Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!"

Transcription

1 Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid - The German exam is the only valid version! All answers must be written on the regular exam sheets (which are in German).

2

3 Sessionsprüfung Regelungstechnik I Page Question (Modeling, Linarization) 8 Points g (Gravitation) x Water Droplet h Wind (System Input) Figure : Water Droplet Hovering in the Air In this exercise, a linearized model of the system presented in Figure has to be created. The System is a water droplet which is kept from falling by air streaming in vertically upwards direction. The droplet has a diameter of d = 2 [mm] and the water has a density of ρ W = [kg/m 3 ]. The droplet is accelerated by the gravitational force towards the ground with 9.8 [m/s 2 ]. Evaporation effects as well as movements of the droplet in horizontal direction can be neglected. The velocity of the air flow (wind) s(t) at ground level (at h = ) can be arbitrarily adjusted and is denoted with u(t) [m/s]. The magnitude of the flow velocity decreases as the height h [m] increases, the functional relationship is as follows: s(t) = u(t) α h [m/s], where α = [/s] is given as a constant parameter. The vertical position of the droplet is denoted by x. The variable h denotes the general coordinate in vertical direction. The force that pushes the water droplet upwards can be modeled as follows: F a = 2 ρ L c a A v 2 [N], where A [m 2 ] is the area of the droplet, c a =.445 [ ] is the drag coefficient, ρ L =.2 [kg/m 3 ] is the density of air, and v denotes the velocity with which the air flows towards the droplet. a) (3 points) Choose the state vector z(t) = [x(t), ẋ(t)] T and derive the nonlinear state space description of the form dz(t) dt = f(z(t), u(t)), w(t) = g(z(t), u(t)). Use the variable names z (t), z 2 (t), u(t), and w(t). b) (2 points) Calculate the flow velocity u e at ground level, which is necessary for the droplet to stay at the positon z,e = 3 [m] in a state of equilibrium.

4 Page 2 Sessionsprüfung Regelungstechnik I c) (3 points) Linearize the system equations around this equilibrium point (no normalization is required). Express the system equations in the standard form (state space description with the matrices {A, b, c, d}). Express the matrices in general form, i.e. use the variables of the system and do not use their corresponding numerical values.

5 Sessionsprüfung Regelungstechnik I Page 3 Question 2 (Frequency domain, time domain) 8 Points The open-loop transfer functions (loop gain) L (s), L 2 (s), L 3 (s), L 4 (s) of 4 control systems are given (see table for solution). Furthermore, the Nyquist plots (see below the diagrams A, B, C and D; plotted for positive frequencies only) of these transfer functions, and the resulting step responses (see on the next page the step responses to 4) of the corresponding closed loop systems are given. Assign the correct Nyquist plot and the correct step response to each of the open loop transfer functions. Use the table provided on the solution page of this question for your solution. You do not need to justify your answers..5 Nyquist Plot A.5 Nyquist Plot B Im Im Re Re.5 Nyquist Plot C.5 Nyquist Plot D Im Im Re Re

6 Page 4 Sessionsprüfung Regelungstechnik I 2 Step Response 2 Step Response 2 Amplitude [ ].5 Amplitude [ ] Time [s] Time [s] 2 Step Response 3 2 Step Response 4 Amplitude [ ].5 Amplitude [ ] Time [s] Time [s]

7 Sessionsprüfung Regelungstechnik I Page 5 Question 3 (Controller Synthesis) 8 Points The department of modeling at your company has created a very accurate model of a system to be controlled. The corresponding dynamics are given by the following transfer function: P (s) = (s + 2 ) (s + 3 ) Your job is to control this system. All tasks of this question can be solved independently of each other. a) (3 points) You have to design a PI-controller C P I (s) = k p ( + T i s ), with the following specifications on the contol system: The crossover frequency must be at ω c =.85 [rad/s]. The phase marigin must be 45. Calculate the values of the parameters {k p, T i } which lead to a control system that fulfills these specifications. b) (3 points) Your colleague suggests the following PD-controller as an alternative controller: ( 5 C P D (s) = 2 s + ) 3 Your boss says she wants the one controller which leads to a faster rise time t 9 in a step response analysis. Which controller do you suggest? c) (2 points) Another colleague suggests two different P-controllers C P (s) = k p with the following specifications Controller Crossover Frequency ω c Phase Margin C P, = k p,.85 [rad/s] 45 C P,2 = k p,2.5 [rad/s] 45 Your boss supports his intentions as she wants to keep the structures of the controllers as simple as possible. What do you think of your colleagues suggestions? Justify your answer.

