Exercise 1 (A Nonminimum Phase System)


 Bernice Leonard
 2 years ago
 Views:
Transcription
1 Prof. Dr. E. Frazzoli Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system, we need to increase the bandwidth of the system and thus design a system with a crossover frequency ω c as high as possible. However, we are constrained by the presence of a RHP zero at z = 2, which limits the crossover frequency. The common rule of thumb used for control design is to assume that ω c <.5z. Thus, for our problem: ω c < rad/sec. So we will add gain along with a lead compensator to try to achieve a crossover frequency of rad/sec while increasing the phase margin. We use a lead compensator of form: At ω =, the phase of P (s) is: C lead = K + s a + s. b ( ) P (jω) ω= = arctan arctan 2 = = ( 4 ) ( ) arctan 2 We need a phase margin of around 45, so we need the compensator to provide: 45 + ( 8 ( )) = To get a condition for a and b we calculate the phase shift at the maximal phase shift frequency ab. C lead (j ab ab ab) = arctan( a ) arctan( b ) b = arctan( a ) arctan( ) b ( ) = 2 arctan b/a π/2, ( b/a ) Now, = 2 arctan 9. Therefore, we want b/a = Since, ω c = = ab, we get: a 29 b = 348 a (arctan( ) = π/2 arctan(x)) x Now, we need to find K: At the crossover frequency ω c, L(s) ωc= =. Therefore, we want:
2 + (/2) 2 + (/29) = 5 + (/4) 2 + (/2) 2 K 2 + (/348) 2 That is, we want K Therefore, our lead compensator is: C lead (s) = s/29 + s/348. But now we need a lag element to compensate for the increased gain of the system at high frequencies due to the lead compensator and to minimize the steady state error. A good rule of thumb so that the lag compensator does not interfere with the phase margin and crossover frequency determined by the lead compensator is to choose its zero to be a decade below the crossover: C lag = s + s + b. The low frequency gain of the system is around = To have low frequency gain of 2, we choose a lag ratio of 2/32.8 = 6.: b =.6 Therefore, our final compensator is: ( + s/29) (s + ) C(s) = C lead (s)c lag (s) = 6.56 ( + s/348) (s +.6) The Bode plot of the resulting Loop transfer function is: 6 Gm = 6.5 db (at 25 rad/s), Pm = 4. deg (at 99.7 rad/s) The step response of the resulting closedloop system is given below. Notice the undershoot caused by the RHP zero. 2
3 Imaginary Axis (seconds  ) Amplitude Exercise 2 (A Dynamic Compensator) The root locus and Bode plot of the system P (s) are given below: 5 Gm = Inf db (at Inf rad/s), Pm = 84.3 deg (at.995 rad/s) 6 Root Locus Real Axis (seconds  ) The root locus plot shows that the closedloop system is stable for all positive gains, i.e., it is stable for C(s) = K, for K >. We take K = for our baseline stabilizing controller, so that the Bode plot above shows us the performance of the system with K =.. The steadystate error following ramp inputs must not exceed 2%. For this requirement to hold, the system must be of Type, i.e., L(s) = C(s)P (s) must have one pole at the origin (i.e., an integrator). Furthermore, the Bodeplot gain of L(s) must be such that e ss.2 for a ramp input. Since P (s) already contains a pole at the origin, there is no need yet to change the structure of the compensator C(s) = K. However, choosing C(s) = would not satisfy the max error requirement because we need the compensator gain to be greater than than /.2, i.e., K 5. From our previous experience we know that a ramp is a input of order m = and that our system is of order γ = since it has a pole in the origin. We can find the steadyerror as follows (also using table ): ess = lim s + L(s) e st s m = lim s sl(s) = lim s Ks s(.s+) = K The Bode plot for K = 5 is shown in the figure below. It is obtained by translating the magnitude plot of the previous Bode plot up by log 5. The phase plot remains unchanged: 3
4 m = m = m = 2 m = 3 γ = +L() γ = L() γ = 2 L() γ = 3 L() Table : Steadystate tracking error depending on the Type γ and the order m of the mth order unit ramp given as reference. Gm = Inf db (at Inf rad/s), Pm = 25.2 deg (at 2.3 rad/s) The error in response to sinusoidal inputs up to 5 rad/sec should not exceed about 5%. The error in response to sinusoidal inputs can be evaluated using the frequency response. The transfer function from the reference input to the error is given by the sensitivity S(s) = + L(s). In particular, the magnitude of the error in response to a unitamplitude sinusoid of frequency ω is given by S(jω) = / + L(jω). Hence, the specification can be written as + L(jω) /.5 = 2, for all ω 5 rad/s. In order to satisfy this equation, L(jω) should be large with respect to, hence we approximate this condition as L(jω) 2, for all ω 5 rad/s. From the magnitude Bode plot of L(s) above, we can see that the system with C(s) = 5 does indeed hit the obstacle imposed by this magnitude condition: at 5 rad/s, the magnitude of L(s) is only about 9. Thus, we need to further increase the gain by at least a factor of 2/9, so let us take C(s) = 5, giving us this new Bode plot: 4
