Exam. 135 minutes, 15 minutes reading time
|
|
- Jade Simpson
- 5 years ago
- Views:
Transcription
1 Exam August 15, 2017 Control Systems I ( L) Prof Emilio Frazzoli Exam Exam Duration: 135 minutes, 15 minutes reading time Number of Problems: 44 Number of Points: 52 Permitted aids: Important: 40 pages (20 sheets) of A4 notes, Appendix D of the book Lino Guzzella: Analysis and Synthesis of Single-Input Single-Output Control Systems, simple calculators will be provided upon request. Questions must be answered on the provided answer sheet; answers given in the booklet will not be considered. There exist multiple versions of the exam, where the order of the answers has been permuted randomly. There are two types of questions: 1. One-best-answer type questions: One unique correct answer has to be marked. The question is worth one point for a correct answer and zero points otherwise. Giving multiple answers to a question will invalidate the answer. These questions are marked 2. True / false type questions: All true statements have to be marked and multiple statements can be true. If all statements are selected correctly, the full number of points is allocated; for one incorrect answer half the number of points; and otherwise zero points. These questions are marked Mark all correct statements. (2 Points). No negative points will be given for incorrect answers. Partial points (Teilpunkte) will not be awarded. You do not need to justify your answers; your calculations will not be considered or graded. Use only the provided paper for your calculations; additional paper is available from the supervisors. Use pens producing a dark, solid and permanent line. The use of pencils is not allowed. Good luck!
2
3 1 Systems and Control Architecture Box 1: Questions 1, 2 A system is described by the equation y(t) = f(u(t)) where f is the function describing the system, u(t) is an input and y(t) is an output signal. u(t) f y(t) Question 1 Mark all correct statements. (2 Points) Which of these systems is not causal? A y(t) = u 2 (t) + 10 B y(t σ) = 2e u(t) σ > 0 C y(t) = u(t + σ) σ > 0 D y(t σ) = 1 σ > 0 Question 2 Which one of these systems is linear? A y(t) = u(t σ) + u(t + σ) σ > 0 B y(t) = 1 C y(t) = (u(t)) 2 D y(t) = au(t) + b a, b R, a 0 Question 3 Your friend and you want to design a control system for an unstable plant that has to operate under various disturbances. A perfect model of the plant is at hand and you have perfect knowledge about it, however you do not have perfect knowledge of the disturbances that might be affecting the plant. Your friend proposes the design of a feedback controller to stabilize the plant. What do you think? A Both a feedforward and feedback controller designs would work. Hence we will choose whichever is easier to implement. B Since we know everything about the plant, we should use the feedforward controller as in this case we will achieve the perfect reference tracking. C I agree with him, we need a feedback controller.
4 2 System Modeling and Analysis Box 2: Questions 4, 5 You are enthusiastic about control systems and decide to learn more about it by building a seesaw as shown in the figure below. You connect two motors with propellers on either side of the seesaw to control the angle α of the seesaw. The propellers have angular velocities ω 1 and ω 2 as shown in the figure and have a distance l from the center of the seesaw. An additional point mass with weight m w and distance to the center of the seesaw l w is present. ω 1 m w, l w ω 2 α α Question 4 Identify which signals are the input and output of the system. A Input: α; Output: ω 1, ω 2 B Input: ω 1, ω 2 ; Output: α C Input: α; Output: ω 1, ω 2 D Input: ω 1 ; Output: ω 2 Question 5 Let J be the angular inertia (including the point mass m w ) of the system. Furthermore, let F 1, F 2 be the forces generated by the propeller on the respective sides of the seesaw. Hint: E kin = 1 2 J α2, E pot = mgh, where m is a mass, g is earths gravitation and h represents height. Then looking at the kinetic and potential energy reservoirs of the system d dt (reservoir content) = inflows outflows, results in which equation of motion: A B α = α = l(f2 F1) mwglw cos α 2J mwglw cos α J C D α = α = (F2 F1)+mwg cos α J l(f2 F1) mwglw cos α J
5 Question 6 You have to design a speed sensor for an electric motor of a new locomotive. First, you need to normalize the physical state variable of the system, i.e. the motor speed z(t) = ω(t), such that the normalized state signal has its maximum value at 1 under regular operating conditions. The gearing is chosen, such that the electric motor has the highest efficiency at the maximum train speed of 200kmph. The gear transmission ratio is γ = 14rpm/kmph. From the data sheet of the electric motor you see that it can reach a speed of 4 000rpm without load. How do you choose the normalizing value z 0? A 2800 rpm. B 200 kmph. C 4000 rpm. D 286 kmph.