8 Page 6 Sessionsprüfung Regelungstechnik I Question 4 (Laplace-Transformation) Points The following subtasks a), b) and c) can be solved independently. a) The two systems Σ and Σ 2 are connected in series. u(t) x(t) y(t) Σ Σ 2 Figure 2: System overview. The output x(t) of Σ is characterized by the following differential equation: ẍ(t) = 4 ẋ(t) 4 x(t) + u(t) with ẋ(t) = x(t) = u(t) =, t. The transfer function of Σ 2 (s) = Y (s) X(s) is given by: Σ 2 (s) = 3 s s + i) ( point) Determine the transfer function of the entire system Σ a (s) = Y (s) ii) iii) U(s). (2 points) The system Σ a (s) is subjected to a step excitation u(t) = h(t), calculate the time domain response y(t). (2 points) Illustrate the time response of ii) in the associated template on the solution page qualitatively. By doing that, also think about the following characteristics: What is the system s static gain? Does the system response overshoot? b) Consider the block diagram in figure 3. x x 2 x 3 y s u 5 s s 2 2 Figure 3: Block diagram. i) ( point) Determine the associated state space description A, B, C, D. ii) (2 points) Determine the transfer function Σ b = Y (s) U(s). c) The time response of another time-invariant SISO system is given as: y(t) = ( e (t T ) cos (ω(t T ))) h(t T ) with T = ms, ω = π 3 rad/s. i) (2 points) Calculate the transfer function of the system Σ c (s).

9 Sessionsprüfung Regelungstechnik I Page 7 Question 5 (Stabilization / Performance & Robustness) 9 Points You would like to develop a controller for a plant with the following state space representation of its model. [ d 3 dt x(t) = y(t) = [ 4 4 ] [ ] a + 3 x(t) + u(t), x() = (a) a ] x(t). (b) The parameter a specifies the actuator. The larger you choose a in the permissible interval 2 a 2 the more expensive is the actuator. Figure 4 shows the set-up of the control system. The control system is used in an environment where it is disturbed by a noise signal n(t) with a frequency of ω n 3 rad /s. r(t) C(s) u(t) P (s) y(t) n Figure 4: Control system with input and output signals. Remark: Solution of a) is required for the solutions of the subsequent questions b)-d). But the questions b)-e) can be partly solved indepedently from each other. a) (2 points) Determine the transfer function P (s) of the plant with the input signal u(t), output signal y(t) and state vector x(t) in function of the actuator parameter a. Determine also the pole(s) and zero(s) of the plant. b) (3 points) In which range the cross over frequency ω c should be selected such that an appropriate controller C(s) may be designed? Which cross over frequency do you choose if at the same time the actuator costs have to be minimized? Determine also the corresponding actuator parameter a that minimizes the actuator costs. Important remark: Based on a special offer you decide to purchase an actuator with a =. Use this value to solve the following questions c) to e). c) (2 points) You would like to stabilize the control system with a P-controller C(s) = k p. Justify why it is surely possible to stabilize the control system for negative values of k p in the range 2 < k p <.3. d) ( point) You use a P-controller according to point c) that stabilizes the control system. What is the amplification of a high frequent sensor noise (ω n ) at the output of the control system? e) ( point) Determine the steady state error of the control system for k p =.3.

