Proportional, Integral & Derivative Control Design. Raktim Bhattacharya


 Norah Sharp
 8 months ago
 Views:
Transcription
1 AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University
2 roportional Control
3 Proportional Control r + e u y C P y C(s) = K, constant gain u(t) = Ke(t) Use Routh s criterion to determine range for K Verify if the system is stabilizable with C(s) = K AERO 422, nstructor: Raktim Bhattacharya 3 / 3
4 Second order system r + e u y C P y A P (s) = (τ s+)(τ 2 s+), C(s) = K Characteristic equation: zeros of + P C τ τ 2 s 2 + (τ + τ 2 )s + AK + = Open loop poles at /τ, /τ 2 Closed loop poles at p = τ + τ 2 + τ 2 2 τ τ 2 + τ AK τ τ 2 2 τ τ 2 p 2 = τ + τ 2 τ 2 2 τ τ 2 + τ AK τ τ 2 2 τ τ 2 AERO 422, nstructor: Raktim Bhattacharya 4 / 3
5 Routh Stability Criterion r + e u y C P y P (s) = 5 A (τ s+)(τ 2 s+), C(s) = K Root Locus Range for stabilization K > s 2 τ τ 2 AK s τ + τ 2 s AK maginary Axis (seconds ) Real Axis (seconds ) Poles move towards each other till K = K = (τ τ 2 ) 2 4τ τ 2 A K > K, poles are purely imaginary K > K, ω n, ζ AERO 422, nstructor: Raktim Bhattacharya 5 / 3
6 Proportional Control Steady State Error r + e u y C P y System Transfer Function G yr (s) := P (s) = Corresponding ODE Steady state A (τ s + )(τ 2 s + ), C(s) = K p P C + P C = AK p τ τ 2 s 2 + (τ + τ 2 )s + AK p + τ τ 2 ÿ + (τ + τ 2 )ẏ + (AK p + )y = AK p r ÿ = ẏ = = y(t) = AK p + AK p r(t) AERO 422, nstructor: Raktim Bhattacharya 6 / 3
7 Proportional Control Summary Root Locus Step Response 8 7 Amplitude maginary Axis (seconds ) 5 5 Mp Time Real Axis (seconds ) ζ (a) Step Response (b) Root Locus (c) M p vs ζ K > K, ω n, ζ t r = 8 ω n, t s = 46 σ Effect of Proportional Control Reduces rise time Good! ncreases overshoot Bad! Large gain small steady state error Amplifies noise and disturbances Bad! Look at frequency characteristics of all the transfer functions (later) AERO 422, nstructor: Raktim Bhattacharya 7 / 3
8 ntegral Control
9 ntegral Control r + e u y C P y C(s) = K /s t u(t) = K e(t) Use Routh s criterion to determine range for K Verify if the system is stabilizable with C(s) = K /s AERO 422, nstructor: Raktim Bhattacharya 9 / 3
10 Second Order System r + e u y C P y P (s) = A (τ s+)(τ 2 s+), C(s) = K /s Characteristic equation: τ τ 2 s 3 + (τ + τ 2 )s 2 + s + AK Root Locus 2 maginary Axis (seconds ) Real Axis (seconds ) s 3 τ τ 2 s 2 τ + τ 2 AK s τ +τ 2 AK τ τ 2 τ +τ 2 s AK < K < τ + τ 2 Aτ τ 2 AERO 422, nstructor: Raktim Bhattacharya / 3
11 ntegral Control Steady State Error r + e u y C P y System P (s) = Transfer Function G yr (s) := Corresponding ODE A (τ s + )(τ 2 s + ), C(s) = K /s P C + P C = AK τ τ 2 s 3 + (τ + τ 2 )s 2 + AK τ τ 2 y + (τ + τ 2 )ÿ + AK y = AK r Steady state y = ÿ = = y(t) = r(t) AERO 422, nstructor: Raktim Bhattacharya / 3
12 ntegral Control Summary r + e u y C P y maginary Axis (seconds ) Root Locus Real Axis (seconds ) K = ζ = M p K = ω n = t r K > τ +τ 2 Aτ τ 2 unstable Zero steady state error Needs anti windup mechanism for saturated actuators discussed later Good disturbance rejection property read book AERO 422, nstructor: Raktim Bhattacharya 2 / 3
13 Control
14 Proportionalntegral Control r + e u y C P y System P (s) = ( A (τ s + )(τ 2 s + ), C(s) = K + ) T s T is called the integral, or reset time /T is reset rate, related to speed of response u(t) is a mixture of two signals t ) u(t) = K (e(t) + T e(t)dτ t u(t) even when e(t) =, because of integral action AERO 422, nstructor: Raktim Bhattacharya 4 / 3
15 Proportionalntegral Control Steady State Error r + e u y C P y System P (s) = ( A (τ s + )(τ 2 s + ), C(s) = K + ) T s Transfer Function G yr (s) := P C + P C = AK(T s + ) T τ τ 2 s 3 + T (τ + τ 2 )s 2 + T ( + AK)s + AK Corresponding ODE ( ) y + ( )ÿ + ( )ẏ + AKy = AKr + ( )ṙ Steady state y = ÿ = ẏ = ṙ = = y(t) = r(t) AERO 422, nstructor: Raktim Bhattacharya 5 / 3
16 Proportionalntegral Control ndependent control over two of the three terms r + e u y C P y System P (s) = ( A (τ s + )(τ 2 s + ), C(s) = K + ) T s Transfer Function G yr (s) := P C + P C = AK(T s + ) T τ τ 2 s 3 + T (τ + τ 2 )s 2 + T ( + AK)s + AK Characteristic Equation T τ τ 2 s 3 + T (τ + τ 2 )s 2 + T ( + AK)s + AK = AERO 422, nstructor: Raktim Bhattacharya 6 / 3
17 Proportionalntegral Control Two tuning variables r + e u y C P y Routh s Table Constraints s 3 T τ τ 2 T (AK + ) s 2 T (τ + τ 2 ) AK s T + AKT AKτ + AKτ 2 τ +τ 2 s AK K >, T > AK τ τ 2 + AK τ + τ 2 AERO 422, nstructor: Raktim Bhattacharya 7 / 3
18 Derivative Control
19 Derivative Control r + e u y C P y C(s) = K D s u(t) = K D ė(t) Use Routh s criterion to determine range for K D Verify if the system is stabilizable with C(s) = K D s Almost never used by itself usually augmented by proportional control Derivative control is not causal depends on future values ė e(t + t) e(t) t Knows the slope known from future values of e(t) e(t) = t, ė = e(t) = t 2, ė = 2t e(t) = sin(t), ė = cos(t) = sin(t + π/2) AERO 422, nstructor: Raktim Bhattacharya 9 / 3
