6.1 Sketch the zdomain root locus and find the critical gain for the following systems K., the closedloop characteristic equation is K + z 0.


 Ashley Austin
 3 years ago
 Views:
Transcription
1 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) To sketch the zdomain root locus by hand, we apply the root locus rules. The plots can more easily be obtained using the MATLAB command rlocus. For Gz K () z 4., the closedloop characteristic equation is K + z.4 At z, we obtain the critical gain, K cr.4 (see also the root locus plot) Root Locus.8.6 Imaginary Axis System: g Gain:.4 Pole: Damping:.755 Overshoot (%): 99.8 Frequency (rad/sec): (ii) For Gz Real Axis Figure 6 Root locus for Problem 6.(i) K () ( z+ 9. )( z 9. ), the closedloop characteristic equation isk + z.8 On the unit circle, z and we obtain K cr.8
2 Root Locus.8 System: g Gain:.8 Pole: 8.3e7 + i Damping: .79 Overshoot (%): Frequency (rad/sec): 57 Imaginary Axis Real Axis Figure 6 Root locus for Problem 6.(ii) Kz (iii) For G() z, the closedloop characteristic equation is ( z. )( z ) K z + z. z +.. On the unit circle, z, K and the critical gain is K cr.4 Root Locus Imaginary Axis System: g Gain:.4 Pole:  Damping: .94 Overshoot (%): Frequency (rad/sec): Real Axis Figure 63 Root locus for Problem 6.(iii) (iv) For G() z Kz ( + 9. ) ( z. )( z 8. ), the closedloop characteristic equation is
3 z z K z +.9 K. On the unit circle, z.9 K +.6 and we obtain K cr Root Locus Imaginary Axis.5 System: g Gain:.934 Pole: i Damping:.39 Overshoot (%): 99 Frequency (rad/sec): Real Axis Figure 64 Root locus for Problem 6.(iv)
4 6.5 Design proportional controllers for the systems of Problem 6. to meet the following specifications where possible. If the design specification cannot be met, explain why and suggest a more appropriate controller. (a) A damping ratio of.7. (b) A steadystate error of % due to a unit step. (c) A steadystate error of % due to a unit ramp. K (i) Gz () z 4. (a) A damping ratio of.7 This is a first order system and does not have an oscillatory response. Hence, the design specification cannot be met with proportional control.
5 A PI controller would give a system with the desired damping ratio but it is probably preferable to use proportional control to obtain a good time response with the appropriate time constant unless the steadystate error is required to be zero. (b) A steadystate error of % due to a unit step The position error constant is K p K G( ) 9.4 e( )% Hence, the gain for % error due to step is K 5.4. Unfortunately, the critical gain for the system is.4 and a gain of 5.4 would make the system unstable and the design specifications cannot be met with proportional control. Using a PD controller would allow us to increase the gain without causing instability. Alternatively, we could use a PI controller to reduce the error due to a step to zero, which would meet the desired specifications. (c) A steadystate error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steadystate error due to ramp. (ii) K G ( z) ( z +.9)( z.9) (a) A damping ratio of.7 The closedloop characteristic polynomial is ζωnt z.8+ K z cos ω T e z + e ζωnt ( ) d For a gain K>.8, we have imaginary poles so that the angle of the pole is ω T ω T ζ ω T.5 π / d n The square magnitude of the poles is ζω K.8 e n T e.4599 The required gain is K n Note that we only need the value of the product ω n T to solve the problem rather than each of the two values. Hence, the result can be verified with MATLAB using a unity sampling period and the command rlocus.
