State Regulator. Advanced Control. design of controllers using pole placement and LQ design rules


 Marsha Nash
 3 years ago
 Views:
Transcription
1 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 space description Markus Kottmann, Date September 23, 29
2 Advanced Control, State Regulator MSE Preliminary Remarks The path to be followed in this and the next document is the following: The first point to be discussed concerns the design of regulators which make use of the state vector x to control a plant. It is assumed that the plant is controllable. The next point concerns the estimation of the state vector x in case that it is not completely measured. To design such estimators (observers) it is assumed that the plant is observable. The third theme concerns the cooperation between the regulator and the observer. Instead of using the true statevector, the regulator then utilizes the estimated statevector. Assumed that the two separate designs of regulator and observer met their specific requirements are these requirements still fulfilled if the regulator and the observer are used together? The assumptions about the plant to be controlled are as follows: Preferably, the plant is controllable and observable. If the plant lacks one or both of the abovementioned properties (see the examples in Fig. 1), then the noncontrollable or nonobservable parts can be removed from the model our methods will work for the remaining model which is controllable and observable. a) b) U 1 s+1 2 s+1 c) U X 1 X 1 X 2 Y X 1 1 Y U + s+1 + X 2 s 2 s+1 1 s 2 1 s 2 X 2 Y Figure 1: Examples of systems that are not controllable If the removed parts are e.g. unstable or badly damped then the control system will not be very successful. In that case, additional actuators or sensors are needed. The plant is assumed to be a lowpass system with a feedthrough matrix D =. The case D is not difficult it just produces larger equations and diagrams. In the sequel, the plant often is given as a SISOsystem. Of course, that is not a restriction. In principle, it is easier to control a plant with more sensors and more actuators. 2 Büchi Roland, Markus Kottmann September 29, 29
3 MSE Advanced Control, State Regulator 1 Feedback of the Full State Vector Figure 2: Control structure with feedforward term and full state feedback In Fig. 2, the control system has two components: the feedback part K, and the feedforward part K vf. The state feedback K is used to determine the dynamics of the closedloop system, while the prefilter K vf is used to guarantee the static gain (y = r ) 1. Note that the error signal e = r y which is used in classical output feedback does not appear in this control structure. In the feedback signal u R, the state variables x 1 (t), x 2 (t)... x n (t) are weighted by the coefficients k 1, k 2,..., k n. x 1 u R = k 1 x 1 + k 2 x k n x n = [ ] x 2 k 1 k 2... k n. = Kx x n The control signal u is The matrix equations are u = w u R = K vf r K x ẋ = A x + B (K vf r Kx) = A x BKx + BK vf r ẋ = (A BK) x + BK vf r and y = C x The whole system can be described by a new set of matrices A g, B b, C g A g = A BK, B g = BK vf, C g = C 1 The prefilter, as used here, is just a gain. More sophisticated prefilters might be used to shape the dynamics of Y (s)/r(s). September 29, 29 Büchi Roland, Markus Kottmann 3
4 Advanced Control, State Regulator MSE The I/Odescription becomes G(s) = Y (s) R(s) = C g(si A g ) 1 B g = C(sI A + BK) 1 BK vf For the moment we assume that K is known. Then the static gain can be adjusted by setting G() = 1 or K vf = (C( A + BK) 1 B) 1, resp. This method is sensitive to parameter variations, so in section 4 we will discuss an alternative which includes integral action. Sections 2 and 3 deal with the question how to calculate K. 2 Pole Placement According to Definition B of controllability, the controllability of (A, B) ensures that the eigenvalues λ 1... λ n of A g = A BK can be placed arbitrarily by proper choice of K. The eigenvalues of A g are given by the roots of det(λi A + BK). To achieve some desired eigenvalues λ 1... λ n the following polynomials must coincide: det(λi A + BK) = (λ λ 1 )(λ λ 2 )... (λ λ n ). The question which arises is about the choice of good pole locations. Some considerations are: Obviously, the poles must be chosen in the left half plane for stability reasons. Transients of the signals should decay towards zero faster than some C e σ 1. The transients should be well damped, e.g. ζ > 1/ 2. Forcing transients to decay fast towards needs energy. Therefore, an upper bound σ 2 > σ is meaningful. ζ = 1/ (2) Im s Re ζ = 1/ (2) σ 2 σ 1 Figure 3: Qualitative sketch of region of good pole locations While the placement of poles is simple for low order systems, it gets somehow unsatisfying when handling systems of order e.g. 4 or higher. 4 Büchi Roland, Markus Kottmann September 29, 29
