Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design


 Phoebe Daniel
 10 months ago
 Views:
Transcription
1 AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University
2
3 Routh s Stability Criterion Given characteristic equation Factor out any roots at the origin and multiply by  if needed D G (s) = s n + a s n + + a n s + a n, How to determine stability? (without using MATLAB :) Necessary condition: All coefficients of characteristic polynomial be positive, ie a i > 0 Any coefficient missing ( = 0) or negative, then poles are outside LHP Necessary and Sufficient condition: System is stable iff all the elements in the first column of Routh array are positive AERO 422, Instructor: Raktim Bhattacharya 3 / 28
4 Routh s Stability Criterion Routh Array where Row n s n a 2 a 4 Row n s n a a 3 a 5 Row n 2 s n 2 b b 2 b 3 Row n 3 s n 3 c c 2 c 3 Row 2 s 2 Row s Row 0 s 0 [ ] a2 det a a 3 b = a [ ] a a det 3 b b 2 c = b [ ] a4 det a a 5 b 2 = a [ ] a a det 5 b b 3 c 2 = b There are two special cases! See webappendix of textbook AERO 422, Instructor: Raktim Bhattacharya 4 / 28
5 Stabilizing Gain Example One parameter r + e u y C P y design parameter Given P (s) = s+ s(s )(s+6) and C(s) = K Characteristic Equation Numerator of + P C No pole zero cancellation s 3 + 5s 2 + (K 6)s + K = 0 Routh s Table Stabilizing Gains s 3 K 6 s 2 5 K s 4K 30 5 s 0 K 4K 30 > 0, K > 0 5 K > 75, K > 0 AERO 422, Instructor: Raktim Bhattacharya 5 / 28
6 Stabilizing Gain Example 2 Two parameters r + e u y C P y Given P (s) = (s+)(s+2), C(s) = K + K I s Characteristic Equation Numerator of + P C No pole zero cancellation Routh s Table s 3 + 3s 2 + (2 + K)s + K I = 0 s K s 2 3 K I s 6+3K K I 3 s 0 K I Stabilizing Gains K > K I /3 2, K I > 0 AERO 422, Instructor: Raktim Bhattacharya 6 / 28
7 Stabilizing Gain Example Two parameters (contd) r + e u y C P y Stabilizing Gains K > K I /3 2, K I > 0 Given P (s) = (s+)(s+2), C(s) = K + K I s Step Response 4 K=,KI= K=0,KI=5 2 Amplitude Time (seconds) What about performance? AERO 422, Instructor: Raktim Bhattacharya 7 / 28
8 Step Response Time Domain Performance Specification 5 Step Response Amplitude Time Second Order System: poles = σ ± jω d, ω n = σ 2 + ω 2 d, ζ = σ/ω n M p = e πζ/ ζ 2 t r = 8 t s = 46 ω n σ AERO 422, Instructor: Raktim Bhattacharya 8 / 28
9 Step Response Time Domain Performance Specification Second Order Systems Desired Location of Poles M p = e πζ/ ζ 2 t r = 8 t s = 46 ω n σ ω n 8/t r ζ ζ(m p ) σ 46/t s Adjust K, K I to satisfy additional performance related constraints Controller gain tuning AERO 422, Instructor: Raktim Bhattacharya 9 / 28
10 Routh s Stability Criterion Conditions that ensure Re p i < α, for α > 0 Given polynomial s n + a s n + + a n s + a n = 0 Modify Routh s stability criterion to ensure Re p i < α, for α > 0 Replace s := q α and substitute in polynomial to get q n + b q n + + b n q + b n = 0 Apply Routh s criterion to the polynomial in q AERO 422, Instructor: Raktim Bhattacharya 0 / 28
11 Routh s Stability Criterion Example r + e u y C P y Given P (s) = s 2 +4s+ and controller C(s) = K + K 2 /s Find range of values for K, K 2 such that all poles are left of α Characteristic equation: s 3 + 4s 2 + (K + )s + K 2 Substitute s := q α, with α = Apply Routh s criterion q 3 + q 2 + (K 4)q K + K AERO 422, Instructor: Raktim Bhattacharya / 28
12 Routh s Stability Criterion Example (contd) Polynomial in q q 3 + q 2 + (K 4)q K + K Routh s Table s 3 K 4 s 2 K 2 K + 2 s 2K K s 0 K 2 K Inequalities 2K K 2 6 > 0, K 2 K + 2 > 0 AERO 422, Instructor: Raktim Bhattacharya 2 / 28
13 Routh s Stability Criterion Example (contd) K K K K (a) 2K K 2 6 > 0, K 2 K + 2 > 0, Poles left of (b) K K 2 /4 + > 0, K 2 > 0, Poles left of 0 AERO 422, Instructor: Raktim Bhattacharya 3 / 28
14
15 Disturbance Rejection d r + u + + e y C P y A Let P (s) := (s/p +)(s/p 2 +), and C(s) := K Total response Y (s) = = P C + P C R(s) + P + P C D(s) AK (s/p + )(s/p 2 + ) + AK R(s) + A (s/p + )(s/p 2 + ) + AK D(s) Steady state value with feedback lim sy (s) = AK s 0 + AK ( ) lim sr(s) + s 0 A + AK ( ) lim sd(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 5 / 28
16 Robustness to Plant Uncertainty r + e u y C P y Transfer function from reference to output G yr (s) = P C + P C = AK (s/p + )(s/p 2 + ) + AK Suppose A A + δa What is the effect on T (s) = G yr (s)? steady state gain AERO 422, Instructor: Raktim Bhattacharya 6 / 28
17 Robustness to Plant Uncertainty (contd) T ss = δt ss = dt ss da δa AK + AK K = ( + AK) 2 δa ( ) ( AK = + AK + AK ) δa A = δt ss T ss = δa + AK A AERO 422, Instructor: Raktim Bhattacharya 7 / 28
18