6 Box 3: Questions 7, 8, 9, 10, 11 You have passed the Control Systems 1 exam and now you are very bored. To do something more exciting, you decide to build a jet-kart, as shown in the figure below. However, before the fun starts you need to model it and design a controller for it. Assume that the cart moves in one direction only (1D motion). To control your vehicle you use thrust from the jet engine (this is your control input) and you are interested in controlling the kart s position. Assume that there are only three kinds of forces acting on the kart: Thrust force F T H (t) = k T H T (t) where T (t) is the thrust from the jet engine (can be both positive or negative) and K th is a constant. High velocity drag force F DH = k DH v 2 (t) where k DH is a constant and v(t) is the linear velocity of the vehicle. Viscous drag force F D = k D v(t) where k D is a constant and v(t) is the velocity. Question 7 Let m is the mass of the vehicle and x(t) its position. Which differential equation models your system? A mẍ k DH ẋ 2 k D ẋ = k T H T (t) B mẍ + k DH ẋ 2 + k D ẋ = k T H T (t) C mẍ k DH ẋ 2 k D ẋ = k T H T (t) D mẍ + k DH ẋ 2 + k D ẋ = 0 Question 8 If you represent your differential equation of the jet kart in state space representation ẋ = f(x, u), y = g(x, u), what is the dimension of the state vector, i.e. if x R n, what is n? A 3 B Cannot be determined from the information given C 1 D 2
7 Question 9 Assume that your jet kart is in a standstill and you d like to accelerate, so you hit the throttle. Unfortunately, something went wrong and instead of producing a steady thrust, the thrust produced by the jet engine is oscillatory. Jet engine s thrust can be described with T (t) = sin(ωt), where ω = 1 rad/s. Other parameters are m = 1000 kg, K TH = 100, k DH = 0 Ns 2 /m 2 and k D = 100 Ns/m. What is the value of jet-kart s velocity response at time t = 20 s? A v(20) 0.11 m/s B v(20) m/s C v(20) m/s D v(20) m/s Question 10 Assume that the jet kart is moving at the constant velocity v = 100 m/s and you turn off the jet engine to slow down. How much time until you can drive through the 30 Zone here in Zurich? (i.e. how much time until you reach a velocity of v = 30 km/h?). For your calculations use m = 1000 kg, k DH = 0 Ns 2 /m 2 and k D = 100 Ns/m. A t 2.5 s B t 250 s C I will never reach that speed since since my jet engine is off D t 25 s Question 11 Assuming that k DH = 0 Ns 2 /m 2 and the system s output is the position x(t) of the jet kart. Is the jet kart BIBO stable? A Cannot be determined from the data given B Yes C No Question 12 Consider a linear time-invariant SISO system. stability criteria. Pick a correct logical relation between different A Asymptotically stable BIBO stable B Asymptotically stable = BIBO stable C Asymptotically stable BIBO stable = Lyapunov stable D Asymptotically stable = BIBO stable Lyapunov stable
8 Question 13 Recall the definition of Lyapunov stability from the lecture: A system is Lyapunov stable if for any bounded initial condition and zero input, the state remains bounded. Let the discrete update equation of state x R for k = 0, 1, 2,... be given by x k+1 = ax k + bu k For which of the following intervals of a is the state x Lyapunov stable? A a [ 0.5, 0.5] B a [ 1, 2] C a (, 0.5] D a (, 0] Question 14 For which values of a is the state BIBO stable (assuming b 0)? A a ( 1, 2) B a (, 0) C a [ 1, 1] D a ( 1, 1) Question 15 Which of the following transfer functions is not causal? A g(s) = s2 +2 s B g(s) = 1000 C g(s) = s3 s 3 +s+2 D g(s) = s+2 s 2 +4s+3 Box 4: Question 16 Consider the state-space system below, where a is a real parameter: ẋ(t) = y(t) = [ 2 [ ] 1 0 x(t) a ] x(t) [ ] a u(t) 1 Question 16 For which values of a is the state-space realization minimal? A a {0, 10} B a = {0, 10} C a = {0, 5, 10} D a = {0, 5} E a {0, 5, 10} F a {0, 5}
