Digital Control: Summary # 7


 Lizbeth Jenkins
 4 years ago
 Views:
Transcription
1 Digital Control: Summary # 7 Proportional, integral and derivative control where K i is controller parameter (gain). It defines the ratio of the control change to the control error. Note that e(k) 0 u(k) u(k ) (4) This means that when the error is ero, the controller output does not change. It is possible to write the k th value for the controller output as follows u() u(0) + K i e() vu(2) u(0) + K i e(2) u(0) + K i e() + K i e(2). k u(k) u(0) + K i e(j) (5) j The output of the controller is proportional to all past errors. This means that the controller takes the history into account. Fig.. Block diagram of an integral controller The proportional controller discussed previously can be used to control the steady state error and the transient. It can be also used to reduces the steady state error; however, an infinite gain is needed to drive the steady state error to ero. Proportional controller: controller output is proportional to the error. Integral controller: change of the controller output is proportional to the error. In other words, the controller output is proportional to the integral of the error. The integral effort is what drives the error to ero. It is possible to write in the sdomain: U(s) K i E(s) () E(s) s U(s) (2) K i where U(s) is the output of the controller and E(s) is the input of the controller, K i is the controller gain. Integral control law Figure shows the block diagram of the controller where e(k) is the error, i.e., the difference between input and output of the system. The error is the input signal of the controller. u(k) is the output of the controller. The integral control law can be expressed as follows: u(k) u(k ) + K i e(k) (3) Find the domain transfer function of the proportional controller u(k) u(k ) + K i e(k) (6) U() U() + K i E() (7) U() K i E() (8) K i (9) Steady state error with integral control In order to illustrate the effect of proportional control on steady state error we consider the following example. Consider the following system: G() 0.3 (0) The steady state error to a unit step input and unity feedback (with proportional controller with gain K ) is given by: e( ) lim( ) () (2) 88 (3) Steady state error to unit step with an integral controller e( ) lim( ) + Ki ( )( 0.3) (4)
2 Digital Controls, spring 208 Summary Step Response Amplitude Closed loop step response with unity feedback only (sec).5 Step Response Amplitude First order system with negative a Closed loop response with integral action Time (sec) Fig. 2. Step response for the closed loop system of (0). Clearly the steady state error is ero with the integral controller Fig. 3. Block diagram with integral controller in the presence of disturbance (top) and block diagram of PI controller (bottom) e( ) lim + Ki ( )( 0.3) (5) e( ) lim ( )( 0.3) ( )( 0.3) + K i 0 (6) Thus, with integral controller, for a system of type 0 and a constant input, the steady state error is ero and does not depend on K i. This means that any value of K i that satisfies stability can be used to drive the steady state error to ero. An illustration is shown in figure 2 for system (0), where the closed loop step response with and without integral controller is shown. Steady state error due to disturbance It is possible to consider the steady state error due to a step disturbance as shown in figure 3 top. Recall that the steady state error in the domain is expressed as E() R() + C()G() D()G() + C()G() G()D() e d ( ) lim( ) + G()C() (7) (8) For a first order system and a unit step disturbance, we can write e d ( ) lim( ) Ki ( 0.3)( ) 0 (9) Therefore, the steady state error to a constant disturbance for any value of K i is 0. This means you can pick any value of K i as long as the system is stable. Proportional + integral Proportional +integral controller combines the advantages of proportional controller and those of integral controller. P control: improves the transient response. I control: ero steady state error. The controller has two terms: u p (k) K p e(k) u i (k) u i (k ) + K i e(k) The PI controller adds these terms together: u(k) u p (k) + u i (k) K p e(k) + u i (k ) + K i e(k) u(k ) u p (k ) + u i (k ) K p e(k ) + u i (k ) u(k) u(k ) K p e(k) K p e(k ) + K i e(k) Thus the final form for the PI controller is or u(k) u(k ) + K p e(k) K p e(k ) + K i e(k) (20) The transfer function of the PI controller is U() E() K p + K i U() E() (K i + K p ) K p (2) (22) Equation (2) emphasies on the components of the controller and equation (22) emphasies on the eros and poles of the controller. Steady state error with PI control: e( ) lim( ) + K p+k i K p (23) PI controller by pole placement One of the most traditional ways to designing PI controllers is by using pole placement techniques. Consider the 2
3 Digital Controls, spring 208 Summary 7 block diagram of figure 3 bottom with G() 0.3 The closed loop transfer function is given by G cl () C()G() + C()G() ((K p+k i) K p) ( )( 0.3) ( )( 0.3)+((K p+k i) K p) ( )( 0.3) (K p + K i ) K p (K p + K i ) K p The characteristic polynomial is (24) (25) (26) Q() (K p + K i ) K p (27) The modeled characteristic polynomial is Q() 2 2 cos(ω d T )e ζωnt + e 2ζωnT (28) From the desired specifications, find the desired characteristic polynomial. Solve for K p, K i : equate the coefficients of the desired polynomial and the modeled characteristic polynomial and solve for K p and K i. Verify the results: check that the closed loop poles are inside the unit circle. Let G() 0.3 (29) Design a PI controller so that: T s s, M p % 0%. Take T 0.s. First T s s ζω n 8 (30) and ζ ln(m p) π 2 + ln 2 M p 9 (3) and ω n 3.56rad/s. The desired characteristic polynomial is obtained: from which we get: Finally: Q() (32) 0.3 K p 0.2 (33) (K p + K i ) (34) K p 0.2 (35) K i.58 (36) In addition to pole placement, other techniques such as graphical tools and empirical methods can be used to design PI controllers. Proportional and derivative controller Integral controllers have the ability to drive the steady state error to ero. While in proportional control, the control action is based on the current error, in integral control, the controller action is based on the previous errors. The derivative term is different where the control action is based on the rate of change of the derivative. The idea in the derivative controller is that the controller should react immediately to