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


 Rodger Victor Harvey
 9 months ago
 Views:
Transcription
1
2
3
4
5
6
7
8
9
10
11
12
13 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 Total You have 2 hours to complete this exam. INSTRUCTIONS 2. This is a closed book exam. You may use one note sheet. 3. Calculators are allowed. 4. Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example, Problem 1.2 Continued. NO credit will be given to solutions that do not meet this requirement. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. 6. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. 1
14 Problem 1: (25 Points) 1. (13 points) Consider the feedback control system in Figure 1. Figure 1: Feedback control system. (a) (7 points) Determine the system type. (b) (3 points) Determine the position, velocity, and acceleration error constants. 2
15 (c) (3 points) For each of the following command inputs, determine the steadystate value of e(t): r 1 (t) = 10 u o (t) r 2 (t) = 10 t u o (t) r 3 (t) = 10 t2 2 u o(t). 3
16 2. (12 points) Figure 2 shows the block diagram of a nonunity feedback system. Figure 2: Nonunity feedback system. (a) (9 points) Determine the system type. (b) (3 points) Determine the value of the nonzero, finite, error constant. 4
17 Problem 2: (25 Points) Figure 3 shows the block diagram of a feedback system that regulates the angular displacement θ(t) of a servomotor. Figure 3: Servomotor system. 1. (7 points) In response to the input θ r (t) = πu o (t), the closedloop system generates the zerostate where τ is a positive constant. θ(t) = C 0 + C 1 e t/τ cos(ω o t + θ), (a) (3 points) Without directly calculating θ(t), what must be the value of the constant C 0? Justify your answer with one or two sentences. (b) (4 points) Determine τ and ω o in terms of the controller gains K 1 and K 2. 5
18 2. (4 points) Given that the closedloop system is BIBO stable, determine the velocity error constant in terms of the controller gains K 1 and K (4 points) Determine the sensitivity of the velocity error constant with respect to the controller gain K 1. 6
19 4. (10 points) Choose the controller gains K 1 and K 2 that achieve the following design specifications: Peak overshoot of 16.3% Error velocity constant of 20 sec 1. 7
20 Problem 3: (25 Points) The root locus for the closedloop system in Figure 4 is shown in Figure 5 as the compensator gain K is varied from zero towards infinity. Figure 4: Closedloop system with proportional control gain K. 1.5 Root Locus Imaginary Axis Real Axis Figure 5: Root locus as the proportional gain K is varied from zero towards infinity. 1. (7 points) Estimate the value of proportional gain K for which the closedloop poles breakin to the real axis. 8
21 2. (18 points) Once again consider the control system in Figure 4, and whose root locus is shown again in Figure 6. Ignoring the effect of the zeros, an engineer chooses the value of K so that the location of the complex conjugate poles yields a settling time 9.2 sec. 1.5 Root Locus Imaginary Axis Real Axis Figure 6: Root locus as the proportional gain K is varied from zero towards infinity. (a) (8 points) What value of gain K did the engineer choose? (b) (5 points) For the chosen value of K, estimate the percent overshoot of the closedloop unitstep response. (c) (5 points) For the chosen value of K, estimate the risetime of the closedloop unitstep response. 9
22 Problem 4: (25 Points) 1. (18 points) Consider the closedloop system in Figure 7 where Gp(s) = s s 2 (s + 6) Figure 7: Closedloop system with proportional control gain K. Neatly sketch by hand root locus in Figure 8 as the proportional gain is varied from zero towards infinity. Be sure to specify breakaway and breakin points, the gain at the breakaway and breakin points, asymptotes, arrival and departure angles, and imaginaryaxis crossings, if any. Root Locus 3 2 Imaginary Axis Real Axis Figure 8: Root locus as the proportional gain K is varied from zero towards infinity. 10
23 11
24 2. (7 points) Write an mfile that constructs the root locus for the system in part 1, draws lines of constant ζ = 0.9 and ω n = 3 rad/s on the plot, and allows the user to determine the gain for a desired pole position by selecting a point in the graphics window. 12
25
If 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 informationProblem Weight Score Total 100
EE 350 EXAM IV 15 December 2010 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 information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More 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 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 informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More informationSchool of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See
More informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
More informationSECTION 5: ROOT LOCUS ANALYSIS
SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path
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 informationProblems XO («) splane. splane *~8 X 5. id) X splane. splane. * Xtg) FIGURE P8.1. jplane. JO) k JO)
Problems 1. For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4] *~8 XO
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 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 informationEE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation
EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginallystable
More informationD G 2 H + + D 2
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.302 Feedback Systems Final Exam May 21, 2007 180 minutes Johnson Ice Rink 1. This examination consists
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 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 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 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 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 informationEE C128 / ME C134 Midterm Fall 2014
