Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.


 Owen Armstrong
 2 years ago
 Views:
Transcription
1 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 θ D dc amplifier Control Transformer Prof. Dr. Y. Samim Ünlüsoy Prof. Dr. Y. Samim Ünlüsoy 1
2 COURSE OUTLINE I. INTRODUCTION & BASIC CONCEPTS II. MODELING DYNAMIC SYSTEMS III. CONTROL SYSTEM COMPONENTS IV. STABILITY V. TRANSIENT RESPONSE VI. STEADY STATE RESPONSE CH VI VII. DISTURBANCE REJECTION VIII. BASIC CONTROL ACTIONS & CONTROLLERS IX. FREQUENCY RESPONSE ANALYSIS X. SENSITIVITY ANALYSIS XI. ROOT LOCUS ANALYSIS Prof. Dr. Y. Samim Ünlüsoy 2
3 STEADY STATE RESPONSE  OBJECTIVES In this chapter : Open loop transfer functions of feedback control systems will be classified. Steady state error of feedback control systems due to step, ramp, and parabolic inputs will be investigated. Selection of controller parameters for a specified steady state error and the measures to be taken to reduce steady state error will be examined. Prof. Dr. Y. Samim Ünlüsoy 3
4 STEADY STATE RESPONSE Steady State Response is the response of a system as time goes to infinity. It is particularly important since it provides an indication of the accuracy of a control system when its output is compared with the desired d input. If they do not agree exactly, then a steady state error exists. Prof. Dr. Y. Samim Ünlüsoy 4
5 STEADY STATE RESPONSE The Steady State Response of a system is judged by the steady state error due to step, ramp, and parabolic (acceleration) inputs. These inputs may be associated with the ability of a control system to : Position itself relative to a stationary target, t Follow a target moving at constant speed, and Track an object that is accelerating. Prof. Dr. Y. Samim Ünlüsoy 5
6 STEADY STATE RESPONSE Note that for the steady state response to exist, the system must be stable. Therefore before going into steady state analysis it would be good practise to check the stability of the system. Prof. Dr. Y. Samim Ünlüsoy 6
7 STEADY STATE RESPONSE The steady state response and error can be obtained by using the final value theorem. The final value theorem : The final value of a time signal can be found from the Laplace transform of the signal in the sdomain. lim y(t)=lim sy(s) t s o Prof. Dr. Y. Samim Ünlüsoy 7
8 STEADY STATE RESPONSE The final value theorem : In the case of the output of a system C(s)=G(s)R(s) )R( ) lim c(t)=lim sg(s)r(s) t s 0 Prof. Dr. Y. Samim Ünlüsoy 8
9 OPEN LOOP TRANSFER FUNCTION The open loop transfer function of a general feedback control system is given by : B(s) = R(s) G(s)H(s) R(s) + E a (s) C(s) G(s)  Remove B(s) H(s) Prof. Dr. Y. Samim Ünlüsoy 9
10 OPEN LOOP TRANSFER FUNCTION The open loop transfer function of a unity feedback control system is given by : B(s) = R(s) G(s) R(s) + E(s) C(s) G(s) Remove  B(s) Prof. Dr. Y. Samim Ünlüsoy 10
11 STEADY STATE ERROR Nise Ch , 7.5, Dorf & Bishop Section 4.5, 5.7 The Laplace transform of actuating error, (s), for a general closed loop system : E a R(s) + E a (s) C(s) G(s)  B(s) H(s) E (s)=r(s)h(s)c(s) a =R(s)H(s)G(s)E (s) a 1 E(s)= a R(s) 1+G(s)H(s) Prof. Dr. Y. Samim Ünlüsoy 11
12 R(s) E(s)= a 1+G(s)H(s) STEADY STATE ERROR H(s)=1 R(s) E(s)= 1+G(s) R(s) + E(s) G(s) C(s)  Note that the actuating and actual errors will be identical only for a unity feedback system, i.e. with an ideal sensor. For a nonunity feedback system, the actual error may not be zero when the actuating error is zero. Prof. Dr. Y. Samim Ünlüsoy 12
13 STEADY STATE RESPONSE Here, with a systematic approach, it will be shown that the steady state error for a unity feedback system due to a certain type of input depends on the type of its open loop transfer function. For a nonunity feedback system, the analysis is somewhat more involved* and will not be covered here. * See Nise, Section 7.6, and Kuo, pp Prof. Dr. Y. Samim Ünlüsoy 13
14 SYSTEM TYPE Nise 7.3, Dorf & Bishop 5.7 The open loop transfer function of a unity feedback control system can be written in the general form : K ( 1+T s) p P p=1 T(s)= M Q 2 N s s s ( 1+τms) 1+2ξ q + 2 m=1 q=1 ωn q ω nq A unity feedback control system with this open loop transfer function is called a type N system. s N K : N poles at the origin (free integrators), : open loop gain. Prof. Dr. Y. Samim Ünlüsoy 14
15 SYSTEM TYPE EXAMPLE 1a R(s) s 2 s ( s+3 ) s 3 ( s+2 ) _ C(s) G(s)= 2 s s+1 s+1 ( s+3) s ( s+2) s( s+2)( s+3) = N=1 : type 1 system Prof. Dr. Y. Samim Ünlüsoy 15
