MAS107 Control Theory Exam Solutions 2008
|
|
- Mary Mitchell
- 6 years ago
- Views:
Transcription
1 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 instead of the gain. I. BODE PLOT A. Write down the zeroes and poles of the following transfer function 0.s G(s) ( + s 0 )( + s 2 0 ) 3 This transfer function consists of four parts The gain constant 0. The derivator s, which is equal to a zero at location s 0 The pole ( + s 0 ), which is equal to a pole at s The pole ( + s 0 ), which is equal to a pole at s B. Draw the asymptotic Bode plots (only gain) for each component of the transfer function above. Use frequencies from 0 0 to 0 5 rad/sec. Add all the components together to get an asymptotic approximation for the Bode plot of G(s). Finally, draw a more accurate Bode-plot based on the asymptotic approximation. In the figure above, note that the gain 0. becomes a constant of 20dB in the Bode amplitude plot. Note also that the derivator s has a gain value of 0dB at frequency 0 0 and increases by 20dB per decade frequency. The individual components are drawn in red color, the total asymptotic plot is shown in green color, while the black curve shows the more accurate Bode plot. The black circles are drawn 3dB below the asymptotic green curve. The black circles in the phase plot above are drawn at 45 o, 0 o and 45 o. II. NYQUIST PLOT A. Draw the Nyquist plot for the following transfer function when 4. G(s) (s + ) 2 We see directly from this transfer function that the phase will go from 0 o to 80 o. Hence, the Nyquist plot will stay in two quadrants only. First, we replace s by jω and multiply by the complex conjugate in the denominator G(jω) (jω + )(jω + ) ω 2 + 2jω + ω 2 + j2ω (( ω 2 ) j2ω) (( ω 2 ) + j2ω)(( ω 2 ) j2ω) ( ω 2 ) 2 + 4ω 2 (( ω2 ) j2ω) Next, let us set 4 and choose a few frequencies ω and calculate the frequency response (complex numbers) ω Re Im
2 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION This gives us approximately the following Nyquist plot, where the red curve is from the table above, while the green curve is the mirror of the red curve about the real axis. D. Based on the Nyquist plot from question 2A, estimate the phase margin graphically (no calculations required). B. Calculate amplitude and phase for the system in question 2A at the following frequencies: 0.5,, 2 and 4 rad/sec. The amplitude A in decibel and phase P are given by the following formulas. ( ) A 20log 0 Re(G(jω))2 + Im(G(jω)) 2 ( ) Im(G(jω)) P atan Re(G(jω)) We then get the following results ω Re Im A P o o o o C. Is the system in question 2A stable for a unity-feedback controller? Yes, the system is stable because N 0 (the number of encirclements of -) and P 0 (the number of open-loop unstable poles). The Nyquist criterion says Z P N 0, which is stable when Z 0. A unit circle is drawn in the Nyquist plot, see the figure above. The phase margin is given by the angle between the negative real axis and the cross-point between the unit circle and the Nyquist curve. From the figure this seems to occur approximately at the black dot representing the frequency ω 2. However, the table in question 2B says that the amplitude A.92dB at ω 2. Hence, the Nyquist curve will cross the unit circle slightly before ω 2 (remember that an amplitude of 0dB represents the unit circle). The phase margin at ω 2 equals 80 o 26.9 o 53. o. Since the cross-point occurs slightly before ω 2, the phase margin is estimated to be approximately 60 o. III. TIME AND FREQUENCY RESPONSE Given the control system in the figure below with the following plant model: G(s) s(s + 2) A. Describe a method to analytically calculate percentage overshoot and settling time when a unit step is applied to the reference input R(s). Hint: Open/closed transfer functions. First, the closed-loop transfer function is calculated. G(s) + G(s) s(s+ 2) + s(s+ 2) s 2 + 2s + ω 2 s 2 + 2ζωs + ω 2
3 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION From the equations above we see that the natural frequency ω and the damping factor ζ equal ω 2 2 ζ 2ω From the exam formula sheet, we then have the following relationships between percentage overshoot and settling time. ζ 2 π 2 %OS 00/exp T s 4 ζω 5.66 ζ B. When a settling time less than.0 second is required, calculate the range of values for the parameter. T s 4 ζω 5.66 ( ) Hence, must be equal to 32 or larger to achieve a settling time of.0 seconds or less. C. When a peak time less than 0.2 second is required, calculate the range of values for the parameter. The formula for peak time was specified as follows. T p T s π ω ζ ( ) () T p Hence, must be equal to or larger to achieve a peak time of 0.2 seconds or less. as there is a steady-state control error. Hence, this type of controller is typically used to achieve zero steady-state error for Type 0 systems when a step is applied as the reference input. The integral action appears at low frequencies in the Bode plot, and hence the PI-controller can be tuned such that it does not worsen bandwidth performance or stability margins. PID-Controller: The PID-controller is the same as the PIcontroller with an added parameter T d which is the derivative time constant. The derivative action reacts on changes in the control error. Hence, the derivative action prevents the system response to change too rapidly. A mechanical equivalent of the derivative action is a damper mechanism. In the Bode plot, the derivative action increases both the amplitude and the phase. Hence, the derivative action can be used to improve system bandwidth as well as stability margins. Lag-Compensator: The lag compensator is very similar to the PI-controller. The main difference is the low