Module 5: Design of Sampled Data Control Systems Lecture Note 8


 Meredith Clark
 1 years ago
 Views:
Transcription
1 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. In laglead compensator the lag part precedes the lead part. A continuous time laglead compensator is given by The corner frequencies are C(s) = K +τ s +α τ s α τ, τ, τ 2, +τ 2 s +α 2 τ 2 s where, α >, α 2 < α 2 τ 2. The frequency response is shown in Figure. 2 2 Figure : Frequency response of a laglead compensator In a nutshell, I. Kar
2 If it is not specified which type of compensator has to be designed, one should first check the PM and BW of the uncompensated system with adjustable gain K. If the BW is smaller than the acceptable BW one may go for lead compensator. If the BW is large, lead compensator may not be useful since it provides high frequency amplification. One may go for a lag compensator when BW is large provided the open loop system is stable. If the lag compensator results in a too low BW (slow speed of response), a laglead compensator may be used.. Laglead compensator design Consider the following system with transfer function G(s) = s(+.s)(+.2s) Design a laglead compensator C(s) such that the phase margin of the compensated system is at least 45 o at gain crossover frequency around rad/sec and the velocity error constant K v is 3. The laglead compensator is given by C(s) = K +τ s +α τ s +τ 2 s +α 2 τ 2 s where, α >, α 2 < When s, C(s) K. K v = lim s sg(s)c(s) = C() = 3 Thus K = 3. Bode plot of the modified system KG(s) is shown in Figure 2. The gain crossover frequency and phase margin of KG(s) are found out to be 9.77 rad/sec and 7.2 o respectively. Since the PM of the uncompensated system with K is negative, we need a lead compensator to compensate for the negative PM and achieve the desired phase margin. However, we know that introduction of a lead compensator will eventually increase the gain crossover frequency to maintain the low frequency gain. Thus the gain crossover frequency of the system cascaded with a lead compensator is likely to be much above the specified one, since the gain crossover frequency of the uncompensated I. Kar 2
3 Gm = 6.2 db (at 7.7 rad/sec), Pm = 7.2 deg (at 9.77 rad/sec) System: untitled 225 : : Figure 2: Frequency response of the uncompensated system of Example system with K is already 9.77 rad/sec. Thus a laglead compensator is required to compensate for both. We design the lead part first. From Figure 2, it is seen that at rad/sec the phase angle of the system is 98 o. Since the new ω g should be rad/sec, the required additional phase at ω g, to maintain the specified PM, is 45 (8 98) = 63 o. With safety margin 2 o, ( ) sin(65 o ) α 2 = =.5 +sin(65 o ) And = τ 2 α2 which gives τ 2 =.45. However, introducing this compensator will actually increase the gain crossover frequency where the phase characteristic will be different than the designed one. This can be seen from Figure 3. The gain crossover frequency is increased to 23.2 rad/sec. At rad/sec, the phase angle is 34 o and gain is 2.6 db. To make this as the actual gain crossover frequency, lag part I. Kar 3
4 Gm =.834 db (at 24.4 rad/sec), Pm = 2.38 deg (at 23.2 rad/sec) 5 5 System: untitled : : System: untitled : : Figure 3: Frequency response of the system in Example with only a lead compensator should provide an attenuation of 2.6 db at high frequencies. At high frequencies the magnitude of the lag compensator part is /α. Thus, 2log α = 2.6 which gives α = Now, /τ should be placed much below the new gain crossover frequency to retain the desired PM. Let /τ be.25. Thus τ = 4 The overall compensator is C(s) = 3 +4s +7.8s +.45s +.225s The frequency response of the system after introducing the above compensator is shown in Figure 4, which shows that the desired performance criteria are met. Example 2: Now let us consider that the system as described in the previous example is subject to a I. Kar 4
5 Gm = 3.3 db (at 24. rad/sec), Pm = 45.3 deg (at rad/sec) Figure 4: Frequency response of the system in Example with a laglead compensator sampled data control system with sampling time T =. sec. We would use MATLAB to derive the plant transfer function wplane. Use the below commands. >> s=tf( s ); >> gc=/(s*(+.*s)*(+.2*s)); >> gz=c2d(gc,., zoh ); You would get The bilinear transformation G z (z) =.5824z z z 3.974z z.223 z = +wt/2 wt/2 = (+.5w) (.5w) will transfer G z (z) into wplane. Use the below commands >> aug=[.,]; >> gwss = bilin(ss(gz),, S_Tust,aug) >> gw=tf(gwss) I. Kar 5