20 Approximate Derivative Control Approximation over Frequency range C(s) = K D s u(t) = K D ė(t) Not implementable in applications Approximate as C(s) = K Ds, α >> pole at far left s/α + K D s, for small s What is the effect of this approximation? Look at a step response AERO 422, nstructor: Raktim Bhattacharya 2 / 3
21 Approximate Derivative Control Effect of the Approximation 5 Step Response Amplitude Time e(t) = r(t) y(t) = y(t) ė(t) = ẏ(t) AERO 422, nstructor: Raktim Bhattacharya 2 / 3
22 Approximate Derivative Control Effect of the Approximation (contd) 6 4 True alp= alp= Error Derivative Time AERO 422, nstructor: Raktim Bhattacharya 22 / 3
23 ProportionalDerivative Control r + e u y C P y Controller Structure ( ) s C(s) = K + T D s/α + s = K P + K D s/α + Tune K P and K D to get desired response Use Routh s table to determine range for stable values Derivative control increases damping reduces overshoot AERO 422, nstructor: Raktim Bhattacharya 23 / 3
24 Control
25 Control Basic dea Controller Structure C(s) = K P + K s + K s D s/α + Tune K P, K and K D to get desired response Use Routh s table to determine range for stable values Proportional term increases ω n and decreases ζ mproves rise time Needs large gain to reduce steadystate error ntegral term increases ω n and decreases ζ Zero steadystate May make the system unstable Derivative control increases damping reduces overshoot and oscillations AERO 422, nstructor: Raktim Bhattacharya 25 / 3
26 Control Example System P (s) = s 2 + 5s + Study behavior of this system with various control strategies Proportional (P) Proportional ntegral () Proportional Derivative () Proportional ntegral and Derivative () AERO 422, nstructor: Raktim Bhattacharya 26 / 3
27 Control Example: P Amplitude Amplitude Amplitude Amplitude 2 Open Loop K= Time (seconds) 2 Open Loop K= Time (seconds) 2 Open Loop K= Time (seconds) 2 Open Loop K= Time (seconds) AERO 422, nstructor: Raktim Bhattacharya 27 / 3
28 Control Example: Step Response 4 2 P Amplitude Time (seconds) AERO 422, nstructor: Raktim Bhattacharya 28 / 3
29 Control Example: (contd) Amplitude Amplitude Amplitude Amplitude 5 5 Open Loop Kp= Kp=, Ki = Time (seconds) 5 5 Open Loop Kp=2 Kp=2, Ki = Time (seconds) 2 Open Loop Kp= Kp=, Ki = Time (seconds) 2 Open Loop Kp=5 Kp=5, Ki = Time (seconds) AERO 422, nstructor: Raktim Bhattacharya 29 / 3
30 Control Example: Step Response 5 Open loop P D Amplitude Time (seconds) AERO 422, nstructor: Raktim Bhattacharya 3 / 3
31 Control Example: Step Response 5 Open loop P Amplitude Time (seconds) AERO 422, nstructor: Raktim Bhattacharya 3 / 3
Raktim 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 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 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 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 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 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 informationControl 2. Proportional and Integral control
Control 2 Proportional and Integral control 1 Disturbance rejection in Proportional Control Θ i =5 + _ Controller K P =20 Motor K=2.45 Θ o Consider first the case where the motor steadystate gain = 2.45
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 informationTuning PI controllers in nonlinear uncertain closedloop systems with interval analysis
Tuning PI controllers in nonlinear uncertain closedloop systems with interval analysis J. Alexandre dit Sandretto, A. Chapoutot and O. Mullier U2IS, ENSTA ParisTech SYNCOP April 11, 2015 Closedloop
More informationME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II
ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and
More 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 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 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 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 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 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 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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #24 Wednesday, March 10, 2004 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Remedies We next turn to the question
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 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 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 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 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 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 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 informationOutline. Classical Control. Lecture 2
