Lecture 7:Time Response Pole-Zero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
|
|
- Noel Burke
- 6 years ago
- Views:
Transcription
1 Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26
2 First-Order Specs: Step : Pole Real inputs contain randomness (for example, noise): unpredictable. Test inputs help to predict response quality under other inputs. step inputs are useful to simulate sudden changes (startup, jump loading, etc) sinusoidal inputs are used to test frequency response (more later) impulses are used to simulate shock conditions ramp inputs are used to simulate transitions between setpoints 2 / 26
3 First-Order Specs: Step : Pole z 2 Im G(s) = p 5 p 4 z 1 p2 p 1 z 1 p 4 p 5 (s z 1 )(s z 1 )(s z 2 ) (s p 5 )(s p 5 )(s p 2 ) 2 (s p 1 )(s p 4 )(s p 4 )(s p 5 )(s p 5 ) Re 3 / 26
4 First-Order First-Order Specs: Step : Pole First-order systems are described by the I/O differential equation or in TF form τċ+c = r(t) C(s) R(s) = 1 τs+1 The number τ is called the time constant of the system. The TF has one real pole at s = 1 τ. Step Applying the unit step input R(s) = 1 s gives the response c(t) = 1 e t τ 4 / 26
5 First-Order Specs: Step : Pole (Stable) first-order systems always track the step command (the error approaches zero asymptotically). The time it takes for the output to reach approx. 98% of its steady value is called settling time, also referred to as 2 % settling time. It equals 4 time constants. 5 / 26
6 First-Order Specs: Step : Pole First-order systems cannot track a ramp command. The tracking error is always equal to the time constant. 6 / 26
7 First-Order Specs: Step : Pole The standard form of a second-order system is: C(s) R(s) = w 2 n s 2 +2ζw n s+w 2 n where w n and ζ are called the natural frequency (in rad/s) and damping ratio (dimensionless), respectively. The response will depend on the nature of the poles: Underdamped: 0 < ζ < 1 Marginally Stable: ζ = 0 Critically Damped: ζ = 1 Overdamped: ζ > 1 7 / 26
8 Case 1: Underdamped (0 < ζ < 1) First-Order Specs: Step : Pole The poles are complex conjugates. The response to a unit step input is given by ( ) c(t) = 1 e ζw nt 1 ζ sin w d t+tan ζ 2 where w d is the damped natural frequency: w d = w n 1 ζ 2 The damped natural frequency is the one obtained by counting the cycles per unit time in an experimental (and underdamped) response and converting the result to radians per second. ζ 8 / 26
9 Underdamped Case First-Order Specs: Step : Pole Poles: ζw n ± w d j cos β = ζ ζw n β w n Im w d = w n 1 ζ 2 0 Re w d 9 / 26
10 Marginally Stable and Overdamped Cases First-Order Specs: Step : Pole Marginally Stable Case (ζ = 0) When ζ = 0 the poles lie on the imaginary axis and the response to a step input is c(t) = 1 cosw n t that is, the oscillation does not die out. The system oscillates with its natural frequency w n. Critically Damped Case (ζ = 1) When ζ = 1 we get a real pole of multiplicity two at s = w n. The response is given by c(t) = 1 e w nt (1+w n t). No oscillations occur for a step input. 10 / 26
11 Overdamped Case (ζ > 1) First-Order Specs: Step : Pole When ζ > 1 the poles lie are real and unequal. The response features two decaying exponential terms (no oscillations): c(t) = 1+ where the poles are w n 2 ζ 2 1 ( e s 1 t s 1 e s2t s 2 s 1,2 = (ζ ± ζ 2 1)w n ) 11 / 26
12 Normalized First-Order Specs: Step : Pole 12 / 26
13 First-Order Specs: Step : Pole Many practical design cases call for time-domain specifications Step response specs are often demanded (worst-case, drastic situation) Common specifications 1. Rise time (T r ) 2. Peak time (T p ) 3. Percent overshoot (P.O.) 4. Settling time (T s ) 13 / 26
14 Specs: Step First-Order Specs: Step : Pole FV Setpoint 1.02*F.V 0.98*F.V Second Order Step Characteristics Overshoot 0 0 Tr Tp Tset 2% Time (sec) Setpoint Steady Error Final Value 14 / 26
15 First-Order Specs: Step 2nd-Order : Pole 2 % settling time: t s = 4 ζw n Rise time: t r = π β w d Peak time: t p = π w d Peak value: M pt = (e ( ζ 1 ζ 2)π +1)y( ) Percent overshoot: P.O.= 100e ( ζ 1 ζ 2)π 15 / 26
16 First-Order Specs: Step 2nd-Order : Pole 16 / 26
17 First-Order Specs: Step Pole Locations and : Pole 17 / 26
18 First-Order Specs: Step : Pole The above design formulas are valid when no zeros are present. In general, zeros spoil the response by increasing the overshoot. If zeros are found on the r.h.p., the system is said to be nonminimum phase, and the response is worsened. Zeros may be neglected in some cases (more about this later in this lecture) 18 / 26
19 : Pole First-Order Specs: Step : Pole When more than two poles are present, or when two distinct real poles are present, it can be difficult to predict the response without computer simulation Frequently, one or two poles are dominant. Remember that each single real pole gives rise to a term of the form a s(s+a) when a step input is applied. If we invert the term, the contribution of the pole is 1 e at. So if the time constant of the pole (1/a) is very small, the transient is very short and does not contribute much to the overall response. On the other hand, poles with large time constants are the ones dictating the response. We say that such poles are dominant. The same applies to complex conjugate pairs, where the time 1 constant taken into consideration is ζw n. 19 / 26
