1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I

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1 MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant a, find the ranges of the controller gains K P and K I so that the closed-loop system is stable. (The conditions on K P and K I may depend on a.) (b) (6 points) Set K I =, i.e., let C(s) be a P controller. Find the range of K P so that the closed-loop system is stable for all a. The condition on K P must not depend on a. Part (a) represents the scenario where the designer knows the value of the parameter a. Part (b) represents the scenario where the designer does not know the value of the parameter a, only knows that a, and tries to design a controller that works for all a. Solution. (a) The characteristic equation of the closed-loop system is ( + K P + K ) I s (s + 3)(s + )(s + a) =, which we can rewrite as s(s + 5s + 6)(s + a) + K P s + K I = s 4 + (a + 5)s 3 + (5a + 6)s + (6a + K P )s + K I =. The corresponding Routh array is For stability we must have s 4 : 5a + 6 K I s 3 : a + 5 6a + K P s : 5a + 6 6a+K P a+5 K I s : 6a + K P (a+5)k I 5a+6 6a+K P a+5 s : K I. 5a + 6 6a + K P a + 5 > 6a + K P (a + 5)K I 5a + 6 6a+K P a+5 > K I >,

2 which are equivalent to K P < 5(a + 5a + 6) < K I < (6a + K P ) {5(a + 5a + 6) K P } (a + 5). (b) When K I =, the characteristic equation of the closed-loop system is which we can rewrite as + K P (s + 3)(s + )(s + a) =, (s + 5s + 6)(s + a) + K P = s 3 + (a + 5)s + (5a + 6)s + (6a + K P ) =. The corresponding Routh array is For stability we must have which are equivalent to s 3 : 5a + 6 s : a + 5 6a + K P s : 5a + 6 6a+K P a+5 s : 6a + K P. 5a + 6 6a + K P > a + 5 6a + K P >, 6a < K P < 5(a + 5a + 6). () Since a, the lower bound 6a in () is less than or equal to and the upper bound 5(a +5a+6) in () is greater than or equal to 3. Therefore, in order for K P satisfy the condition in () for all a, K P must be in the range < K P < 3. Problem. (9 points) Consider the closed-loop system in Problem with G(s) = (s + 3)(s + )(s + a) and the PI controller C(s) = K P + K I s. Let K I = K P. Find the locus of the closed-loop roots with respect to K P in following cases: (a) (3 points) < a < (b) (3 points) a = (c) (3 points) < a <

3 Solution. (a) The characteristic equation of the closed-loop system is + K P (s + ) s(s + 3)(s + )(s + a) =. There are four branches to the locus, starting at s = 3, s =, s = a, and s =, respectively. The real-axis segments s 3, s, and a s are parts of the locus. There are 3 asymptotes centered at ( 3 a) ( ) = 4 a and at the angles ±6 and Therefore, the locus of the closed-loop roots with respect to K P is illustrated in Figure. 4 Root Locus Figure : The root locus of the closed-loop system with respect to K P when < a < (This plot is drawn with a =.5). (b) The characteristic equation of the closed-loop system is + K P s(s + 3)(s + ) =. There are three branches to the locus, starting at s = 3, s =, and s =, respectively. The real-axis segments s 3 and s are parts of the locus. There are 3 asymptotes centered at 3 = 5 and at the angles 3 3 ±6 and 8. Therefore, the locus of the closed-loop roots with respect to K P is illustrated in Figure. (c) The characteristic equation of the closed-loop system is + K P (s + ) s(s + 3)(s + )(s + a) =. There are four branches to the locus, starting at s = 3, s =, s = a, and s =, respectively. The real-axis segments s 3, s a, and s are parts of the locus. There are 3 asymptotes centered at ( 3 a) ( ) = 4 a and at the angles ±6 and Therefore, the locus of the closed-loop roots with respect to K P is illustrated in Figure 3.

