ME 375 Final Examination Thursday, May 7, 2015 SOLUTION
|
|
- Lorraine Gibson
- 6 years ago
- Views:
Transcription
1 ME 375 Final Examination Thursday, May 7, 2015 SOLUTION
2 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 electro-mechanical equations for an electric motor (ignoring armature inductance and armature damping) are given by: J!ω K T i a τ L i a + K T ω e i t where ω is the rotation rate of the motor, J is the mass moment of inertia of the armature, K T is the motor constant, i a is the armature current, τ L is the applied motor torque, e i ( t) is the input voltage applied to the armature and is the armature resistance. An electric motor governed by the above equations is used to propel a cart along a horizontal roadway. In this application, a disk attached to the output shaft of the motor drives one of the rear wheels of the cart through a no-slip frictional contact. The cart moves with no slipping of the wheels on the roadway and experiences a drag force of f D cv to the right as it moves to the left, where v is the speed of the cart to the left. Let M represent the total mass of the motor and cart. Ignore the mass of all of the wheels. For this problem: a) Draw free body diagrams of the driven rear wheel and of the total system. From these develop the differential equation of motion of the cart in terms of the speed v. b) Develop the transfer function relating the input voltage to the cart speed, G( s) V ( s) / E i ( s). c) The motor is given a step input voltage e i (t) e 0 h(t). Determine the steady-state speed attained by the cart as it moves along the roadway.
3 SOLUTION r F D O F M Oy O O x Let W be the radius of the rear wheels. From FBD of the wheel: f f M O F M W f W 0 f F M From the FBD of the cart/wheel: F f F D M!v F M τ L M!v + cv r (1) Taking LT of (1) gives: T L (s) r( Ms + c)v (s) (2) Taking LT of the motor equations gives: JsΩ K T I a T L (3) E i I a + K T Ω (4) For kinematics, we have: ω W r ω Ω W r Ω W W v W ω W V W Ω W rω Ω V / r (5) where ω W is the angular speed of the driven wheel. Combining above gives: (4) I a E i K T Ω (3),(5) & (6) Js V r K T ( E i r K T V ) r 2 ( Ms + c)v E i K T V / r r ( Ms + c )V JsV K T ( J + r 2 M )s + r 2 2 c + K T V K T re i J + Mr 2 And: G(s) V (s) E i (s) K T r / ( J + Mr 2 )s + r 2 c + K 2 T / From FVT: v ss lim sv (s) s 0 lim s e 0 G(s) K G(0) T re 0 s 0 s r 2 2 c + K T!v + r2 c + K T 2 / v K T re i (t) / (6)
4 POBLEM 2 (25%) plant ( s) K ( s) Y ( s) Consider the proportional feedback control system shown above. The plant transfer function s has a static gain of 10, has no zeros and has the three poles shown below. a) Determine the plant transfer function s. b) For the closed-loop root locus with K 0 : i. Determine the K asymptotes and their real-axis intersections. ii. Determine the breakaway/breakin (re-entry) points on the real axis. iii. Determine the departure angles from the open-loop poles. iv. Determine the values of K, if any, at which the root locus crosses into the right-half plane (unstable). v. Sketch the root locus on the plot provided below. vi. For what range of values for the gain K, if any, would the closed-loop response have a 2% settling time t s < 0.5AND a percent overshoot of %OS < 4% for a step input? K θ K 0 p2 p 2 11/ 3 p 3 K 0 θ p2 60 K 0 60 p K
5 SOLUTION A (s) ( s p 1 ) s p 2 A ( s +1 ) s + 5 Since G(0) 10, we have: ( s p 3 ) A ( s +1) ( s j) ( s j) (0) 10 A 29 A 290 A s 3 +11s s + 29 N(s) D(s) ule #2: The L has starting points at the three open-loop poles and end points at infinity. ule #4: The L exists on the real axis to the left of the real pole p 1 1. ule #5: The asymptotes have real axis intercepts and angles of: p i z i 1+ ( j) + ( 5 2 j) 0 σ 0 11 N P N z θ k ( 2k 1) ( 2k 1) 180 ( 2k 1)60 60,180,300 N P N z 3 0 ule #6: BA/E points on the real axis are found from: N dd ds D dn A( 3s s + 39) 0 ds s 1,2 22 ± 222 (4)(3)(39) (2)(3) 22 ± 4 6 3, 13 3 ule #7: From the general equation: ( s z 1 ) ( s p 1 ) + ( s p 2 ) 180 we can write the following for OL pole p 1 : j +1 +θ p2 + ( p 2 p 3 ) p 2 p 1 +θ p2 + ( j j) +θ p2 + ( 4 j) j 180 tan θ p2 + tan θ p θ p2 63.4
6 ule #8: The characteristic equation for the closed-loop system is: 0 D(s) + KN(s) s 3 +11s s K Substituting s jω into the CLCE gives: 3 +11( jω ) ( jω ) K 0 jω jω 3 11ω 2 + j(39ω ) K K 11ω 2 + j( 39 ω 2 )ω Imag : ( 39 ω 2 )ω 0 ω 2 0,39 eal : K 11ω 2 0 K ω ,1.38 For t s ζω n ζω n 8. It is not possible to have all of the roots to the left of 8. Therefore, it is not possible to meet the settling time criteria.
