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


 Jack Carr
 1 years ago
 Views:
Transcription
1
2 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 No Friday Tutorial 10 Apr BREAK 6 17 Apr Root Locus 7 24 Apr Root Locus 2 No Wed/Thu Tut 8 1 May Bode Plots Assign 2 due 9 8 May Bode Plots May State Space Design Assign 3 Due May State Space Design May Advanced Control/Review June Review 2 Assign 4 Due Amme 3500 : Review Slide 2
3 4 Assignments (40%) Assignment 1 : Due Week 4 (5%) Assignment 2 : Due Week 6 (10%) Assignment 3 : Due Week 10 (10%) Assignment 4 : Due Week 13 (15%) Final Exam (60%)* * Note you are expected to pass the exam to pass this course Slide 3
4 To introduce the methods used for the analysis and design of feedback controllers for linear time invariant (LTI) systems Slide 4
5 System 1 (e.g. Controller) System 2 (e.g. Process) System 1 affects system 2, which affects system 1, which affects system 2. Slide 5
6 Vehicle Control and Design (land, sea, air, space) understanding and controlling how the system responds to external disturbances Biomedical (cardiac system, dialysis machine) design and control of systems that interact with the human body. Manufacturing Processes controlled conditions for highperformance materials, pharmaceuticals, microsystems. Biological feedback systems that regulate pressures, concentrations, balance, etc Slide 6
7 Design the dynamics Sluggish systems become quick to respond Unstable systems become stable and predictable Robustness Reject disturbances acting on the system Same response with large variations in the system Slide 7
8 Instability: any feedback loop allows for the possibility of instability. The question of stability has long been central in control theory Measurement noise gets feed back into the actual system response. Slide 8
9 Many control systems can be characterised by these components Plant Disturbance Reference r(t) +  Error e(t) Control Control Signal u(t) Actuator Process Output y(t) Feedback Sensor Sensor Noise Slide 9
10 A block diagram is made up of signals, systems, summing junctions and pickoff points Slide 10
11 A system model is one or more equations that describe the relationship between the system variables often the input(s) and output(s) of the system For physical systems, these equations are derived from study of the physical properties of the system such as mechanics, fluids, electrical, thermodynamics, etc. Slide 11
12 Forcevelocity, forcedisplacement, and impedance translational relationships for springs, viscous dampers, and mass Slide 12
13 Torqueangular velocity, torqueangular displacement, and impedance rotational relationships for springs, viscous dampers, and inertia Slide 13
14 Voltagecurrent, voltagecharge, and impedance relationships for capacitors, resistors, and inductors Slide 14
15 Starting with an impulse response, h(t), and an input, u(t), find y(t) u(t), h(t) convolution y(t) L L 1 U(s), H(s) Multiplication, algebraic manipulation Y(s) Slide 15
16 We normally use tables of Laplace transforms rather than solving the Laplace equations directly This greatly simplifies the transformation process Slide 16
17 The Laplace Transform is a linear transformation between functions in the t domain and s domain Slide 17
18 Second order systems are quite common and are generally written in the following standard form Many systems of interest are of higher order Slide 18
19 We can relate the natural frequency and damping ratio to the s plane θ Slide 19
20 Damping ratio determines the characteristics of the system response Slide 20
21 Rise time, settling time and peak time yield information about the speed of response of the transient response This can help a designer determine if the speed and nature of the response is appropriate Slide 21
22 For a second order system with no finite zeros, the transient response parameters are approximated by Rise time : Overshoot : Settling Time (2%) : Slide 22
23 Im(s) Find allowable region in the splane for the poles of a transfer function to meet the requirements sin 1 ζ ω n t r 0.6 sec Re(s) M p 10% σ t s 3 sec Slide 23
24 This table shows the relationship between input, system type, error constants and steadystate error Slide 24
25 We can easily find the root locations for a second order system What about for a general, possibly higher order, control system? Poles exist when the characteristic equation (denominator) is zero Slide 25
26 The location of the roots, and hence the nature of the system performance, are a function of the system gain K In order to solve for this system performance, we must factor the denominator for specific values of K We define the root locus as the path of the closedloop poles as the system parameter varies from 0 to Slide 26
27 The CL roots move from the OL towards the zeros Additional poles move towards infinity along well defined asymptotes Slide 27
28 We saw that the steady state response for an LTI system excited by a sinusoid with unit amplitude and frequency ω 0 will also exhibit a sinusoidal output of frequency ω 0 with magnitude M(ω 0 ) and a phase φ (ω 0 ) where Notice that both the magnitude and the phase of the response on dependent on the frequency of the input ω 0 Slide 28
29 We often plot the magnitude and phase of the system response as a function of the input frequency The magnitude is normally plotted in db=20logm(ω) vs. log(ω) The phase is plotted in degrees vs. log(ω) The resulting graph is called the Bode plot Slide 29
30 Normalized and scaled Bode plots for a. G(s) = s; b. G(s) = 1/s; c. G(s) = (s + a); d. G(s) = 1/(s + a) Slide 30
31 The complete threeterm controller is described by R(s) E(s) C(s) +  K p +K i /s+k d s G(s) Slide 31
32 The ideal derivative compensator effectively adds a pure differentiator to the forward path of the control system This is effectively equivalent to an additional zero As you should by now be aware, the location of the open loop poles and zeros affects the root locus and hence the transient response of the closed loop system Slide 32
33 Consider a simple second order system whose root locus looks like this (roots 1, 2) Adding a zero to this system drastically changes the shape of the root locus The position of the zero will also change the shape and hence the nature of the transient response Zero at 3 Zero at 5 Slide 33
34 The Bode plot for a PD controller looks like this The stabilizing effect is seen by the increase in phase at frequencies above the break frequency However, the magnitude grows with increasing frequency and will tend to amplify high frequency noise Slide 34
