45283438 Numerical Computing With Simulink Volume I

ot100_granfm3.qxp ot100_granfm3.qxp

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Numerical Computing with Simulink, Volume I Creating Simulations


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Numerical Computing with Simulink, Volume I Creating Simulations

Richard J. Gran Mathematical Analysis Company Norfolk, Massachuse Massachusetts tts

Society for Industrial and Applied Mathematics • Philadelphia


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Copyright © 2007 by the Society for Industrial and Applied Mathematics. 10 9 8 7 6 5 4 3 2 1 All rights reserved. reserved. Printed in the United States of America. No part of this book may be reproduced, stored, or transmitted in any manner without the written permission of the publisher.. For information, write to the Society for Industrial and Applied Mathematics, publisher 3600 Market Street, 6th floor, Philadelphia, PA 19104-2688 19104-2688 USA. Trademarked names may be used in this book without the inclusion of a trademark symbol. These names are used in an editorial context only; no infringement of trademark is intended. Maple is a trademark of Maplesoft, Waterloo, Ontario, Canada. MATLAB, Simulink, Real Time Workshop, SimHydraulics, Stateflow, and Handle Graphics are registered trademarks of The MathWorks, Inc. For MATLAB product information, please contact The MathWorks, Inc., 3 Apple Hill Drive, Natick, MA 01760-2098 USA, 508-6477000, Fax: 508-647-7101, [email protected], www www.mathworks.com .mathworks.com. Figure 1.12 is used with permission of Marcus Orvando, president of Erwin Sattler Clocks of America. Figure 1.16 was taken by Michael Reeve on January 30, 2004, and its source is the Wikipedia article “Foucault Pendulum.” Per Wikipedia, permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, Texts, and no Back-Cover Texts. Texts. Subject to disclaimers. Figure 2.7 is taken from the Wikipedia article “Thermostat.” Per Wikipedia, permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant I nvariant Sections, no Front-Cover Texts, Texts, and no Back-Cover Texts. Texts. Subject to disclaimers.

Library of Congress Cataloging-in-Publication Data Gran, Richard J., 1940Numerical computing with Simulink / Richard J. Gran. v. cm. Includes bibliographical references and index. Contents: v. 1. Creating simulations ISBN 978-0-898716-37-5 (alk. paper) 1. Numerical analysis--Data processing. 2. SIMULINK. I. Title. QA297.G676 2007 518.0285--dc22 2007061803

is a registered trademark.


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To Dr. Richard A. Scheuing: A Friend and Mentor 

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Contents

List of Figures

xi

List of Tables

xvii

Preface

xix

1

Introduction to Simulink 1.1 Using a Picture to Write a Program . . . . . . . . . . . . . . . . . . . . . 1.2 Example 1: Galileo Drops Two Objects from the Leaning Tower of Pisa . 1.3 Example 2: Modeling a Pendulum and the Escapement of a Clock . . . . 1.3.1 History of Pendulum Clocks . . . . . . . . . . . . . . . . . . . . . 1.3.2 A Simulation Model for the Clock . . . . . . . . . . . . . . . . . . 1.4 Example 3: Complex Rotations—The Foucault Pendulum . . . . . . . . . 1.4.1 Forces from Rotations . . . . . . . . . . . . . . . . . . . . . . . . 1.4.2 Foucault Pendulum Dynamics . . . . . . . . . . . . . . . . . . . . 1.5 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 5 17 18 20 24 25 26 31 32

2

Linear Differential Equations, Matrix Algebra, and Control Systems 2.1 Linear Differential Equations: Linear Algebra . . . . . . . . . . . . . . . 2.1.1 Solving a Differential Equation at Discrete Time Steps . . . . . . . 2.1.2 Linear Differential Equations in Simulink . . . . . . . . . . . . . . 2.2 Laplace Transforms for Linear Differential Equations . . . . . . . . . . . 2.3 Linear Feedback Control . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 What Is a Control System? . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Control Systems and Linear Differential Equations . . . . . . . . . 2.4 Linearization and the Control of Linear Systems . . . . . . . . . . . . . . 2.4.1 L inearization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Eigenvalues and the Response of a Linear System . . . . . . . . . . 2.5 Poles and the Roots of the Characteristic Polynomial . . . . . . . . . . . . 2.5.1 Feedback of the Position of the Mass in the Spring-Mass Model . . 2.5.2 Feedback of the Velocity of the Mass in the Spring-Mass Model . . 2.5.3 Comparing Position and Rate Feedback . . . . . . . . . . . . . . . 2.5.4 The Structure of a Control System: Transfer Functions . . . . . . .

33 33 36 38 40 43 44 48 49 49 51 55 56 57 61 62

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2.6 Transfer Functions: Bode Plots . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 The Bode Plot for Continuous Time Systems . . . . . . . . . . . . 2.6.2 Calculating the Bode Plot for Continuous Time Systems . . . . . . 2.7 PD Control, PID Control, and Full State Feedback . . . . . . . . . . . . . 2.7.1 PD Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2 PID Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.3 Full State Feedback . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.4 Getting Derivatives for PID Control or Full State Feedback . . . . 2.8 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

64 65 65 69 69 71 72 74 78 78

3

Nonlinear Differential Equations 81 3.1 The Lorenz Attractor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.1.1 Linear Operating Points: Why the Lorenz Attractor Is Chaotic . . . 83 3.2 Differential Equation Solvers in MATLAB and Simulink . . . . . . . . . 86 3.3 Tables, Interpolation, and Curve Fitting in Simulink . . . . . . . . . . . . 87 3.3.1 T he Simple Lookup Table . . . . . . . . . . . . . . . . . . . . . . 88 3.3.2 Interpolation: Fitting a Polynomial to the Data and Using the Result in Simulink . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.3.3 Using Real Data in the Model: From Workspace and File . . . . . . 92 3.4 Rotations in Three Dimensions: Euler Rotations, Axis-Angle Representations, Direction Cosines, and the Quaternion . . . . . . . . . . 94 3.4.1 E uler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 3.4.2 Direction Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 3.4.3 A xis-Angle Rotations . . . . . . . . . . . . . . . . . . . . . . . . . 99 3.4.4 The Quaternion Representation . . . . . . . . . . . . . . . . . . . . 99 3.5 Modeling the Motion of a Satellite in Orbit . . . . . . . . . . . . . . . . . 105 3.5.1 Creating an Attitude Error When Using Direction Cosines . . . . . 107 3.5.2 Creating an Attitude Error Using Quaternion Representations . . . . 109 3.5.3 The Complete Spacecraft Model . . . . . . . . . . . . . . . . . . . 109 3.6 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

4

Digital Signal Processing in Simulink 115 4.1 Difference Equations, Fibonacci Numbers, and z-Transforms . . . . . . . 116 4.1.1 The z-Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 4.1.2 Fibonacci (Again) Using z-Transforms . . . . . . . . . . . . . . . . 120 4.2 Digital Sequences, Digital Filters, and Signal Processing . . . . . . . . . . 121 4.2.1 Digital Filters, Using z-Transforms, and Discrete Transfer Functions 121 4.2.2 Simulink Experiments: Filtering a Sinusoidal Signal and Aliasing . 123 4.2.3 The Simulink Digital Library . . . . . . . . . . . . . . . . . . . . . 128 4.3 Matrix Algebra and Discrete Systems . . . . . . . . . . . . . . . . . . . . 130 4.4 The Bode Plot for Discrete Time Systems . . . . . . . . . . . . . . . . . . 135 4.5 Digital Filter Design: Sampling Analog Signals, the Sampling Theorem, and Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 4.5.1 Sampling and Reconstructing Analog Signals . . . . . . . . . . . . 137

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4.5.2 Analog Prototypes of Digital Filters: The Butterworth Filter . . . . 142 4.6 The Signal Processing Blockset . . . . . . . . . . . . . . . . . . . . . . . 145 4.6.1 Fundamentals of the Signal Processing Blockset: Analog Filters . . 146 4.6.2 Creating Digital Filters from Analog Filters . . . . . . . . . . . . . 148 4.6.3 Digital Signal Processing . . . . . . . . . . . . . . . . . . . . . . . 149 4.6.4 Implementing Digital Filters: Structures and Limited Precision . . . 153 4.6.5 Batch Filtering Operations, Buffers, and Frames . . . . . . . . . . . 160 4.7 The Phase-Locked Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 4.8 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 5

Random Numbers, White Noise, and Stochastic Processes 173 5.1 Modeling with Random Variables in Simulink: Monte Carlo Simulations . 173 5.1.1 Monte Carlo Analysis and the Central Limit Theorem . . . . . . . . 174 5.1.2 Simulating a Rayleigh Distributed Random Variable . . . . . . . . 176 5.2 Stochastic Processes and White Noise . . . . . . . . . . . . . . . . . . . . 177 5.2.1 The Random Walk Process . . . . . . . . . . . . . . . . . . . . . . 178 5.2.2 Brownian Motion and White Noise . . . . . . . . . . . . . . . . . . 180 5.3 Simulating a System with White Noise Inputs Using the Weiner Process . 184 5.3.1 White Noise and a Spring-Mass-Damper System . . . . . . . . . . 184 5.3.2 Noisy Continuous and Discrete Time Systems: The Covariance Matrix . . . . . . . . . . . . . . . . . . . . . . . . 186 5.3.3 Discrete Time Equivalent of a Continuous Stochastic Process . . . . 189 5.3.4 Modeling a Specified Power Spectral Density: 1 /f Noise . . . . . . 194 5.4 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

6

Modeling a Partial Differential Equation in Simulink 201 6.1 The Heat Equation: Partial Differential Equations in Simulink . . . . . . . 202 6.1.1 Finite Dimensional Models . . . . . . . . . . . . . . . . . . . . . . 202 6.1.2 An Electrical Analogy of the Heat Equation . . . . . . . . . . . . . 203 6.2 Converting the Finite Model into Equations for Simulation with Simulink 205 6.2.1 Using Kirchhoff’s Law to Get the Equations . . . . . . . . . . . . . 206 6.2.2 The State-Space Model . . . . . . . . . . . . . . . . . . . . . . . . 208 6.3 Partial Differential Equations for Vibration . . . . . . . . . . . . . . . . . 212 6.4 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

7

Stateflow: A Tool for Creating and Coding State Diagrams, Complex Logic, Event Driven Actions, and Finite State Machines 215 7.1 Properties of Stateflow: Building a Simple Model . . . . . . . . . . . . . 216 7.1.1 Stateflow Semantics . . . . . . . . . . . . . . . . . . . . . . . . . . 218 7.1.2 Making the Simple Stateflow Chart Do Something . . . . . . . . . 221 7.1.3 Following Stateflow’s Semantics Using the Debugger . . . . . . . . 223 7.2 Using Stateflow: A Controller for Home Heating . . . . . . . . . . . . . . 225 7.2.1 Creating a Model of the System and an Executable Specification . . 225

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7.2.2 Stateflow’s Action Language Types . . . . . . . . . . . . . . . . . . 229 7.2.3 The Heating Controller Layout . . . . . . . . . . . . . . . . . . . . 230 7.2.4 Adding the User Actions, the Digital Clock, and the Stateflow Chart to the Simulink Model of the Home Heating System . . . . . . . . . 231 7.2.5 Some Comments on Creating the GUI . . . . . . . . . . . . . . . . 239 7.3 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 8

Physical Modeling: SimPowerSystems and SimMechanics 241 8.1 SimPowerSystems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 8.1.1 How the SimPowerSystems Blockset Works: Modeling a Nonlinear Resistor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 8.1.2 Using the Nonlinear Resistor Block . . . . . . . . . . . . . . . . . 248 8.2 Modeling an Electric Train Moving on a Rail . . . . . . . . . . . . . . . . 251 8.3 SimMechanics: A Tool for Modeling Mechanical Linkages and Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 8.3.1 Modeling a Pendulum with SimMechanics . . . . . . . . . . . . . . 257 8.3.2 Modeling the Clock: Simulink and SimMechanics Together . . . . 260 8.4 More Complex Models in SimMechanics and SimPowerSystems . . . . . 262 8.5 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

9

Putting Everything Together: Using Simulink in a System Design Process 269 9.1 Specifications Development and Capture . . . . . . . . . . . . . . . . . . 270 9.1.1 Modeling and Analysis: Converting the Specifications into an “Executable Specification” . . . . . . . . . . . . . . . . . . . . . . . . 271 9.2 Modeling the System to Incorporate the Specifications: Lunar Module Rotation Using Time Optimal Control . . . . . . . . . . . . . . . . . . . 272 9.2.1 From Specification to Control Algorithm . . . . . . . . . . . . . . . 273 9.3 Design of System Components to Meet Specifications: Modify the Design to Accommodate Computer Limitations . . . . . . . . . . . . . . . . . . . 276 9.3.1 Final Lunar Module Control System Executable Specification . . . 279 9.3.2 The Control System Logic: Using Stateflow . . . . . . . . . . . . . 283 9.4 Verification and Validation of the Design . . . . . . . . . . . . . . . . . . 285 9.5 The Final Step: Creating Embedded Code . . . . . . . . . . . . . . . . . 286 9.6 Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

10

Conclusion: Thoughts about Broad-Based Knowledge

289

Bibliography

291

Index

295

List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11

1.12 1.13 1.14 1.15

1.16 1.17 1.18 1.19 1.20

2.1 2.2 2.3

Simulink library browser. . . . . . . . . . . . . . . . . . . . . . . . . . Leaning Tower of Pisa. . . . . . . . . . . . . . . . . . . . . . . . . . . . Integrator block dialog. . . . . . . . . . . . . . . . . . . . . . . . . . . . Simulink model for the Leaning Tower of Pisa experiment. . . . . . . . . Scope block graph for the Leaning Tower simulation. . . . . . . . . . . . Leaning Tower Simulink model with air drag added. . . . . . . . . . . . Scope dialog showing how to change the number of axes in the Scope plot. Leaning Tower simulation results when air drag is included. . . . . . . . Adding the second object to the Leaning Tower simulation by vectorizing the air drag. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simulating results. The Leaning Tower Simulink model with a heavy and light object. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Right clicking anywhere in the diagram opens a pull-down menu that allows changes to the colors of the blocks, bold vector line styles, and other model annotations. . . . . . . . . . . . . . . . . . . . . . . . . . . Components of a pendulum clock. . . . . . . . . . . . . . . . . . . . . . Simulink model of the clock. . . . . . . . . . . . . . . . . . . . . . . . . Blocks in the subsystem “Escapement Model.” . . . . . . . . . . . . . . Results of the clock simulation for times around 416 sec. Upper figure is the pendulum angle, and the lower figure is the acceleration applied to the pendulum by the escapement. . . . . . . . . . . . . . . . . . . . . . Foucault pendulum at the Panthéon in Paris. . . . . . . . . . . . . . . . . Axes used to model the Foucault pendulum. . . . . . . . . . . . . . . . . The Simulink model of the Foucault pendulum. . . . . . . . . . . . . . . Simulation results from the Foucault pendulum Simulink model using the default solver. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simulation results for the Foucault pendulum using a tighter tolerance for the solver. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 5 8 9 10 13 13 14 15 16

17 19 21 22

24 25 27 29 29 30

Using Simulink to compare discrete and continuous time state-space versions of the pendulum. . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Outputs asseen inthe Scope blockswhensimulating the model inFigure2.1. 40 Linear pendulum model using Transfer Function and Zero-Pole-Gain blocks from the Simulink Continuous Library. . . . . . . . . . . . . . . 43 xi

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List of Figures 2.4

2.5 2.6

2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2.27 2.28 2.29

Home heating system Simulink model (Thermo_NCS). This model describes the house temperature using a single first order differential equation and assumes that a bimetal thermostat controls the temperature. . . . Subsystem “House” contains the single differential equation that models the house. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Results of simulating the home heating system model. The top figure shows the outdoor and indoor temperatures, and the bottom figure shows the cost of the heat over the 24-hour simulation. . . . . . . . . . . . . . . A typical home-heating thermostat [50]. . . . . . . . . . . . . . . . . . . Hysteresis curve for the thermostat in the Simulink model. . . . . . . . . Model of a spring-mass-damper system using Simulink primitives. . . . Model of a spring-mass-damper system using the state-space model and Simulink’s automatic vectorization. . . . . . . . . . . . . . . . . . . . . Changing the value of the damping ratio from 0 .1 to 0.5 in the model of Figure 2.10. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Generic components of a control system. . . . . . . . . . . . . . . . . . A simple control that feeds back the position of the mass in a springmass-damper system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Feed back the velocity of the mass instead of the position to damp the oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Time response of the mass (position at the top and velocity at the bottom) for the velocity feedback controller. . . . . . . . . . . . . . . . . . . . . Using the state-space model for the spring-mass-damper control system. . Root locus for the spring-mass-damper system. Velocity gain varying from 0 to 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Root locus for the position feedback control. Gain changes only the frequency of oscillation. . . . . . . . . . . . . . . . . . . . . . . . . . . One method of determining the velocity of the mass is to differentiate the position using a linear system that approximates the derivative. . . . . . Simulation result from the Simulink model. . . . . . . . . . . . . . . . . Control System Toolbox interface to Simulink. GUIs allow you to select inputs and outputs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . LTI Viewer results from the LTI Viewer. . . . . . . . . . . . . . . . . . . Using proportional-plus derivative (PD) control for the spring-massdamper. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Response of the spring-mass-damper with the PD controller. There are no oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proportional, integral, and derivative (PID) control in the spring-massdamper system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . PID control of the spring-mass system. Response to a unit step has zero error. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Getting speed from measurements of the mass position and force applied to the mass. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Simulink model to compare three methods for deriving the velocity of the mass. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Methods for deriving rate are selected using the “multiport switch.” . . .

45 46

47 47 48 52 52 53 55 56 57 58 59 60 61 64 64 67 68 70 71 72 73 76 76 77

List of Figures 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13

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Simulink model and results for the Lorenz attractor. . . . . . . . . . . . 83 Exploringthe Lorenzattractor using thelinearizationtoolfromthe control toolbox. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Eigenvalues of the linearized Lorenz attractor as a function of time. . . . 85 Pseudotabulated data for the external temperature in the home heating model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Home heating systemSimulink modelwith tabulatedoutside temperature added. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Dialog to add the tabulated input. . . . . . . . . . . . . . . . . . . . . . 90 Real data must be in a MATLAB array for use in the “From Workspace” block. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Simulink model of the home heating system using measured outdoor temperatures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 From workspacedialog block that uses themeasured outdoor temperature array. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Using real data from measurements in the Simulink model. . . . . . . . . 94 Single axis rotation in three dimensions. . . . . . . . . . . . . . . . . . . 95 Using quaternions to compute the attitude from three simultaneous body axis rotations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Simulink block to convert quaternions into a direction cosine. . . . . . . 103 Combining the quaternion block with the body rotational acceleration. . . 104 Electrical circuit of a motor with the mechanical equations. . . . . . . . . 106 Controlling the rotation of a spacecraft using reaction wheels. . . . . . . 110 Matrixconcatenation blockscreate thematrix Q used to achieve a desired value for q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Simulating the Fibonacci sequence in Simulink. . . . . . . . . . . . . . . 117 Fibonacci sequence graph generated by the Simulink model. . . . . . . . 118 Simulink model for computing the golden ratio from the Fibonacci sequence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Generating the data needed to compute a digital filter transfer function. . 123 Adding callbacks to a Simulink model uses Simulink’s model properties dialog, an option under the Simulink window’s file menu. . . . . . . . . 125 Sampling a sine wave at two different rates, illustrating the effect of aliasing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Changing the numerator and denominator in the Digital Filter block changes the filter icon. . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Using the Digital Filter block to simulate the Fibonacci sequence. . . . . 130 State-space models for discrete time simulations in Simulink. . . . . . . 131 A numerical experiment in Simulink: Does f k+1 f k −1 − f k2 = ±1? . . . . 134 13 iterations of the Fibonacci sequence show that f k +1 f k−1 − f k2 = ±1 (so far). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 The Bode plot for the discrete filter model developed using the Control System Toolbox interface with Simulink. . . . . . . . . . . . . . . . . . 137 Illustrating the steps in the proof of the sampling theorem. . . . . . . . . 139

xiv

List of Figures 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.30 4.31 4.32 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

