Step Response Analysis. Frequency Response, Relation Between Model Descriptions


 Nathan Young
 3 years ago
 Views:
Transcription
1 Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control
2 Content. Step Response Analysis 2. Frequency Response 3. Relation between Model Descriptions
3 Step Response Analysis
4 Step Response From the last lecture, we know that if the input u(t) is a step, then the output in the Laplace domain is Y (s) = G(s)U(s) = G(s) s It is possible to do an inverse transform of Y (s) to get y(t), but is it possible to claim things about y(t) by only studying Y (s)? We will study how the poles affects the step response. (The zeros will be discussed later). 2
5 Initial and Final Value Theorem Let F (s) be the Laplace transformation of f (t), i.e., F (s) = L(f (t))(s). Given that the limits below exist, it holds that: Initial value theorem Final value theorem lim t f (t) = lim s + sf (s) lim t + f (t) = lim s sf (s) For a step response we have that: lim y(t) = lim sy (s) = lim sg(s) t + s s s = G() When can we NOT apply the Final value theorem? 3
6 First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t) Re One pole in s = /T Step response: Y (s) = G(s) s = K s( + st ) G(s) = L K + st 5 5 t y(t) = K ( e t/t ), t 4
7 First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t) Re 5 5 t Final value: G(s) = K + st lim y(t) = lim sy (s) = lim s K t + s s s( + st ) = K 4
8 First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t) Re T is called the timeconstant: G(s) = K + st 5 5 t y(t ) = K( e T /T ) = K( e ).63K i.e., T is the time it takes for the step response to reach 63% of its final value 4
9 First Order System Singularity Chart Step Response (K=) Im.5.5 T = T = 2T = 5 y(t) Re 5 5 t Derivative at zero: lim ẏ(t) = t G(s) = lim s sy (s) = lim s + K + st s + s 2 K s( + st ) = K T 4
10 Second Order System With Real Poles Singularity Chart Step Response (K=) Im.5.5 T = T = 2 y(t) Re G(s) = K ( + st )( + st 2 ) 5 5 t Poles in s = /T and s = /T 2. Step response: ( ) K Te t/t T 2e t/t 2 T y(t) = T 2 T T 2 K ( e t/t ) t T e t/t T = T 2 = T 5
11 Second Order System With Real Poles Singularity Chart Step Response (K=) Im.5.5 T = T = 2 y(t) Re 5 5 t G(s) = K ( + st )( + st 2 ) Final value: lim = lim sk sy (s) = lim t + s s s( + st )( + st 2 ) = K 5
12 Second Order System With Real Poles Singularity Chart Step Response (K=) Im.5.5 T = T = 2 y(t) Re 5 5 t Derivative at zero: lim ẏ(t) = t G(s) = lim s sy (s) = lim s + K ( + st )( + st 2 ) s + s 2 K s( + st )( + st 2 ) = 5
13 Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Relative damping ζ, related to the angle ϕ ζ = cos(ϕ) Singularity Chart Im.5 ω ϕ.5 Re 6
14 Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Inverse transformation for step response yields: y(t) = K = K ( ( ( ) ) ζ 2 e ζωt sin ω ζ2 t + arccos ζ ( ζ 2 e ζωt sin ω ζ2 t + arcsin( ζ )) ) 2, t 6
15 Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Inverse transformation for step response yields: y(t) = K = K ( ( ( ) ) ζ 2 e ζωt sin ω ζ2 t + arccos ζ ( ζ 2 e ζωt sin ω ζ2 t + arcsin( ζ )) ) 2, t Exercise: Check of correct starting point of step response. y() = K = K ( ( ( ζ 2 e sin ω ζ 2 + arcsin( ) ) ζ 2 ) ) ζ 2 ζ 2 = Step Response t 6
16 Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Singularity Chart Step Response (K=) Im ω =.5 ω = ω =.5 y(t).5 Re 5 5 t 6
17 Second Order System With Complex Poles Kω 2 G(s) = s 2 + 2ζω s + ω 2, < ζ < Im Singularity Chart ζ =.3 ζ =.7 ζ =.9 Re y(t).5.5 Step Response (K=) 5 5 t 6
18 Frequency Response
19 Sinusoidal Input Given a transfer function G(s), what happens if we let the input be u(t) = sin(ωt)?.5 y(t) u(t) t t 7
20 Sinusoidal Input It can be shown that if the input is u(t) = sin(ωt), the output 2 will be where y(t) = A sin(ωt + ϕ) A = G(iω) ϕ = arg G(iω) So if we determine a and ϕ for different frequencies ω, we have a description of the transfer function. 2 after the transient has decayed 8
21 Bode Plot Idea: Plot G(iω) and arg G(iω) for different frequencies ω. Magnitude (abs) Phase (deg) Frequency (rad/s) 9
22 Sinusoidal InputOutput: example with frequency sweep (chirp) Resonance frequency of industrial robot IRB2 visible in data.
