线性系统理论来linear system theory and d答案

1、Linear System Theory and DesignSA01010048 LING QING2.1 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 newoutput so that the system in Fig 2.19(b) is linear?Figure 2.

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:y = a * uHere a is a constant. It is a memoryless system. Easy to testify that it is a linear system. The input-output relation in Fi

3、g 2.1(b) can be described as:y = a * u + bHere a and b are all constants. Testify whether it has the property of additivity. Let:y1 = a * u1 + by2 = a * u2 + bthen:( y1 + y2 ) = a * (u1 + u2 ) + 2 * bSo it does not has the property of additivity, therefore, is not a linear system. But we can introdu

4、ce a new output so that it is linear. Let:z = y - bz = a * uz is the new output introduced. Easy to testify that it is a linear system. The input-output relation in Fig 2.1(c) can be described as:y = a(u) * ua(u) is a function of input u. Choose two different input, get the outputs:y1 = a1 * u11Line

5、ar System Theory and DesignSA01010048 LING QINGy2 = a2 * u2Assure:a1 ¹ a2then:( y1 + y2 ) = a1 * u1 + a2 * u2So 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 byg(t) = 2w sin 2w (t - t0 )2w (t - t0 )for

6、all t, where w and to 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

7、output at time ts, excited by 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 inthe real world.2.

8、3 Consider a system whose input u and output y are related byfort £ ay(t) = (P u)(t) := ìu(t)íafort > a0îwhere a is a fixed constant. The system is called a truncation operator, which chops off theinput after time a. Is the system linear? Is it time-invariant? Is it causal?Tra

9、nslation: 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: Consider the input-output relation at any time t, t<=a:y = uEasy to testify that it is linear.Consider the input-output relation at any time t, t>a:y = 0Easy to testify that it is linear. So for any t

10、ime, 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<a, so It decides the system output y(t), t>=to:2Linear System Theory and DesignSA01010048 LING QINGfort0 £ t £ aforotherty(t) = ìu(t)í

11、;0îShift the initial time to to+T. Let to+T>a , then input is u(t-T), t>=to+T. System output:y'(t) = 0Suppose 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 curre

12、nt input u(t) exclusively. So it is acausal system.2.4 The input and output of an initially relaxed system can be denoted by y=Hu, where H is some mathematical operator. Show that if the system is causal, thenPa y = Pa Hu = Pa HPa uwhere Pa is the truncation operator defined in Problem 2.3. Is it tr

13、ue 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=HPau 是正确的吗？Answer: Notice y=Hu, so:Pa y = Pa HuDefine the initial time 0, since the system is causal, output y begins in time 0. If a<=0,then u=Hu. Add operation PaH in both left and right

14、 of the equation:Pa Hu = Pa HPa uIf a>0, we can divide u to 2 parts:for0 £ t £ ap(t) = ìu(t)í0forothertîfort > aq(t) = ì u(t)íî0forothertu(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 i

15、s to say, system output from 0 to a is decided only by p(t). Since PaHu chops off Hu after time a, easy to conclude PaHu=PaHp(t). Notice that p(t)=Pau, also we have:Pa Hu = Pa HPa uIt means under any condition, the following equation is correct:Pa y = Pa Hu = Pa HPa uPaHu=HPau is false. Consider a d

16、elay 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 3.3Linear System Theory and DesignSA01010048 LING QING2.5 Consider a system with input u and output y. Three experiments are performed on the system using the inputs

17、 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 followingstatements are correct if x(0)<>0?1.2.3.If u3=u1+u2, then y3=y1+y2.If u3=0.5(u1+u2), then y3=0.5(y1+y2). If u3=u1-u2,

18、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，下面哪种说法是正确的？Answer: A linear system has the superposition property:a1x1 (t0 ) + a 2 x2 (t0 )ü ® a y (t) + a y (t), t ³ týa u (t) +

19、a u (t), t ³ t1 12 200 þ1 12 2In case 1:a1 = 1a 2 = 1a1x1 (t0 ) + a 2 x2 (t0 ) = 2x(0) ¹ x(0)So y3<>y1+y2.In case 2:a1 = 0.5a 2 = 0.5a1x1 (t0 ) + a 2 x2 (t0 ) = x(0)So y3=0.5(y1+y2).In case 3:a1 = 1a 2 = -1a1x1 (t0 ) + a 2 x2 (t0 ) = 0 ¹ x(0)So y3<>y1-y2.2.6 Consider