EE C128 / ME C134 Midterm Fall 2014 October 16, 2014 Your PRINTED FULL NAME Your STUDENT ID NUMBER Number of additional sheets 1. No computers, no tablets, no connected device (phone etc.) 2. Pocket calculator
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 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
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 informationEE C128 / ME C134 Final Exam Fall 2014
EE C128 / ME C134 Final Exam Fall 2014 December 19, 2014 Your PRINTED FULL NAME Your STUDENT ID NUMBER Number of additional sheets 1. No computers, no tablets, no connected device (phone etc.) 2. Pocket
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 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 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 informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
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 informationEECS C128/ ME C134 Final Wed. Dec. 14, am. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C128/ ME C134 Final Wed. Dec. 14, 211 8111 am Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth 1 points total. Problem Points Score 1 16 2 12
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 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 informationProblem Weight Total 100
EE 350 Problem Set 3 Cover Sheet Fall 2016 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: Submission deadlines: Turn in the written solutions by 4:00 pm on Tuesday September
More informationSan Jose State University Department of Electrical Engineering. Exam 2 Solution. EE 098MIT 6.002x Fall 2012
San Jose State University Department of Electrical Engineering Exam Solution EE 98MIT 6.x Fall 1 losed Book, losed Notes, and no electronic devices. Instructions: There are six problems. Interpretation
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
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 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 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 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 informationMA 113 Calculus I Fall 2009 Exam 3 November 17, 2009
MA 113 Calculus I Fall 2009 Exam 3 November 17, 2009 Answer all of the questions 17 and two of the questions 810. Please indicate which problem is not to be graded by crossing through its number in the
More informationEECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8 am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total
More 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 informationEE402  Discrete Time Systems Spring Lecture 10
EE402  Discrete Time Systems Spring 208 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 0.. Root Locus For continuous time systems the root locus diagram illustrates the location of roots/poles of a closed
More informationCONTROL * ~ SYSTEMS ENGINEERING
CONTROL * ~ SYSTEMS ENGINEERING H Fourth Edition NormanS. Nise California State Polytechnic University, Pomona JOHN WILEY& SONS, INC. Contents 1. Introduction 1 1.1 Introduction, 2 1.2 A History of Control
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 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 informationGEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM. COURSE: ECE 3084A (Prof. Michaels)
GEORGIA INSTITUTE OF TECHNOLOGY SCHOOL of ELECTRICAL & COMPUTER ENGINEERING FINAL EXAM DATE: 30Apr14 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
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 informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More 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 informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More 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 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 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 informationSECTION 8: ROOTLOCUS ANALYSIS. ESE 499 Feedback Control Systems
SECTION 8: ROOTLOCUS ANALYSIS ESE 499 Feedback Control Systems 2 Introduction Introduction 3 Consider a general feedback system: Closedloop transfer function is KKKK ss TT ss = 1 + KKKK ss HH ss GG ss
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 informationRoot locus Analysis. P.S. Gandhi Mechanical Engineering IIT Bombay. Acknowledgements: Mr Chaitanya, SYSCON 07
Root locus Analysis P.S. Gandhi Mechanical Engineering IIT Bombay Acknowledgements: Mr Chaitanya, SYSCON 07 Recap R(t) + _ k p + k s d 1 s( s+ a) C(t) For the above system the closed loop transfer function
More informationMA 113 Calculus I Fall 2016 Exam 3 Tuesday, November 15, True/False 1 T F 2 T F 3 T F 4 T F 5 T F. Name: Section:
MA 113 Calculus I Fall 2016 Exam 3 Tuesday, November 15, 2016 Name: Section: Last 4 digits of student ID #: This exam has five true/false questions (two points each), ten multiple choice questions (five
More informationMA 113 Calculus I Fall 2015 Exam 3 Tuesday, 17 November Multiple Choice Answers. Question
MA 11 Calculus I Fall 2015 Exam Tuesday, 17 November 2015 Name: Section: Last 4 digits of student ID #: This exam has ten multiple choice questions (five points each) and five free response questions (ten
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 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 information6.1 Sketch the zdomain root locus and find the critical gain for the following systems K., the closedloop characteristic equation is K + z 0.
6. Sketch the zdomain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More informationLab # 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 informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus
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 informationMODERN CONTROL SYSTEMS
MODERN CONTROL SYSTEMS Lecture 1 Root Locu Emam Fathy Department of Electrical and Control Engineering email: emfmz@aat.edu http://www.aat.edu/cv.php?dip_unit=346&er=68525 1 Introduction What i root locu?
More informationAutomatic Control A. A.A. 2016/2017 July 7, Corso di Laurea Magistrale in Ingegneria Meccanica. Prof. Luca Bascetta.
