120 Minutes Page 1. Reading-up-time

Transkript

1 120 Minutes Page 1 Reading-up-time For reviewing purposes of the problem statements, there is a reading-up-time of 10 minutes prior to the official examination time. During this period it is not allowed to start solving the problems. This means explicitly that during the entire reading-up-time no writing utensils, e.g. pen, pencil, etc. at all are allowed to be kept on the table. Furthermore the use of carried documents, e.g. books, (electronic) translater, (electronic) dictionaries, etc. is strictly forbidden. When the supervisor refers to the end of the reading-up-time and thus the beginning of the official examination time, you are allowed to take your utensils and documents. Please then, begin with filling in the complete information on the titlepage and on page 3. Good Luck! LAST NAME FIRST NAME MATRIKEL-NO. TABLE-NO. Klausurunterlagen Ich versichere hiermit, dass ich sämtliche für die Durchführung der Klausur vorgesehenen Unterlagen erhalten, und dass ich meine Arbeit ohne fremde Hilfe und ohne Verwendung unerlaubter Hilfsmittel und sonstiger unlauterer Mittel angefertigt habe. Ich weiß, dass ein Bekanntwerden solcher Umstände auch nachträglich zum Ausschluss von der Prüfung führt. Ich versichere weiter, dass ich sämtliche mir überlassenen Arbeitsunterlagen sowie meine Lösung vollständig zurück gegeben habe. Die Abgabe meiner Arbeit wurde in der Teilnehmerliste von Aufsichtsführenden schriftlich vermerkt. Duisburg, den (Unterschrift der/des Studierenden) Falls Klausurunterlagen vorzeitig abgegeben: Uhr

2 Bewertungstabelle Aufgabe 1 Aufgabe 2 Aufgabe 3 Aufgabe 4 Gesamtpunktzahl Angehobene Punktzahl % Bewertung gem. PO in Ziffern (Datum und Unterschrift 1. Prüfer, Univ.-Prof. Dr.-Ing. Dirk Söffker) (Datum und Unterschrift 2. Prüfer, Dr.-Ing. Yan Liu) (Datum und Unterschrift des für die Prüfung verantwortlichen Prüfers, Söffker) Fachnote gemäß Prüfungsordnung: 1,0 1,3 1,7 2,0 2,3 2,7 3,0 3,3 3,7 4,0 5,0 sehr gut gut befriedigend ausreichend mangelhaft Bemerkung:

3 Page 3 Attention: Give your answers to ALL problems directly below the questions in the exam question sheet. You are NOT allowed to use a pencil and also NOT red color (red color is used for corrections). This exam is taken by me as a mandatory (Pflichtfach) elective (Wahlfach) prerequsite (Auflage) subject (cross ONE option according to your own situation). Maximum achievable points: 100 Minimum points for the grade 1,0: 95% Minimum points for the grade 4,0: 50%

4 Problem 1 (30 Points) Page 4 1a) (3 Points) State the difference between a SISO and a MIMO control system with respect to the number of actuators/inputs. A SISO system has one input/actuator. (theoretically) arbitrary (at least one). The number of actuators in MIMO systems is For the following tasks, a system description is given with A = 0 0 1, B = 0, and C = [ ] with a 0 and b 0. a a 3 b It is noted, that no direct transmission between the inputs and the outputs/measurements is given. 1b) (5 Points) Calculate the transfer function matrix of the system depending on the parameters a and b. G(s) = C(sI A) 1 B +D 1 [ ] s(s+3) a s G(s) = a s(s+3) s 0 s 2 (s+3) sa a as a+as s 2 b 2bs 2 G(s) = s 3 +3s 2 as a 1c) (4 Points) State the Rosenbrock matrix of the system depending on the parameters a and b. Determine the invariant zeros, the transmission zeros, and the decoupling zeros of the system. P(s) = s s 1 0 a a s+3 b det(p(s)) = 2bs 2! = 0 Invariant zero: s 0 = 0 Transmission zero: s 0 = 0 Decoupling zeros: none

