Automatic Control EEE 2002 Tutorial Exercise IV
|
|
- Donald Hawkins
- 5 years ago
- Views:
Transcription
1 Automatic Control EEE Tutorial Exercise IV k A second order system is given by G() s =. as + bs + c k'. Write the transfer function as: G() s =. s + ζω ns + ωn () G s = as k = + bs + c s + k / a b s a + c a, hence c b k' = k / a, ω b n =, ζω a n = ζ = a a a c. For k = and ζ =., ω = rad s and define a transfer function and use the command damp to find the damping factor and natural frequency. >> kp=; z=.; wn=; >> num=kp; den=[ *z*wn wn^]; >> g=tf(num,den) Transfer function: s^ + s + >> [z, wn]=damp(g) z =.. wn =.. 3. Based on the previous answer predict the behaviour of the system for a (unit) step input. Since the damping factor is positive and less than I expect a stable but initially oscillatory behaviour. The frequency of these oscillations is ω d = ωn ζ =. =.33rad / s. The overshoot is going to be:
2 >> mp=exp(-(z/sqrt(-z^))*pi) mp =.63 The peak time is: >> wd=wn*sqrt(-z^); tp=pi/wd tp =.7 And finally the settling time: >> ts=3/(z*wn) ts =.. Find the step response of that system using Matlab AND Simulink and hence crosscheck your previous answer. Also plot the system s error. >> step(g).. System: g Peak amplitude:.6 Overshoot (%): 6.3 At time (sec):.79 System: g Settling : Check the properties of the settling time and notice the discrepancy!????
3 . Use the command step as [y,t]=step(sys), where sys is the transfer function of the system given in the previous question. Using these two vectors find the overshoot and peak time. kp=; z=.; wn=; num=kp; den=[ *z*wn wn^]; g=tf(num,den); [y,t]=step(g); yss=y(); mp=max(y) overshoot=*(mp-yss)/yss k=find(y==max(y)); tp=t(k) mp =.6 overshoot = 6.3 tp = Using Matlab find the exact value of the steady state error using Matlab AND Simulink. >> [y,t]=step(g); >> yss=y(); err_ss=-yss err_ss =.999 See also Simulink model, once you run the simulation type: >> plot(error_scope(:,),error_scope(:,))
4 >> error_scope(,) ans = Assume a system with k = and ω = rad s. Use an m-file to calculate and hence plot the pole location in the s-plane for [.,.] ζ (use a step size of.). kp=; wn=; cnt=; for z=.:.:.; den=[ *z*wn wn^]; r(cnt,:)=roots(den); cnt=cnt+; hold on plot(real(r(:,)),imag(r(:,)),'b.') plot(real(r(:,)),imag(r(:,)),'g.')
5 8. Assume a system with ζ =. and ω = rad s. Use an m-file to calculate and hence plot the pole location in the s-plane for [., ] k (use a step size of.). As kp is not in the CE no change will be seen in the pole location. 9. Assume a system with k = and ζ =.. Use an m-file to calculate and hence plot the pole location in the s-plane for [.,] ω (use a step size of ). n kp=; z=.; cnt=; for wn=.::; den=[ *z*wn wn^]; r(cnt,:)=roots(den); cnt=cnt+; hold on plot(real(r(:,)),imag(r(:,)),'b.') plot(real(r(:,)),imag(r(:,)),'g.') Find the unit step response and discuss the results (in connection to your answer given in the previous question) for: a. k =., k = and k = (keep ζ =., ω = rad s ) num=.; num=; num3=; z=., wn=; den=[ *z*wn wn^]; g=tf(num,den); g=tf(num,den); g3=tf(num3,den); subplot(3,,); step(g)
6 subplot(3,,); step(g) subplot(3,,3); step(g3) x b. ζ =., ζ =, ζ =. (keep k = and ω = rad s ) num=; wn=; z=., z=; z3=.; den=[ *z*wn wn^]; den=[ *z*wn wn^]; den3=[ *z3*wn wn^]; g=tf(num,den); g=tf(num,den); g3=tf(num,den3); subplot(3,,); step(g)
7 subplot(3,,); step(g) subplot(3,,3); step(g3) c. ω =.rad / s, ω = rad / s, ω = rad s (keep ζ =., k = ) n n num=; z=.; wn=., wn=; wn3=; den=[ *z*wn wn^]; den=[ *z*wn wn^]; den3=[ *z*wn3 wn3^]; g=tf(num,den); g=tf(num,den); g3=tf(num,den3);
8 subplot(3,,); step(g) subplot(3,,); step(g) subplot(3,,3); step(g3) 3 x x Comments on these responses during the lectures!. Using the specific formula plot the overshoot versus the damping factor. cnt=; for z=.:.: mp(cnt)=exp(-(z/sqrt(-z^))*pi); cnt=cnt+;
9 plot(.:.:,mp,'.') Using the specific formula plot the overshoot versus the natural frequency. The overshoot is not a function of the natural frequency! 3. Using the specific formula plot the peak time versus the damping factor for wn=rad/s. cnt=; wn=; for z=.:.: wd=wn*sqrt(-z^) tp(cnt)=pi/wd; cnt=cnt+; plot(.:.:,tp,'.')
