Dr Ian R. Manchester


 Alexandrina Wood
 3 years ago
 Views:
Transcription
1
2 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 2 Due 8 Bode Plots 9 Bode Plots 2 10 State Space Modeling Assign 3 Due 11 State Space Design Techniques 12 Advanced Control Topics 13 Review Assign 4 Due Dr. Ian R. Manchester Amme 3500 : Introduction Slide 2
3 Review mathematical justification for feedback over open loop control Introduce the PID controller Performance specification: steadystate error Introduction to parameter sensitivity aka robustness Slide 3
4 r(t) Desired Value +  error Controller Control Signal Plant Output Feedback Signal Error: e(t) = r(t) y(t) (Assuming perfect measurement) Slide 4
5 Perfect feedforward controller: G n Output and error: Only zero when: (Perfect Knowledge) (No load or disturbance) Slide 5
6 reference r disturbance d output controller plant A sensor unity feedback y = GK 1+ KG r G 1+ GK d Reference to Output Transfer Function Disturbance to Output Transfer Function Slide 6
7 Equivalent Open Loop Block Diagram r d y = GK 1+ KG r G 1+ GK d Reference to Output Transfer Function Disturbance to Output Transfer Function Slide 7
8 Simplest possible feedback control useful for thermostats, similarly simple problems Often results in oscillations Rapid changes in control input are often infeasible Nonlinear: can be very difficult to analyse Slide 8
9 A very large number (at least 95%) of feedback controllers used in practice have the following simple form: u(t) = K P e + K I t e(τ)dτ + K 0 D e This is called a PID Proportional (P) term feeds back on the current error Integral (I) feedback can eliminate steady state error derivative (D) term can improve dynamic response Slide 9
10 Suppose we just use the P term, i.e. proportional control R(s) E(s) C(s) +  K G(s) We have seen how this form of feedback is able to minimize the effect of disturbances by increasing K This may adversely affect other performance, in particular overshoot Slide 10
11 k = 5 N m s/rad b = 6 N m s/rad * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Let s compute an example of a torsional massspringdamper system with proportional control J = 1 N m, k = 5 N m/rad and b = 6 N m s/rad Slide 11
12 Recall the differential equation The transfer function is J θ + b θ + kθ = τ The closed loop response is R(s) E(s) C(s) +  K G(s) Slide 12
13 K=5 K=100 K=10 K=1000 Slide 13
14 Proportional (P) control acts like a spring pulling the output towards the reference. A very stiff spring (large gain K) can result in highly oscillatory response. Why not add damping? This is PD control R(s) E(s) U(s) C(s) K p +K d s G(s) +  u(t) = K P e + K D e Slide 14
15 D term adds damping and differentiates the reference Slide 15
16 Same torsional setup as before P control: K=100 PD control: K=100+20s Note the anticipatory kick at the beginning, and the damping of oscillations Slide 16
17 We have looked at transient response for second order systems This can be defined by Rise time, T r Settling time, T s Peak time, T p Percentage overshoot. These can all be specified by PD control c final =steady state response * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 17
18 Why use integral control? (The I part in PID) Another important characteristic of systems in general is their steady state performance input Steady state error Steadystate error is the difference between the reference input and output for a particular input as t time output Slide 18
19 Boeing and BAE systems Slide 19
20 Real systems (with nonlinearities): friction/ stiction, backlash, deadzones, etc. Sensor does not see error Actuator produces no output for command Linear systems: Caused by intrinsic dynamics or disturbance inputs SS error can be examined analytically We can eliminate it through integral action Slide 20
21 x d x m b F(s) ms 2 +bs X(s) f x d x m b X d (s) +  E(s) kc F(s) ms 2 +bs X(s) k c Slide 21
22 x d x k m b F(s) ms 2 +bs+k X(s) f x d x k m b X d (s) +  E(s) k c F(s) ms 2 +bs+k X(s) k c Slide 22
23 m x x d g m x x d b f b k c G(s) = mg/s G(s) F(s) ms 2 +bs X(s) X d (s) +  k c ms 2 +bs X(s) Slide 23
24 Consider a unity feedback system. Transfer function T(s) R(s) E(s) C(s) +  G(s) What if we rearrange this to find E(s)? R(s) T(s) C(s)  + E(s) Slide 24
25 We can now find E(s) in terms of the input R (s) and the closed loop transfer function T(s) Or Although we can use E(s) to find e(t) and compute the error in time, we are often interested in the steady state error Slide 25
26 Start with LT for the derivative: Take the limit as s 0 Slide 26
27 The Final Value Theorem can be used to find the steady state value without taking inverse Laplace transforms This only applies if the system is stable Slide 27
28 * N.S. Nise (2004) Control Systems Engineering Wiley & Sons We ll now look at the steady state errors to some common input functions (polynomials) Slide 28
29 Position Velocity Acceleration Slide 29
