Control of Manufacturing Processes


 Lydia Underwood
 1 years ago
 Views:
Transcription
1 Control of Manufacturing Processes Subject Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004
2 The Position Servo Problem, reference position NC Control Robots Injection Molding Screw Forming Press Displacement. Controller Actuator Load d DC motor Hydraulic cylinder Mass Spring Damper position 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 2
3 DC Motor Based Position Servo Reference θ r () Controller u Power Amplifier I DC Motor Load Measureme Transducer θ) Now Measure θ not Ω θ r +  e u I Controller Amplifier Motor/Load θ 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 3
4 Motor Transfer Function I (s) Gp(s) Ω (s) Ω (s) = Gp(s) I (s) G p (s) = K t Js + b = K t / b (J / b)s +1 = K m s +1 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 4
5 Position Servo Block Diagram θ r + e u Ω θ K K m c 1/s τms Controller Position Motor/Load Transducer Encoder θ = 1 K m s s +1 u θ θ r = u = K c (θ r θ) K c K m s s + K ck m 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 5
6 Position Servo Block Diagram θ r + e u Ω θ K K m c 1/s τms Controller Position Motor/Load Transducer Encoder θ = 1 s K m s +1 u u = K c (θ r θ) θ θ r = K c K m s s + K ck m 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 6
7 Position Servo Transfer Function θ θ r = K c K m s s + K ck m Using the canonical variable definitions for a 2nd order system θ θ r = ω n 2 s 2 + 2ζω n s + ω n 2 ω n 2 = K ck m 2ζω n = 1 ζ = 1 1 2ω n 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 7
8 General 2 nd Order System Time Response Ω(t) =Ω SS (1 Be ζω n t sin(ω d t + φ)) A sinusoid of frequency ω d with a magnitude envelope of e ζω n t 1 B = 1 ζ 2 ω d = ω n 1 ζ 2 1 ζ 2 φ = tan 1 ζ 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 8
9 Second Order Step Response /22/04 Lecture 19 D.E. Hardt, all rights reserved 9
10 Overshoot and Damping ζ=0.5 ζ=0.2 ζ= ζ= /22/04 Lecture 19 D.E. Hardt, all rights reserved 10
11 Overshoot and Damping /22/04 Lecture 19 D.E. Hardt, all rights reserved 11
12 Step Response as a Function of Controller Gain 2 θ θ r 1.5 Kc = 1 Kc = 5 Kc = t s 0 ω n t 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 12
13 Key Features of Response Settling Time Is Invariant Overshoot Increases with Gain Error is always Zero! 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 13
14 Settling Time Basic form of Oscillatory Response: y(t) = Ae ζω n t sin(ω d t +φ) exponential envelope sinusoid of frequency ω d Time constant of envelope = 1/ζω n Time to fully decay? 4/ζω n And from above 2ζω n = 1 ζω n = 1 2 = constant 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 14
15 SteadyState State Error θ = θ r ω n 2 s 2 + 2ζω n s + ω n 2 L 1 θ + 2ζω n &θ + ω n2 θ = ω n2 θ r all derivatives 0 ω n 2 θ = ω n 2 θ r θ = θ r Independent of Controller Gain K c 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 15
16 Zero Error for Velocity Servo Add Integrator in Controller Instead of Measurement G c (s) = K c s Ω r (s) + E (s)  K c /s Controller U (s) G p (s) Motor Ω (s) Tachometer 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 16
17 ClosedLoop Transfer Function T(s) G p K c s 1+G p K c s and subs tituting for G p (s): T(s) = KcKm s s + KcKm Same form as Position Transfer Function Thus same properties 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 17
18 Step Response of Integral Controller as a Function of Gain Ω Ω r Kc = 1 Kc = 5 Kc = t s 0 ω n t 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 18
19 Velocity Servo has First Order ClosedLoop Dynamics Better Response and Error with Gain Never Zero?error Conclusions Position Servo has 2nd Order Closed Loop Dynamics Zero error Fixed Settling time Oscillatory Response as Gain Increases 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 19
20 Conclusions Zero Error Can be Achieved with Integrator BUT AT A PRICE! We Need More Options Root Locus for Higher Order Systems 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 20
