Improving a Heart Rate Controller for a Cardiac Pacemaker. Connor Morrow

Size: px
Start display at page:

Download "Improving a Heart Rate Controller for a Cardiac Pacemaker. Connor Morrow"

Transcription

1 Improving a Heart Rate Controller for a Cardiac Pacemaker Connor Morrow 03/13/2018

2 1 In the paper Intelligent Heart Rate Controller for a Cardiac Pacemaker, J. Yadav, A. Rani, and G. Garg detail different control methods for coming up with a controller to improve on the response of a pacemaker. Using a previously established transfer function for the heart and pacemaker mechanics, they design a controller through three different methods: Ziegler-Nechols tuning, Tyreus-Luyben tuning, and fuzzy logic. This example will look into how they developed their controllers while also putting forth a few state space controllers that will improve upon their methods. 1 Pacemaker System To start, let s look at the system that the controller will be integrated into. S. C. Biswas et al came up with a transfer function to model the cardiovascular system in the paper Mathematical Model of Cardiovascular System by Transfer Function Method. The transfer functions that model the heart and the pacemaker are: 169 G H = s s G P = 8 s + 8 The block diagram that includes those transfer functions as well as the controller looks like: 2 Their Controllers Before building our own state space controller, lets first look at how they built their controllers. This will be a quick overview on the methods that they used, as well as looking at the results that they came up with. 2.1 Ziegler-Nichols and Tyreus-Luyben Tuning Ziegler-Nichols and Tyresus-Luyben tuning are ways of forming a PID controller by looking at the critical gain and period of the open loop system. One way to find critical gain would be to use the root locus of the open loop system to find the K value that lies at s = 0 on the real axis. Once that is found, we look at the system after being excited and find the period between the peaks of its continual oscilation.

3 2 From the root locus, we see that the gain associated with marginal stability is: K cr = Now looking at the response we can find the period of the oscillations. P cr = 0.49 sec

4 3 The controller formula for a PID controller is as follows: G c (s) = K p + K i s + K ds It can also be written in a form using time constants as opposed to gains. G c (s) = K p (1 + 1 T i s + T ds) Both the Ziegler-Nichols and Tyreus-Luyben tuning methods use the critical gain and the period to come up with the time constants. The formulas for the time constants are listed in the following table. This gives us the controllers Table 1: Tuning Method Gains Method K p T i T d Ziegler-Nichols PID 0.6K cr 0.5P cr 0.125P cr Tyreus-Luyben PID K cr / P cr P cr /6.3 G ZN = 0.032s s s G T L = 0.14s s s We can now see what the step response of the closed loop system would look like when implementing these controllers. We will be looking at trying to achieve a heart rate beats per minute of 65, 75, and 85.

5 4 These responses are not identical to the ones generated in the paper, as the paper did not explicitly declare their gains. However, they do list what rise time, settling time, and percent maximum overshoot they were able to achieve, which is what we will be designing our controller to do better than. Table 2: Ziegler-Nichols Results Heart Rate Settling Time Rise Time PMO % % % Table 3: Tyreus-Luyben Results Heart Rate Settling Time Rise Time PMO % % % 2.2 Fuzzy Logic Controller Fuzzy logic controllers are built on the fuzzy logic mathematical system. This refers to the ability of logic control that can interpret more than true and false. It is also able to handle partially true. Creating a fuzzy logic controller involves a process of fuzzification of the data, processing that data through a fuzzy logic engine, and finally defuzzifying that information. The paper briefly goes over this process, but doesn t explicitly say how it was performed. Once again, however, they detail the response characteristics of this method, which we will compare our controller to. Heart Rate Settling Time Rise Time PMO % % % 3 State Space Controller Now that we have had an opportunity to look at how they designed their controller, let s begin building a state space controller. In order to do so, we will first create the state space equations. 3.1 State Space Equations Our state variables will come from open loop transfer function of G H and G P. ( G = G = G = G H G P 169 s s ) ( 8 s s s )