6 Root Locus Imaginary Axis System: g Gain:.856 Pole: .54e7 +.4i Damping:.7 Overshoot (%): 4.59 Frequency.7 (rad/sec): Real Axis Figure 6.5 Root locus for Problem 6.5 (ii) and the gain for a damping ratio of.7. (b) A steadystate error of % due to a unit step. The steadystate error is given by e ( )% + K p The error constant for the system is K p 9 K G() ( +.9)(.9) We solve for the gain K The specification can be met because the gain is less than the critical gain. (c) A steadystate error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steadystate error due to ramp. (iii) K z G ( z) ( z.)( z ) (a) A damping ratio of.7 The closedloop characteristic polynomial is
7 z + ζωnt ζωnt ( K.) z +. z cos( ω T ) e z + e Equating coefficients, we have e ζωn T ω T n cos..447 ln(. ) ( ω T ) cos ω T ζ cos(.496.5). 685 d K. cos n ( ω T ) d e ζω n T d K We obtain approximately the same answer using the MALAB command rlocus. Root Locus Imaginary Axis System: g Gain:.595 Pole: i Damping:.698 Overshoot (%): 4.66 Frequency (rad/sec): Real Axis Figure 6.6 Root locus for Problem 6.5 (iii) and the gain for a damping ratio of.7. (b) A steadystate error of % due to a unit step. The steadystate error is zero since the system is type I and the specification can be met for any stable gain. The gain can be selected as in Part (a) to obtain a satisfactory transient response. (c) A steadystate error of % due to a unit ramp. The system is type and can track a ramp input. The velocity error constant is K v ( z ) G( z) T K 8T z T K (.) The sampling period is typically sufficiently small to result in a small gain that is less than the critical value and the specifications can be met.
8 Kz ( + 9. ) (iv) Gz () ( z. )( z 8. ) (a) A damping ratio of.7 cos The closedloop characteristic polynomial is z + K z K z cos ω dt Equating coefficients, we have e ζω ω T ln n n T ζωnt ζωnt ( ) ( ) e z + e.6 +.9K (.6 +.9K ). 4 ( ω T ) cos ω nt ζ cos ln(.6.9k ) d + (.5.4) ( ln.6 +.9K.5.4) (.6. K ) ζω T ( ω T ) e cos ( ) K cos + d n 9 This is a nonlinear equation in K and must be solved numerically. By trial and error, we obtain gain value K.434. Approximately, the same value can be obtained using the MATLAB command rlocus..5 Root Locus.5 System: g Gain:.4 Pole: i Damping:.699 Overshoot (%): 4.64 Frequency (rad/sec): Imaginary Axis Real Axis Figure 6.7 Root locus for Problem 6.5 (iv) and the gain for a damping ratio of.7.
9 (b) A steadystate error of % due to a unit step. The steadystate error is given by e ( )% + K p The error constant for the system is K p 9 K( +.9) G() (.)(.8) We solve for the gain K The specification can be met because the gain is less than the critical gain. (c) A steadystate error of % due to a unit ramp. The system is type and cannot track a ramp input. Hence, the design specification cannot be met with proportional control. A PI controller is needed for the system to become type and have a finite steadystate error due to ramp.
10 6.8 Design a digital PID controller (with T.) for the plant G( s) e s + 5s by applying the ZieglerNichols tuning rules of Table 5.. The ZieglerNichols tuning rules shown in Table 5. can be applied by considering K, τ, L5. We obtain K p.4, T i, T d.5. Thus, by applying (6.3) we have. z + z.4z C z) z. z + ( 4z z
11 6. Use direct control design for the system of Problem 5.7 (with T.), namely, design a controller for the transfer function G ( s) ( s + )( s + 5) to obtain (i) zero steadystate error due to step, (ii) a settling time of less than s, and (iii) an undamped natural frequency of 5 rad/s. Obtain the discretized and the analog output. Then, apply the designed controller to the system G ( s) ( s + )( s + 5)(.s + ) and obtain the discretized and the analog output in order to verify the robustness of the control system. First, we find the discretized process transfer function: ( z ) G( s) G ZAS ( z) Z s.47( z +.889) ( z.948)( z.665) Then, the desired analog characteristic polynomial is s ζω + ωn ns T s 4 ζω n where, according to the specifications, ω n 5 and, which implies that ζ.4. Thus, by taking into account that a zero steadystate error is required, the desired closedloop transfer function is
12 G cl ( s) s 5 + 4s + 5 Then, the desired closedloop transfer function is obtained by using z e st, namely, G cl ( z).9 z By applying (6.43) we have z +.469z C ( z).576( z.948)( z.665)( z + ) ( z )( z +.889)( z.5694) The obtained discretized and analog closedloop system output are shown in Figure 66 and Figure 67 while the corresponding control variable is plotted in Figure 68. If the same controller is applied to the system G ( s) ( s + )( s + 5)(.s + ) the process output obtained is that in Figure 69. It can be seen that the additional lag causes a more significant overshoot and an increment of the settling time..4. process output time [s] Figure 66 Discretized process output for Problem 6.