5.10

Simulation illustrating the sampling theorem. The input signal is 15 sinusoidal signals with frequencies less than 500 Hz. . . . . . . . . . . . 140 Specification of a unity gain low pass filter requires four pieces of information. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 Butterworth filter for D/A conversion using the results of the M-file “butterworthncs.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Using the Signal Processing Blockset to design an analog filter. . . . . . 147 Moving average FIR filter simulation. . . . . . . . . . . . . . . . . . . . 150 Pole-zero plot and filter properties for the moving average filter. . . . . . 152 The Simulink model for the band-pass filter with no computational limitations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 The band-pass filter design created by the Signal Processing Blockset digital filter block. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Band-pass filter simulation results (with and without signal in the pass band). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Fixed-point implementation of a band-pass filter using the Signal Processing Blockset. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Running the Simulink model to determine the minima, maxima, and scaling for all of the fixed-point calculations in the Band-pass filter. . . . 161 Illustrating the use of buffers. Reconstruction of a sampled signal using the FFT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 Using the FFT and the sampling theorem to interpolate a faster sampled version of a sampled analog signal. . . . . . . . . . . . . . . . . . . . . 163 The analog signal (sampled at 1 kHz, top) and the reconstructed signal (sampled at 8 kHz, bottom). . . . . . . . . . . . . . . . . . . . . . . . . 165 Results of interpolating a sampled signal using FFTs and the sampling theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Phase-locked loop (PLL). A nonlinear feedback control for tracking frequency and phase of a sinusoidal signal. . . . . . . . . . . . . . . . . . . 167 Voltage controlled oscillator subsystem in the PLL model. . . . . . . . . 168 Details of the integrator modulo 1 subsystem in Figure 4.30. . . . . . . . 168 Phase-locked loop simulation results. . . . . . . . . . . . . . . . . . . . 169 Monte Carlo simulation demonstrating the central limit theorem. . . . . . 175 Result of the Monte Carlo simulation of the central limit theorem. . . . . 175 A Simulink model that generates Rayleigh random variables. . . . . . . . 177 Simulink model that generates nine samples of a random walk. . . . . . . 179 Nine samples of a random walk process. . . . . . . . . . . . . . . . . . . 179 Simulink model for the spring-mass-damper system. . . . . . . . . . . . 185 Motions of 10 masses with white noise force excitations. . . . . . . . . . 185 White noise block in the Simulink library (masked subsystem). . . . . . 186 Continuous linearsystem covariance matrixcalculationin Simulink. The result is the covariance matrix that can be used for simulating with a covariance equivalent discrete system. . . . . . . . . . . . . . . . . . . . 191 Noise response continuous time simulation and equivalent discrete systems at three different sample times. . . . . . . . . . . . . . . . . . . . . 192

List of Figures 5.11 5.12 5.13

6.1 6.2 6.3 6.4 6.5 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 8.1 8.2 8.3 8.4 8.5 8.6

xv

Simulation results from the four simulations with white noise inputs. . . 193 Simulation of the fractal noise process that has a spectrum proportional to 1 /f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 Power spectral density estimators in the signal processing blockset used to compute the sampled PSD of the 1 /f noise process. . . . . . . . . . . 198 An electrical model of the thermodynamics of a house and its heating system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 Simulink model for the two-room house dynamics. . . . . . . . . . . . . 210 Simulation results when the heat is off. . . . . . . . . . . . . . . . . . . 211 Simulation results when the heat is on continuously. . . . . . . . . . . . 211 Using PID control to maintain constant room temperatures (70 deg F). . . 211 First Stateflow diagram (in the chart) uses a manual switch to create the event. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218 A Stateflow diagram that contains most of the stateflow semantics. . . . . 219 The Simulink model for the simple timer example. . . . . . . . . . . . . 222 Modified Stateflow chart provides the timer outputs “Start,” “On,” and “Trip.” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Outputs resulting from running the simple timer example. . . . . . . . . 224 PID controller for the home-heating example developed in Chapter 6. . . 227 Response of the home heating system with the external temperature from Chapter 3 as input. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 A tentative design for the new home heating controller we want to develop. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 The complete simulation of the home heating controller. . . . . . . . . . 231 Graphical user interface (GUI) that implements the prototype design for the controller. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 Interactionsbetween theGUI andStateflow usechanges in thegain values in three gain blocks in the subsystem called “User_Selections.” . . . . . 233 The first of the two parallel states in the state chart for the heating controller. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 The second of the two parallel states in the home heating controller state chart. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 The final version of the PID control for the home heating controller. . . . 239 A simple electrical circuit model using Simulink and SimPowerSystems. 242 The electrical circuit time response. Current in the resistor and voltage across the capacitor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 43 The SimPowerSystems library in the Simulink browser and the elements sublibrary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 A Simulink and SimPowerSystems subsystem model for a nonlinear or time varying resistor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 Masking the subsystem with an icon that portrays what the variable resistor does makes it easy to identify in an electrical circuit. . . . . . . . . 248 Circuit diagram of a simulation with the nonlinear resistor block. . . . . 249

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List of Figures 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19 8.20 8.21 9.1

9.2 9.3 9.4 9.5 9.6 9.7 9.8

Using the signal builder block in Simulink to define the values for the time varying resistance. . . . . . . . . . . . . . . . . . . . . . . . . . . 250 Using SimPowerSystems: Results from the simple RL network simulation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 Simulink and SimPowerSystems model of a train and its dynamics. . . . 252 Complete model of the train moving along a track. . . . . . . . . . . . . 254 The Simulink blocks that compute the rail resistance as a function of the train location. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Simulation results. Rail resistances, currents in the rails, and current to the train. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Modeling a pendulum using SimMechanics. . . . . . . . . . . . . . . . . 258 Coordinate system for the pendulum. . . . . . . . . . . . . . . . . . . . 258 Simulating the pendulum motion using SimMechanics and its viewer. . . 260 Adding Simulink blocks to a SimMechanics model to simulate the clock as in Chapter 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 Time history of the pendulum motion from the clock simulation. . . . . . 261 SimMechanics model Sim_Mechanics_Vibration. . . . . . . . . 263 String model (the subsystem in the model above). It consists of 20 identical “spring mass” revolute and prismatic joint elements. . . . . . . . . . 263 String simulation. Plucked 15 cm from the left, at the center, and 15 cm from the right. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Using SimPowerSystems to build a model for a four-room house. . . . . 264 The start of the specification capture process. Gather existing designs and simulations and create empty subsystems for pieces that need to be developed. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 The physics of the phase plane logic and the actions that need to be developed. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 Complete simulation of the lunar module digital autopilot. . . . . . . . . 279 Simulating a counter that ticks at 625 microsec for the lunar module Simulink model. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 The blocks in the jet-on-time counter subsystem in Figure 9.4. . . . . . . 281 Lunar module graph of the switch curves and the phase plane motion for the yaw axis. (The graph uses a MATLAB function block in thesimulation.)282 Stateflow logic that determines the location of the state in the phase plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 Simulink blocks that check different model attributes for model verification. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

List of Tables

3.1

4.1 4.2 4.3

5.1

Computation time for the Lorenz model with different solvers and tolerances. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

Using low pass filters. Simulating various filters with various sample times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 Comparison of the error in converting a digital signal to an analog signal using different low pass filters. . . . . . . . . . . . . . . . . . . . . . . . 148 Specification for the band-pass filter and the actual values achieved in the design. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Simulink computation times for the four white noise simulations. . . . . 193

xvii

Preface Simulation has gradually become the preferred method for designing complex systems. The aerospace industry provided the impetus for this design approach because most aerospace systems are at best difficult or at worst impossible to test. I started my career working on the lunar module flight control system. Testing this system in space was extremely expensive andvery difficult. In theentire developmentcycle, there was, in fact, only oneunmannedtest in earth orbit. All of the designs, by necessity, required simulations. Overtime, thisapproach has become pervasive in many industries, particularly the automotive industry. Even in product developments as mundane (seemingly) as disk drives for computers, simulation has become the dominant method of design. A digital simulation of the environment that a system “lives” in is crucial in this modern simulation based design process. The accurate representation of this environment allows the designer to see how well the system being designed performs. It allows the designer to verify that the design meets all of the performance specifications. Moreover, as the simulation develops, the designer always can see how the performance requirements pass through to the various subsystems in the design. This refinement of the specification allows the model to, in essence, become the specification. The simulation environment can also allow the designer of embedded digital signal processing and control algorithms to test the code directly as it is developed to ensure that it satisfies the specification. The last, and newest, step in this process is the ability to use the simulation model automatically to generate the computer code for computers in the system. This can eliminate most of the “hand coding” that is required by the current practice. (In most companies this process consists of handing a stack of papers with the written specification to a team of computer programmers for coding.) This approach eliminates two potential sources of error: the errors introduced by the conversion of the design into a written specification and the errors introduced by the manual coding. Simulink ® is a remarkable tool that fills theniche fora robust, accurate,and easilyused simulation tool. It provides a visual way of constructing a simulation of a complex system that evolves in time (the designer can optimize the system by optimizing the mathematical model’s response). This “model-based design” approach provides accurate simulation as various numerical constants in the model are changed. It allows the user to tune these constants so that the simulation accurately portrays the real world. It then allows sharing of the model among many different users so they can begin to design various components or parts of the system. At the same time, it provides a means for understanding design problems whose solutions will make the system operate in a desired way. Finally, the resulting design’s embedded software is rapidly convertible into embedded code that is xix

xx

Preface

directly usable for testing in a stand-alone computer. All of this can be accomplished using one easily understood graphical environment. A major thesis of this book is that Simulink can form the basis for an integrated design environment. As your use of the tool evolves, the way this can happen should become more apparent. Because simulation is such an important adjunct to numerical analysis, the book can be a supplement for a course on numerical computing. It fits very well with an earlier SIAM book, Numerical Computing with MATLAB by Cleve Moler [29] (which we refer to so frequently that its title is abbreviated NCM in the text), and it has been written with the same eclectic mix of practice and mathematics that makes Cleve’s book so much fun to read. I hope that this work favorably compares with—and comes close to matching the quality of—Cleve’s work. As is the case with Numerical Computing with MATLAB , this book is based on a collection of 88 model files that are in the library Numerical Computing with Simulink (which we abbreviate as NCS), which can be downloaded from the Mathematical Analysis Co. web page (www.math-analysis.com). This library of SIMULINK models and MATLAB® M-files forms an essential part of the book and, through extensions that the reader is encouraged to peruse, an essential part of the exercises. The book will take you on a tour of the Simulink environment, showing you how to develop a system model and then how to execute the design steps to make the model into a functioning design laboratory. Along the way, you will be introduced to the mathematics of systems, including difference equations and z-transforms, ordinary differential equations (both linear and nonlinear), Laplace transforms, numerical methods for solving differential equations, and methods for simulating complex systems from several different disciplines. The mathematics of simulation is not complete without a discussion of random variables and random processes for doing Monte Carlo simulations. Toward this end, we introduce and develop the techniques for modeling random processes with predetermined statistical properties. The mathematics for this type of simulation begins with “white noise,” which is a very difficult entity to pin down. We introduce the concept of Brownian motion and show the connection between this process and the fictitious, but useful, white noise process. The simulation of the Brownian process in Simulink is developed; it is then used to form more (statistically) complex processes. We review and show how to simulate and analyze randomprocesses, andwe formulate anduse thepowerspectraldensity of a process. We introduce other tools, in addition to Simulink, from The MathWorks. Each of the tools is an expansion into a different “domain of knowledge.” The first tool is the Signal Processing Blockset that extends Simulink into the domain of signal processing (both analog and digital). The second tool, Stateflow® , expands Simulink to include state charts and signal flow for modeling event driven systems (i.e., systems where the actions start at times that are asynchronous with the computer’s clock). This tool naturally extends Simulink into the traditional realm of finite state machines. The third tool, SimPowerSystems, extends Simulink into the realm of physical modeling and in particular into the realm of electrical circuits including power systems, motor drives, power generation equipment including power electronics, and three-phase power transmission lines. The last tool, SimMechanics, is also a physical modeling tool; it develops models of mechanical systems such as robots, linkages, etc. These tools can all work together in the common Simulink environment, but they all have their own method for displaying a picture of the underlying system.

Preface

xxi

Both MATLAB and Simulink are available in student versions. All of the examples in this book use the tools in the student version supplemented with the Control System Toolbox, the Signal Processing blockset, SimMechanics, and SimPowerSystems, which can be purchased separately. If you are not a student but wish to learn how to use Simulink using this book, the MathWorks can provide a demonstration copy of MATLAB, Simulink, and the add-ons; contact them at http://www.mathworks.com. This book not only introduces Simulink and discusses many useful tricks to help develop models but it also shows how it is possible for Simulink to be the basis of a complete design process. I hope the material is not too theoretical and that you find the models we will work with are fun, instructive, and useful in the future. As is always the case, innumerable people have contributed to the genesis and development of this book. I thank Cleve Moler for sharing an early manuscript of his wonderful book and The MathWorks for many years of interesting and stimulating work. In particular, I wish to thank the MathWorks staff with whom I have had many long and interesting discussions: Rob Aberg, Paul Barnard, Jason Ghidella, Ned Gully, Loren Shure, and Mark Ullman, and The MathWorks’s CEO, Jack Little. These interesting and productive discussions helped me create the various examples and models described in this book. Words are never enough to thank ones family. My wife, and soulmate, Jean Frova Gran, provides support, joy, affection, and love. My daughter Kathleen keeps my thoughts focused on the future. Her hard work and dedication to the achievement of her goals, despite many small-business frustrations, is laudable. My extended family includes two stepgrandchildren, Ian and Ila, who provide me with insight into the joy brought by the process of learning about this unbelievable world we live in. My stepdaughters Elizabeth and Juliet have a sense of humor and a joie de vivre that are inspirational. Finally, I need to say “thank you” to my sister Cora and her wonderful husband, Arthur, who have always been loving and supportive in more ways than can be stated; Cora is, as the self-proclaimed “matriarch” of the Grans, much more to me than a mere sister. Dr. Richard J. Gran Mathematical Analysis Co. www.math-analysis.com Norfolk, Massachusetts, August 2006

Chapter 1

Introduction to Simulink

The best way to follow the details of this book is to have a copy of Simulink and work the examples as you read the text. The examples described in the book are all in the Numerical Computing with Simulink (NCS) library that is available as a download from The MathWorks (www.mathworks.com). The files are located in an ftp directory on the site. As you work your way through this text, the Simulink and MATLAB files you will run are in boldface Courier type. When you need to run a model or M-file, the text shows the exact MATLAB command you need to type. After you download the NCS library, place it on the MATLAB path using the “Set Path” command under the MATLAB file menu. Alternatively, you can save the NCS files and use the MATLAB directory browser (at the top of the MATLAB window) or the cd command to change to the directory that contains the NCS library.

1.1 Using a Picture to Write a Program In order to understand the Simulink models that we use to illustrate the mathematical principles, this chapter provides the reader with a brief introduction to Simulink and its modeling syntax. In my experience, the more familiarity a user has with the modern click-and-drag techniquefor interfacing with a computer, theeasier it is to build (and understand) a Simulink model. In essence, a Simulink model provides a visual depiction of the signal flow in a system. Conceptually, signal flow diagrams indicate that a “block” in the diagram (as represented by the figure at left) causes an operation on the input, and the result of that operation is the output (which is then possibly the input to the next block). The operation can be as simple as a “gain block,” where the output is simply the product of the input and a constant (called the gain). Signal flow in the diagram splits, so several different blocks have the same input. Then the signal can be coalesced into a single signal through addition (subtraction) and multiplication (division). The diagram illustrates this; the “output” y is y K1 u K2 u (or y (K1 K2 )u). The power of this type of diagram is that K1 or K2 may represent an =

=

−

1

−

2

Chapter 1. Introduction to Simulink

Figure 1.1. Simulink library browser.

operator which might be a matrix, a difference equation, a differential equation, integration, or any of a large number of nonlinear operations. The signals in a block diagram (y , u, and any intermediate terms not explicitly named) are usually functions of time. However, any independent variable may be used as the underlying simulation variable. (For example, it may be the spatial dimension x , or it might simply be interpreted as an integer representing the subscript for a sequence of values.) The Simulink environment consists of a library of elementary building blocks that are available to build a model and a window where the user builds the model. Figure 1.1 shows the Simulink library browser. Open this browser by typing simulink

at the MATLAB command line (or alternatively by clicking on the Simulink icon

that

appears below the menu at the top of the MATLAB window). In either case, the result is that you will open the Simulink browser and a window for building the model. Thebrowser consists of a catalog of 16 types of blocks. Thecatalogis a grouping of the blocks into categories that make it easier for the user to find the appropriate mathematical

1.1. Using a Picture to Write a Program

3

element, function, operation, or grouping of elements. A quick look at the list shows that the first element in the catalog is the “Commonly Used Blocks.” This, as the name implies, is redundant in that it contains blocks from other elements in the catalog. The second set of elements in the library is denoted “Continuous.” This element contains the operator blocks needed to build continuous time ordinary differential equations and perform numerical integration. Similarly, the library element “Discrete” contains the operators needed to develop a discrete time (sampled data) system. All math operations are in the “Math Operations” catalog element. To understand how to build a Simulink model, let us build a simple system. In many applications, simple trigonometric transformations are required to resolve forces into different coordinates. For example, if an object is moving in two dimensions with a vector velocity v at an angle θ relative to the x -axis of a Cartesian coordinate system (see figure), then the velocity along the x -axis is v cos(θ ) and along the y -axis is v sin(θ ).

A Simulink model that will compute this is in the figure at the right. If you do not want to build this model from scratch, then type Simplemodel

at the MATLAB command line. However, if this is the first time you will be using Simulink, you should go through the exercise of building it. Remember that the model building uses a click-and-drag approach. As we discussed above, the trigonometric functionsin theblocks areoperators in thesense that they operate on the input signal v to create the output. To build the model, open Simulink and use the empty model window that opens or, if the empty window is not open, click on the blank paper icon at the upper left of the Simulink library browser (the circled icon in the Simulink library browser, Figure 1.1). The empty model window that is open will appear with the name “Untitled.” All of the mathematical functions needed for building this model are in the Math Operations section of the browser. This section of the browser is opened by clicking on the words “Math Operations” in the browser or by clicking on the + symbol at the “Math Operations” icon in the right side of the browser window. (The browser is set up to work the same as the browser in Windows.) Math Operations contains an icon forall of the math operations available foruse in Simulink models (in alphabetic order). We will need the Trigonometric Function block. To insert this block in the model, click on it and drag it into the “untitled” model window. We will need two of these blocks, so you can drag it twice. Alternatively, after the first block is in the model window, you can right click on the block in the diagram (which will create a copy of the block) and then (still holding down the right click button on the mouse) drag the copy to another location in the diagram.

4


We will assume that the velocity v is a MATLAB variable that is in the MATLAB workspace. To bring this variable into the Simulink model, a block from the Sources library is required. To open this library, locate the Sources Library in the Simulink Browser. This will open a large number of Simulink blocks that provide sources (or inputs) for the model. The “Constant” block allows you to specify a value in MATLAB (created using the MATLAB command window), so click on it and drag it into the untitled model window. To make this input the velocity v , double click on the Constant block to open the “Block Parameters” window. In general, almost all Simulink blocks have some form of dialog that sits (virtually) behind the block; open these dialogs by double clicking the block. As we go through the steps in building a model, we will investigate some of the dialogs. It is a good idea to spend some time familiarizing yourself with the contents of these dialogs by browsing through the library one block at a time and opening each of their dialogs to see what is possible. For now, in the “Block Parameters” window change the Constant value to v , and click OK to close the “Block Parameters” window. We are now ready to create the signal flow in the diagram. Look carefully at the icons in the model. You will see arrowheads at either the input or output (or both) for each of the blocks. You can connect blocks manually or automatically. To connect them manually, click the mouse on the arrowhead at the output of the constant block, drag the line to the input of the Trigonometric Function block (either one), and release the mouse button when the icon changes from a plus sign to a double plus sign. This should leave a line connecting the input to the trig block. To connect the second trig block, start at the input arrow of this block and drag the line to the point on the previous connection that is closest to the block. This will take a little getting used to, so try it a couple of times. The last part of the modeling task is to change the sin function (the default for the trigonometric function from the library) to a cos. This block has a dialog behind it that allows changes, so double click on the block that will become the cosine and use the pull down menu in the dialog called “Function” to select cos. Two additional boxes in the dialog allow you to specify the output type (auto, real, or complex) and the sample time for the signal. We can ignore these advanced functions for now. Try to move the blocks around so they look like the blocks in the Simplemodel figure above. As you were connecting the first two blocks, a message from Simulink should have appeared, telling you that there is an automatic way of connecting the blocks. Do this by clicking (and thereby highlighting) the block that is the input and, while holding down the control (Ctrl) key, clicking on the block that this input will go to. Simulink will then create (and neatly route) a line to connect the two blocks. This simple model introduces you to click and drag icon-based mathematical models. It is simple, obviously too simple for the work needed to create it. It would be far easier to do this in MATLAB. Simulink provides a visual representation of the calculations involved in a process. It is easier to follow complex calculations visually, particularly if they involve flow in time. Toward this end, it is very possible to build a diagram that obscures the flow rather than enhances it. We call such a diagram a “spaghetti Simulink model” (like spaghetti code). To avoid this, it is important to make the connections clear and the model visually easy to navigate. Simulink allows you to do this by dragging the icons around the diagram to make it look better. It is a good idea to get used to this right from the start. Therefore, spend a few minutes manipulating the diagram with the mouse. In addition, blocks move (nudged

1.2. Example 1: Galileo Drops Two Objects from the Leaning Tower of Pisa

5

up or down and back and forth) with the arrow keys. Simply highlight the block and press the arrow key to move it one grid point. To flip a block, highlight it and use Ctrl + i, and to rotate a block, highlight it and use Ctrl + r. Once again, try these steps. If you want to save the example you created, click the floppy disk icon at the top of the model window and save the model in any directory you want. (Be careful not to overwrite the model “Simplemodel” in the NCS directory.)

