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1、 2.1 Consider the memoryless system with characteristics shown in Fig 2.19, in which u denotes the input and y the output. Which of them is a linear system? Is it possible to introduce a new output so that the system in Fig 2.19(b) is linear?Figure 2.19Translation: 考虑具有图2.19中表示的特性的无记忆系统。其中u表示输入，y表示输
2、出。下面哪一个是线性系统？可以找到一个新的输出，使得图2.19(b)中的系统是线性的吗？Answer: The input-output relation in Fig 2.1(a) can be described as: Here a is a constant. It is a memoryless system. Easy to testify that it is a linear system. The input-output relation in Fig 2.1(b) can be described as: Here a and b are all constants. T
3、estify whether it has the property of additivity. Let: then: So it does not has the property of additivity, therefore, is not a linear system. But we can introduce a new output so that it is linear. Let:z is the new output introduced. Easy to testify that it is a linear system. The input-output rela
4、tion in Fig 2.1(c) can be described as:a(u) is a function of input u. Choose two different input, get the outputs: Assure: then: So it does not has the property of additivity, therefore, is not a linear system.2.2 The impulse response of an ideal lowpass filter is given by for all t, where w and to
5、are constants. Is the ideal lowpass filter causal? Is is possible to built the filter in the real world?Translation: 理想低通滤波器的冲激响应如式所示。对于所有的t，w和to，都是常数。理想低通滤波器是因果的吗？现实世界中有可能构造这种滤波器吗？Answer: Consider two different time: ts and tr, ts tr, the value of g(ts-tr) denotes the output at time ts, excited by
6、the impulse input at time tr. It indicates that the system output at time ts is dependent on future input at time tr. In other words, the system is not causal. We know that all physical system should be causal, so it is impossible to built the filter in the real world.2.3 Consider a system whose inp
7、ut u and output y are related by where a is a fixed constant. The system is called a truncation operator, which chops off the input after time a. Is the system linear? Is it time-invariant? Is it causal?Translation: 考虑具有如式所示输入输出关系的系统，a是一个确定的常数。这个系统称作截断器。它截断时间a之后的输入。这个系统是线性的吗？它是定常的吗？是因果的吗？Answer: Con
8、sider the input-output relation at any time t, ta: Easy to testify that it is linear. So for any time, the system is linear. Consider whether it is time-invariable. Define the initial time of input to, system input isu(t), t=to. Let to=to: Shift the initial time to to+T. Let to+Ta , then input is u(
9、t-T), t=to+T. System output: Suppose that u(t) is not equal to 0, y(t) is not equal to y(t-T). According to the definition, this system is not time-invariant. For any time t, system output y(t) is decided by current input u(t) exclusively. So it is a causal system.2.4 The input and output of an init
10、ially relaxed system can be denoted by y=Hu, where H is some mathematical operator. Show that if the system is causal, then where Pa is the truncation operator defined in Problem 2.3. Is it true PaHu=HPau?Translation: 一个初始松弛系统的输入输出可以描述为：y=Hu，这里H是某种数学运算，说明假如系统是因果性的，有如式所示的关系。这里Pa是题2.3中定义的截断函数。PaHu=HPa
11、u是正确的吗？Answer: Notice y=Hu, so: Define the initial time 0, since the system is causal, output y begins in time 0. If a0, we can divide u to 2 parts: u(t)=p(t)+q(t). Pay attention that the system is casual, so the output excited by q(t) cant affect that of p(t). It is to say, system output from 0 to
12、a is decided only by p(t). Since PaHu chops offHu after time a, easy to conclude PaHu=PaHp(t). Notice that p(t)=Pau, also we have: It means under any condition, the following equation is correct:PaHu=HPau is false. Consider a delay operator H, Hu(t)=u(t-2), and a=1, u(t) is a step input begins at ti
13、me 0, then PaHu covers from 1 to 2, but HPau covers from 1 to 3. 2.5 Consider a system with input u and output y. Three experiments are performed on the system using the inputs u1(t), u2(t) and u3(t) for t=0. In each case, the initial state x(0) at time t=0 is the same. The corresponding outputs are
14、 denoted by y1,y2 and y3. Which of the following statements are correct if x(0)0?1. If u3=u1+u2, then y3=y1+y2.2. If u3=0.5(u1+u2), then y3=0.5(y1+y2).3. If u3=u1-u2, then y3=y1-y2.Translation:考虑具有输入u输出y的系统。在此系统上进行三次实验，输入分别为u1(t), u2(t)和u3(t)，t=0。每种情况下，零时刻的初态x(0)都是相同的。相应的输出表示为y1,y2和y3。在x(0)不等于零的情况下，
15、下面哪种说法是正确的？Answer: A linear system has the superposition property: In case 1: So y3y1+y2. In case 2: So y3=0.5(y1+y2). In case 3: So y3y1-y2.2.6 Consider a system whose input and output are related by for all t. Show that the system satisfies the homogeneity property but not the additivity property.
