ECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27


 Kimberly Daniel
 3 years ago
 Views:
Transcription
1 1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability
2 Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system where G ol (s) is the open loop transfer function. 2/27
3 Feedback System (cont.) The closed loop transfer function is Y (s) U(s) = G ol(s) 1 + G ol (s) =: G cl(s). (1) If the open loop transfer function is written in terms of numerator and denominator polynomials, G ol (s) = n ol(s) d ol (s) (2) we have G cl (s) = n cl(s) d cl (s) = n ol (s) d ol (s) + n ol (s), (3) 3/27
4 Feedback System (cont.) so that n cl (s) = n ol (s) (4) and Equations (4) and (5) show that d cl (s) = d ol (s) + n ol (s). (5) a) the zeros of the unity feedback system are identical to the zeros of the open loop system, b) the poles of the closed loop system in general differ from those of the open loop system and are given by the roots of d cl (s) = 0. 4/27
5 Feedback System (cont.) In view of the above d cl (s) = d ol (s) + n ol (s) (6) is called the closed loop characteristic polynomial. It follows that the closed loop system is stable if and only if all roots of d cl (s) = 0 (7) called closed loop characteristic roots lie in the open left half of the complex plane (LHP), or equivalently d cl (s) is a Hurwitz polynomial. 5/27
6 Feedback System (cont.) Example G ol (s) = K(s 1) (s + 1) K(s 2) G cl (s) = (s + 1) + K(s 2) K(s 2) = (1 + K)s + (1 2K) (8) (9a) (9b) 6/27
7 Feedback System (cont.) If K = 1, d cl (s) = 2s 1 = 0 s = 1 /2, (10) but if K = 1 /4 d cl (s) = 5 /4 s + 1 /2 = 0 s = 2 /5. (11) This example shows that the closed loop may be stable or unstable, depending on the value of K, even though the open loop is stable. 7/27
8 Feedback System (cont.) Example G ol (s) = G cl (s) = K s 2 K s 2 + K Thus the open loop system is unstable but the feedback system is stable for K > 2. (12) (13) 8/27
9 Control System In a control system containing plant and controller in unity feedback we have G ol (s) r + e C(s) u P(s) y Figure 2: Unity feedback system with a controller and a plant 9/27
10 Control System (cont.) G ol (s) = P(s) C(s) (14) G cl (s) = P(s) C(s) 1 + P(s) C(s). (15) Writing the transfer functions in terms of numerator and denominator polynomials G ol (s) = n P(s) n C (s) d P (s) d C (s) }{{}}{{} P(s) C(s) (16) G cl (s) = n P (s) n C (s) d P (s) d C (s) + n P (s) n C (s). (17) 10/27
11 Control System (cont.) Therefore the closed loop characteristic polynomial is given by d cl (s) = d P (s) d C (s) + n P (s) n C (s). (18) The roots of d cl (s) = 0 are the closed loop characteristic roots and a controller C(s) stabilizes a plant P(s) if and only if the roots lie in the LHP, or d cl (s) is Hurwitz. 11/27
12 Control System (cont.) Example + C i (s) = K i P(s) = 1 s 1 Figure 3: A gain controller with an unstable plant 12/27
13 Control System (cont.) The controller C 1 (s) = K 1 = 2 stabilizes the closed loop system since d cl (s) = s 1 + K 1 = s + 1 has an LHP root at s = 1. The controller C 2 (s) = K 2 = 1 /2 does not stabilize the closed loop system since d cl (s) = s 1 + K 2 = s 1 /2 has an RHP root at s = 1 /2. 13/27
14 Control System (cont.) Example (Routh Criterion) Consider the following system. r K + P(s) Figure 4: A gain controller K with plant P(s) 14/27
15 Control System (cont.) Find the range of stabilizing K. a) P(s) = s 1 s(s + 1) 1 b) P(s) = c) P(s) = s(s + 1)(s + 2) 1 s(s + 1)(s + 2)(s + 3) d) P(s) = s + 1 s 2 (s + 2) e) P(s) = s + 2 s 2 (s + 1) f ) P(s) = s 1 (s + 1)(s 2) 15/27
16 Control System (cont.) Solution. In each case apply the Routh criterion to the characteristic polynomial of the closed loop system: a) d cl (s) = s(s + 1) + K(s 1) = s 2 + (K + 1)s K. Stabilizing range : 1 < K < 0. b) d cl (s) = s(s + 1)(s + 2) + K Stabilizing range : 0 < K < 6. = s s s + K. (19) (20) 16/27
17 Control System (cont.) c) d cl (s) = s(s + 1)(s + 2)(s + 3) + K = s s s s + K. Stabilizing range : 0 < K < 10. d) d cl (s) = s 2 (s + 2) + K(s + 1) = s s 2 + K s + K. Stabilizing range : 0 < K. e) d cl (s) = s 2 (s + 1) + K(s + 2) Stabilizing range : does not exist. = s 3 + s 2 + K s + 2K. (21) (22) (23) 17/27
