Boise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain


 Denis Goodman
 1 years ago
 Views:
Transcription
1 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: Stabilizing Network Servoamplifier + servomotor i s (+0.5s) o Part A: Draw the Bode plots of the uncompensated openloop transfer function and conclude that the closedloop system is unstable. Can this system be stabilized by a network consisting of a preamplifier with gain K? Part B: Design a stabilizing network so that:. The compensated closedloop system is stable with a phase margin (PM) greater than 5 o and a gain margin (GM) greater than 0 db.. The lowfrequency gain of the compensated openloop system is equal to that of the uncompensated system. 3. The bandwidth of the compensated openloop system is twice that of the uncompensated system. Solution: Part A: Stability Analysis of the Uncompensated System: The magnitude and phase of the uncompensated openloop transfer function are given by: G(jω) = = o (jω) = ( + jω/) ω 80 o + (ω/) tan ω/ 0.5 ω + (ω/) 80 o tan ω = G(ω) φ(ω) The Bode plots (magnitude and phase) of the uncompensated openloop transfer function are shown in solid line on the next page. The closedloop system is unstable because the openloop transfer function has negative phase margin at the crossover frequency ω c = / rad/s. Because of this negative phase margin, the system can never be stabilized by a preamplifier with a real gain K. The Bode magnitude plot would be shifted upwards and the croossover frequency would increase. Clearly, the phase will always be less than 80 o resulting in a negative phase margin.
2 6 db/oct 0 /8 / 6 db/oct / ω (rad/s) MAGNITUDE (db) 36 db/oct 8 db/oct deg/oct 5 deg/oct /8 / / ω (rad/s) PHASE (deg) 90 5 deg/oct 5 deg/oct deg/oct 70 5 deg/oct
3 Part B: StabilizingNetwork Design Stabilizing Network Servoamplifier + servomotor i + K +st α +s α T 0.5 s (+0.5s) o Preamplifier Lead Network We need to add some phase lead to the uncompensated openloop transfer function in order to obtain positive phase margin. To realize the stabilizing network, we select a preamplifier followed by a lead network. The overall compensated openloop transfer function has the following form: G(s) = K α + st + sαt 0.5 s ( + s/) = K α + s/ω s/ω s ( + s/) The lowfrequency approximation of the compensated openloop transfer function should equal that of the uncompensated system. Therefore, G(s) = K α + st + sαt 0.5 s ( + s/) Therefore, the preamplifier gain K should be selected such that = Kα 0.5 s = 0.5 s = Kα = K = α From the previous Bode plots, the bandwidth of the uncompensated openloop transfer function is equal to the zerodb crossover frequency, that is ω B = / rad/s. The new openloop bandwidth should be twice that of the uncompensated bandwith, that is, ω B = = rad/s As a rule of thumb, we want the compensated openloop transfer function to cross the zerodb line with a slope of 6 db/oct (equivalently, 0 db/dec). Using the previous Bode plots, the phase margin of the uncompensated openloop transfer function at ω c = rad/s is equal to 5 o (that is, it is 5 o below the 80 o line). Since we want a minimum phase margin of 5 o, the phase lead to be added at the crossover frequency ω c = rad/s should be at least 5 o + 5 o = 60 o. Recalling the relationship between the maximum phase of a lead network and α, we should have tan α α 60o α α tan 60o = 3 α 3 α + α α α α α + 0 = α 7 8 = or α = 3.9 Since α <, we choose α = 6 = = K = 6 3
4 The maximum phase occurs at ω c = T α = rad/s = T α = T 6 = = T = ω = = T rad/s ω = = = rad/s αt /6 Finally, the openloop transfer function of the compensated system can be expressed as G(s) = 6 + s 6 + s/ 0.5 s ( + s/) = 0.5( + s) s ( + s/) The magnitude and phase of the compensated openloop transfer function are given by: G(jω) = 0.5( + jω) (jω) ( + jω/) = (ω) tan ω ω 80 o [ + (ω/) ] tan ω/ = ω ω ( + ω /6 tan ω 80 o tan ω = G(ω) φ(ω) The Bode plots (magnitude and phase) of the compensated openloop transfer function are shown in dashed line next to the uncompensated Bode plots. Clearly, the system is now stable with a positive phase margin at the crossover frequency ω c = rad/s. We can either use the Bode straightline approximations to compute the gain and phase margins of the compensated system, or we can do exact calculations using a calculator. We will do both below. Phase Margin: Assuming that the crossover frequency is exactly ω c = rad/s, we read an approximate phase margin of PM = 5 o on the graph. An exact computation yields φ(ω c ) = tan ω c 80 o tan ω c = tan 80 o tan PM = 80 o + φ(ω c ) = 80 o 3. o = 7.9 o = 3. o The exact crossover frequency is such that G(ω) = ω ω ( + ω /6) = This expression can be reduced to the following polynomial equation: ω 8 + 3ω ω 56ω 6 = 0 which has ω c = rad/s as a solution. The exact phase margin at this frequency is φ(ω c ) = tan ω c 80 o tan ω c φ(ω c ) = 3.8 o PM = 80 o + φ(ω c ) = 80 o 3.8 o = 8. o = tan o tan