frequency characteristics. While the PI-controller goes to infinity gain at low frequencies, the lag compensator has a limited gain at low frequencies. Hence, the lag compensator can reduce steady-state errors, but not completely remove them. Lead-Compensator: The lead compensator is very similar to a PD controller. The main difference is the high frequency characteristics. While the PD controller goes to infinity gain at high frequencies, the lead compensator has a limited gain at high frequencies. This limited gain is a benefit in systems which contain high frequency measurement noise. B. The Nyquist curve for a system crosses the real axis at -2, circles the point - once and the system is stable in closed loop. How many poles in the right half plane does the system have? Explain your answer. IV. SYSTEM UNDERSTANDING A. Explain the properties of the following controllers: P, PI, PID, lag and lead compensator. Write maximum one page. P-Controller: The P-controller is the simplest controller and contains a single parameter usually denoted p. This parameter affects only the gain of the system, while the total phase is unaffected. Hence, a typical method to tune this controller is to find the frequency ω c where the desired phase margin in the phase plot occurs, and then adjust the gain parameter p such that the amplitude curve crosses the 0dB line at the same frequency ω c. Systems which are open-loop stable (no poles in the right half-plane) have an upper limit on the gain parameter p for stability, while systems which are open-loop unstable have a lower limit on the gain parameter. The P-controller can typically improve system properties such as rise time, settling time, peak time and steady-state error. These improvements usually result in a worse performance in terms of percentage overshoot (stability margin). PI-Controller: The PI-controller is the same as the P- controller with an added parameter T i which is the integral time constant. The integral action will never settle as long Let us use the Nyquist curve above as a concrete example for this problem. The Nyquist curve above is a perfect circle with a radius of 2 and is generated by the following system G(s) 2(s + ) s
4 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION The Nyquist curve crosses the real axis at -2, the curve circles the point - once. The problem text states that the system is stable in closed-loop. Hence, Z P N 0. Since N, we need also that P. In other words, the open-loop system has one pole in the right half-plane. We can also see that the example transfer function above also has a single pole in the right half-plane (at s ). C. For the system in question 4B two different P-controllers are applied. The first P-controller has a gain of 2.0, while the other controller has a gain of 0.4. Are the two new control systems stable or unstable. Explain your answers. The first P-controller will double the radius of the example circle to 4. The Nyquist curve will still encircle the point once. Hence, the control system will be stable. The P- controller with gain 0.4 will change the radius of the example circle to 0.8. The Nyquist curve will no longer encircle the point. Hence, the P-controller with a gain of 0.4 will be unstable for this system. Note, that this system behavior is the opposite of what we see for an open-loop stable system. For an open-loop unstable system, a large P-controller gain will stabilize the system, while small P-controller gains will reduce the stability margins. V. STEADY-STATE ERRORS AND COMPENSATOR G p (s) 0.2 s(s + 2) The controller equals G c 0. A. Calculate the steady-state error when applying a unity step at the input R(s). The system above is a Type system (because of the integrator). Hence, the steady-state error for a unity step is equal to 0. B. Calculate the steady-state error when applying a ramp at the input R(s). The control error equals e(s) + G c (s)g p (s) + 2 s(s+2) s(s + 2) s(s + 2) + 2 Now we can use the final value theorem to determine the steady-state error (knowing that the Laplace transform of a ramp equals s ). 2 [ ] lim s 0 s 2 s s(s + 2) s(s + 2) + 2 lim s 0 [ s + 2 s(s + 2) + 2 ] 2 2 C. Design a lag compensator for the system such that the steady-state error for a ramp is reduced by a factor of 0 without influencing the system s bandwidth to a large degree. The Bode plot of the open-loop transfer function is given below. First, we multiply the system with the gain of 0. From the Bode plot we see that the cross-over frequency w c Hence, to make sure that we do not influence the system bandwidth to a large degree, we choose the zero of the lag compensator equal to 0.0, ie. T 0.0 or T 00. The high frequency gain of the lag compensator needs to compensate for the initial multiplication of 0. Hence, α 0, or αt The lag compensator and the initial gain of 0 are summarized as follows and illustrated in the figure below. Note from the figure that the net gain of the lag compensator and the gain of 0 equals 0dB at high frequencies. This is an important characteristic too avoid affecting the overall system bandwidth as specified in the problem definition. G c (s) 0 s + T α s + αt s s D. eep the lag compensator from question 5C and add a lead compensator such that the bandwidth is increased by a factor 3. We want to increase the cross-over frequency from ω c 0. to ω c 0.3. From the Bode plot in Question 5C, we see that the amplitude at ω c 0.3 equals approximately 0dB. Hence, the lead compensator must contribute with a gain of