9 Question 17 You have derived the nonlinear dynamics of a simple seesaw system to be α = 0.5 cos(α) + 2u, where α(t) denotes the seesaw angle and u(t) denotes the input used to turn the seesaw. You linearize the equation around the equilibrium point α e = 0 to yield ẍ = 2u, where ẍ represents the linearized angular acceleration. For which angle α does linearized angular acceleration deviate not more then ε = 0.2 rad s 2 nonlinear (correct) acceleration? from the A α < ẍ α < ε B α < C α < D α < 60.87
10 3 Frequency Domain Box 5: Questions 18, 19, 20, 21, 22 To keep your annoying younger sibling away, you gave them the seesaw with propellers on its endpoint in the picture below to play with. However, your younger sibling is not an expert in control systems so they need help with the controller design. The seesaw is fitted with two propellers that can produce thrust, a spring and a damper (both located in the seesaw s joint in the middle). You can assume that the joint is located exactly in the middle of the seesaw. Parameters describing the seesaw are its moment of inertia around the center of mass I [kgm 2 ], spring constant k [N m/rad] and damping coefficient d [N ms/rad]. α F 1 α F 2 Question 18 For which values of the spring parameters I, d and k is the seesaw asymptotically stable? A d 0 and k > 2 B I > 0, d > 0 and k > 0 C d 0 and I > 2 k D d 2I 0 Question 19 Assume that I = 1 [kgm 2 ], k = 1 Nm/rad, l = 1 [m] and d = 2 [Nms/rad] and that the input to the system is a difference between forces F 1 and F 2 : u = F 1 F 2 (note that this is a single input and not two inputs). In this question you can observe the output α of the system. What is the value of system s transfer function evaluated at s = i where i = 1? A g(i) = i 2 B g(i) = 1 C g(i) = 1 2i D g(i) = 1 2 Question 20 Unlike in Question 19, you can now observe the output α of the system. What is the value of systems s transfer function evaluated at s = 2i using the same numerical parameter values as in Question 19? A g(2i) = 1 4i+3 B g(2i) = 1 i 1 C g(2i) = 1 4i 3 D g(2i) = 2i 4i 3
11 Question 21 Assume that I = 0.1 [kgm 2 ] and d = 2 [Nms/rad]. For which values of spring constant k you expect to see oscillations in the time response to a step input? A k < 10 B k > 10 C k 10 D k 10 Question 22 Assume that I = 1 [kgm 2 ], k = 1 Nm/rad and d = 2 [Nms/rad]. What is the mode associated with the eigenvalue 1? A B 1 2 [ 1 1 [ 1 0] ] [ 0 C 1] [ ] D Box 6: Question 23 Given an LTI system ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) and the general solution y(t) = Ce At x(0) + C t 0 e A(t τ) Bu(τ)dτ + Du(t). Let the input of the system be an exponential function of time u(t) = e st. In addition the initial condition is zero and there is no feedthrough. x(0) = 0 D = 0 Question 23 Mark all correct statements. (2 Points) What function is the output y(t) converging to, in case the system is asymptotically stable and what function, in case it is unstable? Hint: Determine the steady-state response. A y unstable (t) = (C(sI A) 1 B)e st. B y stable (t) = C(sI A) 1 e At B + (C(sI A) 1 B)e st. C y unstable (t) = C(sI A) 1 e At B + (C(sI A) 1 B)e st. D y stable (t) = (C(sI A) 1 B)e st.
12 Magnitude (db) Magnitude (db) Phase(deg) Phase(deg) Frequency (rad/sec) Frequency (rad/sec) System 1 System 2 Magnitude (db) Magnitude (db) Phase(deg) Phase(deg) Frequency (rad/sec) Frequency (rad/sec) System 3 System 4 Figure 1: Bode plots of four different systems.
13 Box 7: Questions 24, 25 Consider the Bode plots of four different systems are shown in figure 1. Question 24 Consider the open-loop transfer function L(s) = 1000 is the Bode plot corresponding to it? s s+1/2 2 (s+1000). Which of the plots in Figure 1 A System 1 B System 4 C System 2 D System 3 Question 25 Mark all correct statements. (2 Points) Now assume that the Bode plots shown in in Figure 1 show open loop gains of four different systems and that the plant has the transfer function P (s) = 2 s. 2 A System 4 has a nonminimumphase zero. B Out of the four systems shown, system 1 is with the best realizable controller for P (s) = 2 s 2. C The response of system 2 to a unit step is unstable. D System 3 has a nonminimumphase zero. E System 1 has the best phase margin. F Out of the four systems shown, system 3 is with the best realizable controller for P (s) = 2 s 2.
