MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007


 Gwendolyn Burke
 4 years ago
 Views:
Transcription
1 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 m =5.kg, b =.767N sec/m, k = 4 N/m; m =.87 kg, b =8.9 N sec/m, k = 85 N/m. The system model schematic is shown again below for your convenience. Eigenvalues and Eigenvectors in state space representation The eigenvalues of A matrix are poles. Hence, the imaginary parts of the poles are the damped oscillation frequencies (ω d = ω n ζ ), and the real parts are ζω n. The eigenvectors of A matrix represent the modes of the system. Each element of a certain eigenvector indicates the way that state variables change when the system is excited with the corresponding damped natural frequency. For example, if an eigenvector is [ ] T, then the two state variables move in opposite way and the ratio of their magnitudes is. The real and imaginary parts of the eigenvalues represent decay rate (or settling time) and natural frequency. For the complex eigenvector, you may interpret that the real and imaginary part correspond to cos and sin components of the response. Since cos θ = sin(θ + π/) and exp {jπ/} = j, the real part represents the magnitude of cos(ωt) motion, and the imaginary part represents the magnitude of sin(ωt) motion. To give you an easy example, let s consider the.4 Tower system without damping. Then the system matrix is A = , the eigenvectors are j.354 j.354 j.6 j v =, v =, v 3 =, v 4 =. j.3 j.3 j.73 j
2 and the eigenvalues are {±j.583, ±j3.487}. Note that the eigenvalues are purely imaginary because there is no damping. Let s look at v 3 corresponding to the lower natural frequency (rad/s). The first ( j.6) and third ( j.73) elements are the displacement of the tower and slider, respectively. So they move along the same direction in sin fashion (because they are imaginary) without phase delay. The second (.47) and fourth (.987) elements are velocity of the tower and the slider, respectively. They change with the same sign in cos fashion (because they are real) without phase delay. Since velocity is a time derivative of displacement, it totally makes sense that the first and third elements are purely imaginary while the second and fourth ones d are pure real, because (sin ωt) cos(ωt). dt Therefore at the lower natural frequency, the tower and slider move in phase with each other and they oscillate in sin(ωt) fashion. Their velocities oscillate also in phase with each other but in cos(ωt) fashion. Therefore the velocity has phase delay of π/ with respect to the displacement. (You would get the same conclusion from v 4, which corresponds to (rad/s).) If you look at v and v, then you would notice that the slider and tower displacements and velocities move in opposite direction. But you still have the same phase delay between the displacement and the velocity (one goes like sin, the other like cos.) We are now ready to attack the given problem, i.e. the tower and slider with damping. a) Eigenvectors and eigenvalues of matrix A. Answer: From A matrix given by (k + k )/m (b + b )/m k /m b /m A = k /m b /m k /m b /m =, the eigenvectors and eigenvalues are computed by Matlab.
3 The eigenvectors are.39 + j j j j.635 v =, v =,.39 j j j j j j.4 v 3 =, v4 =,.88 + j j and the eigenvalues are {.4 ± j.6, ± j3.858}. b) Justify the following statement: The.4 Tower has two modes of oscillation, one slow and one fast. Compute the damped and natural frequencies of oscillation of the two modes. Answer: We have 4 poles at s =.4 ± j.6, ± j Thus we have two damped natural frequencies (3.858,.6 (rad/s)) and two modes associated with these natural frequencies. Keep in mind that the imaginary part of the pole is ω d = ω n ζ and the real part is ζω n.(or you can use the geometrical relation that cos θ = ζ, where θ is the angle to the pole from the negative real axis.) Mode : ω d =3.858 (rad/s), ζ =.747, and ω n =4.48 (rad/s) Mode : ω d =.6 (rad/s), ζ =.44, and ω n =.36 (rad/s). c) Justify the following statement: In the slow mode the tower and slider mass oscillate in phase, while in the fast mode the tower and slider mass oscillate out of phase. Answer: The eigenvectors v 3 and v 4 of the slow mode are given by.67 + j j j j.4 v 3 =, v 4 = j j Let s look at the position of the tower(.67 ± j.74) and the slider (.88 ± j.659). The real parts have the same sign and the imaginary part does too. Also note that the imaginary part dominates because it is bigger then the real part. Hence, they move a kind of in phase. (Not exactly in phase because the real part is not zero.) 3