18 Tracking Step Inputs In this section we discuss the problem of designing a controller to track step inputs. Consider first the system r + e C(s) u P(s) y Figure 5: A unity feedback system where P(s) = n P(s) d P (s), C(s) = n C (s) d C (s). (24) 18/27
19 Tracking Step Inputs (cont.) The error transfer function is given by 1 E(s) = R(s) 1 + P(s) C(s) (25a) d P (s) d C (s) = d P (s) d C (s) + n P (s) n C (s) R(s) (25b) = d P(s) d C (s) R(s). d cl (s) (25c) 19/27
20 Tracking Step Inputs (cont.) Suppose now that r(t) is a step input of height R 0 R(s) = R 0 s (26) and E(s) = d P(s) d C (s) d cl (s) R 0 s. (27) Thus the forced response e(t) = L 1 (E(s)) consists of weighted exponential terms e λ i t, λ i being the closed loop characteristic roots and a constant term E ss which is the steady state error. 20/27
21 Tracking Step Inputs (cont.) Since the closed loop must be stable the λ i must have negative real parts and so E ss is the residue of the pole at s = 0 in a partial fraction expansion of E(s). Thus, by usual partial fraction formula, lim e(t) = E ss = d P(0) d C (0) R 0 t d cl (0) (28) is the steady state error. 21/27
22 Tracking Step Inputs (cont.) Example Consider the plant P(s) = 1 and controller C(s) = K. As seen s 2 before the closed loop is stable if and only if K > 2. For a unit step input the steady state error is E = K = 2 K 2 (29) 22/27
23 Tracking Step Inputs (cont.) E 2 K Figure 6: which shows that the steady state error can be reduced to zero only with infinite gain. 23/27
24 Tracking Step Inputs (cont.) Example Consider the system r + K s z s 2 + a s + b y Figure 7: A gain controller subject to a unit step input r(t). 24/27
25 Tracking Step Inputs (cont.) The closed loop characteristic polynomial d cl (s) = (s 2 + a s + b) + K(s z) (30a) = s 2 + (a + K)s + (b K z) (30b) and closed loop stability is equivalent to K + a > 0 b K z > 0 (31a) (31b) so that a < K < b z. (32) 25/27
26 Tracking Step Inputs (cont.) The steady state error to a unit step is E ss = d C (0) d P (0) d cl (0) = 1 b b K z. (33) Assuming that the range in (32) is nonempty, that is b z we can sketch the function (33). > a (34) 26/27
27 Tracking Step Inputs (cont.) E 1 a b (b a z) Figure 8: b z K We note that in this example the steady state error has an infimum b or a lower bound of, over the stabilizing range of controller b a z gains. 27/27
ECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : RouthHurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
More informationTest 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010
Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the
More informationAutomatic Control Systems (FCS) Lecture 8 Steady State Error
Automatic Control Systems (FCS) Lecture 8 Steady State Error Introduction Any physical control system inherently suffers steadystate error in response to certain types of inputs. A system may have no
More informationECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8
Learning Objectives ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8 Dr. Oishi oishi@unm.edu November 2, 203 State the phase and gain properties of a root locus Sketch a root locus, by
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design
AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University Routh s Stability
More informationME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II
ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and
More informationSome special cases
Lecture Notes on Control Systems/D. Ghose/2012 87 11.3.1 Some special cases Routh table is easy to form in most cases, but there could be some cases when we need to do some extra work. Case 1: The first
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Stability RouthHurwitz stability criterion Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
More informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More information100 (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 informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More informationRoot Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus  1
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position
More informationFundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc.
Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc. Electrical Engineering Department University of Indonesia 2 Steady State Error How well can
More informationLecture 4 Classical Control Overview II. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 4 Classical Control Overview II Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Stability Analysis through Transfer Function Dr. Radhakant
More informationRoot locus Analysis. P.S. Gandhi Mechanical Engineering IIT Bombay. Acknowledgements: Mr Chaitanya, SYSCON 07
Root locus Analysis P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: Mr Chaitanya, SYSCON 07 Recap R(t) + _ k p + k s d 1 s( s+ a) C(t) For the above system the closed loop transfer function
More informationI Stable, marginally stable, & unstable linear systems. I Relationship between pole locations and stability. I RouthHurwitz criterion
EE C128 / ME C134 Feedback Control Systems Lecture Chapter 6 Stability Lecture abstract Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley Topics
More informationI What is root locus. I System analysis via root locus. I How to plot root locus. Root locus (RL) I Uses the poles and zeros of the OL TF
EE C28 / ME C34 Feedback Control Systems Lecture Chapter 8 Root Locus Techniques Lecture abstract Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley
More information(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 informationSECTION 5: ROOT LOCUS ANALYSIS
SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path
More informationClass 12 Root Locus part II
Class 12 Root Locus part II Revising (from part I): Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K Comple plane jω ais 0 real ais Thus,
More informationLecture 1 Root Locus
Root Locus ELEC304Alper Erdogan 1 1 Lecture 1 Root Locus What is RootLocus? : A graphical representation of closed loop poles as a system parameter varied. Based on RootLocus graph we can choose the
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 informationSteady State Errors. Recall the closedloop transfer function of the system, is
Steady State Errors Outline What is steadystate error? Steadystate error in unity feedback systems Type Number Steadystate error in nonunity feedback systems Steadystate error due to disturbance inputs
More informationOutline. Control systems. Lecture4 Stability. V. Sankaranarayanan. V. Sankaranarayanan Control system
Outline Control systems Lecture4 Stability V. Sankaranarayanan Outline Outline 1 Outline Outline 1 2 Concept of Stability Zero State Response: The zerostate response is due to the input only; all the
More informationControl Systems Engineering ( Chapter 6. Stability ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 6. Stability ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : How to determine
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #22 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Friday, March 5, 24 More General Effects of Open Loop Poles
More informationSECTION 4: STEADY STATE ERROR
SECTION 4: STEADY STATE ERROR MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Steady State Error Introduction 3 Consider a simple unity feedback system The error is the difference between
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 informationEE 380. Linear Control Systems. Lecture 10
EE 380 Linear Control Systems Lecture 10 Professor Jeffrey Schiano Department of Electrical Engineering Lecture 10. 1 Lecture 10 Topics Stability Definitions Methods for Determining Stability Lecture 10.
More informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
More informationPlan of the Lecture. Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control
Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic:
More informationReview: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control
Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control Goal: understand the difference between openloop and closedloop (feedback)
More informationSchool of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
More informationLecture 15 Nyquist Criterion and Diagram
Lecture Notes of Control Systems I  ME 41/Analysis and Synthesis of Linear Control System  ME86 Lecture 15 Nyquist Criterion and Diagram Department of Mechanical Engineering, University Of Saskatchewan,
More informationTransient Response of a SecondOrder System
Transient Response of a SecondOrder System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a wellbehaved closedloop
More informationHomework 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12: Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain
More informationSoftware Engineering/Mechatronics 3DX4. Slides 6: Stability
Software Engineering/Mechatronics 3DX4 Slides 6: Stability Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on lecture notes by P. Taylor and M. Lawford, and Control
More information7.1 Introduction. Apago PDF Enhancer. Definition and Test Inputs. 340 Chapter 7 SteadyState Errors
340 Chapter 7 SteadyState Errors 7. Introduction In Chapter, we saw that control systems analysis and design focus on three specifications: () transient response, (2) stability, and (3) steadystate errors,
More informationMethods for analysis and control of. Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.lag.ensieg.inpg.fr/sename Lead Lag 17th March
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationRouthHurwitz Lecture RouthHurwitz Stability test
ECE 35 RouthHurwitz Leture RouthHurwitz Staility test AStolp /3/6, //9, /6/ Denominator of transfer funtion or signal: s n s n s n 3 s n 3 a s a Usually of the Closedloop transfer funtion denominator
More informationVALLIAMMAI 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 informationFeedback Control Systems (FCS)
Feedback Control Sytem (FCS) Lecture1920 RouthHerwitz Stability Criterion Dr. Imtiaz Huain email: imtiaz.huain@faculty.muet.edu.pk URL :http://imtiazhuainkalwar.weebly.com/ Stability of Higher Order
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition Readings: Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Jacopo Tani Institute for Dynamic Systems and Control
More 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 informationLast week: analysis of pinionrack w velocity feedback
Last week: analysis of pinionrack w velocity feedback Calculation of the steady state error Transfer function: V (s) V ref (s) = 0.362K s +2+0.362K Step input: V ref (s) = s Output: V (s) = s 0.362K s
More informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
More information12.7 Steady State Error
Lecture Notes on Control Systems/D. Ghose/01 106 1.7 Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there
More informationSTABILITY OF CLOSEDLOOP CONTOL SYSTEMS
CHBE320 LECTURE X STABILITY OF CLOSEDLOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 101 Road Map of the Lecture X Stability of closedloop control
More informationLecture 10: Proportional, Integral and Derivative Actions
MCE441: Intr. Linear Control Systems Lecture 10: Proportional, Integral and Derivative Actions Stability Concepts BIBO Stability and The RouthHurwitz Criterion Dorf, Sections 6.1, 6.2, 7.6 Cleveland State
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #14 Wednesday, February 5, 2003 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Chapter 7 Synthesis of SISO Controllers
More informationSystems 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 informationChapter 7 : Root Locus Technique
Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci
More informationSystems 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 informationChemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University
Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this
More informationControl Systems. Frequency Method Nyquist Analysis.
