Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...


 Cameron Ramsey
 3 years ago
 Views:
Transcription
1 Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Given and family names solutions Student ID number Signature Duration: 2 minutes Number of exercises/questions: 5 Maximum score: 32 pts. All solutions must be written (either in English or Italian) within the available blank space (and not on additional paper sheets) It is forbidden to use any electronic device, apart from a non programmable and non graphic calculator Books and lecture material are not allowed It is not allowed to leave the class room within the first 3 minutes
2 This page has been intentionally left blank (to be used only if necessary). 2
3 Question (for all students) The unit step responses of the output of three closedloop systems are shown in the following: Output.6 Output.6 Output Time (s) Time (s) Time (s) Knowing that the three loop transfer functions are the following ones: L (s) = s, L 2 (s) = s ( +.s), L 3 (s) = 3 + 3s, match each loop transfer function with the corresponding closedloop response. Solution: The first two step responses have zero steady state error, whilst the third does not. The sole transfer function with type is the third one. Therefore, the last step response corresponds to L 3. The first two step responses differ from the settling time and from the presence/absence of overshoots. L surely produces a smooth response and hence corresponds to the second response. Finally, the first response corresponds to L 2. 3
4 Exercise 2 (for all students) Consider the closedloop system below, where G (s) = s/3 + s/3. w R(s) u G(s) d + y + + n Design the controller R (s) that satisfies the following specifications: a null steadystate error for a step reference, i.e. e = when w (t) = step (t); a phase margin of at least 6 deg, i.e. ψ m 6 deg; a crossover frequency between and rad/s, i.e. ω c ; attenuation of d (t) on the output of at least 3 db in the bandwidth ω 3 rad/s; attenuation of n (t) on the output of at least 4 db in the bandwidth ω rad/s. Finally, propose a suitable sampling time for the digital implementation of the control law. Solution: The static requirement on the steadystate error is satisfied with R (s) = µ R. We can then set s R (s) = /s, obtaining L (s) = s/3 s + s/3 which however does not satisfy all the other requirements. By selecting: which corresponds to L (s) = s R (s) = s one obtains the Bode diagrams reported in the next page. ( + s/) ( s/3) ( + s/3) ( + s/3) 2 ( + s/3) ( + s/) ( + s/3) ( + s/3) 2 4
5 Magnitude (db) Bode Diagram 35 Phase (deg) Frequency (rad/s) The resulting loop transfer function now satisfies all the requirements with ω c = 32 rad/s, ψ m = 59.9 deg. As for the sampling frequency for the digital implementation we can select ω N = ω c = 32 rad/s which corresponds to T s = π/ω N ms (which corresponds to a phase margin lag of 9 deg). 5
6 Exercise 3 (for all students) Consider the following transfer function G (s) = s (s + ) (s + 2) (s + 3) Using the root locus method discuss whether it is possible or not to design a proportional controller ρ R which stabilises the closedloop system. If so, determine the admissible range of ρ such that the closedloop system is asymptotically stable. Solution: The direct and inverse loci of ρg (s) are sketched in the following: 2 Root Locus 2 Root Locus 5.5 Imaginary Axis (seconds  ) Imaginary Axis (seconds  ) Real Axis (seconds  ) Real Axis (seconds  ) It is therefore possible to implement such a controller. The range of ρ such that the closedloop system is asymptotically stable is of the kind ρ min < ρ < ρ max, where ρ max = 2 3 = 6 while for ρ min < we can evaluate the inverse locus in ρ min = =
7 Exercise 4.A (for students of Automatic Control A, only) Consider a servomechanism with K = Nm/A (torque constant), J m =.2 kgm 2 (motor inertia), n = (reduction ratio), J l = 2 kgm 2 (load inertia). A current impulse exhibits the behaviour shown in the following figure. 5 Impulse Response Amplitude Time (seconds) Based on the acquired data, estimate the the elasticity of the transmission K el and propose a proper tuning for the velocity controller that maximises the damping of closedloop poles. Solution: The oscillation has a period of 2 ms and hence corresponds to ω p = 34.6 rad/s. The inertia ratio is ρ = J lr /J m = and finally the locked frequency results ω z = ω p / + ρ = rad/s. Finally the elasticity of the transmission can be computed as K el = ω 2 zj lr = 987 Nm/rad. As for the tuning of the PI velocity controller we can select T iv = /ω z = 4.5 ms and K pv =.7ω z / (J m + J lr ) =.622 A/rad. 7
8 Exercise 4.B (for students of Automatic Control B, only) The following controller R (s) = +.s + s has been design for a given LTI system in order to obtain a crossover frequency of ω c = rad/s. Design a suitable sampling time T s, compute the corresponding discretisation using the Tustin (bilinear) method, and finally write the corresponding pseudocode (difference equation, u (k) =... ). Solution: In order to meet the Shannon s criterion we can set ω N = ω c and T s = π/ω N =.34 s. In order to obtain the discrete time realisation of the controller we need to substitute s with 2 T s z z +, obtaining R (z) = +. 2 z T s z + z (z ).6369z z + 2 = = = z z (z ) z z T s z + As for the implementation, we have u (k) (.7286z ) = e (k) ( z ) and therefore u (k) =.7286u (k ) +.222e (k) e (k ) 8