1.2 Example 1: Galileo Drops Two Objects from the Leaning Tower of Pisa To illustrate a Simulink model that solves a differential equation, let us build a model that simulates the “experiment” that is attributed to Galileo. In the experiment, he went to the upper level of theLeaning Tower of Pisa (Figure 1.21 ) and dropped a heavy object and a light

Figure 1.2. Leaning Tower of Pisa. 1

This photograph is from my personal collection. It was taken during a visit to Pisa in May of 2006. If Galileo performed the experiment, he most likely would have used the upper parapet at the left of the picture. (Because of the offset at the top, it would have been difficult to use the very top of the tower.)

6


object. The experiment demonstrated that the objects both hit the ground at the same time. Is it true that this would happen? Let us see (this model is in the NCS library, where it is called Leaningtower—you can open it from the library, but if this is your first experience with Simulink, use the following instructions to create the model instead). Thanks to Newton, we know how to model the acceleration due to gravity. Denote the acceleration at the surface of the earth by g (32.2 ft per sec per sec). Since the force on the object is mg and Newton’s first law says that F = ma,

we have m

d 2 x

= −mg. dt 2 From this, it follows immediately that the mass of the object does not enter into the calculation of the position (and velocity), and the underlying equation of motion is

d 2 x dt 2

= −g.

We assume that Galileo dropped the balls from the initial location of 150 ft, so the initial velocity is dxdt (0) = 0 and the initial position is x( 0) = 150. Even though this formulation of the equation of motion explicitly shows that Galileo’s experiment has to work, we continue because we are interested in seeing if the experimental results would in fact have shown that the objects would touch the ground at the same time. Initially we model just the motion as it would be in a vacuum. A little later in this section, we will add air drag to the model. The model of any physical device follows this evolutionary path. Understanding the physical environment allows the developer to understand the ramifications of different assumptions and then add refinements (such as air drag) to the model. The equation above is probably the first differential equation anyone solves. It has a solution (in the foot-pound-second [fps] system) given by dx(t) dt x(t)

= −32.2t, 2 = −16.1t +

150.

Since theexperiment stops when theobjects hit theground, the final time forthis experiment is when x(t) = 0. From the analytic solution, we find that this time is t =



150

16.1

.

Go into MATLAB and compute this value. MATLAB’s full precision for the square root gives >> format long >> sqrt(150/16.1) ans = 3.05233847833680


7

We will return to this calculation after we use Simulink to model and solve this equation because it will highlight a feature of Simulink that is numerically very important. Now, to build the Simulink model of the motion we need to create the left-hand side of the differential equation and then integrate the result twice (once to get the velocity and then again to get the position). Five types of Simulink blocks are required to complete this model: • Three input blocks are used to input the constant g into the equation and to provide the initial position and to provide a value to test the position against to determine whether the object has reached the ground. • Two integrator blocks are needed to compute the velocity and position from the acceleration. • One block will perform the logic test to see when the object hits the ground. • One block will receive the signal from the logic block to stop the simulation. • One block will create a graph of the position and velocity as a function of time. These blocks are in sections of the Simulink library browser as follows: • The inputs are in the Sources library. • The integrator is in the Continuous library. • The block to test to see if the object has hit the ground is a Relational Operation, and it is in the Logic and Bit Operations library. • The block to stop the simulation is in the Sinks library. • The graph is a Scope block and is in the Sinks library. So let us start building the model. As we did above, open up the Simulink browser and a new untitled model window. Accumulate all of the blocks in the window before we start to make connections. Thus, open the Sources, Continuous, Logic and Bit Operations, and Sinks libraries one by one, and grab the icons for a Constant, an Integrator, a Relational Operator, and the Stop and drag them into the untitled model window. The integrator block in the Continuous library uses the Laplace operator 1s to denote integration. This is a bit of an abuse of notation, since strictly speaking this operator multiplies the input only if it is the Laplace transform of the time signal. However, Simulink had its origins with control system designers, and consequently the integration symbol remains in the form that would appear in a control system block diagram. In fact, the integrator symbol invokes C-coded numerical integration algorithms that are the same as those in the MATLAB ODE suite to perform the integration. The C-coded solvers are compiled into linked libraries that are invoked automatically by Simulink when the simulation is executed. (Simulink handles all of the inputs, outputs, initial conditions, etc. that are needed.) We will talk a little more about the ordinary differential equation solvers later, but if the reader is interested in more details about the numerical integration solvers, consult Chapter 7 of NCM [29].

8


Figure 1.3. Integrator block dialog.

Make copies of the integrator and the constant block (since we need two integrators and three constants); as before, either copy it by right clicking or drag a second block into the model from the browser. Make the connections in themodel. Next, change theConstant block values by double clicking to open its dialog box. Change the first of these to make the value – g , change the second to make the value 0 (which will be used to test when the position of the object is at the “ground”), and then change the third so its value is 150 (the initial position of the object at the top of the Leaning Tower of Pisa). Modify the integrator block to bring the initial conditions into the model. When you double click the integrator block, the “Block Parameters” menu, shown in Figure 1.3, appears. In this menu, you can change the way the numerical integration handles the integral. The options are as follows: • Force an integration reset (tothe initial condition) when an external reset event occurs. This reset can be increasing as a function of time (i.e., whenever the reset signal is increasing, the integrator is reset), it can be decreasing, or it can be either (i.e., the integration will take place only if the reset signal is unchanging), and finally, the reset can be forced when the reset signal crosses a preset threshold. • The initial conditions can be internal or external. If it is internal, there is a box called “Initial condition” (as shown above) for the user to specify the value of the initial


9

Figure 1.4. Simulink model for the Leaning Tower of Pisa experiment.

condition. When it is set to external, an additional input to the integrator appears on the integrator icon. We will use this external initial condition to input the height of the leaning tower. • The output of the integrator can be limited (i.e., it can be forced to stay constant if the integration exceeds some minimum or maximum value). Other options are available that will be discussed as we encounter the need for them. Set one of the integrators to have an external initial condition, and leave the other at its default value. (The default is internal with an initial condition of 0.) Then connect the blocks, as we did before, so the model looks like Figure 1.4. The default time for the simulation is 10 sec, which is ample time for the object to hit the ground. Before we start the simulation, we need to specify a value for g (we used the value −g in the dialog for the Constant, and it must be created). To do this, in MATLAB type the command g = 32.2;

Now start the simulation by clicking the right pointing “play” arrow (highlighted with a circle in Figure 1.4). The simulation will run. Double click on the Scope icon and then click the binoculars (at the top of the Scope plot window) to scale the plot properly. The result should look like Figure 1.5.

10


Figure 1.5. Scope block graph for the Leaning Tower simulation.

Notice that the simulation stopped at a time just past 3 sec. We can see how accurately Simulink determined that time by zooming in on the plot. If you put the mouse pointer over the zero crossing point and click, the plot scale will change by about a factor of three. If you do this some number of times, you will see that the crossing time is about 3.0523. The plot will not resolve any finer than that so we cannot see the zero crossing time any better than this. The default for Simulink is to use the variable step differential equation solver ode45. This variable step solver forces the integration to take as long a step as it can consistent with an accuracy of one part in 1000 (the default relative tolerance). Simulink also sends (by default) the time points at which the integration occurred back to MATLAB. The time values are in a vector in MATLAB called tout. To see what the zero crossing time was, get the last entry in tout by typing tout(end)

To get the full precision of the calculation, type format long

The answer should be 3.05233847833684. Remember that the stop time computed above with MATLAB (using the solution of the equation) was 3.05233847833680, a difference of 4 in the last decimal place (i.e., a difference of 4 × 10 14 , which is a little more than the computation tolerance variable in MATLAB called eps). How can the Simulink numerical integration compute this so accurately? −


11

This remarkable accuracy in the numeric calculation of the stopping time is because of a unique feature of Simulink. The variable step integration algorithms that are used to solve the differential equations in a Simulink model will compute the solution at time steps that are as big as possible consistent with the relative tolerance specified for the algorithm. If, in the process of computing the solution, a discontinuity occurs between any two times, the solver forces an interpolation to determine the time at which the discontinuity occurred. Only certain Simulink blocks cause the solver to interpolate. These are as follows: • Abs—the absolute value block. • Backlash—a nonlinear block that simulates a mechanical system’s backlash. • Dead zone—a nonlinear system in which outputs do not change until the absolute value of the input exceeds a fixed value (the variable called deadzone). • Hit crossing—a block that looks at any signal in the Simulink diagram to see if it matches a particular value (the hit). • Integrator—one of the options in the integrator is to reset the integration. This reset value triggers the zero crossing algorithm. The integrator also triggers the zero crossing algorithm if the integration is limited. • MinMax—this block computes the minimum or maximum of a signal. When an input to this block exceeds the maximum or is below the minimum, the interpolation algorithm calculates the new maximum or minimum accurately. • Relational operator—this operation, in thisproblem, detects the crossing of the ground (zero crossing in this case) of the object. • Saturation—a signal entering this block is limited to be between an upper and lower bound (the saturation values). The solver interpolates the time at which the signal crosses the saturation values. • Sign—this block is +1 or −1, depending on whether the signal is positive or negative. The solver interpolates the time at which the signal changes sign. • Step—this is an input function that jumps from one value (usually 0) to another value (usually 1) at a specified time. • Subsystem—a subsystem is a block that contains other blocks. Their use makes diagrams easier to read. They also allow simulations to have blocks that go on and off. Blocks inside a subsystem do not appear at the level of the system, but you can viewthem by double clicking the subsystem block. Subsystems comeon (areenabled) using a signal in the Simulink model that connects to an icon on the subsystem. If you place an Enable block in a subsystem, the icon automatically appears at the top level of the subsystem. Whenever the system enable or disable occurs, the solver forces a time step. • Switch—there is a manual switch that the user can “throw” at any time. The solver computes the time the switch state changes.

12


Now that we have simulated the basic motion of Galileo’s experiment, let us add the effects of air resistance to the model and see if, in fact, the objects hit the ground at the same time. Air drag creates a force that is proportional to the square of the speed of the object. To first order, an object with cross sectional area A moving at a velocity v in air with density ρ will experience a drag force given by f D rag

1 =

2

ρv 2 A.

(In the model so far we have assumed that the force from gravity is downward and is therefore negative, but the air drag force is up so it is positive.) In the fps unit system we have been using, the value of ρ is 0.0765 lbf/ft 3 . Let us assume that the objects are spherical and that their diameters are 0.1 and 0.05 ft each. (If they are made of the same material, the large object will then be 8 times heavier than the small object.) The acceleration of the objects is now due to the force from the air drag combined with the force from gravity, so the differential equation we need to model is now d 2 x dt 2

1 =

2

ρA

2

  dx

dt

−

g.

To get the air drag force for each object we need to know their cross-sectional areas. Since the cross-sectional area is π r 2 , the acceleration from the air drag for the large object is ρ 1 ρπ r 2 (0.1)2 π 0.005πρ and the acceleration for the small object is 12 ρπ (0.05)2 2 2 .00125πρ (they differ by a factor of 4). To create the new model, we need to add the air drag terms. Thus, we need a way to add the force from the air resistance to the force from gravity. This uses a Sum block from the Math library. This block is a small circle (or a rectangle—my personal preference is the circle since it makes it easy for the eye to distinguish summations from the other icons that are mostly rectangular) with two inputs, (each of which has a small plus sign on it). So click and drag this block into the leaning tower model. To get the square of the velocity, we need to take the output of the first integration (the integration that creates the velocity from the acceleration) and multiply it by itself. The block that does this is the Product block in the Math library. Drag a copy of this block into the model also. We will start by creating the model for the heavy object alone. Connect the output of the integrator to the Product block as we did before (using either of the two inputs, it does not matter which one), and then grab the second input to the Product block and drag it until it touches the line that you just connected (the line from the integrator). This will cause the two inputs to the Product block to be the same thing (the velocity), so the output is the square of the velocity. Notice how easy it is to create a nonlinear differential equation in Simulink. The next step is to multiply the velocity squared by 0 .005πρ or 0.003825. Use the Gain block from the Math library to do this. Click and drag a copy of this block into the model then double click on this Gain block and change the value in the dialog to be 0.003825. Figure 1.6 shows the resulting Simulink model, which is in the NCS library. Open it with the MATLAB command: =

=

=

Leaningtower2


13

Figure 1.6. Leaning Tower Simulink model with air drag added.

Figure 1.7. Scope dialog showing how to change the number of axes in the Scope plot.

Note that we changed the Scope block to view both the position and the velocity of the objects. This was done by double clicking the Scope icon and then selecting the “Scope Parameters” dialog by double clicking the second menu item at the top of theScope window. (This is the icon to the right of the printer icon; it is the icon for a tabbed dialog.) The dialog that opens looks like Figure 1.7. Note that the number of axes in the Scope is the result of changing the value in the “Number of Axes” box. We have made it two, which causes the Scope block to have two inputs. We then connected the first input to the Velocity and the second to the Position lines in the model.

14

Chapter 1. Introduction to Simulink Velocity 0

-20 -40 -60 -80

0

0.5

1

1.5

2

2.5

3

3.5

2

2.5

3

3.5

Position 150 100 50 0 -50

0

0.5

1

1.5 Time

Figure 1.8. Leaning Tower simulation results when air drag is included.

If you run this model, the results in the Scope block will resemble the graphs in Figure 1.8. Notice that the air drag causes the object to hit the ground later than we determined when there was no drag (3.35085405691268 seconds instead of 3.05233847833684, about 0.3 seconds later). Now let us compare the light and heavy object. We could do this by going back to the model and changing the cross-sectional area to that of the smaller object. However, there is a feature of Simulink that makes this type of calculation extremely easy. The feature is the automatic “vectorization” of the model. To exploit this feature, open the Gain block in the model (by double clicking) and enter both the values for the large and the small object. Do this using MATLAB’s vector notation. Thus the Gain block dialog will have the values [0.003825 0.000945]. (Include the square brackets, which make the value in the dialog equal to the MATLAB vector, as shown in Figure 1.9.) After running this model with the 2-vector for the drag coefficients, the Scope shows plots for the position and velocity of both of the objects (see Figure 1.10). The heavier object hits the ground well after the light object. The velocity curve in the top graph shows why. The speed of the heavier object comes to a maximum value of about


15

Figure 1.9. Adding the second object to the Leaning Tower simulation by vectorizing the air drag.

70 ft/sec, whereas the lighter object ends up at about 90 ft/sec. This difference is from the increased air drag force on the object because the heavier object has a larger cross-sectional area. This difference is too large for Galileo not to have noticed, so it does lead us to suspect that he never really did the purported experiment. (It is not clear from Galileo’s writings that he understood the effect of drag, but if he did, he could have avoided this by making the objects the same size.) In fact, it is very probable that Galileo relied on the thought experiment he talks about in his 1638 book on the Copernican theory, Two New Sciences [4]. In this thought experiment he argued that in a vacuum, by tying the two objects together with a short rope, if the heavier object were to hit the ground first, at some point the rope would become taut and the light object would begin dragging the heavier. That implies that the combined object would hit the ground a little later than would have been the case had no rope been connecting them. This meant that the combined object (which is heavier than either one) would hit the ground a little later than the single heavy object. However, since the combined weight of the two objects is greater than the single heavy object, this violates the premise that the heavier object must hit the ground first. This deduction was the reason Galileo believed that the Aristotelian statement that the heavy object falls faster was wrong, and that prompted him to perform many experiments with inclined planes to try to understand the dynamics of falling bodies [14]. Since this is the first Simulink model with any substance that we have built, let us use it to illustrate some other features of Simulink. First, let us clean up the model and annotate it. It is a good idea to do this as you create the model so that when you return to the model later (or give the model to someone else to use), it is clear what is simulated. The first thing to do is to assign a name to the

16

Chapter 1. Introduction to Simulink Velocity 0

-50

-100

0

0.5

1

1.5

2

2.5

3

3.5

2

2.5

3

3.5

Position 150 100 50 0 -50

0

0.5

1

1.5 Time

Figure 1.10. Simulating results. The Leaning Tower Simulink model with a heavy and light object.

variables in the diagram, and the second is to name all of the blocks so their functions are clear. It is also helpful to add color to the diagram to indicate the generic function of the blocks. To add a variable name onto a line in the Simulink model, simply double click on the line. This causes an insertion point to appear on the line where you can type any text. The annotation is arbitrary. To annotate the model you created, enter the names in the model, or if you have not created a model, openthe modelLeaningtower3 by typing Leaningtower3 at the MATLAB command line. This will open the model with the annotation shown in Figure 1.11. Simulink automatically changed from a scalar to a vector in the model when we made the gain a MATLAB vector. After making the change, to see that the variables in the model are vectors corresponding to the two different drag accelerations you must right click the mouse anywhere in the model window and select Format from the menu; this will open the submenu shown as part of Figure 1.11. In this menu, we selected “Wide Nonscalar Lines” and “Signal Dimensions” (note that they are checked). After you do this, all of the vector signals in the diagram have a wider line, and the signal dimensions (vectors of size 2 here) are on the signal lines in the model (as shown in Figure 1.11). To change the color of a block, highlight the block and

1.3. Example 2: Modeling a Pendulum and the Escapement of a Clock

17

Figure 1.11. Right clicking anywhere in the diagram opens a pull-down menu that allows changes to the colors of the blocks, bold vector line styles, and other model annotations.

right click. Select “Background Color” from the resulting pop-up menu and then move the mouse pointer to the right to select the desired color. There are other features that Simulink supports which you can discover on your own by playing with the model. As we work our way through the exercises and examples in this book, you will learn a lot more. In the next section, we continue creating some simple models by simulating a pendulum clock. For the interested reader, [30] has a very nice discussion of how Galileo came to realize that objects with different masses had the same motions. The Leaning Tower experiment that showed that the masses would actually move the same had to wait for the invention of the vacuum pump at the end of the 17th century.

1.3 Example 2: Modeling a Pendulum and the Escapement of a Clock The next step in our introduction to Simulink is to investigate a pendulum clock and the mechanism that allows clocks to work: the escapement. This example introduces control systems and how they are simulated. It also continues to discuss both linear and nonlinear differential equations.

18


The escapement mechanism is one of the earliest examples of a feedback control system. An escapement provides the sustaining force for the motion of the clock while ensuring that the pendulum and the clock movement stay in synchronization. Many famous mathematicians and physicists analyzed clocks and early forms of escapements during the 17th and 18th centuries. We start by describing the escapement mechanism fora mechanical pendulum clock; we then show how to build a Simulink model of the clock.

1.3.1 History of Pendulum Clocks Humanshave been building timekeepers that usethe sunor water since at least 1400 BC.The Egyptian clepsydra (from the Greek kleptei, to steal, and hydor , water) was a water-filled bucket with a hole. The water level was a measure of the elapsed time. Sundials provided daylight time keeping. The first mechanical clocks appeared in the 13th century, with the earliest known device dating to about 1290. Who invented the first clock is lost to history, but the main development that made these devices possible is the escapement, a gear with steep sloping teeth. A gear tooth “escapes” from the escapement, causing the gear to rotate until the next tooth engages the opposite side of the escapement. The rotation of the gear registers a count by the clock. At the same time the tooth escapes, the driver imparts energy (through the escaping tooth) into the pendulum, which overcomes the friction losses at the pivot. The escapement thus serves two purposes: it provides the energy to keep the clock in motion, and it provides the means for counting the number of oscillations. Christiaan Huygens invented an escapement similar to that shown in Figure 1.12 in 1656. This type of mechanism was responsible for the emergence of accurate mechanical clocks in the eighteenth century. The mechanism relies on the escapement (the parts labeled 1 and 2 in Figure 1.12).2 With this mechanism, he was able to build a clock that was accurate to within 1 sec per day. Previous clock mechanisms did not have the regulation capability of the Huygens escapement, so they would slow down as the springs unwound or weights that drove the clock dropped down. Earlier clocks required daily resetting to ensure that they were accurate to only the nearest hour. The pendulum mechanism provides the timing for the clock and the escapement provides the forces and connections that • provide energy to the pendulum so the clock does not stop because of friction; • regulate the speed of the gear motions to the natural period of the pendulum so that the timing of the clock is accurate, depending only on the pendulum motion; • decouple the pendulum and the drive; • provide the feedback to ensure that the force from the driver (springs or weights) and wear in the various mechanical parts do not adversely affect the time displayed by the clock. 2

This copyright figure is from a document titled “How a Clock Works” available from Erwin Sattler Clocks of America. You can download the document from the web page http://www.sattlerclocks.com/ck.php. I would like to thank Marcus Orvando, president of Sattler Clocks of America, for permission to use this diagram.


