Systems Analysis and Control


 Todd Carter
 4 years ago
 Views:
Transcription
1 Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 8: Response Characteristics
2 Overview In this Lecture, you will learn: Characteristics of the Response Stability Real Poles SteadyState Error Rise Time Settling Time Complex Poles Complex Pole Locations Damped/Natural Frequency Damping and Damping Ratio M. Peet Lecture 8: Control Systems 2 / 22
3 Feedback Control Recall the Feedback Interconnection  u(s) + K(s) G(s) y(s) Feedback: Controller: u i = K(u y) Plant: y = Gu i The output signal is ŷ(s), ŷ(s) = Ĝ(s) ˆK(s) 1 + Ĝ(s) ˆK(s)û(s) M. Peet Lecture 8: Control Systems 3 / 22
4 Controlling the Inverted Pendulum Model Open Loop Transfer Function 1 Ĝ(s) = Js 2 Mgl 2 Controller: Static Gain: ˆK(s) = K Input: Impulse: û(s) = 1. Closed Loop: Lower Feedback ŷ(s) = Ĝ(s) ˆK(s) = 1 + Ĝ(s) ˆK(s)û(s) K Js 2 Mgl K Js 2 Mgl 2 = K Js 2 Mgl 2 + K M. Peet Lecture 8: Control Systems 4 / 22
5 Controlling the Inverted Pendulum Model Closed Loop Impulse Response: Lower Feedback K ŷ(s) = Js 2 Mgl 2 + K Traits: Infinite Oscillations Oscillates about 0. Amplitude Impulse Response Open Loop Impulse Response: Time (sec) 18 x 105 Impulse Response ŷ(s) = 1 2J 1 J Mgl s Mgl 2J 1 s + Mgl 2J Amplitude Unstable! Time (sec) M. Peet Lecture 8: Control Systems 5 / 22
6 Controlling the Suspension System Open Loop Transfer Function: Set m c = m w = g = c = K 1 = K 2 = 1. x 1 m c Ĝ(s) = s 2 + s + 1 s 4 + 2s 3 + 3s 2 + s + 1 x 2 m w Controller: Static Gain: ˆK(s) = k u Closed Loop: Lower Feedback Ĝ(s) ˆK(s) ŷ(s) = 1 + Ĝ(s) ˆK(s)û(s) = k(s 2 + s + 1) s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) M. Peet Lecture 8: Control Systems 6 / 22
7 Controlling the Suspension Problem Effect of changing the Feedback, k Closed Loop Step Response: ŷ(s) = k(s 2 + s + 1) 1 s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) s High k: Overshot the target Quick Response Closer to desired value of f Low k: Slow Response No overshoot Final value is farther from 1. Questions: Which Traits are important? How to predict the behaviour? Figure: Step Response for different k M. Peet Lecture 8: Control Systems 7 / 22
8 Stability The most basic property is Stability: Definition 1. A system, G is Stable if there exists a K > 0 such that kgukl2 KkukL2 Note: Although this is the true definition for systems defined by transfer functions, it is rarely used. Bounded input means bounded output. Stable is y(t) 0 when u(t) 0. M. Peet Lecture 8: Control Systems 8 / 22
9 Stability Definition 2. The Closed Right HalfPlace, CRHP is the set of complex numbers with nonnegative real part. {s C : Real(s) 0} Im(s) CRHP Re(s) Im(s) Figure: Unstable Re(s) Theorem 3. A system G is stable if and only if it s transfer function Ĝ has no poles in the Closed Right Half Plane. Check stability by checking poles. x is a pole o is a zero M. Peet Lecture 8: Control Systems 9 / 22
10 Predicting SteadyState Error Definition 4. SteadyState Error for a stable system is the final difference between input and output. e ss = lim t u(t) y(t) Usually measured using the step response. Since u(t) = 1, e ss = 1 lim t y(t) Figure: Suspension Response for k = 1 M. Peet Lecture 8: Control Systems 10 / 22
11 Predicting SteadyState Error Recall: For any system G, by partial fraction expansion: So which means and hence ŷ(s) = Ĝ(s)1 s = r 0 s + r 1 s p y(t) = r 0 1(t) + r 1 e p1t r n e pnt lim t y(t) = r 0 e ss = 1 r 0 r n s p n M. Peet Lecture 8: Control Systems 11 / 22
12 Predicting SteadyState Error The steadystate error is given by r 0. e ss = 1 r 0 Recall: The residue at s = 0 is r 0 and is found as r 0 = Ĝ(s) s=0 = lim s 0 Ĝ(s) Thus the steadystate error is e ss = 1 lim s 0 Ĝ(s) This can be generalized to find the limit of any signal: Theorem 5 (Final Value Theorem). lim y(t) = lim sŷ(s) t s 0 Assumes the limit exists (Stability) Can be used to find response to other inputs Ramp, impulse, etc. M. Peet Lecture 8: Control Systems 12 / 22
13 Predicting SteadyState Error Numerical Example Ĝ(s) = The steadystate response is k(s 2 + s + 1) s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) y ss = lim s 0 sŷ(s) = lim s 0 Ĝ(s) = lim s 0 k(s 2 + s + 1) s 4 + 2s 3 + (3 + k)s 2 + (1 + k)s + (1 + k) = k 1 + k The steadystate error is When k = 0, e ss = 1 As k, e ss = 0 e ss = 1 y ss = 1 = k k 1 + k M. Peet Lecture 8: Control Systems 13 / 22
14 Dynamic Response Characteristics Two Types of Response By now, you know that motion is dominated by the poles! Simplify the response by considering response of each pole. Allows quantitative analysis Figure: Real Pole Figure: Complex Pair of Poles We start with Real Poles M. Peet Lecture 8: Control Systems 14 / 22