14 Imaginary Axis Corrected 0.2 Nyquist Diagram Real Axis Figure 2: Nyquist plot of second-order system Question 26 You are given the Nyquist plot of a second-order open-loop transfer function L(s) shown in figure 2. What is the closed-loop system s steady-state error e to a step input of magnitude 1? A 1 6 B 0 C 4 3 D 5 6 Question 27 You are given the plant transfer function P (s) = k jω+2 e jωt with time delay T. Use the 1st order Padé approximation to calculate the frequency at the angle at which the gain margin is usually calculated. Choose the correct calculation below A 3 arctan( ω T 2 ) = π B arctan( ω 2 ) 2 arctan( ω T 2 ) = π C arctan( ω 2 ) + 2 arctan( ω T 2 ) = π D arctan( ω 2 ) 2 arctan( ω T 2 ) = 0
15 4 Controller Design Question 28 Mark all correct statements. (2 Points) Take a system with open-loop gain L(s) = 1 s and consider its positive root-locus curve (k > 0). 2 A For k = 0 both poles of the closed-loop system are identical. B A controller C(s) = k (s+2) will stabilize the closed-loop system for k > 0. C Increasing k to a sufficiently large number will stabilize the resulting closed-loop system and bring satisfactory system performance. D A controller C(s) = s+2 will stabilize the closed-loop system for k > 0. E For k > 0 the absolute value of both poles of the closed-loop system is 2 k. F For k > 0 the absolute value of both poles of the closed-loop system is k. k Question 29 Let P (s) = 1 (s+5) 2, which statement about the controller design for P (s) is correct? A There is no need to apply feedback control to the plant because it is stable and so it will reject any disturbances. B Any P I-Controller with sufficiently high gain can stabilize the plant. C The plant has two stable poles which is why any P -Controller with arbitrary high gain can stabilize the plant and result in sufficiently good control performance. D A P -Controller with a sufficiently small gain is a possible first approach for the design. Box 8: Question 30 Consider the system described by the following block diagram r + e C(s) P (s) y where P (s) = 1 (s + 4)(s + 0.5). We want to design the controller C(s) to satisfy the following specifications: The closed-loop system is stable. The phase margin is about 50 degrees. The cross-over frequency is at about 20 rad/sec. Question 30 Only one of the following controllers satisfies these specifications. Which one is it?
16 A C(s) = 110 s+16 s+25 B C(s) = 816 s+10 s+40 C C(s) = 225 s+33 s+12 D C(s) = 435 s+5 s(s+20) E C(s) = 315 s+27 s(s+15) Question 31 Mark all correct statements. (2 Points) In many cases, second-order system approximations are used to infer useful information about a control system. Which of the following systems should not be approximated as a second-order system? A g(s) = 4 s+1 e st B g(s) = 1 (s+1) 3 C g(s) = 3 (s+1)(s+2) 2 D g(s) = 3 (s+1) 2 (s 1) E g(s) = 0.5 s 1 (s+2) 4 Question 32 The phase of the frequency response of a control system can be decreased by applying a lead element. A True. B False. Box 9: Question 33 Let α(t) = 0.5cos(α(t)) + 2u(t) denote the dynamics of a system. equilibrium point α e = 0 yields the linearized system equations Linearization around the ẍ(t) = 2u(t) Question 33 What is the correct transfer function g(s) of the linearized system if y(t) = x(t)? A g(s) = 2 s 2 B g(s) = 1 2s 2 C g(s) = 2 s D g(s) = 2 s 2 Question 34 Mark all correct statements. (2 Points) Which statements about the transfer function g(s) = 2 s are correct? B The transfer function is proper. A A = 0 0 0, B = 2 3, C = [ ] C The transfer function has two zeros., D = 0 is a state space representation of the transfer function. D The transfer function is strictly proper.