6 5 Looking at G as the transfer function between the input, R, and the output, Y, we get: We can now set up our state variables Y R = 1352 s s Y Ÿ Ẏ = 1352R... Y = 166.4Ẏ 20.8Ÿ R x 1 = Y x 2 = Ẏ = x 1 x 3 = Ÿ = x 2 This gives us the state matrix equation x x 1 0 x 2 = x R x x Y = [ ] x 1 x 2 This gives us our state space matrices that we will enter into Matlab A = 0 0 1, B = 0, C = [ ], D = Before going too much further, we should also realize that state space controllers generally include a steady state error. In order to rectify that, we will include an integrator loop outside of the main control loop. This will produce new state space matrices that look like the following: [A i ] = [ ] [A] {0}, [B [C] 0 i ] = x 3 { } [B], [C 0 i ] = [[C] 0] We will also need the closed loop A and B matrices. They will include the controller matrix, K, that we will derive in the next sections, with K i being the controller gain added because of the integrator loop. [A icl ] = [ ] [A] [B][K] [B]Ki, [B [C] 0 icl ] = { } {0} 1 Those matrices can all be entered into Matlab to handle the calculations. The closed loop matrices should be included in the Matlab code after the controller K is made. With that out of the way, we can begin the actual design process of the controller.

7 6 3.2 Designing a Controller to Outperform the Tuning Methods The controller will be designed in order to model a second order system. That means we can find a ζ and ω n that will meet our design criteria to come up with the controller. This controller will be designed to outperform the the previously discussed tuning methods. The stated PMO and rise time for the Ziegler-Nichols controller was about 23% and sec respectively, whereas the Tyreus-Luyben method achieved a PMO of 10.9% and a rise time of sec. To out perform both controllers, we will design one that will achieve a PMO below 10% and have a rise time below 0.07 sec. From our graphs detailing the relationship between PMO and ζ, for a PMO below 10% we are going to want ζ = 0.6. From there we can calculate ω n from the following equations. ζ = 0.6 t r = ζ ω n ω n = 32.9 We can find the roots for the second order system created with these parameters to be used to determine our controller matrix s 2 + 2ζω n s + ω 2 n = 0 s = 19.7 ± 26.3i Since the the actual system is third order, we will add another arbitrary root for our matrix. This root must be at least five times greater than the second order roots, in order to not have an impact on the system. We will choose Also, as established earlier, we will need a root to compensate for the integrator, which we will choose to be The K matrix is determined by using the place command in Matlab, placing the four roots in A i and B i. One thing to remember is that we will need the gain for the integrator to be the negative of the one that Matlab generates. This gives us the controller matrix: K = [ ] Creating the closed loop A and B matrices through the method discussed in the previous section, and using the ss command in Matlab to generate the state space closed loop equation, we can view the step response of the system. We will only looking at 65bpm to make sure we have achieved our goal.

8 7 This step response yields a PMO of 8.27% and a rise time of sec, which means we ve achieved our goal. 3.3 Designing a Controller to Outperform the Fuzzy Logic Controller Designing our second controller will be nearly identical to the previous method, just with different PMO and rise time goals. For this one, we want to beat a PMO of 2.1% and a rise time of 0.27 sec, which was achieved by the fuzzy logic controller. We will use a PMO of 2% and a rise time of 0.25 sec, which yields the following The step response to that system looks like ζ = 0.9 ω n = 12.2 K = [ ]

9 8 This response has no overshoot and has a rise time of sec, which is once again successful. 3.4 Designing the Optimal Controller For fun, let s now design a controller to beat the tuning methods and the fuzzy logic method at once. We will be trying to achieve a PMO less than 2.1% and a rise time of sec. ζ = 0.8 ω n = 40 K = [ ]

10 9 Once again, state space controllers make it easy to achieve the characteristics that we are looking for. The PMO for this system is 1.18% and the rise time is 0.07 sec. 4 Matlab Code Working with state space matrices can be very tedious, however Matlab makes things a lot easier. The following is the Matlab code, which shows how the tuned controllers were recreated and how the state space controllers were made.

11

12

13

14 13 References [1] S. C. Biswas, A. Das, P. Guha Mathematical Model of Cardiovascular System by Transfer Function Method Calcutta Medical Journal, 4, 2006 [2] J. Yadav, A. Rani, G. Garg Intelligent Heart Rate Controller for Cardiac Pacemaker International Journal of Computer Applications, 7, 2011

6.1 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.