5 MSE Advanced Control, State Regulator 3 Optimal Control An alternative way to determine K makes use of the fundamental idea of optimality. Changing the value of K results in a different behavior of the control system. If the behavior of the control system is judged quantitatively, it is convenient to have a scalar performance index (or cost function) J = f(k). Normally, better performance corresponds to smaller values of J. The optimal value of K then can be found by minimizing J, e.g. by varying K systematically. In control theory, systems that are adjusted to provide a minimum performance index are called optimal control systems. The function f often is defined based on some intermediary variables such as the state vector x or the input u J = f(k) = t f g(x, u)dt where, of course, x and u depend on K. Since it is useless to look at the values of x and u at some specific time only, the index J is formulated as an integral over time. For theoretical considerations, it is convenient to set the final time t f to infinity. Quadratic terms for the function g are prevalent on one hand these are meaningful measures, and on the other hand they often yield nice analytical results. Figure 4: Regulator problem Consider the following regulator problem, see Fig. 4. The initial state of the system is x() = x. The intention is to take x towards zero as fast as possible by using state feedback. If the norm of x is taken, x = x x x 2 n = [ ] x 1 x 2... x n x 1 x 2. x n = xt x September 29, 29 Büchi Roland, Markus Kottmann 5
6 Advanced Control, State Regulator MSE and integrated over time we get the performance index J = t f x T x dt. In the following derivation the time horizon is set to infinity, and a more general weighting scheme for the state variables is used with a symmetric, positive semidefinite matrix Q: J = x T Qx dt (1) Note that the general form of the performance index incorporates a term containing u to weigh the control energy. This will be discussed in section 3.1. To obtain the minimum value of J, we follow [1] and postulate the existence of an exact differential with a constant symmetric matrix P such that d dt (xt P x) = x T Qx where P is to be determined. Applying the product rule for differentiation, d dt (xt P x) = ẋ T P x + x T and substituting ẋ = (A BK) x, we get }{{} H P }{{} x + x T P ẋ d dt (xt P x) = (Hx) T P x + x T P (Hx) = x T H T P x + x T P Hx = x T (H T P + P H) x = x T Qx }{{} Q which is the exact differential we are seeking. Substituting that expression in (1) results in the performance index J = d dt (xt P x)dt = x T P x = ( xt ()P x()) = x T ()P x() In the evaluation of the limit at t =, we have assumed that the system is stable and hence x( ) =, as desired. Therefore to minimize the performance index J, we consider the two equations: J = The design steps are then as follows: (x T Qx)dt = x T ()P x() and (H T P + P H) = Q 1. Determine the matrix P, where H and Q are known. 2. Minimize J = x T ()P x() by adjusting one ore more unspecified system parameters. 6 Büchi Roland, Markus Kottmann September 29, 29
7 MSE Advanced Control, State Regulator 3.1 Linear Quadratic Regulator (LQR) In a more general case, we also consider the control energy of u. A method suitable for computer calculation is stated without proof in the following. Consider the uncompensated MIMO system ẋ = Ax + Bu with feedback The performance index is u = Kx. J = (x T Qx + u T Ru)dt (2) With Q and R both positive definite, it can be shown that (2) is minimized if K = R 1 B T P To calculate the symmetric n n matrix P, an algebraic matrix Riccati (ARE) equation must be solved: A T P + P A P BR 1 B T P = Q. Since the ARE is a quadratic equation, it has multiple solutions. Equation (2) demands for the (only) solution P which is positive definite. The resulting optimal controller is called the linear quadratic regulator (MATLAB command: lqr). The condition about Q can be relaxed: if Q = C T C is positive semidefinite and if (A, C) is observable, then P and K can be calculated in the same way. The reason is that x must be observable through the virtual output y = Cx. 3.2 Performance of the Linear Quadratic Regulator If the control loop is opened at the input u of the plant (see Fig. 4) then the openloop transfer function is given by G (s) = K(sI A) 1 B. It can be shown, that G (jω) has at least a phase margin of ±6 and a gain margin of [.5... [ on each channel. This makes the LQregulator a considerably robust controller. 