15 Step Response Characteristics Real Poles Consider a real pole step response: ŷ(s) = r 1 s p s = r r p s p p s y(t) = r p ( e pt 1 ) Assume stable, so p < 0 Cases: p > 0 implies y(t) p < 0 implies y(t) r p SteadyState Error: e ss = 1 r p M. Peet Lecture 8: Control Systems 15 / 22
16 Step Response Characteristics Rise Time Besides the final value: How quickly will the system respond? Definition 6. The rise time, T r, is the time it takes to go from.1 to.9 of the final value. t 1 when y(t 1 ) =.1 r p is found as.1 = e pt1 1 ln(1.1) = pt 1 ln.9 t 1 = p Likewise for y(t 2 ) =.9 r p t 2 = ln.1 p = 2.31 p =.11 p we get Thus rise time for a Single Pole is: T r = t 2 t 1 = 2.31 p.11 p = 2.2 p M. Peet Lecture 8: Control Systems 16 / 22
17 Step Response Characteristics Settling Time Will it stay there: How fast does it converge? More important for complex poles. Definition 7. The Settling Time, T s, is the time it takes to reach and stay within.99 of the final value. The time at y(t s ) =.99 r p is found as.99 = e pts 1 ln(.01) = pt s ln.01 T s = p = 4.6 p The settling time for a Single Pole is: T s = 4.6 p M. Peet Lecture 8: Control Systems 17 / 22
18 Solution for Complex Poles ŷ(s) = ω 2 d + σ2 s 2 + 2σs + ω 2 d + σ2 1 s = ω 2 n s 2 + 2ζω n s + ω 2 n The poles are at s = σ ± ω d ı and s = 0. The solution is: y(t) = 1 e (cos(ω σt d t) σ ) sin(ω d t) ω d = 1 e σt ω n ω d sin (ω d t + φ) Where σ = ζω n, ω d = ω n 1 ζ2 and φ = tan 1 ( ωd ζω n ). The result is oscillation with an Exponential Envelope. Envelope decays at rate σ Speed of oscillation is ω d, the Damped Frequency 1 s = k 1 s + k 2 s 2 + 2ζω n s + ωn 2 + r 2 s M. Peet Lecture 8: Control Systems 18 / 22
19 Damping We use several adjectives to describe exponential decay: Undamped Oscillation continues forever, σ = 0 Underdamped Oscillation continues for many cycles. Damped Critically Damped No oscillation or overshoot. ω = 0 M. Peet Lecture 8: Control Systems 19 / 22
20 Step Response Characteristics Damping Ratio Besides ω, there is another way to measure oscillation Im(s) Definition 8. The Natural Frequency of a pole at p = σ + ıω d is ω n = σ 2 + ω 2 d. ω n Re(s) 1 1 for ŷ(s) = s 2 +as+b s, ω n = b. Radius of the pole in complex plane. Frequency of least damping. Also known as resonant frequency M. Peet Lecture 8: Control Systems 20 / 22
21 Step Response Characteristics Damping Ratio Besides σ, there are other ways to measure damping Definition 9. The Damping Ratio of a pole at p = σ + ıω is ζ = σ ω n. 1 1 for ŷ(s) = s 2 +as+b s, ζ = The amount the amplitude decreases per oscillation. a 2ω n. The angle that the pole makes in the complex plane. M. Peet Lecture 8: Control Systems 21 / 22
22 Summary What have we learned today? Characteristics of the Response Real Poles SteadyState Error Rise Time Settling Time Complex Poles Complex Pole Locations Damped/Natural Frequency Damping and Damping Ratio Continued in Next Lecture M. Peet Lecture 8: Control Systems 22 / 22
Systems 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 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 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 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12: Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain
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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 13: Root Locus Continued Overview In this Lecture, you will learn: Review Definition of Root Locus Points on the Real Axis
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 informationReview: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system
Plan of the Lecture Review: transient and steadystate response; DC gain and the FVT Today s topic: systemmodeling diagrams; prototype 2ndorder system Plan of the Lecture Review: transient and steadystate
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 Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
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 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 informationControl Systems, Lecture04
Control Systems, Lecture04 İbrahim Beklan Küçükdemiral Yıldız Teknik Üniversitesi 2015 1 / 53 Transfer Functions The output response of a system is the sum of two responses: the forced response and the
More informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
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 openloop system (a system without feedbac) and a closedloop