17 Question 35 Consider the four pole/zero diagrams in figure 3 indication the pole, zero locations of 4 transfer functions: Im Im +3 Im Im 1 Re 1 3 Re 6 2 Re Re (a) (b) (c) (d) Figure 3: Pole/zero plots of four different transfer functions. Your colleague asks you to order these transfer functions with increasing t 90 response time to a unit step input. What is the correct ordering? A c, d, b, a B c, d, a, b C d, c, a, b D d, b, c, a Box 10: Questions 36, 37 Let α(t) = 0.5cos(α(t))+2u(t) be the dynamics of a system. Linearization around the equilibrium point α e = 0 yields the linearized system equations Furthermore let y(t) = x(t). ẍ(t) = 2u(t) (1) Question 36 Mark all correct statements. (2 Points) Mark all correct statements. A The system is stabilizable. B The system is detectable. C The system is fully observable. D The system is fully controllable. Question 37 Mark all correct statements. (2 Points) Let L(s) denote the loop gain, S(s) the sensitivity and T (s) the complementary sensitivity of the system in equation 1. A The sensitivity is asymptotically stable. B The poles of the sensitivity and the complementary sensitivity are identical. C The magnitude of the open loop gain response to a unit step input is bounded. D The complementary sensitivity is BIBOstable.
18 Question 38 Your colleague has a system with one unstable open-loop pole. He used mathematica to find a stabilizing controller but has unfortunately mixed up his scripts. Figure 4 shows the plots that he presents to you. Help him to assign the correct Nyquist plots to the respective step responses. A 1 c, 2 b, 3 a, 4 d B 1 a, 2 b, 3 d, 4 c C 1 b, 2 c, 3 d, 4 a D 1 c, 2 b, 3 d, 4 a
19 Nyquist plot of system Step response a Nyquist plot of system Step response b Nyquist plot of system Step response c Nyquist plot of system Step response d. Figure 4: Bode plots of four different systems.
20 Im(jω) Im(jω) Re(jω) Re(jω) Im(jω) 0.2 Nyquist plot A -1.5 Im(jω) 1.0 Nyquist plot B Re(jω) Re(jω) Nyquist plot C Nyquist plot D Figure 5: Figure illustrating different possible Nyquist diagrams for a lead compensator C lead (s) Question 39 Your colleague has designed a lead compensator. C lead = s/2 + 1 s/8 + 1 She has given you several Nyquist plots of and asks you which one corresponds to the lead compensator transfer function she designed. Pick the correct Nyquist plot in Figure 5 corresponding to the above lead compensator. A Nyquist plot D B Nyquist plot C C Nyquist plot A D Nyquist plot B
21 Amplitude Corrected 0.7 Step Response Time (seconds) Figure 6: Step response of plant P (s) Question 40 You are tasked with designing a lag compensator C lag = k s/a+1 s/b+1 to increase the gain of your system at low frequencies. Currently you have L(0) = 5, you wish to achieve L(0) = 200. Furthermore, you know that your crossover frequency is ω c = 100 rad/s and that you do not want your lag compensator to substantially influence the system s phase margin. A C lag (s) = 40 s/ s/10+1 B C lag (s) = 40 s/10+1 s/ C C lag (s) = s/2.5+1 s/100+1 D C lag (s) = 40 s/100+1 s/2.5+1 Question 41 Your coworker designed a controller for plant P (s), however the system does not track a step response as shown in figure 6. Which of the following controllers might ensure a correct tracking of the step input with zero steady state error (e = 0)? A PD B C lead (s) = s/a+1 s/b+1 with 0 < a < b C PI D P k (s+p)(s+r) s+q. Choose the correct coeffi- Question 42 You are given a Padé approximation of the form e st cients: A k = T, p = 2 T ± j 2 T, r = 2 T j 2 T, q = 6 T B k = 0.5T, p = 1 T, r = 3 T, q = 3 T C k = 0.5T, j 1 T, q = 3 T D k = 0.5T, j 2 T, q = 3 T p = 2 T ± j 1 T, r = 2 T p = 2 T ± j 2 T, r = 2 T
22 Phase [ ] Amplitude [db] Corrected Frequency [rad/s] Figure 7: Bode plot of transfer function L(s) Question 43 You are given the Bode plot of the open-loop gain L(s) in Figure 7. Which magnitude can the multiplicative model disturbance W 2 (s) reach at the crossover frequency ω c while still ensuring stability? A 16.9 db B 6.29 db Question 44 Which of the following statements is true about the Anti-Windup scheme? A The Anti-windup reduces the total error multiplied by a gain parameter when the input saturates. B The Anti-windup reduces the proportional error term multiplied by a gain parameter when the input saturates. C The Anti-windup restricts the input to the maximal input when the input saturates. D The Anti-windup reduces the integral error multiplied by a gain parameter when the input saturates.