4 State Regulator Including Integral Part In this section we discuss the problem of designing a compensator that provides an asymptotic tracking of a constant reference input r(t) = r with zero steadystate error. From classical control it is known that this can be achieved if the openloop is at least of type 1, i.e. if it has at least one open integrator. This idea is formalized here by introducing an internal model of the reference input in the compensator. September 29, 29 Büchi Roland, Markus Kottmann 7
8 Advanced Control, State Regulator MSE Figure 5: Block diagram of a regulator with integral part In Fig. 5 the tracking error e is defined as and its time derivative yields e = y r, ė = ẏ (note the sign!) {}}{ ṙ = ẏ = Cẋ Defining two intermediate variables, z and w, as a new system results: ˆx z = ẋ and w = u, Â [{}} ]{ [{}} ]{ [{}}{ ė C e = ż A z ˆx ] + ˆB {[ }}{ B ] û {}}{ w (3) If the system (Â, ˆB) is controllable, then some feedback gain ˆK which stabilizes (3) can be designed e.g. by pole placement or using the LQR method û = ˆK ˆx. Since the error e is a state variable it converges towards zero, i.e. the requirement of an asymptotic tracking with zero steadystate error is fulfilled. Splitting the feedback term yields [ ] û = w = ˆK e ˆx = [K 1 K 2 ] = K z 1 e K 2 z and by integration the final control law u(t) = K 1 t e(τ)dτ K 2 x(t) which is illustrated in Fig Büchi Roland, Markus Kottmann September 29, 29
9 Bibliography [1] Richard C. Dorf and Robert H. Bishop. Modern Control Systems. Pearson, 28. [2] William S. Levine, editor. The Control Handbook. The Electrical Engineering Handbook Series. CRC Press/IEEE Press, September 29, 29 Büchi Roland, Markus Kottmann 9
10 Advanced Control, State Regulator MSE Solved Exercises 1 Pole Placement Given is a system in state space description [ ] [ ] 1 A =, B = a K s, C = [ c 1 ], D = a = 4, K s = 8, c 1 = 6.4 For each of the following cases, find a controller u = k x + k vf u such that the closed loop system has a unitary static gain and the poles p 1 and p 2 : 1. p 1 = p 2 = p 1,2 = 15 ± 15j 3. p 1 = p 2 = 5 1 Büchi Roland, Markus Kottmann September 29, 29
11 MSE Advanced Control, State Regulator 2 Optimal Control 1 Consider the control system shown below. The state vector is x = [x 1 openloop system is unstable, therefore state feedback is introduced. x 2 ] T. The Figure 6: Block diagram of a LQ regulator The performance index is J = (x T Qx)dt. Additional conditions are: Q = I and k 1 = k 2 = k and u < 1 at initial conditions x() = 1. Find the plant s matrices A, B, C and D. 2. Find the matrices H and P. 3. Find k by minimizing the performance index J. [ 1 ] September 29, 29 Büchi Roland, Markus Kottmann 11
12 Advanced Control, State Regulator MSE 3 Optimal Control 2 Given is a scalar system ẋ = ax + bu with a simplified algebraic Riccati equation: 2ap 1 r b2 p 2 + q = 1. Find a controller u = kx by minimizing the performance index J J = (qx 2 + ru 2 )dt, (q, r > ) 2. Find a controller u = kx, which stabilizes the system using minimal control energy (q = ). 12 Büchi Roland, Markus Kottmann September 29, 29
13 MSE Advanced Control, State Regulator 4 LQRegulator of an Inverted Pendulum Consider an inverted pendulum, where the input signal u(t) is a force, applied to the cart, and the output signal y(t) is the position of the cart. The state vector [x 1, x 2, x 3, x 4 ] T is assumed to be [y, ẏ, θ, θ] T. The parameters are given by M = 1kg, m =.1kg, g = 1m/s 2, l = 1m. Figure 7: inverted pendulum Thus, the following system description can be found: 1 A = , B = 1.11, C = [ 1 ], D = Compute some optimal control regulators using Matlab command lqr and plot step responses of the closedloop systems at different weights of control energy r. Find also suitable prefilters K vf. 2. For the same weights r of the control energy, plot the Bode diagram of the closedloop systems and discuss the results. 3. Discuss the robustness of the system. September 29, 29 Büchi Roland, Markus Kottmann 13