20、a system whose input and output are related byìu 2 (t) / u(t - 1)ifu(t - 1) ¹ 0y(t) = íî0ifu(t - 1) = 0for all t.Show that the system satisfies the homogeneity property but not the additivity property. 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 i

21、nitially relaxed, system input:p(t) = au(t)a is any real constant. Then system output q(t):4Linear System Theory and DesignSA01010048 LING QINGì p 2 (t) / p(t - 1)ifp(t - 1) ¹ 0q(t) = íî0ifp(t - 1) = 0ìau (t) / u(t - 1)ifu(t - 1) ¹ 02=íî0ifu(t - 1) = 0So it sa

22、tisfies 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:r(1) = m2 (1) / m(0) = 4s(1) = n 2 (1) / n(0) = -9y(1) = m(1) + n(1)2 /m(0) + n(0) = 0¹ r(1) + s(1）So the sys

23、tem 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.Translation: 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

24、齐次性。Answer: Any rational number a can be denoted by:a = m / nHere m and n are both integer. Firstly, prove that if system input-output can be described as following:x ® ythen:mx ® myEasy to conclude it from additivity.Secondly, prove that if a system input-output can be described as follow

25、ing:x ® ythen:x / n ® y / nSuppose:x / n ® uUsing additivity:n * (x / n) = x ® nuSo:y = nuu = y / n5Linear System Theory and DesignSA01010048 LING QINGIt is to say that:x / n ® y / nThen:x * m / n ® y * m / n ax ® ayIt is the property of homogeneity.2.8 Let g(t,T)=

26、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:x = t + Ty = t - TSo:x + yx - yt =T =22Then:g(t,T ) = g( x + y , x - y )22= g( x + y + a, x - y + a)22= g( x + y + - x + y , x - y + - x + y )2= g( y,0)22

27、2So:¶g (t,T ) = ¶g( y,0) = 0¶x¶xIt 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-stateresponse excited by the input u(t) shown in Fig2.20(b)?Fig2.206Linear System Theory and DesignSA01010048 LING QINGTr

28、anslation: 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) and u(t):t0 £ t £ 1g (t) = ìí2 - t1 £ t £ 2î0 £ t £ 1u(t) = ì 1í- 1 1 £ t £ 2îthen y(t) equals to the convolution integral:ty(

29、t) = ò g(r)u(t - r)dr0If 0=<t=<1, 0=<r=<1, 0<=t-r<=1:ty(t) = ò rdr =0t 22If 1<=t<=2:t -11ty(t) = ò g(r)u(t - r)dr + ò g(r)u(t - r)dr + ò g(r)u(t - r)drt -101= y1 (t) + y2 (t) + y3 (t)Calculate integral separately:t -1y1 (t) = ò g(r)u(t - r)dr00 &

30、#163; r £ 11 £ t - r £ 2t -1= ò - rdr =0- (t - 1)221y2 (t) = ò g(r)u(t - r)drt -10 £ r £ 10 £ t - r £ 111(t - 1)2= ò rdr = 2 -2t -1ty3 (t) = ò g(r)u(t - r)dr11 £ r £ 20 £ t - r £ 1t= ò (2 - r)dr = 2(t - 1) -t 2 - 121y(t)

31、 = y (t) + y (t) + y (t) = - 3 t 2 + 4t - 212327Linear System Theory and DesignSA01010048 LING QING2.10 Consider a system described by····y+ 2 y- 3y = u- uWhat 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

32、 the Laplace transform to system input-output equation, supposing that the System is initial relaxed:s 2Y (s) + 2sY (s) - 3Y (s) = sY (s) - Y (s)System transfer function:s - 1G(s) = U (s) =1=s + 3s 2 + 2s - 3Y (s)Impulse response:g(t) = L-1G(s) = L-11 = e-3ts + 32.11 Let y(t) be the unit-step respon

33、se of a linear time-invariant system. Show that the impulseresponse of the system equals dy(t)/dt.Translation: y(t)是线性定常系统的阶跃响应。说明系统的冲激响应等于 dy(t)/dt.Answer: Let m(t) be the impulse response, and system transfer function is G(s):Y (s) = G(s) * 1s(s)M (s) = Y (s) * sSo:m(t) = dy(t) / dt2.12 Consider a