30 For a step input (R(s)=1/s), we find: If lim s 0 G(s) is finite, we define to be the position error constant. If there is a pole at the origin then steadystate error is zero sr(s) 1+ G(s) s(1/s) 1+ G(s) e( ) = lim s 0 = lim s 0 = 1 1+ lim s 0 G(s) Slide 30
31 For a general G(s) The denominator must be zero for zero steady state error. Therefore n 1 Implies at least one pole must be at the origin In order to have zero steady state error, G(s) must approach as s approaches 0 Slide 31
32 For a system with no poles at the origin there will be some constant error to a step input For one or more poles at the origin, the error will be integrated away and will reach zero in the steadystate Input e( ) Slide 32
33 For a ramp input (R(s)=1/s 2 ), we find We can define the velocity error constant, K v, such that Slide 33
34 Once again, in order to have zero steady state error, sg(s) must approach as s approaches 0 Implies at least two poles must be at the origin Slide 34
35 Not surprisingly, for a parabolic input (R(s)=1/s 3 ), we find We can define the acceleration error constant, K a, such that Slide 35
36 Only one is useful for a given system: K step = lim s 0 G(s) K v = lim s 0 sg(s) K a = lim s 0 s 2 G(s) (book calls this K p, don t confuse with proportional control gain!!) They can be determined from the OL transfer function using the formulae presented Only for unity feedback, stable systems. Slide 36
37 For unity negative feedback, the steadystate errors are dependent on the number of pure integrations in the forward path We define system type to be the number of poles n in the forward path (either from controller or plant) A system with n = 0 is Type 0. If n = 1 or n = 2, the corresponding system type is Type 1 or Type 2 respectively * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 37
38 This table shows the relationship between input, system type, error constants and steadystate error. Note: only for unity feedback cases! * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 38
39 This just scratches the surface: a very important result in control theory is the internal model principle: For any feedback system to give zero error for a particular class of reference or disturbance inputs, the system itself must be capable of internally reproducing all signals in this class. This can be interpreted like so: it is necessary for the system to internally estimate the state of the reference or disturbance If you do advanced control, you will learn about state observers and Kalman filters, which explicitly perform this task Slide 39
40 * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Find the steady state error for inputs of r(t)=5u(t), r(t) =5tu(t) and r(t)=5t 2 u(t), where u(t) refers to step input First verify that the system is stable Yes System type? Type 1 Slide 40
41 For step r(t)=5u(t) For ramp r(t)=5tu(t) For parabola r(t)=5t 2 u(t) Slide 41
42 Just as damping ratio ζ, settling time T s, peak time T p, and percent overshoot are used as specifications for transient response, The position constant, K step, velocity constant, K v, and acceleration constant, K a, can be used as specifications for steadystate error Slide 42
43 u(t) = K P e + K I t e(τ)dτ + K 0 D e Intuitively: if the controller continues to see an error signal over a long time, then the integral term builds up, and increases control signal Slowly it builds up to the level required to completely kill the error Slide 43
44 For a PD controller, if the error and its derivative are both zero, then u = K P e + K D e = 0 So, a zerosteadystateerror response to a step must result in zero control input. This is not possible if there is some force (e.g. a spring, or gravity, or wind) pulling the system away from the reference Slide 44
45 Given this control system, find K so that there is 10% error in the steady state response to a ramp input System Stable? Yes System Type? Type 1 * N.S. Nise (2004) Control Systems Engineering Wiley & Sons Slide 45
46 For LTI systems we have assumed that the parameters of the system are perfectly known However, there are instances in which the parameters of a system may change due to, for example: changes in ambient temperature, Altitude wear Manufacturing tolerances It is important to understand the degree to which changes in system parameters affect system performance. This is called sensitivity analysis Slide 46
47 Sensitivity is defined as the normalised ratio of the change in the transfer function T to the change in the plant G: S(s) = T(s) /T(s) G(s) /G(s) = T(s) G(s) Note that S is itself a transfer function: G(s) T(s) The sensitivity to plant changes is dynamic the feedback system may be more sensitive to plant changes when subjected to some inputs than others Slide 47
48 T(s) = Y(s) R(s) = G(s) G n (s) S = T(s) G(s) G(s) T(s) T(s) G(s) = 1 G n (s) S = 1 G n 1 G n =1 Slide 48
49 r d T(s) = Y(s) R(s) = T(s) G(s) = = KG 1+ KG K 1+ KG K 2 G 1+ KG K ( 1+ KG) 2 ( ) 2 S = = T /T G /G = 1 1+ GK With very high gains: K G(1+ KG) (1+ KG) 2 GK lim S = 0 K Slide 49
