback control law,uðtÞ¼KxðtÞ.Then,we may substi-tute the u(Á)to obtainxððkþ1ÞT mÀT sÞ¼A mÀ1xðkT mÞþNX mÀ2i¼0A i BKxðkT mÀT sÞþX mÀ2i¼0A i EdðkT mÞandxððkþ1ÞT mÞ¼A m xðkT mÞþNX mÀ1i¼1A iþh"#BKxðkT mÀT sÞþðNÀhÞBKxððkþ1ÞT mÀT sÞþX mÀ1i¼0A i EdðkT mÞ:Substitute xððkþ1ÞT mÀT sÞinto the above equation. Thenxððkþ1ÞT mÞ¼½A mþðNÀhÞBKA mÀ1 xðkT mÞþNX mÀ1i¼1A iþðNÀhÞNBKX mÀ2i¼0A iþh"#ÂBKxðkT mÀT sÞþðNÀhÞBKX mÀ2i¼0A i"þX mÀ1i¼0A i#EdðkT mÞ:If we let^x½k ¼xðkT mÀT sÞxðkT mÞ!and d½k ¼dðkT mÞ,then the above equations can be written as^x½kþ1 ¼Èðm,hÞ^x½k þE m d½k ,whereIn this case,m¼T m=T s!2,and h¼0,1,...,d max.Èðm,hÞ¼NX mÀ2i¼0A i BK A mÀ1NX mÀ1i¼1A iþðNÀhÞNBKX mÀ2i¼0A iþh!BK A mþðNÀhÞBKA mÀ1 266664377775E m¼X mÀ2i¼0A i EðNÀhÞBKX mÀ2i¼0A i EþX mÀ1i¼0A i E266664377775:Disturbance attenuation for network control systems1451For the case of m ¼1,namely no packet dropout,thefollowing dynamic equation is derived^x½k þ1 ¼Èð1,h Þ^x ½k þE 1d ½k ,whereÈð1,h Þ¼0IhBK A þðN Àh ÞBK!,E 1¼0E!:For the the case of aperiodic packet dropouts,one may look at the delay and packet dropout pattern (figure 2)of the NCS as a succession of ramps of various lengths (T m 1þh 1,T m 2þh 2,...,T m k þh k ,...).Therefore,the NCS along with a typical aperiodic delay and packet dropout pattern can be seen as a dynamical system switching among the dynamics with different periodic delay and packet dropout pattern Èðm ,h Þ,for m ¼1,...,D max and h ¼0,1,2,...,d max .This observation leads to modelling the NCS as a switched system namely as^x½k þ1 ¼Èðm ,h Þ^x ½k þE m d ½k z ½k ¼C 0ÂÃ^x½k ,)ð4ÞwhereHere Èðm ,h Þ2f Èð1,0Þ,Èð1,1Þ,...,Èð1,D max Þ,Èð2,0Þ,...,ÈðD max ,0Þ,...,ÈðD max ,d max Þg ,where D max corresponds to the maximum number of successively dropped packets,and d max is the maximum access delay.For notational simplicity,let us denote the index of all the subsystems by q ¼m þh ÂD max ,and call the collec-tion f 1,2,...,D max Âðd max þ1Þg the mode set Q ,q 2Q .Therefore,we rewrite (4)as^x½k þ1 ¼Èq ^x ½k þE q d ½k z ½k ¼C 0ÂÃ^x½k :)ð5ÞAssociate (5)with a class of piecewise constant functionsof time :Z þ!Q ,called switching signals.Note that each switching signal corresponds to a (maybe aperio-dic)delay and packet dropout pattern.In order to study the effects of bounded uncertain access delay and packet dropouts on the NCSs’stability and performance,one needs to consider all possible delay and packet dropout patterns which corresponds to considering the arbitrary switching signals for (5).Therefore,the stability and performance problems of the NCS are equivalent to the stability and performance problems of the switched system (5)with arbitrary switching.To illustrate,we consider the following example.Example 1:Consider the continuous-time integrator with disturbances as the plant_xðt Þ¼0100"#x ðt Þþ01"#u ðt Þþ0:10:1"#d ðt Þz ðt Þ¼11ÂÃx ðt Þ:Assume that the sampling period T s is 0.1second.The controller reads the receiving buffer every T ¼0:01s ,i.e.N ¼T s =T ¼10.It is assumed that the sensor only tries to send the new sampled state value during the first 0.02s of each sampling period T s ,from which we may obtain that the maximum delay (if successfully arrived)is max ¼0:02s and d max ¼0:02=T ¼2.Èðm ,h Þ¼0I hBK A þðN Àh ÞBK!,m ¼1N Xm À2i ¼0A i BK A m À1N X m À1i ¼1A i þðN Àh ÞNBK X m À2i ¼0A i þh !BK A m þðN Àh ÞBKAm À1266664377775,m !28>>>>>>>>>><>>>>>>>>>>:E m ¼0E !,m ¼1Xm À2i ¼0A i E ðN Àh ÞBKX m À2i ¼0A i E þX m À1i ¼0A i E 266664377775,m !2:8>>>>>>>>>><>>>>>>>>>>:1452H.Lin and P.J.AntsaklisAlso assume that at most three successive packet-dropouts can occur,namely D max ¼4.Therefore,the above NCS can be modelled as an arbitrary switching system with D max Âðd max þ1Þ¼12modes.The state matrices for each mode can be determined by substituting the following valuesA ¼e A cT s ¼10:101"#,B ¼ðTe A ct B c dt¼0:000050:01"#E ¼ðT se A ct E c dt ¼0:1050:1"#,K ¼À2À1ÂÃinto the expressions for Èðm ,h Þand E m in (4)for all possible values of m 2f 1,2,3,4g and h ¼f 0,1,2g .For instance,the mode corresponding to the case of two successive packet dropouts (m ¼3)and the third packet arriving with delay 0:02s (h ¼2),i.e.,the eleventh mode (2ÂD max þ3¼11),can be described by^x ½k þ1 ¼À0:0220À0:01101:00000:2000À0:4000À0:200001:0000À0:1020À0:05100:99920:2994À0:4047À0:2023À0:16000:8880266664377775^x ½k þ0:22000:20000:23990:1736266664377775d ½kz ½k ¼1100ÂÃ^x½k :&Remarks:Similar techniques were used in Bauer et al.