13 6.3 Design a deadbeat controller for the system of Problem 5.7 to obtain perfect tracking of a unit step in minimum finite time. Obtain the analog output for the system and compare your design to that obtained in Problem 5.7. Apply then the controller to the process G ( s) ( s + )( s + 5)(.s + ) in order to verify the robustness of the control system. We consider the same discretized process transfer function of Problem 6., but in this case we set G cl ( z) z. Thus, by applying (6.43) we have C ( z) ( z.948)( z.665) ( z )( z +.889) The corresponding discretized and analog output are shown in Figure 63, the control variable in Figure 63. We observe that the deadbeat controller causes wide intersampling oscillations and requires a much higher control effort. It results in an increased overshoot without significantly decreasing the settling time. When the system G ( s) ( s + )( s + 5)(.s + ) is considered, the process output obtained is plotted in Error! Reference source not found...5 process output time [s] Figure 63 Discretized and analog output for Problem 6.3
14 3 control variable time [s] Figure 63 Control variable for Problem 6.3. process output time [s] Figure 63 Analog output for Problem 6.3 with uncertainty
15 6.8 To examine the effect of the sampling period on the relative stability and transient response of a digital control system, consider the system G ( s) ( s + )( s + 5) a) Obtain the transfer function of the system, the root locus and the critical gain for T.,.5,. s. Using the MATLAB commands zpk and cd, we obtain:
16 >> gzpk([],[,5],) Zero/pole/gain: (s+) (s+5) >> gdcd(g,.) Zero/pole/gain: 4.93e5 (z+.98) (z.99) (z.95) Sampling time:. >> gdcd(g,.5) Zero/pole/gain:.37 (z+.949) (z.95) (z.7788) Sampling time:.5 >> gd3cd(g,.) Zero/pole/gain:.47 (z+.889) (z.948) (z.665) Sampling time:. Using the command >> [Gm,PM]margin(gd) We obtain the gain margin, which is the critical gain value. The values obtain are summarized in the following table. Sampling Period Zero Pole Pole K cr e b) Obtain the step response for each system at a gain of 5.
17 .4 Step Response. Amplitude Time (sec) Figure 64 Step response for sampling period T. (blue), T.5 (green), and T. s (red). c) Discuss the effect of the sampling period on the transient response and relative stability of the system based on your results from (a) and (b). The transfer functions for the different sampling period with generally faster poles for slower sampling. Faster poles are poles that are closer to the origin. The systems with slower sampling does not monitor the process as closely as ones with faster sampling. As a result, the step response of the system is more oscillatory.
ECSE 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 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 informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More 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 closedloop
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 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 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 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 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 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 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 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, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
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 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 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 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 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 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 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 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 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 information1 x(k +1)=(Φ LH) x(k) = T 1 x 2 (k) x1 (0) 1 T x 2(0) T x 1 (0) x 2 (0) x(1) = x(2) = x(3) =
567 This is often referred to as Þnite settling time or deadbeat design because the dynamics will settle in a Þnite number of sample periods. This estimator always drives the error to zero in time 2T or
More informationProportional plus Integral (PI) Controller
Proportional plus Integral (PI) Controller 1. A pole is placed at the origin 2. This causes the system type to increase by 1 and as a result the error is reduced to zero. 3. Originally a point A is on
More 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 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 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 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 informationFeedback Control of Linear SISO systems. Process Dynamics and Control
Feedback Control of Linear SISO systems Process Dynamics and Control 1 OpenLoop Process The study of dynamics was limited to openloop systems Observe process behavior as a result of specific input signals
More informationChapter 12. Feedback Control Characteristics of Feedback Systems
Chapter 1 Feedbac Control Feedbac control allows a system dynamic response to be modified without changing any system components. Below, we show an openloop system (a system without feedbac) and a closedloop
More informationDepartment of Electronics and Instrumentation Engineering M. E CONTROL AND INSTRUMENTATION ENGINEERING CL7101 CONTROL SYSTEM DESIGN Unit I BASICS AND ROOTLOCUS DESIGN PARTA (2 marks) 1. What are the