23
24
25
26
27
28
29
30
31 Answer sheet: student number please encode your student number, and write your first and last name below Firstname and lastname: How to select answer B : Answer B chosen. Corrected, answer B chosen. Double corrected, answer B chosen No choice made. Answers must be given exclusively on this sheet; answers given on the other sheets will not be counted. 1 Systems and Control Architecture Q1: A B C D Q2: A B C D Q3: A B C 2 System Modeling and Analysis Q4: A B C D Q5: A B C D Q6: A B C D Q7: A B C D Q8: A B C D Q9: A B C D Q10: A B C D Q11: A B C Q12: A B C D Q13: A B C D Q14: A B C D Q15: A B C D Q16: A B C D E F Q17: A B C D 3 Frequency Domain Q18: A B C D Q19: A B C D Q20: A B C D Q21: A B C D Q22: A B C D Q23: A B C D Q24: A B C D Q25: A B C D E F Q26: A B C D Q27: A B C D 4 Controller Design Q28: A B C D E F Q29: A B C D Q30: A B C D E Q31: A B C D E Q32: A B Q33: A B C D Q34: A B C D Q35: A B C D Q36: A B C D Q37: A B C D Q38: A B C D
32 Q39: A B C D Q40: A B C D Q41: A B C D Q42: A B C D Q43: A B Q44: A B C D
Exam. 135 minutes + 15 minutes reading time
Exam January 23, 27 Control Systems I (5-59-L) Prof. Emilio Frazzoli Exam Exam Duration: 35 minutes + 5 minutes reading time Number of Problems: 45 Number of Points: 53 Permitted aids: Important: 4 pages
More informationExam. 135 minutes, 15 minutes reading time
Exam August 6, 208 Control Systems II (5-0590-00) Dr. Jacopo Tani Exam Exam Duration: 35 minutes, 5 minutes reading time Number of Problems: 35 Number of Points: 47 Permitted aids: 0 pages (5 sheets) A4.
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 9. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -
More informationPrüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 5. 2. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -
More informationControls Problems for Qualifying Exam - Spring 2014
Controls Problems for Qualifying Exam - Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationQuiz. Good luck! Signals & Systems ( ) Number of Problems: 19. Number of Points: 19. Permitted aids:
Quiz November th, Signals & Systems (--) Prof. R. D Andrea Quiz Exam Duration: Min Number of Problems: Number of Points: Permitted aids: Important: None Questions must be answered on the provided answer
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More informationLecture 1: Feedback Control Loop
Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure
More informationControl Systems I. Lecture 2: Modeling. Suggested Readings: Åström & Murray Ch. 2-3, Guzzella Ch Emilio Frazzoli
Control Systems I Lecture 2: Modeling Suggested Readings: Åström & Murray Ch. 2-3, Guzzella Ch. 2-3 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich September 29, 2017 E. Frazzoli
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 5-059-00 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationControl Systems I. Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback. Readings: Emilio Frazzoli
Control Systems I Lecture 4: Diagonalization, Modal Analysis, Intro to Feedback Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH)
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationEECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8- am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationControl Systems I. Lecture 2: Modeling and Linearization. Suggested Readings: Åström & Murray Ch Jacopo Tani
Control Systems I Lecture 2: Modeling and Linearization Suggested Readings: Åström & Murray Ch. 2-3 Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich September 28, 2018 J. Tani, E.
More informationECEN 420 LINEAR CONTROL SYSTEMS. Lecture 6 Mathematical Representation of Physical Systems II 1/67
1/67 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 6 Mathematical Representation of Physical Systems II State Variable Models for Dynamic Systems u 1 u 2 u ṙ. Internal Variables x 1, x 2 x n y 1 y 2. y m Figure
More informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationr + - FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of hand-written notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationControl Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/2017
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 13: Stability Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 13: Stability p.1/20 Outline Input-Output
More informationDesign Methods for Control Systems
Design Methods for Control Systems Maarten Steinbuch TU/e Gjerrit Meinsma UT Dutch Institute of Systems and Control Winter term 2002-2003 Schedule November 25 MSt December 2 MSt Homework # 1 December 9
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 5-8 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationThe loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)
Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationsc Control Systems Design Q.1, Sem.1, Ac. Yr. 2010/11
sc46 - Control Systems Design Q Sem Ac Yr / Mock Exam originally given November 5 9 Notes: Please be reminded that only an A4 paper with formulas may be used during the exam no other material is to be
More informationFeedback Control of Linear SISO systems. Process Dynamics and Control
Feedback Control of Linear SISO systems Process Dynamics and Control 1 Open-Loop Process The study of dynamics was limited to open-loop systems Observe process behavior as a result of specific input signals
More informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pick-and-place robot to move the link of a robot between two positions. Usually
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 6: Generalized and Controller Design Overview In this Lecture, you will learn: Generalized? What about changing OTHER parameters
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequency-response design method Goal: wrap up lead and
More informationME 132, Fall 2017, UC Berkeley, A. Packard 334 # 6 # 7 # 13 # 15 # 14
ME 132, Fall 2017, UC Berkeley, A. Packard 334 30.3 Fall 2017 Final # 1 # 2 # 3 # 4 # 5 # 6 # 7 # 8 NAME 20 15 20 15 15 18 15 20 # 9 # 10 # 11 # 12 # 13 # 14 # 15 # 16 18 12 12 15 12 20 18 15 Facts: 1.