Outline Outline Outline Review of Material from Lecture 2 New Stuff  Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New
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 informationIntegrator Windup
3.5.2. Integrator Windup 3.5.2.1. Definition So far we have mainly been concerned with linear behaviour, as is often the case with analysis and design of control systems. There is, however, one nonlinear
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
More informationControl Systems II. ETH, MAVT, IDSC, Lecture 4 17/03/2017. G. Ducard
Control Systems II ETH, MAVT, IDSC, Lecture 4 17/03/2017 Lecture plan: Control Systems II, IDSC, 2017 SISO Control Design 24.02 Lecture 1 Recalls, Introductory case study 03.03 Lecture 2 Cascaded Control
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 informationChemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University
Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this
More informationEECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8 am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total
More 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 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 informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More 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 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 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 informationFall 線性系統 Linear Systems. Chapter 08 State Feedback & State Estimators (SISO) FengLi Lian. NTUEE Sep07 Jan08
Fall 2007 線性系統 Linear Systems Chapter 08 State Feedback & State Estimators (SISO) FengLi Lian NTUEE Sep07 Jan08 Materials used in these lecture notes are adopted from Linear System Theory & Design, 3rd.
More informationCDS 101/110a: Lecture 102 Control Systems Implementation
CDS 101/110a: Lecture 102 Control Systems Implementation Richard M. Murray 5 December 2012 Goals Provide an overview of the key principles, concepts and tools from control theory  Classical control 
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 informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Norms for Signals and Systems
. AERO 632: Design of Advance Flight Control System Norms for Signals and. Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Norms for Signals ...
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 informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
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 informationSystem Types in Feedback Control with Saturating Actuators
System Types in Feedback Control with Saturating Actuators Yongsoon Eun, Pierre T. Kabamba, and Semyon M. Meerkov Department of Electrical Engineering and Computer Science, University of Michigan, Ann
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 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 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 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 informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
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 informationHomework 11 Solution  AME 30315, Spring 2015
1 Homework 11 Solution  AME 30315, Spring 2015 Problem 1 [10/10 pts] R +  K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
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 information1 (30 pts) Dominant Pole
EECS C8/ME C34 Fall Problem Set 9 Solutions (3 pts) Dominant Pole For the following transfer function: Y (s) U(s) = (s + )(s + ) a) Give state space description of the system in parallel form (ẋ = Ax +
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationControl Lab. Thermal Plant. Chriss Grimholt
Control Lab Thermal Plant Chriss Grimholt Process System Engineering Department of Chemical Engineering Norwegian University of Science and Technology October 3, 23 C. Grimholt (NTNU) Thermal Plant October
More informationChapter 7 : Root Locus Technique
Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real
More informationGoodwin, Graebe, Salgado, Prentice Hall Chapter 11. Chapter 11. Dealing with Constraints
Chapter 11 Dealing with Constraints Topics to be covered An ubiquitous problem in control is that all real actuators have limited authority. This implies that they are constrained in amplitude and/or rate
More informationPitch Rate CAS Design Project
Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability
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 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 informationHOMEWORK 4: MATH 265: SOLUTIONS. y p = cos(ω 0t) 9 ω 2 0
HOMEWORK 4: MATH 265: SOLUTIONS. Find the solution to the initial value problems y + 9y = cos(ωt) with y(0) = 0, y (0) = 0 (account for all ω > 0). Draw a plot of the solution when ω = and when ω = 3.