5 5 Gm = Inf db (at Inf rad/s), Pm = 6.8 deg (at 33.2 rad/s) System: untitled : 5.35 : The crossover frequency should be at about 5 rad/sec to meet bandwidth requirements while limiting the response to highfrequency noise. With the current design C(s) = 5, the crossover frequency occurs at approximately ω c = 33.2 rad/s, and at 5 rad/sec the magnitude of the frequency response is about.4. Hence, in order to make sure that the crossover frequency is exactly at 5 rad/s, we need to multiply K by a factor of /.4 = 2.5. We thus even further increase K to K = 288, satisfying the first three requirements. 4. The phase margin should be about 5, and the ratio of the break frequencies of C should not exceed 5 to limit noise effects. While the choice of C(s) = 288 satisfies requirements 3, increasing the gain makes the phase margin very small. We now have a phase margin of only 2. We can use a lead compensator to increase the phase at crossover frequency, i.e., consider a compensator of the form C(s) = K s a + s b +. The maximum phase lead we can get is limited by the constraint on the ratio α = b/a of the break frequencies of the compensator imposed by our requirement. Using the straightline approximation, we know that the phase lead will be degrees one decade below the zero break frequency, and will increase linearly with log ω at a rate of 45 degrees/decade until one decade below the pole break frequency, at which point it will remain flat and eventually decrease. In other words, an estimate of the maximum phase lead is φ max = 45 log (5) = (More accurate calculations, yield a max phase lead of about 4.) In other words, a lead compensator, if used correctly, could increase the phase margin to the 455 degrees range, as desired. How do we get the maximum phase lead contribution to the phase margin? We can achieve this by placing the midpoint (on a log scale, i.e., the geometric mean) of the compensator zero/pole break frequencies exactly at the desired crossover frequency. Doing the math, we get a b = a α = 5rad/s, i.e., a = 5/ 5 = rad/s, and b = 5a = 5 5 =.8 rad/s. The resulting Bode plot is shown below. The phase at 5 rad/s is as desired. However, the crossover frequency has moved to the right as an effect of the increased gain at high frequencies due to the lead compensator. 5
6 Amplitude In order to recover the correct crossover frequency, and the desired phase margin, we can lower the gain (i.e., shift the magnitude Bode plot downwards). Fortunately, we have some room to reduce the gain and still meet requirements () and (2), since we artificially increased the gain in step (3). For C(s) = 288(s/ )/(s/.8 + ), the magnitude of the L(s) at 5 rad/s is about 2.5, so let s reduce the gain by a factor of 2.5: C(s) = 5 s/ s/.8 + = 575 s s +.8 Note that this compensator satisfies the error specifications () and (2), since K is still greater than the minimum values we obtained from those requirements. If this had not been the case, we could have added a lag compensator at low frequencies. The corresponding Bodeplot is: Gm = Inf db (at Inf rad/s), Pm = 53 deg (at 5.3 rad/s) The step response of the resulting closedloop system with designed C(s) is: Exercise 3 (Time Delays) a) We analyatically derive the magnitude and phase of :e jt dω = cos(t d ω) j sin(t d ω). 6
7 Magnitude: e jtdω = cos(t d ω) j sin(t d ω) = (cos(t d ω)) 2 + ( sin(t d ω)) 2 = cos 2 (T d ω) + sin 2 (T d ω) = Phase: ( ) e jtdω sin(td ω) = arctan cos(t d ω) ( ) sin(td ω) = arctan cos(t d ω) = arctan (tan(t d ω)) = T d ω Below are linear plots of both the magnitude and phase for T d = : On a logarithmic scale, it looks like this:
8 Amplitude b). The first, second, and third order Padé approximations for T d = are: st order: e s 2 s 2 + s 2nd order: e s s2 6s + 2 s 2 + 6s + 2 3rd order: e s s3 + 2s 2 6s + 2 s 3 + 2s 2 + 6s The corresponding Bode plots of e s and its first through third order Padé approximations are: Exact First order Second order Third order The corresponding step responses are:.5 Exact First order Second order Third order c). The Bode plots are as follows: 8
9 Amplitude Amplitude # No delay 432 Exact delay of. sec First order approx of. sec delay No delay Exact delay of. sec First order approx of. sec delay The step responses are as follows: # 33.6 No delay Exact delay of. sec No delay Exact delay of. sec First order approx of. sec delay d) As we saw from the plots above, while the designed compensator can handle small delays, the system becomes destabilized for larger time delays. The first order Padé approximation for G d (s) = e T ds is D(s) = 2/T d s 2/T d + s. This is an allpass filter with a RHP zero equal to 2/T d. As the timedelay T d increases, this RHP zero approaches the origin and further and further limits the crossover frequency of the system, thus decreasing the phase without changing the gain crossover. Thus, a large enough time delay can destabilize the system. Remember our rule of thumb that the crossover frequency w c <.5z, where z is a RHP zero. Thus, for T d =., w c < rad/sec, while in the previous exercise we assumed we could reach a crossover frequency of w c = 5 rad/sec. As time delay is an important consideration for many control applications, it can thus also be useful to consider the time delay margin along with the gain and phase margins. You can use Matlab s allmargin to calculate all these margins for a given loop function. For the plant P (s) and C(s) designed in this solution set for exercise 2, the time delay margin claculated by Matlab is.84 seconds. Therefore, it is no surprise that the system was unable to handle a delay of. seconds. 9