19

Figure 1.12. Components of a pendulum clock.

Before we go into the details of the design of the escapement, let us look at how it works. The escapement is on a shaft attached to the clock case through a bushing called a back cock. The pendulum mount does not connect to this shaft. The pendulum suspension allows the pendulum to swing with minimal friction unencumbered by the escapement. As the pendulum swings, it alternately engages the left and right side of the escapement anchor (1). In the process a force from the escapement wheel (2) goes to the pendulum through the crutch (4) and the crutch pin (5). The crutch assembly also forces the escapement to rotate in synchronism with the pendulum. The pendulum itself (3) moves back and forth with a period that is very closely determined by its length. (Once again we return to Galileo, who demonstrated this in 1582.) The pendulum includes a nut at its bottom (not shown), whose adjustment changes the effective length and thereby adjusts the period of the pendulum. The escapement wheel (2) has a link to all of the gears in the gear train (they are not all shown in Figure 1.12) that turns the hands (8) of the clock. For the remainder of this description, we focus on the escapement anchor (1) and its two sides. On the two sides of the anchor are triangular teeth; as the anchor rotates, one of the two sides of each tooth alternately enters and exits the escapement gear, hence the names entrance and exit pallets. If you look carefully at the anchor in Figure 1.12, you

20


will see them: they are triangular pins that pass through each end of the anchor. As the pendulum moves to the left, it forces the anchor to rotate and ultimately disengage the right exit pallet face from the escapement wheel (2). When it does, the exit pallet on the opposite side of the anchor engages the next tooth of the escapement wheel (9), and the escapement gear moves one tooth. If the pendulum is set to swing at a period of 1 sec (as is the case for the mechanism in Figure 1.12), the result of the motion of the escapement gear is to move the second hand on the clock by one second (which implies that there are 60 teeth on the escapement gear). When the right face of the anchor engages the gear, the anchor and the escapement wheel interact. The escapement wheel, which is torqued by the drive weight (7), applies a force back through the anchor (1) and then, through the crutch and crutch pin (4 and 5), to the pendulum. The alternate release and reengagement of the anchor and escape wheel provides the energy to the pendulum that compensates for the energy lost to friction. As a by-product of this interplay of the pendulum and the escapement wheel, we hear the distinctive and comforting “tick-tock” sound that a clock makes.

1.3.2

A Simulation Model for the Clock

This Simulink model of the clock is in the NCS library, and you can open it using the command Clocksim. The clock model has two parts: the model of the pendulum, and the model of the escapement. The pendulum model comes from the application of Newton’s laws by balancing the force due to the acceleration with the forces created by gravity. Let us assume the following: • The pendulum angle is θ . • The mass of the pendulum is m. • The length of the pendulum is l . • The inertia of the pendulum is J . (If we assume that the mass of the pendulum is concentrated at the end of the shaft, the inertia is given by J ml 2 .) =

• The damping force coefficient is c (Newtons/radians/sec). Then the equation of motion of the pendulum is J

d 2 θ(t) dt 2

=

c

dθ(t)

−

dt

−

mgl sin(θ(t)).

Substituting ml 2 for J gives d 2 θ(t) dt 2

c dθ(t) =

−

ml 2

dt

g −

l

sin(θ(t)).

These dynamics are in the pendulum clock Simulink model shown in Figure 1.13. This model uses Simulink primitive blocks for integrators, summing junctions, gains, and


21

Pend. Angle

Damping Accel.

dampcoef

Display Variables PendulumAngleForce

Escapement Model

Force

0.1148

c/(m*len^2) 1 s

Thet2dot Escapement Accel.

Force from escapment over the pendulum mass

Initial Pend. Angle

Units of

dampcoef = 0.01 g = 9.8 plength = 0.2485

Thetadot

1 xo s 0.2

2

acceleration m/s

Gravity Accel.

1/plength 1/length

g*sin(u)

Pend. Angle

Gravity Force g*sin(Theta)

Figure 1.13. Simulink model of the clock.

the sine function (from the Continuous and the Math Operations libraries). We also used a new concept, called a subsystem, to create a block called “Escapement Model.” Inside this block are Simulink primitives that model the escapement. Open the subsystem by double clicking, or view it using the model browser. Let us follow the signal flow in the model. We start at the output of the summing junction, where the angular acceleration of the pendulum, Theta2dot, is the result of subtracting the three terms: • the damping acceleration mlc 2 θ ˙ from the equation above ( dampcoef*Thetadot), • the gravity acceleration from the equation above ( g/plength*sin(Theta)), • the acceleration created by the escapement that we develop next. The Simulink model creates the variables representing the angular position and velocity, Theta and Thetadot , by integrating the angular acceleration Theta2dot. A pendulum clock only sustains its oscillation if the pendulum starts at an angle that causes the escapement to operate. Consequently, an initial condition is used in the last integrator (we use 0.2 radians, about 11.5 degrees) to start the clock. A feature of Simulink is the ability to create a subsystem to simplify the model. Typically, a subsystemis a Simulinkmodel thatgroups the equations for a particular common function into a single block. The details of the block are in the subsystem and are not visible until the subsystem block is opened (by double clicking or browsing), but its functionality is captured so that the user can see the interactions. The clock’s single subsystem is the model for the escapement. This “Escapement Model” block has the equations shown in Figure 1.14. These nonlinear escapement mechanism equations consist of the logic needed to model the force created by the escapement.

22

Chapter 1. Introduction to Simulink Clock Escapement Mode l

Restore Sign 1 Pendulum Angle

|u|

<=

Abs Penhigh 0.15

Relational Operator1

1 Force

>=

Penlow 0.1

Product1

Relational Operator2

du/dt

<

Derivative 0

Relational Operator3 penlow1

Creates a force that is applied when the pe ndulum is at the end of its swing

Figure 1.14. Blocks in the subsystem “Escapement Model.”

The escapement works in thesame wayas youwould push a swing. A small accurately timed push occurs when the exit pallet leaves the escapement. The push is always in the direction of the pendulum’s motion and it occurs just after the pendulum reaches its maximum or minimum swing (depending on the sign of the angle). The attributes of the escapement model are as follows: • Since the pendulum motion is symmetric, the calculations use positive angles (with the absolute value function block). The force applied has the correct sign because we multiply it by +1 or −1, depending on the sign of the angle. (The block called “Restore Sign” does this.) • The pendulum angle at which the escapement engages (applies an acceleration) is denoted by Penlow . (As can be seen, it is given a value of 0.1 radians.) • The angle of the pendulum at which the escapement disengages (jumps from one gear tooth to the next) is denoted by Penhigh (with a value of 0.15 radians). • The escapement applies the force only when the pendulum is accelerating (i.e., when the pendulum is moving toward the center of its swing). We check to see if the derivative of the pendulum angle is positive or not, and multiply the force by 1 when it is and by 0 when it is not in the “Relational Operator 3” block.


23

The plot at right shows the force applied by the escapement model (the Y axis) vs. the pendulum angle (X axis). Notice that the angle at which the force engages is 0.1 and –0.1 radians, and the angle at which the force stops is 0.15 and –0.15 radians as required. We used the Math Operations and Logic libraries to develop this model, so let us go through the logic systematically. The absolute value of the pendulum angle is the first operation. This block is the first one in the Math Operations library. At the conclusion of the logic, the direction of the force comes from the sign of the pendulum angle. In the model this is done by multiplying the force created (which will always be positive) by the sign of the pendulum angle. When the absolute value of the pendulum angle is less than or equal to 0.15, the “Relational Operator1” block is one; otherwise it is zero. When the angle is greater than or equal to 0.1 the “Relational Operator2” block is one; otherwise it is zero. The last part of the logic insures that the only time the force is applied is when the pendulum angle is getting smaller (in the absolute sense). This is accomplished by checking that the derivative of the pendulum angle is negative. (The derivative block is in the Continuous library). The product of all of these terms causes the output to be one if and only if the pendulum angle is between 0.1 and 0.15 and the pendulum angle is decreasing, or the pendulum angle is between −0.1 and −0.15 and the pendulum angle is increasing. The simulation is set up to run for 1000 sec, and the output in the Scope block shows the pendulum angle and the escapement acceleration. After you run the model, you can reveal the details of the plot by using the mouse to encircle an area on the plot to zoom in on. If you do this (it was done here between 412 and 421 sec), the plot resembles Figure 1.15. The early part of the simulation shows that the pendulum angle grows from its initial value (of 0.2 radians) to 0.2415 radians. This stable value for the maximum pendulum angle is a consequence of the escapement. At this point, numerical experiments with the clock can investigate different questions. For example, the fact that the escapement force is determined by some device that converts potentialenergy into kinetic energy (weights, a spring, a pneumatic devicethat uses variations in air pressure, etc.) means that over its operation the force that the escapement applies to the pendulum will decrease. To understand the effect of this decreasing force on the time keeping of the clock, simulate the escapement with different values of Gain in the Gain block labeled “Force from escapement over the pendulum mass.” Evaluate the change, if any, in the period of swing of the pendulum, and the amplitude of the swing when you change this value (see Example 1.2).

24


Figure 1.15. Results of theclocksimulation fortimesaround 416 sec. Upper figure is the pendulum angle, and the lower figure is the acceleration applied to the pendulum by the escapement.

1.4 Example 3: Complex Rotations—The Foucault Pendulum Léon Foucault, in 1848, noticed that a pendulum mounted in the chuck of a rotating lathe always moved back and forth in the same plane even if the chuck rotated. This gave him the idea that he could demonstrate that the earth rotated by setting in motion a long pendulum and observing the path of motion over the period of a day. In a series of experiments with gradually longer pendulums, he became convinced that his concept was valid. In February of 1851, he unveiled a public demonstration for the Paris Exposition, where, at the Eglise Ste.-Geneviève in Paris, he set up a pendulum that was 67 meters long. The original pendulum, shown in Figure 1.163 , has been back at the Panthéon since 1995. The Foucault pendulum both illustrates the mechanics of a body moving in a rotating frame andis thesimplest example of thecomplex motions that result from gyroscopic forces. We will investigate these forces in detail in Chapter 3. For this introduction to Simulink, we have looked at Galileo’s experiment, the pendulum for use in a clock, and now the Foucault pendulum. The dynamics are not too complex, but they do require that the full three dimensions be included. (However, one of these motions, the vertical direction, may be ignored.) The model we create can also be used as a simple example of the Coriolis effect that causes weather patterns to move from west to east (in the northern hemisphere). 3

The original church is now a public building called the Panthéon on the Place du Panthéon in Paris. This photgraph is from the Wikipedia web site [51].

1.4. Example 3: Complex Rotations—The Foucault Pendulum

25

Figure 1.16. Foucault pendulum at the Panthéon in Paris.

1.4.1

Forces from Rotations

A particle moving with constant speed along a circular path is always changing its direction of travel; it is acceleratingall of thetime. Theforces associated with this change in direction are the centrifugal and centripetal forces. The figure at right illustrates the motion and the vectors associatedwith the location of the particle as it moves over an interval t . The vectors in the diagram are the radial location at the two times t + t ; the unit vectors for the radial motion, ur (t) ; and the direction perpendicular to the radial motion, uθ (t). To restrict the particle to motions with a constant radius, we assume that r (t) = r ur (t) , where r is the radius (a scalar constant). Differentiating this to get the linear velocity of the particle gives v

(t) =

d r (t) dt

=r

d ur (t) dt

.

r (t) From the figure above, the derivative of the unit vector ( d dt ) is given by calculating the difference in this unit vector over the time t . Since t is small, the directions of the unit vector uθ over this interval can be assumed constant. In addition, the magnitude of the perturbation |uθ (t + t) − uθ (t)| is the small arc length θ . r (t) Putting these together gives the derivative d dt as u

u

d ur (t) dt

= lim

t →0

ur

(t + t) − ur (t) t

= lim

t →0

ur

(t) + uθ θ − ur (t) t

= lim

t →0

θ uθ

t

=

uθ

ω.

Here ω is the angular velocity of the particle’s motion. Therefore we have v(t) = uθ ωr or, equivalently, that the magnitude of the velocity (the particle’s speed) is ωr .

26


To get the relationship between the angular acceleration and the particle’s speed we differentiate the vector velocity using the fact that the radius is constant and the angular velocity and unit vector perpendicular to the radial motion are changing with time. Thus the acceleration of the particle is a

(t) =

d v(t) dt

dω

=

uθ

dt

r

+

d uθ dt

ωr .

Following steps similar to those we used above for differentiating ur , the derivative of uθ is − ur ω, where the minus sign comes from the fact that the difference in uθ at the times t and t+t points inward (along the radial direction this is negative). Using these values, we get the acceleration of the particle as (in this expression, α = dω ) dt a

(t) =

αr − ur ω2 r.

uθ

This acceleration has two components: tangential and radial. The tangential component is the instantaneous linear acceleration and its magnitude is αr . The radial component is 2 the centrifugal acceleration and its magnitude is ω2 r = vr . These two results should be familiar from introductory physics, but we show them here to remindyou that motions under rotations need to account for both the vector magnitude and its directions.

1.4.2 Foucault Pendulum Dynamics When there are no forces on an object, its trajectory in inertial space is a straight line. If we view this motionin a rotating coordinate system, themotion will appearcurved. Because the curve is not thetrue motion of theobject butis a consequence of theobserver’s motion, to get the equations of motion for the observer we use a fictitious force, called the Coriolis force. We have indirectly seen this kind of fictitious force in the centrifugal acceleration above. Following the derivation in Section 1.4.1, if a three-dimensional vector b is rotating, then its derivative in an inertial (stationary) frame is 

d b 

 dt  Inertial



=

d b  

dt  Rotating

+ ω × b| Inertial .

Using this to get the velocity of a rotating vector gives v Inertial = this result again to differentiate the inertial velocity gives a Inertial

=

d dt

v Rotating + ω × r .

Applying

(v Rotating + ω × r) Rotating + ω × v Rotating + ω × (ω × r).

The derivative of the first term on the right must account for the fact that the angular velocity and the radius vector are both time varying. Thus, we have a Inertial

= a Rotating +

dω dt

× r + 2ω × v Rotating + ω × (ω × r).

This equation gives the acceleration in the rotating coordinate system as a Rotating

= a Inertial − 2ω × v Rotating − ω × (ω × r) −

dω dt

× r.

Since the force on the body in the rotating frame is the mass times the acceleration in the rotating frame and the force on the body in the inertial frame is the mass times the inertial


27

z

x

Figure 1.17. Axes used to model the Foucault pendulum.

acceleration, the “Coriolis force” that accounts for the perceived motion in the rotating frame is dω FCoriolis = −2mω × v Rotating − mω × (ω × r) − m × r. dt We can now model the Foucault pendulum. Figure 1.17 shows the coordinate system we use. The y coordinate comes out of the paper. The origin of these coordinates is the rest coordinates of the pendulum, and the z axis goes through the point of suspension of the pendulum. We assume that the pendulum length is L. In the figure, the Earth is rotating at the angular velocity , the pendulum is at the   latitude λ, and we denote the coordinates of the pendulum with the vector x, y, z . Since the pendulum is rotating at the Earth’s rate, the Coriolis forces on the pendulum are C

2m(v × ) = −2m

= −



dx , dt

dy , dt

dz dt



×



 cos λ,

0,

−

 sin λ



.

In the vertical direction, the tension in the wire holding the pendulum is approximately constant (equal to the weight of the pendulum bob mg), so the velocity along the vertical is essentially zero. Therefore, the Coriolis forces (from the cross product) are 2m



dy  sin λ, dt

−

dx  sin λ, dt

dy  cos λ dt



.

Now, when the pendulum moves, there is a restoring force from gravity. This force is proportional to mg sin(θ), where θ is the pendulum angle, as we saw for the pendulum y above. Forthe pendulum, theangle is Lx along the x -axis and L along the y -axis. We assume that the pendulum motion is small so the acceleration is mgθ . Thus the accelerations on the

28


Foucault pendulum are d 2 x dt 2

2 sin(λ)

=

2

d y dt 2

2 sin(λ)

= −

dy dt

dx dt

(2πf P )2 x,

−

−

(2πf P )2 y.

We ignore the accelerations in the vertical direction since to first order they are zero. g The term (2πf P )2 = L is the square of the period of the pendulum, exactly as was used in the clock example above. An interesting aspect of this problem is the magnitudes of the coefficients in the differential equations. The rate of rotation of the earth is essentially once every 24 hours (it is actually a little less), so 

∼ =

2π 24 ∗ 60 ∗ 60

=

2π 86, 400

=

0.00007272205.

The coefficient in the model depends on the latitude. Let us assume we are at 45 degrees north latitude, so the coefficients are 2 sin(λ) = 0.00010284. If we assume that the pendulum length is such that the period of the pendulum swing is 1 minute, we obtain that the other coefficient in the differential equation is 2

(2πf P )

=

  2π 60

2 =

0.01096622.

These coefficients differby about four ordersof magnitude, which will make these equations difficult to solve numerically. As we will see, Simulink allows the user to accommodate this so-called stiffness without difficulty. We cannow build a Simulink model forthe Foucault pendulum using these differential equations. First, though, let us compute how long the simulation should be set up to run. The period of the total motion of the Foucault pendulum depends upon our latitude. This T E arth 2π = period is  sin , where T E arth is the time for one rotation of the earth. (We assume (λ) sin(λ) the earth rotates in 24 hours so, for a latitude of 45 degrees, the period is 1.2219e+005 sec.) We will make this time the simulation stop time. Figure 1.18 shows the model for the Foucault pendulum. Once again, to gain familiarity with Simulink and to ensure you know how to build a model like this, you should create this model yourself. We call this model Foucault_Pendulum in the NCS library. By now, you shouldbe findingit reasonably easy to picture what theSimulinkdiagram portrays. Remember that the picture represents the mathematics and, because the signal flow represents the various parts of the mathematics, it is important when the model reinforces this flow. In this picture, you should be able to clearly distinguish the x and y components of the pendulum motion. We couldemphasize this by coloring the two second order differential equation blocksets with a different color. (The models that are in the NCS library use this annotation.) When you run this model, the Scope block creates the oscillation over one period, as shown in Figure 1.19. As can be seen from the plots, the motions in the east-west and northsouth directions are out of phase by 90 degrees and essentially create a circular motion. The initial condition in the model sets the pendulum at 1 ft to the north and at exactly zero along


29

Pendulm Freq Squared Earth Rate at Latitude (Lat)

2*OmegaE*sin(Lat)

OmegaP^2 Pend. Accel.

1 s

Coriolis Accel (y)

Integrate y Accel.

dy/dt

y Axis (E-W) Motion

1 s

dy/dt

Integrate y Vel.

2*OmegaE*sin(Lat)

x Axis (N-S) Motion

Pendulum Freq Squared

Coriolis Accel. (x)

Pend. Accel.

OmegaP^2

Earth Rate at Latitude 1 s dx/dt

Integrate x Accel. 1 s

dx/dt

Integrate x Vel.

Figure 1.18. The Simulink model of the Foucault pendulum.

y Axis (E-W) Motion 1

) t e e F D ( 0.5 y n i m u l 0 u d n e P f o -0.5 n o i t o M

-1 0

2

4

6

8

10

12

14

Time x 10

4

x Axis (N-S) Motion 1

) t e e F ( x 0.5 n i m u l 0 u d n e P f o n -0.5 o i t o M -1 0

2

4

6

8 Time

10

12

14 x 10

4

Figure 1.19. Simulation results from the Foucault pendulum Simulink model using the default solver.

Scope

30

Chapter 1. Introduction to Simulink y Axis (E-W) Motion 1

0.5

0

-0.5

-1 0

2

4

6

8

10

12

14 x 10

4

x Ax is (N-S) Motion 1

0.5

0

-0.5

-1 0

2

4

6

8 Time

10

12

14 x 10

4

Figure 1.20. Simulation results for the Foucault pendulum using a tighter tolerance for the solver.

the east-west line. If for some reason the pendulum starts with a motion that is out of the plane, the pendulum will precess during its motion sweeping out a more elliptic path. The parameters in the model come from code used by the pre-load function (in the “Model Properties” dialog under “Edit” in the pull-down menu at the top of the Simulink model). You can experiment with different latitudes for the pendulum by changing the value of Lat (by simply typing the new value of Lat at the MATLAB command line). Let us use this result to explore the numerical integration tools in Simulink. Chapter 7 of Cleve Moler’s Numerical Computing with MATLAB contains a discussion of the differential equation solvers in MATLAB. The solvers in Simulink are identical, except that they are linked libraries that Simulink automatically uses when the model runs (because you clicked the “run” button). The results in Figure 1.19 use the default solver in Simulink (the default is ode45 with a relative tolerance of 1e-3). If you look carefully at this figure, you will notice that the amplitude of the pendulum motion is decreasing over time. There is, however, no physical reason for this. We have not included any pendulum damping in the model, and the terms that couple the x and y motions should not introduce damping. To see if this effect is a consequence of the numerical solver, let us go into the model and change the solver to make it more accurate. Open the “Configuration Parameters” dialog under the Simulation pull-down menu (or use Ctrl + E), and in the dialog change the relative tolerance to 1e-5. If you run the simulation, the result appears as shown in Figure 1.20.