50 We can see a pattern in many of the sensitivity relations of feedback systems Y(s) R(s) = G(s)K(s) 1+ G(s)K(s) Response to reference command: (should = 1) Y(s) N(s) = G(s)K(s) 1+ G(s)K(s) Response to measurement noise: (should = 0) Y(s) D(s) = 1 1+ G(s)K(s) Response to disturbance at output: (should = 0) dy(s) /Y(s) dg(s) /G(s) = 1 1+ G(s)K(s) Sensitivity to model uncertainty: (should = 0) Note also that 1 1+ G(s)K(s) + G(s)K(s) 1+ G(s)K(s) =1 Slide 50
51 Since they show up so much, they have been given names: The Sensitivity Function S(s) = The Complementary Sensitivity Function T(s) = 1 1+ G(s)K(s) G(s)K(s) 1+ G(s)K(s) Complementary because S+T=1 for all s Many advanced control techniques are based on shaping these transfer functions Slide 51
52 PID: u(t) = K P e + K I t 0 e(τ)dτ + K D e U(s) E(s) = K(s) = K P + K I s + K D s Error e(t) Slope of e(t) ~ derivative control term Area under curve ~ integral control term e(t) ~ proportional control Time t Slide 52
53 Slide 53
54 Same torsional setup as before P control: K=100 PD control: K(s)=100+10s PID control: K(s)=80+60/s+6s Note the zero steadystate error for PID Slide 54
55 PID with Different spring constants: k =5 (nominal), 7.5, and 2.5 Nm/rad Slide 55
56 r(t) G d G n d(t) Slide 56
57 Nise Sections (Steady state errors) Franklin & Powell Section Åström & Murray (free online) Chapter 10 Slide 57
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 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 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 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 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 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
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 informationAnalysis and Design of Control Systems in the Time Domain
Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
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 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 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 informationPart IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 5. An Introduction to Feedback Control Systems
Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL Glenn Vinnicombe HANDOUT 5 An Introduction to Feedback Control Systems ē(s) ȳ(s) Σ K(s) G(s) z(s) H(s) z(s) = H(s)G(s)K(s) L(s) ē(s)=
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 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 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 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 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 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 Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance
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 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 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 information06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance.
Chapter 06 Feedback 06 Feedback Control System Characteristics The role of error signals to characterize feedback control system performance. Lesson of the Course Fondamenti di Controlli Automatici of
More 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 informationBASIC PROPERTIES OF FEEDBACK
ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)
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 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 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 informationController Design using Root Locus
Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
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 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 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 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 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 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 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 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 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 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 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 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 informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More 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 informationRoot Locus Design Example #4
Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is
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 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 informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
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 informationLecture 1: Feedback Control Loop
Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure
More information6.1 Sketch the zdomain root locus and find the critical gain for the following systems K., the closedloop characteristic equation is K + z 0.