More 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 informationRoot Locus. Motivation Sketching Root Locus Examples. School of Mechanical Engineering Purdue University. ME375 Root Locus  1
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: D 0.09 Position
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 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 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 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 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 informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.04A Systems and Controls Spring 2013 Problem Set #4 Posted: Thursday, Mar. 7, 13 Due: Thursday, Mar. 14, 13 1. Sketch the Root
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 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 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 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 information1 An Overview and Brief History of Feedback Control 1. 2 Dynamic Models 23. Contents. Preface. xiii
Contents 1 An Overview and Brief History of Feedback Control 1 A Perspective on Feedback Control 1 Chapter Overview 2 1.1 A Simple Feedback System 3 1.2 A First Analysis of Feedback 6 1.3 Feedback System
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 informationRoot Locus Design Example #4
Root Locus Design Example #4 A. Introduction The plant model represents a linearization of the heading dynamics of a 25, ton tanker ship under empty load conditions. The reference input signal R(s) is
More informationa. Closedloop system; b. equivalent transfer function Then the CLTF () T is s the poles of () T are s from a contribution of a
Root Locus Simple definition Locus of points on the s plane that represents the poles of a system as one or more parameter vary. RL and its relation to poles of a closed loop system RL and its relation
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 informationController Design using Root Locus
Chapter 4 Controller Design using Root Locus 4. PD Control Root locus is a useful tool to design different types of controllers. Below, we will illustrate the design of proportional derivative controllers
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 informationSolutions for Tutorial 10 Stability Analysis
Solutions for Tutorial 1 Stability Analysis 1.1 In this question, you will analyze the series of three isothermal CSTR s show in Figure 1.1. The model for each reactor is the same at presented in Textbook
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 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 informationDIGITAL CONTROLLER DESIGN
ECE4540/5540: Digital Control Systems 5 DIGITAL CONTROLLER DESIGN 5.: Direct digital design: Steadystate accuracy We have spent quite a bit of time discussing digital hybrid system analysis, and some
More 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 informationControl of SingleInput SingleOutput Systems
Control of SingleInput SingleOutput Systems Dimitrios HristuVarsakelis 1 and William S. Levine 2 1 Department of Applied Informatics, University of Macedonia, Thessaloniki, 546, Greece dcv@uom.gr 2
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: SteadyState Error Unit 7: Root Locus Techniques
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 informationLecture 25: Tue Nov 27, 2018
Lecture 25: Tue Nov 27, 2018 Reminder: Lab 3 moved to Tuesday Dec 4 Lecture: review timedomain characteristics of 2ndorder systems intro to control: feedback openloop vs closedloop control intro to
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 information] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command prefilter [ 0.
EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965372 Problem (Analysis of a Feedback System) Consider the feedback system
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
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 informationAnswers to multiple choice questions
Answers to multiple choice questions Chapter 2 M2.1 (b) M2.2 (a) M2.3 (d) M2.4 (b) M2.5 (a) M2.6 (b) M2.7 (b) M2.8 (c) M2.9 (a) M2.10 (b) Chapter 3 M3.1 (b) M3.2 (d) M3.3 (d) M3.4 (d) M3.5 (c) M3.6 (c)
More informationImproving a Heart Rate Controller for a Cardiac Pacemaker. Connor Morrow
Improving a Heart Rate Controller for a Cardiac Pacemaker Connor Morrow 03/13/2018 1 In the paper Intelligent Heart Rate Controller for a Cardiac Pacemaker, J. Yadav, A. Rani, and G. Garg detail different
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 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 informationControl Systems. Design of State Feedback Control.
Control Systems Design of State Feedback Control chibum@seoultech.ac.kr Outline Design of State feedback control Dominant pole design Symmetric root locus (linear quadratic regulation) 2 Selection of closedloop
More information2.010 Fall 2000 Solution of Homework Assignment 8
2.1 Fall 2 Solution of Homework Assignment 8 1. Root Locus Analysis of Hydraulic Servomechanism. The block diagram of the controlled hydraulic servomechanism is shown in Fig. 1 e r e error + i Σ C(s) P(s)
More informationRoot Locus Design Example #3
Root Locus Design Example #3 A. Introduction The system represents a linear model for vertical motion of an underwater vehicle at zero forward speed. The vehicle is assumed to have zero pitch and roll
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 informationNotes for ECE320. Winter by R. Throne
Notes for ECE3 Winter 45 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................