1.5. Further Reading

31

Notice that this result is what we expect; there is no discernable damping in the motion and the period is exactly what we predicted. The need for the tighter tolerance in the calculation is because of the order of magnitude difference of five in the coefficients in the model, as we noted above. In general, experimentation is required with the differential equation solver values in order to insure that the solution is correct. The best way to do this is to try different solvers and different tolerances to see if they create discernable differences in the solutions. We return to the solvers as we look at other simulations. In particular, we will look at some very stiff systems and the use of stiff solvers in Chapter 3. In Chapter 3, we will also explore the more complex motions of objects undergoing gyroscopic forces when the motion takes place in three dimensions. First, however, we will look at how Simulink creates simulation models for linear systems, how Simulink handles vectors and matrices, and, last, when to use Simulink as part of the process of designing control systems.

1.5 Further Reading Galileo showed that a falling body’s acceleration is independent of its mass by making detailed measurements with an inclined plane, as was described in his 1632 Dialogue Concerning the Two Chief World Systems: Ptolemaic and Copernican [14]. Recent and old versions of his devices are in the Institute and Museum of the History of Science in Florence, Italy (http://www.imss.firenze.it/), and you can view a digital version of the original manuscript of Galileo’s “Notes on Motion” (Folios 33 to 196) from this site. The location for the manuscript is http://www.imss.fi.it/ms72/index.htm. An excellent translation of the Galileo dialogue describing the inclined plane measurements that he made is in The World of Mathematics, Volume 2, edited by James R. Newman [30]. There is a wonderful essay by John H. Lienhard from the University of Houston (at http://www.uh.edu/engines/epi1307.htm) that describes the clock as the first device ever engineered. The essay ties the clock back to Galileo and traces its history from the early writings of Francis Bacon to Huygens and Hooke. The document that you can download from Sattler Clocks (referred to in footnote 2) has much more detail on the mechanisms involved in the escapement (and in particular the escapement in Figure 1.12 that was invented by George Graham in 1720) and the other very interesting features of a pendulum clock. They also have extensive pictures of some elegantly designed and beautiful clocks. I used the clock example to develop an analysis of the clock escapement in the IEEE Control Systems Society Magazine [38]. Shortly after the publication of this article, a set of articles also appeared in the same magazine on several interesting early control systems, including the clock [3], [19]. In a recent Science article [37] the authors demonstrated that moths maintain control during their flight because their antennae vibrate. This vibration, like the motion of the Foucault pendulum, precesses as the moth rotates, giving the moth a signal that tells it what changes it needs to make to maintain its orientation. The authors demonstrated this by cutting the antennae from moths and observing that their flight was unstable. They subsequently reattached the antennae, and the moths were again able to fly stably. You should carefully review The MathWorks’s User’s Manual for Simulink [42]. Simulink models can get very large and take a long time to run. The data generated by

32


the model can also overwhelm the computer’s memory. Managing the use of memory can improve the performance of your models. Reference [41] has a good discussion of the methods available for managing the memory in a Simulink model.

Exercises 1.1 Use the Leaning Tower of Pisa model as the starting point to add a horizontal component to the velocity at the instant the objects start. Include the effect of air drag. Also, add an input to the model to simulate different wind speeds. Experiment with this model to see how these do or do not affect the experiment. 1.2 In the clock simulation, increase the friction force to see how large it must be before the escapement ceases working. Perform some numerical experiments with the model to see if there is any effect on the accuracy because of the increased friction. When you evaluated the change in the period and amplitude of swing of the pendulum as you changed the Gain, you should not have seen a significant change. What change in the model will allow you to see the effect of lower forces applied by the escapement? Modify the model to look at this, and experiment with different escapement forces. What is the effect? Why is the effect so small? How does the escapement ameliorate these forces? Look at Figure 1.12 carefully. The Sattler Company has a spring mounted on thependulumthat transmitsthe force from thecrutch pinto thependulum. Does the Simulink model have this mechanism? How would you go about adding this mechanism to the model? Is there a good reason for the Sattler clock to use this spring? 1.3 Experiment with the different Simulink solvers (one by one) to see what effect they have on the simulation of the Foucault pendulum. Which ones work well? Try changing the tolerances used for each of the solvers to see if they stop working. 1.4 Create a Simulink model of a mass suspended on a spring under the influence of gravity. Add a force that is proportional to the velocity of the mass to model the damping. Pick some parameter values for the mass, the spring constant, the damping, and, of course, gravity, and simulate the system. Explore what happens when you increase the value of the damping from zero. Can you explain the behavior you are seeing in terms of the solution of the underlying differential equation? (If you cannot, we will see why in the next chapter.)

Chapter 2

Linear Differential Equations, Matrix Algebra, and Control Systems We have seen in Chapter 1 how Simulink allows vectorization of the parameters in a model (using MATLAB notation for the vectors) with the resulting model simultaneously simulating multiple situations. In a similar manner, Simulink allows matrix algebra using blocks. In this chapter, we will exploit this capability to build easy to understand models using vectors, matrices, and related computational tools. In fact, it is this capability that, when used properly, contributes to the readability and ease of use of many models. To make it easy to understand these capabilities, we will spend some time talking about the solution to linear differential equations using matrix techniques.

2.1 Linear Differential Equations: Linear Algebra The general form of an nth order linear differential equation is an+1

d n y(t) dt n

+

an

d n−1 y(t) dt n−1

+

an−1


+ · · · + a2

dy(t) dt

+

a1 y

=

u(t).

In addition to the equation, we must specify the values (at t = 0) of the dependent variable y(t) and its derivatives up to the order n − 1. Specifying the initial conditions creates a class of differential equation called an initial value problem. The coefficients in the equation can be functions of t , but for now, we will assume that they are constants. One way to solve this equation is to use a test solution of the form y(t) = ept , where p is an unknown. Substituting this for y(t) into the differential equation gives a polynomial equation of order n for p. The solution of this polynomial gives n values of p that, with the initial conditions and the input u(t), provide the solution y(t). This may be familiar to you from a course on differential equations. However, there is a better way of understanding the solution to this equation, and it relies on matrix algebra. In the differential equation above, let us build an n-vector x (the vector will be n × 1) that consists of y and its derivatives as follows: x

(t)

=



y(t)

dy(t)

d 2 y(t)

dt

dt 2 33

···


T



.

34

Chapter 2. Linear Differential Equations, Matrix Algebra, and Control Systems

Using this vector, the differential equation becomes a first order vector-matrix differential equation. The equation comes about quite naturally because all of the derivates of the elements in the vector x(t) are linear combinations of elements in x(t). Thus, any linear differential equation becomes d x(t)

= Ax(t) + bu(t),

dt

where A and b are the matrices

A

   =  

−

0 0 .. . 0 a1 an+1

−

1 0 .. . 0 a2 an+1

−

0 1 .. . 0 a3

··· ··· ··· ··· ···

an+1

−

0 0 .. . 1 an an+1

   ,  

b

   =  

0 0 0 .. . 1 a n+1

    . 

If this is the first time you have encountered this equation, you should verify that the matrix equation and the original differential equation are the same (see Problem 2.1). We will solve this equation in two steps. First, we assume that u(t) ≡ 0 and get the solution; then we will let u(t) be nonzero and derive the solution. We will do this using matrix algebra. When u(t) ≡ 0 we have d x(t) dt with

= Ax(t) x(0) = x0 .

If A in this equation were a scalar (say a ), then the solution would be x(t) = eat x( 0). This comes immediately by substituting this value for x into the differential equation and using d at the fact that dt e x0 = ae at x0 . Can we expand the idea of an exponential to a matrix? To answer this we need to think about what the exponential function is. When we say that a function eat exists, what we really mean is that the infinite series e

at

= 1 + at +

a 2 t 2

2!

+

a 3 t 3

3!

+···+

a n t n n!

+···

is convergent and well behaved. Therefore, we can differentiate this by taking the derivative of each term in the series. Can we do the same thing with a matrix? Let’s define the matrix exponential (analogous to the scalar exponential) as eAt = I + At +

A

2 2

t

2!

+

A

3 3

t

3!

+···+

An t n

n!

+···.

This series converges because we can always find a matrix, T, which will convert the matrix A to a diagonal or Jordan form. For now, let us assume that the A matrix is diagonalizable. That is, there exists a matrix T such that TAT−1 = , a diagonal matrix. (See Chapter 10 of Numerical Computing with MATLAB [29] for a discussion of eigenvalues and the diagonalization of a matrix in MATLAB.)

2.1. Linear Differential Equations: Linear Algebra

−1

T

35

Multiplying both sides of the definition of the matrix exponential by on the right gives 2

Te

−1

At

T

−1

= I + TAT

t +

3

−1 2

TA T

t

+

2!

n

−1 3

TA T

t

+···+

3!

T

on the left and

−1 n

TA T

t

n!

+···.

If we insert the identity matrix in the form of T−1 T between each of the powers of A in this series, we get Te

A t

−1

T

−1

= I + TAT

−1

t +

TA(T

−1 2

T)AT

t

2!

−1

+

TA(T

−1

T)A(T

−1 3

T)AT

3!

t

+···.

Now, by grouping slightly differently we get Te

At

−1

T

−1

= I + TAT

t +

(TAT−1 )(TAT−1 )t 2

2!

+

(TAT−1 )(TAT−1 )(TAT−1 )t 3

3!

+···.

Therefore, the infinite series is now the sum of an infinite number of diagonal matrices whose n diagonal elements are the infinite series that sums to the exponential eλi t for each eigenvalue λi of the matrix A. The steps for this are as follows: Te

At

−1

T

= I + t + λ1 t

e  0  = 0  ... 0

2 t 2

2!

+

0

3 t 3

3!

+···+

0 0

···

λ3 t

···

0 0 0

.. .

.. .

···

.. .

0

0

···

eλn t

eλ2 t

0

e

···

n t n n!

+···= e

t

   . 

Therefore, the matrix exponential exists and is absolutely convergent. This constructive proof of the existence of the matrix exponential also shows how to compute it. To construct it, first compute the diagonal matrix above from each of the eigenvalues of the matrix and then multiply the result on the right by T and on the left by T−1 . MATLAB allows construction of a simple M-file that will generate the solution matrix this way. The code is expm_ncs. There is a built-in matrix exponential function in MATLABthat performsthe calculation, but sincethe codeis not visible, the M-file expm_ncs is inthe NCS library. Open this fileand followthe steps ituses toform thematrix exponential. We are only halfway to solving the linear differential equation we set out to solve. So now, let us remove the restriction that u(t) ≡ 0. Before we show what happens when we do this, there are a couple of facts about the matrix exponential that you need to know. (Spend a few seconds verifying these facts, using the hint below each of the statements; see Exercise 2.2.) Fact 1: The derivative of eAt is AeAt . Show this by differentiating the infinite series term by term. Fact 2: The inverse of eAt is e−At .

36


Show this by taking the derivative of the product eAt e−At (using the first fact) and show that the result is zero. Then show that this means that the product is a constant equal to the identity matrix. Now, return to the differential equation d xdt (t) = Ax(t) + bu(t) and let u(t) be nonzero. Conceptually, the input u(t) is like a change to the initial conditions of the differential equation at every time step. Thus, it seems reasonable that the solution to the equation should have the form eAt c(t), where c(t) is a “pseudoinitial condition” that is a function of time that is to be determined. Substituting this assumed solution into the differential equation gives d dt

eAt c(t) = AeAt c(t) + bu(t).





However, d dt

eAt c(t) = AeAt c(t) + eAt





When this result is used, we get AeAt c(t) +

d dt

c(t).

d c(t) = AeAt c(t) + bu(t) eAt dt or d c(t) = bu(t). eAt dt

Multiplying by e−At gives the following for c(t) : c(t) =



t

e−Aτ bu(τ)dτ + x0 .

0

The fact that the constant of integration is x0 comes from letting t = 0 in the solution. Therefore the solution of the differential equation is At

At

x(t) = e c(t) = e x0 +



t

eA(t −τ) bu(τ)dτ.

0

We call the integral in this expression the convolution integral.

2.1.1 Solving a Differential Equation at Discrete Time Steps In our discussion of digital filters in Chapter 4, we will need to understand how to convert a continuous time linear system into an equivalent discrete time (digital) representation. One method uses the solution above. Assume that the analog-to-digital (A/D) converter at the input samples u(t) to make it piecewise constant (i.e., the analog input is sampled at times kt and held constant over the interval t) . We call this a “zero order hold” sampler. Under this assumption, the solution above becomes a difference equation by following these steps: • Assume that the initial time, t 0 , is the sample kt and the current time, t , is (k + 1)t . • Assume that u(t) is sampled so that its value is constant over the sample interval, i.e., u(t) = u(kt), kt ≤ t < (k + 1)t , which we denote by uk .


37

• Denote the value of the vector x(kt) by xk . With these assumptions, the solution becomes the difference equation xk +1 =

e

At

xk +



(k +1)t

eA( (k+1)t −τ) dτ buk

kt

= (t)xk +  (t)uk .



t

The matrix (t) is eAt , and the matrix (t) = 0 eAτ dτ comes from the integral by changing the integration variable (k + 1)t − τ = τ  (which makes dτ = −dτ  ):  (t) =



(k+1)t

kt

e

A( (k +1)t −τ )

dτ b

=



t



eAτ dτ  b.

0

Notice that this difference equation is also a vector matrix equation. It corresponds to the solution of the differential equation at the times kt , and this discrete time solution is the same as the continuous time solution when the input u(t) is a constant. We can now solve any linear differential equation we encounter using linear algebra. In fact, we have a whole bunch of ways of solving the equation. So let us explore them. Before we do, there are some names that you should be aware of that are associated with the solution method that we just went through. We call the vector x(t) in the differential equation the “state vector” and the differential equation the “state-space” model. In the solution, the matrix  = eAt is the transition matrix. If you check the Simulink Continuous library, you will find the state-space icon that allows you to put a linear differential equation in state-space form in a model. Similarly, in the list of digital filters in the Discrete library, the eighth block is the “Discrete State-Space.” It allows you to enter the difference equation above. In order to match the continuous and discrete models, we need to be able to compute for any A and b the values of (t) and (t). MATLAB can do this. The NCS library contains a code called c2d_ncs that will create these matrices. The inputs to c2d_ncs are A, b, and the sample time t ; the outputs are phi and gamma. The M-file, shown below, uses the matrix exponential expm in MATLAB. This code is similar to the code called c2d that is part of the control system toolbox in MATLAB. function [Phi, Gamma] = c2d_ncs(a, b, deltat) % C2D_NCS Converts the continuous time state space model to a discrete % time state space model that is equivalent at the sample times under % the assumption that the input is constant over the sample time. % % [Phi, Gamma] = C2D(A,B,deltat) converts: % . % x = Ax + Bu % into the discrete-time state-space system: % % x[k+1] = Phi * x[k] + Gamma * u[k] [ma,na] [mb,nb]

= size(a) = size(b)

38

Chapter Chapter 2. Linear Linear Differe Differential ntial Equations, Equations, Matrix Matrix Algebra, Algebra, and Control Control Systems Systems

if ma˜=na ma˜=na error( error(’Th ’The e matrix matrix a must must be square square’) ’) end if mb˜=ma mb˜=ma erro error( r(’T ’The he matr matrix ix b must must have have the the same same numb number er of rows rows as the the matr matrix ix a’) a’) end augmented augmented_matr _matrix ix exp_a xp_au ug Phi Gamma

= = = =

[[a b];zeros( b];zeros(nb,na nb,na+nb)] +nb)]; ; expm xpm(au (augme gmente nted_ma d_mat trix rix*de *delta ltat); t); exp_aug( 1:na, 1:na ); exp_aug( 1:na, na+1:na+nb );

The cute trick in this code is to build the augmented matrix



A

b

0nxn

0nxm



, where the

zeros at the bottom have the same number of rows as A and the same number of columns as the concatena concatenation tion of A and b. Computi Computing ng the matrix matrix exponenti exponential al of this matrix matrix automati automaticall cally y t t At generates the integral 0 e bdt . (See Exercise Exercise 2.3, where you will will verify verify this yourself. yourself. Hint: Use the fact that the integral of the bottom n rows in the matrix is a constant.)



2.1.2 2.1.2 Linear Linear Differ Differentia entiall Equation Equationss in Simulink Simulink We can now go to Simulink Simulink and solve some linear differe differential ntial equation equations. s. In the clock example, the equation for the pendulum was d 2 θ dt 2

=−

c dθ ml 2

d t

−

g l

(θ ) + 0.1148 u(t). sin(θ)

(θ ) ≈ θ . (Thi To make this linear, assume that the pendulum angle is small so that sin (θ) (Thiss comes from the fact that the lim θ →0 sinθ (θ) = 1.) The parameter parameter values values in the model were such that the linear differential equation becomes ¨ + 0.01 θ ˙ + 39.4366 θ = 0.1148 u(t). θ

Using the state-space form we developed above gives d x(t) dt

=



0

1

−39.4366

−0.01



(t ) + x(t)



0 0.1148



u(t).

If we convert this, using the c2d_ncs M-file, into a discrete system with a sample time of 0.01 seconds, the matrices  and  are phi = 0.99 0.9980 8028 2888 8836 3633 3385 85 -0.3 -0.394 9408 0871 7138 3885 8513 130 0

gamma = 0.00 0.0099 9992 9292 9288 8874 7448 48 0.99 0.9979 7928 2895 9543 4345 4511 11

0.00 0.0000 0005 0573 7379 7922 2261 61 0.00 0.0011 1147 4718 1882 8234 3479 79


39

x' = Ax+Bu y = Cx+Du Scope2

State-Space 0 Scope

Constant y(n)=Cx(n)+Du(n) x(n+1)=Ax(n)+Bu(n) Discrete State-Space

Scope1

Figure 2.1. Using Simulink to compare discrete and continuous time state-space versions of the pendulum.

We can now compare the solutions in Simulink for the continuous version and the discreteversion discreteversion using using the state-sp state-space ace blocks blocks for each. each. The NCS library library containsthe containsthe Simulink Simulink model State_Space, whic which h you you can can open open (or (or buil build d your yourse self lf usin using g Figu Figure re 2.1 2.1 as the the guid guide) e).. The state-space block requires values for two additional matrices (called C and D). They define define the output output of the block. block. Specifi Specifical cally ly,, the output output of the block, block, denote denoted d by y, is Because we defined defined the state state x with the output y as the first entry (and y = Cx + Du. Because the derivatives of y as the subsequent entries), the output and its derivatives are always availabl availablee as a simple simp matri product. If we are interested interested only in the value of y , then  le matrix  x product. the matrix c = 1 0 , and since the input u does not appear in the output, D = 0 (a scalar). These values, along with the values of A and b, are set when the model opens using the model preload function callback callback in the model properties. We also have set the initial displacement of the pendulum to be 1 and the initial velocity of the pendulum to be 0. This is done in the block dialogs by setting the initial condition vector to be [1 0]. As an exerci exercise, se, let us see how the two solution solutionss compar compare. e. We set up the model so that a sum block creates the difference between the two solutions and sends the differenc ferencee to a Scope. Scope. We also also set the model model up to run for 1000 1000 sec, sec, and to compute compute the solution to the continuous time model with the ode45 algorithm with the relative tolerance set to10−10 . The Simulink model that compares the state-space models in discrete and continuous form are in Figure 2.1; Figure 2.2 contains plots that show the solution for the continuous model and the difference between the continuous solution and the discrete solution. Notice that the largest difference is about 3 × 10−8 , which is almost entirely due to the conversion from continuous to discrete form. This is because in the Configuration Parameters dialog (under the Simulation menu) we set the relative tolerance for the ode45 solver to 10 −10 . The ability ability to simulate simulate a continuou continuouss time linear linear differe differentia ntiall equation equation using the iterativ iterativee discrete version is obviously very useful when it comes to designing a digital filter. We will return to this later when we show details about the design of digital filters. For now, we need to spend some time on a detour that shows how to get the frequency response of a continuous time system. To do this we need to introduce and investigate the properties of the Laplace transform.

