Systems Analysis and Control


 Eleanore Rose
 10 months ago
 Views:
Transcription
1 Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain
2 Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering Phase Margin Lag Compensators Change in steadystate error M. Peet Lecture 24: Control Systems 2 / 27
3 Phase and Gain Margin Recall: 3 indicators Gain Margin Phase Margin Bandwidth From Phase Margin and ClosedLoop Bandwidth: Percent Overshoot Peak Time Rise Time Settling Time Percent Overshoot: ζ is from Φ M only %OS = e ( πζ 1 ζ ) 2 M. Peet Lecture 24: Control Systems 3 / 27
4 Phase and Gain Margin Given ζ, we need ω BW to find T r, T s and T p T s = f 1(ζ) ω BW T p = f 2(ζ) ω BW T r = f 3(ζ) ω BW M. Peet Lecture 24: Control Systems 4 / 27
5 Phase and Gain Margin This is all analysis. Design Problem: Achieve 1% Overshoot T r = 2s T s = 1s Step 1: Translate into Φ M and ω BW constraints. Get desired ζ from 1% Overshoot ζ = =.57 ( ln ) %OS 1 π 2 + ln 2 ( %OS 1 ) M. Peet Lecture 24: Control Systems 5 / 27
6 Phase and Gain Margin The desired ζ yields a desired phase margin. Φ M,desired = ζdes 1 = 57 The toughest constraint will be Rise Time: T r < 2s ω BW = f 3(ζ) T r > 2.12 = M. Peet Lecture 24: Control Systems 6 / 27
7 Phase and Gain Margin We can also look at settling time: T s < 1s ω BW = f 1(ζ) T s > 8 1 =.8 Therefore: We want Phase margin of Φ M = 57 Bandwidth of ω BW > 1 M. Peet Lecture 24: Control Systems 7 / 27
8 Lead Compensators So far we only know how to find Φ M and ω BW, but not influence them. Φ M = 35 at ω c = 2.13 ω BW = 3.7 Question: How can we increase Φ M and decrease ω BW? Magnitude (db) Bode Diagram Gm = 1.2 db (at 4.24 rad/s), Pm = 35 deg (at 2.13 rad/s) 15 9 Phase (deg) Answer: Increase gain at ω > ω BW,desired So that G(ıωBW,desired ) = 7dB. Increase phase by 3 at crossover frequency Frequency (rad/s) Lead Compensation M. Peet Lecture 24: Control Systems 8 / 27
9 Lead Compensators Question: How do we design our lead compensator? In root locus, we had pole placement. Look at the Two Lead Compensators: D(s) = k s + 1 s and D 2 (s) = k s + 1 s Bode Diagram 4 3 Magnitude (db) Phase (deg) Frequency (rad/sec) What are the differences? M. Peet Lecture 24: Control Systems 9 / 27
10 Lead Compensators Consider the generalized form of lead compensation α determines how much phase is added α < 1 for lead compensation α > 1 for lag compensation T determines where the phase is added. D(s) = k T s + 1 αt s + 1 We want extra phase at the crossover frequency Note that magnitude is also added at high frequency. Could increase the crossover frequency. Changes the phase margin. M. Peet Lecture 24: Control Systems 1 / 27
11 Lead Compensators Question: Where is the phase added? Find the point of Maximum Phase. The Phase contribution of the lead compensator is Φ(ω) = D(ıω) = T (ıω + 1) (αt ıω + 1) = tan 1 (T ω) tan 1 (αt ω) To find peak phase contribution, set D(ıω) ω =. Which means Dividing by 1 α, we get Φ ω = (T ω) 2 T (αt ω) 2 αt = 1 + (αt ω) 2 (1 + (T ω) 2 )α = 1 α + (α 1)(αT 2 ω 2 ) = 1 αt 2 ω 2 = or ω max = 1 T α M. Peet Lecture 24: Control Systems 11 / 27
12 Lead Compensators So we get that the maximum phase contribution is at ω max = 1 T 1 T α. Convert to a log ω Bode plot: log ω max = 1 2 The maximum occurs at the average of log 1 T [ log 1 T + log 1 ] αt and log 1 αt. 9 Phase (deg) Frequency (rad/sec) log(1/t) log(1/ αt) ( log(1/ T) + log(1/ αt) ) 2 M. Peet Lecture 24: Control Systems 12 / 27
13 Lead Compensators Reconsider our example: T = 1 α = 1 1 D(s) = k s + 1 s Phase (deg) Frequency (rad/sec) ω = 1 ω = 1 ω max = Maximum Phase occurs at ω = 1 T α = M. Peet Lecture 24: Control Systems 13 / 27
14 Lead Compensators Reconsider our example: T = 1 α = 1 1 D(s) = k s + 1 s Phase (deg) Frequency (rad/sec) ω = 1 ω = 1 ω max =1 Maximum Phase occurs at ω = 1 T α = 1 M. Peet Lecture 24: Control Systems 14 / 27
15 Lead Compensators So if we want to add phase at ω c, then we need This is not definitive T α = 1 ω c Depends on how much phase we want to add. Now consider the case where we want to add 3 of phase margin. Question: How much phase does a lead compensator add? We already know the frequency of peak phase. Make this the crossover frequency Independent of T! Φ max = tan 1 1 α tan 1 ( αω) ( ) 1 α = sin α M. Peet Lecture 24: Control Systems 15 / 27
16 Lead Compensators We can plot Φ max vs. 1 α. For 3 phase, we want 1 α = 3. Lets add 3 of phase at ω c = 2.13 T = 1 3 = ω c α 2.13 =.814 M. Peet Lecture 24: Control Systems 16 / 27
17 Lead Compensators This gives us a lead compensator: D(s) = T s + 1 αt s + 1 =.814s s Bode Diagram Magnitude (db) Phase (deg) Frequency (rad/s) M. Peet Lecture 24: Control Systems 17 / 27