Frequency Method Nyquist Analysis chibum@seoultech.ac.kr Outline Polar plots Nyquist plots Factors of polar plots PolarNyquist Plots Polar plot: he locus of the magnitude of ω vs. the phase of ω on polar
More informationProblems XO («) splane. splane *~8 X 5. id) X splane. splane. * Xtg) FIGURE P8.1. jplane. JO) k JO)
Problems 1. For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4] *~8 XO
More informationChapter 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 Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
More informationAnalysis of Stability &
INC 34 Feedback Control Sytem Analyi of Stability & SteadyState Error S Wonga arawan.won@kmutt.ac.th Summary from previou cla Firtorder & econd order ytem repone τ ωn ζω ω n n.8.6.4. ζ ζ. ζ.5 ζ ζ.5 ct.8.6.4...4.6.8..4.6.8
More informationSECTION 7: STEADYSTATE ERROR. ESE 499 Feedback Control Systems
SECTION 7: STEADYSTATE ERROR ESE 499 Feedback Control Systems 2 Introduction SteadyState Error Introduction 3 Consider a simple unityfeedback system The error is the difference between the reference
More informationME 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 informationTime Response of Systems
Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) splane Time response p =0 s p =0,p 2 =0 s 2 t p =
More information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationModule 07 Control Systems Design & Analysis via RootLocus Method
Module 07 Control Systems Design & Analysis via RootLocus Method Ahmad F. Taha EE 3413: Analysis and Desgin of Control Systems Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ taha March
More informationMethods for analysis and control of dynamical systems Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.gipsalab.fr/ o.sename 5th February 2015 Outline
More informationDidier HENRION henrion
POLYNOMIAL METHODS FOR ROBUST CONTROL PART I.1 ROBUST STABILITY ANALYSIS: SINGLE PARAMETER UNCERTAINTY Didier HENRION www.laas.fr/ henrion henrion@laas.fr Pont Neuf over river Garonne in Toulouse June
More informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
More informationRemember that : Definition :
Stability This lecture we will concentrate on How to determine the stability of a system represented as a transfer function How to determine the stability of a system represented in statespace How to
More informationEECE 460 : Control System Design
EECE 460 : Control System Design SISO Pole Placement Guy A. Dumont UBC EECE January 2011 Guy A. Dumont (UBC EECE) EECE 460: Pole Placement January 2011 1 / 29 Contents 1 Preview 2 Polynomial Pole Placement
More informationControl Systems. Root Locus & Pole Assignment. L. Lanari
Control Systems Root Locus & Pole Assignment L. Lanari Outline rootlocus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS  Root
More informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Boundedinput boundedoutput (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationBangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory
Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system
More informationS I X SOLUTIONS TO CASE STUDIES CHALLENGES. Antenna Control: Stability Design via Gain K s s s+76.39K. T(s) =
S I X Stability SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Stability Design via Gain From the antenna control challenge of Chapter 5, Make a Routh table: 76.39K s 3 +151.32s 2 +198s+76.39K s
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationSystems 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 informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE 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 informationECE 388 Automatic Control
Controllability and State Feedback Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage:
More informationCHAPTER # 9 ROOT LOCUS ANALYSES
F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closedloop system is closely related to the location of the closedloop poles. If the system
More informationLecture 7:Time Response PoleZero 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 FirstOrder Specs: Step : Pole Real inputs contain
More informationControl System (ECE411) Lectures 13 & 14
Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 SteadyState Error Analysis Remark: For a unity feedback system
More informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More informationDigital Control: Summary # 7
Digital Control: Summary # 7 Proportional, integral and derivative control where K i is controller parameter (gain). It defines the ratio of the control change to the control error. Note that e(k) 0 u(k)
More informationECEN 326 Electronic Circuits
ECEN 326 Electronic Circuits Stability Dr. Aydın İlker Karşılayan Texas A&M University Department of Electrical and Computer Engineering Ideal Configuration V i Σ V ε a(s) V o V fb f a(s) = V o V ε (s)
More informationSECTION 8: ROOTLOCUS ANALYSIS. ESE 499 Feedback Control Systems
SECTION 8: ROOTLOCUS ANALYSIS ESE 499 Feedback Control Systems 2 Introduction Introduction 3 Consider a general feedback system: Closedloop transfer function is KKKK ss TT ss = 1 + KKKK ss HH ss GG ss
More informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More informationCourse roadmap. ME451: Control Systems. Example of Laplace transform. Lecture 2 Laplace transform. Laplace transform
ME45: Control Systems Lecture 2 Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Transfer function Models for systems electrical mechanical electromechanical Block
More information