Classify a transfer function to see which order or ramp it can follow and with which expected error.


 Dustin Butler
 2 years ago
 Views:
Transcription
1 Dr. J. Tani, Prof. Dr. E. Frazzoli Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: Learning objectives: The student can 30th November 208 Classify a transfer function to see which order or ramp it can follow and with which expected error. Understand time domain specifications: steadystatevalue, overshoot, rise time and settling time. Calculate these values for a second order system. Apply the dominant pole approximation to simplify a system as secondorder and understand the utility of being able to use the dominant pole approximation. Remember conditions on S(s), T (s) to ensure noise rejection and allow for accurate reference tracking. Apply frequency domain specifications to assess the robustness of a control system. Background So far we have looked at analysis issues (how to determine closed loop stability). We will now concentrate on control synthesis, i.e. how to design a feedback control system with a desired behavior. The closed loop behavior can be given using Time Domain Specifications Frequency Domain specifications Steady State Error We derive the conditions that a unit ramp of order m can be tracked, i.e., a reference of the form r(t) := m! tm. Assuming that S(s) is stable we can use the final value theorem. Furthermore we write L(s) = s ˆL(s) with ˆL(s) having no integrators in the origin and γ being the Type of γ L(s). We can summarize the steady state error for all possible cases in table. γ = 0 + ˆL(0) m = 0 m = m = 2 m = 3 γ = 0 ˆL(0) γ = ˆL(0) γ = ˆL(0) Table : Steadystate tracking error depending on the Type γ and the order m of the mth order unit ramp given as reference.
2 Time Domain Specifications Many closed loop systems have a dominant pole pair such that their step responses are approximated well by a secondorder system with the same pole configuration: T (s) = ω n s 2 + 2ζω n s + ω 2 n y(t) = e σt cos(ω n ζ 2 ) While in class you have seen how to compute many important features of the closed loop time response looking at the property of the closed loop system, it is possible to directly consider the open loop transfer function L(s). (derivation in Analysis and Synthesis of SingleInput Single Output Control Systems, L. Guzzella Chapter 0.3. We can in fact relate the rise time or T 90 (time to reach the 90 % of the steady state output) and the percentage of overshoot M p (with respect to the steady state value) using the following approximations T 90 =.7 ω c M p = 7 ϕ 7 where ϕ and ω c are respectively the open loop phase margin and the crossover frequency. Frequency Domain specifications Frequency domain specifications relate requirements such as disturbance/noise rejection, reference tracking and stability to the frequency response of the open loop transfer function L(s). Noise and Disturbance Rejection Recall that S(s) is the transfer function from disturbance/reference to error T (s) is the transfer function from noise to output. Reference tracking/disturbance rejection and noise rejection can be met simultaneously if S(s) and T (s) are both small but S(s) + T (s) = + L(s) + L(s) + L(s) = Fortunately it generally happens that disturbances and reference have a low frequency content while noise is most concentrated at high frequencies. A practical requirements is to make S(jω) at low frequencies and T (jω) at high frequencies. Furthermore, the crossover frequency is close to the bandwidth ω b of T (s). Although this is not a fixed dependency, in most designs ω b will be very close to ω c. So the previous requirements can be summarized by the following rules of thumb: ω b ω c < 0. ω noise ω b ω c > 0 ω d where the noise frequency spectrum N(jω) > 0 db only for ω > ω noise. the disturbance frequency spectrum D(jω) > 0 db only for ω < ω d.
