PD, PI, PID Compensation. M. Sami Fadali Professor of Electrical Engineering University of Nevada


 Pauline Kelley Henry
 4 years ago
 Views:
Transcription
1 PD, PI, PID Compensation M. Sami Fadali Professor of Electrical Engineering University of Nevada 1
2 Outline PD compensation. PI compensation. PID compensation. 2
3 PD Control L= loop gain s cl = desired closedloop pole location. Find a controller C s.t. CL s cl = ±180 or its odd multiple (angle condition). Controller Angle for s cl θ c = ±180 L s cl Zero location a = ω d tan θ c + ζω n 3
4 Remarks Graphical procedures are no longer needed. CAD procedure to obtain the design parameters for specified ζ& ω n (s cl known).» evalfr(l,scl) % Evaluate L at scl» polyval( num,scl) % Evaluate num at scl The complex value and its angle can also be evaluated using any hand calculator. 4
5 Procedure:MATLAB or Calculator 1. Calculate π L s cl»theta = pi angle(evalfr(l, scl) ) 2. Calculate zero location using a = ω d + ζω tan θ n c 3. Calculate new loop gain (with zero)» Lc = tf( [1, a],1)*l 4. Calculate gain (magnitude condition)» K = 1/abs( evalfr( Lc, scl)) 5. Check time response of PDcompensated system. Modify the design to meet the desired specifications if necessary (MATLAB). 5
6 Procedure 5.2: Given e & ζ 1. Obtain error constant K e from e and determine a system parameter K f that remains free after K e is fixed for the system with PD control. 2 Rewrite the closedloop characteristic equation of the PD controlled system as 1 + K f G f s = 0 6
7 Procedure 5.2 (Cont.) 3. Obtain K f corresponding to the desired closedloop pole location. As in Procedure 5.1, K f can be obtained by clicking on the MATLAB root locus plot or applying the magnitude condition using MATLAB or a calculator. 4. Calculate the free parameter from the gain K f. 5. Check the time response of the PD compensated system and modify the design to meet the desired specifications if necessary. 7
8 Example 5.5 Use a CAD package to design a PD controller for the type I system 1 G s = s s + 4 to meet the following specifications (a) ζ = 0.7 & ω n = 10 rad/s. (b) ζ = 0.7 & e = 4% due to a unit ramp. 8
9 (a) ζ = 0.7 & ω n = 10 rad/s MATLAB Calculate pole location & corresponding gain >> scl = 10*exp( j*( piacos(0.7) ) ) scl = I >> g=zpk([ ],[0,4],1); >> theta=pi ( angle( evalfr( g, scl) ) ) theta =
10 (a) Design (cont.)» a = 10 * sqrt(10.7^2)/ tan(theta) + 7 a = » k =1/abs(evalfr(tf( [1, a],1)*g,scl)) % k = Gain at scl: K = 10 10
11 RL of Uncompensated System 8 Root Locus Imaginary Axis Real Axis 11
12 RL of PDcompensated System 8 Root Locus 0.7 Imaginary Axis System: gcomp Gain: 10 Pole: i Damping: Overshoot (%): 4.56 Frequency (rad/sec): Real Axis 12
13 (b) ζ = 0.7, e % = 4% for unit ramp K v = Lim s 0 = Ka sc s G s = Lim s 0 sk 4 = 100 e % = 25 Ka = 100, a = 100/K s + a s s + 4 Closedloop characteristic equation with PD 1 + K s+a s s+4 = 0 13
14 (b) Design (cont.) 1 + K s+a = 0 s s+4 Assume that K varies with K a fixed, then s s Ks + Ka = 0, Ka = 100 s 1 + K s 2 + 4s = 0 RL= circle centered at the origin 14
15 RL of PDcompensated System K a fixed 15
16 (b) Design Cont. Desired location: intersection of root locus with ζ = 0.7 radial line. K = 10, a = 10 (same values as in Example 5.3). MATLAB command to obtain the gain» k = 1/abs( evalfr( tf([1,10],[ 1, 4, 0]), scl) ) k = Time responses of two designs are identical. 16
17 PI Control Integral control to improve e (increases type by one) worse transient response or instability. Add proportional control controller has a pole and a zero. Transfer function of proportionalplusintegral (PI) controller C s = K p + K i s = K s + a p, a = K i s K p 17
18 PI Remarks Used in cascade compensation (integral term in the feedback path is equivalent to a differentiator in the forward path) PI design for a plant transfer function G s is the same as PD design of G s /s. A better design is often possible by "almost canceling" the controller zero and the controller pole (negligible effect on time response). 18
19 Procedure Design a proportional controller for the system to meet the transient response specifications, i.e. place the dominant closedloop poles at s cl = ζω n ± jω d. 2. Add a PI controller with the zero location a such that φ = 3 5 (small angle) a = ζ+ ω n 1 ζ 2 tan φ or a = ζω n Tune the gain of the system to moved the closedloop pole closer to s cl. 19
20 Comments Use PI control only if Pcontrol meets the transient response but not the steadystate error specifications. Otherwise, use another control. Use polezero diagram to prove zero location formula. 20
21 Polezero Diagram of PI Controller s cl tan 180 θ p = ω d ζω n j d z s cl a p s cl a n tan 180 θ z = ω d ζω n a 21
22 Proof From Figure tan 180 θ p = ω d ζω n = tan 180 θ z = ω d ζω n a = x = a/ω n 1 ζ2 ζ 1 ζ2 ζ x Controller angle at s cl φ = θ z θ p = 180 θ p 180 θ z 22
23 Proof (Cont.) Trig. Identity tan A B = tan φ = 1 ζ 2 ζ x 1 ζ2 ζ 1 1 ζ2 ζ x Solve for x = a ω n = ζ+ Multiplying by ω n gives a 1 ζ 2 ζ 1 tan A tan B 1+tan A tan B = 1 ζ 2 tan φ x 1 ζ2 1 ζx 23