5 Gain Margin: From the Bode plots, the compensated openloop transfer function has a phase of 80 o at ω o = rad/s. The actual phase at ω o = rad/s is φ(ω o ) = tan ω o 80 o tan ω o The approximate gain margin at this frequency is GM (db) = 0 log G(ω o ) = tan 80 o tan = 0 log G(ω o ) = ( ) = db = 83. o Using the fact that there is a maximum error of 6 db at the double pole ω o = rad/s, we can conclude that the gain margin is truly GM = 6 = 8 db An exact computation yields G(ω o ) = ωo ωo( + ωo/6) = ( + = 0.5 /6) GM = 0 log G(ω o ) = 0 log 0.5 = 8.0 db To be more accurate, we need to compute the exact frequency at which the phase is equal to 80 o. Using a calculator, and, by trial and error: By interpolation, ω o ω o (rad/s) φ(ω o ) (deg) 68.5 o 76 o 78 o 79.5 o 80.8 o = 3.7 rad/s. We check that φ(ω o ) = tan ω o 80 o tan ω o φ(ω o ) = o = 80 o = tan o 3.7 tan G(ω o ) = ωo ωo( + ωo/6) = ( = 0.3 /6) GM = 0 log G(ω o ) = 0 log 0.3 = 6.9 db Therefore, all of the criteria have been met and this design is satisfactory. Exercise #: Redo the design using decade straightline approximations. Exercise #: Redo the design using a PD controller of the form G c (s) = K p + K d s = K p ( + sk d /K p ) = ( + sk). (Notice that K p must be equal to using Criterion #.) Design for an approximate gain K that will meet the design objectives. There is no need for a preamplifier in this case. 5
ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationActive Control? Contact : Website : Teaching
Active Control? Contact : bmokrani@ulb.ac.be Website : http://scmero.ulb.ac.be Teaching Active Control? Disturbances System Measurement Control Controler. Regulator.,,, Aims of an Active Control Disturbances
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 informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More 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 informationCHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION
CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationLecture 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.
More informationModule 5: Design of Sampled Data Control Systems Lecture Note 8
Module 5: Design of Sampled Data Control Systems Lecture Note 8 Laglead Compensator When a single lead or lag compensator cannot guarantee the specified design criteria, a laglead compensator is used.
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
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 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 informationMAE 143B  Homework 9
MAE 143B  Homework 9 7.1 a) We have stable firstorder poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straightline
More informationFrequency Response Analysis
Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
More 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 informationAsymptotic Bode Plot & LeadLag Compensator
Asymptotic Bode Plot & LeadLag Compensator. Introduction Consider a general transfer function Ang Man Shun 20225 G(s = n k=0 a ks k m k=0 b ks k = A n k=0 (s z k m k=0 (s p k m > n When s =, transfer
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
More information8.1.6 Quadratic pole response: resonance
8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Secondorder denominator, of the form 1+a 1 s + a s v 1 (s) + C R Twopole lowpass filter example v (s) with
More informationEE 4343/ Control System Design Project LECTURE 10
Copyright S. Ikenaga 998 All rights reserved EE 4343/5329  Control System Design Project LECTURE EE 4343/5329 Homepage EE 4343/5329 Course Outline Design of Phaselead and Phaselag compensators using
More informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
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
More informationECE382/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
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationECE382/ME482 Spring 2005 Homework 8 Solution December 11,
ECE382/ME482 Spring 25 Homework 8 Solution December 11, 27 1 Solution to HW8 P1.21 We are given a system with open loop transfer function G(s) = K s(s/2 + 1)(s/6 + 1) and unity negative feedback. We are
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
More informationECE 388 Automatic Control