3 Unstable Poles Consider the plant P (s) = a s, a > 0 with one unstable pole π+ = a. The closed loop L(s) = k p P (s) can be stabilized only if the crossover frequency is larger than π + since for stability, we need to encircle the point. Using Nyquist, this is possible with a gain k p smaller than a. Informally, we can say that the controller must be faster than the fastest unstable pole, leading to the rule of thumb ω c > 2 π + being π + the fastest unstable pole. Non Minimum Phase Zeros Using Root Locus we can see that a simple proportional gain will move the closed loop poles to the open loop zeros. In case of non minimum phase zeros this is equivalent to eventually reach closed loop instability for k = k critic. This shows that the loop gain and hence the bandwidth is limited and particularly upper bounded by the non minimum phase zeros. This consideration applies to general plants but for a detailed derivation, the interested reader is reminded to Analysis and Synthesis of SingleInput Single Output Control Systems, L. Guzzella (Chapter 9.5). As a rule of thumb we choose ω c < 0.5 ζ + with ζ + being the slowest non minimum phase zero. Plant Delay Delay can be approximated as a first order system using Taylor approximation e st st Thus its effect can be approximated as a non minimum phase zero at the frequency ω delay = /T. We apply the same rule of thumb and we get ω c < 0.5 ω delay = (4) 2T Plant Uncertainty Given that the nominal closed loop plant is stable we would like to account for modeling errors as well. This means that the actual loop shape is in a range of possible plants: L real (s) = L(s) + L(s)W 2 (s) with < (5) being W 2 (s) a multiplicative uncertainty weight which quantifies the uncertainty about the plant at each frequency. E.g. if W 2 (s) = 0 at the specific complex frequency s, we restore L real (s) = L(s). Then we would like that a stable design based on L(s) is still stable for all possible plants. Alternatively, using Nyquist, we might ask that the number of encirclements of the point does not change for all L real (s). Since L(s)W 2 (s) draws a circle of possible plants around L(s), this is equivalent to ask that the distance of L(s) from the point is larger than this radius or L(jω) ( ) = L(jω) + > L(jω)W 2 (jω) T (jω) < W 2 (jω) (6) This requirement can be fulfilled if ω b ω c < ω 2 being ω 2 the first frequency where W 2 (jω) > 0 db (percentage uncertainty is larger than ). The applied rule of thumb is to choose ω c < 0.5 ω 2 () (2) (3) (7) (8)
4 Exercise (Noise Amplification and Bandwidth) Consider the following first order system G(s) = σ s + (9) a) Design a controller such that the closed loop step response is as in figure Desired step response y(t) t[s] Figure : Desired step response for the closed loop system. constant is approximately equal to 0.s. As shown in the plot, the time b) What would you do to get a faster response? c) Derive the transfer function from the input to the noise. d) According to the two previous answers, which limitation does the noise impose in terms of a faster system response? Assume that the noise has the following frequency content: N(jω) = 0.00jω + (0) Can you think to a practical solution and as a consequence, what would be a requirement for the feasibility of a good control system? Exercise 2 (Constraints on controller design) Please use the following constraints on the crossover frequency ω c (outlined at the end of chapter 0 of the accompanying book) to answer the following questions max{0 ω d, 2 ω π } < ω c < min{0.5 ω 2, 0.5 ω delay, 0.5 ω ζ, 0. ω n },
5 where ω d refers to the relevant disturbance frequency, ω π refers to the maximal frequency of unstable poles, ω 2 is the earliest point at which the uncertainty reaches 00%, i.e. W 2 (s) =, ω delay = T delay is the frequency of the delay, ω ζ is the minimal frequency of all nonminimum phase zeros and ω n is the frequency of the noise. All these frequencies impose design constraints on a potential controller. The task in this exercise is then to determine whether such a system can or cannot fulfill the design constraints summarized above. You are given a system plant defined by the following equations: ẋ (t) = x 2 (t) ẋ 2 (t) = x (t).5 x 2 (t) + u(t 0.0) y(t) = 4 x (t) + x 2 (t). Disturbances occur mainly in the frequency range 0 rad/s to 0. rad/s. The present noise has high frequency of ω n = 6000 rad/s. Furthermore, the transfer function of uncertainty W 2 (s) is: W 2 (s) = (s + ). 0 The goal now is to design a controller (not part of this exercise) which always reduces all disturbances d(t) by at least a factor of 0. a) What conditions must the crossover frequency satisfy in order to achieve disturbance and noise rejection? b) Derive the plant transfer function P (s). c) Derive the frequency constraints eventually introduced by nonminimum phase zeros and unstable poles. d) Which are the constraints on the bandwidth imposed by the delay and model uncertainty? e) Find the range of admissible frequencies according to the previously identified constraints. Exercise 3 (Homework: Implementation Considerations) You are in charge of designing a controller for a small electronic motor whose plant P (s) has the usual second order transfer function: P (s) = ω 2 n s 2 + 2ζω n s + ω 2 n () with the parameters being: ω n = 2 rad/s ζ = 0.9 After presenting your system model and controller to the board, some concerns are raised. Address them individually using theoretical and considerations and graphical illustrations and if necessary propose a way to solve the issue. a) An engineer form the electrical engineering department is worried that you did not consider the dynamics of the sensor in your system model. The sensor has the transfer function P Sensor = 0 s+0.