16、Translation:考虑输入输出关系如式的系统,证明系统满足齐次性,但是不满足可加性.Answer: Suppose the system is initially relaxed,system input:a is any real constant. Then system output q(t): So it satisfies the homogeneity property. If the system satisfies the additivity property, consider system input m(t) and n(t), m(0)=1, m(1)=2; n
17、(0)=-1, n(1)=3. Then system outputs at time 1 are: So the system does not satisfy the additivity property.2.7 Show that if the additivity property holds, then the homogeneity property holds for all rational numbers a . Thus if a system has “continuity” property, then additivity implies homogeneity.T
18、ranslation:说明系统如果具有可加性，那么对所有有理数a具有齐次性。因而对具有某种连续性质的系统，可加性导致齐次性。Answer: Any rational number a can be denoted by: Here m and n are both integer. Firstly, prove that if system input-output can be described as following: then: Easy to conclude it from additivity. Secondly, prove that if a system input-ou
19、tput can be described as following: then: Suppose: Using additivity: So: It is to say that: Then: It is the property of homogeneity.2.8 Let g(t,T)=g(t+a,T+a) for all t,T and a. Show that g(t,T) depends only on t-T. Translation:设对于所有的t,T 和a，g(t,T)=g(t+a,T+a)。说明g(t,T)仅依赖于t-T。Answer: Define: So: Then:
20、So: It proves that g(t,T) depends only on t-T.2.9 Consider a system with impulse response as shown in Fig2.20(a). What is the zero-state response excited by the input u(t) shown in Fig2.20(b)?Fig2.20Translation: 考虑冲激响应如图2.20(a)所示的系统，由如图2.20(b)所示输入u(t)激励的零状态响应是什么？Answer: Write out the function of g(t
21、) and u(t): then y(t) equals to the convolution integral: If 0=t=1, 0=r=1, 0=t-r=1: If 1=t=2: Calculate integral separately:2.10 Consider a system described by What are the transfer function and the impulse response of the system?Translation:考虑如式所描述的系统，它的传递函数和冲激响应是什么？Answer: Applying the Laplace tra
22、nsform to system input-output equation, supposing that the System is initial relaxed: System transfer function: Impulse response:2.11 Let y(t) be the unit-step response of a linear time-invariant system. Show that the impulseresponse of the system equals dy(t)/dt.Translation:y(t)是线性定常系统的单位阶跃响应。说明系统的
23、冲激响应等于dy(t)/dt.Answer: Let m(t)be the impulse response, and system transfer function is G(s): So:2.12 Consider a two-input and two-output system described bywhere Nij and Dij are polynomials of p:=d/dt. What is the transfer matrix of the system?Translation:考虑如式描述的两输入两输出系统，Nij和Dij是p:=d/dt的多项式。系统的传递矩阵
24、是什么？Answer: For any polynomial of p, N(p), its Laplace transform is N(s). Applying Laplace transform to the input-output equation: Write to the form of matrix:= So the transfer function matrix is:= By the premising that the matrix inverse: exists.2.11 Consider the feedback systems shows in Fig2.5. S
25、how that the unit-step responses of the positive-feedback system are as shown in Fig2.21(a) for a=1 and in Fig2.21(b) for a=0.5. Show also that the unit-step responses of the negative-feedback system are as shown in Fig2.21(c) and 2.21(d), respectively, for a=1 and a=0.5. Fig 2.21Translation: 考虑图2.5
26、中所示反馈系统。说明正反馈系统的单位阶跃响应，当a=1时，如图2.21(a)所示。当a=0.5时，如图2.21(b)所示。说明负反馈系统的单位阶跃响应如图2.21(c)和2.21(b)所示，相应地，对a=1和a=0.5。Answer: Firstly, consider the positive-feedback system. Its impulse response is:Using convolution integral:When input is unit-step signal: Easy to draw the response curve, for a=1 and a=0.5,