9 Exercise 5.A (for students of Automatic Control A, only) A closedloop system is characterised by the following loop transfer function L (s) = s ( +.s) compute the crossover frequency and the corresponding phase margin. Then design a suitable fistorder analog antialiasing filter which contribute in a phase margin lag not higher than 5 deg and draw the corresponding circuit. Solution: The crossover frequency is approximately ω c = rad/s while the corresponding phase margin is of 84.3 deg. A first order antialiasing filter is a transfer function of the form: H (s) = + st Therefore, the parameter T > must be tuned in order to have a phase lag not higher than 5 deg at rad/s and therefore: H (j) = atant 5, T <.875 s We can select T =. s, while a possible analog implementation is the following one C V R R R v out which has the following transfer function H (s) = C = µf.. One could select, for example, R = kω and + src 9
10 Exercise 5.B (for students of Automatic Control B, only) Given the following continuous time signal y (t) = + sin (t) + sin (7t), answer to the following questions: find the maximum sampling time T s for which there is no aliasing effect; assuming T s =.5 s, find the frequency of the aliasing harmonics; Solution: The highest frequency harmonic happen at ω max = 7 rad/s, therefore in order to satisfy the Shannon s criterion, T s < π =.4488 s. ωmax Clearly T s =.5 s does not satisfy the Shannon s criterion. Therefore, an aliasing harmonic will be present on the acquired of signal. The corresponding sampling frequency is ω s = 2π/T s = 4π and therefore the aliasing harmonic is at 7 ω s /2 =.768 rad/s.
Automatic 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 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 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 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 informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More 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 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 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 informationIndex. Index. More information. in this web service Cambridge University Press
Atype elements, 4 7, 18, 31, 168, 198, 202, 219, 220, 222, 225 Atype variables. See Across variable ac current, 172, 251 ac induction motor, 251 Acceleration rotational, 30 translational, 16 Accumulator,
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 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 informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More 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 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 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 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 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 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 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 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 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 informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
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 information6.302 Feedback Systems
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.302 Feedback Systems Fall Term 2005 Issued : November 18, 2005 Lab 2 Series Compensation in Practice Due
More informationExam. 135 minutes + 15 minutes reading time
Exam January 23, 27 Control Systems I (559L) Prof. Emilio Frazzoli Exam Exam Duration: 35 minutes + 5 minutes reading time Number of Problems: 45 Number of Points: 53 Permitted aids: Important: 4 pages
More informationAndrea Zanchettin Automatic Control AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Spring Semester, Linear systems (frequency domain)
1 AUTOMATIC CONTROL Andrea M. Zanchettin, PhD Spring Semester, 2018 Linear systems (frequency domain) 2 Motivations Consider an LTI system Thanks to the Lagrange s formula we can compute the motion of
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 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 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 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 informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 9. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
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 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 informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationDepartment of Electronics and Instrumentation Engineering M. E CONTROL AND INSTRUMENTATION ENGINEERING CL7101 CONTROL SYSTEM DESIGN Unit I BASICS AND ROOTLOCUS DESIGN PARTA (2 marks) 1. What are the
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 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 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 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 informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
More 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 informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
More 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 informationECE Circuit Theory. Final Examination. December 5, 2008
ECE 212 H1F Pg 1 of 12 ECE 212  Circuit Theory Final Examination December 5, 2008 1. Policy: closed book, calculators allowed. Show all work. 2. Work in the provided space. 3. The exam has 3 problems
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 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 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 505900 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 informationAndrea Zanchettin Automatic Control 1 AUTOMATIC CONTROL. Andrea M. Zanchettin, PhD Spring Semester, Discrete time linear systems
Andrea Zanchettin Automatic Control 1 AUTOMATIC CONTROL Andrea M. Zanchettin, PhD Spring Semester, 2018 Discrete time linear systems Andrea Zanchettin Automatic Control 2 Discrete time linear systems Modern
