Due Wednesday, February 6th EE/MFS 599 HW #5

Save this PDF as:

Size: px
Start display at page:

Transcription

1 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)] W(s) + - G c (s) G(s) Y(s) a) Let q= (type zero system) and plot the OPEN-LOOP step response for K=, and. Ans:.4 EE/MFS 599: Open-Loop Step respons for H(s) = /[(s+)(s +9)]

2 EE/MFS 599: Open-Loop Step respons for H(s) = /[(s+)(s +9)] EE/MFS 599: Open-Loop Step respons for H(s) = /[(s+)(s +9)] b) Measure the settling time and overshoot. Does changing the value of the gain alter the dynamics of the system (i.e., is the settling time faster, does the overshoot change)? Ans: The settling time (%) is about 4 seconds and is the same for all three responses. The overshoot is zero. No, changing the gain of the open-loop system does not alter the dynamics c) Now, consider the closed-loop system and let our controller Gc(s) be a simple proportional gain K. Use the matlab rlocus() command to plot the closed-loop poles of this system as we vary K from. to. Do these poles ever go unstable (cross into the RHP)?

3 Ans: The following is the result of the rlocus() command in Matlab 3 Root Locus Imaginary Axis Real Axis As we can see, the poles never cross the jω-axis and the closed-loop system is stable for all values of K d) Look at the root locus. Let s be the pole that corresponds to a settling time of.8 seconds and a damping ratio of.77 (i.e., about 5% overshoot). Is the pole s on your root locus? If so, find the value of the gain K which produced these poles by solving the equation, +KG(s)= Ans: The pole s is -5+j5 and is on the root locus. The value of K which produces this pole is K = -/G(s) = 4. e) Find the static error and velocity coefficients (Kp and Kv) and calculate what the theorectical steady-state error should be for a step and ramp input Ans: Kp = 4/9 and Kv = zero. Therefore, the steady-state error for a step is /(+4/9) = 9/5 =.8 and e ss for a ramp is. f) Plot the closed-loop step and ramp response for Gc(s) = K in part d). Does the settling time and overshoot for the step response agree with what you expected? Ans: Yes (see plot below)

4 .9 EE/MFS 599: Closed-Loop Step respons for H(s) = 4/[(s+)(s +9)] EE/MFS 599: Closed-Loop ramp response for H(s) = 4/[(s+)(s +9)] g) Use your plots to find the steady-state error for each input and compare to the theoretical answers. Can your system follow a ramp? Ans: for the closed-loop step response, y ss =.8 thus the steady-state error is -.8 =.8 which agrees with our theory. The ramp error is clearly headed toward infinity which also agrees with our theory.

5 h) We can decrease the steady-state error due to a step by increasing the value of the gain K. Try using a K ten times greater and again find the step response and measure the steady-state error and compare to the new value of /(+Kp). What happened to the overshoot when you increased the gain by a factor of? Ans: As can bee seen from the plot below, the steady-state error due to a step is much smaller when we increase the gain by a factor of. Theoretically, the value is /( +4/9) = 9/49 =.5. Next, consider the closed-loop system when q= (type system) a) Now, consider the closed-loop system and let our controller Gc(s) be a simple proportional gain K. Use the matlab rlocus() command to plot the closed-loop poles of this system as we vary K from. to. Do these poles ever go unstable (cross into the RHP)? Ans: Yes, the closed-loop system goes unstable as shown by the root locus plot below: Root Locus 5 5 Imaginary Axis Real Axis b) Look at the root locus. Let s be the pole that corresponds to a settling time of seconds and a damping ratio of.77. Is the pole s on your root locus? The desired dominant closed-loop pole is s= -4+j4 which is not on the root locus c) If no, we have to design a PD compensator to force the root locus to pass thru s. Let Gpd(s) = Kp + Kds = K(s+z) where K=Kd and z=kp/kd. Find values of K and z to make s a closed-loop pole of the system Ans: K(s+z) = -/G(s) = -9.+j5.6. Thus K = 5.6/4 = 3.9 and z = -9./K +4 =.64. Thus, Gpd(s) = 3.9(s+.64)=3.9s+6.4. So, Kp =6.4 and Kd = 3.9 d) Find the static error and velocity coefficients (Kp and Kv) and calculate what the theorectical steady-state error should be for a step and ramp input Ans: Kp is now infinite and Kv = 6.4*/9=7.. Thus, e ss for a step is zero and e ss for a ramp is /7. =.46. e) Plot the closed-loop step and ramp response for Gpd(s) = K(s+z) in part c). Does the settling time and overshoot for the step response agree with what you expected? If not, what have you inserted in the system that could be influencing these values? Ans: Recall Mohannad stated in class that a ζ =.77 yields an overshoot of 4.3%. The settling time should be -4/ζω n = second. As can be seen from the step response, the overshoot is much greater than expected (8%) and the settling time is slightly larger (.35 seconds). This is because we have inserted a zero at -.64 which interferes with the dominant poles at -+j.