10 Page 8 Sessionsprüfung Regelungstechnik I Question 6 (Bode-Diagram/Nyquist Criterion) Points The following subtasks a) and b) can be solved independently. a) The bode-diagram of a critically damped plant with the transfer function P (s) was measured. Figure 5: Bode-diagram of the plant with the corresponding transfer function P (s). i) (2 points) Determine the transfer function of the plant P (s) with the aid of the measured bode-diagram in figure 5. ii) ( point) The system outlined in figure 6 is controlled using a proportional controller with the transfer function C (s) = k p =. The control system is subjected to a disturbance (unit step) on input w. Will the output of the system return to the original value, without a steady state error? Assume that the system was at equilibrium with r = and y =, prior to the disturbance. r + - C (s) + + w P (s) y Figure 6: Control system for the plant P (s). iii) iv) ( point) Determine the phase margin in case the plant P (s) of exercise i) is controlled by the controller C (s) = k p = according to the structure presented in figure 6 (w = ). ( point) Which structural changes would you consider to apply to C (s) firstly, in order to eliminate the steady state error?

11 Sessionsprüfung Regelungstechnik I Page 9 b) Another control system consists of the plant P 2 (s) and the controller C 2 (s). The transfer function P 2 (s) is known and the Nyquist-diagram of P 2 (s) is given in figure 8. P 2 (s) = s (s + ) 3 (2) Figure 8: Nyquist-diagram of the plant P 2 (s) Figure 7: Block-diagram of the control system. r + - C 2 (s) P 2 (s) y i) (2 points) The Bode-diagram of the controller C 2 (s) is given. Illustrate the according Nyquist-diagram as accurately as possible in the associated template on the solution page. It is not demanded to derive the exact transfer function. Figure 9: Bode-diagram of the controller C 2 (s). ii) (3 points) Assume now that a simple P-controller C 2 (s) = k p is used. Make use of the Nyquist-criterion in order to calculate the range of k p that leads to an asymptotically stable system.

12 Page Sessionsprüfung Regelungstechnik I Question 7 (System Analysis) 7 Points Consider the model of a geosynchronous satellite as shown in Figure. The gravitational force of the earth is approximately equal to F g = MG r 2 m, where M > is the mass of the earth and G > is its gravitational constant. The mass of the satellite is described by m >. The distance from the satellite to the earth s center of mass is expressed with the radius r >. The centrifugal force counteracts to the gravitational force and is given by (3) F z = m r ω 2, (4) where ω is the rotational velocity of the satellite around the earth. F t ω m F z F g r M Figure : Geosynchronous satellite Assume a thruster with the force F t R is mounted tangential onto the satellite. As a result, the equations of motion for the geosynchronous satellite yield r = r ω 2 MG r 2, ω = F t r m. The linearization about the equilibria, r >, ω >, F t, =, leads to the system of linear equations, i.e. ω MG 2 r r x 3 ω ṙ = x + F t with x = r. (7) r m ω (5) (6)

13 Sessionsprüfung Regelungstechnik I Page a) ( point) State the system matrices {A, b, c, d} for the system of linear equations (7) with the assumption that the radius r and the rotational velocity ω are measured. b) (2 points) In the sense of Lyapunov, is the given system stable, asymptotically stable or unstable? Justify your answer mathematically. c) Given the tangential thruster, is the system controllable? i) (2 points) Make a point about the controllability of the system and justify your answer mathematically. ii) iii) ( point) Assume that the satellite has slightly approached the earth. Is the tangential thruster able to bring the satellite back onto its original orbit? ( point) If yes, does the thruster have to accelerate or decelerate the satellite? If no, why is the thruster not able to do so? d) ( point) How would the system matrices look like, if the tangential thruster was replaced by a radial thruster F r R that acts in the same direction than the centrifugal force? e) (2 points) Which statements hold about the Lyapunov stability, the observability, and the controllability of the new system?