6 to find out the transfer function in wplane, as G w (w) =.756w3.636w 2.75w w w w Since the velocity error constant criterion will produce the same controller dcgain K, the gain of the laglead compensator is designed to be 3. The Bode plot of the uncompensated system with K = 3 is shown in Figure 5. 6 Gm =.6 db (at 5.44 rad/sec), Pm = 44 deg (at.4 rad/sec) System: untitled : : Figure 5: Bode plot of the uncompensated system for Example 2 From Figure 5, it is seen that at rad/sec the phase angle of the system is 39 = 22 o. Thus a huge phase lead (86 o ) is required if we want to acieve a PM of 45 o which is not possible with a single lead compensator. Let us lower the PM requirement to a minimum of 2 o at ω g = rad/sec. Since the new ω g should be rad/sec, the required additional phase at ω g, to maintain the specified PM, is 2 (8 22) = 6 o. With safety margin 5 o, ( ) sin(66 o ) α 2 = =.45 +sin(66 o ) I. Kar 6
7 And = τ 2 α2 which gives τ 2 =.47. However, introducing this compensator will actually increase the gain crossover frequency where the phase characteristic will be different than the designed one. This can be seen from Figure 6. 5 Gm =.8 db (at 4.2 rad/sec), Pm = Inf System: untitled : : Figure 6: Frequency response of the system in Example 2 with only a lead compensator Also, as seen from Figure 6, the GM of the system is negative. Thus we need a lag compensator to lower the magnitude at rad/sec. At rad/sec, the magnitude is 4.2 db. To make this as the actual gain crossover frequency, lag part should provide an attenuation of 4.2 db at high frequencies. Thus, 2log α = 4.2 which gives α = 5.. Now, /τ should be placed much below the new gain crossover frequency to retain the desired PM. Let /τ be / =. Thus τ = I. Kar 7
8 The overall compensator is C(w) = 3 ( )( ) +w +.47w +5.w +.25w The frequency response of the system after introducing the above compensator is shown in Figure 7, which shows that the desired performance criteria are met. 8 Gm = 2.87 db (at 3.6 rad/sec), Pm = 2 deg (at rad/sec) Figure 7: Frequency response of the system in Example 2 with a laglead compensator Reconverting the controller in zdomain, we get ( )( ).235z z 5.93 C(z) = 3 z.986 z I. Kar 8
LINEAR 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #11 Wednesday, January 28, 2004 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Relative Stability: Stability
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 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 informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #15 Friday, February 6, 2004 Dr. Ian C. Bruce Room: CRL229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca (3) CohenCoon Reaction Curve Method
More informationIMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION. K. M. Yanev A. Obok Opok
IMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION K. M. Yanev A. Obok Opok Considering marginal control systems, a useful technique, contributing to the method of multistage compensation is suggested. A
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 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 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 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 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 informationDynamic Compensation using root locus method
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 9 Dynamic Compensation using root locus method [] (Final00)For the system shown in the
More informationME 475/591 Control Systems Final Exam Fall '99
ME 475/591 Control Systems Final Exam Fall '99 Closed book closed notes portion of exam. Answer 5 of the 6 questions below (20 points total) 1) What is a phase margin? Under ideal circumstances, what does
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 informationSystems Engineering and Control
Cork Institute of Technology Bachelor of Engineering (Honours) in Mechanical Engineering  Award (NFQ Level 8) Autumn 2007 Systems Engineering and Control (Time: 3 Hours) Answer any FIVE Questions Examiners:
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 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 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
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 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 informationStability and Robustness 1
Lecture 2 Stability and Robustness This lecture discusses the role of stability in feedback design. The emphasis is notonyes/notestsforstability,butratheronhowtomeasurethedistanceto instability. The small
More informationTopic # Feedback Control Systems
Topic #19 16.31 Feedback Control Systems Stengel Chapter 6 Question: how well do the large gain and phase margins discussed for LQR map over to DOFB using LQR and LQE (called LQG)? Fall 2010 16.30/31 19
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 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 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 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 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 informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Intro Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /5/27 Outline Closed Loop Transfer
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 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 informationReturn Difference Function and ClosedLoop Roots SingleInput/SingleOutput Control Systems