7 POBLEM 3 (25%) A plant with a transfer function s 1 s s + 1 is to be controlled with unity PD feedback control where the controller has a transfer function G C ( s) K P + K D s. a) Derive the closed-loop transfer function and closed-loop characteristic equation for this control system. b) Determine the gain values K P and K D corresponding to a 2% settling time of t s 4 and a %OS 4 for a step input. c) If K P 4, what value of the gain K D produces the smallest possible 2% settling time for a step input? SOLUTION The CL transfer function and characteristic equation are given by: G CL (s) G C 1+ G C D CL (s) 1+ G C 1+ K P + K D s s s +1 0 s 2 + ( 1+ K D )s + K P 0 respectively. The general form of the characteristic equation for a second-order system is: s 2 + 2ζω n s + ω n 2 0 For %OS 4 andt s 4, we have: %OS 100e ζπ / 1 ζ 2 ζ + π 2 ln %OS / 100 ln 2 %OS / 100 t s ζω ζω n 1 ω n n Balancing coefficients in the CLCE s: K P ω n K D 2ζω n K D 2ζω n 1 2 ( 1) π 2 ln 0.04 ln The smallest possible settling time occurs when ζ 1. Since ω n K P 4 2, we can write: K D 2ζω n
8 POBLEM 4 PAT A (10%) Consider the electrical circuit shown below. Develop an expression for the impedance V i ( s) I ( s) Z s. Express your final answer as a single fraction. a b i v i (t) + - L d e C SOLUTION Inductor be and capacitor de are connected in series. That connection is in parallel with resistor bd. In turn, that is in series with resistor ab. Therefore: Z be Ls and: Z de 1 Cs Z bd Z bde Z bd Z be + Z de + 1 Ls +1/ Cs 1 + Cs LCs 2 +1 LCs Cs LCs 2 + LCs 2 Z bde + LCs Cs which leads to: LCs 2 + Z(s) Z a + Z bde + LCs 2 + Cs +1 2LCs Cs + 2 LCs 2 + Cs +1
9 POBLEM 4 (continued) PAT B (10%) s plant K ( s) Y ( s) The closed-loop transfer function for the above system has the following characteristic equation: D CL (s) s 4 + 5Ks 3 + (3+ K)s 2 + 6s K 0 a) Determine the plant transfer function (s). b) Determine the magnitude of the steady-state response of the closed-loop system to a unit harmonic input r(t) sin5t for K 1. SOLUTION 0 s 4 + 5Ks 3 + (3+ K)s 2 + 6s K s 4 + 3s 2 + 6s K 5s 3 + s s 3 + s K s 4 + 3s 2 + 6s K (s) The CL transfer function is written as: For K 1: G CL (s) K (s) K 5s 3 + s K (s) s 4 + 3s 2 + 6s K 5s 3 + s s 3 + s 2 +1 G CL (s) s 4 + 5s 3 + 4s 2 + 6s + 5 Magnitude of steady-state response for harmonic input: G CL (5 j) 5(5 j) 3 + (5 j) 2 +1 (5 j) 4 + 5(5 j) 3 + 4(5 j) 2 + 6(5 j) j j
10 POBLEM 4 (continued) PAT C (5%) G 1 (s) s Y ( s) G 2 (s) Consider the feedback control system above, with: s 2 +1 G 1 (s) s 3 + 3s +1 G 2 (s) s s + 4 Determine the single input-output differential equation governing the output y(t) for an input r(t). What is the order of the closed-loop system? SOLUTION From block diagram: Y G 1 G 2 Y Y G 1 (s 2 +1) / (s 3 + 3s +1) 1+ G 1 G 2 1+ s(s 2 +1) / (s 3 + 3s +1)(s + 4) (s 2 +1)(s + 4) (s 3 + 3s +1)(s + 4) + s(s 2 +1) s 3 + 4s 2 + s + 4 s 4 + 5s 3 + 3s 2 +14s + 4 ( s 4 + 5s 3 + 3s 2 +14s + 4)Y ( s 3 + 4s 2 + s + 4) Performing an inverse LT on the above:!!!! y + 5!!! y + 3!!y +14!y + 4y!!! r (t) + 4!! r(t) +!r(t) + 4r(t) Fourth-order system.