(2001),in which a NCS with bounded packet dropout was modelled as a polytopic uncertain linear time-variant system.In Bauer et al.(2001),it is assumed that the plant and the controller are well synchronized,while in this paper we do not have the synchronization assumption.In addition,we also consider uncertain access delay in our switched NCS model.Now we have modelled the NCS with uncertain access delay and packet dropout effects as a switched system (5)with arbitrary switching between its N ¼D max Âðd max þ1Þmodes.In the following sections we will study the asymptotic stability and disturbance attenua-tion properties of such NCSs within the framework of switched systems.For notational simplicity we willwrite ^xas x in the sequel. 4.Stability analysisThe effects of the uncertain access delay and packet dropouts on the persistent disturbance attenuation property,namely the l 1induced norm from d ½k to z ½k for the NCSs (5)will now be investigated.It is assumed that the disturbance d ½k is contained in the l 1unit ball,i.e.,D ¼f d :k d k l 1 1g .The l 1induced norm from d ½k to z ½k is defined asinf ¼inf f :k z ½k k l 1 ,8d ½k ,k d ½k k l 1 1g :The first problem we need to answer isProblem 1:Under what conditions the l 1induced norm from d ½k to z ½k for the NCSs with bounded uncertain access delay and packet dropouts is finite?The answer to Problem 1is equivalent to the condition for the arbitrarily switching system (4)to have a finite l 1induced gain.In Lin and Antsaklis (2003),it is shown that a necessary and sufficient condition for an arbitra-rily switching system (5)to have a finite inf is that the corresponding autonomous switched system x ½k þ1 ¼È x ½k be asymptotically stable under arbitrary switch-ing signals.Therefore,Problem 1is transformed into a stability analysis problem for autonomous switched systems under arbitrary switching,which has been studied extensively in the literature,and is typically being dealt with by constructing a common Lyapunov function;see the survey papers by Liberzon and Morse (1999),Decarlo et al.(2000),Lin an Antsaklis (2005),the recent book by Liberzon (2003)and the references cited therein.Various attempts have been made,for example in Narendra and Balakrishnan (1994),Shorten and Narendra (1998),Liberzon et al.(1999)and Liberzon and Tempo (2003)to find a common quadratic Lyapunov function for the family of systems,ensuring the asymptotic stability of switched systems for all switching signals.In Liberzon et al.(1999)and Agrachev and Liberzon (2001),Lie algebra conditions were given which imply the existence of a common quadratic Lyapunov function.It is worth pointing out that a converse Lyapunov theorem was derived in Dayawansa and Martin (1999)for the globally asympto-tic stability of arbitrary switching systems.This converse Lyapunov theorem justifies the common Lyapunov method which was pursued in the literature.However,most of the work has been restricted to the case of quad-ratic Lyapunov function which only gives sufficient stability test criteria.In the present paper a necessary and sufficient condition for asymptotic stability of arbi-trarily switching systems is given thus improving the sufficient only conditions found in the literature.Disturbance attenuation for network control systems1453。