14 Advanced Control, State Regulator MSE Solutions 1 Solution Pole Placement 14 Büchi Roland, Markus Kottmann September 29, 29
15 MSE Advanced Control, State Regulator 2 Solution Optimal Control (1) September 29, 29 Büchi Roland, Markus Kottmann 15
16 Advanced Control, State Regulator MSE 3 Solution Optimal Control (2) 16 Büchi Roland, Markus Kottmann September 29, 29
17 MSE Advanced Control, State Regulator 4 Solution LQRegulator of an Inverted Pendulum September 29, 29 Büchi Roland, Markus Kottmann 17
18 Advanced Control, State Regulator MSE Step Response Amplitude r =.1 r = 1 r = Time (sec) Bode Diagram Magnitude (db) 51 r =.1 r = 1 r = Phase (deg) Frequency (rad/sec) 1 Nyquist Diagram Openloop r =.1 r = 1 r = Büchi Roland, Markus Kottmann September 29, 29
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 informationCourse Outline. Higher Order Poles: Example. Higher Order Poles. Amme 3500 : System Dynamics & Control. State Space Design. 1 G(s) = s(s + 2)(s +10)
Amme 35 : System Dynamics Control State Space Design Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the splane
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 informationLinear State Feedback Controller Design
Assignment For EE5101  Linear Systems Sem I AY2010/2011 Linear State Feedback Controller Design Phang Swee King A0033585A Email: king@nus.edu.sg NGS/ECE Dept. Faculty of Engineering National University
More information5. Observerbased Controller Design
EE635  Control System Theory 5. Observerbased Controller Design Jitkomut Songsiri state feedback poleplacement design regulation and tracking state observer feedback observer design LQR and LQG 51
More informationEE C128 / ME C134 Final Exam Fall 2014
EE C128 / ME C134 Final Exam Fall 2014 December 19, 2014 Your PRINTED FULL NAME Your STUDENT ID NUMBER Number of additional sheets 1. No computers, no tablets, no connected device (phone etc.) 2. Pocket
More informationSuppose that we have a specific single stage dynamic system governed by the following equation:
Dynamic Optimisation Discrete Dynamic Systems A single stage example Suppose that we have a specific single stage dynamic system governed by the following equation: x 1 = ax 0 + bu 0, x 0 = x i (1) where
More informationEL 625 Lecture 10. Pole Placement and Observer Design. ẋ = Ax (1)
EL 625 Lecture 0 EL 625 Lecture 0 Pole Placement and Observer Design Pole Placement Consider the system ẋ Ax () The solution to this system is x(t) e At x(0) (2) If the eigenvalues of A all lie in the
More informationCONTROL DESIGN FOR SET POINT TRACKING
Chapter 5 CONTROL DESIGN FOR SET POINT TRACKING In this chapter, we extend the pole placement, observerbased output feedback design to solve tracking problems. By tracking we mean that the output is commanded
More informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationMODERN CONTROL DESIGN
CHAPTER 8 MODERN CONTROL DESIGN The classical design techniques of Chapters 6 and 7 are based on the rootlocus and frequency response that utilize only the plant output for feedback with a dynamic controller
More informationTopic # Feedback Control
Topic #5 6.3 Feedback Control StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or,evenifwecanchangethepolelocations. Where do we put the poles? Linear
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 informationChapter 3. State Feedback  Pole Placement. Motivation
Chapter 3 State Feedback  Pole Placement Motivation Whereas classical control theory is based on output feedback, this course mainly deals with control system design by state feedback. This modelbased
More information1 (30 pts) Dominant Pole
EECS C8/ME C34 Fall Problem Set 9 Solutions (3 pts) Dominant Pole For the following transfer function: Y (s) U(s) = (s + )(s + ) a) Give state space description of the system in parallel form (ẋ = Ax +
More informationTopic # Feedback Control Systems
Topic #17 16.31 Feedback Control Systems Deterministic LQR Optimal control and the Riccati equation Weight Selection Fall 2007 16.31 17 1 Linear Quadratic Regulator (LQR) Have seen the solutions to the
More information6 OUTPUT FEEDBACK DESIGN
6 OUTPUT FEEDBACK DESIGN When the whole sate vector is not available for feedback, i.e, we can measure only y = Cx. 6.1 Review of observer design Recall from the first class in linear systems that a simple
More informationControl Systems. Design of State Feedback Control.