19 Analysis of Steady State Error r + e u y C P y E(s) = + P C R(s) = e ss = lim s s 0 + L(s) = lim s 0 + L(s) s k Investigate e ss 0 for various values of k s k+ Value of k r(t) 0 (t) e ss = constant = Type 0 t e ss = constant = Type I 2 t 2 /2! e ss = constant = Type II AERO 422, Instructor: Raktim Bhattacharya 9 / 28
20 Steady State Error Type Zero r + e u y C P y Type 0 System Constant steady state error to step reference k = 0 e ss = lim s 0 + L(s) s k = lim s 0 + L(s) = + K p Position Error Constant K p K p = lim L(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 20 / 28
21 Steady State Error Type One r + e u y C P y Type System Constant steady state error to ramp reference k = e ss = lim s 0 + L(s) = lim s 0 + L(s) s k s = lim s 0 sl(s) = K v Velocity Error Constant K v K v = lim sl(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 2 / 28
22 Steady State Error Type Two r + e u y C P y Type 2 System Constant steady state error to parabolic reference k = 2 e ss = lim s 0 + L(s) = lim s 0 + L(s) s k s 2 = lim s 0 s 2 L(s) = K a Acceleration Error Constant K a K a = lim s 2 L(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 22 / 28
23 Steady State Error Summary r + e u y C P y Various Constants K p = lim s 0 L(s) Steady State Errors K v = lim s 0 sl(s) K a = lim s 0 s 2 L(s) Step Ramp Parabola Type 0 +K p Type I 0 K v Type II 0 0 K a AERO 422, Instructor: Raktim Bhattacharya 23 / 28
24 Steady State Error Summary (contd) r + e u y C P y Quickly identify ability to track polynomials Robustness property higher type tracks lower order polynomials Can be extended to study G yd and other transfer functions AERO 422, Instructor: Raktim Bhattacharya 24 / 28
25 Truxal s Formula Let T (s) := G yr (s) be given by T (s) = K Πm i= (s z i) Π n i= (s p i) Most common case: e ss to step is zero = Type I system DC gain lim s 0 T (s) = System error E(s) = R(s)( T (s)) System error due to ramp e ss = lim se(s) = lim s( T (s)) s 0 s 0 s 2 = lim T (s) s 0 s Using L Hopital s rule dt e ss = lim s 0 ds = K v for type I systems AERO 422, Instructor: Raktim Bhattacharya 25 / 28
26 Truxal s Formula Contd Let T (s) := G yr (s) be given by Using L Hopital s rule T (s) = K Πm i= (s z i) Π n i= (s p i) dt e ss = lim s 0 ds = K v for type I systems K v is related to the slope of T (s) at origin AERO 422, Instructor: Raktim Bhattacharya 26 / 28
27 Truxal s Formula Contd Let T (s) := G yr (s) be given by Rewrite Truxal s Formula T (s) = K Πm i= (s z i) Π n i= (s p i) dt e ss = lim T(0) = s 0 ds T d = lim log T (s) s 0 ds d = lim s 0 ds ( ) m n log(k) + log(s z i ) log(s p i ) K v = m i= i= z i n p i= i i= AERO 422, Instructor: Raktim Bhattacharya 27 / 28
28 Truxal s Formula Design Implication K v = m i= z i n p i= i Observe effect of pole/zero location on /K v Useful for design of dynamic compensators Example Third order type I system has closedloop poles 2 ± 2j, 0 The system has one zero Where should it be for K v = 0? AERO 422, Instructor: Raktim Bhattacharya 28 / 28
Proportional, Integral & Derivative Control Design. Raktim Bhattacharya
AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University
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 informationChapter 7  Solved Problems
Chapter 7  Solved Problems Solved Problem 7.1. A continuous time system has transfer function G o (s) given by G o (s) = B o(s) A o (s) = 2 (s 1)(s + 2) = 2 s 2 + s 2 (1) Find a controller of minimal
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 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 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 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 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 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 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 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 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 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 informationPD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada
PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closedloop
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 informationChapter 6 SteadyState Analysis of ContinuousTime Systems
Chapter 6 SteadyState Analysis of ContinuousTime Systems 6.1 INTRODUCTION One of the objectives of a control systems engineer is to minimize the steadystate error of the closedloop system response
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 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 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 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 informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