14 Page 2 Sessionsprüfung Regelungstechnik I Question 8 (Multiple-Choice) 8 Points Decide whether the following statements are true or false and check the corresponding check box with an X ( ) on the solution page of this question. You are not required to justify your answers. All questions are equally weighted ( Point). There will be a reduction of one point for a wrong answer. Unanswered questions will get points. The minimum sum for all questions is points. a) The differential equation δẋ = δx + 4 δu is the linearization of the non-linear system ẋ = 4 x + + 3x + u 2 around the equilibrium point {x e =, u e = 2}. b) A constant signal u(t) = at the input of a system with the transfer function Σ(s) = produces for t a constant output signal of 2. s s 2 +6s+5 c) An unstable system with the transfer function s (s 2) C(s) = k p (k p R). can be stabilized by a P-controller d) The state space representation {A, b, c, d} of a second order system has the transfer function P (s) = s+3. The system is completely controllable and observable. s 2 +s 6 e) The following state space model {A, b, c, d} represents a realization for a system with the transfer function Σ(s) = s s 3 : [ ] [ ] 2 A =, b =, 3 2 c = [ ], D = [ ] f) The Matlab instruction P = zpk(,[-i +i 3],-2) defines in Matlab a system with the transfer function P (s) = 2s (s 3)(s 2 2s+2). g) The system with the transfer function Σ(s) = s+7 + s 4 has no system zeros. h) The transfer function of a closed loop system from the reference signal to the output y is T (s) = s+ (complementary sensitivity). The loop gain of the control system is s 2 +4s+ L(s) = s+ s(s+3). Be aware of this fact!

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!

Übersetzungshilfe / 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 information

Exam. 135 minutes + 15 minutes reading time

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 information

MAS107 Control Theory Exam Solutions 2008

MAS107 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 information

Represent this system in terms of a block diagram consisting only of. g From Newton s law: 2 : θ sin θ 9 θ ` T

Represent this system in terms of a block diagram consisting only of. g From Newton s law: 2 : θ sin θ 9 θ ` T Exercise (Block diagram decomposition). Consider a system P that maps each input to the solutions of 9 4 ` 3 9 Represent this system in terms of a block diagram consisting only of integrator systems, represented

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Dr 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 information

Richiami di Controlli Automatici

Richiami 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 information

FEEDBACK CONTROL SYSTEMS

FEEDBACK CONTROL SYSTEMS FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control

More information

Automatic 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 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 information

Lecture 1: Feedback Control Loop

Lecture 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 information

Control 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 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 information

Control Systems II. ETH, MAVT, IDSC, Lecture 4 17/03/2017. G. Ducard

Control 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 information

EECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.

EECS 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 information

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =

100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) = 1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot

More information

Analysis and Design of Control Systems in the Time Domain

Analysis and Design of Control Systems in the Time Domain Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.

More information

x(t) = x(t h), x(t) 2 R ), where is the time delay, the transfer function for such a e s Figure 1: Simple Time Delay Block Diagram e i! =1 \e i!t =!

x(t) = x(t h), x(t) 2 R ), where is the time delay, the transfer function for such a e s Figure 1: Simple Time Delay Block Diagram e i! =1 \e i!t =! 1 Time-Delay Systems 1.1 Introduction Recitation Notes: Time Delays and Nyquist Plots Review In control systems a challenging area is operating in the presence of delays. Delays can be attributed to acquiring

More information

Frequency domain analysis

Frequency 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 information

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.

More information

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year

More information

Goodwin, Graebe, Salgado, Prentice Hall Chapter 11. Chapter 11. Dealing with Constraints

Goodwin, 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 information

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.

6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0. 6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)

More information

Implementation of a Communication Satellite Orbit Controller Design Using State Space Techniques

Implementation of a Communication Satellite Orbit Controller Design Using State Space Techniques ASEAN J Sci Technol Dev, 29(), 29 49 Implementation of a Communication Satellite Orbit Controller Design Using State Space Techniques M T Hla *, Y M Lae 2, S L Kyaw 3 and M N Zaw 4 Department of Electronic

More information

Feedback Control of Linear SISO systems. Process Dynamics and Control

Feedback 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 information

Pole placement control: state space and polynomial approaches Lecture 2

Pole 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

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques [] For the following system, Design a compensator such

More information

EECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.