6.1 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 information

EE3CL4: Introduction to Linear Control Systems

EE3CL4: 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 information

EE C128 / ME C134 Fall 2014 HW 6.2 Solutions. HW 6.2 Solutions

EE 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 information

MASSACHUSETTS 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 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 information

EEE 184 Project: Option 1

EEE 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 information

ME 475/591 Control Systems Final Exam Fall '99

ME 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 information

(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:

(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

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur

VALLIAMMAI 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 information

Due Wednesday, February 6th EE/MFS 599 HW #5

Due 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 information

Outline. Classical Control. Lecture 5

Outline. 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 information

ECE-320: Linear Control Systems Homework 8. 1) For one of the rectilinear systems in lab, I found the following state variable representations:

ECE-320: Linear Control Systems Homework 8. 1) For one of the rectilinear systems in lab, I found the following state variable representations: ECE-30: Linear Control Systems Homework 8 Due: Thursday May 6, 00 at the beginning of class ) For one of the rectilinear systems in lab, I found the following state variable representations: 0 0 q q+ 74.805.6469

More information

MASSACHUSETTS 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 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 information

Iterative Feedback Tuning

Iterative Feedback Tuning Iterative Feedback Tuning Michel Gevers CESAME - UCL Louvain-la-Neuve Belgium Collaboration : H. Hjalmarsson, S. Gunnarsson, O. Lequin, E. Bosmans, L. Triest, M. Mossberg Outline Problem formulation Iterative

More information

Control 2. Proportional and Integral control

Control 2. Proportional and Integral control Control 2 Proportional and Integral control 1 Disturbance rejection in Proportional Control Θ i =5 + _ Controller K P =20 Motor K=2.45 Θ o Consider first the case where the motor steadystate gain = 2.45

More information

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007

MASSACHUSETTS 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 information

Software Engineering 3DX3. Slides 8: Root Locus Techniques

Software 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 information

KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING

KINGS 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 information

Systems Analysis and Control

Systems 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 information

Root Locus Design Example #4

Root 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 information

Chapter 2. Classical Control System Design. Dutch Institute of Systems and Control

Chapter 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 information

Chapter 6 - Solved Problems

Chapter 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 Z-N tuning strategy for the nominal

More information

CHAPTER 10: STABILITY &TUNING

CHAPTER 10: STABILITY &TUNING When I complete this chapter, I want to be able to do the following. Determine the stability of a process without control Determine the stability of a closed-loop feedback control system Use these approaches

More information

K c < K u K c = K u K c > K u step 4 Calculate and implement PID parameters using the the Ziegler-Nichols tuning tables: 30

K c < K u K c = K u K c > K u step 4 Calculate and implement PID parameters using the the Ziegler-Nichols tuning tables: 30 1.5 QUANTITIVE PID TUNING METHODS Tuning PID parameters is not a trivial task in general. Various tuning methods have been proposed for dierent model descriptions and performance criteria. 1.5.1 CONTINUOUS

More information

Notes for ECE-320. Winter by R. Throne

Notes for ECE-320. Winter by R. Throne Notes for ECE-3 Winter 4-5 by R. Throne Contents Table of Laplace Transforms 5 Laplace Transform Review 6. Poles and Zeros.................................... 6. Proper and Strictly Proper Transfer Functions...................

More information

R a) Compare open loop and closed loop control systems. b) Clearly bring out, from basics, Force-current and Force-Voltage analogies.

R 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 information

Control Systems. University Questions

Control Systems. University Questions University Questions UNIT-1 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 information

Control Systems I Lecture 10: System Specifications

Control Systems I Lecture 10: System Specifications Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture

More information

MAS107 Control Theory Exam Solutions 2008

MAS107 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 information

Multivariable Control Laboratory experiment 2 The Quadruple Tank 1

Multivariable Control Laboratory experiment 2 The Quadruple Tank 1 Multivariable Control Laboratory experiment 2 The Quadruple Tank 1 Department of Automatic Control Lund Institute of Technology 1. Introduction The aim of this laboratory exercise is to study some different

More information

Controller Design using Root Locus

Controller 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 information

Course roadmap. Step response for 2nd-order system. Step response for 2nd-order system

Course 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 information

Controls Problems for Qualifying Exam - Spring 2014

Controls 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 information

Laboratory 11 Control Systems Laboratory ECE3557. State Feedback Controller for Position Control of a Flexible Joint

Laboratory 11 Control Systems Laboratory ECE3557. State Feedback Controller for Position Control of a Flexible Joint Laboratory 11 State Feedback Controller for Position Control of a Flexible Joint 11.1 Objective The objective of this laboratory is to design a full state feedback controller for endpoint position control

More information

Root Locus Design Example #3

Root 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 information

PID Control. Objectives

PID Control. Objectives PID Control Objectives The objective of this lab is to study basic design issues for proportional-integral-derivative control laws. Emphasis is placed on transient responses and steady-state errors. The