Control Systems Design of State Feedback Control chibum@seoultech.ac.kr Outline Design of State feedback control Dominant pole design Symmetric root locus (linear quadratic regulation) 2 Selection of closedloop
More informationQuadratic Stability of Dynamical Systems. Raktim Bhattacharya Aerospace Engineering, Texas A&M University
.. Quadratic Stability of Dynamical Systems Raktim Bhattacharya Aerospace Engineering, Texas A&M University Quadratic Lyapunov Functions Quadratic Stability Dynamical system is quadratically stable if
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 splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationZeros and zero dynamics
CHAPTER 4 Zeros and zero dynamics 41 Zero dynamics for SISO systems Consider a linear system defined by a strictly proper scalar transfer function that does not have any common zero and pole: g(s) =α p(s)
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 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 8111 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 informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the
More informationHomework Solution # 3
ECSE 644 Optimal Control Feb, 4 Due: Feb 17, 4 (Tuesday) Homework Solution # 3 1 (5%) Consider the discrete nonlinear control system in Homework # For the optimal control and trajectory that you have found
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 informationExam. 135 minutes, 15 minutes reading time
Exam August 6, 208 Control Systems II (5059000) 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 informationControl System Design
ELEC4410 Control System Design Lecture 19: Feedback from Estimated States and DiscreteTime Control Design Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science
More informationState Feedback and State Estimators Linear System Theory and Design, Chapter 8.
1 Linear System Theory and Design, http://zitompul.wordpress.com 2 0 1 4 2 Homework 7: State Estimators (a) For the same system as discussed in previous slides, design another closedloop state estimator,
More informationPole placement control: state space and polynomial approaches Lecture 2
: state space and polynomial approaches Lecture 2 : a state O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.fr www.gipsalab.fr/ o.sename based November 21, 2017 Outline : a state
More informationẋ n = f n (x 1,...,x n,u 1,...,u m ) (5) y 1 = g 1 (x 1,...,x n,u 1,...,u m ) (6) y p = g p (x 1,...,x n,u 1,...,u m ) (7)
EEE582 Topical Outline A.A. Rodriguez Fall 2007 GWC 352, 9653712 The following represents a detailed topical outline of the course. It attempts to highlight most of the key concepts to be covered and
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 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 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 58 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 informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationEE221A Linear System Theory Final Exam
EE221A Linear System Theory Final Exam Professor C. Tomlin Department of Electrical Engineering and Computer Sciences, UC Berkeley Fall 2016 12/16/16, 811am Your answers must be supported by analysis,
More informationChapter 8 Stabilization: State Feedback 8. Introduction: Stabilization One reason feedback control systems are designed is to stabilize systems that m
Lectures on Dynamic Systems and Control Mohammed Dahleh Munther A. Dahleh George Verghese Department of Electrical Engineering and Computer Science Massachuasetts Institute of echnology c Chapter 8 Stabilization:
More informationChapter 3. LQ, LQG and Control System Design. Dutch Institute of Systems and Control
Chapter 3 LQ, LQG and Control System H 2 Design Overview LQ optimization state feedback LQG optimization output feedback H 2 optimization nonstochastic version of LQG Application to feedback system design
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 informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationTopic # Feedback Control Systems
Topic #19 16.31 Feedback Control Systems Stengel Chapter 6 Question: how well do the large gain and phase margins discussed for LQR map over to DOFB using LQR and LQE (called LQG)? Fall 2010 16.30/31 19
More information1. 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 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 informationExtensions and applications of LQ
Extensions and applications of LQ 1 Discrete time systems 2 Assigning closed loop pole location 3 Frequency shaping LQ Regulator for Discrete Time Systems Consider the discrete time system: x(k + 1) =
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 informationIntro. Computer Control Systems: F9
Intro. Computer Control Systems: F9 Statefeedback control and observers Dave Zachariah Dept. Information Technology, Div. Systems and Control 1 / 21 dave.zachariah@it.uu.se F8: Quiz! 2 / 21 dave.zachariah@it.uu.se
More informationCoordinated Tracking Control of Multiple Laboratory Helicopters: Centralized and DeCentralized Design Approaches
Coordinated Tracking Control of Multiple Laboratory Helicopters: Centralized and DeCentralized Design Approaches Hugh H. T. Liu University of Toronto, Toronto, Ontario, M3H 5T6, Canada Sebastian Nowotny
More informationHere represents the impulse (or delta) function. is an diagonal matrix of intensities, and is an diagonal matrix of intensities.