21 Outline Root Locus Interpretation of Servo Crossing the Line: 3rd Order Systems The effect of Different Controllers : PI Zeros and Integrators Summary of InProcess Control Application to Material Control: Roll Bending Application to Output Control: Welding 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 21
22 Root Locus Interpretation θ r + e u Ω θ K K c m 1/s τms Controller Motor/Load Position Transducer θ θ r = K c K m s s + K ck m Characteristic Equation: s s + K c K m = 0 or s 2 + 2ζω n s +ω n 2 = 0 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 22
23 Root Locus Interpretation s 2 + 2ζω n s +ω n 2 = 0 Solution s 1,s 2 = ζω n ±ω n ζ 2 1 ζ>1 roots are real ζ<1 roots are complex ζ=1 roots repeat for K m = =1 s 1,s 2 = 1 2 ± 1 4K c 4 K c <0.25 roots are real K c >0.25 roots are complex K c =0.25 roots repeat 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 23
24 Root Locus Interpretation s 1,s 2 = 1 2 ± 1 4K c 4 Im K c s 1,s 2 ζ ω n 0 0, , , , ? Re 11/2 1.5 ± j ± j2.95 1/ /22/04 Lecture 19 D.E. Hardt, all rights reserved 24
25 Root Locus Interpretation Recall Ω(t) = Ω SS (1 Be ζω n t sin(ω d t + φ)) s 1,s 2 = ζω n ± ω n ζ 2 1 Im for ζ < 1 ω d = ζω n ± jω n 1 ζ 2 ζω n = ζω n ± jω d Re 1 Real part Imaginary Part 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 25
26 Root Locus Interpretation For any root on the root locus: Im ω d = (ζω n ) 2 + ( 1 ζ 2 ω ) 2 n = ω n = cos(β) = ζω n ω n β = ζ Re ζω n The root location tells us: Settling time and natural frequency 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 26
27 SPlane  Response Relationships Same ζ Im Different ζω n Re 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 27
28 SPlane  Response Relationships Greater angle β Im Lower ζ More Overshoot and Oscillation Re 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 28
29 And for the Servo Root Locus K=1 Im K=0.5 K=0 Re K=0.25 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 29
30 Conclusions: Root Locus Locus of All Possible Roots of CE as K c changes Locations(s) Indicate Characteristic Response Change in System will change Root Locus What are System  Locus Relationships? Go to Higher Order systems 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 30
31 The Limit to Feedback: Instability θ r + e u K K Ω θ c m 1/s τms Controller Motor/Load Position Transducer K c K m s( s+1) a s+a feedback filter K=1; τ=1/a 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 31
32 System Closed Loop Roots K m K c s( s+1) a s+a θ θ r = K c K m (s + a) s 3 + ( 1/ + a)s 2 + (a / )s + K ck m a Roots of this C.E? Look at Root Locus
33 Open  Loop Roots? G(s) H(s) T(s) = G(s) 1+ G(s)H(s) Thus the C.E. = 1+GH(s) = 0 For Our System GH(s) = K c K m a s( s +1)(s + a) s( s +1)(s + a) = 0 s 1, s 2, s 3 =[1/, 0, a] 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 33
34 Root locus? Start at Open Loop Poles Lie on Real Axis to left of odd root starting from + + Are Attracted by Zeros End at Zeros or Infinity Repelled by Poles 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 34
35 Root Locus Without Filter Same System as Before Flows to zeros at 1/ 0 Balanced Repulsion from Poles
36 Root Locus with Filter a 1/ 0 starts at open loop roots real part lies to the left of odd number of roots 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 36
37 Effect of the Filter Pole? Poles repel the root locus a 1/ 0
38 Effect of the Filter Pole roots repel the root locus and closer is stronger a 1/ 0
39 Instability Crossover to positive real roots a 1/ 0 Conclusion: At some value of K c the closedloop system will go unstable 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 39
40 The Remedy?: Basic rules: roots repel the root locus zeros attract the root locus α 1/ 0
41 Zeros Applied Here a α 1/ 0 if the zero is closer it wins!
42 Applied Here Basic rules: roots repel the root locus zeros attract the root locus α a 1/ 0 if the pole is closer it wins!