Exercise 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
More informationStudio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter
Studio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC LowPass
More informationCDS 101/110 Homework #7 Solution
Amplitude Amplitude CDS / Homework #7 Solution Problem (CDS, CDS ): (5 points) From (.), k i = a = a( a)2 P (a) Note that the above equation is unbounded, so it does not make sense to talk about maximum
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servooriented
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 informationSchool of Mechanical Engineering Purdue University
Case Study ME375 Frequency Response  1 Case Study SUPPORT POWER WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power wire, which is suspended from a catenary. During
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 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 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 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 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 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 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 informationControl Systems 2. Lecture 4: Sensitivity function limits. Roy Smith
Control Systems 2 Lecture 4: Sensitivity function limits Roy Smith 2017314 4.1 Inputoutput controllability Control design questions: 1. How well can the plant be controlled? 2. What control structure
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 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 informationProcess Control & Instrumentation (CH 3040)
Firstorder systems Process Control & Instrumentation (CH 3040) Arun K. Tangirala Department of Chemical Engineering, IIT Madras January  April 010 Lectures: Mon, Tue, Wed, Fri Extra class: Thu A firstorder
More informationPositioning Servo Design Example
Positioning Servo Design Example 1 Goal. The goal in this design example is to design a control system that will be used in a pickandplace robot to move the link of a robot between two positions. Usually
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 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 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 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 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 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 informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design
AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University Routh s Stability
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 informationReglerteknik: Exercises
Reglerteknik: Exercises Exercises, Hints, Answers Liten reglerteknisk ordlista Introduktion till Control System Toolbox ver. 5 This version: January 3, 25 AUTOMATIC CONTROL REGLERTEKNIK LINKÖPINGS UNIVERSITET
More informationSTABILITY OF CLOSEDLOOP CONTOL SYSTEMS
CHBE320 LECTURE X STABILITY OF CLOSEDLOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 101 Road Map of the Lecture X Stability of closedloop control
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 informationProfessor Fearing EE C128 / ME C134 Problem Set 4 Solution Fall 2010 Jansen Sheng and Wenjie Chen, UC Berkeley. control input. error Controller D(s)
Professor Fearing EE C18 / ME C13 Problem Set Solution Fall 1 Jansen Sheng and Wenjie Chen, UC Berkeley reference input r(t) + Σ error e(t) Controller D(s) grid 8 pixels control input u(t) plant G(s) output
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 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 informationRobust and Optimal Control, Spring A: SISO Feedback Control A.1 Internal Stability and Youla Parameterization
Robust and Optimal Control, Spring 2015 Instructor: Prof. Masayuki Fujita (S5303B) A: SISO Feedback Control A.1 Internal Stability and Youla Parameterization A.2 Sensitivity and Feedback Performance A.3
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 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 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 informationExam. 135 minutes, 15 minutes reading time
Exam August 15, 2017 Control Systems I (151059100L) Prof Emilio Frazzoli Exam Exam Duration: 135 minutes, 15 minutes reading time Number of Problems: 44 Number of Points: 52 Permitted aids: Important:
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 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 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 information