10 . Using the specific formula plot the peak time versus the natural frequency for z=.rad/s Using the specific formula plot the settling time (% and % in the same graph) versus the natural frequency for z=.rad/s: cnt=; z=.; for wn=.:.: ts(cnt)=3/z/wn; ts(cnt)=/z/wn; cnt=cnt+; plot(.:.:,ts,'r.'), hold on plot(.:.:,ts,'b.') Using the specific formula plot the settling time (% and % in the same graph) versus the damping factor for wn=rad/s:
11 cnt=; wn=; for z=.:.: ts(cnt)=3/z/wn; ts(cnt)=/z/wn; cnt=cnt+; plot(.:.:,ts,'r.'), hold on plot(.:.:,ts,'b.')
Lab # 4 Time Response Analysis
Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an
More information6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.
6. Sketch the z-domain 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 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 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 open-loop system (a system without feedbac) and a closed-loop
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System
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 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 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 informationMatlab Controller Design. 1. Control system toolbox 2. Functions for model analysis 3. Linear system simulation 4. Biochemical reactor linearization
Matlab Controller Design. Control system toolbox 2. Functions for model analysis 3. Linear system simulation 4. Biochemical reactor linearization Control System Toolbox Provides algorithms and tools for
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 informationSOLUTIONS TO CASE STUDIES CHALLENGES
F O U R Time Response SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Open-Loop Response The forward transfer function for angular velocity is, G(s) = ω 0(s) V P (s) = 24 (s+50)(s+.32) a. ω 0 (t)
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 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 informationChapter 7: Time Domain Analysis
Chapter 7: Time Domain Analysis Samantha Ramirez Preview Questions How do the system parameters affect the response? How are the parameters linked to the system poles or eigenvalues? How can Laplace transforms
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, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
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 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 informationStep Response of First-Order Systems
INTRODUCTION This tutorial discusses the response of a first-order system to a unit step function input. In particular, it addresses the time constant and how that affects the speed of the system s response.