18 Lead Compensators Add this to the original plot to get 4 Bode Diagram Gm = 8.79 db (at 6.15 rad/s), Pm = 38.6 deg (at 3.44 rad/s) 2 Magnitude (db) Phase (deg) Frequency (rad/s) New Phase margin is New ω c = 4.79, new ω BW = 9. Crossover frequency is increased, which reduces the phase margin! M. Peet Lecture 24: Control Systems 18 / 27
19 Lead Compensators To avoid changing the crossover frequency, we control the gain. D(s) = k T s + 1 αt s + 1 We want D(ω c ) = 1. Since we have T = 1 ω c α, this yields 1 α ı + 1 D(ıω c ) = k αı + 1 Thus to preserve the crossover frequency, we set α + 1 k = 1 α = + 1 = M. Peet Lecture 24: Control Systems 19 / 27
20 Lead Compensators Apply the final controller to get 4 Bode Diagram Gm = 13.6 db (at 6.15 rad/s), Pm = 65 deg (at 2.13 rad/s) Magnitude (db) Phase (deg) Frequency (rad/s) New Phase margin is 65. New ω c = 2.13, new ω BW = 4. New phase margin is on target  bandwidth increased slightly. M. Peet Lecture 24: Control Systems 2 / 27
21 Lag Compensators What about steadystate error? Want to increase G() No effect on Φ M or ω BW Bode Diagram 3 2 Magnitude (db) Phase (deg) Frequency (rad/sec) Ignore the lead compensator. Increase G() by 15dB. No change at ω c or ω BW e ss = G() M. Peet Lecture 24: Control Systems 21 / 27
22 Lag Compensators Question: How do we increase G() without changing ω BW? We want to reduce gain at ω =. Want no effect on Φ M or G() Look at the Two Lag Compensators: D(s) = 2 log D() = 15 2 log D(ıω BW ) = s s 1 s and D 2 (s) = s Bode Diagram Magnitude (db) Phase (deg) Frequency (rad/sec) M. Peet Lecture 24: Control Systems 22 / 27
23 Lag Compensators Lets use the form: α determines how much phase is added D(s) = T s + 1 αt s + 1 α > 1 for lag compensation As before, min phase is given by ω = 1 T α Center this point at low frequency Change in magnitude at high frequency is 2 log α(t s + 1) 2 log(αt s + 1) If T is large, this is just. At low frequency, gain is 2 log α M. Peet Lecture 24: Control Systems 23 / 27
24 Lag Compensators For our problem, set T =.1 and then we want so α = 1.75 = log α = 15 D(s) = s s Bode Diagram 15 Magnitude (db) Phase (deg) Frequency (rad/sec) M. Peet Lecture 24: Control Systems 24 / 27
25 Lag Compensators The system with lag compensation: 15 Bode Diagram Gm = 1.2 db (at 4.23 rad/sec), Pm = 34.8 deg (at 2.13 rad/sec) 1 Magnitude (db) Phase (deg) Frequency (rad/sec) Bandwidth: ω BW = 3.7 Phase Margin: Φ M = 35 M. Peet Lecture 24: Control Systems 25 / 27
26 Lag Compensators Combine the lead and lag compensators: 15 Bode Diagram Gm = 12.4 db (at 6.92 rad/sec), Pm = 51.1 deg (at 2.78 rad/sec) 1 Magnitude (db) Phase (deg) Frequency (rad/sec) Φ M = 51 ω BW = 4.78 M. Peet Lecture 24: Control Systems 26 / 27
27 Summary You Have Learned: Classical Control Systems Exam Material: Root Locus Drawing Asymptotes, Break Points, etc. Compensation Gain, LeadLag, Notch Filters Bode and Frequency Response What is frequency response? Drawing Bode Plot Closed Loop Dynamics Compensation Nyquist Plot Drawing and Concepts Stability Margins M. Peet Lecture 24: Control Systems 27 / 27
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
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 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 informationELECTRONICS & 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 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 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 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 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 informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More 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 informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
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 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 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 informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
More 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 informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
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 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 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 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 STAFF NAME: Mr. P.NARASIMMAN BRANCH : ECE Mr.K.R.VENKATESAN YEAR : II SEMESTER
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 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 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 informationECE382/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.