27、 respectively, as Fig 2.21(a) and Fig2.21(b) shown. Secondly, consider the negative-feedback system. Its impulse response is:Using convolution integral:When input is unit-step signal: Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(c) and Fig2.21(d) shown.2.14 Draw an o
28、p-amp circuit diagram for2.15 Find state equations to describe the pendulum system in Fig 2.22. The systems are useful to model one- or two-link robotic manipulators. If , and are very small, can you consider the two systems as linear?Translation:试找出图2.22所示单摆系统的状态方程。这个系统对研究一个或两个连接的机器人操作臂很有用。假如角度都很小时
29、，能否考虑系统为线性？Answer: For Fig2.22(a), the application of Newtons law to the linear movements yields:Assuming and to be small, we can use the approximation =, =1. By retaining only the linear terms in and , we obtain and:Select state variables as , and output For Fig2.22(b), the application of Newtons l
30、aw to the linear movements yields: Assuming , and , to be small, we can use the approximation =, =, =1, =1. By retaining only the linear terms in , and , , we obtain , and: Select state variables as , and output :+2.17 The soft landing phase of a lunar module descending on the moon can be modeled as
31、 shown in Fig2.24. The thrust generated is assumed to be proportional to the derivation of m, where m is the mass of the module. Then the system can be described by Where g is the gravity constant on the lunar surface. Define state variables of the system as:, , , Find a state-space equation to desc
32、ribe the system. Translation:登月舱降落在月球时，软着陆阶段的模型如图2.24所示。产生的冲激力与m的微分成正比。系统可以描述如式所示形式。g是月球表面的重力加速度常数。定义状态变量如式所示，试图找出系统的状态空间方程描述。Answer: The system is not linear, so we can linearize it.Suppose: So:= Define state variables as:, , , Then:=+=2.19 Find a state equation to describe the network shown in Fig
33、2.26.Find also its transfer function.Translation:试写出描述图2.26所示网络的状态方程，以与它的传递函数。Answer: Select state variables as: Voltage of left capacitor: Voltage of right capacitor: Current of inductorApplying Kirchhoffs current law:=From the upper equations, we get:They can be combined in matrix form as:=+Use MA
34、TLAB to compute transfer function. We type:A=-1,0,0;0,0,-1;0,1,-1;B=1;1;0;C=0,1,0;D=0;N1,D1=ss2tf(A,B,C,D,1)Which yields:N1 = 0 1.0000 2.0000 1.0000D1 = 1.0000 2.0000 2.0000 1.0000So the transfer function is:2.20 Find a state equation to describe the network shown in Fig2.2. Compute also its transfe
35、r matrix.Translation:试写出描述图2.2所示网络的状态方程，计算它的传递函数矩阵。Answer: Select state variables as Fig 2.2 shown. Applying Kirchhoffs current law:=From the upper equations, we get:They can be combined in matrix form as:=+Applying Laplace Transform to upper equations:2.18 Find the transfer functions from to and fr
36、om to of the hydraulic tank system shown in Fig2.25. Does the transfer function from to equal the product of the two transfer functions? Is this also true for the system shown in Fig2.14?Translation:试写出图2.25所示水箱系统从到的传递函数和从到的传递函数。从到的传递函数等于两个传递函数的乘积吗？这对图2.14所示系统也是正确的吗？Answer: Write out the equation about , and : Applying Laplace transform: So: But it is not true for Fig2.14, because of the loading problem in the two tanks.17 / 17