35 For compensation using passive components, a pole and zero will result If the pole position is selected such that it is to the left of the zero, the resulting compensator will behave like an ideal derivative compensator The name Lead Compensation reflects the fact that this compensator imparts a phase lead Slide 35
36 First find a point in the splane that we d like to have on the root locus Place the lead pole at 20 and solve for the position of the zero x θ 1 =120 θ 2 =112.4 θ 3 =20.2 τηερεφορε θ c =72.6 Slide 36
37 The Bode plot for a Lead compensator looks like this The frequency of the phase increase can be designed to meet a particular phase margin requirement The high frequency magnitude is now limited Slide 37
38 If we rewrite the transfer function for the integral compensator we find This is simply a pole at the origin and a zero at some other position to be selected based on our design requirements normally close to the origin to minimize the angular contribution of the compensator Slide 38
39 Slide 39
40 The Bode plot for a PI controller looks like this The break frequency is usually located at a frequency substantially lower than the crossover frequency to minimize the effect on the phase margin Slide 40
41 As with the lead compensation, using passive components results in a pole and zero If the pole position is selected such that it is to the right of the zero near the origin, the resulting compensator will behave like an ideal integral compensator although it will not increase the system type The name Lag Compensation reflects the fact that this compensator imparts a phase lag Slide 41
42 Slide 42
43 The Bode plot for a Lag compensator looks like this This compensator effectively raises the magnitude for low frequencies The effect of the phase lag can be minimized by careful selection of the centre frequency Slide 43
44 System equations are described in matrix form. Most fundamental are the state variables: A set of parameters that completely describes the current state of the system. For example position and velocity of a moving body Using measurements of the outputs, the system state can be computed  observers Using the system state, control strategies can be devised to achieve desired performance. Pole placement Optimal controllers Attractive for multiinput multioutput (MIMO) systems Slide 44
45 Using the state space approach, we represent a system by a set of n firstorder differential equations: The output of the system is expressed as: x  state vector y  output vector u  input vector A  state matrix B  input matrix C  output matrix D  feedthrough or feedforward matrix (often zero) Slide 45
46 We can draw a block diagram describing the general State Space Model Slide 46
47 We can represent a general state space system as a Block Diagram. If we feedback the state variables, we end up with n controllable parameters. State feedback with the control input u=kx +r. * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 47
48 We can then control the pole locations by finding appropriate values for K This allows us to select the position of all the closed loop system roots during our design. There are a number of methods for selecting and designing controllers in state space, including pole placement and optimal control methods via the Linear Quadratic Regulator algorithm. Slide 48
49 Setting u=kx+r yields Rearranging the state equation and taking LT yields Select values of K so that the eigenvalues (root locations) of (ABK) are at a particular location Slide 49
50 Controllability and Observability are fundamental concepts. For output feedback controllers, the separation principle tells us that we can design an observer and a controller separately and the combined controller is stabilizing. Slide 50
51 You should be familiar with the concepts reviewed in this lecture Modelling of dynamic systems Specification of second order systems Root Locus Bode Plots Design and properties of PID (and variants), Lead and Lag controllers State Space Modelling and Design Slide 51
52 You will not be required to find roots of polynomials higher than second order. You will be provided with a selected set of equations you may require for solving the problems In order to prepare I would suggest that you Review the assignment questions, making sure you understand the material covered this semester Look over previous years exams Slide 52
53 In order to understand system performance, we must be able to model these systems The study of control provides us with a process for analysing, understanding and deisgning for the behaviour of a system Slide 53
Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus
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 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 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 informationSchool of Engineering Faculty of Built Environment, Engineering, Technology & Design
Module Name and Code : ENG60803 Real Time Instrumentation Semester and Year : Semester 5/6, Year 3 Lecture Number/ Week : Lecture 3, Week 3 Learning Outcome (s) : LO5 Module Coordinator/Tutor : Dr. Phang
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 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 informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
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 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 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 informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More 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 informationFATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY
FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY Senkottai Village, Madurai Sivagangai Main Road, Madurai  625 020. An ISO 9001:2008 Certified Institution DEPARTMENT OF ELECTRONICS AND COMMUNICATION
More informationIndex. Index. More information. in this web service Cambridge University Press
Atype elements, 4 7, 18, 31, 168, 198, 202, 219, 220, 222, 225 Atype variables. See Across variable ac current, 172, 251 ac induction motor, 251 Acceleration rotational, 30 translational, 16 Accumulator,
More informationDr. Ian R. Manchester
Dr 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
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationPrüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013 Problem Set #4 Posted: Thursday, Mar. 7, 13 Due: Thursday, Mar. 14, 13 1. Sketch the Root
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationIntroduction to Root Locus. What is root locus?