EECS 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 information

Autonomous Mobile Robot Design

Autonomous 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 information

Lecture 12. Upcoming labs: Final Exam on 12/21/2015 (Monday)10:30-12:30

Lecture 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 information

1 Mathematics. 1.1 Determine the one-sided Laplace transform of the following signals. + 2y = σ(t) dt 2 + 3dy dt. , where A is a constant.

1 Mathematics. 1.1 Determine the one-sided Laplace transform of the following signals. + 2y = σ(t) dt 2 + 3dy dt. , where A is a constant. Mathematics. Determine the one-sided Laplace transform of the following signals. {, t < a) u(t) =, where A is a constant. A, t {, t < b) u(t) =, where A is a constant. At, t c) u(t) = e 2t for t. d) u(t)

More information

Homework 7 - Solutions

Homework 7 - Solutions Homework 7 - Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the

More information

D G 2 H + + D 2

D 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 information

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications: 1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.

More information

Today (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10

Today (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 information

Index. Index. More information. in this web service Cambridge University Press

Index. Index. More information.  in this web service Cambridge University Press A-type elements, 4 7, 18, 31, 168, 198, 202, 219, 220, 222, 225 A-type variables. See Across variable ac current, 172, 251 ac induction motor, 251 Acceleration rotational, 30 translational, 16 Accumulator,

More information

Modeling and Control Overview

Modeling and Control Overview Modeling and Control Overview D R. T A R E K A. T U T U N J I A D V A N C E D C O N T R O L S Y S T E M S M E C H A T R O N I C S E N G I N E E R I N G D E P A R T M E N T P H I L A D E L P H I A U N I

More information

AN INTRODUCTION TO THE CONTROL THEORY

AN INTRODUCTION TO THE CONTROL THEORY Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter

More information

RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS. Ryszard Gessing

RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS. Ryszard Gessing RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS Ryszard Gessing Politechnika Śl aska Instytut Automatyki, ul. Akademicka 16, 44-101 Gliwice, Poland, fax: +4832 372127, email: gessing@ia.gliwice.edu.pl

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance

More information

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions EE C28 / ME C34 Fall 24 HW 8 - Solutions HW 8 - Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot

More information

FREQUENCY-RESPONSE DESIGN

FREQUENCY-RESPONSE DESIGN ECE45/55: Feedback Control Systems. 9 FREQUENCY-RESPONSE DESIGN 9.: PD and lead compensation networks The frequency-response methods we have seen so far largely tell us about stability and stability margins

More information

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral

More information

1. Find the solution of the following uncontrolled linear system. 2 α 1 1

1. Find the solution of the following uncontrolled linear system. 2 α 1 1 Appendix B Revision Problems 1. Find the solution of the following uncontrolled linear system 0 1 1 ẋ = x, x(0) =. 2 3 1 Class test, August 1998 2. Given the linear system described by 2 α 1 1 ẋ = x +

More information

H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )

H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at ) .7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a

More information

Lecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion

Lecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 First-Order Specs: Step : Pole Real inputs contain

More information

Mechatronics Engineering. Li Wen

Mechatronics Engineering. Li Wen Mechatronics Engineering Li Wen Bio-inspired robot-dc motor drive Unstable system Mirko Kovac,EPFL Modeling and simulation of the control system Problems 1. Why we establish mathematical model of the control

More information

ECE317 : Feedback and Control

ECE317 : Feedback and Control ECE317 : Feedback and Control Lecture : Steady-state error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace

More information

Introduction to Feedback Control

Introduction to Feedback Control Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering

More information

Transfer function and linearization

Transfer function and linearization Transfer function and linearization Daniele Carnevale Dipartimento di Ing. Civile ed Ing. Informatica (DICII), University of Rome Tor Vergata Corso di Controlli Automatici, A.A. 24-25 Testo del corso:

More information

Fall 線性系統 Linear Systems. Chapter 08 State Feedback & State Estimators (SISO) Feng-Li Lian. NTU-EE Sep07 Jan08

Fall 線性系統 Linear Systems. Chapter 08 State Feedback & State Estimators (SISO) Feng-Li Lian. NTU-EE Sep07 Jan08 Fall 2007 線性系統 Linear Systems Chapter 08 State Feedback & State Estimators (SISO) Feng-Li Lian NTU-EE Sep07 Jan08 Materials used in these lecture notes are adopted from Linear System Theory & Design, 3rd.