any change in the error. The derivative control law has the form: u d (k) K d (e(k) e(k )) (37) where K d is the derivative control gain. It defines the importance given to the rate of change of the error. From equation (37), it is clear that the derivative action cannot react to a constant error. In practice, derivative action is never alone. It is used with proportional control and sometimes with integral control. The transfer function of the derivative action is given by U d () K d (E() E()) (38) U d E () K d (39) where K d is the controller gain. Proportional + derivative The transfer function of proportional + derivative controller is given by U E () K p + K d (K p + K d ) K d (40) Proportional derivative controllers have predictive ability and can be used effectively to reduce overshoot. Proportional + derivative + integral actions Proportionalintegralderivative controllers combine three different actions, three parameters K p, K i, K d are associated each with each action. The output of the controller is u(k) u p (k) + u i (k) + u d (k) (4) k K p e(k) + K i e(i) + K d (e(k) e(k )) i0 The transfer function of a PID controller is given by U() E() K p + K i + K d (42) The block diagram of a PID controller is shown in figure 4. The controller adds two poles to the system: one at the origin and the other one at. The controller transfer function can be written as C() (K p + K i + K d ) 2 (K p + 2K d ) + K d ( ) (43) There exist different methods to design PID controllers: Pole placement methods 3
4 Digital Controls, spring 208 Summary 7 The closed loop characteristic polynomial is given by Q() (K p +K i +K d ) 2 3(K p +2K d )+3K d (50) By equating the two polynomials, we obtain T erm P ID Desired 3 2 3(K p + K d + K i ).3 3(K p + 2K d ) K d (5) The table above shows a comparison between the two polynomials in terms of their coefficients. The next step is to equate the coefficients and solve for the gains. Finally, by solving we get: K p 3 K i 0.06 (52) K d Fig. 4. Block diagram of a PID control Graphical methods such as root locus Empirical methods (powerful in practice when it is difficult to obtain an algebraic model of the system) PID control using pole placement Pole placement means we chose the poles of the closed loop system. The chosen poles translate the desired characteristics of the system. The method is very similar to the PI design algorithm discussed previously. Consider the open loop transfer function given by G() 3 2 (44) Note that the open loop system is unstable. The desired poles locations are p,2 ± j0. (45) p (46) The desired characteristic polynomial is given by Q() ( p )( p 2 )( p 3 ) ( + j0.)( j0.)( 0.3)(47) (48) The closed loop transfer function is given by G cl () C()G() + C()G() (49) where C() is the transfer function of the PID controller given by (43). 4
5 Digital Controls, spring 208 Summary 7 x2 % Initial liquid level yx % Another variable for the liquid level f.75 % Numerical value of the disturbance T. % Sampling time r5 % desired liquid level, i.e., reference Kp2 % Proportionality gain Ki % Integral gain ep0 % previous sample error for time0:t:0 % Proportional u(kp)*(ry) % The proportional control law only Fig. 5. Tank system yy+t*(uf) % Open loop transfer function plot(time, y, b+ ) hold on end Reference Level With PI action 2.5 With P action only time(s) Fig. 6. Closed loop system response with P and PI controller This example shows a simple implementation of the P and PI controllers to control the liquid level of figure 5. The closed loop system response is shown in figure 6. Sample code (for P only) can be found below. 5
6.1 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 informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
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 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 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 informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More 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 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 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 informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More 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 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 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 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 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 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 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 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 informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
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 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 informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More 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 information7.2 Controller tuning from specified characteristic polynomial
192 Finn Haugen: PID Control 7.2 Controller tuning from specified characteristic polynomial 7.2.1 Introduction The subsequent sections explain controller tuning based on specifications of the characteristic
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 informationEE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO
EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total
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 informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More 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 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 informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control
More informationINTRODUCTION TO 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 informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
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 information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationEE 422G  Signals and Systems Laboratory
EE 4G  Signals and Systems Laboratory Lab 9 PID Control Kevin D. Donohue Department of Electrical and Computer Engineering University of Kentucky Lexington, KY 40506 April, 04 Objectives: Identify the
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationEEL2216 Control Theory CT1: PID Controller Design
EEL6 Control Theory CT: PID Controller Design. Objectives (i) To design proportionalintegralderivative (PID) controller for closed loop control. (ii) To evaluate the performance of different controllers