Introduction to Root Locus What is root locus? A graphical representation of the closed loop poles as a system parameter (Gain K) is varied Method of analysis and design for stability and transient response
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
More informationKINGS 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 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 informationCourse Outline. Closed Loop Stability. Stability. Amme 3500 : System Dynamics & Control. Nyquist Stability. Dr. Dunant Halim
Amme 3 : System Dynamics & Control Nyquist Stability Dr. Dunant Halim Course Outline Week Date Content Assignment Notes 1 5 Mar Introduction 2 12 Mar Frequency Domain Modelling 3 19 Mar System Response
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 informationVALLIAMMAI 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 informationControl Systems. EC / EE / IN. For
Control Systems For EC / EE / IN By www.thegateacademy.com Syllabus Syllabus for Control Systems Basic Control System Components; Block Diagrammatic Description, Reduction of Block Diagrams. Open Loop
More informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
More informationINTRODUCTION 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 lineartimeinvariant
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 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 informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
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 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 informationLecture 12. Upcoming labs: Final Exam on 12/21/2015 (Monday)10:3012:30
289 Upcoming labs: Lecture 12 Lab 20: Internal model control (finish up) Lab 22: Force or Torque control experiments [Integrative] (23 sessions) Final Exam on 12/21/2015 (Monday)10:3012:30 Today: Recap
More informationEE3CL4: 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 6: Generalized and Controller Design Overview In this Lecture, you will learn: Generalized? What about changing OTHER parameters
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servooriented
More informationINSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad ELECTRICAL AND ELECTRONICS ENGINEERING TUTORIAL QUESTION BANK
Course Name Course Code Class Branch INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad 500 043 ELECTRICAL AND ELECTRONICS ENGINEERING TUTORIAL QUESTION BAN : CONTROL SYSTEMS : A50 : III B. Tech
More informationCHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System
CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages
More informationAutomatic Control (TSRT15): Lecture 7
Automatic Control (TSRT15): Lecture 7 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13282226 Office: Bhouse extrance 2527 Outline 2 Feedforward
More informationI Laplace transform. I Transfer function. I Conversion between systems in time, frequencydomain, and transfer
EE C128 / ME C134 Feedback Control Systems Lecture Chapter 2 Modeling in the Frequency Domain Alexandre Bayen Department of Electrical Engineering & Computer Science University of California Berkeley Lecture
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 informationContents. PART I METHODS AND CONCEPTS 2. Transfer Function Approach Frequency Domain Representations... 42
Contents Preface.............................................. xiii 1. Introduction......................................... 1 1.1 Continuous and Discrete Control Systems................. 4 1.2 OpenLoop
More informationChapter 9: Controller design
Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer function from disturbances to the output 9.2.2. Feedback
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 informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad  500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III
More informationControl System. Contents
Contents Chapter Topic Page Chapter Chapter Chapter3 Chapter4 Introduction Transfer Function, Block Diagrams and Signal Flow Graphs Mathematical Modeling Control System 35 Time Response Analysis of
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 informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationAcceleration Feedback
Acceleration Feedback Mechanical Engineer Modeling & Simulation Electro Mechanics Electrical Electronics Engineer Sensors Actuators Computer Systems Engineer Embedded Control Controls Engineer Mechatronic
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 informationNote. Design via State Space
Note Design via State Space Reference: Norman S. Nise, Sections 3.5, 3.6, 7.8, 12.1, 12.2, and 12.8 of Control Systems Engineering, 7 th Edition, John Wiley & Sons, INC., 2014 Department of Mechanical
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
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 informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationExam. 135 minutes + 15 minutes reading time
Exam January 23, 27 Control Systems I (559L) 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 informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
More information1 An Overview and Brief History of Feedback Control 1. 2 Dynamic Models 23. Contents. Preface. xiii
Contents 1 An Overview and Brief History of Feedback Control 1 A Perspective on Feedback Control 1 Chapter Overview 2 1.1 A Simple Feedback System 3 1.2 A First Analysis of Feedback 6 1.3 Feedback System
More informationEC CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING
EC 2255  CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING 1. What is meant by a system? It is an arrangement of physical components related in such a manner as to form an entire unit. 2. List the two types
More informationIntroduction to Controls
EE 474 Review Exam 1 Name Answer each of the questions. Show your work. Note were essaytype answers are requested. Answer with complete sentences. Incomplete sentences will count heavily against the grade.