19 KALMAN FILTER 19.1 Introduction In the previous section, we derived the linear quadratic regulator as an optimal solution for the fullstate feedback control problem. The inherent assumption was that
More information1 Steady State Error (30 pts)
Professor Fearing EECS C28/ME C34 Problem Set Fall 2 Steady State Error (3 pts) Given the following continuous time (CT) system ] ẋ = A x + B u = x + 2 7 ] u(t), y = ] x () a) Given error e(t) = r(t) y(t)
More informationIntermediate Process Control CHE576 Lecture Notes # 2
Intermediate Process Control CHE576 Lecture Notes # 2 B. Huang Department of Chemical & Materials Engineering University of Alberta, Edmonton, Alberta, Canada February 4, 2008 2 Chapter 2 Introduction
More informationDenis ARZELIER arzelier
COURSE ON LMI OPTIMIZATION WITH APPLICATIONS IN CONTROL PART II.2 LMIs IN SYSTEMS CONTROL STATESPACE METHODS PERFORMANCE ANALYSIS and SYNTHESIS Denis ARZELIER www.laas.fr/ arzelier arzelier@laas.fr 15
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 informationRobotics. Control Theory. Marc Toussaint U Stuttgart
Robotics Control Theory Topics in control theory, optimal control, HJB equation, infinite horizon case, LinearQuadratic optimal control, Riccati equations (differential, algebraic, discretetime), controllability,
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 informationControl Systems Design, SC4026. SC4026 Fall 2010, dr. A. Abate, DCSC, TU Delft
Control Systems Design, SC426 SC426 Fall 2, dr A Abate, DCSC, TU Delft Lecture 5 Controllable Canonical and Observable Canonical Forms Stabilization by State Feedback State Estimation, Observer Design
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequencydomain analysis and control design (15 pt) Given is a
More informationTransfer 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. 2425 Testo del corso:
More informationCDS 101/110a: Lecture 81 Frequency Domain Design
CDS 11/11a: Lecture 81 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 informationRobust Control 5 Nominal Controller Design Continued
Robust Control 5 Nominal Controller Design Continued Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 4/14/2003 Outline he LQR Problem A Generalization to LQR MinMax
More informationTRACKING AND DISTURBANCE REJECTION
TRACKING AND DISTURBANCE REJECTION Sadegh Bolouki Lecture slides for ECE 515 University of Illinois, UrbanaChampaign Fall 2016 S. Bolouki (UIUC) 1 / 13 General objective: The output to track a reference
More informationEEE582 Homework Problems
EEE582 Homework Problems HW. Write a statespace realization of the linearized model for the cruise control system around speeds v = 4 (Section.3, http://tsakalis.faculty.asu.edu/notes/models.pdf). Use
More informationLecture 4 Continuous time linear quadratic regulator
EE363 Winter 200809 Lecture 4 Continuous time linear quadratic regulator continuoustime LQR problem dynamic programming solution Hamiltonian system and two point boundary value problem infinite horizon
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 informationChapter 9 Observers, Modelbased Controllers 9. Introduction In here we deal with the general case where only a subset of the states, or linear combin
Lectures on Dynamic Systems and Control Mohammed Dahleh Munther A. Dahleh George Verghese Department of Electrical Engineering and Computer Science Massachuasetts Institute of Technology c Chapter 9 Observers,
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 DMAVT ETH Zürich October 13, 2017 E. Frazzoli (ETH)