5 1d) (3 Points) For which parameters a and b is the system described by A,B,C fully controllable? Page 5 Q S = [ B AB A 2 B ] 0 0 b Q S = 0 b 3b b 3b ab+9b Requirement: det(q S )! 0 for full controllability For b 0 the system is fully controllable. 1e) (4 Points) Assuming full observability, a state observer should be designed for the previously given system A,B,C with λ 1 = 1, λ 2 = 2, and λ 3 = 1. For this subtask it can be assumed that a = 1. Calculate the gains of the observer matrix L = [ l 1 l 2 l 3 ] T. λ 1 2l 1 det(λi (A LC)) = det 0 λ 2l λ+3+2l 3 Characteristic polynomial: λ 3 +λ 2 (3+2l 3 )+λ(2l 2 +2l 1 1)+2l 2 1 comparing the coefficients with: λ 3 +2λ 2 λ 2 leads to the results: l 1 = 0.5, l 2 = 0.5, and l 3 = f) (6 Points) Assume b = 1. Calculate the gain matrix K = [ ] k 1 k 2 k 3 (depending on a) by using the pole placement method to realize state feedback control. The desired eigenvalues of the closed-loop system are given as λ 1 = 1, λ 2 = 2, and λ 3 = 1. λ 1 0 det(λi (A BK)) = det 0 λ 1 k 1 a k 2 a λ+k 3 +3 Characteristic polynomial: λ 3 +λ 2 (k 3 +3)+λ(k 2 a)+k 1 a comparing the coefficients with: λ 3 +2λ 2 λ 2 leads to the results: k 1 = 2+a, k 2 = 1+a, and k 3 = 1.

6 Problem 2 (20 Points) Given is the standard description of a linear MIMO system. The system described by b A = 2 1 1, B = b 2, C = 3 5 9, and D = Page 6 should be controlled using the state feedback matrix K = [ k 1 k 2 k 3 ]. 2a) (2 Points) Is it necessary for practical purposes to build an observer for the given system to realize full state feedback control? State reason. No, All states are measured. Rank C = 3 = n 2b) (3 Points) For which b 1, b 2 (b 1, b 2 0) is the system fully controllable? Q s = [ B AB A 2 B ] = Rank Q B! = n = 3 det Q B 0 b 1 0 b b 2 b 2 2b 1 b 2 4b 2 0 2b 1 +3b 2 6b 1 3b 2 Full controllability for 12b b 2 1b 2 3b 1 b b or 3(2b 1 +b 2 )(2b 1 b 2 )(b 1 1.5b 2 ) 0 b 2 2b 1 b 2 2b 1 b 1 1.5b 2

7 2c) (1 Point) For task 2b) is it possible to design the controller eigenvalues arbitrarily? Page 7 Yes, if the conclusion for full controllability is fulfilled. 2d) (5 Points) A modified system is given with A = a 0 a and B = [ 0 0 b ] T with b 0. Calculate k 1, k 2, and k 3 so that the eigenvalues of the closed-loop system are λ 1 = 3, λ 2 = 1, and λ 3 = 2. Characteristic equation of the closed-loop system: det(λi A + BK)=0 λ 4 0 λi A+BK = 3 λ 1 bk 1 a bk 2 λ a+bk 3 = λ 3 +(bk 3 a)λ 2 +(bk 2 12)λ+4bk 1 12bk 3 +8a = 0 λ 1 = 3 4bk 1 3bk 2 3bk 3 a+9 = 0 λ 2 = 1 4bk 1 +bk 2 11bk 3 +7a 11 = 0 λ 3 = 2 4bk 1 +2bk 2 8bk 3 +4a 16 = 0 = k 1 = (2a+3)/2b k 2 = 5/b k 3 = a/b

8 2e) (1 Point) State the equations to calculate left and right eigenvectors. Page 8 Left eigenvectors w i : w T i A = λ i w T i Right eigenvectors v i : Av i = λ i v i 2f) (5 Points) [ ] [ ] 0 1 b1 A modified system is given with A = and B =. Is the system stabilizable a 2 b [ ] 2 x (use Hautus criterion with x i = i1 as the left eigenvector of A)? State conditions. 1 λ 1 = 1 a+1, λ 2 = 1+ a+1 The eigenvalues λ 1 is stable for all values of a, but the stability of λ 2 depends on the value of a. Therefore it is necessary to check the controllability of λ 2. If it is fully controllable, the system is stabilizable. Using Hautus criterion, the controllability of the eigenvalue is guaranteed if x T 2B 0. x T 2(λ 2 I A)=0 [ ] a x T 2 = 0 a a+1+1 [ ] a+1+1 x 2 = 1 x T 2B = [ a ][ ] b 1 = b b 1 a+1+b1 +b 2 2 The system is stabilizable, = if b 1 a+1+b1 +b 2 0.

9 2g) (3 Points) Page 9 An optimal controller should be realized using standard routines solving Algebraic Riccati Equation. The system matrices are given with A, B; additionally Re{λ i (A)} < 0 is known. After long discussions the weighting matrices have been chosen as Q = , R = [ r 0 0 r ], r > 0. Which statement(s) is/are true (multiple choices possible: wrong answers are subtracted)? The open-loop system is not boundary stable. The design process depends not on A, B. If A, B is not fully controllable, the controlled system can still be stable. The chosen design process (LQR) is principally always establishing stable systems dynamics, if A, B is fully controllable. The chosen design parameters always guarantees perfect dynamical behavior with respect to stablity and dynamical properties.