4 At the fast mode, the eigenvectors are,.39 + j j j j.635 v =, v =..39 j j Let s look at the first (.39 + j.37) and third (.39 j.376) elements in the same context. The first element is dominated by the real part, and the third one is dominated by the imaginary part. Thus we expect that the displacement of the tower and the slider have a phase delay about π/, which means that they are out of phase (or not in phase). (You would get the same conclusion from v, whose damped frequency is negative of the damped frequency of v.) Therefore, the tower and slider oscillate in phase in the slow mode, while out of phase in the fast mode. d) Which mode has been excited by the impulse input? Compare the damped frequency of oscillation of the modethat you think has been excited with the frequency of oscillation that you measure from the impulse response simulation. Answer: x Impulse Response tower displacement Amplitude 4 6 System: tower Time (sec):.874 Amplitude:.63 System: tower Time (sec):.9 Amplitude: Time (sec) π The measured damped frequency ω d = = (rad/s). It (.9.874) is very close to the higher mode frequency. We conclude that the second mode is predominantly excited by the impulse. e) Is the phase relationship between the tower and tower responses consistent with the oscillation mode that the system is in? Answer: 4
5 Impulse Response tower displacement slider displacement Amplitude System: tower Time (sec):.967 Amplitude:.3 System: tower Time (sec):.655 Amplitude: Time (sec) π The measured damped frequency: ω d = =.384 (rad/s). ( ) It is very close to the higher mode frequency and the second (faster) mode is excited. The displacement of the slider is delayed with respect to the displacement of the tower. Note that the position of the tower (blue) and the slider (green) are out of phase by about π/ as we expected in (c).. Nise chapter, problem 4 (page 777). Answer: G(s) = =. (s +)(s +3)(s +7) s 3 +s +3s + Please refer to the notes (Lecture 8) or chapter 3.5 in the Nise textbook about converting a transfer function to state space representation. q = q + u 3 [ ] y = q With a state variable feedback loop, A BK = (k + ) (k + 3) (k 3 + ) det(si (A BK)) = s 3 +(k 3 + )s +(k + 3)s + k 3 + =. The desired %OS is 5% and a settling time is.75 (s). From these desired values, ln(%os/) ζ = =.569, π +ln (%OS/) 5
6 and 4 ω n = =.373. (.75)ζ The desired characteristic equation is (s + p)(s +ζω n s + ω n )=(s + p)(s +.666s ) The third pole should be times as far from the imaginary axis as the dominant pole pair s = ± j8.83. Thus we choose p =53.3. The characteristic equation should be identical to the desired one, s 3 +( p)s + ( p)s p = s 3 +(k 3 + )s +(k + 3)s + k +. Thus k 3 + = p, k + 3 = p, k + = p. Using p = 53.3, we find k = 565.6, k = 643.9, and k 3 =53.. The beauty of the state space representation and state feedback is that you can observe and control individual state variables (e.g. position, velocity, acceleration, etc.) of systems. In complex systems, the state space representation may give you easier understanding of the system dynamics and better. 3. Nise chapter 9, problem (page 674). Answer: a. G(s) = s(s +)(s +4) G(jω)= = = s(s +)(s +4) s=jω s 3 +6s +8s s=jω 6ω j(8ω ω 3 ) = =. jω 3 6ω +8jω 6ω + j( ω 3 +8ω) 36ω 4 +(8ω ω 3 ) Hence, M(ω) =, 36ω 4 +(8ω ω 3 ) and ( ) ω 8 φ(ω) =tan. 6ω s +5 b. G(s) = s(s +)(s +4) G(jω) = 5+ jω 6ω j(8ω ω 3 ) = (5 + jω) 6ω + j( ω 3 +8ω) 36ω 4 +(8ω ω 3 ) = { 3ω + ω (8 ω )} + j { 6ω 3 5(8ω ω 3 )} 36ω 4 +(8ω ω 3 ) = ω (ω + ) + jω( ω 4). 36ω 4 +(8ω ω 3 ) 6
7 Hence, { ω (ω + )} + {ω( ω 4)} M(ω) =, 36ω 4 +(8ω ω 3 ) and ( ) ω 4 φ(ω) =tan. ω(ω + ) (s +3)(s +5) s +8s +5 c. G(s) = = s(s +)(s +4) s 3 +6s +8s G(jω) = Hence, = = (5 ω )+ j8ω = 6ω j(8ω ω 3 ) { (5 ω )+ j8ω } 6ω + j( ω 3 +8ω) 36ω 4 +(8ω ω 3 ) { 6ω (5 ω )+8ω(8ω ω 3 )} + j { 48ω 3 (5 ω )(8ω ω 3 )} 36ω 4 +(8ω ω 3 ) ω ( ω 6) + jω( ω 4 5ω ) 36ω 4 +(8ω ω 3 ) {ω ( ω 6)} + {ω( ω 4 5ω )} M(ω) =, 36ω 4 +(8ω ω 3 ) and ( ) ω 4 5ω φ = tan. ω( ω 6) 4. Nise chapter 9, problem (page 674). Use Matlab to generate the plots. Answer: From the analytical expression in question 3, the magnitude and phase of the system are plotted here without using bode command in Matlab. Note that the magnitude is not plotted in (db). a. G(s) = s(s +)(s +4) Magnitude 5 Phase(degree) 5 5 Frequency (rad/s) Frequency (rad/s) 7
8 s +5 b. G(s) = s(s +)(s +4) Magnitude 4 Frequency (rad/s) Phase(degree) 4 6 s +5 c. G(s) = s(s +)(s +4) 8 Frequency (rad/s) Magnitude 9 Frequency (rad/s) Phase(degree) 95 5 Frequency (rad/s) See the following results generated by built in bode command of Matlab. Note that the magnitude are plotted in (db). a. G(s) = s(s +)(s +4) 8
9 Bode Diagram 5 Magnitude (db) Phase (deg) Frequency (rad/sec) s +5 b. G(s) = s(s +)(s +4) Bode Diagram Magnitude (db) Phase (deg) 35 8 Frequency (rad/sec) s +5 c. G(s) = s(s +)(s +4) 4 Bode Diagram Magnitude (db) Phase (deg) 5 Frequency (rad/sec) 9
10 5. Nise chapter 9, problem 4 (page 674). a. G(s) = s(s +)(s +4) Answer: (Note that this is a color plot) s +5 b. G(s) = s(s +)(s +4) Answer: (Note that this is a color plot) s +5 c. G(s) = s(s +)(s +4) Answer: (Note that this is a color plot)