EE 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 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 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 informationHomework 7 - Solutions
Homework 7 - Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationSoftware Engineering 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 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 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 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 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 s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
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 informationTransient Response of a Second-Order System
Transient Response of a Second-Order System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a well-behaved closed-loop
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 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 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 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 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(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. 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 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 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 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 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 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 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: Steady-State Error Unit 7: Root Locus Techniques
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 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 informationECEN 420 LINEAR CONTROL SYSTEMS. Lecture 6 Mathematical Representation of Physical Systems II 1/67
1/67 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 6 Mathematical Representation of Physical Systems II State Variable Models for Dynamic Systems u 1 u 2 u ṙ. Internal Variables x 1, x 2 x n y 1 y 2. y m Figure
More informationConventional Paper-I-2011 PART-A
Conventional Paper-I-0 PART-A.a Give five properties of static magnetic field intensity. What are the different methods by which it can be calculated? Write a Maxwell s equation relating this in integral
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 informationRotational Systems, Gears, and DC Servo Motors
Rotational Systems Rotational Systems, Gears, and DC Servo Motors Rotational systems behave exactly like translational systems, except that The state (angle) is denoted with rather than x (position) Inertia
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 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 pick-and-place robot to move the link of a robot between two positions. Usually
More informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More 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 informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More 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 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 Steady-state Error and Type 0, Type
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 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 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 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 informationSchool of Engineering Faculty of Built Environment, Engineering, Technology & Design
Module Name and Code : ENG60803 Real Time Instrumentation Semester and Year : Semester 5/6, Year 3 Lecture Number/ Week : Lecture 3, Week 3 Learning Outcome (s) : LO5 Module Co-ordinator/Tutor : Dr. Phang
More informationLab 3: Quanser Hardware and Proportional Control
Lab 3: Quanser Hardware and Proportional Control The worst wheel of the cart makes the most noise. Benjamin Franklin 1 Objectives The goal of this lab is to: 1. familiarize you with Quanser s QuaRC tools
More informationLecture 1 Root Locus
Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the
More informationCHAPTER 7 STEADY-STATE RESPONSE ANALYSES
CHAPTER 7 STEADY-STATE 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 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 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 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 Process Control
Introduction to Process Control For more visit :- www.mpgirnari.in By: M. P. Girnari (SSEC, Bhavnagar) For more visit:- www.mpgirnari.in 1 Contents: Introduction Process control Dynamics Stability The
More informationDesign of a Lead Compensator
Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc - Funded by MHRD The Lecture Contains Standard Forms of
More informationChapter three. Mathematical Modeling of mechanical end electrical systems. Laith Batarseh
Chapter three Mathematical Modeling of mechanical end electrical systems Laith Batarseh 1 Next Previous Mathematical Modeling of mechanical end electrical systems Dynamic system modeling Definition of
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 informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
More 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 informationExam. 135 minutes + 15 minutes reading time
Exam January 23, 27 Control Systems I (5-59-L) 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 6: Generalized and Controller Design Overview In this Lecture, you will learn: Generalized? What about changing OTHER parameters
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 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 informationChapter 2 SDOF Vibration Control 2.1 Transfer Function
Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. Kwang-Chun Ho kwangho@hansung.ac.kr Tel: 02-760-4253 Fax:02-760-4435 Introduction In this lesson, you will learn the following : The
More informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
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 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 linear-time-invariant
More informationRoot Locus Design Example #3
Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll
More informationLab 3: Model based Position Control of a Cart
I. Objective Lab 3: Model based Position Control of a Cart The goal of this lab is to help understand the methodology to design a controller using the given plant dynamics. Specifically, we would do position
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 informationIntroduction to Control (034040) lecture no. 2
Introduction to Control (034040) lecture no. 2 Leonid Mirkin Faculty of Mechanical Engineering Technion IIT Setup: Abstract control problem to begin with y P(s) u where P is a plant u is a control signal
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 informationLinear Control Systems Solution to Assignment #1
Linear Control Systems Solution to Assignment # Instructor: H. Karimi Issued: Mehr 0, 389 Due: Mehr 8, 389 Solution to Exercise. a) Using the superposition property of linear systems we can compute the
More informationJRE SCHOOL OF Engineering
JRE SCHOOL OF Engineering Class Test-1 Examinations September 2014 Subject Name Electromechanical Energy Conversion-II Subject Code EEE -501 Roll No. of Student Max Marks 30 Marks Max Duration 1 hour Date
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 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 informationMechatronics Engineering. Li Wen
Mechatronics Engineering Li Wen Bio-inspired robot-dc motor drive Unstable system Mirko Kovac,EPFL Modeling and simulation of the control system Problems 1. Why we establish mathematical model of the control
More informationSelf-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current.