机器学习题库一、 极大似然1、 ML estimation of exponential model (10)A Gaussian distribution is often used to model data on the real line, but is sometimesinappropriate when the data are often close to zero but constrained to be nonnegative. In such cases one can fit an exponential distribution, whose probability density function is given by()1xb p x e b-=Given N observations x i drawn from such a distribution:(a) Write down the likelihood as a function of the scale parameter b.(b) Write down the derivative of the log likelihood.(c) Give a simple expression for the ML estimate for b.2、换成Poisson 分布：()|,0,1,2,...!x e p x y x θθθ-==()()()()()1111log |log log !log log !N Ni i i i N N i i i i l p x x x x N x θθθθθθ======--⎡⎤=--⎢⎥⎣⎦∑∑∑∑3、二、 贝叶斯假设在考试的多项选择中，考生知道正确答案的概率为p ，猜测答案的概率为1-p ，并且假设考生知道正确答案答对题的概率为1，猜中正确答案的概率为1，其中m 为多选项的数目。

Single Variable Calculus_西北工业大学中国大学mooc课后章节答案期末考试题库2023年1.If f (x) and g (x) are differentiable on (a, b), 【图片】and f (x) > 0, g (x) > 0,x∈(a, b), then when x∈(a, b), we have答案:2.For what values of a and b will 【图片】be differentiable for all values of x?答案:a=-1/2, b=13.The evaluation of integral【图片】(where x>1) is答案:4.Find the derivative of【图片】答案:5.Find the centroid of a thin, flat plate covering the “triangular” region i n thefirst quadrant bounded by they-axis, the parabola【图片】, and the line【图片】.答案:6.If【图片】, find the limit of g(x) as x approaches the indicated value.答案:7.Find the derivative of the function below at x=0,【图片】答案:8.【图片】is答案:-1/329.If f (x) is continuous and F′(x) = f(x), then答案:10.Find the volume of the solid generated by revolving the region bounded bythe curve【图片】and the lines【图片】about【图片】.答案:11.The mean value【图片】that satisfies the Rolle’s Theorem on the function【图片】is答案:12.The critical number of 【图片】is ( )答案:0 and 213.Which statement is true?【图片】答案:A14.If【图片】,then【图片】答案:15.Evaluate【图片】.答案:16.The integtral of【图片】is答案:17.When x approaches infinity, the limit of【图片】is答案:18.The evaluation of integral【图片】is答案:19.If【图片】has continuous second-order derivative, and【图片】, then答案:20.Find the length of the enclosed loop【图片】shown here. The loop starts at【图片】and ends at【图片】.【图片】答案:21.The height of a body moving vertically is given by 【图片】, with s in metersand t in se conds. The body’s maximum height is ( )答案:22.If f (x) is increasing and f(x) > 0, then答案:23. A rock climber is about to haul up 100 N of equipment that has been hangingbeneath her on 40 m of rope that weighs 0.8 newton per meter. How much work will it take? (Hint: Solve for the rope and equipment separately, thenadd.)答案:24.The integral of【图片】is答案:25.Expand【图片】by partial function答案:26.Assume that u is a function of x and v is the derivative of u, then thederivative of arcsin(u) is答案:27.Find the center of mass of a thin plate covering the region bounded below bythe parabola 【图片】and above by the line 【图片】, if the density at the point 【图片】is 【图片】.答案:28.Find the limit【图片】答案:-129.Find the length of the curve【图片】, from【图片】 to【图片】.答案:53/630.Find the volume of the solid generated by revolving the regions bounded bythe curve 【图片】and line 【图片】about the x-axis.答案:31.Find the total area of the shaded region in the following picture.【图片】答案:4/332.The total area between the region 【图片】and the x-axis is答案:33.Which statement is NOT true?答案:34.Calculate【图片】答案:-135.The second derivative of the function y=secx is ( )答案:36.If gas in a cylinder is maintained at a constant temperature T, the pressure Pis related to the volume V by a formula of the form 【图片】in which a, b, n, and R are constants. Then【图片】答案:37.If【图片】then【图片】.答案:38.Calculate 【图片】The limit is ( )答案:139.Find the tangent to the folium of descartes 【图片】at the point (3,3)答案:x+y=640.Let 【图片】The tangent line to the graph of g(x) at (0,0) is ( ).