11 6. For the following transfer function (s + ) G(s) = (s +)(s +4s + 4 ) sketch the Bode asymptotic magnitude and phase plots, using appropriate corrections, and compare with the exact result using Matlab. Answer: ( s ) (s + ) G(s) = = ( + ) (s +)(s +4s + 4 ) 4 s 4s (s +) The system has one zero at s =, one real pole at s =, and one pole pair whose natural frequency ω n = (rad/s) and damping ratio ζ = 4//ω n =.. Hence, the system has three break frequencies:,, (rad/s). The magnitude at the low frequency asymptote is log 3 = 6 (db). The correction log ζ = 7.96 (db). TheBodeasymptoticplot:
12 The exact Bode plot by Matlab: 6 Bode Diagram Magnitude (db) Phase (deg) Frequency (rad/sec)
MASSACHUSETTS 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 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 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 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 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 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 2: Drawing Bode Plots, Part 2 Overview In this Lecture, you will learn: Simple Plots Real Zeros Real Poles Complex
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 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 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 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More informationChapter 8: Frequency Domain Analysis
Chapter 8: Frequency Domain Analysis Samantha Ramirez Preview Questions 1. What is the steadystate response of a linear system excited by a cyclic or oscillatory input? 2. How does one characterize the
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 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 informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment 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 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 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 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 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 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 information8.1.6 Quadratic pole response: resonance
8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Secondorder denominator, of the form 1+a 1 s + a s v 1 (s) + C R Twopole lowpass filter example v (s) with
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
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 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 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 informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More 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 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 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 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 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 7: Time Domain Analysis
Chapter 7: Time Domain Analysis Samantha Ramirez Preview Questions How do the system parameters affect the response? How are the parameters linked to the system poles or eigenvalues? How can Laplace transforms
More informationDigital Control: Part 2. ENGI 7825: Control Systems II Andrew Vardy
Digital Control: Part 2 ENGI 7825: Control Systems II Andrew Vardy Mapping the splane onto the zplane We re almost ready to design a controller for a DT system, however we will have to consider where
More informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
More 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 informationMAE 143B  Homework 9
MAE 143B  Homework 9 7.1 a) We have stable firstorder poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straightline
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 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 information2.010 Fall 2000 Solution of Homework Assignment 7
. Fall Solution of Homework Assignment 7. Control of Hydraulic Servomechanism. We return to the Hydraulic Servomechanism of Problem in Homework Assignment 6 with additional data which permits quantitative
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
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 informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More 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 informationK(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s
321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify
More informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
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 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 informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
More informationResponse to a pure sinusoid
Harvard University Division of Engineering and Applied Sciences ES 145/215  INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid
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 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 informationFrequency Response of Linear Time Invariant Systems
ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z
More informationagree w/input bond => + sign disagree w/input bond =>  sign
1 ME 344 REVIEW FOR FINAL EXAM LOCATION: CPE 2.204 M. D. BRYANT DATE: Wednesday, May 7, 2008 9noon Finals week office hours: May 6, 47 pm Permitted at final exam: 1 sheet of formulas & calculator I.