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 15: Root Locus Part 4 Overview In this Lecture, you will learn: Which Poles go to Zeroes? Arrival Angles Picking Points? Calculating
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 informationLecture 25: Tue Nov 27, 2018
Lecture 25: Tue Nov 27, 2018 Reminder: Lab 3 moved to Tuesday Dec 4 Lecture: review timedomain characteristics of 2ndorder systems intro to control: feedback openloop vs closedloop control intro to
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 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 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 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 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 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 informationTransient Response of a SecondOrder System
Transient Response of a SecondOrder System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a wellbehaved closedloop
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 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 information16.31 Homework 2 Solution
16.31 Homework Solution Prof. S. R. Hall Issued: September, 6 Due: September 9, 6 Problem 1. (Dominant Pole Locations) [FPE 3.36 (a),(c),(d), page 161]. Consider the second order system ωn H(s) = (s/p
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 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 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 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 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 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 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 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 informationSchool of Mechanical Engineering Purdue University. ME375 Dynamic Response  1
Dynamic Response of Linear Systems Linear System Response Superposition Principle Responses to Specific Inputs Dynamic Response of f1 1st to Order Systems Characteristic Equation  Free Response Stable
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 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 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 informationProblem Value Score Total 100/105
RULES This is a closed book, closed notes test. You are, however, allowed one piece of paper (front side only) for notes and definitions, but no sample problems. The top half is the same as from the first
More 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 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 informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
More 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 information12.7 Steady State Error
Lecture Notes on Control Systems/D. Ghose/01 106 1.7 Steady State Error For first order systems we have noticed an overall improvement in performance in terms of rise time and settling time. But there
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 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 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 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 informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control DMAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
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 informationHomework 11 Solution  AME 30315, Spring 2015
1 Homework 11 Solution  AME 30315, Spring 2015 Problem 1 [10/10 pts] R +  K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
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 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 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 informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
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: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
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 informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More 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 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 informationModeling and Experimentation: MassSpringDamper System Dynamics
Modeling and Experimentation: MassSpringDamper System Dynamics Prof. R.G. Longoria Department of Mechanical Engineering The University of Texas at Austin July 20, 2014 Overview 1 This lab is meant to
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 informationForced Mechanical Vibrations
Forced Mechanical Vibrations Today we use methods for solving nonhomogeneous second order linear differential equations to study the behavior of mechanical systems.. Forcing: Transient and Steady State