More informationSYSTEMTEORI  KALMAN FILTER VS LQ CONTROL
SYSTEMTEORI  KALMAN FILTER VS LQ CONTROL 1. Optimal regulator with noisy measurement Consider the following system: ẋ = Ax + Bu + w, x(0) = x 0 where w(t) is white noise with Ew(t) = 0, and x 0 is a stochastic
More informationTopic # Feedback Control Systems
Topic #20 16.31 Feedback Control Systems Closedloop system analysis Bounded Gain Theorem Robust Stability Fall 2007 16.31 20 1 SISO Performance Objectives Basic setup: d i d o r u y G c (s) G(s) n control
More informationTopic # Feedback Control Systems
Topic #14 16.31 Feedback Control Systems StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or, even if we can change the pole locations. Where do we
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 informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More informationPOLE PLACEMENT. Sadegh Bolouki. Lecture slides for ECE 515. University of Illinois, UrbanaChampaign. Fall S. Bolouki (UIUC) 1 / 19
POLE PLACEMENT Sadegh Bolouki Lecture slides for ECE 515 University of Illinois, UrbanaChampaign Fall 2016 S. Bolouki (UIUC) 1 / 19 Outline 1 State Feedback 2 Observer 3 Observer Feedback 4 Reduced Order
More informationSYSTEMTEORI  ÖVNING 5: FEEDBACK, POLE ASSIGNMENT AND OBSERVER
SYSTEMTEORI  ÖVNING 5: FEEDBACK, POLE ASSIGNMENT AND OBSERVER Exercise 54 Consider the system: ẍ aẋ bx u where u is the input and x the output signal (a): Determine a state space realization (b): Is the
More informationAustralian Journal of Basic and Applied Sciences, 3(4): , 2009 ISSN Modern Control Design of Power System
Australian Journal of Basic and Applied Sciences, 3(4): 42674273, 29 ISSN 99878 Modern Control Design of Power System Atef Saleh Othman AlMashakbeh Tafila Technical University, Electrical Engineering
More informationTopic # /31 Feedback Control Systems. Analysis of Nonlinear Systems Lyapunov Stability Analysis
Topic # 16.30/31 Feedback Control Systems Analysis of Nonlinear Systems Lyapunov Stability Analysis Fall 010 16.30/31 Lyapunov Stability Analysis Very general method to prove (or disprove) stability of
More informationAutomatic Control II Computer exercise 3. LQG Design
Uppsala University Information Technology Systems and Control HN,FS,KN 200010 Last revised by HR August 16, 2017 Automatic Control II Computer exercise 3 LQG Design Preparations: Read Chapters 5 and 9
More informationCDS 101: Lecture 51 Reachability and State Space Feedback
CDS 11: Lecture 51 Reachability and State Space Feedback Richard M. Murray 23 October 26 Goals: Define reachability of a control system Give tests for reachability of linear systems and apply to examples
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 informationIntroduction to Nonlinear Control Lecture # 4 Passivity
p. 1/6 Introduction to Nonlinear Control Lecture # 4 Passivity È p. 2/6 Memoryless Functions ¹ y È Ý Ù È È È È u (b) µ power inflow = uy Resistor is passive if uy 0 p. 3/6 y y y u u u (a) (b) (c) Passive
More informationLecture 2: Discretetime Linear Quadratic Optimal Control
ME 33, U Berkeley, Spring 04 Xu hen Lecture : Discretetime Linear Quadratic Optimal ontrol Big picture Example onvergence of finitetime LQ solutions Big picture previously: dynamic programming and finitehorizon
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 505900 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 informationMassachusetts Institute of Technology. Department of Electrical Engineering and Computer Science : MULTIVARIABLE CONTROL SYSTEMS by A.