Lead Compensator and PID Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage: http://ece388.cankaya.edu.tr
More informationLINEAR CONTROL SYSTEMS. Ali Karimpour Associate Professor Ferdowsi University of Mashhad
LINEAR CONTROL SYSTEMS Ali Karimpour Associate Professor Ferdowsi University of Mashhad Controller design in the frequency domain Topics to be covered include: Lag controller design 2 Dr. Ali Karimpour
More informationEngraving Machine Example
Engraving Machine Example MCE44  Fall 8 Dr. Richter November 24, 28 Basic Design The Xaxis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional
More informationPM diagram of the Transfer Function and its use in the Design of Controllers
PM diagram of the Transfer Function and its use in the Design of Controllers Santiago Garrido, Luis Moreno Abstract This paper presents the graphical chromatic representation of the phase and the magnitude
More informationNyquist Criterion For Stability of Closed Loop System
Nyquist Criterion For Stability of Closed Loop System Prof. N. Puri ECE Department, Rutgers University Nyquist Theorem Given a closed loop system: r(t) + KG(s) = K N(s) c(t) H(s) = KG(s) +KG(s) = KN(s)
More informationH(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
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 informationK(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s
321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify
More information2ndorder filters. EE 230 secondorder filters 1
ndorder filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth, first, or secondorder polyinomials in the numerator. Use two reactive
More informationAnalysis and Design of Analog Integrated Circuits Lecture 12. Feedback
Analysis and Design of Analog Integrated Circuits Lecture 12 Feedback Michael H. Perrott March 11, 2012 Copyright 2012 by Michael H. Perrott All rights reserved. Open Loop Versus Closed Loop Amplifier
More informationThe 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
More informationECEN 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)
More informationProfessor Fearing EE C128 / ME C134 Problem Set 7 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley
Professor Fearing EE C8 / ME C34 Problem Set 7 Solution Fall Jansen Sheng and Wenjie Chen, UC Berkeley. 35 pts Lag compensation. For open loop plant Gs ss+5s+8 a Find compensator gain Ds k such that the
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
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 informationSKEE 3143 CONTROL SYSTEM DESIGN. CHAPTER 3 Compensator Design Using the Bode Plot
SKEE 3143 CONTROL SYSTEM DESIGN CHAPTER 3 Compenator Deign Uing the Bode Plot 1 Chapter Outline 3.1 Introduc4on Re viit to Frequency Repone, ploang frequency repone, bode plot tability analyi. 3.2 Gain
More informationESE319 Introduction to Microelectronics. Feedback Basics
Feedback Basics Feedback concept Feedback in emitter follower Stability Onepole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability
More informationESE319 Introduction to Microelectronics. Feedback Basics
Feedback Basics Stability Feedback concept Feedback in emitter follower Onepole feedback and root locus Frequency dependent feedback and root locus Gain and phase margins Conditions for closed loop stability
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 informationTheory of Machines and Automatic Control Winter 2018/2019
Theory of Machines and Automatic Control Winter 2018/2019 Lecturer: Sebastian Korczak, PhD, Eng. Institute of Machine Design Fundamentals  Department of Mechanics http://www.ipbm.simr.pw.edu.pl/ Lecture
More informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10
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 informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More 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 informationSolution: K m = R 1 = 10. From the original circuit, Z L1 = jωl 1 = j10 Ω. For the scaled circuit, L 1 = jk m ωl 1 = j10 10 = j100 Ω, Z L
Problem 9.9 Circuit (b) in Fig. P9.9 is a scaled version of circuit (a). The scaling process may have involved magnitude or frequency scaling, or both simultaneously. If R = kω gets scaled to R = kω, supply
More informationThe Frequencyresponse Design Method
Chapter 6 The Frequencyresponse Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)
More informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationLecture 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,
More informationControl Systems. Control Systems Design LeadLag Compensator.