16 SYSTEM TYPE EXAMPLE 1b R(s) + _ s 2 s+1 2 ( s+3 ) s 3 ( s+2) C(s) 1 ( s+1 ) s+1 G(s)= = s( 2) s+1 ( 3) s+1 s s+1 s Open loop gain : 1/18 Prof. Dr. Y. Samim Ünlüsoy 16
17 E(s)= R(s) 1+G(s) STEADY STATE ERROR Using the final value theorem : e =lim e(t) ss t =lim sr(s) lims 0sE(s)=lim s 0 1+G(s) Note that t the steady state t error depends on the input and the open loop transfer function of the system. Prof. Dr. Y. Samim Ünlüsoy 17
18 STEADY STATE ERROR Step Input Step Input : r(t)=r R(s)= R s R s s e ss =lims 0 1+G(s) R = 1+lim G(s) s 0 Prof. Dr. Y. Samim Ünlüsoy 18
19 R e ss = 1+lims 0 G(s) STEADY STATE ERROR Define the position error constant Then K =lim s s 0 G(s) e = ss R 1+K s Prof. Dr. Y. Samim Ünlüsoy 19
20 K s =G(0) R e ss = 1+K s STATE ( 1 )( 2 ) ( m ) ( τ )( τ ) ( τ ) K T s +1 T s T s +1 G(s)= N s 1s+1 2s+1... ns+1 STEADY STATE ERROR K s = K for type 0 systems for type 1 or higher systems e = ss R 1+K 0 for type 0 systems for type 1 or higher systems Prof. Dr. Y. Samim Ünlüsoy 20
21 STEADY STATE ERROR Ramp Input Ramp Input : r(t)=rt { } L t n! = s n n+1 R R(s)= s 2 R s 2 s e ss =lims 0 1+G(s) =lim s 0 s 0 R s+sg(s) R = lim sg(s) Prof. Dr. Y. Samim Ünlüsoy 21
22 R e ss = lims 0 sg(s) STEADY STATE ERROR Define the velocity error constant Then K =lim v s 0 sg(s) e = ss R K v Prof. Dr. Y. Samim Ünlüsoy 22
23 K v =lims 0sG(s) ss R ( 1 )( 2 ) ( m ) ( τ )( τ ) ( τ ) K T s +1 T s T s +1 G(s)= N s 1s+1 2s+1... ns+1 e = K STEADY v STATE ERROR v 0 K = K for type 0 systems for type 1 systems for type 2 or higher systems R for type 0 systems e = ss for type 1 systems K for type 2 or higher systems K 0 Prof. Dr. Y. Samim Ünlüsoy 23
24 STEADY STATE ERROR Acceleration Input Parabolic (acceleration) Input : R 2 R r(t)= t s 3 2 s e ss =lim s 0 1+G(s) { } n! L t n = s n+1 R =lims 0 s 2 +s 2 G(s) R(s)= R 3 s R = lim s 2 G(s) s 0 Prof. Dr. Y. Samim Ünlüsoy 24
25 R e = ss 2 lim s 0s G(s) STEADY STATE ERROR Define the acceleration error constant K =lim 2 a s 0s G(s) Then e ss R = K a Prof. Dr. Y. Samim Ünlüsoy 25
26 2 K a =lims 0s G(s) ss R ( 1 )( 2 ) ( m ) ( τ )( τ ) ( τ ) K T s +1 T s T s +1 G(s)= N s 1s+1 2s+1... ns+1 e = K STEADY a STATE ERROR a 0 K = K for type 0 and 1 systems for type 2 systems for type 3 or higher systems R for type 0 and 1 systems e = ss for type 2 systems K for type 3 or higher systems K 0 Prof. Dr. Y. Samim Ünlüsoy 26
27 STEADY STATE ERROR  Summary System Type Step Input Ramp Input Parabolic Input R 0 1+K 1 0 R K R K Prof. Dr. Y. Samim Ünlüsoy 27
28 STEADY STATE ERROR Multiple Inputs For linear systems, the steady state error for the simultaneous application of two or more inputs will be the superposition p of the steady state errors due to each input applied separately. Prof. Dr. Y. Samim Ünlüsoy 28
29 STEADY STATE ERROR For example, for an input of 3t r(t)=1+2t the steady state error is given as the superposition of the steady responses to each of the inputs e ss = K K K s v a Prof. Dr. Y. Samim Ünlüsoy 29
30 STEADY STATE ERROR Therefore the steady state error of the system subjected to the composite input will be : e ss = Ks Kv Ka = + + = 1+K 0 0 for type 0systems = K 0 = for type1systems = + + = 1+ K K for type 2 systems = + + = 0 1+ for type 3 or higher systems Prof. Dr. Y. Samim Ünlüsoy 30
31 STEADY STATE ERROR  Observations It is observed that : i) When e ss is finite, increasing the open loop gain decreases the steady state error. Step : R 1+K Ramp : Acceleration : R K K : open loop gain Prof. Dr. Y. Samim Ünlüsoy 31
32 STEADY STATE ERROR  Observations ii) As the type of the system is increased, e ss decreases. Therefore one may attempt to improve the steady state response by including an integrator in the controller. This, however, may cause stability problems which become critical for type 3 or higher systems. Prof. Dr. Y. Samim Ünlüsoy 32
33 STEADY STATE ERROR  Observations iii) The approach to the minimization of the steady state error in control systems may thus be : Determine the maximum possible open loop gain and check the maximum allowable open loop gain that will not result in instability. Choose the smaller of the two open loop gain values. Prof. Dr. Y. Samim Ünlüsoy 33
34 STEADY STATE ERROR Example 1a Determine the steady state error for r(t)=2+3t R(s) + _ Amplifier gain 10 ss ( ) C(s) G(s)= = = s(s+5) s(5)(0.2s+1) s(0.2s+1) N=1 : type 1 system, K=200 Prof. Dr. Y. Samim Ünlüsoy 34