More informationCourse roadmap. Step response for 2nd-order system. Step response for 2nd-order system
ME45: Control Systems Lecture Time response of nd-order systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
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 informationControl Systems, Lecture04
Control Systems, Lecture04 İbrahim Beklan Küçükdemiral Yıldız Teknik Üniversitesi 2015 1 / 53 Transfer Functions The output response of a system is the sum of two responses: the forced response and the
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 informationChapter #4 EEE Automatic Control
Spring 008 EEE 00 Chapter #4 EEE 00 Automatic Control Root Locu Chapter 4 /4 Spring 008 EEE 00 Introduction Repone depend on ytem and controller parameter > Cloed loop pole location depend on ytem and
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 informationState Feedback Controller for Position Control of a Flexible Link
Laboratory 12 Control Systems Laboratory ECE3557 Laboratory 12 State Feedback Controller for Position Control of a Flexible Link 12.1 Objective The objective of this laboratory is to design a full state
More information12.7 Steady State Error
Lecture Notes on Control Systems/D. Ghose/01 106 1.7 Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there
More informationControl 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationLecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 First-Order Specs: Step : Pole Real inputs contain
More informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
More 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 informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informationEEL2216 Control Theory CT1: PID Controller Design
EEL6 Control Theory CT: PID Controller Design. Objectives (i) To design proportional-integral-derivative (PID) controller for closed loop control. (ii) To evaluate the performance of different controllers
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 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 informationLaboratory handouts, ME 340
Laboratory handouts, ME 340 This document contains summary theory, solved exercises, prelab assignments, lab instructions, and report assignments for Lab 4. 2014-2016 Harry Dankowicz, unless otherwise
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 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 informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329 - Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phase-lead and Phase-lag compensators using
More informationYTÜ Mechanical Engineering Department
YTÜ Mechanical Engineering Department Lecture of Special Laboratory of Machine Theory, System Dynamics and Control Division Coupled Tank 1 Level Control with using Feedforward PI Controller Lab Report
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA EXPERIMENT NO : CS II/ TITLE : FAMILIARIZATION
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 informationECE-320: Linear Control Systems Homework 8. 1) For one of the rectilinear systems in lab, I found the following state variable representations:
ECE-30: Linear Control Systems Homework 8 Due: Thursday May 6, 00 at the beginning of class ) For one of the rectilinear systems in lab, I found the following state variable representations: 0 0 q q+ 74.805.6469
More information16.31 Homework 2 Solution
16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p
More informationCALIFORNIA INSTITUTE OF TECHNOLOGY Control and Dynamical Systems
CDS 101 1. Åström and Murray, Exercise 1.3 2. Åström and Murray, Exercise 1.4 3. Åström and Murray, Exercise 2.6, parts (a) and (b) CDS 110a 1. Åström and Murray, Exercise 1.4 2. Åström and Murray, Exercise
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 informationLab Experiment 2: Performance of First order and second order systems
Lab Experiment 2: Performance of First order and second order systems Objective: The objective of this exercise will be to study the performance characteristics of first and second order systems using
More informationNotes for ECE-320. Winter by R. Throne
Notes for ECE-3 Winter 4-5 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
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 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 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 informationYTÜ Mechanical Engineering Department
YTÜ Mechanical Engineering Department Lecture of Special Laboratory of Machine Theory, System Dynamics and Control Division Coupled Tank 1 Level Control with using Feedforward PI Controller Lab Date: Lab
More informationOKLAHOMA STATE UNIVERSITY
OKLAHOMA STATE UNIVERSITY ECEN 4413 - Automatic Control Systems Matlab Lecture 1 Introduction and Control Basics Presented by Moayed Daneshyari 1 What is Matlab? Invented by Cleve Moler in late 1970s to
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 informationController Design using Root Locus
Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers
More informationFeedback Control part 2
Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open- and closed-loop control Everything before chapter 7 are open-loop systems Transient response Design criteria Translate criteria
More informationPitch Rate CAS Design Project
Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability
More 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 open-loop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the state-variable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationProblem Value Score Total 100/105
RULES This is a closed book, closed notes test. You are, however, allowed one piece of paper (front side only) for notes and definitions, but no sample problems. The top half is the same as from the first
More informationQuanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual
Quanser NI-ELVIS 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 informationMultivariable Control Laboratory experiment 2 The Quadruple Tank 1
Multivariable Control Laboratory experiment 2 The Quadruple Tank 1 Department of Automatic Control Lund Institute of Technology 1. Introduction The aim of this laboratory exercise is to study some different
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 informationChapter 6 Steady-State Analysis of Continuous-Time Systems
Chapter 6 Steady-State Analysis of Continuous-Time Systems 6.1 INTRODUCTION One of the objectives of a control systems engineer is to minimize the steady-state error of the closed-loop system response
More informationMAE 143B - Homework 7
MAE 143B - Homework 7 6.7 Multiplying the first ODE by m u and subtracting the product of the second ODE with m s, we get m s m u (ẍ s ẍ i ) + m u b s (ẋ s ẋ u ) + m u k s (x s x u ) + m s b s (ẋ s ẋ u
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 D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationDigital Control: Summary # 7
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)
More informationME 375 System Modeling and Analysis. Homework 11 Solution. Out: 18 November 2011 Due: 30 November 2011 = + +
Out: 8 November Due: 3 November Problem : You are given the following system: Gs () =. s + s+ a) Using Lalace and Inverse Lalace, calculate the unit ste resonse of this system (assume zero initial conditions).