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 informationSecond Order Systems
Second Order Systems independent energy storage elements => Resonance: inertance & capacitance trade energy, kinetic to potential Example: Automobile Suspension x z vertical motions suspension spring shock
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #4 Monday, January 13, 2003 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Impulse and Step Responses of ContinuousTime
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More 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 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 informationLab # 4 Time Response Analysis
Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an
More 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 informationLab Experiment 2: Performance of First order and second order systems
Lab Experiment 2: Performance of First order and second order systems Objective: The objective of this exercise will be to study the performance characteristics of first and second order systems using
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 informationSecond Order and Higher Order Systems
Second Order and Higher Order Systems 1. Second Order System In this section, we shall obtain the response of a typical secondorder control system to a step input. In terms of damping ratio and natural
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 informationControl Systems Design
ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:
More informationGoals for today 2.004
Goals for today Block diagrams revisited Block diagram components Block diagram cascade Summing and pickoff junctions Feedback topology Negative vs positive feedback Example of a system with feedback Derivation
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 informationC(s) R(s) 1 C(s) C(s) C(s) = s  T. Ts + 1 = 1 s  1. s + (1 T) Taking the inverse Laplace transform of Equation (5 2), we obtain
analyses of the step response, ramp response, and impulse response of the secondorder systems are presented. Section 5 4 discusses the transientresponse analysis of higherorder systems. Section 5 5 gives
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 informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
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 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 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 informationDigital Control: Summary # 7
Digital Control: Summary # 7 Proportional, integral and derivative control where K i is controller parameter (gain). It defines the ratio of the control change to the control error. Note that e(k) 0 u(k)
More informationDynamic Behavior. Chapter 5
1 Dynamic Behavior In analyzing process dynamic and process control systems, it is important to know how the process responds to changes in the process inputs. A number of standard types of input changes
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 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 informationIdentification Methods for Structural Systems. Prof. Dr. Eleni Chatzi System Stability  26 March, 2014
Prof. Dr. Eleni Chatzi System Stability  26 March, 24 Fundamentals Overview System Stability Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can
More informationTransient Stability Analysis with PowerWorld Simulator
Transient Stability Analysis with PowerWorld Simulator T11: Single Machine Infinite Bus Modeling (SMIB) 21 South First Street Champaign, Illinois 6182 +1 (217) 384.633 support@powerworld.com http://www.powerworld.com
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 informationVibrations: Second Order Systems with One Degree of Freedom, Free Response
Single Degree of Freedom System 1.003J/1.053J Dynamics and Control I, Spring 007 Professor Thomas Peacock 5//007 Lecture 0 Vibrations: Second Order Systems with One Degree of Freedom, Free Response Single
More informationBasic Procedures for Common Problems
Basic Procedures for Common Problems ECHE 550, Fall 2002 Steady State Multivariable Modeling and Control 1 Determine what variables are available to manipulate (inputs, u) and what variables are available
More information8. Introduction and Chapter Objectives
Real Analog  Circuits Chapter 8: Second Order Circuits 8. Introduction and Chapter Objectives Second order systems are, by definition, systems whose inputoutput relationship is a second order differential
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 information