10 Page 10 Problem 3 (25 Points) The mechanical system of a crane bridge with adjustable gripper length h(t) is shown below. The differential equations for the position of the trolley s K (t) and the gripper θ(t) (θ(t) h(t) φ(t)) can be described by s K (t) = m G(g ḧ(t)) θ(t)+ 1 F(t) and m K h(t) m K θ(t) = (m K +m G )(g ḧ(t)) θ(t) 1 F(t). m K h(t) m K Trolley m K s K (t) F(t) s G s φ(t) h(t) z G m G z Gripper 3a) (5 Points) Because of a misoperation the length of the gripper is not adjustable (h(t) is constant with h(t) = H). Derive for this case a model description in the form M q(t)+d q(t)+kq(t) = Vf(t) with q(t) = [ sk (t) θ(t) ], and also the state space model with F(t) as input, s K (t) as output, and the state vector x(t) = s K (t) ṡ K (t) θ(t) θ(t).

11 Page 11 h(t) = H ḧ(t) = 0 [ ] 0 m Gg 1 0 m + K H 0 1][ sk θ 0 (m K +m K )g m K H [ ] sk = θ 1 m K 1 m K F or: [ ] [ mk H 0 0 mg g + 0 m K H][ sk θ 0 (m K +m G )g ][ sk ] = θ [ ] H F H State space representation: m G g m A = K H , 0 0 (m K +m G )g 0 m K H B = 0 1 m K 0 1 m K, C = [ ], D = 0. ẋ = Ax+Bu y = Cx+Du

12 3b) (4 Points) Assume A = Page 12. Determine the characteristic polynomial of the system and check the asymptotic stability of the system using the Hurwitz criterion. Characteristic polynomial: λ 4 +15λ 3 +3λ 2 +10λ+1 i) All coefficents a i > 0. ii) H = H 1 = 15 > 0 H 2 = 35 > 0 H 3 = 125 > 0 H 4 = H 3 > 0 The system is asymptotic stable.

13 Page 13 3c) (6 Points) Assume measurements are available with C = The matrices A, B are given as A = and B = 0. An output feedback is realized by u(t) = K y y(t) + Vw(t) with K y = [1 0 1] and V = [3 2 1]. Determine the transfer function matrix G w (s) of the closed loop system. Assume x 0 = [ ] T. G w (s) = C(sI A+BK y C) 1 BV s = s 3 s s 0 [ ] s 2 s s 2 s 2 s 2 = s s s s 2 s 2 s = 1 s s s s s s s

14 3d) (2 Points) A system is given with A = [ ] , B = Determine the pole(s) and invariant zero(s) of the system. [ ] 2, C = [ 1 1 ], and D = 0. 1 Page 14 G(s) = C(sI A) 1 B +D = [ 1 1 ] 1 2 s 1 s s+1 = 3(s+1) s 2 1 3(s+1) = (s+1)(s 1) [ ] The pole of the system: s = 1 The invariant zero of the system: s 0 = 1 3e) (2 Points) Assume a system has three poles s 1 = 1, s 2 = 2, s 3 = 3, one transmission zero s o1 = 0, one output decoupling zero s o2 = 1, and no input decoupling zero. Is the system fully controllable? Is the system fully observable? State also the minimum order number of the system. State reasons. The system has no input decoupling zeros. fully controllable The system has one output decoupling zero. not fully observable Minimum order number = 4.

15 3f) (6 Points) A system is given with A = 1 0 0, B = 1, C = [ c 0 1 ], and D = a b For which values of a, b, and c is the system Page 15 i) asymptotic stable? ii) fully controllable? iii) fully observable? i) The eigenvalues of the system can be calculated by det(λi A) =! 0 λ 1 0 det(λi A) = det 1 λ 0 =! λ a (λ 2 1)(λ a) = 0 λ 1 = 1, λ 2 = 1, λ 1 = a. Because there is a positive eigenvalue λ 1 = 1, the system is NOT asymptotic stable. ii) Using Kalman criterion to check the controllability: Q S = [ B AB A 2 B ] = b ab 2 a 2 b 2a 1 The system is fully controllable if det(q S ) 0 det(q S ) = 2a a 2 b+b+1 0 iii) Using Kalman criterion to check the observability: C c 0 1 Q B = CA = 1 c 2 a CA 2 c a 2 2a 1 a 2 The system is fully observable if det(q B ) 0 det(q B ) = a 2 c 2 +2ac c 2 +4c 3 0

