Systems Analysis and Control


 Eleanore Rose
 3 years 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
Systems 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationBoise 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:
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 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 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 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 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
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 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 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 13: Root Locus Continued Overview In this Lecture, you will learn: Review Definition of Root Locus Points on the Real Axis
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 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 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 informationEE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions
EE C28 / ME C34 Fall 24 HW 6.2 Solutions. PI Controller For the system G = K (s+)(s+3)(s+8) HW 6.2 Solutions in negative feedback operating at a damping ratio of., we are going to design a PI controller
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 6: Generalized and Controller Design Overview In this Lecture, you will learn: Generalized? What about changing OTHER parameters
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
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 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 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 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 informationCDS 101/110a: Lecture 81 Frequency Domain Design
CDS 11/11a: Lecture 81 Frequency Domain Design Richard M. Murray 17 November 28 Goals: Describe canonical control design problem and standard performance measures Show how to use loop shaping to achieve
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 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 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 information] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command prefilter [ 0.
EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965372 Problem (Analysis of a Feedback System) Consider the feedback system
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 informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
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 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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 15: Root Locus Part 4 Overview In this Lecture, you will learn: Which Poles go to Zeroes? Arrival Angles Picking Points? Calculating
More informationME 375 EXAM #1 Friday, March 13, 2015 SOLUTION
ME 375 EXAM #1 Friday, March 13, 2015 SOLUTION PROBLEM 1 A system is made up of a homogeneous disk (of mass m and outer radius R), particle A (of mass m) and particle B (of mass m). The disk is pinned
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 informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12: Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain
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 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 informationEE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO
EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total
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 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 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 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 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 informationMEM 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
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 informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e at, an exponential function s + a sin wt, a sine fun
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 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 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 informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
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 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 informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More 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 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 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 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 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 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 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 informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More 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 informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationControls Problems for Qualifying Exam  Spring 2014
Controls Problems for Qualifying Exam  Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationDiscrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture
Discrete Systems Mark Cannon Hilary Term 22  Lecture 4 Step response and pole locations 4  Review Definition of transform: U() = Z{u k } = u k k k= Discrete transfer function: Y () U() = G() = Z{g k},
More 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 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 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 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 informationDesign of a Lead Compensator
Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc  Funded by MHRD The Lecture Contains Standard Forms of
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 informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
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 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 informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More informationEE402  Discrete Time Systems Spring Lecture 10
EE402  Discrete Time Systems Spring 208 Lecturer: Asst. Prof. M. Mert Ankarali Lecture 0.. Root Locus For continuous time systems the root locus diagram illustrates the location of roots/poles of a closed
More 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 informationDesign via Frequency Response
4th Edition E L E V E N Design via Frequency Response SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Gain Design a. The required phase margin for 25% overshoot (ζ = 0.404), found from Eq. (10.73),
More information