More information

EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO

EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO EE 380 EXAM II 3 November 2011 Last Name (Print): First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score 1 25 2 25 3 25 4 25 Total

More information

ECSE 4962 Control Systems Design. A Brief Tutorial on Control Design

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 information

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING. MSc SYSTEMS ENGINEERING AND ENGINEERING MANAGEMENT SEMESTER 2 EXAMINATION 2015/2016

UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING. MSc SYSTEMS ENGINEERING AND ENGINEERING MANAGEMENT SEMESTER 2 EXAMINATION 2015/2016 TW2 UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING MSc SYSTEMS ENGINEERING AND ENGINEERING MANAGEMENT SEMESTER 2 EXAMINATION 2015/2016 ADVANCED CONTROL TECHNOLOGY MODULE NO: EEM7015 Date: Monday 16 May 2016

More information

Video 5.1 Vijay Kumar and Ani Hsieh

Video 5.1 Vijay Kumar and Ani Hsieh Video 5.1 Vijay Kumar and Ani Hsieh Robo3x-1.1 1 The Purpose of Control Input/Stimulus/ Disturbance System or Plant Output/ Response Understand the Black Box Evaluate the Performance Change the Behavior

More information

Discrete Systems. Step response and pole locations. Mark Cannon. Hilary Term Lecture

Discrete 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 information

Lecture 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 Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This

More information

Transient 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

Transient 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 information

TUNING-RULES FOR FRACTIONAL PID CONTROLLERS

TUNING-RULES FOR FRACTIONAL PID CONTROLLERS TUNING-RULES FOR FRACTIONAL PID CONTROLLERS Duarte Valério, José Sá da Costa Technical Univ. of Lisbon Instituto Superior Técnico Department of Mechanical Engineering GCAR Av. Rovisco Pais, 49- Lisboa,

More information

Homework 7 - Solutions

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 information

1 An Overview and Brief History of Feedback Control 1. 2 Dynamic Models 23. Contents. Preface. xiii

1 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 information

Bangladesh University of Engineering and Technology. EEE 402: Control System I Laboratory

Bangladesh 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 information

If you need more room, use the backs of the pages and indicate that you have done so.

If 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 information

State Feedback Controller for Position Control of a Flexible Link

State Feedback Controller for Position Control of a Flexible Link Laboratory 12 Control Systems Laboratory ECE3557 Laboratory 12 State Feedback Controller for Position Control of a Flexible Link 12.1 Objective The objective of this laboratory is to design a full state

More information

IMPROVED TECHNIQUE OF MULTI-STAGE COMPENSATION. K. M. Yanev A. Obok Opok

IMPROVED TECHNIQUE OF MULTI-STAGE COMPENSATION. K. M. Yanev A. Obok Opok IMPROVED TECHNIQUE OF MULTI-STAGE COMPENSATION K. M. Yanev A. Obok Opok Considering marginal control systems, a useful technique, contributing to the method of multi-stage compensation is suggested. A

More information

Pitch Rate CAS Design Project

Pitch Rate CAS Design Project Pitch Rate CAS Design Project Washington University in St. Louis MAE 433 Control Systems Bob Rowe 4.4.7 Design Project Part 2 This is the second part of an ongoing project to design a control and stability

More information

Chapter 5 HW Solution

Chapter 5 HW Solution Chapter 5 HW Solution Review Questions. 1, 6. As usual, I think these are just a matter of text lookup. 1. Name the four components of a block diagram for a linear, time-invariant system. Let s see, I

More information

EE C128 / ME C134 Fall 2014 HW 9 Solutions. HW 9 Solutions. 10(s + 3) s(s + 2)(s + 5) G(s) =

EE 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 open-loop plant, HW 9 Solutions G(s) = 1(s + 3) s(s + 2)(s + 5) design the state-variable feedback controller u = Kx + r, where K = [k 1 k 2 k 3 ] is the feedback

More information

Inverted Pendulum: State-Space Methods for Controller Design

Inverted Pendulum: State-Space Methods for Controller Design 1 de 12 18/10/2015 22:45 Tips Effects TIPS ABOUT BASICS HARDWARE INDEX NEXT INTRODUCTION CRUISE CONTROL MOTOR SPEED MOTOR POSITION SYSTEM MODELING ANALYSIS Inverted Pendulum: State-Space Methods for Controller