More information

Recitation 11: Time delays

Recitation 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 information

Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis

Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis J. Alexandre dit Sandretto, A. Chapoutot and O. Mullier U2IS, ENSTA ParisTech SYNCOP April 11, 2015 Closed-loop

More information

Inverted Pendulum. Objectives

Inverted Pendulum. Objectives Inverted Pendulum Objectives The objective of this lab is to experiment with the stabilization of an unstable system. The inverted pendulum problem is taken as an example and the animation program gives

More information

Professor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley

Professor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the

More information

ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop system when the gain K changes from 0 to. H(s) H ( s) = ( s)

ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop system when the gain K changes from 0 to. H(s) H ( s) = ( s) C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s) - H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closed-loop system when the gain K changes from 0 to 1+ K G ( s)

More information

An Introduction to Control Systems

An Introduction to Control Systems An Introduction to Control Systems Signals and Systems: 3C1 Control Systems Handout 1 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie November 21, 2012 Recall the concept of a

More information

University of Utah Electrical & Computer Engineering Department ECE 3510 Lab 9 Inverted Pendulum

University of Utah Electrical & Computer Engineering Department ECE 3510 Lab 9 Inverted Pendulum University of Utah Electrical & Computer Engineering Department ECE 3510 Lab 9 Inverted Pendulum p1 ECE 3510 Lab 9, Inverted Pendulum M. Bodson, A. Stolp, 4/2/13 rev, 4/9/13 Objectives The objective of

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real

More information

Systems Analysis and Control

Systems Analysis and Control Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles

More information

PID Control. Objectives

PID Control. Objectives PID Control Objectives The objective of this lab is to study basic design issues for proportional-integral-derivative control laws. Emphasis is placed on transient responses and steady-state errors. The

More information

Proportional, Integral & Derivative Control Design. Raktim Bhattacharya

Proportional, Integral & Derivative Control Design. Raktim Bhattacharya AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University

More information

INTRODUCTION TO DIGITAL CONTROL

INTRODUCTION TO DIGITAL CONTROL ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a linear-time-invariant

More information

06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance.

06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Chapter 06 Feedback 06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Lesson of the Course Fondamenti di Controlli Automatici of

More information

Tradeoffs and Limits of Performance

Tradeoffs and Limits of Performance Chapter 9 Tradeoffs and Limits of Performance 9. Introduction Fundamental limits of feedback systems will be investigated in this chapter. We begin in Section 9.2 by discussing the basic feedback loop

More information

Loop shaping exercise

Loop shaping exercise Loop shaping exercise Excerpt 1 from Controlli Automatici - Esercizi di Sintesi, L. Lanari, G. Oriolo, EUROMA - La Goliardica, 1997. It s a generic book with some typical problems in control, not a collection

More information

Learn2Control Laboratory

Learn2Control Laboratory Learn2Control Laboratory Version 3.2 Summer Term 2014 1 This Script is for use in the scope of the Process Control lab. It is in no way claimed to be in any scientific way complete or unique. Errors should

More information

CHAPTER 10: STABILITY &TUNING

CHAPTER 10: STABILITY &TUNING When I complete this chapter, I want to be able to do the following. Determine the stability of a process without control Determine the stability of a closed-loop feedback control system Use these approaches