Spectral Properties of Linear Quadratic Regulators Robert Stengel Optimal Control and Estimation MAE 546 Princeton University, 2018! Stability margins of singleinput/singleoutput (SISO) systems! Characterizations
More information6.302 Feedback Systems Recitation 17: Black s Formula Revisited, and Lead Compensation Prof. Joel L. Dawson
Recitation 7: Black s Formula Revisited, and Lead Compensation By now, applying Black s Formula to a feedback system is almost a reflex: x(s) G(s) Y(s) H(s) This formula actually lends itself rather naturally
More informationProcedure for sketching bode plots (mentioned on Oct 5 th notes, Pg. 20)
Procedure for sketching bode plots (mentioned on Oct 5 th notes, Pg. 20) 1. Rewrite the transfer function in proper p form. 2. Separate the transfer function into its constituent parts. 3. Draw the Bode
More informationChapter 6  Solved Problems
Chapter 6  Solved Problems Solved Problem 6.. Contributed by  James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the ZN tuning strategy for the nominal
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 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 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 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 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 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 informationD(s) G(s) A control system design definition
R E Compensation D(s) U Plant G(s) Y Figure 7. A control system design definition x x x 2 x 2 U 2 s s 7 2 Y Figure 7.2 A block diagram representing Eq. (7.) in control form z U 2 s z Y 4 z 2 s z 2 3 Figure
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 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 informationModule 6: Deadbeat Response Design Lecture Note 1
Module 6: Deadbeat Response Design Lecture Note 1 1 Design of digital control systems with dead beat response So far we have discussed the design methods which are extensions of continuous time design
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM II LAB EE 693 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA EXPERIMENT NO : CS II/ TITLE : FAMILIARIZATION
More informationDesign Methods for Control Systems
Design Methods for Control Systems Maarten Steinbuch TU/e Gjerrit Meinsma UT Dutch Institute of Systems and Control Winter term 20022003 Schedule November 25 MSt December 2 MSt Homework # 1 December 9
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 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 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 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 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 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 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 informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
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 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 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 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 informationRational Implementation of Distributed Delay Using Extended Bilinear Transformations
Rational Implementation of Distributed Delay Using Extended Bilinear Transformations QingChang Zhong zhongqc@ieee.org, http://come.to/zhongqc School of Electronics University of Glamorgan United Kingdom
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 informationOPTIMAL DESIGN AND ANALYSIS OF A CHEMICAL PROCESS CONTROL SYSTEM
International Journal of Advances in Engineering & Technology, Mar. 23. OPTIMAL DESIGN AND ANALYSIS OF A CHEMICAL PROCESS CONTROL SYSTEM Ashis Kumar Das Faculty of Tech., Uttar Banga Krishi Viswavidyalaya,
More informationCDS 101/110a: Lecture 101 Robust Performance
CDS 11/11a: Lecture 11 Robust Performance Richard M. Murray 1 December 28 Goals: Describe how to represent uncertainty in process dynamics Describe how to analyze a system in the presence of uncertainty
More informationElectronics II. Final Examination
The University of Toledo f6fs_elct7.fm  Electronics II Final Examination Problems Points. 5. 0 3. 5 Total 40 Was the exam fair? yes no The University of Toledo f6fs_elct7.fm  Problem 5 points Given is
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 9. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
More informationDepartment of Electronics and Instrumentation Engineering M. E CONTROL AND INSTRUMENTATION ENGINEERING CL7101 CONTROL SYSTEM DESIGN Unit I BASICS AND ROOTLOCUS DESIGN PARTA (2 marks) 1. What are the
More informationUnit 11  Week 7: Quantitative feedback theory (Part 1/2)
X reviewer3@nptel.iitm.ac.in Courses» Control System Design Announcements Course Ask a Question Progress Mentor FAQ Unit 11  Week 7: Quantitative feedback theory (Part 1/2) Course outline How to access
More information