More informationState Regulator. Advanced Control. design of controllers using pole placement and LQ design rules
Advanced Control State Regulator Scope design of controllers using pole placement and LQ design rules Keywords pole placement, optimal control, LQ regulator, weighting matrixes Prerequisites Contact state
More informationChapter 12. Feedback Control Characteristics of Feedback Systems
Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an openloop system (a system without feedbac) and a closedloop
More informationControl Systems I. Lecture 2: Modeling. Suggested Readings: Åström & Murray Ch. 23, Guzzella Ch Emilio Frazzoli
Control Systems I Lecture 2: Modeling Suggested Readings: Åström & Murray Ch. 23, Guzzella Ch. 23 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich September 29, 2017 E. Frazzoli
More informationToday s goals So far Today 2.004
Today s goals So far Feedback as a means for specifying the dynamic response of a system Root Locus: from the openloop poles/zeros to the closedloop poles Moving the closedloop poles around Today Proportional
More informationEE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation
EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginallystable
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 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 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 informationagree w/input bond => + sign disagree w/input bond =>  sign
1 ME 344 REVIEW FOR FINAL EXAM LOCATION: CPE 2.204 M. D. BRYANT DATE: Wednesday, May 7, 2008 9noon Finals week office hours: May 6, 47 pm Permitted at final exam: 1 sheet of formulas & calculator I.
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 informationVALLIAMMAI ENGINEERING COLLEGE
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK V SEMESTER IC650 CONTROL SYSTEMS Regulation 203 Academic Year 207 8 Prepared
More informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
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 informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 09Dec13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
More informationLecture 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 informationFeedback Control of Linear SISO systems. Process Dynamics and Control
Feedback Control of Linear SISO systems Process Dynamics and Control 1 OpenLoop Process The study of dynamics was limited to openloop systems Observe process behavior as a result of specific input signals
More informationEE 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 informationEE 422G  Signals and Systems Laboratory
EE 4G  Signals and Systems Laboratory Lab 9 PID Control Kevin D. Donohue Department of Electrical and Computer Engineering University of Kentucky Lexington, KY 40506 April, 04 Objectives: Identify the
More informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
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 informationLecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore A Fundamental Problem in Control Systems Poles of open
More informationVideo 5.1 Vijay Kumar and Ani Hsieh
Video 5.1 Vijay Kumar and Ani Hsieh Robo3x1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior
More informationDesign of a Lead Compensator
Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc  Funded by MHRD The Lecture Contains Standard Forms of
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 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 informationRoot Locus Design Example #3
Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll
More informationStep input, ramp input, parabolic input and impulse input signals. 2. What is the initial slope of a step response of a first order system?
IC6501 CONTROL SYSTEM UNITII TIME RESPONSE PARTA 1. What are the standard test signals employed for time domain studies?(or) List the standard test signals used in analysis of control systems? (April
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop 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 closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationUnit 8: Part 2: PD, PID, and Feedback Compensation
Ideal Derivative Compensation (PD) Lead Compensation PID Controller Design Feedback Compensation Physical Realization of Compensation Unit 8: Part 2: PD, PID, and Feedback Compensation Engineering 5821:
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 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 informationLaboratory Exercise 1 DC servo
Laboratory Exercise DC servo PerOlof Källén ø 0,8 POWER SAT. OVL.RESET POS.RESET Moment Reference ø 0,5 ø 0,5 ø 0,5 ø 0,65 ø 0,65 Int ø 0,8 ø 0,8 Σ k Js + d ø 0,8 s ø 0 8 Off Off ø 0,8 Ext. Int. + x0,
More information