More information

Transient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n

Transient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2

More information

Chapter 7 - Solved Problems

Chapter 7 - Solved Problems Chapter 7 - Solved Problems Solved Problem 7.1. A continuous time system has transfer function G o (s) given by G o (s) = B o(s) A o (s) = 2 (s 1)(s + 2) = 2 s 2 + s 2 (1) Find a controller of minimal

More information

Control System Design

Control System Design ELEC ENG 4CL4: Control System Design Notes for Lecture #24 Wednesday, March 10, 2004 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Remedies We next turn to the question

More information

r + - FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic

r + - 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 information

MIMO analysis: loop-at-a-time

MIMO analysis: loop-at-a-time MIMO robustness MIMO analysis: loop-at-a-time y 1 y 2 P (s) + + K 2 (s) r 1 r 2 K 1 (s) Plant: P (s) = 1 s 2 + α 2 s α 2 α(s + 1) α(s + 1) s α 2. (take α = 10 in the following numerical analysis) Controller:

More information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open

More information

EE3CL4: Introduction to Linear Control Systems

EE3CL4: Introduction to Linear Control Systems 1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We

More information

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES

CHAPTER 7 STEADY-STATE RESPONSE ANALYSES CHAPTER 7 STEADY-STATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of

More information

Andrea Zanchettin Automatic Control AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Spring Semester, Linear systems (frequency domain)

Andrea Zanchettin Automatic Control AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Spring Semester, Linear systems (frequency domain) 1 AUTOMATIC CONTROL Andrea M. Zanchettin, PhD Spring Semester, 2018 Linear systems (frequency domain) 2 Motivations Consider an LTI system Thanks to the Lagrange s formula we can compute the motion of

More information

Analysis of SISO Control Loops

Analysis 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 information

BIBO STABILITY AND ASYMPTOTIC STABILITY

BIBO 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 information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS STAFF NAME: Mr. P.NARASIMMAN BRANCH : ECE Mr.K.R.VENKATESAN YEAR : II SEMESTER

More information

Lecture: Sampling. Automatic Control 2. Sampling. Prof. Alberto Bemporad. University of Trento. Academic year

Lecture: Sampling. Automatic Control 2. Sampling. Prof. Alberto Bemporad. University of Trento. Academic year Automatic Control 2 Sampling Prof. Alberto Bemporad University of rento Academic year 2010-2011 Prof. Alberto Bemporad (University of rento) Automatic Control 2 Academic year 2010-2011 1 / 31 ime-discretization

More information

Control of Electromechanical Systems

Control of Electromechanical Systems Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance

More information

CDS 101/110a: Lecture 10-2 Control Systems Implementation

CDS 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 information

Control Systems Design

Control 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 information

Circular Orbits. Slide Pearson Education, Inc.

Circular Orbits. Slide Pearson Education, Inc. Circular Orbits The figure shows a perfectly smooth, spherical, airless planet with one tower of height h. A projectile is launched parallel to the ground with speed v 0. If v 0 is very small, as in trajectory

More information

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closed-loop behavior what we want it to be. To review: - G c (s) G(s) H(s) you are here! plant For

More information

Singular Value Decomposition Analysis

Singular Value Decomposition Analysis Singular Value Decomposition Analysis Singular Value Decomposition Analysis Introduction Introduce a linear algebra tool: singular values of a matrix Motivation Why do we need singular values in MIMO control

More information

Stabilizing the dual inverted pendulum

Stabilizing the dual inverted pendulum Stabilizing the dual inverted pendulum Taylor W. Barton Massachusetts Institute of Technology, Cambridge, MA 02139 USA (e-mail: tbarton@mit.edu) Abstract: A classical control approach to stabilizing a

More information

AA/EE/ME 548: Problem Session Notes #5

AA/EE/ME 548: Problem Session Notes #5 AA/EE/ME 548: Problem Session Notes #5 Review of Nyquist and Bode Plots. Nyquist Stability Criterion. LQG/LTR Method Tuesday, March 2, 203 Outline:. A review of Bode plots. 2. A review of Nyquist plots