4 Root Locus Figure : The root locus of the closed-loop system with respect to K P when a =. 4 Root Locus Figure 3: The root locus of the closed-loop system with respect to K P when < a < (This plot is drawn with a =.5). Problem 3. (5 points) Consider a closed-loop system with a plant G(s) = e εs s where < ε < is a constant. Sketch (qualitatively) the Nyquist plot of G(s). Then consider G(s) in feedback with a constant gain K and determine the range of K for which the closed-loop system is stable. Supporting material and hint: (i) Denote φ(ω) = εω + arctan ω π. (ii) Note that φ (ω) = ε + + ω.

5 (iii) Observe that φ() = π, denote by ω (ε) the positive solution of φ(ω) = π, and denote by ω n (ε) the solutions of φ(ω) = (n + )π for n =,,... (iv) Do not calculate ω (ε), ω (ε), ω (ε),..., but do use them in your answer. Solution. In order to draw the Nyquist plot of G(s), evaluate G(s) at s = jω for ω so that G(jω) = e jεω jω = ω + e j(π arctan ω) e jεω = ω + ejφ(ω), from which it follows that ) G(j) =, ) G(jω) =, and ω + 3) G(jω) = φ(ω). To investigate φ(ω), we first look into the φ (ω). It is clear that 4) φ () = ε >, 5) φ (ω) is monotonically decreasing as ω increases from to, and 6) lim ω φ (ω) = ε <, from which it follows that 7) there exists a unique positive solution ω > of the equation φ (ω) =, 8) φ(ω) is monotonically increasing for ω ω, 9) φ(ω) is monotonically decreasing for ω > ω. Moreover, we know ) φ() = π, ) lim ω φ(ω) = from the definition of φ(ω), ) φ(ω) < π for all ω from the fact that arctan ω < π for all ω. From the arguments above, we can conclude the followings: 3) The Nyquist plot of G(s) starts off at (, ) : from ) 4) The distance from the origin to the Nyquist plot is monotonically decreasing and the Nyquist plot converges to : from ) 5) The Nyquist plot moves counterclockwise first : from 8) 6) The Nyquist plot turns back before it reaches the negative imaginary axis and, then, moves clockwise endlessly. : from 8), ), 9), and ) Finally, using the properties 3), 4), 5), and 6), we can qualitatively draw the Nyquist plot of G(s) which is illustrated in Figure 4 and 5. From the properties of φ(ω) and Figure 4 and 5, it follows that the Nyquist plot of G(s) intersects the negative real axis infinitely many times at, G(jω (ε)), G(jω (ε)), G(jω (ε)), and < G(jω (ε)) = ω (ε) + < G(jω (ε)) = ω (ε) + < G(jω (ε)) = ω (ε) +. <....

6 Nyquist Diagram Figure 4: The Nyquist plot of G(s) (This plot is drawn with ε =.5). Nyquist Diagram Figure 5: The magnified plot of Figure 4 around the origin. Since G(s) has one unstable pole, the closed-loop system is stable if, and only if, #CCW #CW= where the encirclement is around (, j). And, this only happens when < < K K G(jω (ε)) = ω. (ε) + Therefore, the range of K is < K < ω (ε) +. Problem 4. (4 points) (a) (6 points) Sketch the Nyquist plot of the plant G(s) = s + s(s ). and de- 5(s + ) (b) (8 points) Sketch the Nyquist plot of the plant G(s) = (s + )(s 3s + ) termine the stability of a unity feedback system with this plant.