答案:x-axis41.Find the derivative of the function below at x=0, 【图片】答案:It does not exist42.Find【图片】答案:43.The average value of 【图片】over theinterval [【图片】] is答案:44.Find the average rate of change of the function【图片】over the giveninterval [2,3]答案:1945.For【图片】 find the number【图片】 by using the two steps learned in 2.3.答案:0.0546.The linearization of the function 【图片】at x=1 is ( ).答案:47.If and only if x=ln(y),y=e^x.答案:正确48.Find the derivative of the function【图片】答案:49.Find the derivative of the function 【图片】It is ( )答案:50.If f (x) is an antiderivative of【图片】then【图片】答案:51.If f ′(x ) < 0, f ′′(x ) < 0, x∈(a, b), then the graph of f (x) on (a, b) is答案:decreasing and concave down.52.If【图片】, find【图片】.答案:753.At what points are the function【图片】 continuous?答案:Discontinuous at odd integer multiples of , but continuous at all other x.54.On what interval is the function 【图片】continuous?答案:55.On what interval is the function【图片】continuous?答案:56.【图片】【图片】and【图片】答案:0, 357.Suppose that the functionf(x)is second order continuous differentiable, and【图片】,【图片】. Therefore,【图片】答案:58.When x approaches 0, the limit of【图片】is答案:59.Find the area of the surface generated by revolving the curve 【图片】aboutthe x-axis to generate a solid.答案:60.Find the average rate of change of the function【图片】 over the giveninterval [0,2]答案:161.Find the limit of the function【图片】 and is the function continuous at thepoint being approached?答案:The limit is 0 and the function is continuous at62.The integral of [x/(x^2+1)]dx is答案:1/2[ln(x^2+1)]+C63.When x approaches 0, the limit of (1+3x)^(1/x) is答案:e^364.When x approaches infinity, the limit of x^(1/x) is答案:165.When x approaches infinity, for two functions f(x) and g(x), the limit off(x)/g(x) is infinity, and the limit of g(x)/f(x) is 0, thus a relationship between their growth rates can be said that答案:Function f(x) grouws faster than g(x).66. A function f is called a One-to-One function if it never takes on the same valuetwice.答案:正确67.The integtral of [e^(2x+1)]dx is答案:1/2[e^(2x+1)]+C68. A force of 2 N will stretch a rubber band 2 cm (0.02 m). Assuming thatHooke's Law applies, how far will a 4-N force stretch the rubber band?答案:4 cm69.Find the area of the surface generated by revolving the curve【图片】aboutthey-axis.答案:70.Which statement is true?答案:71.Which statement is false?答案:72.Find the integration formula of the solid volume generated by the curve 【图片】, the x-axis, and the line 【图片】revolved about the x-axis by the shell method.答案:73.Find the integration formula of the area of the region bounded above by thecurve 【图片】, below by the curve 【图片】, on the left by 【图片】, and on the right by 【图片】.答案:74.If 【图片】is continuous on [-1,1] and the average value is 2, then 【图片】答案:475. A cubic function is a polynomial of degree 3; that is, it has the form 【图片】，where a≠0. Then ( ) is false.答案:x=1 is critical number when the cubic function has only one criticalnumber.76.The graph of【图片】has ( )asymptotes.答案:377.If 【图片】then答案:78.The average value of【图片】on【图片】is答案:79.If f (x) is continuous on (−1, 1), and【图片】then答案:80.The derivative of the function【图片】 is答案:81.The function 【图片】has ( )答案:A. neither a local maximum nor a local minimum82.Find the derivative of function【图片】答案:83.Find y' , if【图片】答案:84.The derivative of 【图片】is( )答案:85.Let【图片】,Then【图片】答案:18x(x+1)86.At what points, is the function 【图片】continuous?答案:A. Discontinuous only when x= 3 or x= 187.Find the derivative of x(e^x).答案:e^x(x+1)88.The integral of (1/x)dx is答案:ln|x|+C89.Find the area of the surface generated by revolving the curve 【图片】aboutthe y-axis to generate a solid.答案:90.Find the length of the curve【图片】.答案:7ing the trapezoidal rule to estimate the integralwith n=4 steps【图片】答案:0.70500。