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 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 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 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 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
More informationIntroduction to Feedback Control
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More informationProblem 1 (Analysis of a Feedback System  Bode, Root Locus, Nyquist) Consider the feedback system defined by the open loop transfer function 1.
1 EEE480 Final Exam, Spring 2016 A.A. Rodriguez Rules: Calculators permitted, One 8.5 11 sheet, closed notes/books, open minds GWC 352, 9653712 Problem 1 (Analysis of a Feedback System  Bode, Root Locus,
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 informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationDesired Bode plot shape
Desired Bode plot shape 0dB Want high gain Use PI or lag control Low freq ess, type High low freq gain for steady state tracking Low high freq gain for noise attenuation Sufficient PM near ω gc for stability
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
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 informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
More informationEECS C128/ ME C134 Final Wed. Dec. 15, am. Closed book. Two pages of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Wed. Dec. 5, 2 8 am Closed book. Two pages of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 2 2 6 3 4 4 5 6 6 7 8 2 Total
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 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 informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More 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 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 informationNADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni
NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501
More informationECE137B Final Exam. There are 5 problems on this exam and you have 3 hours There are pages 119 in the exam: please make sure all are there.
ECE37B Final Exam There are 5 problems on this exam and you have 3 hours There are pages 9 in the exam: please make sure all are there. Do not open this exam until told to do so Show all work: Credit
More informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
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 informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More informationEE3CL4: 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
More information6.1 Sketch the zdomain root locus and find the critical gain for the following systems K., the closedloop characteristic equation is K + z 0.
6. Sketch the zdomain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance
More 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 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 informationMechanical Systems Part A: StateSpace Systems Lecture AL12
AL: 436433 Mechanical Systems Part A: StateSpace Systems Lecture AL Case study Case study AL: Design of a satellite attitude control system see Franklin, Powell & EmamiNaeini, Ch. 9. Requirements: accurate
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 informationTopic # Feedback Control. StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback
Topic #17 16.31 Feedback Control StateSpace Systems Closedloop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1 Fall
More informationEE Control Systems
Copyright FL Lewis 7 All rights reserved Updated: Moday, November 1, 7 EE 4314  Cotrol Systems Bode Plot Performace Specificatios The Bode Plot was developed by Hedrik Wade Bode i 1938 while he worked
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 3.. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
More informationDESIGN MICROELECTRONICS ELCT 703 (W17) LECTURE 3: OPAMP CMOS CIRCUIT. Dr. Eman Azab Assistant Professor Office: C
MICROELECTRONICS ELCT 703 (W17) LECTURE 3: OPAMP CMOS CIRCUIT DESIGN Dr. Eman Azab Assistant Professor Office: C3.315 Email: eman.azab@guc.edu.eg 1 TWO STAGE CMOS OPAMP It consists of two stages: First
More informationCDS 101/110a: Lecture 102 Control Systems Implementation
CDS 101/110a: Lecture 102 Control Systems Implementation Richard M. Murray 5 December 2012 Goals Provide an overview of the key principles, concepts and tools from control theory  Classical control 
More information1 Mathematics. 1.1 Determine the onesided Laplace transform of the following signals. + 2y = σ(t) dt 2 + 3dy dt. , where A is a constant.