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 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 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 informationECE 388 Automatic Control
Lead Compensator and PID Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage: http://ece388.cankaya.edu.tr
More informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
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 information1 An Overview and Brief History of Feedback Control 1. 2 Dynamic Models 23. Contents. Preface. xiii
Contents 1 An Overview and Brief History of Feedback Control 1 A Perspective on Feedback Control 1 Chapter Overview 2 1.1 A Simple Feedback System 3 1.2 A First Analysis of Feedback 6 1.3 Feedback System
More informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
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: 09Dec13 COURSE: ECE 3084A (Prof. Michaels) NAME: STUDENT #: LAST, FIRST Write your name on the front page
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 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 information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
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 informationQuanser NIELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual
Quanser NIELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control DC Motor Control Trainer (DCMCT) Student Manual Table of Contents 1 Laboratory Objectives1 2 References1 3 DCMCT Plant
More 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 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 informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
More informationADMISSION TEST INDUSTRIAL AUTOMATION ENGINEERING
UNIVERSITÀ DEGLI STUDI DI PAVIA ADMISSION TEST INDUSTRIAL AUTOMATION ENGINEERING September 26, 2016 The candidates are required to answer the following multiple choice test which includes 30 questions;
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More 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 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 informationFrequency Response Analysis
Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions
More informationControl of Electromechanical Systems
Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance
More informationChapter 6  Solved Problems
Chapter 6  Solved Problems Solved Problem 6.. Contributed by  James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the ZN tuning strategy for the nominal
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 information10ES43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A
10ES43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
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 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 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 informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
More informationWind Turbine Control
Wind Turbine Control W. E. Leithead University of Strathclyde, Glasgow Supergen Student Workshop 1 Outline 1. Introduction 2. Control Basics 3. General Control Objectives 4. Constant Speed Pitch Regulated
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 informationMATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 3 PERIODIC FUNCTIONS
MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 3 PERIODIC FUNCTIONS This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals.
More informationExam. 135 minutes, 15 minutes reading time
Exam August 15, 2017 Control Systems I (151059100L) Prof Emilio Frazzoli Exam Exam Duration: 135 minutes, 15 minutes reading time Number of Problems: 44 Number of Points: 52 Permitted aids: Important:
More informationFATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY
FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY Senkottai Village, Madurai Sivagangai Main Road, Madurai  625 020. An ISO 9001:2008 Certified Institution DEPARTMENT OF ELECTRONICS AND COMMUNICATION
More informationEE3CL4: Introduction to Linear Control Systems
1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We
More 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 informationFrequency response. Pavel Máša  XE31EO2. XE31EO2 Lecture11. Pavel Máša  XE31EO2  Frequency response
Frequency response XE3EO2 Lecture Pavel Máša  Frequency response INTRODUCTION Frequency response describe frequency dependence of output to input voltage magnitude ratio and its phase shift as a function
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 informationTime Response of Systems
Chapter 0 Time Response of Systems 0. Some Standard Time Responses Let us try to get some impulse time responses just by inspection: Poles F (s) f(t) splane Time response p =0 s p =0,p 2 =0 s 2 t p =
More informationEC6405  CONTROL SYSTEM ENGINEERING Questions and Answers Unit  I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
More informationLecture 7:Time Response PoleZero 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 FirstOrder Specs: Step : Pole Real inputs contain
More informationBoise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain
Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type servomechanism:
More informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pickandplace robot to move the link of a robot between two positions. Usually
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 informationECE Branch GATE Paper The order of the differential equation + + = is (A) 1 (B) 2
Question 1 Question 20 carry one mark each. 1. The order of the differential equation + + = is (A) 1 (B) 2 (C) 3 (D) 4 2. The Fourier series of a real periodic function has only P. Cosine terms if it is
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 information