More information

Example on Root Locus Sketching and Control Design

Example 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 information

ECEn 483 / ME 431 Case Studies. Randal W. Beard Brigham Young University

ECEn 483 / ME 431 Case Studies. Randal W. Beard Brigham Young University ECEn 483 / ME 431 Case Studies Randal W. Beard Brigham Young University Updated: December 2, 2014 ii Contents 1 Single Link Robot Arm 1 2 Pendulum on a Cart 9 3 Satellite Attitude Control 17 4 UUV Roll

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review

Dr 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 information

Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 5. An Introduction to Feedback Control Systems

Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 5. An Introduction to Feedback Control Systems Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL Glenn Vinnicombe HANDOUT 5 An Introduction to Feedback Control Systems ē(s) ȳ(s) Σ K(s) G(s) z(s) H(s) z(s) = H(s)G(s)K(s) L(s) ē(s)=

More information

Lecture 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 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 information

Design of an Intelligent Control Scheme for Synchronizing Two Coupled Van Der Pol Oscillators

Design of an Intelligent Control Scheme for Synchronizing Two Coupled Van Der Pol Oscillators International Journal of ChemTech Research CODEN (USA): IJCRGG ISSN : 974-49 Vol.6, No., pp 533-548, October 4 CBSE4 [ nd and 3 rd April 4] Challenges in Biochemical Engineering and Biotechnology for Sustainable

More information

BASIC PROPERTIES OF FEEDBACK

BASIC 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 information

CYBER EXPLORATION LABORATORY EXPERIMENTS

CYBER 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 information

Design of a Lead Compensator

Design 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 information

CHAPTER 6 CLOSED LOOP STUDIES

CHAPTER 6 CLOSED LOOP STUDIES 180 CHAPTER 6 CLOSED LOOP STUDIES Improvement of closed-loop performance needs proper tuning of controller parameters that requires process model structure and the estimation of respective parameters which

More information

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus

Dr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Root Locus 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 information

Open Loop Tuning Rules

Open Loop Tuning Rules Open Loop Tuning Rules Based on approximate process models Process Reaction Curve: The process reaction curve is an approximate model of the process, assuming the process behaves as a first order plus

More information

CHAPTER 1 Basic Concepts of Control System. CHAPTER 6 Hydraulic Control System

CHAPTER 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 information

In-Process Control in Thermal Rapid Prototyping

In-Process Control in Thermal Rapid Prototyping In-Process Control in Thermal Rapid Prototyping Nadya Peek in MAS.864, May 2011 May 16, 2011 Abstract This report compares thermal control algorithms for curing carbon fibre and epoxy resin composite parts

More information

7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM

7.4 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 information

7.2 Controller tuning from specified characteristic polynomial

7.2 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

Chapter 12. Feedback Control Characteristics of Feedback Systems

Chapter 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 open-loop system (a system without feedbac) and a closed-loop

More information

Control Systems Design

Control Systems Design ELEC4410 Control Systems Design Lecture 18: State Feedback Tracking and State Estimation Julio H. Braslavsky julio@ee.newcastle.edu.au School of Electrical Engineering and Computer Science Lecture 18:

More information

Outline. Classical Control. Lecture 1

Outline. 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 information

EE C128 / ME C134 Fall 2014 HW 8 - Solutions. HW 8 - Solutions

EE 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

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

CHAPTER 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 information

Introduction to System Identification and Adaptive Control

Introduction to System Identification and Adaptive Control Introduction to System Identification and Adaptive Control A. Khaki Sedigh Control Systems Group Faculty of Electrical and Computer Engineering K. N. Toosi University of Technology May 2009 Introduction

More information

Dynamic 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. 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 information

SECTION 5: ROOT LOCUS ANALYSIS

SECTION 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 information

RamchandraBhosale, Bindu R (Electrical Department, Fr.CRIT,Navi Mumbai,India)

RamchandraBhosale, Bindu R (Electrical Department, Fr.CRIT,Navi Mumbai,India) Indirect Vector Control of Induction motor using Fuzzy Logic Controller RamchandraBhosale, Bindu R (Electrical Department, Fr.CRIT,Navi Mumbai,India) ABSTRACT: AC motors are widely used in industries for

More information

Understanding Exponents Eric Rasmusen September 18, 2018

Understanding Exponents Eric Rasmusen September 18, 2018 Understanding Exponents Eric Rasmusen September 18, 2018 These notes are rather long, but mathematics often has the perverse feature that if someone writes a long explanation, the reader can read it much