Design LeadLag Compensator hibum@seoulteh.a.kr Outline Lead ompensator design in frequeny domain Lead ompensator design steps. Example on lead ompensator design. Frequeny Domain Design Frequeny response
More informationAn Internal Stability Example
An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several polezero cancellations between the controller and
More information6.302 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 closedloop behavior what we want it to be. To review:  G c (s) G(s) H(s) you are here! plant For
More informationFREQUENCY 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
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationMAE 143B  Homework 9
MAE 43B  Homework 9 7.2 2 2 3.8.6.4.2.2 9 8 2 2 3 a) G(s) = (s+)(s+).4.6.8.2.2.4.6.8. Polar plot; red for negative ; no encirclements of, a.s. under unit feedback... 2 2 3. 4 9 2 2 3 h) G(s) = s+ s(s+)..2.4.6.8.2.4
More informationEKT 119 ELECTRIC CIRCUIT II. Chapter 3: Frequency Response of AC Circuit Sem2 2015/2016 Dr. Mohd Rashidi Che Beson
EKT 9 ELECTRIC CIRCUIT II Chapter 3: Frequency Response of AC Circuit Sem 05/06 Dr. Mohd Rashidi Che Beson TRANSFER FUNCTION (TF Frequency response can be obtained by using transfer function. DEFINITION:
More informationRobust fixedorder H Controller Design for Spectral Models by Convex Optimization
Robust fixedorder H Controller Design for Spectral Models by Convex Optimization Alireza Karimi, Gorka Galdos and Roland Longchamp Abstract A new approach for robust fixedorder H controller design by
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationFrequency (rad/s)
. The frequency response of the plant in a unity feedback control systems is shown in Figure. a) What is the static velocity error coefficient K v for the system? b) A lead compensator with a transfer
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
More informationPoles and Zeros and Transfer Functions
Poles and Zeros and Transfer Functions Transfer Function: Considerations: Factorization: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial
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 informationECEN 607 (ESS) OpAmps Stability and Frequency Compensation Techniques. Analog & MixedSignal Center Texas A&M University
ECEN 67 (ESS) OpAmps Stability and Frequency Compensation Techniques Analog & MixedSignal Center Texas A&M University Stability of Linear Systems Harold S. Black, 97 Negative feedback concept Negative
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationFEEDBACK AND STABILITY
FEEDBCK ND STBILITY THE NEGTIVEFEEDBCK LOOP x IN X OUT x S + x IN x OUT Σ Signal source _ β Open loop Closed loop x F Feedback network Output x S input signal x OUT x IN x F feedback signal x IN x S x
More informationMAE 142 Homework #2 (Design Project) SOLUTIONS. (a) The main body s kinematic relationship is: φ θ ψ. + C 3 (ψ) 0 + C 3 (ψ)c 1 (θ)
MAE 42 Homework #2 (Design Project) SOLUTIONS. Top Dynamics (a) The main body s kinematic relationship is: ω b/a ω b/a + ω a /a + ω a /a ψâ 3 + θâ + â 3 ψˆb 3 + θâ + â 3 ψˆb 3 + C 3 (ψ) θâ ψ + C 3 (ψ)c
More informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad  500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III
More informationAutomatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: Student ID number... Signature...
Automatic Control (MSc in Mechanical Engineering) Lecturer: Andrea Zanchettin Date: 29..23 Given and family names......................solutions...................... Student ID number..........................
More 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 informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
More information= rad/sec. We can find the last parameter, T, from ωcg new
EE572 Solution to HW#22. Keep working on your project!! 1. Consider the following system: W(s) + T s =1 msec G lead (z) G zoh (z) 8 ( s+ 4)  a) Design a lead compensator, G lead (z), which meets the following
More informationThe loop shaping paradigm. Lecture 7. Loop analysis of feedback systems (2) Essential specifications (2)
Lecture 7. Loop analysis of feedback systems (2). Loop shaping 2. Performance limitations The loop shaping paradigm. Estimate performance and robustness of the feedback system from the loop transfer L(jω)
More informationLoop shaping exercise
Loop shaping exercise Excerpt 1 from Controlli Automatici  Esercizi di Sintesi, L. Lanari, G. Oriolo, EUROMA  La Goliardica, 1997. It s a generic book with some typical problems in control, not a collection
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 informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationDIGITAL CONTROLLER DESIGN
ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steadystate accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some
More informationChapter 9: Controller design
Chapter 9. Controller Design 9.1. Introduction 9.2. Effect of negative feedback on the network transfer functions 9.2.1. Feedback reduces the transfer function from disturbances to the output 9.2.2. Feedback
More informationSecondorder filters. EE 230 secondorder filters 1
Secondorder filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth, first, or secondorder polynomials in the numerator. Use two
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More 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 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 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 informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationThe FrequencyResponse
6 The FrequencyResponse Design Method A Perspective on the FrequencyResponse Design Method The design of feedback control systems in industry is probably accomplished using frequencyresponse methods
More informationGeneralized Fractional Order Reset Element (GFrORE)
ENOC 217, June 253, 217, Budapest, Hungary Generalized Fractional Order Reset Element (GFrORE) Niranjan Saikumar and Hassan HosseinNia Precision and Microsystems Engineering, 3ME, TU Delft, The Netherlands
More informationVALLIAMMAI ENGINEERING COLLEGE
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK V SEMESTER IC650 CONTROL SYSTEMS Regulation 203 Academic Year 207 8 Prepared
More information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
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 information