線性系統(tǒng)課后答案第2章習題

1、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 input u and output y are related by where a is a fixed constant. The system i

2、s called a truncation operator, which chops off the input after time a. Is the system linear? Is it time-invariant? Is it causal?Translation: 考慮具有如式所示輸入輸出關系的系統(tǒng)，a是一個確定的常數(shù)。這個系統(tǒng)稱作 截斷器。它截斷時間a之后的輸入。這個系統(tǒng)是線性的嗎？它是定常的嗎？是因果 的嗎？Answer: Consider the input-output relation at any time t, ta: Easy to testify that

3、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 is u(t), t=to. Let to=to: Shift the initial time to to+T. Let to+Ta , then input is u(t-T), t=to+T. System output: Suppose that u(t) is not equal to 0, y(t)

4、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 initially relaxed system can be denoted by y=Hu, where H is some mathematic

5、al 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: 一個初始松弛系統(tǒng)的輸入輸出可以描述為：y=Hu，這里H是某種數(shù)學運算，說明假設系統(tǒng)是因果性的，有如式所示的關系。這里Pa是題2.3PaHu=HPau是正確的嗎？Answer: Notice y=Hu, so: Define the initial time 0, since the system is c

6、ausal, 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 a is decided only by p(t). Since PaHu chops off Hu after time a, easy to conclud

7、e 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 time 0, then PaHu covers from 1 to 2, but HPau covers from 1 to 3. 2.5 Consider

8、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 denoted by y1,y2 and y3. Which of the following statements are correct if x(0

9、)0?If u3=u1+u2, then y3=y1+y2.If u3=0.5(u1+u2), then y3=0.5(y1+y2).If u3=u1-u2, then y3=y1-y2.Translation: 考慮具有輸入u輸出y的系統(tǒng)。在此系統(tǒng)上進行三次實驗，輸入分別為u1(t), u2(t) 和u3(t)，t=0。每種情況下，零時刻的初態(tài)x(0)都是相同的。相應的輸出表示為y1,y2 和y3。在x(0)不等于零的情況下，下面哪種說法是正確的？Answer: A linear system has the superposition property: In case 1: So y3y

10、1+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.Translation: 考慮輸入輸出關系如式的系統(tǒng),證明系統(tǒng)滿足齊次性,但是不滿足可加性.Answer: Suppose the system is initiall

11、y 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(0)=-1, n(1)=3. Then system outputs at time 1 are: So the system does not satisfy

12、 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.Translation: 說明系統(tǒng)如果具有可加性，那么對所有有理數(shù)a具有齊次性。因而對具有某種連續(xù)性質(zhì)的系統(tǒng)，可加性導致齊次性。Answer: Any rationa

13、l 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-output can be described as following: then: Suppose: Using additivity: So: It is to

14、 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: 設對于所有的t,T 和 a，g(t,T)=g(t+a,T+a)。說明g(t,T)僅依賴于t-T。Answer: Define: So: Then: So: It proves that g(t,T) depends only on t-T.2.9 Consider a system with impuls

15、e 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: 考慮沖激響應如圖2.20(a)所示的系統(tǒng)，由如圖2.20(b)所示輸入u(t)鼓勵的零狀 態(tài)響應是什么？Answer: Write out the function of g(t) and u(t): then y(t) equals to the convolution integral: If 0=t=1, 0=r=1, 0=

16、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: Answer: Applying the Laplace transform to system input-output equation, supposing that the System is initial relaxed: System transfer

17、 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)是線性定常系統(tǒng)的單位階躍響應。說明系統(tǒng)的沖激響應等于dy(t)/dt.Answer: Let m(t) be the impulse response, and system transfer function is G(s): So:2

18、.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: 考慮如式描述的兩輸入兩輸出系統(tǒng)， Nij和Dij 是p:=d/dt的多項式。系統(tǒng)的傳遞矩陣是什么？Answer: For any polynomial of p, N(p), its Laplace transform is N(s). Applying Laplace tran

19、sform 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. Show that the unit-step responses of the positive-feedback system are as shown in Fig2.21(a)

20、 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 Fig 2.21(c) and 2.21(d), respectively, for a=1 and a=0.5. Fig 2.21Translation: 考慮圖2.5中所示反應系統(tǒng)。說明正反應系統(tǒng)的單位階躍響應，當a=1時，如圖2.21(a)所示。當a=0.5時，如圖2.21(b)所示。說明負反應系統(tǒng)的單位階躍響應如圖2.21(c)和2.21(

21、b)所示，相應地，對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, respectively, as Fig 2.21(a) and Fig 2.21(b) shown. Secondly, consider the negative-fee

22、dback 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 Fig 2.21(d) shown.2.14 Draw an op-amp circuit diagram for 2.15 Find state equations to describe the pendulum system

23、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: 試找出圖2.22所示單擺系統(tǒng)的狀態(tài)方程。這個系統(tǒng)對研究一個或兩個連接的機器人操作臂很有用。假設角度都很小時，能否考慮系統(tǒng)為線性？Answer: For Fig2.22(a), the application of Newtons law to the linear mo

24、vements 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 law to the linear movements yields: Assuming , and , to be small, we can use the

25、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 shown in Fig2.24. The thrust generated is assumed to be proportional to the d

26、erivation 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 describe the system. Translation: 登月艙降落在月球時，軟著陸階段的模型如圖2.24所示。產(chǎn)生的沖激力與m的微分成正比。系統(tǒng)可以描

27、述如式所示形式。g是月球外表的重力加速度常數(shù)。定義狀態(tài)變量如式所示，試圖找出系統(tǒng)的狀態(tài)空間方程描述。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 Fig2.26.Find also its transfer function.Translation: 試寫出描述圖2.26Answer: 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 MATLAB to compute transfer function. We typ