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 informationIMC based automatic tuning method for PID controllers in a Smith predictor configuration
Computers and Chemical Engineering 28 (2004) 281 290 IMC based automatic tuning method for PID controllers in a Smith predictor configuration Ibrahim Kaya Department of Electrical and Electronics Engineering,
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 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 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 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 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 STAFF NAME: Mr. P.NARASIMMAN BRANCH : ECE Mr.K.R.VENKATESAN YEAR : II SEMESTER
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 informationAppendix A MoReRT Controllers Design Demo Software
Appendix A MoReRT Controllers Design Demo Software The use of the proposed ModelReference Robust Tuning (MoReRT) design methodology, described in Chap. 4, to tune a twodegreeoffreedom (2DoF) proportional
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 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 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 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 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 informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More 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 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 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 informationPower System Control
Power System Control Basic Control Engineering Prof. Wonhee Kim School of Energy Systems Engineering, ChungAng University 2 Contents Why feedback? System Modeling in Frequency Domain System Modeling in
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
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 informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
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 informationSingular Value Decomposition Analysis
Singular Value Decomposition Analysis Singular Value Decomposition Analysis Introduction Introduce a linear algebra tool: singular values of a matrix Motivation Why do we need singular values in MIMO control
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 informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More informationLecture 1 Root Locus
Root Locus ELEC304Alper Erdogan 1 1 Lecture 1 Root Locus What is RootLocus? : A graphical representation of closed loop poles as a system parameter varied. Based on RootLocus graph we can choose the
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationRELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS. Ryszard Gessing
RELAY CONTROL WITH PARALLEL COMPENSATOR FOR NONMINIMUM PHASE PLANTS Ryszard Gessing Politechnika Śl aska Instytut Automatyki, ul. Akademicka 16, 44101 Gliwice, Poland, fax: +4832 372127, email: gessing@ia.gliwice.edu.pl
More informationAn Introduction to Linear Matrix Inequalities. Raktim Bhattacharya Aerospace Engineering, Texas A&M University
An Introduction to Linear Matrix Inequalities Raktim Bhattacharya Aerospace Engineering, Texas A&M University Linear Matrix Inequalities What are they? Inequalities involving matrix variables Matrix variables
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 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 informationCONTROL DESIGN FOR SET POINT TRACKING
Chapter 5 CONTROL DESIGN FOR SET POINT TRACKING In this chapter, we extend the pole placement, observerbased output feedback design to solve tracking problems. By tracking we mean that the output is commanded
More informationBasic Properties of Feedback
4 Basic Properties of Feedback A Perspective on the Properties of Feedback A major goal of control design is to use the tools available to keep the error small for any input and in the face of expected
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 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 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 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 informationESE319 Introduction to Microelectronics. Feedback Basics
Feedback Basics Stability Feedback concept Feedback in emitter follower Onepole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability
More informationClass 12 Root Locus part II
Class 12 Root Locus part II Revising (from part I): Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K Comple plane jω ais 0 real ais Thus,
More 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 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 informationLaboratory 11 Control Systems Laboratory ECE3557. State Feedback Controller for Position Control of a Flexible Joint