23 Sinusoidal InputOutput: example with frequency sweep (chirp) Resonance frequency of industrial robot IRB2 visible in data.
24 Bode Plot  Products of Transfer Functions Let G(s) = G (s)g 2 (s)g 3 (s) then log G(iω) = log G (iω) + log G 2 (iω) + log G 3 (iω) arg G(iω) = arg G (iω) + arg G 2 (iω) + arg G 3 (iω) This means that we can construct Bode plots of transfer functions from simple building blocks for which we know the Bode plots.
25 Bode Plot of G(s) = K If G(s) = K then log G(iω) = log( K ) arg G(iω) = (if K >, else + 8 or 8 deg) 2
26 Bode Plot of G(s) = K Magnitude (abs) Phase (deg) K = 4 K = K = Frequency (rad/s) 2
27 Bode Plot of G(s) = s n If G(s) = s n then log G(iω) = n log(ω) arg G(iω) = n π 2 3
28 Bode Plot of G(s) = s n Magnitude (abs) Phase (deg) n = 2 2 n = n = Frequency (rad/s) 3
29 Bode Plot of G(s) = ( + st ) n If then G(s) = ( + st ) n log G(iω) = n log( + ω 2 T 2 ) arg G(iω) = n arg( + iωt ) = n arctan (ωt ) For small ω log G(iω) arg G(iω) For large ω log G(iω) n log(ωt ) arg G(iω) n π 2 4
30 Bode Plot of G(s) = ( + st ) n Magnitude (abs) Phase (deg) n = 2 n = 2 n = T Frequency (rad/s) 4
31 Bode Plot of G(s) = ( + 2ζs/ω + (s/ω ) 2 ) n G(s) = ( + 2ζs/ω + (s/ω ) 2 ) n For small ω log G(iω) arg(iω) For large ω ( ) ω log G(iω) 2n log ω arg G(iω) nπ 5
32 Bode Plot of G(s) = ( + 2ζs/ω + (s/ω ) 2 ) n Magnitude (abs) Phase (deg) 2 ζ =.2 ζ =. ζ = Frequency (rad/s) 5
33 Bode Plot of G(s) = e sl G(s) = e sl Describes a pure time delay with delay L, i.e, y(t) = u(t L) log G(iω) = arg G(iω) = ωl 6
34 Bode Plot of G(s) = e sl Magnitude (abs) Phase (deg) 2 L =. 2 L = 5 L = Frequency (rad/s) 6
35 Bode Plot of G(s) = e sl Same delay may appear as different phase lag for different frequencies! Example Delay.52 sec between input and output (Upper): Period time = 2π 6.28 sec. Delay represents phase lag of deg 6.28 (Lower): Period time = π 3.4 sec. Delay represents phase lag of deg
36 Bode Plot of G(s) = e sl Same delay may appear as different phase lag for different frequencies! Example Delay.52 sec between input and output (Upper): Period time = 2π 6.28 sec. Delay represents phase lag of deg 6.28 (Lower): Period time = π 3.4 sec. Delay represents phase lag of deg
37 Bode Plot of G(s) = e sl Check phase in Bode diagram for e.52s for sin(t) ω =. rad/s sin(2t) ω = 2. rad/s >> s=tf( s ) >> G=exp(.52*s); >> bode(g,.,5) % Bode plot in frequencyrange [... 5] rad/s 6
38 Bode Plot of Composite Transfer Function Example Draw the Bode plot of the transfer function G(s) = (s + 2) s(s + 2) 2 First step, write it as product of sample transfer functions: G(s) = (s + 2) s(s + 2) 2 =.5 s ( +.5s) ( +.5s) 2 Then determine the corner frequencies: w c2 =2 w c =2 (s + 2) {}}{{}}{ G(s) = s(s + 2) 2 =.5 s ( +.5s) ( +.5s) 2 7
39 Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) Frequency (rad/s) 8
40 Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) Frequency (rad/s) 8
41 Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) Frequency (rad/s) 8
42 Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) Frequency (rad/s) 8
43 Bode Plot of Composite Transfer Function G(s) = (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Magnitude (abs) Frequency (rad/s) 8
44 Bode Plot of Composite Transfer Function G(s) = 45 (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Phase (deg) Frequency (rad/s) 9
45 Bode Plot of Composite Transfer Function G(s) = 45 (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Phase (deg) Frequency (rad/s) 9
46 Bode Plot of Composite Transfer Function G(s) = 45 (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Phase (deg) Frequency (rad/s) 9
47 Bode Plot of Composite Transfer Function G(s) = 45 (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Phase (deg) Frequency (rad/s) 9
48 Bode Plot of Composite Transfer Function G(s) = 45 (s + 2) s(s + 2) 2 =.5 s w c =2 {}}{ ( +.5s) w c2 =2 {}}{ ( +.5s) 2 Phase (deg) Frequency (rad/s) 9
49 Nyquist Plot By removing the frequency information, we can plot the transfer function in one plot instead of two..5 Im G(iω) arg G(iω) G(iω).5.5 Re G(iω) 2
50 Nyquist Plot By removing the frequency information, we can plot the transfer function in one plot instead of two..5 Im G(iω) arg G(iω) G(iω).5.5 Re G(iω) Split the transfer function into real and imaginary part: G(s) = G(iω) = + s + iω = + ω 2 i ω + ω 2 Is this the transfer function in the plot above? 2
51 From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω) Re G(iω) 2
52 From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω) Re G(iω) 2
53 From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω) Re G(iω) 2
54 From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω) Re G(iω) 2
55 From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω) Re G(iω) 2
56 From Bode Plot to Nyquist Plot Magnitude (abs) Phase (deg) Frequency (rad/s) 2 2 Frequency (rad/s) Im G(iω) Re G(iω) 2
57 Relation between Model Descriptions
58 Singlecapacitive Processes K st Singularity chart.5 Step response Nyquist plot 2 4 Bode plot
59 Multicapacitive Processes K (st +)(st 2 +).5.5 Singularity chart.5 Step response Nyquist plot Bode plot