7 Solution. (a) Evaluated values of G(s) at s = jω for ω > is G(jω) = jω + ω jω (jω + )(ω j) = ω(ω + 4) = 3ω + j( ω + ) ω(ω + 4) = 3 ω j ω + ω(ω + 4), from which it follows that ) lim ω + G(jω) = j(+ ), ) lim ω G(jω) = j( ), 3) lim ω G(jω) =, 4) G(jω) intersects the imaginary axis at ω =, and 5) G(jω) intersects the real axis at ω = with G(j ) =. Therefore, the Nyquist plot of G(s) is given in Figure 6. 3 Nyquist Diagram Figure 6: The Nyquist plot of G(s) in (a). (b) Evaluated values of G(s) at s = jω for ω is G(jω) = from which it follows that ) G(j) = 5, 5(jω + ) (jω + )( ω j3ω + ) = 5(jω + )( jω + )( ω + + j3ω) (ω + ){( ω + ) + 9ω } 5(ω ω ) = (ω + ){( ω + ) + 9ω } + j 5ω(ω 3) (ω + ){( ω + ) + 9ω },

8 ) lim ω G(jω) =, 3) Re[G(jω)] < and Im[G(jω)] < for sufficiently large ω, 4) G(jω) intersects the imaginary axis at ω =.55 with Im[G(j.55)] >, and 3 5) G(jω) intersects the real axis at ω =, with Re[G(j 3/)] = =.485 <. 7 Therefore, the Nyquist plot of G(s) is given in Figure 7. 4 Nyquist Diagram Figure 7: The Nyquist plot of G(s) in (b). Since G(s) has two unstable poles, the closed-loop system is stable if, and only if, #CCW #CW= = where the encirclement is around (, j). And, from the Nyquist plot in Figure 7, we have #CCW #CW=. Therefore, the closed-loop system is stable. Problem 5. ( points) Sketch the Bode plots of the following systems: (a) (6 points) G(s) = (b) (6 points) G(s) = (s + ) 3 (s.) (s + )(s + ) s + s + (s +.)(s +.s + ) Solution. (a) An approximate Ĝ(jω) is obtained in the following: ) For < ω, log Ĝ(jω) = log G(j) = log = : db/dec Ĝ(jω) = G(j) =. ) For < ω, log Ĝ(jω) = log Ĝ(j ) + ( ) (log ω log ) = 4(log ω + ) : 4 db/dec Ĝ(jω) = Ĝ(j ) ( 9 ) = + 8 = 8.

9 3) For < ω, 4) For < ω, 5) For < ω 3, log Ĝ(jω) = log Ĝ(j) + ( + 3) (log ω log ) = 6 + log ω : db/dec Ĝ(jω) = Ĝ(j) = = 45. log Ĝ(jω) = log Ĝ(j) + ( + 3 ) (log ω log ) = 4 : db/dec Ĝ(jω) = Ĝ(j) 9 = 45 9 = 36. log Ĝ(jω) = log Ĝ(j ) + ( + 3 ) (log ω log ) = 4 (log ω ) : db/dec Ĝ(jω) = Ĝ(j ) 9 = 36 9 = 7. The approximate Ĝ(jω) and the Bode plot of G(s) are given in Figure 8. Magnitude (db) Bode Diagram Frequency (rad/sec) Phase (deg) Frequency (rad/sec) Figure 8: The approximate Ĝ(jω) and the Bode plot of G(s) in (a). (b) The system G(s) can be rewritten as G(s) = s + ζ ω n s + ω n (s +.)(s + ζ ω n s + ω n) where ζ =., ω n =, ζ =.5, and ω n =. An approximate Ĝ(jω) is obtained in the following: ) For < ω, log Ĝ(jω) = log G(j) = log = : db/dec Ĝ(jω) = G(j) =.

10 ) For < ω, 3) For < ω, 4) For < ω, log Ĝ(jω) = log Ĝ(j ) + ( ) (log ω log ) = (log ω + ) : db/dec Ĝ(jω) = Ĝ(j ) 9 = 9 = 9. log Ĝ(jω) = log Ĝ(j) + ( ) (log ω log ) = 6 log ω : 6 db/dec Ĝ(jω) = Ĝ(j) 9 = 9 8 = 7. log Ĝ(jω) = log Ĝ(j) + ( + ) (log ω log ) = 6 (log ω ) : db/dec Ĝ(jω) = Ĝ(j) + 9 = = 9. The approximate Ĝ(jω) and the Bode plot of G(s) are given in Figure 9. Magnitude (db) Bode Diagram Frequency (rad/sec) Phase (deg) Frequency (rad/sec) Figure 9: The approximate Ĝ(jω) and the Bode plot of G(s) in (b). In the Bode plot, the peaks in magnitude and the sharp changes in phase are caused by ζ =. and ζ =.5.