35 STEADY STATE ERROR Example 1b G(s) = 200 s(0.2s+1) N=1 : type 1 system, K=200 e ss = 0 for r(t)=2 (step input) R 3 e ss = = =0.015 for r(t)= 3t (ramp input) K 200 Hence e = =1.5 % ss It is obvious that if you increase the value of the open loop gain K, say by increasing amplifier gain, steady state error will decrease. Prof. Dr. Y. Samim Ünlüsoy 35
36 STEADY STATE ERROR Example 1c Let us check the maximum value of the amplifier gain without causing instability. R(s) + _ Amplifier gain K a 10 ss+5 ( ) C(s) 10K a ( ) C(s) ss+5 = = R(s) 10Ka 1+ ss+5 ( ) 10K a 2 s +5s+10Ka Prof. Dr. Y. Samim Ünlüsoy 36
37 STEADY STATE ERROR Example 1d Thus the characteristic polynomial is given by : 2 D(s)= s +5s+10K a It is obvious that it passes Hurwitz test. Application of Routh s stability criterion will result in only one condition on stability : s K a s s 0 10K a K > 0 a Prof. Dr. Y. Samim Ünlüsoy 37
38 STEADY STATE ERROR Example 1e As an alternative solution, let us calculate the steady state error using the final value theorem C(s) s(s+5) 1000 = = R(s) 1000 s(s+5) s(s+5) r(t)=2+3t 2 3 R(s)= + s s E(s)=R(s)C(s)=R(s) R(s) s(s+5)+1000 s(s+5) s(s +5) 2 3 E(s)= R(s)= s +5s+1000 s +5s s s Prof. Dr. Y. Samim Ünlüsoy 38
39 STEADY STATE ERROR Example 1f Thus with the error expression in Laplace domain : s+5 2s+3 E(s)= 2 s +5s+1000 s e =lim se(s) ss s 0 2s+3 s+5 =lim s s 0 s s 2 +5s = = =1.5 % 1000 It is obvious that this method requires considerably higher effort. Prof. Dr. Y. Samim Ünlüsoy 39
40 STEADY STATE ERROR Example 2a Determine the value of amplifier gain K a such that t the steady state t error for a ramp input r(t)=3t is going to be at most 2 %. R(s) 300 K a + ss+5 ( )( s+2 ) _ Amplifier gain C(s) 300Ka 300Ka 30Ka s(s+2) s+5 s 2 (5)(0.5s+1) 5s+1) 0.2s+1 s(0.5s+1)(0.2s+1) G(s)= = ( ) ( ) ( ) = N=1 : type 1 system, K=30K a Prof. Dr. Y. Samim Ünlüsoy 40
41 STEADY STATE ERROR Example 2b G(s) = 30K a s(0.5s+1)(0.2s+1) N=1 : type 1 system, K=3K ; a R=3 R 3 ess = = 0.02 for r(t)= 3t (ramp input) K 30K 3 Hence Ka = a ( ) Now, the limit on K a with respect to stability should be checked. Prof. Dr. Y. Samim Ünlüsoy 41
42 STEADY STATE ERROR Example 2c Let us check the maximum value of the amplifier gain without causing instability. R(s) + _ K a Amplifier gain 30 ss+5 s+2 ( )( ) C(s) 30K a ( ) C(s) s(s+2) s+5 = = R(s) 30Ka 1+ s(s+2) s+5 ( ) 30K a 3 2 s +7s +10s+30Ka Prof. Dr. Y. Samim Ünlüsoy 42
43 STEADY STATE ERROR Example 2d Thus the characteristic polynomial is given by : 3 2 D(s)= s +7s +10s +30K a It is obvious that it passes Hurwitz test. Application of Routh s stability criterion will result in : s s 2 a > 7 30K a 7030K a 7 s 1 0 s 0 30K a Ka > 7 0 K 0 K 70 a < = Prof. Dr. Y. Samim Ünlüsoy 43
44 STEADY STATE ERROR Example 2e It is obvious that the value of K required a for the specified steady state error will make the system unstable. Therefore, the minimum possible steady state error will be higher than the desired value. Can you determine the minimum possible steady state t error for this system? Prof. Dr. Y. Samim Ünlüsoy 44
Radar 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 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 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 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 informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More informationSteady State Errors. Recall the closedloop transfer function of the system, is
Steady State Errors Outline What is steadystate error? Steadystate error in unity feedback systems Type Number Steadystate error in nonunity feedback systems Steadystate error due to disturbance inputs
More 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 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 informationTest 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010
Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationControl Systems. EC / EE / IN. For
Control Systems For EC / EE / IN By www.thegateacademy.com Syllabus Syllabus for Control Systems Basic Control System Components; Block Diagrammatic Description, Reduction of Block Diagrams. Open Loop
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationLecture 4 Classical Control Overview II. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 4 Classical Control Overview II Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Stability Analysis through Transfer Function Dr. Radhakant
More 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 informationLast week: analysis of pinionrack w velocity feedback