60 Integrating Processes s.5.5 Singularity chart 4 2 Step response Nyquist plot 2 4 Bode plot
61 Oscillative Processes Kω 2 s 2 +2ζω s+ω 2, < ζ < Singularity chart.5 Step response Nyquist plot Bode plot 8 25
62 Delay Processes K st + e sl Step response.5 Nyquist plot Bode plot
63 Process with Inverse Responses sa+ (st +)(st 2 +) Singularity chart Step response Nyquist plot Bode plot
Systems 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 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 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 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 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 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 informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
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 informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, 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 informationExam in Systems Engineering/Process Control
Department of AUTOMATIC CONTROL Exam in Systems Engineering/Process Control 2762 Points and grading All answers must include a clear motivation. Answers may be given in English or Swedish. The total
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 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 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 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 informationReglerteknik: Exercises
Reglerteknik: Exercises Exercises, Hints, Answers Liten reglerteknisk ordlista Introduktion till Control System Toolbox ver. 5 This version: January 3, 25 AUTOMATIC CONTROL REGLERTEKNIK LINKÖPINGS UNIVERSITET
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 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 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 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 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 informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More 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 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 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 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 informationReglerteknik Allmän Kurs. Del 2. Lösningar till Exempelsamling. Läsår 2015/16
Reglerteknik Allmän Kurs Del Lösningar till Exempelsamling Läsår 5/6 Avdelningen för Reglerteknik, KTH, SE 44 Stockholm, SWEDEN AUTOMATIC CONTROL COMMUNICATION SYSTEMS LINKÖPINGS UNIVERSITET Reglerteknik
More information1 Mathematics. 1.1 Determine the onesided Laplace transform of the following signals. + 2y = σ(t) dt 2 + 3dy dt. , where A is a constant.
Mathematics. Determine the onesided Laplace transform of the following signals. {, t < a) u(t) =, where A is a constant. A, t {, t < b) u(t) =, where A is a constant. At, t c) u(t) = e 2t for t. d) u(t)
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 informationChapter Eight. Transfer Functions. 8.1 Frequency Domain Modeling
Chapter Eight Transfer Functions The typical regulator system can frequently be described, in essentials, by differential equations of no more than perhaps the second, third or fourth order. In contrast,
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
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 informationECEN 420 LINEAR CONTROL SYSTEMS. Lecture 2 Laplace Transform I 1/52
1/52 ECEN 420 LINEAR CONTROL SYSTEMS Lecture 2 Laplace Transform I Linear Time Invariant Systems A general LTI system may be described by the linear constant coefficient differential equation: a n d n
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 informationOutline. Classical Control. Lecture 2
Outline Outline Outline Review of Material from Lecture 2 New Stuff  Outline Review of Lecture System Performance Effect of Poles Review of Material from Lecture System Performance Effect of Poles 2 New
More informationCourse roadmap. ME451: Control Systems. Example of Laplace transform. Lecture 2 Laplace transform. Laplace transform
ME45: Control Systems Lecture 2 Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Transfer function Models for systems electrical mechanical electromechanical Block
More informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that 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 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 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 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 informationLecture 9 Infinite Impulse Response Filters
Lecture 9 Infinite Impulse Response Filters Outline 9 Infinite Impulse Response Filters 9 FirstOrder LowPass Filter 93 IIR Filter Design 5 93 CT Butterworth filter design 5 93 Bilinear transform 7 9
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 8: Response Characteristics Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles
More 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 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 informationControl Systems. Control Systems Design LeadLag Compensator.