40

Chapter Chapter 2. Linear Linear Differe Differential ntial Equations, Equations, Matrix Matrix Algebra, Algebra, and Control Control Systems Systems Dynamics

0.2 0.04 0.1

0.02 0

0 -0.02 -0.04

-0.1

399

399.5

400

400.5

401

Time -0.2 0

200

400

600

800

1000

Time

x 10

-8

Scope 2 -- Continuous State -Space Model x 10

-8

1

4 0 -1

2

399 0

399.5

400

400.5

401

401.5

Time

-2

-4 0

200

400

600

800

1000

Time

Scope -- Difference Between Continuous and Discrete Model

Output from from Continuous State-Space Model (Scope2) and the Difference Between the Continuous and Discrete Models (Scope)

Figure 2.2. Outputs as seen in the Scope blocks when simulating the model in Figure 2.1.

2.2 Laplace Laplace Transfo ransforms rms for for Linear Linear Diffe Differen rential tial Equation Equationss Early in his career, Pierre-Simon Laplace (1749–1827) was interested in the stability of the orbits of planets. He developed his eponymous transform to solve the linear differendifferential equations that defined the perturbations away from the nominal orbit of a planet, and consequently there is no single paper devoted only to the transform. In fact, he treated the transform as simply a tool and most times used it without explanation. The transform converts a differential equation into an equivalent algebraic equation whose solution is more manageable. Today, oday, this use is not as important because we have computers, and tools like Simulink, to do this for us. However, However, one aspect of the Laplace transfor transform m is still still extensiv extensively ely used: the ability ability to understand understand the effect effect of a differe differentia ntiall equation equation on a sinusoida sinusoidall input in terms terms of the frequency frequency of the sinusoid. sinusoid. We will make

2.2. Laplace Transforms for Linear Differential Equations

41

extensive use of this in our discussions of electrical systems, mechanical systems, and both analog and digital filtering. The Laplace transform of a function f(t) is ∞

F(s) =



e−st f(t)dt.

0

With this definition, it is easy to build a table of Laplace transforms for different types of functions. The simplest is the the function f(t) = 1 when t ≥ 0 and f(t) = 0 when t < 0. This is the step function because its value steps from 0 to 1 at the time t = 0. Putting this function in the definition, we get ∞

F(s) =



e

−st

dt = −

0

1 s

∞

e

−st

 

1

=

s

0

.

When the value of the integral is evaluated at infinity, the assumption is made that the real part of s is > 0. In a similar way, way, the transform transform of e−at is s +1 a . The Laplace Laplace transforms transforms for the sine and cosine can use this transform and the fact that sin (bt) = eibt +e−ibt 2

eibt −e−ibt 2i

and

cos(bt) = (see Exercise 2.4). We can also also apply apply the Laplac Laplacee transf transform orm to the statestate-spa space ce model model we develo developed ped in the previo previous us sectio section. n. In order order to do this this we need need to know know the Laplac Laplacee transf transform orm of the deriva derivativ tivee df(t) of a function. Thus, if the Laplace transform of a function f(t) is F(s), the derivative dt has the Laplace transform sF(s) − f (0). This follows from the definition and integration by parts as follows: ∞

 0

df (t) dt

∞

e

−st

dt = s



f (t) e −st dt + f(t)e −st

0

∞



0

= sF(s) − f (0).

Since the Laplace transform of the derivative is s times the Laplace transform of the functi function on differ different entiat iated, ed, it follow followss that that the Laplac Laplacee transf transform orm of the integr integral al is 1 /s. From From this this comes the Simulink notation for the integral block in the Continuous library. Let us apply this definition to the state-space model d xdt (t) = Ax(t) + bu(t). The The Laplace transform of the vector x(t) , X(s) , is the Laplace transform of each element, so s X(s) − x(0) = AX(s) + bU(s).

This gives X(s) as −1

X(s) = [s I − A]

−1

x(0) + [s I − A]

bU(s).

t t

(t ) = eA(t −t 0 ) x0 + t eA(t −τ ) bu(τ)dτ we developed If we compare this to the solution x(t) 0 earlier, it is obvious that the Laplace transforms of the two terms in the solution are (using the symbol L to denote the Laplace transform)



((t)) = L (e (e L ((t))

−At

) = [s I − A]−1

and t

L

 t 0

e

A(t −τ )



bu(τ)dτ

= [s I − A]−1 bU(s).

42


The last transform is the convolution theorem; it says that the Laplace transform of the convolution integral is the product of the transforms of the two functions in the integral. Also note that the Laplace transform of the matrix exponential has exactly the same form as the transform of the scalar exponential (except the inverse is used and the Laplace variable is multiplied by the identity matrix). As we noted when we introduced the state-space model, the output (for a scalar input u) is y = Cx + du. One of the most important uses of the Laplace transform is the transfer function of a linear Y(s) system. The transfer function is U(s) when the initial conditions are zero. In the state-space model, if y(t) (t ) is of dimension m (i.e., there are m outputs from the model), then m transfer functions are created. Using the Laplace transforms above gives the transfer function for the state-space model as Y(s) 1 = C[s I − A ]− b + d. U(s) The transfer function of the linear system can specify the properties of the differential equation equation in a Simulink Simulink simulation simulation.. For example, example, the transfer transfer functions functions that result result from taking the Laplace transform of the state-space model for the clock dynamics are (once again using the symbol L to denote the operation of taking the Laplace transform) d x(t)

L

d t

= L



0 −39.4366 y(t) =

1 −0.01



1

0

 

x(t) (t ) +



0 0.1148

 

u(t) ,

x(t)

Moreover, using the derivation above and the fact that the initial condition is zero for the transfer function, we get Y(s) U(s)

=[ 1

=[ 1 =

0]

0]

 

s 39.4366

s + 0.01 −39.4366

1 s



s 2 + 0.01s + 39.4366

0.1148 s2

−1 s + 0.01

+ 0.01s + 39.4366

−1

  

0 0.1148

0 0.1148





.

The Simulink model in Figure 2.3 uses the two transfer-function blocks in Simulink to define the clock example that we developed developed in Section Section 1.3. We opened this from the NCS library with the command Clock_transfer_functions. Notice that the blocks are Transfe Transferr Fcn and Zero-Pole. Zero-Pole. The first uses the transfer transfer function function in exactly the form that we just developed, whereas the second uses the form with the numerator and denominator polynomi polynomials als factored. factored. The numerato numeratorr polynomia polynomials ls roots roots are the zeros zeros of the transfer transfer function function since the roots of the numerator, when substituted for s , result in a value of zero for the

2.3. Linear Feedback Control

43

0.1148 s2 +0.01s+39.4366 Transfer Fcn

Scope

[10000/9]

0 Constant

Scope2

IC

0.1148 (s+0.005+6.27985469577123i)(s+0.005-6.27985469577123i) Zero-Pole

Scope1

Figure 2.3. Linear pendulum model using Transfer Function and Zero-Pole-Gain blocks from the Simulink Continuous Library.

transfer function. Similarly the roots of the denominator are called the poles of the transfer function since setting s to these values causes the transfer function to become infinite. (As poles go, these are truly long poles.) The simulation would result in a value of exactly zero if there were no input since the transfer function assumes that the initial condition condition is zero. No initial condition combined with an input of 0 would mean that everything in the simulation is exactly zero for all time. For this reason, the model uses the Initial Condition (IC) block from the Signal Attributes librar library y in Simuli Simulink nk to force force an initi initial al value value for the input input to the transf transfer er functi functions ons.. This This block block causes only the input u(t) to have an initial condition, not the output. To get the output to have the right value at the start of the simulation is rather delicate. It involves issues about how long the input stays at the value specified by the IC block, the amplitude of the output caused by this initial input, and the desired output amplitude. To ensure that the simulation produces consistent results, the solver for this model was a fixed step solver with a step size of 0.01 sec. The initial condition for the input was set to 10000/9, the value needed to force the output to be 0.2 at t 0. The simulation results then are the same as the state-space and the original clock simulation results shown above. The difficulties associated with getting the initial condition right makes transfer functions difficult to use for initial value problems. For this reason, either the state-space model or the model created using integrators (as in the clock simulation above) is the better choice. =

2.3

Linear Linear Feedb Feedbac ack k Cont Contro roll

One of the more powerful uses for Simulink is the evaluation of the performance of potential control strategies for complex systems. systems. If you are designing an aircraft flight-control system; a spacecraft’s guidance, navigation, or attitude control system; or an automobile engine controller, you often have neither the luxury nor the ability to build the device with the intention intention of trying the control control system and then tinkering tinkering with it until it works. The preferred method for these complex systems is to build a simulation of the device and use

44


the simulation to tinker. This section will introduce the methods for control system design and its mathematics and then will show how to use Simulink to tinker with the design.

2.3.1 What Is a Control System? Early in theindustrial revolution, it becameclearthat some attributes of a mechanical system required regulation. One of the first examples was the control of water flowing through a sluice gate to maintain a constant speed for the water wheel. Although many control ideas were tried, they often operated in an unstable way or hunted excessively, causing the water wheel to alternately speed up and slow down. James Clerk Maxwell analyzed such a device and showed that its stability and performance required the roots of an equation that came from the differential equation (the poles that we discussed in Section 2.2). Unfortunately, the mathematics at the time allowed factoring only of third order and some fourth order polynomials, so it was very difficult to design a system that involved dynamics that had more than three differential equations. Mathematicians found an approach that handled higher order systems, but today the digital computer and programs such as MATLAB and Simulink have made those methods interesting only as historical footnotes. So what exactly is a control system? The easiest way to understand a control systemis to show an example. In the process, we will build a Simulink model that we can experiment with to understand the various design issues and the consequences of incorrect choices. One of the simplest and most commonly encountered control systems is that used to keep the temperature constant for a home heating system. The dynamics for this system can be grossly described by assuming that the entire house is a single thermal mass which has both heat loss (via conduction through the walls, windows, floors, and ceilings) and heat gain from the heating plant (via the radiators, convectors, or air vents, depending on whether the system uses steam, hot water, or hot air). Simple thermodynamics states that the temperature, T , of the house is C

dT dt

=

kA l

(T − T o utside ) + Qin u(t),

where C is the thermal capacity of the house, T is the temperature of the house, T o utside is the air temperature outside the house, k is the average coefficient of thermal conductivity of all of the house surfaces, A is the area of all of the house surfaces, l is the average thickness of the insulation in the house, Qin is the heat from the heat exchanger, u(t), the control signal applied to the furnace, is either zero or one. Most heating appliances in the United States use the English system of units, where heat is in British thermal units (BTUs), temperature is in degrees Fahrenheit, dimensions are in feet, and thermal capacity is defined as the thermal mass of the house times the specific heat capacity of the materials in the house. (The units are BTUs per degree F.) Clearly, a house does not consist of a single material, or a single compartment with a temperature T , or a single thermal mass. For this reason, a more detailed model would have many terms corresponding to different rooms in the house, with different losses correspond-


45

Figure 2.4. Home heating system Simulink model (Thermo_NCS). This model describes the house temperature using a single first order differential equation and assumes that a bimetal thermostat controls the temperature.

ing to the walls, doors, windows, floors, ceilings, etc. We will develop a more complex model like this in Chapter 6. We will use a model that is one of The MathWorks Simulink demonstrations. (This model, sldemo_househeat, comes with Simulink, but we have a version of the model in the NCS library.) Open the model by typing Thermo_NCS at the MATLAB prompt. This opens the model shown in Figure 2.4. The model uses meter-kilogram-second units, so there are conversions from degrees Fahrenheit to Celsius and back in the model. (These are the blocks called F2C and C2F, respectively). One new feature of Simulink that we have not seen before is in the model. It is the multiplexer block (or the “mux” block) that is found in the Signal Routing library. This block converts multiple signals into a single vector signal. In the model, it forces the Scope block to plot two signals together in a single plot. We configured the Scope block in this model to show two separate plot axes; we did this by opening the plot “parameters” dialog by double clicking the icon at the top of the plot that is next to the printer icon and then changing the “Number of axes” to two. We have seen the use of a subsystem in the clock example. Here we use subsystems to group two of the complex models into a visually simplified model by grouping the equations for the house and the thermostat inside subsystem blocks. The house model contains the differential equation above, and the thermostat model uses the “hysteresis” block from the “Discontinuous” library in Simulink. Thehousemodelis a subsystemin theSimulink modelthat implements thedifferential equation above. The subsystem is in Figure 2.5. The thermal capacity of the house is computed from the total air mass, M , in the house and the specific heat, c, of air. The thermal conductivity comes from the equivalent thermal resistance to the flow of heat through the walls, ceilings, glass, etc. All of these

46


1

1/(M*c)

Heater QDot In

1/s

1 Indoor Temp Ti n

1/Mc

1/Req Thermodynamic Model for the House

1/Req 2 Outdoor Temp Tout

Figure 2.5. Subsystem House models the house. “

”

contains the single differential equation that

data and the data for the heating system (which is electric) are loaded from the MATLAB M-file Thermdat_NCS.m in the NCS Library. To see the various data and their functional relationship to the dimensions of the house and the insulation properties, etc., edit and review the data in this file. The model uses a sinusoidal input for the outside temperature that gives a variation of 15 deg F around an average temperature of 50 deg F. The plots of the indoor and outdoor temperatures and the total heating cost for a 24-hour (86,400 sec) period are in Figure 2.6. We can now see the features of this home heating controller (and any other control system for that matter). It consists of a system that needs to be controlled (in this case, the temperature of the house needs to be controlled, and the thermodynamics of the house is the system). The control system needs to measure the variables we are trying to control (in this case the thermostat measures the house temperature), and it needs a device to calculate the difference between the desired and actual variable. It also needs a device (a controller) that converts the desired state into an action that causes the system to move toward the desired condition. In this case, the control is the heater; the controller is the thermostat. (The thermostat plays the dual role of both the temperature sensor, working on the difference between the actual and desired temperature values, and the control in this application.) In order to model the thermostat, we need to look at how this device works. The photograph in Figure 2.74 shows a typical thermostat. The temperature difference is detected using a metal coil (2) made from two dissimilar materials, each with different heat expansion rates (so-called bimetals). The bimetal’s expansion difference translates into a torque that causes a contact (4 and 6) to open and close. A lever (3) sets the tension of the coil and therefore the room temperature set point. Moving the lever makes the contact coil rotate closer or further away from the contact (6). When the temperature drops below the set point, the bimetal rotates and the contacts close, making the heat source come on. When the temperature rises above the set point, the contacts open and the heat goes off. The contact 4 has a magnet that holds it closed (just to the left of the (6) in Figure 2.7). This forces the contacts to stay closed until the temperature of the room rises well past the 4

This figure is from the article on thermostats in Wikipedia, The Free Encyclopedia web site [50].


47

80 70 60

Indoor Temperature (with a 70 deg. F set point)

50 Outdoor Temperature

40 30

0

1

2

3

4

5

6

7

8 x 10

4

Heating Cost ($) 40

30

20

10

0

0

2

4

6

8

10

12

14

Time

16

18 x 10

4

Figure 2.6. Results of simulating the home heating system model. The top figure shows the outdoor and indoor temperatures, and the bottom figure shows the cost of the heat over the 24-hour simulation.

Figure 2.7. A typical home-heating thermostat [50].

48

Chapter 2. Linear Differential Equations, Matrix Algebra, and Control Systems Out

1

In 0

In

Ton

Toff

Tset Figure 2.8. Hysteresis curve for the thermostat in the Simulink model.

point at which the contacts closed. (The force from the magnet is greatest when the contacts are closed because of the close proximity of the magnet and the contact.) This effect, called hysteresis, results in the room temperature ’s rising above the desired value and then falling below the desired value by some amount. Look closely at the room temperature in the plot above; the room temperature rises to about 5 deg above and falls about 5 deg below the desired value. In the next section, we will develop a simple controller that will eliminate this problem. The thermostat’s mechanical motion can be modeled using the built-inhysteresis block in the Simulink discontinuous block library. This block captures the essential features of the thermostat, as can be seen from the input-output map of the block. Figure 2.8 shows the input-output graph. An input-output map of a device shows how the output changes as the input varies. In the thermostat, the temperature is the input, and the output is either 0 (off) or 1 (on). When the temperature is below the set value ( T set ), the thermostat is on and the output of the block is 1; as the temperature increases because of the heat being provided, the temperature inevitably becomes greater than the off temperature ( T off ) and the thermostat turns off (the output becomes 0). As the house cools down, the temperature must fall below the value T off to the value T on before the output changes to 1 again. The difference between T set and T off and T set and T on is set to 5 deg in the model by changing the values in the dialog that opens when you double click on the block. Once you have built the model, try different values for the hysteresis, change the temperature pro file for the outside, and see if you can find an outside temperature below which the house will not stay at the desired set point temperature.

2.3.2 Control Systems and Linear Differential Equations The basic dynamics for the heating system is the first order linear differential equation for the heat flow into and out of the house. These dynamics are linear, but the system is nonlinear because of the thermostat. The controller nonlinearity causes the heat to cycle to

2.4. Linearization and the Control of Linear Systems

49

a temperature that is 5 deg too hot and then fall to a temperature that is 5 deg too cold. This 10-deg swing in the room can cause discomfort for the occupants since the room is either too hot or too cold. What could we do to fix this? The simple answer is to make the controller linear, because if you did so, the temperature would stay almost exactly at the set point. Therefore, we would like to control theheater so that thetemperatureis proportionalto the difference; it should be possible to keep the temperature at the desired value without the 10-deg swing produced by the thermostat. A simple way to achieve this would be to control the heater using a device that makes the heat output proportional to the input. Electronic devices exist that allow this form of control. We will actually build such a controller in Chapter 6, refine it further in Chapter 7, and look at a more complete version in Chapter 8. Another important aspect of a control system is the fact that a properly operating controller causes the system to deviate away from the set point by only a small amount. This means that even a very nonlinear system is linearizable for purposes of control. Let us formalize these ideas and use the results to justify the use of linear systems analysis for control systems.

2.4 Linearization and the Control of Linear Systems In the most general situation, the underlying mathematical model of a system is nonlinear. Nonlinearities occur naturally because of the three dimensional world we live in. For example, a rotation about an axis is described through a coordinate transformation which involves the sine and cosine of the rotational angle. (We will discuss this in Chapter 3.) Another example is the air drag that is a function of the square of the speed; we modeled this when we were exploring the Leaning Tower of Pisa in Chapter 1. Other nonlinear effects are a consequence of the properties of the materials that we encounter. All control system designs use linear differential equations to model the physical systems under control. This is true because a control system is usually trying to keep the system at a particular value or is trying to track an input. In either case, the perturbations away from the desired input are small. We will show an example of this in Section 2.4.2, but first let us see how we go about finding a model for small perturbations. (It is the linear model.)

2.4.1 Linearization As we stated above, the goal of a control system is to make some system attribute constant or to follow a slowly changing command. (Think of the autopilot on an airplane that is trying to keep the altitude of the plane constant.) In other words, the control is trying to keep perturbations small. To make this concept more precise, let us consider a general nonlinear differential equation that models a physical system. This differential equation might come from a Simulink model that you have built. The form of the equation then will be a set of nonlinear differential equations that couple through nonlinear functions as follows: dx i (t) dt

=

f i (x1 , x2 , . . . , xn , t) + g(u1 , u2 , . . . , um , t),

i

=

1, 2, . . . , n .

There are n first order differential equations in this description. The n values of xi (t) are lumped into an n-vector x(t) , the m values of ui (t) are lumped into an m-vector u(t) , and

50


the n functionsf i (x1 , x2 , . . . , xn , t) are lumped into an n-vector f . Doing this changes the way we write the differential equation to the more compact d x(t) dt

=

f (x, t) + g (u,t).

To investigate small perturbations δ x(t) away from some nominal solution xnom (t) of this equation, let x(t) = xnom (t) + δ x(t) and u(t) = unom (t) + δ u(t), and substitute them into the differential equation. The result is d(xnom (t)

+

δ x(t))

dt

f (xnom (t)

=

+

δ x(t),t) + g( unom (t)

If the perturbations are small enough, then a Taylor series gives a away from the nominal as d(xnom (t)) dt

+

d(δx(t)) dt

=

f (xnom (t),t) +

∂ f (xnom (t),t) ∂x

+g (unom (t),t) +

+

δ u(t), t).

first

order perturbation

δ x(t) + · · ·

∂ g (unom (t),t) ∂u

δ u(t)

+ · · ·.

In these equations, the partial derivatives of any vector with respect to another vector are matrices defined as follows:

∂ f ∂x

=

   

∂f 1 ∂x1 .. . ∂f n ∂x1

···

.. . ···

∂f 1 ∂xn .. . ∂f n ∂xn

   . 

The Taylor series terms beyond the first derivatives are small as long as the perturbations are small, so we can neglect them. Using the fact that the nominal values satisfy the differential equation, i.e., that d( xnom (t)) dt

=

f (xnom (t),t) + g (unom (t), t)),

the differential equation for the perturbations is d(δx(t)) dt

=

∂ f (xnom (t),t) ∂x

δ x(t) +

∂ g (unom (t),t) ∂u

δ u(t).