高二英语数学建模方法单选题20题(含答案)1. In a math modeling project, we need to “analyze data”. Which of the following phrases has a similar meaning?A. look up dataB. sort out dataC. examine dataD. collect data答案：C。

“analyze data”是分析数据的意思。

A 选项“look up data”是查找数据；B 选项“sort out data”是整理数据；C 选项“examine data”是检查、分析数据；D 选项“collect data”是收集数据。

所以正确答案是C。

2. When we build a math model, we often use “assumptions”. What does “assumptions” mean in this context?A. factsB. guessesC. resultsD. methods答案：B。

在建立数学模型时，“assumptions”是假设的意思。

A 选项“facts”是事实；B 选项“guesses”是猜测，与假设意思相近；C 选项“results”是结果；D 选项“methods”是方法。

所以正确答案是B。

3. In math modeling, “validate the model” means to _____.A. make the modelB. test the modelC. change the modelD. explain the model答案：B。

“validate the model”在数学建模中是验证模型的意思。

A 选项“make the model”是制作模型；B 选项“test the model”是测试模型，与验证模型意思相近；C 选项“change the model”是改变模型；D 选项“explain the model”是解释模型。

1Additional Exercises for Chapter 41.For each of the following systems,use a quadratic Lyapunov function candidate to show that the origin is asymptotically stable.Then,investigate whether the origin is globally asymptotically stable.(1)˙x 1=−x 1+x 22,˙x 2=−x 2(2)˙x 1=(x 1−x 2)(x 21+x 22−1),˙x 2=(x 1+x 2)(x 21+x 22−1)(3)˙x 1=−x 1+x 21x 2,˙x 2=−x 2+x 1ing V (x )=x 21+x 22,study stability of the origin of the system˙x 1=x 1(k 2−x 21−x 22)+x 2(x 21+x 22+k 2),˙x 2=−x 1(k 2+x 21+x 22)+x 2(k 2−x 21−x 22)when (a)k =0and (b)k =0.ing the variable gradient method,find a Lyapunov function V (x )that shows asymptotic stability ofthe origin of the system˙x 1=x 2,˙x 2=−(x 1+x 2)−sin(x 1+x 2)4.Consider the system˙x 1=x 2,˙x 2=x 1−sat(2x 1+x 2)Show that the origin is asymptotically stable,but not globally asymptotically stable.5.Show that the origin of the following system is unstable.˙x 1=−x 1+x 62,˙x 2=x 32+x 616.Consider the system˙z =−m i =1a i y i ,˙y i =−h (z,y )y i +b i g (z ),i =1,2,...,mwhere z is a scalar,y T =(y 1,...,y m ).The functions h (·,·)and g (·)are continuously differentiable for all (z,y )and satisfy zg (z )>0,∀z =0,h (z,y )>0,∀(z,y )=0,and z0g (σ)dσ→∞as |z |→∞.The constants a i and b i satisfy b i =0and a i /b i >0,∀i =1,2,...,m .Show that the origin is an equilibrium point,and investigate its stability using a Lyapunov function candidate of the formV (z,y )=α z 0g (σ)dσ+mi =1βi y 2i7.Consider the system˙x 1=x 2,˙x 2=−x 1−x 2sat(x 22−x 23),˙x 3=x 3sat(x 22−x 23)where sat(·)is the saturation function.Show that the origin is the unique equilibrium point,and useV (x )=x T x to show that it is globally asymptotically stable.8.The origin x =0is an equilibrium point of the system˙x 1=−kh (x )x 1+x 2,˙x 2=−h (x )x 2−x 31Let D ={x ∈R 2| x 2<1}.Using V (x )=14x 41+12x 22,investigate stability of the origin in each ofthe following cases.(1)k >0,h (x )>0,∀x ∈D ;(2)k >0,h (x )>0,∀x ∈R 2;(3)k >0,h (x )<0,∀x ∈D ;(4)k >0,h (x )=0,∀x ∈D ;(5)k =0,h (x )>0,∀x ∈D ;(6)k =0,h (x )>0,∀x ∈R 2.29.Consider the system˙x 1=−x 1+g (x 3),˙x 2=−g (x 3),˙x 3=−ax 1+bx 2−cg (x 3)where a ,b ,and c are positive constants and g (·)is a locally Lipschitz function that satisfiesg (0)=0and yg (y )>0,∀0<|y |<k,k >0(a)Show that the origin is an isolated equilibrium point.(b)With V (x )=12ax 21+12bx 22+ x 3g (y )dy as a Lyapunov function candidate,show that the origin is asymptotically stable.(c)Suppose yg (y )>0∀y =0.Is the origin globally asymptotically stable?10.Consider the system˙x 1=x 2,˙x 2=−a sin x 1−kx 1−dx 2−cx 3,˙x 3=−x 3+x 2where all coefficients are positive and k >a .Using V (x )=2a x 10sin y dy +kx 21+x 22+px 23with some p >0,show that the origin is globally asymptotically stable.11.Show that the system˙x 1=11+x 3−x 1,˙x 2=x 1−2x 2,˙x 3=x 2−3x 3has a unique equilibrium point in the region x i ≥0,i =1,2,3,and investigate stability of this point using linearization.12.For each of the following systems,use linearization to show that the origin is asymptotically stable.Then,show that the origin is globally asymptotically stable.(1)˙x 1=−x 1+x 2˙x 2=(x 1+x 2)sin x 1−3x 2(2)˙x 1=−x 31+x 2˙x 2=−ax 1−bx 2,a,b >013.Consider the system˙x 1=−x 31+α(t )x 2,˙x 2=−α(t )x 1−x 32where α(t )is a continuous,bounded function.Show that the origin is globally uniformly asymptoticallystable.Is it exponentially stable?14.Consider the system˙x 1=x 2,˙x 2=−x 1−(1+b cos t )x 2Find b ∗>0such that the origin is exponentially stable for all |b |<b ∗.15.Consider the system˙x 1=x 2−g (t )x 1(x 21+x 22),˙x 2=−x 1−g (t )x 2(x 21+x 22)where g (t )is continuously differentiable,bounded,and g (t )≥k >0for all t ≥0.Is the originuniformly asymptotically stable?Is it exponentially stable?16.Consider two systems represented by˙x =f (x )(1)˙x =h (x )f (x )(2)where f :R n →R n and h :R n →R are continuously differentiable,f (0)=0,and h (0)>0.Show that the origin of (1)is exponentially stable if and only if the origin of (2)is exponentially stable.17.Investigate input-to-state stability of the system˙x 1=(x 1−x 2+u )(x 21+x 22−1),˙x 2=(x 1+x 2+u )(x 21+x 22−1)。