Laboratory 11 State Feedback Controller for Position Control of a Flexible Joint 11.1 Objective The objective of this laboratory is to design a full state feedback controller for endpoint position control
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 informationHomework Assignment 3
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full
More informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
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 information6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson
Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closedloop behavior what we want it to be. To review:  G c (s) G(s) H(s) you are here! plant For
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 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 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 informationQuanser NIELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual
Quanser NIELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control DC Motor Control Trainer (DCMCT) Student Manual Table of Contents 1 Laboratory Objectives1 2 References1 3 DCMCT Plant
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 informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
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 informationThe output voltage is given by,
71 The output voltage is given by, = (3.1) The inductor and capacitor values of the Boost converter are derived by having the same assumption as that of the Buck converter. Now the critical value of the
More informationUnit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace
Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, 010 1 Pair Table Unit, Part : Unit,
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 informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
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 informationPole placement control: state space and polynomial approaches Lecture 2
: state space and polynomial approaches Lecture 2 : a state O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.fr www.gipsalab.fr/ o.sename based November 21, 2017 Outline : a state
More informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
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 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 informationProblem Set 5 Solutions 1
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski Problem Set 5 Solutions The problem set deals with Hankel
More informationReview of Linear TimeInvariant Network Analysis
D1 APPENDIX D Review of Linear TimeInvariant Network Analysis Consider a network with input x(t) and output y(t) as shown in Figure D1. If an input x 1 (t) produces an output y 1 (t), and an input x
More informationPower System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur. Module 3 Lecture 8
Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur Module 3 Lecture 8 Welcome to lecture number 8 of module 3. In the previous
More informationAn Internal Stability Example
An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several polezero cancellations between the controller and
More informationSAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015
FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequencydomain analysis and control design (15 pt) Given is a
More informationLecture 9. Welcome back! Coming week labs: Today: Lab 16 System Identification (2 sessions)
232 Welcome back! Coming week labs: Lecture 9 Lab 16 System Identification (2 sessions) Today: Review of Lab 15 System identification (ala ME4232) Time domain Frequency domain 1 Future Labs To develop
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 informationMultivariable Control. Lecture 05. Multivariable Poles and Zeros. John T. Wen. September 14, 2006
Multivariable Control Lecture 05 Multivariable Poles and Zeros John T. Wen September 4, 2006 SISO poles/zeros SISO transfer function: G(s) = n(s) d(s) (no common factors between n(s) and d(s)). Poles:
More informationPole Placement Design
Department of Automatic Control LTH, Lund University 1 Introduction 2 Simple Examples 3 Polynomial Design 4 State Space Design 5 Robustness and Design Rules 6 Model Reduction 7 Oscillatory Systems 8 Summary
More informationSMITH MCMILLAN FORMS
Appendix B SMITH MCMILLAN FORMS B. Introduction Smith McMillan forms correspond to the underlying structures of natural MIMO transferfunction matrices. The key ideas are summarized below. B.2 Polynomial
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
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 information