Design LeadLag Compensator hibum@seoulteh.a.kr Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design. Frequeny Domain Design Frequeny response
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 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 informationNotes on the Periodically Forced Harmonic Oscillator
Notes on the Periodically orced Harmonic Oscillator Warren Weckesser Math 38  Differential Equations 1 The Periodically orced Harmonic Oscillator. By periodically forced harmonic oscillator, we mean the
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 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 informationAnalysis and Design of Control Systems in the Time Domain
Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.
More 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 informationPart IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 5. An Introduction to Feedback Control Systems
Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL Glenn Vinnicombe HANDOUT 5 An Introduction to Feedback Control Systems ē(s) ȳ(s) Σ K(s) G(s) z(s) H(s) z(s) = H(s)G(s)K(s) L(s) ē(s)=
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 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 informationDr. Ian R. Manchester
Dr Ian R. Manchester Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus
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 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 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 informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
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 informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
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 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 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 information9.5 The Transfer Function
Lecture Notes on Control Systems/D. Ghose/2012 0 9.5 The Transfer Function Consider the nth order linear, timeinvariant dynamical system. dy a 0 y + a 1 dt + a d 2 y 2 dt + + a d n y 2 n dt b du 0u +
More informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
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 informationStability and Robustness 1
Lecture 2 Stability and Robustness This lecture discusses the role of stability in feedback design. The emphasis is notonyes/notestsforstability,butratheronhowtomeasurethedistanceto instability. The small
More informationSchool of Mechanical Engineering Purdue University
Case Study ME375 Frequency Response  1 Case Study SUPPORT POWER WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power wire, which is suspended from a catenary. During
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 informationGATE EE Topic wise Questions SIGNALS & SYSTEMS
www.gatehelp.com GATE EE Topic wise Questions YEAR 010 ONE MARK Question. 1 For the system /( s + 1), the approximate time taken for a step response to reach 98% of the final value is (A) 1 s (B) s (C)
More informationPrüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
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 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 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 informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Intro Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /5/27 Outline Closed Loop Transfer
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More informationA Simple PID Control Design for Systems with Time Delay
Industrial Electrical Engineering and Automation CODEN:LUTEDX/(TEIE7266)/116/(2017) A Simple PID Control Design for Systems with Time Delay Mats Lilja Division of Industrial Electrical Engineering and
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 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 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 informationA sufficient condition for the existence of the Fourier transform of f : R C is. f(t) dt <. f(t) = 0 otherwise. dt =
Fourier transform Definition.. Let f : R C. F [ft)] = ˆf : R C defined by The Fourier transform of f is the function F [ft)]ω) = ˆfω) := ft)e iωt dt. The inverse Fourier transform of f is the function
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA EXPERIMENT NO : CS II/ TITLE : FAMILIARIZATION
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the
More informationThe loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)
Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)
More information2.3 Oscillation. The harmonic oscillator equation is the differential equation. d 2 y dt 2 r y (r > 0). Its solutions have the form
2. Oscillation So far, we have used differential equations to describe functions that grow or decay over time. The next most common behavior for a function is to oscillate, meaning that it increases and
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationUnit 2: Modeling in the Frequency Domain Part 2: The Laplace Transform. The Laplace Transform. The need for Laplace
Unit : Modeling in the Frequency Domain Part : Engineering 81: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland January 1, 010 1 Pair Table Unit, Part : Unit,
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 informationPM diagram of the Transfer Function and its use in the Design of Controllers
PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude
More informationAutomatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...
Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number..........................
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance
More information+ + LAPLACE TRANSFORM. Differentiation & Integration of Transforms; Convolution; Partial Fraction Formulas; Systems of DEs; Periodic Functions.
COLOR LAYER red LAPLACE TRANSFORM Differentiation & Integration of Transforms; Convolution; Partial Fraction Formulas; Systems of DEs; Periodic Functions. + Differentiation of Transforms. F (s) e st f(t)
More information