Tikhonov regularizationFrom Wikipedia, the free encyclopediaTikhonov regularization is the most commonly used method of of named for . In , the method is also known as ridge regression . It is related to the for problems.The standard approach to solve an of given as,b Ax =is known as and seeks to minimize the2bAx -where •is the . However, the matrix A may be or yielding a non-unique solution. In order to give preference to a particular solution with desirable properties, the regularization term is included in this minimization:22xb Ax Γ+-for some suitably chosen Tikhonov matrix , Γ. In many cases, this matrix is chosen as the Γ= I , giving preference to solutions with smaller norms. In other cases, operators ., a or a weighted ) may be used to enforce smoothness if the underlying vector is believed to be mostly continuous. This regularizationimproves the conditioning of the problem, thus enabling a numerical solution. An explicit solution, denoted by , is given by:()b A A A xTTT 1ˆ-ΓΓ+=The effect of regularization may be varied via the scale of matrix Γ. For Γ=αI , when α = 0 this reduces to the unregularized least squares solution providedthat (A T A)−1 exists.Contents••••••••Bayesian interpretationAlthough at first the choice of the solution to this regularized problem may look artificial, and indeed the matrix Γseems rather arbitrary, the process can be justified from a . Note that for an ill-posed problem one must necessarily introduce some additional assumptions in order to get a stable solution.Statistically we might assume that we know that x is a random variable with a . For simplicity we take the mean to be zero and assume that each component isindependent with σx. Our data is also subject to errors, and we take the errorsin b to be also with zero mean and standard deviation σb. Under these assumptions the Tikhonov-regularized solution is the solution given the dataand the a priori distribution of x, according to . The Tikhonov matrix is then Γ=αI for Tikhonov factor α = σb/ σx.If the assumption of is replaced by assumptions of and uncorrelatedness of , and still assume zero mean, then the entails that the solution is minimal . Generalized Tikhonov regularizationFor general multivariate normal distributions for x and the data error, one can apply a transformation of the variables to reduce to the case above. Equivalently,one can seek an x to minimize22Q P x x b Ax -+-where we have used 2P x to stand for the weighted norm x T Px (cf. the ). In the Bayesian interpretation P is the inverse of b , x 0 is the of x , and Q is the inverse covariance matrix of x . The Tikhonov matrix is then given as a factorization of the matrix Q = ΓT Γ. the ), and is considered a . This generalized problem can be solved explicitly using the formula()()010Ax b P A QPA A x T T-++-[] Regularization in Hilbert spaceTypically discrete linear ill-conditioned problems result as discretization of , and one can formulate Tikhonov regularization in the original infinite dimensional context. In the above we can interpret A as a on , and x and b as elements in the domain and range of A . The operator ΓΓ+T A A *is then a bounded invertible operator.Relation to singular value decomposition and Wiener filterWith Γ= αI , this least squares solution can be analyzed in a special way viathe . Given the singular value decomposition of AT V U A ∑=with singular values σi , the Tikhonov regularized solution can be expressed asb VDU xT =ˆ where D has diagonal values22ασσ+=i i ii Dand is zero elsewhere. This demonstrates the effect of the Tikhonov parameteron the of the regularized problem. For the generalized case a similar representation can be derived using a . Finally, it is related to the :∑==qi iiT i i v bu f x1ˆσwhere the Wiener weights are 222ασσ+=i i i f and q is the of A .Determination of the Tikhonov factorThe optimal regularization parameter α is usually unknown and often in practical problems is determined by an ad hoc method. A possible approach relies on the Bayesian interpretation described above. Other approaches include the , , , and . proved that the optimal parameter, in the sense of minimizes:()()[]21222ˆTTXIX XX I Tr y X RSSG -+--==αβτwhereis the and τ is the effective number .Using the previous SVD decomposition, we can simplify the above expression:()()21'22221'∑∑==++-=qi iiiqi iiub u ub u y RSS ασα()21'2220∑=++=qi iiiub u RSS RSS ασαand∑∑==++-=+-=qi iqi i i q m m 12221222ασαασστRelation to probabilistic formulationThe probabilistic formulation of an introduces (when all uncertainties are Gaussian) a covariance matrix C M representing the a priori uncertainties on the model parameters, and a covariance matrix C D representing the uncertainties on the observed parameters (see, for instance, Tarantola, 2004 ). In the special case when these two matrices are diagonal and isotropic,and, and, in this case, the equations of inverse theory reduce to theequations above, with α = σD / σM .HistoryTikhonov regularization has been invented independently in many differentcontexts. It became widely known from its application to integral equations from the work of and D. L. Phillips. Some authors use the term Tikhonov-Phillips regularization . The finite dimensional case was expounded by A. E. Hoerl, who took a statistical approach, and by M. Foster, who interpreted this method as a - filter. Following Hoerl, it is known in the statistical literature as ridge regression .[] References•(1943). "Об устойчивости обратных задач [On the stability of inverse problems]". 39 (5): 195–198.•Tychonoff, A. N. (1963). "О решении некорректно поставленных задач и методе регуляризации [Solution of incorrectly formulated problems and the regularization method]". Doklady Akademii Nauk SSSR151:501–504.. Translated in Soviet Mathematics4: 1035–1038. •Tychonoff, A. N.; V. Y. Arsenin (1977). Solution of Ill-posed Problems.Washington: Winston & Sons. .•Hansen, ., 1998, Rank-deficient and Discrete ill-posed problems, SIAM •Hoerl AE, 1962, Application of ridge analysis to regression problems, Chemical Engineering Progress, 58, 54-59.•Foster M, 1961, An application of the Wiener-Kolmogorov smoothing theory to matrix inversion, J. SIAM, 9, 387-392•Phillips DL, 1962, A technique for the numerical solution of certain integral equations of the first kind, J Assoc Comput Mach, 9, 84-97•Tarantola A, 2004, Inverse Problem Theory (), Society for Industrial and Applied Mathematics,•Wahba, G, 1990, Spline Models for Observational Data, Society for Industrial and Applied Mathematics。