6 EE/MFS 599: Closed-Loop PD Compensated Step response for H(s) = /[s(s+)(s +9)] EE/MFS 599: Closed-Loop PD Compensated Ramp response for H(s) = /[s(s+)(s +9)] f) Use your plots to find the steady-state error for each input and compare to the theoretical answers. Can your system follow a ramp? Ans: The ramp response now can follow a ramp and the steady-state error is -y()=-.8564=.436 (instead of.46). The discrepancy is because we have only simulated for seconds. If we allowed the simulation to run longer, we would obtain an identical answer. The steady-state error due to a ramp is zero as predicted.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:

Root Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus - 1

Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position

Homework 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

AN 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

Transient 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

Essence of the Root Locus Technique

Essence of the Root Locus Technique In this chapter we study a method for finding locations of system poles. The method is presented for a very general set-up, namely for the case when the closed-loop

9/9/2011 Classical Control 1

MM11 Root Locus Design Method Reading material: FC pp.270-328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),

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

ECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]

ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative

Root 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

School of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:

Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus - 1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See

Introduction 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

Transient 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

(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.

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)

ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques

CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques [] For the following system, Design a compensator such

CHAPTER 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

100 (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

INTRODUCTION 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

ME 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

Alireza Mousavi Brunel University

Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched

Chapter 7 : Root Locus Technique

Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci

a. 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

Example 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

Root 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

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

Design via Root Locus

Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steady-state error (Sections

Lab # 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

Performance 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

Lecture 1 Root Locus

Root Locus ELEC304-Alper Erdogan 1 1 Lecture 1 Root Locus What is Root-Locus? : A graphical representation of closed loop poles as a system parameter varied. Based on Root-Locus graph we can choose the

EEE 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)

Automatic 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

PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada

PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop

Chapter 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

EEL2216 Control Theory CT1: PID Controller Design

EEL6 Control Theory CT: PID Controller Design. Objectives (i) To design proportional-integral-derivative (PID) controller for closed loop control. (ii) To evaluate the performance of different controllers

Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc.

Fundamental of Control Systems Steady State Error Lecturer: Dr. Wahidin Wahab M.Sc. Aries Subiantoro, ST. MSc. Electrical Engineering Department University of Indonesia 2 Steady State Error How well can

1 (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

KINGS 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

Dynamic 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

EE3CL4: Introduction to Linear Control Systems

1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We

Notes 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...................