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 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 informationCHAPTER 7 STEADYSTATE RESPONSE ANALYSES
CHAPTER 7 STEADYSTATE 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 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 informationTransient Response of a SecondOrder System
Transient Response of a SecondOrder System ECEN 830 Spring 01 1. Introduction In connection with this experiment, you are selecting the gains in your feedback loop to obtain a wellbehaved closedloop
More informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
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 informationR a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies.
SET  1 II B. Tech II Semester Supplementary Examinations Dec 01 1. a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Forcecurrent and ForceVoltage analogies..
More informationLecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore A Fundamental Problem in Control Systems Poles of open
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 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 information9/9/2011 Classical Control 1
MM11 Root Locus Design Method Reading material: FC pp.270328 9/9/2011 Classical Control 1 What have we talked in lecture (MM10)? Lead and lag compensators D(s)=(s+z)/(s+p) with z < p or z > p D(s)=K(Ts+1)/(Ts+1),
More informationBASIC PROPERTIES OF FEEDBACK
ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)
More informationParameter Derivation of Type2 DiscreteTime PhaseLocked Loops Containing Feedback Delays
Parameter Derivation of Type DiscreteTime PhaseLocked Loops Containing Feedback Delays Joey Wilson, Andrew Nelson, and Behrouz FarhangBoroujeny joey.wilson@utah.edu, nelson@math.utah.edu, farhang@ece.utah.edu
More informationHomework 11 Solution  AME 30315, Spring 2015
1 Homework 11 Solution  AME 30315, Spring 2015 Problem 1 [10/10 pts] R +  K G(s) Y Gpsq Θpsq{Ipsq and we are interested in the closedloop pole locations as the parameter k is varied. Θpsq Ipsq k ωn
More informationUnit 8: Part 2: PD, PID, and Feedback Compensation
Ideal Derivative Compensation (PD) Lead Compensation PID Controller Design Feedback Compensation Physical Realization of Compensation Unit 8: Part 2: PD, PID, and Feedback Compensation Engineering 5821:
More informationLab # 4 Time Response Analysis
Islamic University of Gaza Faculty of Engineering Computer Engineering Dep. Feedback Control Systems Lab Eng. Tareq Abu Aisha Lab # 4 Lab # 4 Time Response Analysis What is the Time Response? It is an
More informationIMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION. K. M. Yanev A. Obok Opok
IMPROVED TECHNIQUE OF MULTISTAGE COMPENSATION K. M. Yanev A. Obok Opok Considering marginal control systems, a useful technique, contributing to the method of multistage compensation is suggested. A
More informationDynamic System Response. Dynamic System Response K. Craig 1
Dynamic System Response Dynamic System Response K. Craig 1 Dynamic System Response LTI Behavior vs. NonLTI Behavior Solution of Linear, ConstantCoefficient, Ordinary Differential Equations Classical
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 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 Steadystate Error and Type 0, Type
More informationDepartment of Mechanical Engineering
Department of Mechanical Engineering 2.14 ANALYSIS AND DESIGN OF FEEDBACK CONTROL SYSTEMS Fall Term 23 Problem Set 5: Solutions Problem 1: Nise, Ch. 6, Problem 2. Notice that there are sign changes in
More informationEE451/551: Digital Control. Final Exam Review Fall 2013
EE45/55: Digital Control Final Exam Review Fall 03 Exam Overview The Final Exam will consist of four/five questions for EE45/55 students based on Chapters 7 and a bonus based on Chapters 8 9 (students
More informationChapter 7: Time Domain Analysis
Chapter 7: Time Domain Analysis Samantha Ramirez Preview Questions How do the system parameters affect the response? How are the parameters linked to the system poles or eigenvalues? How can Laplace transforms
More informationSchool of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See
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 informationCascade Control of a Continuous Stirred Tank Reactor (CSTR)
Journal of Applied and Industrial Sciences, 213, 1 (4): 1623, ISSN: 23284595 (PRINT), ISSN: 2328469 (ONLINE) Research Article Cascade Control of a Continuous Stirred Tank Reactor (CSTR) 16 A. O. Ahmed
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, 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 information