24 Controller Angle at s cl = φ < 3 vs. ζ x = a ω n = ζ 10 x 1 ζ2 1 φ = tan 1 ζx ζ 1 ζ2 = tan 1 10 ζ 2 < 3 24
25 Example 5.6 Design a controller for the position control system to perfectly track a ramp input with a dominant pair with ζ = 0.7 & ω n = 4 rad/s. G s = 1 s s
26 Solution To use the PD procedure for PI design, consider G s 1 G i s = = s s 2 s + 10 Procedure 5.1 with modified transfer function. Unstable for all gains (see root locus plot). Controller must provide an angle 249 at the desired closedloop pole location. Zero at Cursor at desired pole location on compensated system RL gives a gain of about
27 RL of System With Integrator 6 Root Locus 5 4 Imaginary Axis Real Axis 27
28 RL of PIcompensated System 28
29 Analytical Design Closedloop characteristic polynomial s s 2 + Ks + Ka = s + α s 2 + 2ζω n s + ω n 2 = s 3 + α + 2ζω n s 2 + 2ζω n α + ω n 2 s + αω n 2 α = 10 2ζω n = = 4.4 K = ω n α + 2ζω n = = a = αω n = K = Values obtained earlier, approximately. 29
30 MATLAB» scl = 4*exp(j*(pi acos( 0.7)) ) scl = i» theta = pi + angle( polyval( [ 1, 10, 0, 0], scl ) ) theta = » a = 4*sqrt(1.7^2)/tan(theta)+ 4*.7 a = » k = abs(polyval( [ 1, 10, 0,0], scl )/ polyval([1,a], scl) ) k =
31 Design I Results Closedloop transfer function for Design I s G cl s = s s s s = s s s + 16 Zero close to the closedloop poles excessive PO (see step response together with the response for a later design: Design II). 31
32 Step Response of PIcompensated System Design I (dotted), Design II (solid). 32
33 Design II: Procedure 5.3 For ζ = 0.7, K & ω n = rad/s. zero location ω a = n ζ + 1 ζ 2 tan φ = Closedloop transfer function 50 s G cl s = s s s s = s s s Gain slightly reduced to improve response / tan
34 Compare Designs I & II Design II dominant poles: ζ = 0.7, ω n = rad/s Time response for Designs I and II Percentage overshoot for Design II (almost polezero cancellation) << Percentage overshoot for Design I (II better) 34
35 PID (proportionalplusintegralplusderivative) Control Improves both transient & steadystate response. K p, K i, K d = proportional, integral, derivative gain, resp. Controller zeros: real or complex conjugate. C s = K p + K i s + K s 2 + 2ζω n s + ω n ds = K d s 2ζω n = K p, ω 2 K n = K i d K d 2 35
36 PID Design Procedures 1. Cancel real or complex conjugate LHP poles. 2. Cancel pole closest to (not on) the imaginary axis then add a PD controller by applying Procedures 5.1 or 5.2 to the reduced transfer function with an added pole at the origin. 3. Follow Procedure 5.3 with the proportional control design step modified to PD design. The PD design meets the transient response specifications and PI control is then added to improve the steadystate response. 36
37 Example 5.7 Design a PID controller for the transfer function to obtain zero e due to ramp, ζ = 0.7 and ω n = 4 rad/s. 1 G s = s s + 1 s
38 Design I Cancel pole at 1 with a zero & add an integrator 1 G i s = s 2 s + 10 Same as transfer function of Example 5.6 Add PD control of Example 5.6 PID controller C s = 50 s + 1 s s 38
39 Design II Design PD control to meet transient response specs. Select ω n = 5 rad/s (anticipate effect of adding PI). Design PD controller using MATLAB (see book: pdcon)» [k,a,scl] = pdcon(0.7, 5, tf(1, [1,11,10,0])) k = a = scl = i 39
40 Design II (Cont.) Obtain PI zero (approx. same as ζω n /10)» b =5 /(0.7 + sqrt(1.49)/tan(3*pi/ 180) ) % b = Transfer function for Design II (reduce gain to 40 to correct for PI) s s C s = 40 s 40
41 Conclusions Compare step responses for Designs I and II: Design I is superior because the zeros in Design II result in excessive overshoot. Plant transfer function favors pole cancellation in this case because the remaining real axis pole is far in the LHP. If the remaining pole is at 3, say, the second design procedure would give better results. MORAL: No easy solutions in design, only recipes to experiment with until satisfactory results are obtained. 41
42 Time response for Design I(dotted) and Design II (solid) 42
43 Example 5.8 Design a PID controller for the transfer function to obtain zero e due to step, ζ = 0.7 and ω n 4 rad/s G s = 1 (s+10)(s 2 +2s+10) 43
44 Solution Complex conjugate pair slows down the time response. Third pole is far in the LHP. Cancel the complex conjugate poles with zeros and add an integrator G s = 1 s(s+10) 44
45 Solution (cont.) Stable for all gains. Closedloop characteristic polynomial s s + K = s 2 + 2ζω n s + ω n 2 Equate coefficients with ζ = 0.7 ω n = 10 2ζ = = PID controller (meets design specs) G s = s2 + 2s + 10 s 45
46 Step Response of PIDcompensated System In practice, polezero cancellation may not occur but near cancellation is sufficient to obtain a satisfactory time response 46
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.
6. Sketch the zdomain root locus and find the critical gain for the following systems K (i) Gz () z 4. (ii) Gz K () ( z+ 9. )( z 9. ) (iii) Gz () Kz ( z. )( z ) (iv) Gz () Kz ( + 9. ) ( z. )( z 8. ) (i)