6 b) Another engineer from the marketing department wants to sell the component as a replacement for the wellknown Manekineko ( ) (Figure 2), so he wants it to wave with a given frequency ω wave (0, 0] depending on the realtime turnover of the shop. He would like to know if the system can fulfill that requirement and if this is not the case which controller would make this possible. Figure 2: ManekNeki lucky, waiving cat product.
7 Exercise 4 (Homework: Time domain specifications) The bode plot of the open loop gain L(jω) of a control system is depicted in figure 3. Magnitude (db) L(s) 4 L(s) Phase (deg) L(s) 4 L(s) Frequency (rad/sec) Figure 3: Bode plot  The relationship of the crossover frequency ω c, phase margin ϕ to time domain specifications. In the plot we show the magnitude and the frequency of both the original plant (blue) and the plant obtained halving the controller gain (orange). Please assume the closed loop system is approximately similar to a secondorder system. Please also note that this approximation is only valid for simple loop gains L(s) (see chapter 0 of [?]). If nonminimum phase zeros or unstable poles exist in P (s), the secondorder system approximation may not yield satisfactory results. a) How large is the static error e ss of the system depicted in figure 3? Please assume a unit step as reference input (r(t) = h(t)). b) Please use the following formulas to answer the next questions. The relationship of the overshoot ˆɛ and phase margin can be approximated as follows M p = 7 ϕ 7 Similarly, the relationship of rise time T 90 and crossover frequency ω c can be approximated as T 90 =.7 ω c
8 Determine the maximum overshoot M p and the rise time T 90 of the closed loop control system. Make use of the approximation formulas for M p and T 90. c) The control gain k p of the controller is reduced by a factor of 4. Redetermine the values M p and T 90. How do the values of M p and T 90 change compared to the previous control gain? d) Sketch the step response of the control system for both cases. e) Approximately which control gain k p is required to achieve a phase margin of 60? Which control gain k p is needed to obtain a crossover frequency ω c =? Exercise 5 (Homework: Static error e ss ) The transfer functions of plant P (s) and of two controllers C (s) and C 2 (s) are provided. P (s) = s + ( C (s) = 2 C 2 (s) = 2 + ) s a) Calculate the open loop transfer functions (L (s), L 2 (s)), the resulting sensitivities (S (s), S 2 (s)) and the complementary sensitivities (T (s), T 2 (s)) using controllers C (s) and C 2 (s), respectively. b) During the calculation to obtain L 2 (s) a zeropole cancellation occurs. Does this pose a problem? Explain. c) Calculate the system response to a step input of height. Only calculate the static response (t ). How large is the static error e ss? d) Calculate the system response of both proposed controllers to a step disturbance d of height. Again only calculate the static state response (t ). How large is the static error e ss? e) Simulate the step responses from c) and d) using Python. For this purpose you can use the control library which can be installed using conda. Run the conda terminal as Administrator and type in the following commands: conda install c pythoncontrol c cyclus slycot control Then you can use the following functions: sys = control.tf(num, den): it returns a transfer function whit numerator coefficients num and denominator coefficients den in decreasing power order. This transfer functions can then be summed and multiplied together. y, t, x = control.matlab.lsim(sys, U, T): it returns the output y, for the internally computed time t and the state evolution x for the specified system sys simulated for the time vector T under the input U.
Control Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
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 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 informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
More 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 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 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 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 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 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 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 informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationExam. 135 minutes + 15 minutes reading time
Exam January 23, 27 Control Systems I (559L) Prof. Emilio Frazzoli Exam Exam Duration: 35 minutes + 5 minutes reading time Number of Problems: 45 Number of Points: 53 Permitted aids: Important: 4 pages
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. 8. 24 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationÜbersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 5. 2. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid 
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 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 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 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 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 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 informationPrüfung Regelungstechnik I (Control Systems I) Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam!