More information

Homework Assignment 3

Homework Assignment 3 ECE382/ME482 Fall 2008 Homework 3 Solution October 20, 2008 1 Homework Assignment 3 Assigned September 30, 2008. Due in lecture October 7, 2008. Note that you must include all of your work to obtain full

More information

Lab 3: Model based Position Control of a Cart

Lab 3: Model based Position Control of a Cart I. Objective Lab 3: Model based Position Control of a Cart The goal of this lab is to help understand the methodology to design a controller using the given plant dynamics. Specifically, we would do position

More information

Feedback Control of Linear SISO systems. Process Dynamics and Control

Feedback Control of Linear SISO systems. Process Dynamics and Control Feedback Control of Linear SISO systems Process Dynamics and Control 1 Open-Loop Process The study of dynamics was limited to open-loop systems Observe process behavior as a result of specific input signals

More information

Systems Analysis and Control

Systems 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

Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis

Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis Tuning PI controllers in non-linear uncertain closed-loop systems with interval analysis J. Alexandre dit Sandretto, A. Chapoutot and O. Mullier U2IS, ENSTA ParisTech SYNCOP April 11, 2015 Closed-loop

More information

Engraving Machine Example

Engraving Machine Example Engraving Machine Example MCE44 - Fall 8 Dr. Richter November 24, 28 Basic Design The X-axis of the engraving machine has the transfer function G(s) = s(s + )(s + 2) In this basic example, we use a proportional

More information

Part II. Advanced PID Design Methods

Part II. Advanced PID Design Methods Part II Advanced PID Design Methods 54 Controller transfer function C(s) = k p (1 + 1 T i s + T d s) (71) Many extensions known to the basic design methods introduced in RT I. Four advanced approaches

More information

Feedback Control part 2

Feedback Control part 2 Overview Feedback Control part EGR 36 April 19, 017 Concepts from EGR 0 Open- and closed-loop control Everything before chapter 7 are open-loop systems Transient response Design criteria Translate criteria

More information

The requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot --- in time domain

The 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 information

Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur. Module 3 Lecture 8

Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur. Module 3 Lecture 8 Power System Operations and Control Prof. S.N. Singh Department of Electrical Engineering Indian Institute of Technology, Kanpur Module 3 Lecture 8 Welcome to lecture number 8 of module 3. In the previous

More information

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual

Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control. DC Motor Control Trainer (DCMCT) Student Manual Quanser NI-ELVIS Trainer (QNET) Series: QNET Experiment #02: DC Motor Position Control DC Motor Control Trainer (DCMCT) Student Manual Table of Contents 1 Laboratory Objectives1 2 References1 3 DCMCT Plant

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 pre-filter [ 0.

] [ 200. ] 3 [ 10 4 s. [ ] s + 10 [ P = s [ 10 8 ] 3. s s (s 1)(s 2) series compensator ] 2. s command pre-filter [ 0. EEE480 Exam 2, Spring 204 A.A. Rodriguez Rules: Calculators permitted, One 8.5 sheet, closed notes/books, open minds GWC 352, 965-372 Problem (Analysis of a Feedback System) Consider the feedback system

More information

ECE 388 Automatic Control

ECE 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 information

1 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) =

1 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 information

Automatic Control (TSRT15): Lecture 4

Automatic Control (TSRT15): Lecture 4 Automatic Control (TSRT15): Lecture 4 Tianshi Chen Division of Automatic Control Dept. of Electrical Engineering Email: tschen@isy.liu.se Phone: 13-282226 Office: B-house extrance 25-27 Review of the last

More information

Chap 8. State Feedback and State Estimators

Chap 8. State Feedback and State Estimators Chap 8. State Feedback and State Estimators Outlines Introduction State feedback Regulation and tracking State estimator Feedback from estimated states State feedback-multivariable case State estimators-multivariable

More information

Lab # 4 Time Response Analysis

Lab # 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 information

Unit 8: Part 2: PD, PID, and Feedback Compensation

Unit 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 information

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad

INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III

More information

PD, 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 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 information

Professional Portfolio Selection Techniques: From Markowitz to Innovative Engineering

Professional Portfolio Selection Techniques: From Markowitz to Innovative Engineering Massachusetts Institute of Technology Sponsor: Electrical Engineering and Computer Science Cosponsor: Science Engineering and Business Club Professional Portfolio Selection Techniques: From Markowitz to

More information