34、 two-input and two-output system described byD11 ( p) y1 (t) + D12 ( p) y2 (t) = N11 ( p)u1 (t) + N12 ( p)u2 (t)D21 ( p) y1 (t) + D22 ( p) y2 (t) = N 21 ( p)u1 (t) + N 22 ( p)u2 (t)where 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， N

35、ij 和 Dij 是 p:=d/dt 的多项式。系统的传递矩阵是什么？Answer: For any polynomial of p, N(p), its Laplace transform is N(s).Applying Laplace transform to the input-output equation:D11 (s)Y1 (s) + D12 (s)Y2 (s) = N11 (s)U1 (s) + N12 (s)U 2 (s)D21 (s)Y1 (s) + D22 (s)Y2 (s) = N 21 (s)U1 (s) + N 22 (s)U 2 (s)8Linear System

36、 Theory and DesignSA01010048LING QINGWrite to the form of matrix:é D11 (s)D12 (s)ùéY1 (s)ùé N11 (s)N12 (s)ùéU1 (s)ùêDD (s)úêY (s)ú = êN (s)N (s)úêU (s)ú(s)ë 2122ûë 2ûë2122ûë 2û(s)&

37、#249;-1 é NéY (s)ùé D(s)D(s)N(s)ùéU (s)ù1111211121ú = êêúêúêúëY2 (s)ûëD21 (s)D22 (s)ûëN 21 (s)N 22 (s)ûëU 2 (s)ûSo the transfer function matrix is:(s)ù-1 é Né D(s)ù(

38、s)D(s)N11121112G(s) = êúêúëD21 (s)D22 (s)ûëN 21 (s)N 22 (s)ûBy the premising that the matrix inverse:(s)ù-1é D(s)D1112êD(s)D (s)úë 2122ûexists.2.11 Consider the feedback systems shows in Fig2.5. Show that the unit-step responses o

39、f 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 中所示反馈系统。说明正反馈系统的阶跃响应，当 a=1 时，如图 2

40、.21(a)所示。当 a=0.5 时，如图 2.21(b)所示。说明负反馈系统的如图 2.21(c)和 2.21(b)所示，相应地，对 a=1 和 a=0.5。阶跃响应9Linear System Theory and DesignSA01010048 LING QINGAnswer: Firstly, consider the positive-feedback system. Its impulse response is:¥g(t) = å aid (t - i)i=1Using convolution integral:¥y(t) = å ai

41、r(t - i)i=1When input is unit-step signal:ny(n) = å aii=1y(t) = y(n)n £ t £ n + 1Easy 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-feedback system. Its impulse response is:¥g(t) = -å(-a)i d

42、(t - i)i=1Using convolution integral:¥y(t) = -å(-a)i r(t - i)i=1When input is unit-step signal:ny(n) = -å(-a)ii=1y(t) = y(n)n £ t £ n + 1Easy 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 op-amp circuit diagram

43、foré- 24ùéù2·x = ê5ú x + ê- 4úu0ëûëûy = 3 10x - 2u2.15 Find state equations to describe the pendulum system in Fig 2.22. The systems are useful toone- or two-link robotic manipulators. If q , q1 and q 2 are very small, can youmconsider

44、 theas linear?10Linear System Theory and DesignSA01010048 LING QINGTranslation: 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，能否考虑系统为线性？Answer: For Fig2.22(a), the application of Newtons law to the linear movements yields:· 2d 2··f cosq - mg = m(l cosq ) = ml(-q sinq - q cos

45、q )dt 2d 2dt 2· 2··u - f sinq = m(l sinq ) = ml(q cosq - q sinq )·Assuming q and q to be small, we can use the approximation sinq =q ,cosq =1.·By retaining only the linear terms in q and q , we obtain f = mg and:··g1q = -q +ulml·Select state variables as x1 =

46、q , x2 = q and output y = q= é01ù + éù1·x0ú xê1/ mlúuê- g / lëûëûy = 10xFor Fig2.22(b), the application of Newtons law to the linear movements yields:d 2f1 cosq1 - f 2 cosq 2 - m1 g = m1(l1 cosq1 )dt 2···= m l (-sinq -2

47、cosq )q 1q 11 111d 2f 2 sinq 2 - f1 sinq1 = m1(l1 sinq1 )dt 2···= m l (cosq -2 sinq )q 1q 11 111d 2f 2 cosq 2 - m2 g = m2(l1 cosq1 + l2 cosq 2 )dt 2······= m l(-sinq -2 cosq ) + m lq 1q 1(-sinq -2 cosqq 2q 2)2 1112 222d 2u - f 2 sinq 2 = m2(l1 sinq1 + l2