Last week: analysis of pinionrack w velocity feedback Calculation of the steady state error Transfer function: V (s) V ref (s) = 0.362K s +2+0.362K Step input: V ref (s) = s Output: V (s) = s 0.362K s
More informationAutomatic Control Systems (FCS) Lecture 8 Steady State Error
Automatic Control Systems (FCS) Lecture 8 Steady State Error Introduction Any physical control system inherently suffers steadystate error in response to certain types of inputs. A system may have no
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design
AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University Routh s Stability
More 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 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 information10ES43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A
10ES43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative
More 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 informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad  500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III
More 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 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 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 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 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 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 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 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 informationBangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory
Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24Jan15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
More 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 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 information7.1 Introduction. Apago PDF Enhancer. Definition and Test Inputs. 340 Chapter 7 SteadyState Errors
340 Chapter 7 SteadyState Errors 7. Introduction In Chapter, we saw that control systems analysis and design focus on three specifications: () transient response, (2) stability, and (3) steadystate errors,
More 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 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 informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
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 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 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 informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More 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 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 informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8
Learning Objectives ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8 Dr. Oishi oishi@unm.edu November 2, 203 State the phase and gain properties of a root locus Sketch a root locus, by
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 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 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 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 informationModule 07 Control Systems Design & Analysis via RootLocus Method
Module 07 Control Systems Design & Analysis via RootLocus Method Ahmad F. Taha EE 3413: Analysis and Desgin of Control Systems Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ taha March
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 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 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 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 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 informationBASIC PROPERTIES OF FEEDBACK
ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)
More information( ) ( = ) = ( ) ( ) ( )
( ) Vρ C st s T t 0 wc Ti s T s Q s (8) K T ( s) Q ( s) + Ti ( s) (0) τs+ τs+ V ρ K and τ wc w T (s)g (s)q (s) + G (s)t(s) i G and G are transfer functions and independent of the inputs, Q and T i. Note
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #18 Basic Control Loop Analysis" April 15, 2004 Revisit Temperature Control Problem τ dy dt + y = u τ = time constant = gain y ss =
More 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 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 informationSome special cases
Lecture Notes on Control Systems/D. Ghose/2012 87 11.3.1 Some special cases Routh table is easy to form in most cases, but there could be some cases when we need to do some extra work. Case 1: The first