Corso di Laurea Magistrale in Ingegneria Meccanica Automatic Control A Prof. Luca Bascetta A.A. 2016/2017 July 7, 2017 Name: Surname: University ID number: Signature: This file consists of 8 pages (including
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 informationCONTROL SYSTEMS ENGINEERING Sixth Edition International Student Version
CONTROL SYSTEMS ENGINEERING Sixth Edition International Student Version Norman S. Nise California State Polytechnic University, Pomona John Wiley fir Sons, Inc. Contents PREFACE, vii 1. INTRODUCTION, 1
More informationPhysics 1252 Exam #3E (MakeUp)
Physics 1252 Exam #3E (MakeUp) Instructions: This is a closedbook, closednotes exam. You are allowed to use a clean printout of your formula sheet, any scientific calculator, and a ruler. Do not write
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 information1. This is a CLOSED BOOK, CLOSED NOTES exam. A list of equations for your reference is provided below.
_ Instructions 1. This is a CLOSED BOOK, CLOSED NOTES exam. A list of equations for your reference is provided below. 2. Begin each problem on the printed sheet and continue on the lined paper provided.
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 informationMath 131 Exam 3 November 29, :008:30 p.m.
Name (Last, First) ID # Signature Lecturer Section # university of massachusetts amherst department of mathematics and statistics Math 131 Exam 3 November 29, 2006 7:008:30 p.m. Instructions Turn off
More informationMethods for analysis and control of. Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.lag.ensieg.inpg.fr/sename Lead Lag 17th March
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 informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
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 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 informationIMPORTANT. Read these directions carefully: You do not need to show work for the Multiple Choice questions.
Physics 208: Electricity and Magnetism Common Exam 3, November 14 th 2016 Print your name neatly: First name: Last name: Sign your name: Please fill in your Student ID number (UIN): _   Your classroom
More informationMA 113 Calculus I Fall 2013 Exam 3 Tuesday, 19 November Multiple Choice Answers. Question
MA 113 Calculus I Fall 2013 Exam 3 Tuesday, 19 November 2013 Name: Section: Last 4 digits of student ID #: This exam has ten multiple choice questions (five points each) and five free response questions
More informationRoot Locus. Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering
Root Locus Signals and Systems: 3C1 Control Systems Handout 3 Dr. David Corrigan Electronic and Electrical Engineering corrigad@tcd.ie Recall, the example of the PI controller car cruise control system.
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 informationMathematics 131 Final Exam 02 May 2013
Mathematics 3 Final Exam 0 May 03 Directions: This exam should consist of twelve multiple choice questions and four handgraded questions. Multiple choice questions are worth five points apiece. The first
More informationME 375 EXAM #1 Friday, March 13, 2015 SOLUTION
ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned
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 informationEE 4443/5329. LAB 3: Control of Industrial Systems. Simulation and Hardware Control (PID Design) The Torsion Disks. (ECP SystemsModel: 205)
EE 4443/539 LAB 3: Control of Industrial Systems Simulation and Hardware Control (PID Design) The Torsion Disks (ECP SystemsModel: 05) Compiled by: Nitin Swamy Email: nswamy@lakeshore.uta.edu Email: okuljaca@lakeshore.uta.edu
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 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 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 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. 20142016 Harry Dankowicz, unless otherwise
More informationMath Fall 2012 Exam 1 UMKC. Name. Student ID. Instructions: (a) The use of laptop or computer is prohibited.
Math  Fall Exam UMKC Name Student ID Instructions: (a) The use of laptop or computer is prohibited. (b) Total time allowed for the exam: 75 min. (c) Calculators may not be shared. (d) For Part (Problems
More informationNAME: ht () 1 2π. Hj0 ( ) dω Find the value of BW for the system having the following impulse response.
University of California at Berkeley Department of Electrical Engineering and Computer Sciences Professor J. M. Kahn, EECS 120, Fall 1998 Final Examination, Wednesday, December 16, 1998, 58 pm NAME: 1.
More information(a) Draw the coordinate system you are using and draw the free body diagram of the block during rotation with constant speed.
4[25 pts.] A block of mass m is placed at the side surface of a cone. The cone can rotate about an axis through its center so that the block can make circular motion. The static friction coefficient between
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 informationDigital Control Systems State Feedback Control
Digital Control Systems State Feedback Control Illustrating the Effects of ClosedLoop Eigenvalue Location and Control Saturation for a Stable OpenLoop System ContinuousTime System Gs () Y() s 1 = =
More informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
More informationComprehensive Exam in Real Analysis Fall 2006 Thursday September 14, :0011:30am INSTRUCTIONS
Exam Packet # Comprehensive Exam in Real Analysis Fall 2006 Thursday September 14, 2006 9:0011:30am Name (please print): Student ID: INSTRUCTIONS (1) The examination is divided into three sections to
More informationMultiple Choice Answers. MA 110 Precalculus Spring 2015 Exam 3 14 April Question
MA 110 Precalculus Spring 2015 Exam 3 14 April 2015 Name: Section: Last 4 digits of student ID #: This exam has ten multiple choice questions (four points each) and five free response questions (seven
More informationMethods for analysis and control of dynamical systems Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.gipsalab.fr/ o.sename 5th February 2015 Outline
More information