20 Pole Criterion First-Order Specs: Step : Pole Begin by finding all poles of the TF and finding their time constants. Remember: For a real pole at s = a, the time constant is 1/a, and for a complex conjugate pair with 0 < ζ < 1, it is 1 ζw n. Sort out the time constants: (high) τ 1,τ 2...,τ n (low). Attempt to find a gap of a factor of eight or more between two consecutive time constants, starting with the highest. If such a gap is found, neglect all poles on the fast side of the gap (small time constants). Careful! when reassembling the TF. You may be inadvertently changing the TF gain (a.k.a. DC gain): Make sure that the responses to a step input have the same final value. If not, correct the reduced TF with an appropriate constant. 20 / 26
21 Pole : Example First-Order Specs: Step : Pole G(s) = 3 s s s s+32 Poles: s 1 = 2, s 2 = 4, s 3,4 = 0.2 ± 1.989i(ζ = 0.1, w n = 2). Time constants for real poles: τ 1 = 0.5 and τ 2 = For the complex poles : τ = 5. The factored TF is: G(s) = 3 (s+2)(s+4)(s s+4). Time constants: 5,0.5,0.25. A gap of a factor of 10 is found, therefore we neglect the fast poles at s 2 = 4 and s 1 = 2. If we simply eliminate the factors (s + 2) and (s + 4), we have a gain mismatch. In order to have the same final value, we correct the gain to obtain: G r (s) = 3/8 s s + 4 Check with the final value theorem: lim s 0 sg r (s) = lim s 0 sg(s). 21 / 26
22 First-Order Specs: Step : Pole For second-order systems, we can predict the effect of a single real zero on the overshoot. Consider G(s) = (w2 n /a)(s + a) s 2 + 2ζw n s + wn 2 If a/ζw n > 8, neglect the zero due to dominance: G r (s) = wn 2 s 2 +2ζwns+wn 2. If not, use the chart: 22 / 26
23 Example First-Order Specs: Step : Pole Sketch the response of the following TF when the input is a step of size 2: 3(s+5) G(s) = (s+23)(s 2 +5s+16) 23 / 26
24 Solution First-Order Specs: Step : Pole 24 / 26
25 Solution First-Order Specs: Step : Pole 25 / 26
26 Matlab Simulation First-Order Specs: Step : Pole c(t) c(t p )= t s =1.3 s c( )= Overshoot=12.76% t 26 / 26
Second 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 second-order control system to a step input. In terms of damping ratio and natural
More informationCourse roadmap. Step response for 2nd-order system. Step response for 2nd-order system
ME45: Control Systems Lecture Time response of nd-order systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
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 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 second-order systems are presented. Section 5 4 discusses the transient-response analysis of higherorder systems. Section 5 5 gives
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More 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 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 informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steady-state Error and Type 0, Type
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More 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 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 informationControl Systems. University Questions
University Questions UNIT-1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More 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 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 informationRoot Locus Design Example #4
Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus
More informationEngraving Machine Example
Engraving Machine Example MCE44 - Fall 8 Dr. Richter November 24, 28 Basic Design The X-axis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional
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 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 informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steady-state error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationControl System (ECE411) Lectures 13 & 14
Control System (ECE411) Lectures 13 & 14, Professor Department of Electrical and Computer Engineering Colorado State University Fall 2016 Steady-State Error Analysis Remark: For a unity feedback system
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 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 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) s-plane Time response p =0 s p =0,p 2 =0 s 2 t p =
More informationProcess Control & Instrumentation (CH 3040)
First-order 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 first-order
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 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback
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 informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steady-state error, and transient response for computer-controlled systems. Transfer functions,
More informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
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 informationCHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS
CHAPTER 5 : REDUCTION OF MULTIPLE SUBSYSTEMS Objectives Students should be able to: Reduce a block diagram of multiple subsystems to a single block representing the transfer function from input to output
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 informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
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 informationEE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation
EE 3CL4: Introduction to Control Systems Lab 4: Lead Compensation Tim Davidson Ext. 27352 davidson@mcmaster.ca Objective To use the root locus technique to design a lead compensator for a marginally-stable
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 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 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 information6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.