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski QParameterization 1 This lecture introduces the socalled
More informationReturn Difference Function and ClosedLoop Roots SingleInput/SingleOutput Control Systems
Spectral Properties of Linear Quadratic Regulators Robert Stengel Optimal Control and Estimation MAE 546 Princeton University, 2018! Stability margins of singleinput/singleoutput (SISO) systems! Characterizations
More informationIntroduction to Modern Control MT 2016
CDT Autonomous and Intelligent Machines & Systems Introduction to Modern Control MT 2016 Alessandro Abate Lecture 2 Firstorder ordinary differential equations (ODE) Solution of a linear ODE Hints to nonlinear
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 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 DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationLQR, Kalman Filter, and LQG. Postgraduate Course, M.Sc. Electrical Engineering Department College of Engineering University of Salahaddin
LQR, Kalman Filter, and LQG Postgraduate Course, M.Sc. Electrical Engineering Department College of Engineering University of Salahaddin May 2015 Linear Quadratic Regulator (LQR) Consider a linear system
More informationControl 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 informationSubject: Optimal Control Assignment1 (Related to Lecture notes 110)
Subject: Optimal Control Assignment (Related to Lecture notes ). Design a oil mug, shown in fig., to hold as much oil possible. The height and radius of the mug should not be more than 6cm. The mug must
More informationECEn 483 / ME 431 Case Studies. Randal W. Beard Brigham Young University
ECEn 483 / ME 431 Case Studies Randal W. Beard Brigham Young University Updated: December 2, 2014 ii Contents 1 Single Link Robot Arm 1 2 Pendulum on a Cart 9 3 Satellite Attitude Control 17 4 UUV Roll
More informationOptimal Polynomial Control for DiscreteTime Systems
1 Optimal Polynomial Control for DiscreteTime Systems Prof Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning this paper should
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control DMAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More information5HC99 Embedded Vision Control. Feedback Control Systems. dr. Dip Goswami Flux Department of Electrical Engineering
5HC99 Embedded Vision Control Feedback Control Systems dr. Dip Goswami d.goswami@tue.nl Flux 04.135 Department of Electrical Engineering 1 Example Feedback control system: regulates the behavior of dynamical
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 20022003 Schedule November 25 MSt December 2 MSt Homework # 1 December 9
More informationCONTROL SYSTEMS, ROBOTICS, AND AUTOMATION Vol. III Controller Design  Boris Lohmann
CONROL SYSEMS, ROBOICS, AND AUOMAION Vol. III Controller Design  Boris Lohmann CONROLLER DESIGN Boris Lohmann Institut für Automatisierungstechnik, Universität Bremen, Germany Keywords: State Feedback
More informationOutline. 1 Linear Quadratic Problem. 2 Constraints. 3 Dynamic Programming Solution. 4 The Infinite Horizon LQ Problem.
Model Predictive Control Short Course Regulation James B. Rawlings Michael J. Risbeck Nishith R. Patel Department of Chemical and Biological Engineering Copyright c 217 by James B. Rawlings Outline 1 Linear
More informationOPTIMAL CONTROL. Sadegh Bolouki. Lecture slides for ECE 515. University of Illinois, UrbanaChampaign. Fall S. Bolouki (UIUC) 1 / 28
OPTIMAL CONTROL Sadegh Bolouki Lecture slides for ECE 515 University of Illinois, UrbanaChampaign Fall 2016 S. Bolouki (UIUC) 1 / 28 (Example from Optimal Control Theory, Kirk) Objective: To get from
More informationSchool of Mechanical Engineering Purdue University. ME375 Feedback Control  1
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationECEEN 5448 Fall 2011 Homework #4 Solutions
ECEEN 5448 Fall 2 Homework #4 Solutions Professor David G. Meyer Novemeber 29, 2. The statespace realization is A = [ [ ; b = ; c = [ which describes, of course, a free mass (in normalized units) with
More informationMTNS 06, Kyoto (July, 2006) Shinji Hara The University of Tokyo, Japan
MTNS 06, Kyoto (July, 2006) Shinji Hara The University of Tokyo, Japan Outline Motivation & Background: H2 Tracking Performance Limits: new paradigm Explicit analytical solutions with examples H2 Regulation
More informationControl of SingleInput SingleOutput Systems
Control of SingleInput SingleOutput Systems Dimitrios HristuVarsakelis 1 and William S. Levine 2 1 Department of Applied Informatics, University of Macedonia, Thessaloniki, 546, Greece dcv@uom.gr 2
More information