More informationModeling and Analysis of Dynamic Systems
Modeling and Analysis of Dynamic Systems by Dr. Guillaume Ducard Fall 2016 Institute for Dynamic Systems and Control ETH Zurich, Switzerland based on script from: Prof. Dr. Lino Guzzella 1/33 Outline 1
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationAutomatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...
Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number..........................
More informationEECS C128/ ME C134 Final Wed. Dec. 14, am. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C128/ ME C134 Final Wed. Dec. 14, 211 81-11 am Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth 1 points total. Problem Points Score 1 16 2 12
More informationIntroduction to Controls
EE 474 Review Exam 1 Name Answer each of the questions. Show your work. Note were essay-type answers are requested. Answer with complete sentences. Incomplete sentences will count heavily against the grade.
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More informationLecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.
ISS0031 Modeling and Identification Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design. Aleksei Tepljakov, Ph.D. September 30, 2015 Linear Dynamic Systems Definition
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationBIBO STABILITY AND ASYMPTOTIC STABILITY
BIBO STABILITY AND ASYMPTOTIC STABILITY FRANCESCO NORI Abstract. In this report with discuss the concepts of bounded-input boundedoutput stability (BIBO) and of Lyapunov stability. Examples are given to
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servo-oriented
More informationExercise 1 (A Non-minimum Phase System)
Prof. Dr. E. Frazzoli 5-59- Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Non-minimum Phase System) To decrease the rise time of the system,
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition Readings: Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Jacopo Tani Institute for Dynamic Systems and Control
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationD G 2 H + + D 2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.302 Feedback Systems Final Exam May 21, 2007 180 minutes Johnson Ice Rink 1. This examination consists
More informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 09-Dec-13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More informationAutonomous Mobile Robot Design
Autonomous Mobile Robot Design Topic: Guidance and Control Introduction and PID Loops Dr. Kostas Alexis (CSE) Autonomous Robot Challenges How do I control where to go? Autonomous Mobile Robot Design Topic:
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:
More informationVideo 5.1 Vijay Kumar and Ani Hsieh
Video 5.1 Vijay Kumar and Ani Hsieh Robo3x-1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior
More informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More informationIntroduction to Process Control
Introduction to Process Control For more visit :- www.mpgirnari.in By: M. P. Girnari (SSEC, Bhavnagar) For more visit:- www.mpgirnari.in 1 Contents: Introduction Process control Dynamics Stability The
More informationControl Systems. Root Locus & Pole Assignment. L. Lanari
Control Systems Root Locus & Pole Assignment L. Lanari Outline root-locus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS - Root
More informationLecture 12. Upcoming labs: Final Exam on 12/21/2015 (Monday)10:30-12:30
289 Upcoming labs: Lecture 12 Lab 20: Internal model control (finish up) Lab 22: Force or Torque control experiments [Integrative] (2-3 sessions) Final Exam on 12/21/2015 (Monday)10:30-12:30 Today: Recap
More informationGoodwin, Graebe, Salgado, Prentice Hall Chapter 11. Chapter 11. Dealing with Constraints
Chapter 11 Dealing with Constraints Topics to be covered An ubiquitous problem in control is that all real actuators have limited authority. This implies that they are constrained in amplitude and/or rate
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R + - E G c (s) G(s) C Figure 1: The standard block diagram
More informationControl Systems II. ETH, MAVT, IDSC, Lecture 4 17/03/2017. G. Ducard
Control Systems II ETH, MAVT, IDSC, Lecture 4 17/03/2017 Lecture plan: Control Systems II, IDSC, 2017 SISO Control Design 24.02 Lecture 1 Recalls, Introductory case study 03.03 Lecture 2 Cascaded Control
More informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationExercise 1 (A Non-minimum Phase System)
Prof. Dr. E. Frazzoli 5-59- Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Non-minimum Phase System) To increase the rise time of the system, we
More informationState Regulator. Advanced Control. design of controllers using pole placement and LQ design rules
Advanced Control State Regulator Scope design of controllers using pole placement and LQ design rules Keywords pole placement, optimal control, LQ regulator, weighting matrixes Prerequisites Contact state
More informationExam in Systems Engineering/Process Control
Department of AUTOMATIC CONTROL Exam in Systems Engineering/Process Control 27-6-2 Points and grading All answers must include a clear motivation. Answers may be given in English or Swedish. The total
More informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
More informationRecitation 11: Time delays
Recitation : Time delays Emilio Frazzoli Laboratory for Information and Decision Systems Massachusetts Institute of Technology November, 00. Introduction and motivation. Delays are incurred when the controller
More informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
More informationAutomatic Control A. A.A. 2016/2017 July 7, Corso di Laurea Magistrale in Ingegneria Meccanica. Prof. Luca Bascetta.