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Intro Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /5/27 Outline Closed Loop Transfer
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More informationEEE 184 Project: Option 1
EEE 184 Project: Option 1 Date: November 16th 2012 Due: December 3rd 2012 Work Alone, show your work, and comment your results. Comments, clarity, and organization are important. Same wrong result or same
More informationUnit 8: Part 2: PD, PID, and Feedback Compensation
Ideal Derivative Compensation (PD) Lead Compensation PID Controller Design Feedback Compensation Physical Realization of Compensation Unit 8: Part 2: PD, PID, and Feedback Compensation Engineering 5821:
More 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 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 informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More 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 information1 Chapter 9: Design via Root Locus
1 Figure 9.1 a. Sample root locus, showing possible design point via gain adjustment (A) and desired design point that cannot be met via simple gain adjustment (B); b. responses from poles at A and B 2
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 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 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 informationCDS 101/110 Homework #7 Solution
Amplitude Amplitude CDS / Homework #7 Solution Problem (CDS, CDS ): (5 points) From (.), k i = a = a( a)2 P (a) Note that the above equation is unbounded, so it does not make sense to talk about maximum
More informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More 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 informationPID Control. Objectives
PID Control Objectives The objective of this lab is to study basic design issues for proportionalintegralderivative control laws. Emphasis is placed on transient responses and steadystate errors. The
More informationFeedback design for the Buck Converter
Feedback design for the Buck Converter Portland State University Department of Electrical and Computer Engineering Portland, Oregon, USA December 30, 2009 Abstract In this paper we explore two compensation
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
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 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 informationTopic # Feedback Control
Topic #5 6.3 Feedback Control StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or,evenifwecanchangethepolelocations. Where do we put the poles? Linear
More 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 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 informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationSoftware Engineering 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 informationDesign of a Lead Compensator
Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc  Funded by MHRD The Lecture Contains Standard Forms of
More 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 informationControl Systems. Design of State Feedback Control.
Control Systems Design of State Feedback Control chibum@seoultech.ac.kr Outline Design of State feedback control Dominant pole design Symmetric root locus (linear quadratic regulation) 2 Selection of closedloop
More informationEL2450: Hybrid and Embedded Control Systems: Homework 1
EL2450: Hybrid and Embedded Control Systems: Homework 1 [To be handed in February 11] Introduction The objective of this homework is to understand the basics of digital control including modelling, controller
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 information] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command prefilter [ 0.
EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965372 Problem (Analysis of a Feedback System) Consider the feedback system
More informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
More 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 informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More informationEssence of the Root Locus Technique
Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general setup, namely for the case when the closedloop
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 informationAutomatic Control (TSRT15): Lecture 7
Automatic Control (TSRT15): Lecture 7 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13282226 Office: Bhouse extrance 2527 Outline 2 Feedforward
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 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 information1 x(k +1)=(Φ LH) x(k) = T 1 x 2 (k) x1 (0) 1 T x 2(0) T x 1 (0) x 2 (0) x(1) = x(2) = x(3) =
567 This is often referred to as Þnite settling time or deadbeat design because the dynamics will settle in a Þnite number of sample periods. This estimator always drives the error to zero in time 2T or
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 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 informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
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 informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
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 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 informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
More informationOptimal Polynomial Control for DiscreteTime Systems
1 Optimal Polynomial Control for DiscreteTime Systems Prof Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning this paper should
More informationChapter 2 SDOF Vibration Control 2.1 Transfer Function
Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:
More 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013 Problem Set #4 Posted: Thursday, Mar. 7, 13 Due: Thursday, Mar. 14, 13 1. Sketch the Root
More informationControl Systems Lab  SC4070 Control techniques
Control Systems Lab  SC4070 Control techniques Dr. Manuel Mazo Jr. Delft Center for Systems and Control (TU Delft) m.mazo@tudelft.nl Tel.:0152788131 TU Delft, February 16, 2015 (slides modified from
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
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 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 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 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationProportional, 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 information