计量经济学章节习题CHAPTER 1THE NATURE OF REGRESSION ANALYSIS 1.1a. please use the above data to compute the inflation rate of each country.(a) These rates (%) are as follows. They are year-over-year, starting with 1974 as there is no data prior to 1973. These rates are, respectively, for Canada, France, Germany, Italy, Japan, UK and US.10.784313.58382 6.84713419.41748 23.17328 0.157700.110360 1084071 11.70483 5.96125217.07317 11.69492 0.244582 0.091278 7.584830 9.567198 4,360056 16.66661 9.5599390.1641790.057621 7.792208 9.563410 3.638814 19.34524 8.171745 0.158120 0.065026 8.950086 9.108159 2.730819 1246883 4.225352 0.083026 0.075908 9.320695 10.60870 4.050633 15.52106 3.685504 0.134583 0.113497 9.971098 13.67925 5A74453 21.30518 7.7014220.178679 0.134986 12.48357 13.27801 6.343714 19.30380 4.8404840.119745 0.163155 10.86449 11.96581 5.314534 16.31300 2.938090 0.085324 0.061606 5.795574 9.487459 3.295572 14.93729 1.732926 0.046122 0.032124 4.282869 7.669323 2.392822 10.61508 2.304609 0.050100 0.043173 4.1069725.827937 2.044791 8.609865 1.958864 0.060115 0.035511 4.128440 2.534965 -0.0954206.110652 0.672430 0.034203 0.018587 4.317181 3.239557 0.191022 4.591440 0.000000 0.041775 0.036496 4.054054 2.725021 1.334604 4.985119 0.763359 0.049290 0.041373 4.951299 3.456592 2.728128 6.591070 2.367424 0.077229 0.048183 4.795050 3.3411032.747253 6.1170213.052729 0.095344 0.054032 5.6088563.157895 3.654189 6.390977 3.231598 0.058704 0.042081 1.537386 2.4052484.9871025.300353 1.552174 0.036966 0.030103 1.789401 2.135231 4.5045054.250559 1.283148 0.015980 0.029936 0.202840 1.602787 2.742947 3.915309 0.760135 0.024803 0.025606 2.159244 1.783265 1.8306645.369128 -0.167645 0.033648 0.028340 1.585205 2.021563 1.498127 3.870652 0.167926 0.024557 0.029528 1.6254881.1889041.6974171.7452831.6764460.0312150.022945b. please plot the inflation rate of each country, with time as lateral axis, and inflation rate as vertical axis in the Plane Rectangular Coordinate System.c. what conclusion can you draw from the inflation history from in above 7 countries?As you can see from this figure, the inflation rate of each ofthe countries has generally declined over the years.(c) As you can see from this figure, the inflation rate of each of the countries has generally declined over the years.d. which country change the largest in inflation rate? Can you give an explaination?(d) As a measure of variability, we can use the standard deviation. These standard deviations are0.036, 0.044, 0.018, 0.062, 0.051, 0.060, and 0.032, respectively, for Canada, France, Germany, Italy, Japan, UK and USA. The highest variability s thus found for Italy and the lowest for Germany.2.12.1 What is the conditional expectation function or the population regression function? Conditional expectation, also called conditional mathematical expectation. For convenience, we discuss two random variables is deduced with the occasion of the eta, assuming they have the density function p (x, y), and the p (y ∣x) under the condition of known factor = x, density function, the conditions of eta to p1 (x) of density function is deduced. Defined under the condition of factor = x, eta conditional mathematical expectation is defined a s: E {eta ∣ factor = x} = ∫ yf (y ∣ x) dy.Under the condition of a given variable Xi,interpreted variable expect trajectory is called populatio n regression line orpopulation regression curve.The correspondingfunct i on：E（Y/Xi）=f(Xi)is called population regression function,PRF.u in regression analysis? What is the difference 2.3 What is the role of the stochastic error termibetween the stochastic error term and the residualu ?iA regression model can never be a completely accurate description of reality. Therefore, there is bound to be some difference between the actual values of the regressand and its values estimated from the chosen model. This difference is simple the stochastic error term,whose various forms are discussed in the chapter. The residual is the sample counterpart of the stochastic error term.2.5What do we mean by a linear regression model?It tells how the mean or average of the sub-population of Y varies with the fixed values of the explanatory variable(s).2.7 Are the following model the linear model? Why they are or not?(a) Taking the natural log, we find that, which becomes a linear regression model.(b). The following transformation, known as the logit transformation, makes this model a linear regression model:(c) a linear regression mdel(d) a nonlinear regression model(e) a nonlinear regression model, as is raised to the third power.BLUE（名词解释及证明）Terms definition and verifications.What is the Gauss-Markov theorem?Under the assumption of Classic Linear Model (CLM), the Ordinary Least Square (OLS) can make the estimators is the smallest deviation among the unbiased linear estimator, which are calledbest linear unbiased estimator(BLUE). Under Assumptions MLR.1 through MLR.5, are thebest linear unbiased estimators(BLUEs) of .What does the BLUE mean?when the standard set of assumptions holds, The best linear unbiased estimator means that the estimators have the characteristics as follows: linear parameter, unbiased estimators of parameters, and effective parameters, namely the unbiased parameters have the smallest deviation. minimum variance unbiased estimators,How to deduce the estimators of OLS?given the partial difference of to the sum of residual square, we get the following two equations:Please give the verification of the BLUE?(1) linear parameterswhere,Sois the linear function of Yi; since it is the weighted function of Yi with the wight of ki. , it isa linear estimator. in the same reason, is a linear estimator.