6. Sketch the zdomain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More 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 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 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 informationLecture 25: Tue Nov 27, 2018
Lecture 25: Tue Nov 27, 2018 Reminder: Lab 3 moved to Tuesday Dec 4 Lecture: review timedomain characteristics of 2ndorder systems intro to control: feedback openloop vs closedloop control intro to
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 informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More information( ) ( = ) = ( ) ( ) ( )
( ) Vρ C st s T t 0 wc Ti s T s Q s (8) K T ( s) Q ( s) + Ti ( s) (0) τs+ τs+ V ρ K and τ wc w T (s)g (s)q (s) + G (s)t(s) i G and G are transfer functions and independent of the inputs, Q and T i. Note
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 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 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 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 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 informationCourse Background. Loosely speaking, control is the process of getting something to do what you want it to do (or not do, as the case may be).
ECE4520/5520: Multivariable Control Systems I. 1 1 Course Background 1.1: From time to frequency domain Loosely speaking, control is the process of getting something to do what you want it to do (or not
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
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 informationChapter 2 SDOF Vibration Control 2.1 Transfer Function
Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:
More informationUltimate State. MEM 355 Performance Enhancement of Dynamical Systems
Ultimate State MEM 355 Performance Enhancement of Dnamical Sstems Harr G. Kwatn Department of Mechanical Engineering & Mechanics Drexel Universit Outline Design Criteria two step process Ultimate state
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 informationAn Introduction to Control Systems
An Introduction to Control Systems Signals and Systems: 3C1 Control Systems Handout 1 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie November 21, 2012 Recall the concept of a
More 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 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 informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
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 informationC(s) R(s) 1 C(s) C(s) C(s) = s  T. Ts + 1 = 1 s  1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain
analyses of the step response, ramp response, and impulse response of the secondorder systems are presented. Section 5 4 discusses the transientresponse analysis of higherorder systems. Section 5 5 gives
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 informationProblem Value Score Total 100/105
RULES This is a closed book, closed notes test. You are, however, allowed one piece of paper (front side only) for notes and definitions, but no sample problems. The top half is the same as from the first
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 informationEEL2216 Control Theory CT1: PID Controller Design
EEL6 Control Theory CT: PID Controller Design. Objectives (i) To design proportionalintegralderivative (PID) controller for closed loop control. (ii) To evaluate the performance of different controllers
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More informationProcess Control J.P. CORRIOU. Reaction and Process Engineering Laboratory University of LorraineCNRS, Nancy (France) Zhejiang University 2016
Process Control J.P. CORRIOU Reaction and Process Engineering Laboratory University of LorraineCNRS, Nancy (France) Zhejiang University 206 J.P. Corriou (LRGP) Process Control Zhejiang University 206
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More information10ES43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A
10ES43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative
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 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 informationIntroduction to Control (034040) lecture no. 2
Introduction to Control (034040) lecture no. 2 Leonid Mirkin Faculty of Mechanical Engineering Technion IIT Setup: Abstract control problem to begin with y P(s) u where P is a plant u is a control signal
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
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 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 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 informationMODELING OF CONTROL SYSTEMS
1 MODELING OF CONTROL SYSTEMS Feb15 Dr. Mohammed Morsy Outline Introduction Differential equations and Linearization of nonlinear mathematical models Transfer function and impulse response function Laplace
More informationEE451/551: Digital Control. Chapter 3: Modeling of Digital Control Systems
EE451/551: Digital Control Chapter 3: Modeling of Digital Control Systems Common Digital Control Configurations AsnotedinCh1 commondigitalcontrolconfigurations As noted in Ch 1, common digital control
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 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 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 informationBasic Procedures for Common Problems
Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available
More informationYTÜ Mechanical Engineering Department
YTÜ Mechanical Engineering Department Lecture of Special Laboratory of Machine Theory, System Dynamics and Control Division Coupled Tank 1 Level Control with using Feedforward PI Controller Lab Date: Lab
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More information