Prüfung Regelungstechnik I (Control Systems I) Prof. Dr. Lino Guzzella 29. 8. 2 Übersetzungshilfe / Translation aid (English) To be returned at the end of the exam! Do not mark up this translation aid
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition Readings: Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Jacopo Tani Institute for Dynamic Systems and Control
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 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 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 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 informationLecture 1: Feedback Control Loop
Lecture : Feedback Control Loop Loop Transfer function The standard feedback control system structure is depicted in Figure. This represend(t) n(t) r(t) e(t) u(t) v(t) η(t) y(t) F (s) C(s) P (s) Figure
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 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 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 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 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 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 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 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 informationChapter 7  Solved Problems
Chapter 7  Solved Problems Solved Problem 7.1. A continuous time system has transfer function G o (s) given by G o (s) = B o(s) A o (s) = 2 (s 1)(s + 2) = 2 s 2 + s 2 (1) Find a controller of minimal
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 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 information9. TwoDegreesofFreedom Design
9. TwoDegreesofFreedom Design In some feedback schemes we have additional degreesoffreedom outside the feedback path. For example, feed forwarding known disturbance signals or reference signals. In
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 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 informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control DMAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
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 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the
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 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 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 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 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 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 informationRecitation 11: Time delays
Recitation : Time delays Emilio Frazzoli Laboratory for Information and Decision Systems Massachusetts Institute of Technology November, 00. Introduction and motivation. Delays are incurred when the controller
More informationUncertainty and Robustness for SISO Systems
Uncertainty and Robustness for SISO Systems ELEC 571L Robust Multivariable Control prepared by: Greg Stewart Outline Nature of uncertainty (models and signals). Physical sources of model uncertainty. Mathematical
More informationChapter 15  Solved Problems
Chapter 5  Solved Problems Solved Problem 5.. Contributed by  Alvaro Liendo, Universidad Tecnica Federico Santa Maria, Consider a plant having a nominal model given by G o (s) = s + 2 The aim of the
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
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 informationTopic # Feedback Control
Topic #4 16.31 Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests. Fall 2007 16.31 4 2 Frequency Stability
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 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 informationRobust fixedorder H Controller Design for Spectral Models by Convex Optimization
Robust fixedorder H Controller Design for Spectral Models by Convex Optimization Alireza Karimi, Gorka Galdos and Roland Longchamp Abstract A new approach for robust fixedorder H controller design by
More 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 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 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 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 informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Guzzella 9.13, Emilio Frazzoli
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Guzzella 9.13, 13.3 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 3, 2017 E. Frazzoli (ETH)
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 informationEE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =
1. Pole Placement Given the following openloop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the statevariable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback
More informationTopic # Feedback Control Systems
Topic #1 16.31 Feedback Control Systems Motivation Basic Linear System Response Fall 2007 16.31 1 1 16.31: Introduction r(t) e(t) d(t) y(t) G c (s) G(s) u(t) Goal: Design a controller G c (s) so that the
More informationTradeoffs and Limits of Performance
Chapter 9 Tradeoffs and Limits of Performance 9. Introduction Fundamental limits of feedback systems will be investigated in this chapter. We begin in Section 9.2 by discussing the basic feedback loop
More informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the splane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
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 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 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 informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationAn Internal Stability Example
An Internal Stability Example Roy Smith 26 April 2015 To illustrate the concept of internal stability we will look at an example where there are several polezero cancellations between the controller and
More 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 informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
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 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 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 informationFEEDBACK CONTROL SYSTEMS
FEEDBAC CONTROL SYSTEMS. Control System Design. Open and ClosedLoop Control Systems 3. Why ClosedLoop Control? 4. Case Study  Speed Control of a DC Motor 5. SteadyState Errors in Unity Feedback Control
More informationControl of Electromechanical Systems
Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance
More informationDesign and Tuning of Fractionalorder PID Controllers for Timedelayed Processes
Design and Tuning of Fractionalorder PID Controllers for Timedelayed Processes Emmanuel Edet Technology and Innovation Centre University of Strathclyde 99 George Street Glasgow, United Kingdom emmanuel.edet@strath.ac.uk
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
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 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 informationChapter 13 Digital Control
Chapter 13 Digital Control Chapter 12 was concerned with building models for systems acting under digital control. We next turn to the question of control itself. Topics to be covered include: why one
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
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 informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
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 informationTopic # Feedback Control Systems
Topic #20 16.31 Feedback Control Systems Closedloop system analysis Bounded Gain Theorem Robust Stability Fall 2007 16.31 20 1 SISO Performance Objectives Basic setup: d i d o r u y G c (s) G(s) n control
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 informationClosedloop system 2/1/2016. Generally MIMO case. Twodegreesoffreedom (2 DOF) control structure. (2 DOF structure) The closed loop equations become
Closedloop system enerally MIMO case Twodegreesoffreedom (2 DOF) control structure (2 DOF structure) 2 The closed loop equations become solving for z gives where is the closed loop transfer function
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
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 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 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 information