48、sinq 2 )dt 2······= m l (cosq -2 sinq ) + m lq 1q 1(cosq -2 sinq )q 2q 22 1112 22211Linear System Theory and DesignSA01010048 LING QING··Assuming q1 , q 2 andq 1 , q 2to be small, we can use the approximation sinq1 =q1,sinq 2 =q 2 , cosq1 =1, cosq 2 =1. By ret

49、aining only the linear terms inq1, q 2 and··q 1 , q 2 , we obtainf 2 = m2 g , f1 = (m1 + m2 )g and:(m1 + m2 )gm2 g··q 1q1 +q 2= -m lm l1 11 1(m1 + m2 )g(m1 + m2 )g··1q 2=q1 -q 2 + mum lm ll1 21 22 2··Select state variables asx1 = q1= q 1, x3 = q 2= q 2x2, x4,a

50、nd outputé y1 ùéq1 ùê y ú = êqú :ë 2 ûë 2 ûé ·ùéê0100000ùé x ùéêêêùú000ê x1 ú1ú2 úúúê·- (m + m )g / m lêúm g / m l0xx 2=

51、 êúêúu1201 121 10+ê ·úê1úê xúêx3 ú3 úêúêêúê ·ú(m + m )g / m l- (m + m )g / m l0x1/ m lë122 2ûë 4 ûë2 2 ûëx 4 û1 212é x1 ù0ù êx

52、0;é y ùé1000ú ê2 ú1êú = ê10û ê x3 úë y2 ûë0êx úë 4 û2.17 The soft landing phase oar module descending on the moon can be med as shownin Fig2.24. The thrust generated is assumed to be proportional to the de

53、rivation of m, wherem is the mass of the module. Then the system can be described by···m y = -k m- mgWhere g is the gravity constant on the lunar surface. Define state variables of the system as:··x1 = y , x2 = y ,x3 = m , y = mFind a state-space equation to describe the sys

54、tem.Translation: 登月舱降落在月球时，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:度常数。定12Linear System Theory and DesignSA01010048LING QING-······y = -gt 2 / 2 + yy = -gt + y-y =

55、-g + y-··m = m0 + mm = mSo:-···-(m0 + m) (-g + y) = - k m - (m0 + m)g-···- m0 g - m g + m0 y = - k m - m0 g - m g-···m0 y = - k mDefine state variables as:-·-·= y ,= y ,= m ,y = mx1x 2x3Then:éùéù·-ê x 

56、0;êúxê1 ú1êúé01000ùé0ù·ú êúêú-ê-êú-0ú êx2 ú + ê- k / m0 ú uêx2 ú = ê0·êúêúêë00úû ê -êëúû1-ú

57、;êúxê x3 ú3êúëûëûéù-ê x1 úêúêú-y = 100 xê2 úêú-ê x3 úêúëû2.19 Find a state equation to describe the network shown in Fig2.26.Find also its transfer function.Tra

58、nslation: 试写出描述图 2.26 所示网络的状态方程，以及它的传递函数。Answer: Select state variables as:x1 : Voltage of left capacitorx2 : Voltage of right capacitorx3 : Current of inductorApplying Kirchhoffs current law:13Linear System Theory and DesignSA01010048LING QING·x3 = (x2 - L x3 ) / R·u = C x1 + x1 R u = C x

59、2 + x3y = x2From the upper equations, we get:··x1 = -x1 / CR + u / C = -x1 + ux2 = -x3 / C + u / C = -x3 + u··x3 = x2 / L - Rx3 / L = x2 - x3y = x2They can be combined in matrix form as:éù·ê x1 úé- 1ê0 ù é x1 ùé1ù0·

60、ú êúê úêú0- 1ú êx2 ú + ê1úux=0êêú2·ê 01- 1ú ê xê0úúêúëû ë 3 ûë ûêë x3 úûé x1 ùy = 0 10êxúê2 úê&

61、#235; x3 úûUseto 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 =0D1 =1.00002.00001.00001.00002.00002.00001.0000So the transfer function is:s 2 + 2s + 1s + 1G(s) =s3 + 2s 2 + 2s + 1s 2 + s + 114Linear System Theory and DesignSA01010048 LING QING2.20 Find a state equation to describ