Inductance Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current. Basis of the electrical circuit element called an
More informationLinear Systems Theory
ME 3253 Linear Systems Theory Review Class Overview and Introduction 1. How to build dynamic system model for physical system? 2. How to analyze the dynamic system? -- Time domain -- Frequency domain (Laplace
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 informationInductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits
Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the timevarying
More informationChapter 7 : Root Locus Technique
Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci
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 informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage 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, Force-current and Force-Voltage analogies..
More informationEE102 Homework 2, 3, and 4 Solutions
EE12 Prof. S. Boyd EE12 Homework 2, 3, and 4 Solutions 7. Some convolution systems. Consider a convolution system, y(t) = + u(t τ)h(τ) dτ, where h is a function called the kernel or impulse response of
More informationFirst and Second Order Circuits. Claudio Talarico, Gonzaga University Spring 2015
First and Second Order Circuits Claudio Talarico, Gonzaga University Spring 2015 Capacitors and Inductors intuition: bucket of charge q = Cv i = C dv dt Resist change of voltage DC open circuit Store voltage
More informationME 230: Kinematics and Dynamics Spring 2014 Section AD. Final Exam Review: Rigid Body Dynamics Practice Problem
ME 230: Kinematics and Dynamics Spring 2014 Section AD Final Exam Review: Rigid Body Dynamics Practice Problem 1. A rigid uniform flat disk of mass m, and radius R is moving in the plane towards a wall
More informationSolutions to Skill-Assessment Exercises
Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
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 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 informationImplementation Issues for the Virtual Spring
Implementation Issues for the Virtual Spring J. S. Freudenberg EECS 461 Embedded Control Systems 1 Introduction One of the tasks in Lab 4 is to attach the haptic wheel to a virtual reference position with
More information2.010 Fall 2000 Solution of Homework Assignment 1
2. Fall 2 Solution of Homework Assignment. Compact Disk Player. This is essentially a reprise of Problems and 2 from the Fall 999 2.3 Homework Assignment 7. t is included here to encourage you to review
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 informationChapter 32. Inductance
Chapter 32 Inductance Joseph Henry 1797 1878 American physicist First director of the Smithsonian Improved design of electromagnet Constructed one of the first motors Discovered self-inductance Unit of
More informationMECHATRONICS ENGINEERING TECHNOLOGY. Modeling a Servo Motor System
Modeling a Servo Motor System Definitions Motor: A device that receives a continuous (Analog) signal and operates continuously in time. Digital Controller: Discretizes the amplitude of the signal and also
More informationIntroduction to Controls
EE 474 Review Exam 1 Name Answer each of the questions. Show your work. Note were essay-type answers are requested. Answer with complete sentences. Incomplete sentences will count heavily against the grade.
More informationPD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada
PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop
More informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24-Jan-15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
More informationPHYSICS 218 Final Exam Fall, 2014
PHYSICS 18 Final Exam Fall, 014 Name: Signature: E-mail: Section Number: No calculators are allowed in the test. Be sure to put a box around your final answers and clearly indicate your work to your grader.
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 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 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 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 information