Chapter 6 Steady-State Analysis of Continuous-Time Systems

Chapter 6 Steady-State Analysis of Continuous-Time Systems 6.1 INTRODUCTION One of the objectives of a control systems engineer is to minimize the steady-state error of the closed-loop system response

School of Mechanical Engineering Purdue University. ME375 Feedback Control - 1

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

ROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop 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 closed-loop system when the gain K changes from 0 to 1+ K G ( s)

Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback

Topic #17 16.31 Feedback Control State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall

Solution for Mechanical Measurement & Control

Solution for Mechanical Measurement & Control December-2015 Index Q.1) a). 2-3 b).3-4 c). 5 d). 6 Q.2) a). 7 b). 7 to 9 c). 10-11 Q.3) a). 11-12 b). 12-13 c). 13 Q.4) a). 14-15 b). 15 (N.A.) Q.5) a). 15

Pitch Rate CAS Design Project

Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability

1 Chapter 9: Design via Root Locus

1 Figure 9.1 a. Sample root locus, showing possible design point via gain adjustment (A) and desired design point that cannot be met via simple gain adjustment (B); b. responses from poles at A and B 2

ECE382/ME482 Spring 2005 Homework 7 Solution April 17, K(s + 0.2) s 2 (s + 2)(s + 5) G(s) =

ECE382/ME482 Spring 25 Homework 7 Solution April 17, 25 1 Solution to HW7 AP9.5 We are given a system with open loop transfer function G(s) = K(s +.2) s 2 (s + 2)(s + 5) (1) and unity negative feedback.

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control DC Motor Control Trainer (DCMCT) Student Manual Table of Contents 1 Laboratory Objectives1 2 References1 3 DCMCT Plant

Lecture 7:Time Response Pole-Zero 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 First-Order Specs: Step : Pole Real inputs contain

EE3CL4: 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

Chapter 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

Feedback Control part 2

Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open- and closed-loop control Everything before chapter 7 are open-loop systems Transient response Design criteria Translate criteria

Lab 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

SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015

FACULTY OF ENGINEERING AND SCIENCE SAMPLE SOLUTION TO EXAM in MAS501 Control Systems 2 Autumn 2015 Lecturer: Michael Ruderman Problem 1: Frequency-domain analysis and control design (15 pt) Given is a

Root 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

ME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II

ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and

FEEDBACK 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

ME 375 System Modeling and Analysis. Homework 11 Solution. Out: 18 November 2011 Due: 30 November 2011 = + +

Out: 8 November Due: 3 November Problem : You are given the following system: Gs () =. s + s+ a) Using Lalace and Inverse Lalace, calculate the unit ste resonse of this system (assume zero initial conditions).

DEPARTMENT 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

APPLICATIONS 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

Analysis 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.

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

Feedback Control of Linear SISO systems. Process Dynamics and Control

Feedback Control of Linear SISO systems Process Dynamics and Control 1 Open-Loop Process The study of dynamics was limited to open-loop systems Observe process behavior as a result of specific input signals

EE Control Systems LECTURE 14

Updated: Tueday, March 3, 999 EE 434 - Control Sytem LECTURE 4 Copyright FL Lewi 999 All right reerved ROOT LOCUS DESIGN TECHNIQUE Suppoe the cloed-loop tranfer function depend on a deign parameter k We

NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni-625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501

H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at )

.7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a

Unit 7: Part 1: Sketching the Root Locus. Root Locus. Vector Representation of Complex Numbers

Root Locus Root Locus Unit 7: Part 1: Sketching the Root Locus Engineering 5821: Control Systems I Faculty of Engineering & Applied Science Memorial University of Newfoundland 1 Root Locus Vector Representation

16.31 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

Homework 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

Laplace Transform Analysis of Signals and Systems

Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.

FREQUENCY-RESPONSE DESIGN

ECE45/55: Feedback Control Systems. 9 FREQUENCY-RESPONSE DESIGN 9.: PD and lead compensation networks The frequency-response methods we have seen so far largely tell us about stability and stability margins

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics

Systems 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

The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location.