16 Page 16 Problem 4 (25 Points) Gripper systems are typical devices applied for manufacturing tasks. They enable the system to handle sensitive system components. For a robot gripper system, as shown in Fig. 4.1, a control algorithm has to be designed. The gripper consists of three fingers with three rotational joints each. The kinematic relations of a finger are illustrated in Fig α 3 (t) α 2 (t) τ 3 (t) α 1 (t) τ 2 (t) Robotiq Fig. 4.1: Gripper system (Source: Robotiq) τ 1 (t) Fig. 4.2: Kinematics of the gripper system In this context α(t) = [ α 1 (t) α 2 (t) α 3 (t) ] T denotes the angles of the rotational joints and τ(t) = [ τ 1 (t) τ 2 (t) τ 3 (t) ] T the torques applied on them by the actuator. For the dynamical behavior of this system a simplified relation given by J a J b J c D a D b D c K a K b K c J b J a J b α(t)+ D b D a D b α(t)+ K b K a K b α(t) = τ(t) (4.1) J c J b J a D c D b D a K c K b K a can be assumed, where J a,b,c denote the inertia, D a,b,c the damping, and K a,b,c the stiffness coefficients of the system.

17 4a) (1 Point) Give the order of the system stated above when described by a system matrix A? Page 17 n = 6 4b) (6 Points) Give the state space representation of the system defined by the equations stated in eq. 4.1 in terms of the matrices A,B,C,D. Therefore use the state vector x(t) = [ α 1 (t) α 1 (t) α 2 (t) α 2 (t) α 3 (t) α 3 (t) ] T. The torques τ i (t), i = 1,2,3 of the actuators have to be applied as system inputs and the angles of the joints α i (t), i = 1,2,3 have to be considered as system outputs. How can the input/output structure be classified (SISO, SIMO, MISO, or MIMO)? For the purpose of simplification, assume J b = J c = 0. A = B = K a /J a D a /J a K b /J a D b /J a K c /J a D c /J a K b /J a D b /J a K a /J a D a /J a K b /J a D b /J a K c /J a D c /J a K b /J a D b /J a K a /J a D a /J a ẋ = Ax+Bu y = Cx+Du /J a /J a /J a MIMO System, C = , D =

18 4c) (2 Points) Page 18 Considering the conditions within a specific working range of the system the dynamic behavior can be simplified as A = (p 1 +2p 3 ) 3/ p , B = 7.886p 3, C = [ ], and D = 0 with p 1, p 2, p 3 > 0. Give the eigenvalues of the system in terms of the parameters p 1, p 2, p 3. What can be concluded concerning the stability [asymptotic stable, stable, unstable] of the system? λ 1 = 1.371, λ 2 = p 2, λ 3 = unstable for p 2 > 0.

19 4d) (2 Points) Page 19 For a similiar system the eigenvalues are obtained as λ 1,2 = 1±2 (p 1 4p 2 ) 1/2, and λ 3 = For which range of parameters p i > 0, i = 1,2 is the system asymptotic stable? λ 2 = 1+ p 1 4p 2 > p 1 4p 2 < 0

20 Page 20 4e) (6 Points) Using a specific type of a differential gear the dynamical behavior of the system changes to A = d 1 d 2 2, B = d 3, C = [ ], and D = 0, with d i > 0, i = 1,2,3. For d 1 = d 2 = d 3 = 2 the eigenvalues of this system are λ 1 = λ 2 = 1, and λ 3 = 2. Which of these eigenvalues are observable, which are controllable? Use suitable criteria to check these properties. Q S = [ A AB A 2 B ] = Q B = C CA CA 2 ( λ1 I A λ 1 = 1 : rank C ( λ3 I A λ 3 = 2 : rank C = 0 not fully observable ) = rank ) = rank, det(q S ) = 26 0, fully controllable = 3 observable = 2 not observable

21 4f) (8 Points) Page 21 For a similiar system the eigenvalues λ 1 = 4, λ 2 = 1, and λ 3 = 2 and the corresponding eigenvectors q 1 = [ ] T, q2 = [ ] T, and q3 = [ ] T are given. The B and C matrices are known as 4f 1 B = f 1 f 2 and C = [ f 1 f 2 1 ]. 1 f 1 Check the range of the parameters f i,i = 1,2 for the system to be fully controllable. Therefor use the Gilbert criterion. For which parameters is the system fully observable? V = 0 1 0, V 1 = B = V 1 B = 5f 1 0.5f f 1 f 2 f f fully controllable, if 5f 1 0.5f f 1 f 2 f f C = CV = [ f 1 f 2 +1 f 1 +2 ] fully observable, if f 1 0 f 2 1 f 1 2