Mathematics. Determine the onesided Laplace transform of the following signals. {, t < a) u(t) =, where A is a constant. A, t {, t < b) u(t) =, where A is a constant. At, t c) u(t) = e 2t for t. d) u(t)
More informationLecture 17 Date:
Lecture 17 Date: 27.10.2016 Feedback and Properties, Types of Feedback Amplifier Stability Gain and Phase Margin Modification Elements of Feedback System: (a) The feed forward amplifier [H(s)] ; (b) A
More informationFrequency Response part 2 (I&N Chap 12)
Frequency Response part 2 (I&N Chap 12) Introduction & TFs Decibel Scale & Bode Plots Resonance Scaling Filter Networks Applications/Design Frequency response; based on slides by J. Yan Slide 3.1 Example
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 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 informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationECE320: Linear Control Systems Homework 8. 1) For one of the rectilinear systems in lab, I found the following state variable representations:
ECE30: Linear Control Systems Homework 8 Due: Thursday May 6, 00 at the beginning of class ) For one of the rectilinear systems in lab, I found the following state variable representations: 0 0 q q+ 74.805.6469
More informationPD, 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 closedloop
More informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
More 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 informationFeedback Control part 2
Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open and closedloop control Everything before chapter 7 are openloop systems Transient response Design criteria Translate criteria
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 Steadystate Error and Type 0, Type
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 informationFeedback design for the Buck Converter
Feedback design for the Buck Converter Portland State University Department of Electrical and Computer Engineering Portland, Oregon, USA December 30, 2009 Abstract In this paper we explore two compensation
More informationSAMPLE 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: Frequencydomain analysis and control design (15 pt) Given is a
More informationControl Systems Design
ELEC4410 Control Systems Design Lecture 13: Stability Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 13: Stability p.1/20 Outline InputOutput
More informationAPPLICATIONS 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
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 informationReglerteknik 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
More informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More information732 Appendix E: Previous EEE480 Exams. Rules: One sheet permitted, calculators permitted. GWC 352,
732 Aedix E: Previous EEE0 Exams EEE0 Exam 2, Srig 2008 A.A. Rodriguez Rules: Oe 8. sheet ermitted, calculators ermitted. GWC 32, 9372 Problem Aalysis of a Feedback System Cosider the feedback system
More informationStep Response for the Transfer Function of a Sensor
Step Response f the Transfer Function of a Sens G(s)=Y(s)/X(s) of a sens with X(s) input and Y(s) output A) First Order Instruments a) First der transfer function G(s)=k/(1+Ts), k=gain, T = time constant
More informationHomework 11 Solution  AME 30315, Spring 2015
1 Homework 11 Solution  AME 30315, Spring 2015 Problem 1 [10/10 pts] R +  K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
More information2.010 Fall 2000 Solution of Homework Assignment 8
2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)
More information376 CHAPTER 6. THE FREQUENCYRESPONSE DESIGN METHOD. D(s) = we get the compensated system with :
376 CHAPTER 6. THE FREQUENCYRESPONSE DESIGN METHOD Therefore by applying the lead compenator with ome gain adjutment : D() =.12 4.5 +1 9 +1 we get the compenated ytem with : PM =65, ω c = 22 rad/ec, o
More informationLecture 8  SISO Loop Design
Lecture 8  SISO Loop Deign Deign approache, given pec Loophaping: inband and outofband pec Fundamental deign limitation for the loop Gorinevky Control Engineering 81 Modern Control Theory Appy reult
More informationStudio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter
Studio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC LowPass
More informationSchool of Mechanical Engineering Purdue University. ME375 Feedback Control  1
Introduction to Feedback Control Control System Design Why Control? OpenLoop vs ClosedLoop (Feedback) Why Use Feedback Control? ClosedLoop Control System Structure Elements of a Feedback Control System
More information9/9/2011 Classical Control 1
MM11 Root Locus Design Method Reading material: FC pp.270328 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),
More informationConventional PaperI Part A. 1. (a) Define intrinsic wave impedance for a medium and derive the equation for intrinsic vy
EEConventional PaperI IES01 www.gateforum.com Conventional PaperI01 Part A 1. (a) Define intrinsic wave impedance for a medium and derive the equation for intrinsic vy impedance for a lossy dielectric
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationCascade Control of a Continuous Stirred Tank Reactor (CSTR)
Journal of Applied and Industrial Sciences, 213, 1 (4): 1623, ISSN: 23284595 (PRINT), ISSN: 2328469 (ONLINE) Research Article Cascade Control of a Continuous Stirred Tank Reactor (CSTR) 16 A. O. Ahmed
More informationEC6405  CONTROL SYSTEM ENGINEERING Questions and Answers Unit  I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
More informationThe Nyquist Stability Test
Handout X: EE24 Fall 2002 The Nyquist Stability Test.0 Introduction With negative feedback, the closedloop transfer function A(s) approaches the reciprocal of the feedback gain, f, as the magnitude of
More informationANALYSIS OF SMALLSIGNAL MODEL OF A PWM DCDC BUCKBOOST CONVERTER IN CCM.
ANALYSIS OF SMALLSIGNAL MODEL OF A PWM DCDC BUCKBOOST CONVERTER IN CCM. A thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in Engineering By Julie J. Lee
More informationAdditional ClosedLoop Frequency Response Material (Second edition, Chapter 14)
Appendix J Additional ClosedLoop Frequency Response Material (Second edition, Chapter 4) APPENDIX CONTENTS J. ClosedLoop Behavior J.2 Bode Stability Criterion J.3 Nyquist Stability Criterion J.4 Gain
More information