6. Sketch the z-domain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More 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 informationTransient Response of a Second-Order System
Transient Response of a Second-Order System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a well-behaved closed-loop
More informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control
More 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 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 informationCHAPTER 7 STEADY-STATE RESPONSE ANALYSES
CHAPTER 7 STEADY-STATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More 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 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 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 informationProblems -X-O («) s-plane. s-plane *~8 -X -5. id) X s-plane. s-plane. -* Xtg) FIGURE P8.1. j-plane. JO) k JO)
Problems 1. For each of the root loci shown in Figure P8.1, tell whether or not the sketch can be a root locus. If the sketch cannot be a root locus, explain why. Give all reasons. [Section: 8.4] *~8 -X-O
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 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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #4 Monday, January 13, 2003 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Impulse and Step Responses of Continuous-Time
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 informationAN INTRODUCTION TO THE CONTROL THEORY
Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter
More informationChapter 12. Feedback Control Characteristics of Feedback Systems
Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an open-loop system (a system without feedbac) and a closed-loop
More 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 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 informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open
More informationChapter 5 HW Solution
Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, time-invariant system. Let s see, I
More informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More 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 input-output relationship is a second order differential
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 informationMAE 143B - Homework 9
MAE 143B - Homework 9 7.1 a) We have stable first-order 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 straight-line
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 informationLaboratory handouts, ME 340
Laboratory handouts, ME 340 This document contains summary theory, solved exercises, prelab assignments, lab instructions, and report assignments for Lab 4. 2014-2016 Harry Dankowicz, unless otherwise
More informationVideo 5.1 Vijay Kumar and Ani Hsieh
Video 5.1 Vijay Kumar and Ani Hsieh Robo3x-1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior
More informationCourse Background. Loosely speaking, control is the process of getting something to do what you want it to do (or not do, as the case may be).
ECE4520/5520: Multivariable Control Systems I. 1 1 Course Background 1.1: From time to frequency domain Loosely speaking, control is the process of getting something to do what you want it to do (or not
More informationAutomatic Control (TSRT15): Lecture 4
Automatic Control (TSRT15): Lecture 4 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13-282226 Office: B-house extrance 25-27 Review of the last
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a linear-time-invariant
More information10ES-43 CONTROL SYSTEMS ( ECE A B&C Section) % of Portions covered Reference Cumulative Chapter. Topic to be covered. Part A
10ES-43 CONTROL SYSTEMS ( ECE A B&C Section) Faculty : Shreyus G & Prashanth V Chapter Title/ Class # Reference Literature Topic to be covered Part A No of Hours:52 % of Portions covered Reference Cumulative
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: E-mail: info@madeeasy.in Ph: 0-4546 CLASS TEST 08-9 ELECTRONICS ENGINEERING
More 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 informationDEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: CONTROL SYSTEMS YEAR / SEM: II / IV UNIT I SYSTEMS AND THEIR REPRESENTATION PARTA [2
More 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 informationa. Closed-loop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s- plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
More informationExercise 1 (A Non-minimum Phase System)
Prof. Dr. E. Frazzoli 5-59- Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Non-minimum Phase System) To decrease the rise time of the system,
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: Steady-State Error Unit 7: Root Locus Techniques
More informationCHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System
CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages
More 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 informationLecture 25: Tue Nov 27, 2018
Lecture 25: Tue Nov 27, 2018 Reminder: Lab 3 moved to Tuesday Dec 4 Lecture: review time-domain characteristics of 2nd-order systems intro to control: feedback open-loop vs closed-loop control intro to
More informationNotes for ECE-320. Winter by R. Throne
Notes for ECE-3 Winter 4-5 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
More informationEE451/551: Digital Control. Final Exam Review Fall 2013
EE45/55: Digital Control Final Exam Review Fall 03 Exam Overview The Final Exam will consist of four/five questions for EE45/55 students based on Chapters 7 and a bonus based on Chapters 8 9 (students
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Routh-Hurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
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 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 informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
More 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 informationHomework Assignment 3
ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? Open-Loop vs Closed-Loop (Feedback) Why Use Feedback Control? Closed-Loop Control System Structure Elements of a Feedback Control System
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.
SET - 1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies..
More informationBangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory
Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system
More informationAnswers to multiple choice questions
Answers to multiple choice questions Chapter 2 M2.1 (b) M2.2 (a) M2.3 (d) M2.4 (b) M2.5 (a) M2.6 (b) M2.7 (b) M2.8 (c) M2.9 (a) M2.10 (b) Chapter 3 M3.1 (b) M3.2 (d) M3.3 (d) M3.4 (d) M3.5 (c) M3.6 (c)
More informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More informationPoles, Zeros and System Response
Time Response After the engineer obtains a mathematical representation of a subsystem, the subsystem is analyzed for its transient and steady state responses to see if these characteristics yield the desired
More informationEE402 - Discrete Time Systems Spring Lecture 10
EE402 - Discrete Time Systems Spring 208 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 0.. Root Locus For continuous time systems the root locus diagram illustrates the location of roots/poles of a closed
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 informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More information