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 informationINSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad ELECTRICAL AND ELECTRONICS ENGINEERING TUTORIAL QUESTION BANK
Course Name Course Code Class Branch INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad 500 043 ELECTRICAL AND ELECTRONICS ENGINEERING TUTORIAL QUESTION BAN : CONTROL SYSTEMS : A50 : III B. Tech
More 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 informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More 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 informationSECTION 4: STEADY STATE ERROR
SECTION 4: STEADY STATE ERROR MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Steady State Error Introduction 3 Consider a simple unity feedback system The error is the difference between
More 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 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 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 informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
More 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 informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More 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 informationEE451/551: Digital Control. Chapter 3: Modeling of Digital Control Systems
EE451/551: Digital Control Chapter 3: Modeling of Digital Control Systems Common Digital Control Configurations AsnotedinCh1 commondigitalcontrolconfigurations As noted in Ch 1, common digital control
More 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 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 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 informationOutline. Control systems. Lecture4 Stability. V. Sankaranarayanan. V. Sankaranarayanan Control system
Outline Control systems Lecture4 Stability V. Sankaranarayanan Outline Outline 1 Outline Outline 1 2 Concept of Stability Zero State Response: The zerostate response is due to the input only; all the
More informationPlan of the Lecture. Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control
Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic:
More informationReview: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control
Plan of the Lecture Review: stability; Routh Hurwitz criterion Today s topic: basic properties and benefits of feedback control Goal: understand the difference between openloop and closedloop (feedback)
More informationMAE143 B  Linear Control  Spring 2018 Midterm, May 3rd
MAE143 B  Linear Control  Spring 2018 Midterm, May 3rd Instructions: 1. This exam is open book. You can consult any printed or written material of your liking. 2. You have 70 minutes. 3. Most questions
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 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 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 informationCourse roadmap. ME451: Control Systems. What is Root Locus? (Review) Characteristic equation & root locus. Lecture 18 Root locus: Sketch of proofs
ME451: Control Systems Modeling Course roadmap Analysis Design Lecture 18 Root locus: Sketch of proofs Dr. Jongeun Choi Department of Mechanical Engineering Michigan State University Laplace transform
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 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 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 informationSECTION 7: STEADYSTATE ERROR. ESE 499 Feedback Control Systems
SECTION 7: STEADYSTATE ERROR ESE 499 Feedback Control Systems 2 Introduction SteadyState Error Introduction 3 Consider a simple unityfeedback system The error is the difference between the reference
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 informationLinear State Feedback Controller Design
Assignment For EE5101  Linear Systems Sem I AY2010/2011 Linear State Feedback Controller Design Phang Swee King A0033585A Email: king@nus.edu.sg NGS/ECE Dept. Faculty of Engineering National University
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
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 informationCONTROL SYSTEMS LECTURE NOTES B.TECH (II YEAR II SEM) ( ) Prepared by: Mrs.P.ANITHA, Associate Professor Mr.V.KIRAN KUMAR, Assistant Professor
LECTURE NOTES B.TECH (II YEAR II SEM) (201718) Prepared by: Mrs.P.ANITHA, Associate Professor Mr.V.KIRAN KUMAR, Assistant Professor Department of Electronics and Communication Engineering MALLA REDDY
More information