5 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION dB (or 3.6) at ω c. From the formula sheet, we have G c (jω c ) β β 3.6 ( ) The formula sheet gives us the final parameter T of the lead compensator. ω c T G c (s) β T β ω c β s + T s + βt 0(s ) s From the Bode plots below, we see that the lead compensator raises the gain with 0dB at ω c 0.3 and that the phase has its maximum phase lift at the same frequency. Taking the Laplace transform and inserting m, d, we can find the transfer function from f(t) to x(t). x(s 2 + s) f(s) x f (s) s B. Calculate the transfer function from the battery voltage to the current in the figure below. V (t) Ri(t) + C i(t)dt VI. MODELLING A. Calculate the transfer function from the force f(t) to the position x(t) for the mechanical system in the figure below. Taking Laplace, we get the following transfer function The spring has no effect, since the force f(t) is the same on both sides of the spring (Newton s st law). The differential equation which describes the motion of the mass are given below. mẍ(t) f(t) dẋ(t) (2) V (s) Ri(s) + Cs i(s) ( i(s) R + ) Cs i(s) RCs + Cs i V (s) Cs RCs + 0.5s 0.5s +
6 MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION C. The figure below shows two watertanks. The height in tank is h (metre) while the height in tank 2 is h 2. The area of tank and 2 are A and A 2 (m 2 ), respectively. The inflow to tank is u (m 3 /sec) while the outflow is k h. The inflow to tank 2 is k h, while the outflow is k 2 h 2. Write down the two differential equations which describe the system. Based on the differential equations, write down the transfer function from u(s) to h2(s). A h dt h 2 A 2 dt Taking Laplace, we get u k h k h k 2 h 2 A sh u k h h (A s + k ) u h u A s + k A 2 sh 2 k h k 2 h 2 h 2 (A 2 s + k 2 ) k h u ( A s ) + ( k ) h 2 u (s) k A 2 s + k 2 A s + k k
Homework 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 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 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 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 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 information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closed-loop system
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 linear-time-invariant
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 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 informationROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop 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 closed-loop 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 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 informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the s-plane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
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 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 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 informationPID controllers. Laith Batarseh. PID controllers
Next Previous 24-Jan-15 Chapter six Laith Batarseh Home End The controller choice is an important step in the control process because this element is responsible of reducing the error (e ss ), rise time
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 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 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 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 informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 Lead-Lag Compensation Techniques [] For the following system, Design a compensator such
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 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 informationClass 13 Frequency domain analysis
Class 13 Frequency domain analysis The frequency response is the output of the system in steady state when the input of the system is sinusoidal Methods of system analysis by the frequency response, as
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage 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, Force-current and Force-Voltage analogies..
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
More informationAnalysis and Design of Control Systems in the Time Domain
Chapter 6 Analysis and Design of Control Systems in the Time Domain 6. Concepts of feedback control Given a system, we can classify it as an open loop or a closed loop depends on the usage of the feedback.
More informationAN INTRODUCTION TO THE CONTROL THEORY
Open-Loop controller An Open-Loop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, non-linear dynamics and parameter
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 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 5-059-00 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 informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unity-feedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
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 hand-written notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
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 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 informationSolutions to Skill-Assessment Exercises
Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More 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 Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral
More informationCHAPTER 7 STEADY-STATE RESPONSE ANALYSES
CHAPTER 7 STEADY-STATE RESPONSE ANALYSES 1. Introduction The steady state error is a measure of system accuracy. These errors arise from the nature of the inputs, system type and from nonlinearities of
More 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 frequency-response design method Goal: wrap up lead and
More information6.1 Sketch the z-domain root locus and find the critical gain for the following systems K., the closed-loop characteristic equation is K + z 0.