43 Making Zeros:The PI controller E(s) K p +K I /s U(s) U(s) E(s) = K p + K I s = K ps + K I s = K p(s +α) s α = K I K p zero integrator 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 43
44 Other Considerations Effect of Parameter Variations Consider 2nd Order System Effect of NonConstant Disturbances Random Inputs 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 44
45 Summary of InProcess Feedback Control: Feedback of Machine State InProcess Lumped Position, Force, Speed Feedback of Material States InProcess Point locations or Distributed forces, temperatures, pressures, etc. Feedback of Output InProcess Direct (or estimated) Geometry Feedback 4/22/04 Lecture 19 D.E. Hardt, all rights reserved 45
Control 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 informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
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 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 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 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 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 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 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 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 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 informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, timeinvariant system. Let s see, I
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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
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 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 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 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 informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
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 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 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 informationFrequency Response of Linear Time Invariant Systems
ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z
More 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 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 informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
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 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 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 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 informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
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 informationDr 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 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 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 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 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 informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the
More 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More 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 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 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 informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More 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 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 information2.010 Fall 2000 Solution of Homework Assignment 7
. Fall Solution of Homework Assignment 7. Control of Hydraulic Servomechanism. We return to the Hydraulic Servomechanism of Problem in Homework Assignment 6 with additional data which permits quantitative
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 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 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 informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
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 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 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 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 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 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 informationReview: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Plan of the Lecture Review: transient and steadystate
More informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More informationTransient Stability Analysis with PowerWorld Simulator
Transient Stability Analysis with PowerWorld Simulator T11: Single Machine Infinite Bus Modeling (SMIB) 21 South First Street Champaign, Illinois 6182 +1 (217) 384.633 support@powerworld.com http://www.powerworld.com
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 informationDigital Control: Part 2. ENGI 7825: Control Systems II Andrew Vardy
Digital Control: Part 2 ENGI 7825: Control Systems II Andrew Vardy Mapping the splane onto the zplane We re almost ready to design a controller for a DT system, however we will have to consider where
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 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 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 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 informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
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 informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pickandplace robot to move the link of a robot between two positions. Usually
More information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More informationTopic # Feedback Control
Topic #5 6.3 Feedback Control StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or,evenifwecanchangethepolelocations. Where do we put the poles? Linear
More 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 informationGoals for today 2.004
Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation
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 informationEEE 188: Digital Control Systems
EEE 88: Digital Control Systems Lecture summary # the controlled variable. Example: cruise control. In feedback control, sensors and measurements play an important role. In discrete time systems, the control
More informationReglerteknik: Exercises
Reglerteknik: Exercises Exercises, Hints, Answers Liten reglerteknisk ordlista Introduktion till Control System Toolbox ver. 5 This version: January 3, 25 AUTOMATIC CONTROL REGLERTEKNIK LINKÖPINGS UNIVERSITET
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 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 informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
More informationProfessional Portfolio Selection Techniques: From Markowitz to Innovative Engineering
Massachusetts Institute of Technology Sponsor: Electrical Engineering and Computer Science Cosponsor: Science Engineering and Business Club Professional Portfolio Selection Techniques: From Markowitz to
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 informationAutomatic Control (TSRT15): Lecture 4
Automatic Control (TSRT15): Lecture 4 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13282226 Office: Bhouse extrance 2527 Review of the last
More informationChapter 7 Control. Part Classical Control. Mobile Robotics  Prof Alonzo Kelly, CMU RI
Chapter 7 Control 7.1 Classical Control Part 1 1 7.1 Classical Control Outline 7.1.1 Introduction 7.1.2 Virtual Spring Damper 7.1.3 Feedback Control 7.1.4 Model Referenced and Feedforward Control Summary
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
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 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 Deartment, Middle East Technical University Armature controlled dc motor Outside θ D outut Inside θ r inut r θ m Gearbox Control Transmitter θ D
More information16.31 Homework 2 Solution
16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p
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 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 informationAnswers to multiple choice questions
Answers to multiple choice questions Chapter 2 M2.1 (b) M2.2 (a) M2.3 (d) M2.4 (b) M2.5 (a) M2.6 (b) M2.7 (b) M2.8 (c) M2.9 (a) M2.10 (b) Chapter 3 M3.1 (b) M3.2 (d) M3.3 (d) M3.4 (d) M3.5 (c) M3.6 (c)
More information2.004 Dynamics and Control II Spring 2008
MT OpenCourseWare http://ocw.mit.edu.004 Dynamics and Control Spring 008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts nstitute of Technology
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 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 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 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 information1 Mathematics. 1.1 Determine the onesided Laplace transform of the following signals. + 2y = σ(t) dt 2 + 3dy dt. , where A is a constant.
Mathematics. Determine the onesided Laplace transform of the following signals. {, t < a) u(t) =, where A is a constant. A, t {, t < b) u(t) =, where A is a constant. At, t c) u(t) = e 2t for t. d) u(t)
More information