Corso di Laurea Magistrale in Ingegneria Meccanica Automatic Control A Prof. Luca Bascetta A.A. 2016/2017 July 7, 2017 Name: Surname: University ID number: Signature: This file consists of 8 pages (including
More informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET - II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationIntroduction to Modern Control MT 2016
CDT Autonomous and Intelligent Machines & Systems Introduction to Modern Control MT 2016 Alessandro Abate Lecture 2 First-order ordinary differential equations (ODE) Solution of a linear ODE Hints to nonlinear
More information6.241 Dynamic Systems and Control
6.241 Dynamic Systems and Control Lecture 24: H2 Synthesis Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology May 4, 2011 E. Frazzoli (MIT) Lecture 24: H 2 Synthesis May
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 2010-2011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 2010-2011
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationModeling and Analysis of Dynamic Systems
Modeling and Analysis of Dynamic Systems Dr. Guillaume Ducard Fall 2017 Institute for Dynamic Systems and Control ETH Zurich, Switzerland G. Ducard c 1 / 54 Outline 1 G. Ducard c 2 / 54 Outline 1 G. Ducard
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationRobust and Optimal Control, Spring A: SISO Feedback Control A.1 Internal Stability and Youla Parameterization
Robust and Optimal Control, Spring 2015 Instructor: Prof. Masayuki Fujita (S5-303B) A: SISO Feedback Control A.1 Internal Stability and Youla Parameterization A.2 Sensitivity and Feedback Performance A.3
More informationManufacturing Equipment Control
QUESTION 1 An electric drive spindle has the following parameters: J m = 2 1 3 kg m 2, R a = 8 Ω, K t =.5 N m/a, K v =.5 V/(rad/s), K a = 2, J s = 4 1 2 kg m 2, and K s =.3. Ignore electrical dynamics
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationCDS 101/110a: Lecture 8-1 Frequency Domain Design
CDS 11/11a: Lecture 8-1 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve
More informationECE 388 Automatic Control
Lead Compensator and PID Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage: http://ece388.cankaya.edu.tr
More informationLecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore
Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore A Fundamental Problem in Control Systems Poles of open
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationCollocated versus non-collocated control [H04Q7]
Collocated versus non-collocated control [H04Q7] Jan Swevers September 2008 0-0 Contents Some concepts of structural dynamics Collocated versus non-collocated control Summary This lecture is based on parts
More informationCDS 101/110a: Lecture 10-2 Control Systems Implementation
CDS 101/110a: Lecture 10-2 Control Systems Implementation Richard M. Murray 5 December 2012 Goals Provide an overview of the key principles, concepts and tools from control theory - Classical control -
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More informationProblem Set 4 Solution 1
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski Problem Set 4 Solution Problem 4. For the SISO feedback
More informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 30-Apr-14 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More informationPole placement control: state space and polynomial approaches Lecture 2
: state space and polynomial approaches Lecture 2 : a state O. Sename 1 1 Gipsa-lab, CNRS-INPG, FRANCE Olivier.Sename@gipsa-lab.fr www.gipsa-lab.fr/ o.sename -based November 21, 2017 Outline : a state
More information