6. Sketch the z-domain 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 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 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 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 21-211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21-211 1 / 39 Feedback
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 informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
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 Phase-lead and Phase-lag compensators using
More informationUNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS
ENG0016 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING BENG (HONS) IN BIOMEDICAL ENGINEERING SEMESTER 1 EXAMINATION 2017/2018 ADVANCED BIOMECHATRONIC SYSTEMS MODULE NO: BME6003 Date: Friday 19 January 2018
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 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 informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 5-8 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
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 informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationDr Ian R. Manchester
Week Content Notes 1 Introduction 2 Frequency Domain Modelling 3 Transient Performance and the s-plane 4 Block Diagrams 5 Feedback System Characteristics Assign 1 Due 6 Root Locus 7 Root Locus 2 Assign
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 informationEC6405 - CONTROL SYSTEM ENGINEERING Questions and Answers Unit - I Control System Modeling Two marks 1. What is control system? A system consists of a number of components connected together to perform
More 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 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 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 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 informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31-3-17 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationTable of Laplacetransform
Appendix Table of Laplacetransform pairs 1(t) f(s) oct), unit impulse at t = 0 a, a constant or step of magnitude a at t = 0 a s t, a ramp function e- at, an exponential function s + a sin wt, a sine fun
More 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R + - E G c (s) G(s) C Figure 1: The standard block diagram
More informationDigital Control: Summary # 7
Digital Control: Summary # 7 Proportional, integral and derivative control where K i is controller parameter (gain). It defines the ratio of the control change to the control error. Note that e(k) 0 u(k)
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 D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
More 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 informationCourse Outline. Closed Loop Stability. Stability. Amme 3500 : System Dynamics & Control. Nyquist Stability. Dr. Dunant Halim
Amme 3 : System Dynamics & Control Nyquist Stability Dr. Dunant Halim Course Outline Week Date Content Assignment Notes 1 5 Mar Introduction 2 12 Mar Frequency Domain Modelling 3 19 Mar System Response
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 informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steady-state error, and transient response for computer-controlled systems. Transfer functions,
More informationThe Frequency-response Design Method
Chapter 6 The Frequency-response 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 informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationFATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY
FATIMA MICHAEL COLLEGE OF ENGINEERING & TECHNOLOGY Senkottai Village, Madurai Sivagangai Main Road, Madurai - 625 020. An ISO 9001:2008 Certified Institution DEPARTMENT OF ELECTRONICS AND COMMUNICATION
More informationCourse roadmap. Step response for 2nd-order system. Step response for 2nd-order system
ME45: Control Systems Lecture Time response of nd-order systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationChapter 2 SDOF Vibration Control 2.1 Transfer Function
Chapter SDOF Vibration Control.1 Transfer Function mx ɺɺ( t) + cxɺ ( t) + kx( t) = F( t) Defines the transfer function as output over input X ( s) 1 = G( s) = (1.39) F( s) ms + cs + k s is a complex number:
More informationFREQUENCY-RESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCY-RESPONSE DESIGN 9.: PD and lead compensation networks The frequency-response methods we have seen so far largely tell us about stability and stability margins
More informationIntroduction to Process Control
Introduction to Process Control For more visit :- www.mpgirnari.in By: M. P. Girnari (SSEC, Bhavnagar) For more visit:- www.mpgirnari.in 1 Contents: Introduction Process control Dynamics Stability The
More informationTest 2 SOLUTIONS. ENGI 5821: Control Systems I. March 15, 2010
Test 2 SOLUTIONS ENGI 5821: Control Systems I March 15, 2010 Total marks: 20 Name: Student #: Answer each question in the space provided or on the back of a page with an indication of where to find the
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 informationEE3CL4: Introduction to Linear Control Systems
1 / 17 EE3CL4: Introduction to Linear Control Systems Section 7: McMaster University Winter 2018 2 / 17 Outline 1 4 / 17 Cascade compensation Throughout this lecture we consider the case of H(s) = 1. We
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steady-state Error and Type 0, Type
More informationCYBER EXPLORATION LABORATORY EXPERIMENTS
CYBER EXPLORATION LABORATORY EXPERIMENTS 1 2 Cyber Exploration oratory Experiments Chapter 2 Experiment 1 Objectives To learn to use MATLAB to: (1) generate polynomial, (2) manipulate polynomials, (3)
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 informationME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II
ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and
More informationThe Frequency-Response
6 The Frequency-Response Design Method A Perspective on the Frequency-Response Design Method The design of feedback control systems in industry is probably accomplished using frequency-response methods
More informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
More 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 informationPD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada
PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1 Outline PD compensation. PI compensation. PID compensation. 2 PD Control L= loop gain s cl = desired closed-loop
More informationStability of Feedback Control Systems: Absolute and Relative
Stability of Feedback Control Systems: Absolute and Relative Dr. Kevin Craig Greenheck Chair in Engineering Design & Professor of Mechanical Engineering Marquette University Stability: Absolute and Relative
More informationEEE 184: Introduction to feedback systems
EEE 84: Introduction to feedback systems Summary 6 8 8 x 7 7 6 Level() 6 5 4 4 5 5 time(s) 4 6 8 Time (seconds) Fig.. Illustration of BIBO stability: stable system (the input is a unit step) Fig.. step)
More information7.2 Controller tuning from specified characteristic polynomial
192 Finn Haugen: PID Control 7.2 Controller tuning from specified characteristic polynomial 7.2.1 Introduction The subsequent sections explain controller tuning based on specifications of the characteristic
More information