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
More informationState Regulator. Advanced Control. design of controllers using pole placement and LQ design rules
Advanced Control State Regulator Scope design of controllers using pole placement and LQ design rules Keywords pole placement, optimal control, LQ regulator, weighting matrixes Prerequisites Contact state
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
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 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 informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servooriented
More information2.004 Dynamics and Control II Spring 2008
MT OpenCourseWare http://ocw.mit.edu.004 Dynamics and Control Spring 008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts nstitute of Technology
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More information2.004 Dynamics and Control II Spring 2008
MIT OpenCourseWare http://ocw.mit.edu 2.004 Dynamics and Control II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 8 I * * Massachusetts
More informationDamped Oscillation Solution
Lecture 19 (Chapter 7): Energy Damping, s 1 OverDamped Oscillation Solution Damped Oscillation Solution The last case has β 2 ω 2 0 > 0. In this case we define another real frequency ω 2 = β 2 ω 2 0. In
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 13: Stability Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 13: Stability p.1/20 Outline InputOutput
More informationDate: 1 April (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.
PH1140: Oscillations and Waves Name: Solutions Conference: Date: 1 April 2005 EXAM #1: D2005 INSTRUCTIONS: (1) The only reference material you may use is one 8½x11 crib sheet and a calculator. (2) Show
More informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
More informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
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 informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
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 informationLaboratory handout 5 Mode shapes and resonance
laboratory handouts, me 34 82 Laboratory handout 5 Mode shapes and resonance In this handout, material and assignments marked as optional can be skipped when preparing for the lab, but may provide a useful
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 informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More informationDate: 31 March (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.
PH1140: Oscillations and Waves Name: SOLUTIONS AT END Conference: Date: 31 March 2005 EXAM #1: D2006 INSTRUCTIONS: (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.
More informationSection 4.9; Section 5.6. June 30, Free Mechanical Vibrations/Couple MassSpring System
Section 4.9; Section 5.6 Free Mechanical Vibrations/Couple MassSpring System June 30, 2009 Today s Session Today s Session A Summary of This Session: Today s Session A Summary of This Session: (1) Free
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 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 informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
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 informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationControl System (ECE411) Lectures 13 & 14
Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 SteadyState Error Analysis Remark: For a unity feedback system
More informationExercises Lecture 15
AM1 Mathematical Analysis 1 Oct. 011 Feb. 01 Date: January 7 Exercises Lecture 15 Harmonic Oscillators In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More informationDifferential Equations: Homework 8
Differential Equations: Homework 8 Alvin Lin January 08  May 08 Section.6 Exercise Find a general solution to the differential equation using the method of variation of parameters. y + y = tan(t) r +
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 informationTopic # Feedback Control
Topic #5 6.3 Feedback Control StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or,evenifwecanchangethepolelocations. Where do we put the poles? Linear
More 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 informationProcess Control & Instrumentation (CH 3040)
Firstorder systems Process Control & Instrumentation (CH 3040) Arun K. Tangirala Department of Chemical Engineering, IIT Madras January  April 010 Lectures: Mon, Tue, Wed, Fri Extra class: Thu A firstorder
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 informationThe Frequencyresponse Design Method
Chapter 6 The Frequencyresponse Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
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 informationECE320: Linear Control Systems Homework 8. 1) For one of the rectilinear systems in lab, I found the following state variable representations:
ECE30: Linear Control Systems Homework 8 Due: Thursday May 6, 00 at the beginning of class ) For one of the rectilinear systems in lab, I found the following state variable representations: 0 0 q q+ 74.805.6469
More 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 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 informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More 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 information