(2) unbiased estimatorssowhat is the variance of ?due towe know , thenassume , and(),(due to the definitionof).(3) efficiency: minimum variance unbiased estimators(a)the OLS estimator of is unbiased estimator.thenwhereso is an unbiased estimator of true variance .(b) minimum varianceassume another linear estimator of true parameter here maybenot equal to ki.let , thenwhen ki (the weight of OLS), the variance of is the variance of OLS .so do .3.1 Please explain the assumption of the first column is equivalent to the second column.[]121212.,()()()i i i i i i i i i i i Y X u Therefore E Y X E X u X X E u X ββββββ=++=++=++，since thes β'are constants and X isnonstochastic.12=sin ce ()i i i X E u X ββ+，is zero by assumption.(2)Givei j cov(u u )0for for=?all i ，j(i j ≠),then[]{}()i j i i j i j cov(YY )Y (Y )Y (Y ) 1(u u ),j from the results in E E E E ??=--??=i j (u )(u )E E =,because the error terms are not correlated by assumption=0，since eachi u has zero mean by assumption.（3）Given []222i i i i i i 2var(u ),var(Y )Y (Y )(u )var(u )i i i X X E E E X σσ==-===by assumption3.6Note that :22i ji jyxxyiix y x y and xyββ==∑∑∑∑Multiplying the two ,we obtain the expression for r2,the squared sample correlation coefficient.3.7 Even though ??.1,yx xyββ=it may still matter (for causality and theory)if Y is regressed on X orX on Y ,since it ij just the product of the two that equals 1.This does not say that ??yx xy ββ=.3.6-3.73.6 Let and are the slop of Y regress to X, and X regress to Y respectively. pleaseexplain , meanwhile r is the correlation coefficient between X and Y..3.7 If in the question 3.6, please explain what the difference between regression of Y to X and regression of X to Y?What is the 10 assumption of classic linear model?1 Parameter is linear(Linear in Parameters),2 random regressor (Random Sampling) 3.the mean of error terms is zero (Zero Conditional Mean 4. Homoskedasticity 5. error term is no-correlation 6. the covariance is zero between error term and regressor 7 the number of observation is less than the number of regressors 8. regressor is various 9. the model is miss specification 10. non multicollinearity(No Perfect Collinearity) What is the 11 standard set of assumptions of classic normallinear model?1 Parameter is linear(Linear in Parameters),2 random regressor (Random Sampling) 3.the mean of error terms is zero (Zero Conditional Mean 4. Homoskedasticity 5. error term is no-correlation 6. the covariance is zero between error term and regressor 7 the number of observation is less than the number of regressors 8. regressor is various 9. the model is miss specification 10. non multicollinearity(No Perfect Collinearity) 11 Normal Sampling Distributions5.1 Please judge what is true, false, or not sure? and explain the reason.5.1 (a) True . The I test is based on variables with a normal distribution.Since the estimators of ,O and fl2 are linen combinations of the error u, which is assumed to be normally distributed under CLRM,these estimators arc also normally distributed.(b) True. So long as E(u1) =0, the OLS estimators are unbiased.No probabilistic assumptions are required to establish unbiasedness.(c) True. In this case the Eq. (1) in App. 3%, Sec. 3A.l, will be absent. This topic is discussed more fully in Chap. 6, Sec. 6.1.(d) True. The p value is the smallest level of significance at which the null hypothesis can be rejected. The terms level of significance and size of the test are synonymous.(e) True. This follows from Eq. (1) of App. 3A, Sec. 3A.1.(f) False. All we can say is that the data at hand does notpermit us to reject the null hypothesis.(g) False. A larger a2 may be counterbalanced by a larger .It is only if the latter is held constant, the statement can be true.(h) False. The conditional mean of a random variable depends on the vaJues taken by another (conditioning) variable. Only if the two variables are independent, that the conditional and unconditional means can be the same.(I) True. This is obvious from Eq. (3.1.7).(j) True. Refer of Eq. (3.5.2). IfX h as no influence on 1, ‘6 will be zero, in whichcase∑∑=2^2i iu y5.3 from the data of 2.6 about the revenue and education level. we get the equation (3.7.3)se = (0.8355) ( ) t = ( ) (9.6536) r 2=0.8944 n=13a. fill the figure in the blank ( ).b. explain the coefficient 0.6416?c. do you refuse the assumption that true slop coefficient is zero? which test you useand why? what is the statistic figure of p value?d. assumed there is no r 2 in the regression, can you get it from the other figure?5.3 (a) se of the slope coefficient is:6536.96417.0=0.0664the ivalue under H0: fl1=0,is: =8351.07347.0=0.8797(b) On average, mean hourly wage goes up by about 64 cents for an additional year of schooling. (c) Here n=13, so df= 11. lithe null hypothesis were true,the estimated (value is 9.6536. The probability of obtaining such a (value is extremely small; the p value is practically zero. Therefore, one can reject the null hypothesis that education has no effect on hourly earnings.(d) The ESS =74.9389; RSS = 8.8454; numerator df= 1, denominator df =11,，F =93.1929. The p value of such an F under the null hypothesis that there is no=2relationship between the two variables is 0.000001, which is extremely small. We can thus reject the null hypothesis with great confidence.Note that the F value is approximately the square of the t value under the same null hypothesis.(e) In the bivariate case, given Ho: β2=0, there is the following relationshipbetween the r value and r2: r 2=)]2([22-+n t t 。