Introduction to the Nyquist criterion The Nyquist criterion relates the stability of a closed system to the open-loop frequency response and open loop pole location. Mapping. If we take a complex number

The Frequency-Response

6 The Frequency-Response Design Method A Perspective on the Frequency-Response Design Method The design of feedback control systems in industry is probably accomplished using frequency-response methods

Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur. Module 3 Lecture 8

Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur Module 3 Lecture 8 Welcome to lecture number 8 of module 3. In the previous

Control System Design

ELEC ENG 4CL4: Control System Design Notes for Lecture #22 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Friday, March 5, 24 More General Effects of Open Loop Poles

Chap 8. State Feedback and State Estimators

Chap 8. State Feedback and State Estimators Outlines Introduction State feedback Regulation and tracking State estimator Feedback from estimated states State feedback-multivariable case State estimators-multivariable

Chapter 9 Tradeoffs and Limits of Performance 9. Introduction Fundamental limits of feedback systems will be investigated in this chapter. We begin in Section 9.2 by discussing the basic feedback loop

Reglerteknik 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

The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain

Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may

MAK 391 System Dynamics & Control. Presentation Topic. The Root Locus Method. Student Number: Group: I-B. Name & Surname: Göksel CANSEVEN

MAK 391 System Dynamics & Control Presentation Topic The Root Locus Method Student Number: 9901.06047 Group: I-B Name & Surname: Göksel CANSEVEN Date: December 2001 The Root-Locus Method Göksel CANSEVEN

PM 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

Control System Design

ELEC4410 Control System Design Lecture 19: Feedback from Estimated States and Discrete-Time Control Design Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science

Lecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore

Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.

Chapter 7 - Solved Problems

Chapter 7 - Solved Problems Solved Problem 7.1. A continuous time system has transfer function G o (s) given by G o (s) = B o(s) A o (s) = 2 (s 1)(s + 2) = 2 s 2 + s 2 (1) Find a controller of minimal

Design of a Heading Autopilot for Mariner Class Ship with Wave Filtering Based on Passive Observer

Design of a Heading Autopilot for Mariner Class Ship with Wave Filtering Based on Passive Observer 1 Mridul Pande, K K Mangrulkar 1, Aerospace Engg Dept DIAT (DU), Pune Email: 1 mridul_pande000@yahoo.com

Solutions to Skill-Assessment Exercises

Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part

FREQUENCY-RESPONSE ANALYSIS

ECE450/550: Feedback Control Systems. 8 FREQUENCY-RESPONSE ANALYSIS 8.: Motivation to study frequency-response methods Advantages and disadvantages to root-locus design approach: ADVANTAGES: Good indicator

CHAPTER 4 STATE FEEDBACK AND OUTPUT FEEDBACK CONTROLLERS

54 CHAPTER 4 STATE FEEDBACK AND OUTPUT FEEDBACK CONTROLLERS 4.1 INTRODUCTION In control theory, a controller is a device which monitors and affects the operational conditions of a given dynamic system.

Class 11 Root Locus part I

Class 11 Root Locus part I Closed loop system G(s) G(s) G(s) Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K We shall assume here K >,

Autonomous Mobile Robot Design

Autonomous Mobile Robot Design Topic: Guidance and Control Introduction and PID Loops Dr. Kostas Alexis (CSE) Autonomous Robot Challenges How do I control where to go? Autonomous Mobile Robot Design Topic:

6.302 Feedback Systems Recitation 16: Compensation Prof. Joel L. Dawson

Bode Obstacle Course is one technique for doing compensation, or designing a feedback system to make the closed-loop behavior what we want it to be. To review: - G c (s) G(s) H(s) you are here! plant For

EE 370L Controls Laboratory. Laboratory Exercise #7 Root Locus. Department of Electrical and Computer Engineering University of Nevada, at Las Vegas

EE 370L Controls Laboratory Laboratory Exercise #7 Root Locus Department of Electrical an Computer Engineering University of Nevaa, at Las Vegas 1. Learning Objectives To emonstrate the concept of error

Proportional, Integral & Derivative Control Design. Raktim Bhattacharya

AERO 422: Active Controls for Aerospace Vehicles Proportional, ntegral & Derivative Control Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University