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Outline. 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

MEM 355 Performance Enhancement of Dynamical Systems MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions

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.

EE 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

Some solutions of the written exam of January 27th, 2014 TEORIA DEI SISTEMI Systems Theory) Prof. C. Manes, Prof. A. Germani Some solutions of the written exam of January 7th, 0 Problem. Consider a feedback control system with unit feedback gain, with the following

Lecture 11. Frequency Response in Discrete Time Control Systems EE42 - Discrete Time Systems Spring 28 Lecturer: Asst. Prof. M. Mert Ankarali Lecture.. Frequency Response in Discrete Time Control Systems Let s assume u[k], y[k], and G(z) represents the input, output,

Delhi 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

Classify a transfer function to see which order or ramp it can follow and with which expected error. Dr. J. Tani, Prof. Dr. E. Frazzoli 5-059-00 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,

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

Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type- servomechanism:

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

Chapter 6 - Solved Problems Chapter 6 - Solved Problems Solved Problem 6.. Contributed by - James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the Z-N tuning strategy for the nominal

STABILITY ANALYSIS TECHNIQUES ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to control-system design: 1 Stability, 2 Steady-state response, 3 Transient response

ECEN 326 Electronic Circuits ECEN 326 Electronic Circuits Stability Dr. Aydın İlker Karşılayan Texas A&M University Department of Electrical and Computer Engineering Ideal Configuration V i Σ V ε a(s) V o V fb f a(s) = V o V ε (s)

EE 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

Control 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

CONTROL SYSTEM STABILITY. CHARACTERISTIC EQUATION: The overall transfer function for a. where A B X Y are polynomials. Substitution into the TF gives: CONTROL SYSTEM STABILITY CHARACTERISTIC EQUATION: The overall transfer function for a feedback control system is: TF = G / [1+GH]. The G and H functions can be put into the form: G(S) = A(S) / B(S) H(S)

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.

Plan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design. Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequency-response design method Goal: wrap up lead and

Radar 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

Control Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:

Control Systems. Root Locus & Pole Assignment. L. Lanari Control Systems Root Locus & Pole Assignment L. Lanari Outline root-locus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS - Root

MAE 143B - Homework 8 Solutions MAE 43B - Homework 8 Solutions P6.4 b) With this system, the root locus simply starts at the pole and ends at the zero. Sketches by hand and matlab are in Figure. In matlab, use zpk to build the system

AMME3500: System Dynamics & Control Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13

Outline. 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?

Procedure for sketching bode plots (mentioned on Oct 5 th notes, Pg. 20) Procedure for sketching bode plots (mentioned on Oct 5 th notes, Pg. 20) 1. Rewrite the transfer function in proper p form. 2. Separate the transfer function into its constituent parts. 3. Draw the Bode

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:

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

FREQUENCY RESPONSE ANALYSIS Closed Loop Frequency Response Closed Loop Frequency Response The Bode plot is generally constructed for an open loop transfer function of a system. In order to draw the Bode plot for a closed loop system, the transfer function has

Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 5. 2. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid -

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

EE302 - Feedback Systems Spring Lecture KG(s)H(s) = KG(s) EE3 - Feedback Systems Spring 19 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 1.. 1.1 Root Locus In control theory, root locus analysis is a graphical analysis method for investigating the change of

Module 07 Control Systems Design & Analysis via Root-Locus Method Module 07 Control Systems Design & Analysis via Root-Locus Method Ahmad F. Taha EE 3413: Analysis and Desgin of Control Systems Email: ahmad.taha@utsa.edu Webpage: http://engineering.utsa.edu/ taha March

Introduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31 Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions 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.