More information

Control Systems I. Lecture 1: Introduction. Suggested Readings: Åström & Murray Ch. 1, Guzzella Ch. 1. Emilio Frazzoli

Control Systems I. Lecture 1: Introduction. Suggested Readings: Åström & Murray Ch. 1, Guzzella Ch. 1. Emilio Frazzoli Control Systems I Lecture 1: Introduction Suggested Readings: Åström & Murray Ch. 1, Guzzella Ch. 1 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich September 22, 2017 E. Frazzoli

More information

Design via Root Locus

Design via Root Locus Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steady-state error (Sections

More information

UNIVERSITY OF MISSOURI-COLUMBIA PHYSICS DEPARTMENT. PART I Qualifying Examination. August 20, 2013, 5:00 p.m. to 8:00 p.m.

UNIVERSITY OF MISSOURI-COLUMBIA PHYSICS DEPARTMENT. PART I Qualifying Examination. August 20, 2013, 5:00 p.m. to 8:00 p.m. UNIVERSITY OF MISSOURI-COLUMBIA PHYSICS DEPARTMENT PART I Qualifying Examination August 20, 2013, 5:00 p.m. to 8:00 p.m. Instructions: The only material you are allowed in the examination room is a writing

More information

School of Mechanical Engineering Purdue University. ME375 Feedback Control - 1

School of Mechanical Engineering Purdue University. ME375 Feedback Control - 1 Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System

More information

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION

ME 375 Final Examination Thursday, May 7, 2015 SOLUTION ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled

More information

Quiz Number 4 PHYSICS April 17, 2009

Quiz Number 4 PHYSICS April 17, 2009 Instructions Write your name, student ID and name of your TA instructor clearly on all sheets and fill your name and student ID on the bubble sheet. Solve all multiple choice questions. No penalty is given

More information

Reglerteknik Allmän Kurs. Del 2. Lösningar till Exempelsamling. Läsår 2015/16

Reglerteknik Allmän Kurs. Del 2. Lösningar till Exempelsamling. Läsår 2015/16 Reglerteknik Allmän Kurs Del Lösningar till Exempelsamling Läsår 5/6 Avdelningen för Reglerteknik, KTH, SE 44 Stockholm, SWEDEN AUTOMATIC CONTROL COMMUNICATION SYSTEMS LINKÖPINGS UNIVERSITET Reglerteknik

More information

MCE/EEC 647/747: Robot Dynamics and Control. Lecture 8: Basic Lyapunov Stability Theory

MCE/EEC 647/747: Robot Dynamics and Control. Lecture 8: Basic Lyapunov Stability Theory MCE/EEC 647/747: Robot Dynamics and Control Lecture 8: Basic Lyapunov Stability Theory Reading: SHV Appendix Mechanical Engineering Hanz Richter, PhD MCE503 p.1/17 Stability in the sense of Lyapunov A

More information

Control Systems II. Gioele Zardini June 3, 2017

Control Systems II. Gioele Zardini June 3, 2017 Control Systems II Gioele Zardini gzardini@studentethzch June 3, 207 Abstract This Skript is made of my notes from the lecture Control Systems II of Dr Gregor Ochsner (literature of Prof Dr Lino Guzzella)

More information

Outline. Classical Control. Lecture 2

Outline. Classical Control. Lecture 2 Outline Outline Outline Review of Material from Lecture 2 New Stuff - Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New

More information

PID controllers, part I

PID controllers, part I Faculty of Mechanical and Power Engineering Dr inŝ. JANUSZ LICHOTA CONTROL SYSTEMS PID controllers, part I Wrocław 2007 CONTENTS Controller s classification PID controller what is it? Typical controller

More information

Frequency Response of Linear Time Invariant Systems

Frequency Response of Linear Time Invariant Systems ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z

More information

Automatic Control (TSRT15): Lecture 1

Automatic Control (TSRT15): Lecture 1 Automatic Control (TSRT15): Lecture 1 Tianshi Chen* Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13-282226 Office: B-house extrance 25-27 * All lecture

More information