Since the matrices in this equation are evaluated at the nominal value, they are time varying, and as a result the equation is linear and in the state-variable form d xdt (t) = Ax(t) + Bu(t) . ∂ f (x

(t),t)

∂ g (u

(t),t)

nom nom In this state equation, the matrix A is , the matrix B is , the ∂x ∂x states are the perturbations in x(t), and the controls are the inputs δ u(t) that create the perturbations. To recap this result, the use of the linear equation is justi fied by the fact that when the control system is working properly, the perturbations are small and the Taylor series is


51

an excellent approximation to the actual dynamics. For this reason, discussions on control systems design generally consider only linear models. In a later chapter, we will show how to use the above linearization to create the linear model directly from Simulink. Before we do this, let us create some Simulink models in various forms for a mass attached to a spring.

2.4.2 Eigenvalues and the Response of a Linear System Since linear differential equations form the basis for control system design, it is important to understand the relationships between the parameters in the state-variable model and the solution of the differential equations. Let us create some Simulink models for a springmass system that is typical of many mechanical systems. A spring stores mechanical energy whenever it is compressed or expanded. The energy that the spring stores returns whenever the spring is not compressed or expanded. The force that the spring applies is typically nonlinear. However, consistent with the discussion above, we can linearize the force to give f = Kx , where x is the displacement. ( K is the proportionality constant for the spring, and it has a negative sign so the force is positive when the spring is compressed and negative when it expands.) The units of K are force per unit displacement (Newtons per meter or pounds per inch, for example). This linear equation by itself describes an ideal spring; there is no friction. When there is Kx + D dx/dt friction, it is usual to add another linear force, a so-called damping force, that is proportional to the speed of the object. The model for this is f = D dx . A spring with a mass attached dt therefore has a model that is very much like the x (t ) M pendulum in Chapter 1. Let us develop an equa Mg tion of motion for the spring-mass system in the figure at the right. In the figure, the motion x(t) is positive down, so the forces from the spring and the damping are negative (upward). The figure contains a diagram that shows the force balance used to get the equation of motion. Using the force balance in the figure along with Newton ’s second law gives M

d 2 x dt 2

= −Kx −

D

dx dt

+

Mg.

We can now easily build a Simulink model for this equation. The model Spring_Mass1 is in the NCS library, but as usual, you should create the model yourself before you open it from the library. The model is in Figure 2.9. Remember that all of the data for the model load from a callback function. We can also use the state-space methods we developed above to convert this equation into a state-space model. Letting x

=



x(t) dx(t) dt



,

52

Chapter 2. Linear Differential Equations, Matrix Algebra, and Control Systems Spring Mass Model Using the Equations of Motion and Basic Simulink Blocks

Velocity of Mass

M*g

1 s

1/M

mg

1 xo s

Integrator

1/mass

Position of Mass

Integrator1

Scope

Damper D

Initial Positi on

0 Spring K

Figure 2.9. Model of a spring-mass-damper system using Simulink primitives.

Integrator

Add Input u(t)

Step

2

K*u Bu

2

1 s

State Derivative = Ax+Bu

2 2

Position and Speed of Mass

2

Matrix A

Ax

0 1 -omegan^2 -2*zeta*omegan [2x2]

Multiply [2x2]

Matrix Multiply

2

2

Figure 2.10. Model of a spring-mass-damper system using the state-space model and Simulink s automatic vectorization. ’

the model is d x(t) dt

=



0 −

1

K M

−

D M



x(t)

+

  0

1 M

u(t).

A Simulink model for this is easy to build. The first, and easiest, way to build the model is to use the State-Space block in the Continuous library (you should try this), but we will use an alternate method that exploits the automatic use of vectors in Simulink. The model (shown in Figure 2.10) is Spring_Mass2 in the NCS library. This model uses a characterization for the parameters of the second order differential equation that make the calculation of the solution easier to understand. The parameters are the undamped natural frequency (denoted by ωn ) and the damping ratio (denoted by ζ ). The values of these parameters are

ωn

=



K

M

and

ζ =

D/M

2ωn

.


53

2

ζ =0.4

ζ = 0.1

ζ = 0.2

1.5

1

ζ = 0.5 0.5

ζ = 0.3

0

-0.5

-1 0

5

10

15

20

25

30

Time

Figure 2.11. Changing the value of the damping ratio from 0.1 to 0.5 in the model of Figure 2.10.

When we insert these values in the state-space differential equation, we get d x(t) dt

=



0 2 −ωn

1 −2ζ ωn



x(t) +

  0

1 M

u(t).

In the model we let u(t) = g , corresponding to the force of gravity applied at t = 0. The simulation shows the motion when ωn = 1 and ζ = 0.1, which is identical to the Spring_Mass1 simulation. When this model opens, MATLAB sets the value for omegan, zeta, and M to 1, 0.1, and 1, respectively. The motion of the mass is a sine wave whose amplitude gradually decreases. We can see what happens if we vary the parameters in this model. First, keep ωn = 1 and let ζ vary between 0 and 1. The derivative is Ax + bu and the input is a function called the step function, which is found in the Sources library. Thus, as an exercise, vary zeta in the model from the value 0.1 to 1 by changing the value of zeta in MATLAB and rerunning the model. For each value of zeta entered, you will see a plot that looks like Figure 2.11. (This figure is an overplot for values of zeta from 0.1 to 0.5 in steps of 0.1.)

54


It should be evident that the bigger we make zeta, the more rapidly the oscillations disappear. (We say the oscillations are damped.) It is important to understand this since it is critical in the design of a control system. The second exercise is to increase the value of ωn . As you do so, you will see that the frequency of the oscillations increases. There is a subtle connection between the actual frequency of the oscillation and the damping ratio. Remember that we showed that the solution of the differential equation in state-space −1 t form depends on the transition matrix given by eAt T e T, where  is the diagonal matrix that contains the eigenvalues of the matrix A. Using the undamped natural frequency and damping ratio parameters in the A matrix, let us see what its eigenvalues are. The eigenvalues come from setting the determinant det(λI A)to 0, so

=

−

det(λI

− A) = det



λ ωn2

−1 λ + 2ζ ω



n

2

2 n

= λ + 2ζ ω λ + ω = 0. n

There are two eigenvalues, the roots of this equation, given (assuming that ς λ1 λ2



2

≤ 1) by

= −ζ ω + i  1 − ζ ω , = −ζ ω − i 1 − ζ ω . n n

n

2

n

Therefore, the diagonal matrix et is et

 − = e

ζ ωn i

+

√ −



1 ζ 2 ωn t

0

0 e

−

ζ ωn i

−

√ −

   −

1 ζ 2 ωn t

.

Since the solution is T−1 et T, the frequency of the oscillations is 1 ζ 2 ωn and the damping comes from the term e−ζ ωn t . Thus, the closer ζ gets to 1, the less oscillatory the response is and the faster the initial conditions disappear. The complete solution of −1 t the differential equation comes from e At T e T. We ask you to complete this in − 1 Exercise 2.5. (You need to compute T and T and finish the multiplication.) Several additional facts about the solution to this spring mass equation are important. First, when ζ >1, the two eigenvalues are real and the solution becomes the sum of two exponentials (i.e., it is no longer oscillatory). Second, if the mass or the damping term were negative for any reason (such as a badly designed control system), the solution will grow without bound. (The term ζ ωn in the solution would be negative, so the solution terms

=

√

2

e(−ζ ωn +i 1−ζ ωn )t would grow without bound.) The solution to the state-space differential equation using Laplace transforms was X(s)

1

1

= [s I − A]− x(0) + [s I − A]− bU(s).

The inverse Laplace transform involves exponentials of the form e−pi t , where pi is one of the roots (poles) of the denominator polynomial det (s I A). This determinant is the same one we get when we compute the eigenvalues of the matrix A, so poles and eigenvalues are the same creature in different clothes. The major difference is in the ease and accuracy of calculation. Poles are the roots of the polynomial; factoring it is inherently prone to

−

2.5. Poles and the Roots of the Characteristic Polynomial

White Noise

55

Dynamics of the Distrurbance num(s) den(s)

Control Gain Signal 1

K

Dynamics of the Controller

Dynamics of the System Being Controlled

num(s)

num(s)

den(s)

den(s)

1 Output

Input Signal

num(s) den(s) Dynamics of the Measurement Device

Figure 2.12. Generic components of a control system.

numeric issues, whereas the eigenvalues of a matrix are computed using numerically robust algorithms (see NCM, Chapter 10, in particular Section 10.5). These are all important aspects for the understanding of the control of a linear system, which we explore next.

2.5 Poles and the Roots of the Characteristic Polynomial Now that we know that poles and eigenvalues are the same, let us use this knowledge. By now, you should be familiar with the notation used in Simulink for modeling linear systems. The idea is to use the transfer function (which is the Laplace transform of the output divided by the Laplace transform of the input) as a multiplicative element inside the block. As a result, you can write the equations in the pictorial form of the block diagram where the multiplications use theblocks (with the transfer function inside) and parallel paths represented by lines that join a sum or difference (which use summing junctions). This idea predates Simulink and has represented control systems for over 50 years. Figure 2.12 shows the general form of a linear control system when the device to be controlled is a linear differential equation. Let us bring each of the elements in this figure into the control system framework we established above; from left to right in the diagram these are as follows: •

The Input Signal is the set point (in the case of the thermostat) which models the input that the control system is trying to maintain (or track).

•

The Control Gain is one or more gains in the system. They ensure the control system performs properly.

•

The Dynamics of the Controller models the control device. (This might not be dynamic, as was the case for the thermostat.)

56

Chapter 2. Linear Differential Equations, Matrix Algebra, and Control Systems •

The Dynamics of the Disturbance is a model for the way that “the world” is acting to cause the system to deviate from the desired performance. Note that the model assumes that the disturbance adds to the controller.

•

The Dynamics of the System Being Controlled —often called the plant dynamics—is the differential equation (or transfer function) of the system that is to be controlled.

•

Notice that the Output is measured, and this measurement compares with the input signal through a difference. The controller forces the system to react as desired, using the difference signal.

Each of these functions exists in any control system. They sometimes use a common device (as was the case for the thermostat, where the difference, the set point, and the measurement are in this single device).

2.5.1 Feedback of the Position of the Mass in the Spring-Mass Model Let us build a simple control system that will attempt to control the spring-mass damper system that we developed in Section 2.4.2. We assume that you are holding the mass at some location below its rest position, and you then let it go. We also assume that the control system will attempt to stop the mass motion as it approaches the rest position, but with no oscillation and as little residual motion as possible. The Simulink simulation in Figure 2.13 is the model. By now, this model should be easy for you to build. Try to build it without looking too closely at the figure. (Use the model Spring_Mass1 from the NCS library as the starting point, or open the complete model from NCS library using the command SpringMass_Control.) As you review the diagram, identify each of the generic control system features that we discussed. This model has the values K 20, D 2, M 10, and initially the gain 0. The initial condition for y has been set up so the mass is initially at rest (at t 0 this requires 2 that d dt y2 0 and dy 0 in the differential equation), so the initial value of y needs to be dt =

=

=

=

=

=

=

m*32.2 mg

Gain*m Step at 1 second

Control Gain (Initially 1 )

1/m 1/mass

1 s

Velocity of Mass

Integrator

1 xo s

Position of Mass

Integrator1

Damper d -32.2/(k/m+Gain) Spring

Initial Condition

k

Figure 2.13. A simple control that feeds back the position of the mass in a springmass-damper system.

Scope

2.5. Poles and the Roots of the Characteristic Polynomial computed. With the gain at 0, the initial position is

−g K . / M

57

(As an exercise, show why this is

so.) This initial condition for y is an external input to the integrator using the dialog for the integrator y . With this initial condition, the mass is at rest when the simulation starts. The first thing you need totry inthis model is tochange the value ofgain and see what happens. If you have done what was asked, i.e., varied the value of gain, you should see that you cannot achieve the desired result (i.e., having the motion stop without an oscillation). All that seems to happen as the value of gain is changed is that the oscillation frequency increases. Why is this and what can be done to achieve the desired performance? The why part of this question is easy: it can be answered by inspection from the block diagram. Look carefully at what the feedback is doing. The gain multiplies the position (after the summing junction), which is exactly what the spring force is doing. Thus, increasing the gain is like changing the restoring force of the spring. In the solution for the spring mass system we developed in the previous section, the result of changing K √ is to change the parameter ωn = K/M . Therefore, changing gain will only change the frequency of oscillation of the mass, and it will never cause the mass to stop without an oscillation.

2.5.2 Feedback of the Velocity of the Mass in the Spring-Mass Model From this discussion and thesolutionwe developedin theprevioussection, it should be clear D/M thatthe onlyway tocause the oscillation to stopquickly isto alter the value of ζ = 2ωn . The most direct way to do this is to change the effective D . Unfortunately, as it is constructed, the control system will not do this. If, however, we were to measure the velocity of the mass and then multiply this by a gain, we will create a force that is proportional to the velocity. This is equivalent to the dy mass damping term D dt that appears in the differential equation. Thus, we will modify the Simulink diagram to allow a feedback from the velocity. The resulting block diagram, shown in Figure 2.14, is the model SpringMass_Vel_Control in the NCS library.

M*32.2 mg

Gain*M Step at 1 second

Control Gain (Initially 1)

1/M 1/mass

1 s

Velocity of Mass

Integrator

1 xo s

Position of Mass

Integrator1

Damper D -32.2/(K/M) Spring

Initial Condition

K

Figure 2.14. Feed back the velocity of the mass instead of the position to damp the oscillations.

Scope

58


Position of Mass -15

-15.2

-15.4

-15.6

-15.8

-16

0

5

10

15

20

25

Time

Figure 2.15. Time response of the mass (position at the top and velocity at the bottom) for the velocity feedback controller.

This model uses a measurement of the velocity of the mass as the feedback variable. If you open this model and play with the gain variable, you will see that the response of the mass to the change in the position of the mass is less oscillatory as the gain is increased. In fact when the gain is set to 1, there is only one cycle of oscillation before the mass settles, and when the gain is 2, there are no longer any oscillations. (See the plot of the response with these different gain values in Figure 2.15.) For now, let us bypass how one would measure the speed and accept the fact that the control objective (that of ensuring that the mass stops with little or no oscillatory motion) is achieved with this control. Instead, let us try to put in place a way of determining why the second of the two control systems achieved the desired result while the first did not. To do this we will use the state-space model (with the built-in state-space block) for the various pieces in the control system. The Simulink model then looks like Figure 2.16.

30


59

Figure 2.16. Using the state-space model for the spring-mass-damper control system.

The state-space model of the spring is d x dt

0 K

  

=

−

1 D

−

M

dt

x+

M

y dy

0 1

    

u(t),

M

1 0

= measured

0 1

x.

dy

Moreover, from the model, we see that u(t) = Gain(Stepinput − dt ) − 32.2. If we substitute this into the state-space model, we get the state-space model of the closed loop system as

d x dt

=

  

0

1

      x+

−

K M

−

D M

0 1

M

dy

dt

=[ 0

which reduces to the model (using the fact that d x dt

=



0 K

−

M

1 −

Gain Stepinput −

D + Gain M

dt



− 32.2,

1] x,

  0 0

1 M

0 1

   x+

dy

1



x=



0 0

0 1 M

x)

(Gain Stepinput ) − 32.2.

M

This illustrates clearly that the gain term is equivalent to a change in the damping term D since the gain adds to D. The eigenvalues of the state matrix in this equation are the roots of the polynomial det(λI − A) = det



λ K/M

−1 λ + Gain + D/M



2 = λ + (Gain + D/M)λ + K/M,

and they clearly change as the gain changes. Let us draw a plot that will allow us to see how the roots change.

60

Chapter 2. Linear Differential Equations, Matrix Algebra, and Control Systems 1.5

1

s e u 0.5 l a v n e g i E e h t f 0 o t r a p y r a n i g a -0.5 m I

-1

-1.5 -2.5

-2

-1.5 -1 Real part of the Eigenvalues

-0.5

0

Figure 2.17. Root locus for the spring-mass-damper system. Velocity gain varying from 0 to 3.

If we let the gain vary from 0 to 3 in steps of 0.1, and we use the parameter values in the model above (i.e., M = 10; D = 2; K = 20), the eigenvalues change as shown in Figure 2.17. Remember that the response of the system is a sum of exponentials, where each of the exponents is an eigenvalue. Thus, as long as the eigenvalues are complex, the solution will be oscillatory. When the eigenvalues are all real and negative, on the other hand, the solution will asymptotically decay with no oscillations. The plot we created allows us to select a gain that will ensure that this happens (and meet the requirements that the control system had to satisfy). This plot, called a root locus by control engineers, is the most useful way of understanding the effect on a control system s response when the gain is changed. Before computers, the rules for creating a root locus by hand were quite elaborate. Fortunately, today we can do this easily using MATLAB (and Simulink). For example, the code that created the plot is “

”

’

ev =[]; for Gain = 0:0.1:3 ev = [ev eig(A-[0 0;0 Gain])]; end plot(real(ev),imag(ev),’x’) axis(’square’) grid


61

4

3

2 s e u l a v n 1 e g i E e h t f 0 o t r a p y r a n -1 i g a m I

-2

-3

-4

- 0.1 8

- 0.1 6

- 0.1 4 -0 .12 - 0. 1 - 0.0 8 - 0.0 6 Real part of the Eigenvalues

- 0.0 4

-0 .0 2

0

Figure 2.18. Root locus for the position feedback control. Gain changes only the frequency of oscillation.

2.5.3 Comparing Position and Rate Feedback In contrast to this root locus, let us look at the locus for the first control system we developed. In this system, the gain multiplied the position of the mass, and the control system model in state-space form is d x dt

=



0 K

−

1 D

M

−

M

0 1

   x+

Gain(Stepinput − y) − 32.2,

M y = [ 1 0] x.

As we did above, multiplying the matrices gives d x dt

=



0 −

K + Gain M

1 D − M

0 1

   x+

y=[ 1

M 0] x.

Therefore, the eigenvalues of the matrix



0

1

Gain − K +M

D − M

will be the root locus. Figure 2.18 shows the plot.



(GainStepinput ) − 32.2,

62


As we observed before, this control does nothing but change the frequency because the only change is the imaginary part of the eigenvalues.

2.5.4 The Structure of a Control System: Transfer Functions A generic control system has several essential components. These are the plant, which we will denote by G(s); the measurement, which we will denote by H(s); and the control dynamics (which might be only a gain), which we denote by K (s). The structure of the control system is in the figure below. U (s)

E (s)

+

C (s)

K (s)

G(s)

Y (s)

H (s) It is a simple matter to write the equation for the transfer function of the closed loop control system. If wedenote thedifference between theinput on theleftand themeasurement as E(s), then E(s) = U(s) − H(s)Y(s), Y(s) = K(s)G(s)E(s), which we solve for the transfer function T(s)

=

Y(s) U(s)

Y(s) U(s)

=

to give K(s)G(s)

1 + K(s)G(s)H(s)

.

Let us take this equation one step further by replacing each of the transfer functions with their respective numerators and denominators. Thus, if the transfer functions are G(s)

=

n1 (s) d 1 (s)

then T(s)

=

,

K(s)

=

K

n2 (s) d 2 (s)

,

H (s)

=

n3 (s) d 3 (s)

n1 (s)n2 (s)d 3 (s) d 1 (s)d 2 (s)d 3 (s)

+

Kn 1 (s)n2 (s)n3 (s)

,

.

The inverse Laplace transform of this transfer function comes from the partial fraction expansion into a sum of terms involving the roots of the denominator polynomial. This expansion is n  aj T(s) = a0 + . s + pj j =1

The denominator term s + pj is the j th factor of the denominator polynomial (i.e., −pj is the j th root of the polynomial d 1 (s)d 2 (s)d 3 (s) + Kn1 (s)n2 (s)n3 (s)). These “poles” of the closed loop transfer function obviously determine the response of the system. Notice


63

that the denominator polynomial is determined from the numerator and denominator of K(s)G(s)H(s) ; i.e., the poles are determined by the numerator, denominator, and gain terms of the individual blocks in the system. (These result from the open loop transfer functions of the system, the transfer function that would result if there were no feedback.) We could compute the root locus as the gain K varies using this polynomial equation, but this is not numerically robust. As was pointed out above, factoring polynomials is not a good idea. However, the use of Laplace transforms in block diagrams does allow some control system features to be readily developed. For example, in the control system model we developed above, how would we create the measurement of the velocity of the mass? One way would be to measure the position and differentiate it. A block that would do that df would contain a single s (since the Laplace transform of the derivative dt was sF(s), where F(s) is the Laplace transform of f(t)). If one were to try to build an analog device that would do this (using an electronic or mechanical device, for example), it becomes obvious rather quickly that this cannot be easily done. A simple idea would be to use a coil of wire and a magnet attached to the mass to develop a signal that is proportional to the speed of the mass. This comes about naturally because the voltage across the coil is the rate of change of the magnetic flux, which comes from the coil inductance and the coil motion. The electrical model for this is (where P is a constant that depends on the geometry of the coil and magnet) dφ dt

=

d(Li

−

dt

P y)

=

L

di dt

−

P

dy dt

= −iR .