More informationIndex. INDEX_p /15/02 3:08 PM Page 765
INDEX_p.765770 11/15/02 3:08 PM Page 765 Index N A Adaptive control, 144 Adiabatic reactors, 465 Algorithm, control, 5 Allpass factorization, 257 Allpass, frequency response, 225 Amplitude, 216 Amplitude
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 informationDERIVATIVE FREE OUTPUT FEEDBACK ADAPTIVE CONTROL
DERIVATIVE FREE OUTPUT FEEDBACK ADAPTIVE CONTROL Tansel YUCELEN, * Kilsoo KIM, and Anthony J. CALISE Georgia Institute of Technology, Yucelen Atlanta, * GA 30332, USA * tansel@gatech.edu AIAA Guidance,
More informationControl System Design
ELEC4410 Control System Design Lecture 19: Feedback from Estimated States and DiscreteTime Control Design Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science
More informationChapter 8. Feedback Controllers. Figure 8.1 Schematic diagram for a stirredtank blending system.
Feedback Controllers Figure 8.1 Schematic diagram for a stirredtank blending system. 1 Basic Control Modes Next we consider the three basic control modes starting with the simplest mode, proportional
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 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 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 informationECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =
ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.
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 informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
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 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 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 informationLecture 7: Antiwindup and friction compensation
Lecture 7: Antiwindup and friction compensation Compensation for saturations (antiwindup) Friction models Friction compensation Material Lecture slides Course Outline Lecture 13 Lecture 26 Lecture
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 informationChapter Eleven. Frequency Domain Design Sensitivity Functions
Feedback Systems by Astrom and Murray, v2.11b http://www.cds.caltech.edu/~murray/fbswiki Chapter Eleven Frequency Domain Design Sensitivity improvements in one frequency range must be paid for with sensitivity
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationRoot locus 5. tw4 = 450. Root Locus S51 S O L U T I O N S
Root Locus S51 S O L U T I O N S Root locus 5 Note: All references to Figures and Equations whose numbers are not preceded by an "S" refer to the textbook. (a) Rule 2 is all that is required to find the
More informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationCONTROL OF OSCILLATIONS IN MANUFACTURING NETWORKS
PHYSCON 2009, Catania, Italy, September, 1 September, 4 2009 CONTROL OF OSCILLATIONS IN MANUFACTURING NETWORKS Alexander Y. Pogromsky Department of Mechanical Engineering Eindhoven University of Technology
More informationApplication Note #3413
Application Note #3413 Manual Tuning Methods Tuning the controller seems to be a difficult task to some users; however, after getting familiar with the theories and tricks behind it, one might find the
More informationState Feedback and State Estimators Linear System Theory and Design, Chapter 8.
1 Linear System Theory and Design, http://zitompul.wordpress.com 2 0 1 4 2 Homework 7: State Estimators (a) For the same system as discussed in previous slides, design another closedloop state estimator,
More informationDistributed RealTime Control Systems
Distributed RealTime Control Systems Chapter 9 Discrete PID Control 1 Computer Control 2 Approximation of Continuous Time Controllers Design Strategy: Design a continuous time controller C c (s) and then
More informationEL2520 Control Theory and Practice
EL2520 Control Theory and Practice Lecture 8: Linear quadratic control Mikael Johansson School of Electrical Engineering KTH, Stockholm, Sweden Linear quadratic control Allows to compute the controller
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 informationUniversity of Science and Technology, Sudan Department of Chemical Engineering.
ISO 91:28 Certified Volume 3, Issue 6, November 214 Design and Decoupling of Control System for a Continuous Stirred Tank Reactor (CSTR) Georgeous, N.B *1 and Gasmalseed, G.A, Abdalla, B.K (12) University
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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #22 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Friday, March 5, 24 More General Effects of Open Loop Poles
More informationRecitation 11: Time delays
Recitation : Time delays Emilio Frazzoli Laboratory for Information and Decision Systems Massachusetts Institute of Technology November, 00. Introduction and motivation. Delays are incurred when the controller
More informationLecture 13: Internal Model Principle and Repetitive Control
ME 233, UC Berkeley, Spring 2014 Xu Chen Lecture 13: Internal Model Principle and Repetitive Control Big picture review of integral control in PID design example: 0 Es) C s) Ds) + + P s) Y s) where P s)
More informationLecture 7:Time Response PoleZero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 FirstOrder Specs: Step : Pole Real inputs contain
More 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 information