More 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 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 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 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 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 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 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 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 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 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 informationECSE 4962 Control Systems Design. A Brief Tutorial on Control Design
ECSE 4962 Control Systems Design A Brief Tutorial on Control Design Instructor: Professor John T. Wen TA: Ben Potsaid http://www.cat.rpi.edu/~wen/ecse4962s04/ Don t Wait Until The Last Minute! You got
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More 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 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 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 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 information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More informationSECTION 5: ROOT LOCUS ANALYSIS
SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems 2 Introduction Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path
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 informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 OpenLoop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
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 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 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 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 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 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 informationDesign of a Lead Compensator
Design of a Lead Compensator Dr. Bishakh Bhattacharya Professor, Department of Mechanical Engineering IIT Kanpur Joint Initiative of IITs and IISc  Funded by MHRD The Lecture Contains Standard Forms of
More information1 Chapter 9: Design via Root Locus
1 Figure 9.1 a. Sample root locus, showing possible design point via gain adjustment (A) and desired design point that cannot be met via simple gain adjustment (B); b. responses from poles at A and B 2
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 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 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 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 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 information2.004 Dynamics and Control II Spring 2008
MT OpenCourseWare http://ocw.mit.edu.004 Dynamics and Control Spring 008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Massachusetts nstitute of Technology
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 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 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 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 informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
More informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
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 informationCHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System
CHAPTER 1 Basic Concepts of Control System 1. What is open loop control systems and closed loop control systems? Compare open loop control system with closed loop control system. Write down major advantages
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 informationMethods for analysis and control of dynamical systems Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.gipsalab.fr/ o.sename 5th February 2015 Outline
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 informationControl of Electromechanical Systems
Control of Electromechanical Systems November 3, 27 Exercise Consider the feedback control scheme of the motor speed ω in Fig., where the torque actuation includes a time constant τ A =. s and a disturbance
More 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 informationCourse roadmap. ME451: Control Systems. What is Root Locus? (Review) Characteristic equation & root locus. Lecture 18 Root locus: Sketch of proofs
ME451: Control Systems Modeling Course roadmap Analysis Design Lecture 18 Root locus: Sketch of proofs Dr. Jongeun Choi Department of Mechanical Engineering Michigan State University Laplace transform
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 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 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 informationMethods for analysis and control of. Lecture 4: The root locus design method
Methods for analysis and control of Lecture 4: The root locus design method O. Sename 1 1 Gipsalab, CNRSINPG, FRANCE Olivier.Sename@gipsalab.inpg.fr www.lag.ensieg.inpg.fr/sename Lead Lag 17th March
More informationAN INTRODUCTION TO THE CONTROL THEORY
OpenLoop controller An OpenLoop (OL) controller is characterized by no direct connection between the output of the system and its input; therefore external disturbance, nonlinear dynamics and parameter