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:
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 informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationEE 4314 Lab 1 Handout Control Systems Simulation with MATLAB and SIMULINK Spring Lab Information
EE 4314 Lab 1 Handout Control Systems Simulation with MATLAB and SIMULINK Spring 2013 1. Lab Information This is a take-home lab assignment. There is no experiment for this lab. You will study the tutorial
More informationCDS 101/110: Lecture 3.1 Linear Systems
CDS /: Lecture 3. Linear Systems Goals for Today: Describe and motivate linear system models: Summarize properties, examples, and tools Joel Burdick (substituting for Richard Murray) jwb@robotics.caltech.edu,
More informationEE 370L Controls Laboratory. Laboratory Exercise #7 Root Locus. Department of Electrical and Computer Engineering University of Nevada, at Las Vegas
EE 370L Controls Laboratory Laboratory Exercise #7 Root Locus Department of Electrical an Computer Engineering University of Nevaa, at Las Vegas 1. Learning Objectives To emonstrate the concept of error
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 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 informationOverview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piece-wise linear approximations First-order terms Second-order terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 5-059-00 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control
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 informationThe control of a gantry
The control of a gantry AAE 364L In this experiment we will design a controller for a gantry or crane. Without a controller the pendulum of crane will swing for a long time. The idea is to use control
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, time-invariant system. Let s see, I
More information9/9/2011 Classical Control 1
MM11 Root Locus Design Method Reading material: FC pp.270-328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),
More informationAN INTRODUCTION TO THE CONTROL THEORY
Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter
More informationMEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 2017) Lecture DOF Modeling of Shock Absorbers. This lecture covers:
MEG6007: Advanced Dynamics -Principles and Computational Methods (Fall, 207) Lecture 4. 2-DOF Modeling of Shock Absorbers This lecture covers: Revisit 2-DOF Spring-Mass System Understand the den Hartog
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 informationHomework 11 Solution - AME 30315, Spring 2015
1 Homework 11 Solution - AME 30315, Spring 2015 Problem 1 [10/10 pts] R + - K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closed-loop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
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 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback
More informationImproving a Heart Rate Controller for a Cardiac Pacemaker. Connor Morrow
Improving a Heart Rate Controller for a Cardiac Pacemaker Connor Morrow 03/13/2018 1 In the paper Intelligent Heart Rate Controller for a Cardiac Pacemaker, J. Yadav, A. Rani, and G. Garg detail different
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 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 informationRecursive, Infinite Impulse Response (IIR) Digital Filters:
Recursive, Infinite Impulse Response (IIR) Digital Filters: Filters defined by Laplace Domain transfer functions (analog devices) can be easily converted to Z domain transfer functions (digital, sampled
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 informationAutomatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...
Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number..........................
More informationAppendix 3B MATLAB Functions for Modeling and Time-domain analysis
Appendix 3B MATLAB Functions for Modeling and Time-domain analysis MATLAB control system Toolbox contain the following functions for the time-domain response step impulse initial lsim gensig damp ltiview
More informationu (t t ) + e ζωn (t tw )
LINEAR CIRCUITS LABORATORY OSCILLATIONS AND DAMPING EFFECT PART I TRANSIENT RESPONSE TO A SQUARE PULSE Transfer Function F(S) = ω n 2 S 2 + 2ζω n S + ω n 2 F(S) = S 2 + 3 RC ( RC) 2 S + 1 RC ( ) 2 where
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus
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 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 informationControl Systems. University Questions
University Questions UNIT-1 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 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: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONICS ENGINEERING
More information