More informationCHAPTER # 9 ROOT LOCUS ANALYSES
F K א CHAPTER # 9 ROOT LOCUS ANALYSES 1. Introduction The basic characteristic of the transient response of a closedloop system is closely related to the location of the closedloop poles. If the system
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 informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More informationBangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory
Bangladesh University of Engineering and Technology Electrical and Electronic Engineering Department EEE 402: Control System I Laboratory Experiment No. 4 a) Effect of input waveform, loop gain, and system
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 informationMeeting Design Specs using Root Locus
Meeting Design Specs using Root Locus So far, we have Lead compensators which cancel a pole and move it left, speeding up the root locus. PID compensators which add a zero at s=0 and add zero, one, or
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 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 informationTopic # Feedback Control
Topic #5 6.3 Feedback Control StateSpace Systems Fullstate Feedback Control How do we change the poles of the statespace system? Or,evenifwecanchangethepolelocations. Where do we put the poles? Linear
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 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 informationDesign via Root Locus
Design via Root Locus I 9 Chapter Learning Outcomes J After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steadystate error (Sections
More information9. TwoDegreesofFreedom Design
9. TwoDegreesofFreedom Design In some feedback schemes we have additional degreesoffreedom outside the feedback path. For example, feed forwarding known disturbance signals or reference signals. In
More 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 informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationControl Systems. Root Locus & Pole Assignment. L. Lanari
Control Systems Root Locus & Pole Assignment L. Lanari Outline rootlocus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS  Root
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : RouthHurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
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 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 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 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 informationIntroduction to Root Locus. What is root locus?
Introduction to Root Locus What is root locus? A graphical representation of the closed loop poles as a system parameter (Gain K) is varied Method of analysis and design for stability and transient response
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 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 informationDesign via Root Locus
Design via Root Locus 9 Chapter Learning Outcomes After completing this chapter the student will be able to: Use the root locus to design cascade compensators to improve the steadystate error (Sections
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 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 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 informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
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 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 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 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 informationEEE 184 Project: Option 1
EEE 184 Project: Option 1 Date: November 16th 2012 Due: December 3rd 2012 Work Alone, show your work, and comment your results. Comments, clarity, and organization are important. Same wrong result or same
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 informationRoot Locus Contents. Root locus, sketching algorithm. Root locus, examples. Root locus, proofs. Root locus, control examples
Root Locu Content Root locu, ketching algorithm Root locu, example Root locu, proof Root locu, control example Root locu, influence of zero and pole Root locu, lead lag controller deign 9 Spring ME45 
More informationRoot Locus Design. MEM 355 Performance Enhancement of Dynamical Systems
Root Locus Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline The root locus design method is an iterative,
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 informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationStep input, ramp input, parabolic input and impulse input signals. 2. What is the initial slope of a step response of a first order system?
IC6501 CONTROL SYSTEM UNITII TIME RESPONSE PARTA 1. What are the standard test signals employed for time domain studies?(or) List the standard test signals used in analysis of control systems? (April
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 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 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 information