Let the measurement be the voltage iR ; then the Laplace transform of this equation R/Ls = P , which, when gives the measurement-block transfer function as H(s) = RI(s) X(s) s +R/L added to the simulation diagram of the mass damper, gives the E (s) C (s) + ( ) U s Y (s) K G(s) model at right. (G(s) is the Laplace transform of the massdamper dynamics.) R s This exercise shows that L P some measurements come with s + R L additional dynamics. In this case, the measurement came with the dynamics associated with the inductor and resistor in the electrical circuit. The analysis of the control system must account for these to ensure that the system is stable despite the added dynamics. Once again, Simulink can rapidly analyze the effect of adding the dynamics. With a fewmousestrokes, themodelchanges to include thedynamicsof themeasurement(theycan be either a transfer function or differential equation). The resulting Simulink model with the response of the system is in Figure 2.19. (Instead of the model SpringMas_Vel_Contol, use SpringMass_Control_Sensor_Dynamics.) In the modified model, the transfer function block was used, and the values of the parameters in the model were set to make P(R/L) = 1 and R/L = 100. Note that the response of the control system with the value of the Gain = 100 (Figure 2.20) is almost identical in the two models, indicating that the sensor design does notchange the result. Once again, Simulink allowed us rapidly to answer the question of the effect of the additional sensor dynamics on the stability and response of the system.

64


Figure 2.19. One method of determining the velocity of the mass is to differentiate the position using a linear system that approximates the derivative.

Figure 2.20. Simulation result from the Simulink model.

2.6

Transfer Functions: Bode Plots

It is possible to build a simulation in Simulink that uses a sinusoidal input. You could use any of the simulations we have created so far. If the simulations are linear, all of the outputs will be sinusoids at the same frequency as the input. In steady state, the outputs will have different amplitudes and will cross through zero at times that are different from the input. Therefore, the amplitude and phase of the sinusoid at the various frequencies describe the sinusoidal response of the linear system. In 1938, Bell Labs scientist Hendrik Wade Bode (pronounced Boh-dee ) demonstrated that a plot of the frequency response of a linear system contains all of the information needed to understand what feedback around “

”

2.6. Transfer Functions: Bode Plots

65

the system would do. His Bode plots are still a mainstay of control system design. Let us see how to compute a Bode plot and how Simulink (and the Control System Toolbox in MATLAB) can do the work.

2.6.1 The Bode Plot for Continuous Time Systems Remember that the Laplace transform for an exponential gives the Laplace transform of a sine wave (see Problem 2.4). Using the approach in this problem, the Laplace transform of sin(ωt) is 1 1 eiωt − e−iωt ω 2i 2i L . = − = 2 s − iω s + iω s + ω2 2i





Now, if we want to know the response of a systemwith transfer function H(s) to a sinusoidal input, we use the fact that the transfer function is the ratio of the Laplace transform of the output divided by the Laplace transform of the input. Thus, if the input U(s) is the Laplace transform above, the output Y(s) is Y(s) = H(s)

ω s 2 + ω2

= H(s)

1 2i



1 s + iω

−

1 s − iω



.

The partial fraction expansion of the right-hand side above gives np

Y(s) =

 k =1

H (−pk ) s + pk

+

1 2i



H (−iω) s + iω

+

H(iω) s − iω



,

where the expansion is around the n poles (pk ) of the transfer function H(s) and the two imaginary poles from the sinusoidal input. The inverse Laplace transform of this has two parts: the transientresponse fromthe first termabove and the sinusoidalsteady state response from the second term above. The inverse transform gives the response of the system as np

y(t) =



αk H (−pk )e−pk t + |H(iω)| sin(ωt + φ(ω)).

k =1

The αk are constants that depend on the initial conditions and the proof of this assertion is Exercise 2.6. The Bode plot is a plot of 20 log10 |H(iω)| and φ(ω) versus ω on semilog axes. These show what the sinusoidal steady state response of the system is because the output is a sinusoid with amplitude |H(iω)| and phase φ(ω). Rapid and accurate calculation of the magnitude and the phase of H(s) uses the statespace representation of the system. The calculations areeasy to do in MATLAB, and there is a strong connection between Simulink and these calculations. The next section shows this.

2.6.2 Calculating the Bode Plot for Continuous Time Systems In Section 2.2, we showed that the transfer function for a linear system in state-space form is Y(s) = C[s I − A]−1 b. We now know that the Bode plot comes from this transfer function U(s)

66


iω and then computing the magnitude and phase of the resulting complex by letting s variable. Thus, the Bode plot contains the two terms =

Bode amplitude Bode phase

=

=

tan

20log10 1

−





C (iω I

Im(C (iω I Re(C (iω I

A)

−

−

A)

−

A)

1

−

1

b)

1

b)

−

−

 

b ,

.

It should be obvious that the calculation of these terms in MATLAB is trivial. The code would look something like this: function [mag, phase] = bode(a,b,c,omega) % This function computes the magnitude and phase of a % transfer function when the system dynamics are in % state-space form. The inputs are: % a - the A matrix of the system % b - the B matrix of the system % c - the C matrix of the system % omega vector of values for the frequency (rad/sec) for iw = sqrt(-1)*omega h = c*(iw-a)/b; hmag = abs(h); hphase = phase(h); mag = [mag hmag]; phase = [phase hphase]; end

This code segment is not terribly ef ficient, and it is prone to errors because it manipulates the matrix a in an unscaled form. In the Control Systems Toolbox, there is a built in function that does the calculation in a far more ef ficient, and therefore faster, form. The function is called bode, and it returns both the magnitude and the phase. The MATLAB command logspace generatesa vectorof equally spacedvalues foromegaso that a semilog plot of the magnitude versus the log base 10 of the frequency is a smooth curve. (To run the following examples you will need the Control Systems Toolbox.) The Control Systems Toolboxuses a MATLAB objectcalled an “lti object” that allows the calculations of the various linear system attributes such as the poles, zeros, Bode plot, and other plots in a simple and seamless way. The user may enter the lti object data using state-space models or transferfunctions (factored into the poles and zerosor as polynomials). It also allows the system to be discrete (i.e., in the form of a z-transform) or continuous (i.e., in the form of a Laplace transform). The rules for manipulating the lti object are part of the object’s de finition and the overloading of the various operations. “Overloading” anoperationisawayofrede fining themeaningof themathoperators “+ or ”, “* or / ”, so they are meaningful operations for the objects. For the lti object, transfer functions, pole-zero representations, and state-space representations are interchangeable. The “+ or ” adds or subtracts the transfer functions (with the automatic changes in the state-space model that the additions or subtractions imply), and the “* or / ” operators are multiplication and division of the transfer functions. (For these operations the order of the transfer functions increases, so the resulting polynomials change their order and the state-space model changes dimensions.) −

−

2.6. Transfer Functions: Bode Plots

67

Figure 2.21. Control System Toolbox interface to Simulink. GUIs allow you to select inputs and outputs.

It is beyond the scope of this book to discuss how to build a MATLAB object, but if the reader is interested, the documentation that comes with MATLAB shows how to create and use objects. The example shows how to create a polynomial object and how to then overload all of MATLAB s operations so that the appropriate rules for the polynomial arithmetic are invoked whenever the user types the symbols +, −, *, /, 2, etc. at the MATLAB command line. Let us use the model Spring_Mass2 that we created in Section 2.5 to illustrate the use of Simulink to get the lti model and the Bode plots. Open the model by typing Spring_Mass2 at the command line after changing to the NCS directory. We now want to invoke the connections between Simulink and MATLAB that the Control Systems Toolbox allows. Under the Tools menu in the model, select Control Design and the submenu Linear Analysis. This will invoke the linearization tool and the GUI shown at the left in Figure 2.21 will appear. This GUI allows you to select where you want the inputs and outputs to be. Any signal in the model can be selected for either of these; just go to the model, right click on the line that has the desired signal, and then select from the resulting dialog Linearization Points, and then from the submenu whether the point is an input, an output, or possibly a combination of the two. You need to select one input and one output, and when you are done, the GUI will resemble the GUI at the right in Figure 2.21. Now, click the Linearize button on the GUI and watch. A new LTI window will open and the comment LTI viewer is being launched will appear. The step response of the output will appear in the LTI Viewer window (see Figure 2.22). To see other types of plots, right click anywhere in the window and select the submenu Plot Types and then Bode. The result will be the Bode plots for the two outputs from the model shown below. (There are two outputs because the Simulink model has a 2-vector output that has both the position and velocity of the mass.) From the LTI Viewer window, you can export the lti object to MATLAB for further analysis. To do this, select from the File menu in the viewer the Export command, and then, in the LTI Viewer Export GUI that comes up, select the model to export. The GUI ’

ˆ

“

“

”

”

“

”

“

”

“

“

”

”

68


Figure 2.22. LTI Viewer results from the LTI Viewer.

gives a tentative name of sys to the lti model, so if you select this it will create the object in MATLAB. To export now, click the Export to workspace button. Nowgo to the MATLAB command line and type sys. You will see that MATLAB now contains the lti object sys, and it is a state-space model. If you want to convert this to a transfer function, type tf(sys). All of the MATLAB commands and the resulting answers are in the MATLAB code segment below. “

”

>> sys a = Spring_Mass/ Spring_Mass/

Spring_Mass/ 0 -1

b = Spring_Mass/ Spring_Mass/

Spring_Mass/ 0 1

Spring_Mass/ 1 -0.2

2.7. PD Control, PID Control, and Full State Feedback

69

c = Spring_Mass/ Spring_Mass/

Spring_Mass/ 1 0

Spring_Mass/ 0 1

d = Spring_Mass/ Spring_Mass/

Spring_Mass/ 0 0

Continuous-time model. >> tf(sys) Transfer function from input "Spring_Mass/Step (1)" to output... 1 Spring_Mass/Integrator (pout 1, ch 1): ----------------sˆ2 + 0.2 s + 1

Spring_Mass/Integrator (pout 1, ch 2):

s ----------------sˆ2 + 0.2 s + 1

Other plot types and results are available with the LTI Viewer. Spend some time exploring them and work with some of the problems at the end of this chapter.

2.7 2.7.1

PD Control, PID Control, and Full State Feedback PD Control

The simple control systems that we have investigated so far were mostly second order systems (i.e., systems described by second order differential equations). There are good reasons for thoroughly exploring this class of system, because any mechanical system that has a moving mass creates a second order differential equation via Newton ’s law of motion 2 f (f = ma implies d dt x2 = m ). We saw in the spring-mass example that if the position and derivative (speed) were the feedback variables any desired response is possible by appropriate choices of gains. Control engineers call this PD control (P for position and D for derivative). The simplest way to see that this control will allow the system to be set arbitrarily is to use the state-space formulation of the system. In state-space form the second order differential equation is d x dt

=

y=

 

0 −a 1 0

1 −b 0 1



   x+

0 a

u,

x.

The first component of the vector measurement y is the position, and the second component

70

Chapter 2. Linear Differential Equations, Matrix Algebra, and Control Systems Position Control Gain

P

x' = Ax+Bu y = Cx+Du Step at 1 second

Position

State-Space Model of Spring

D

Position and Velocity

Velocity

Velocity Control Gain

Figure 2.23. Using proportional-plus derivative (PD) control for the spring-massdamper.

is the derivative (because of the de finition of the state x). Therefore, the PD control is u

=



P

D

  y

0 D

P 0

=



x + uin ,

where the input to the closed loop system is uin . Substituting this back into the state-space model above gives d x dt

=

=

 

0 −a

1 −b

0 a(P − 1)

    0 a

x+

1

aD

−

b

0 D

P 0



x + uin

x + uin .

It is clear from the matrix that by selecting P and D the entire closed loop system can respond in any desired way. This makes PD control the simplest and easiest to use in a mechanical system, and as such, it should be the first choice for a simple design. There is, however, one problem with a simple PD controller. Most systems require that the response in steady state should have some predictable value. (When the controller is trying to track the input, the desired value is the steady state value of uin .) In the above system, we denote the steady state value of x by S. S comes from setting the derivative of x to zero. (The name steady state comes from the fact that nothing in the system is changing; hence the derivative of all of the states must be zero.) Therefore, S

= −



0 a(P − 1)

−1

1 aD

−

b



uin .

Figure 2.23 is the Simulink model PD_Control from the NCS library. Simulating this model with a , b, P , and D equal to 100, 2, 1, and 0.2, respectively, shows that the steady state value of the output is 0.5 when the input steady state value is 1 (see Figure 2.24).


71

Position 0.8

0.6

0.4

0.2

0

0

2

4

6

8

10

12

14

16

18

20

12

14

16

18

20

Velocity 3

2

1

0

-1

0

2

4

6

8

10 Time

Figure 2.24. Response of the spring-mass-damper with the PD controller. There are no oscillations.

It is obvious that the steady state value of neither component of x is equal to uin . We now address the question, How does one force the steady state value of one of the components to match uin ?

2.7.2

PID Control

We will use the block diagram of the PD control system we developed above to see how it is possible to force one of the state components to be equal to the input in steady state. If the control system were only trying to improve the response with no attempt to track the input (the step), the response we got above would be fine. Assume that we want the response to track the input precisely (for example, this control might be for the steering system in an automobile so that the angle of the steering wheel will always be proportionally to the angle of the front tires). In this case, the steady state response is not adequate. One possibility would be to add a gain in front of the step input that doubles the input. Then the value of the output would be 1 when the step is in steady state, but it is easy to see that any change in the values of a , b, P , or D would change this relationship. So what can we do?

72


Integral Gain 1 s

Position and Velocity

I

Integrator1 x' = Ax+Bu y = Cx+Du

P Step at 1 second

Position Gain1

Position

State-Space Model of Spring

D Velocity (Derivative) (Derivative) Gai n

Figure 2.25. Proportional, integral, and derivative (PID) control in the springmass-damper system.

Think about the error between the input and the output as measured by the difference between the value of uin and the position measured in the system. (This is the output from the difference block in the diagram above.) If we were to insert an integrator into the model at this point, then, were the steady state error at this point S , the output of the integrator would be St . The input to the second order spring-m spring-mass ass system system is now St , so it begins to move away from the steady state value of 0.5. As it moves, the feedback causes the output to get get clos closer er to the the stea steady dy stat statee valu valuee of the the inpu input. t. In fact fact,, when when the the outp output ut matc matche hess the the inpu inputt in steady state, the difference will be exactly zero and the integrator stops. The result will be that the input and the output match exactly and the states will not change values because the integrator output will be an unchanging constant. Let us do this in the model. Add an integrator from the continuous library library in Simulink so it integrates the output of the difference block (the difference between the input and the measured position of the mass). Multiply the integral by a gain I and then add the result to the difference (see Figure Figure 2.25). Simulate the system and verify that the output and input match. (This model is called PID_Control in the NCS library.) While you are doing the simulation of this model, look carefully at the difference between the input and the output (add another Scope block displaying the difference if you want to make the difference clear), and verify that we have achieved zero steady state error, as we set out to do. Also, experiment with different values for the integral gain I (the model uses I 7) to see what effect this has on the time it takes to reach steady state. The result of simulating the PID controller with the nominal values is in Figure 2.26. The figure clearly shows that the steady state value of the mass position is exactly one, and the error between the desired and actual position is zero. =

2.7.3 2.7.3

Full Full State State Feedb Feedback ack

Most models used in textbooks to illustrate control concepts are second order. The reasons for this are that •

second order systems are easy to visualize and manipulate;

•

second order systems are the basic building blocks of mechanical systems through Newton’s equations of motion.


73

Position 1.5

1

0.5

0

0

1

2

3

4

5

6

7

8

9

10 10

6

7

8

9

10 10

Velocity 5 4 3 2 1 0 -1

0

1

2

3

4

5 Time

Figure 2.26. PID control of the spring-mass system. Response to a unit step has zero error. error.

Unfortunately, the real world is not restricted to second order systems. What do we do if we want the response of a higher order system to match some desired response? The PD contro controlle llerr we develo developed ped used used feedba feedback ck of both both the positi position on and the deriva derivativ tivee of the output, which turned out to be the full state of the model when we used the output y dy and its derivativ derivative, e, dt , as the the stat state. e. In othe otherr word wordss the the full full stat statee of the the syst system em was was assu assume med d to be available (i.e., measurements of the entire state were assumed to exist). This generalizes very easily. easily. We have seen that any linear system given by a differential differential equation of the form d n y(t) dt n

+

an−1


+

an−2


+ · · · + a0 y(t) =

bu(t)

has the vector-matrix (state-space) form

d x(t) (t ) dt

=

   

0 0 .. . 0 −a0

1 0 .. . 0 −a1

0 1 .. . 0 −a2

··· ··· ··· ··· ···

0 0 .. . 1 −an−1

   x(t)  (t )

+

0  0   ...  u(t).  0  b

74


If the entire state is available for control (meaning that y and all of its derivatives up to the (n − 1)st are available), then the controller can be u(t) = K1 x1 + K2 x2 + · · · + Kn xn = [ K1

K2

···

Kn ]x = Kx.

When used in the state-space equation, the result is

  d (t)  =  dt x

0 0 .. . 0 −a0

1 0 .. . 0 −a1

0 1 .. . 0 −a2

··· ··· ··· ··· ···

   

0 0 .. . 1 −an−1

x

0  0  . (t) +  ..   0 

Kx.

b

Since K is a row vector, the product of the b vector and K is an n × n matrix that, when added to the state matrix, results in

  d (t)  =  dt

0 0 .. . 0

1 0 .. . 0

0 1 .. . 0

bK1 − a0

bK2 − a1

bK3 − a2

x

0 0 .. . 1

··· ··· ··· ··· ···

bKn − an−1

   

x(t).

Now every one of the original coef ficients in the differential equation has an additive term from the gain matrix, and as a result, the gains change the entire last row of the state matrix. This means that any desired desired response is possible possible for the entire entire system. This is “full state feedback.” There are some subtleties associated with full state feedback having to do with statespace models that are not in the form we used here (i.e., where the state is not the output and all of its derivatives). The result we showed is still true.

2.7.4 2.7.4 Getting Getting Derivat Derivative ivess for PID Control Control or or Full State State Feedbac Feedbackk The assumption in all of the discussions above was that some device measures all of the derivatives of the output. What happens when this is not so? Three methods are available to create measurements for a full state feedback implementation. The first rst appr approa oach ch to crea creati ting ng the the deri deriva vati tive ve of a vari variab able le is to use use a line linear ar syst system em that approximates the derivative. One possibility is to use the transfer function H d d (s) =

αs s+α

=α−

α2 s+α

.

This is not exactly the derivative, but it is close. (It is the derivative for all motions that are 1 faster than approximately t max max = α .) The second approach is to use a digital approximation approximation to the derivative. Remember that the derivative is approximately dx dt

≈

x(kt + t) − x(kt) t

=

1 t

(xk+1 − xk ).


75

This digital approximation is reasonably good if the sample time is fast relative to the rate of change in the position. In the third approach, the fact that the derivative of the output (or even linear combinations of the state of the system) is available through related measurements. One example of this would be a measurement of the force at the support point of a spring-mass system. The force measured would be the sum of the gravity force ( mg), the spring force ( Kx ), and ). Thus, this measurement, the damping force (D dx measurement, combined with a measurement of the dt position, can compute the velocity. velocity. If the force measurement is f Meas M eas , the acceleration of the mass is the result of subtracting the three accelerations applied to the mass. They are •

•

•

the accelerat acceleration ion due to the spring spring (obtaine (obtained d using using the position position measurem measurement ent multipli multiplied ed by the spring constant), gravity acceleration, the acceleration from the damping (obtained as feedback by integrating this acceleration estimate). Thus a pseudomeasurement of velocity is d 2 x d t 2

 

f Meas =

m

Meas

−

g

K −

m

xMeas

D dx −

m dt

where dx d t

 

Meas

d 2 x

=

   dt 2

 

,

Meas

dt .

Meas

In this implementation, we have an implicit feedback loop because we integrate the pseudomeasurem measurement ent of accelera acceleration tion (the left-hand left-hand side of this equation) equation) for use on the right dx (as dt Meas ). This approach uses a priori values for the spring constant and the force of gravity (which means that in the actual control system they need to remain constant over time), but despite this, control systems are often built using this type of computation to provide a pseudomeasur pseudomeasuremen ementt of the velocity velocity (or other states). states). Using Using this approach, approach, you can recompute the gains so that the feedback explicitly uses the measured values of force and displacement, displacement, thereby eliminating the explicit computation of the derivative. Once again, this requires that the parameters in the model be constant or, if they are not, that the gains change as the parameters vary. These These equati equations ons can be assemb assembled led into into the Simuli Simulink nk diagra diagram m in Figure Figure 2.27; 2.27; the linear linear system that results derives the rate using no derivatives (actually, it uses only integration). Theinitialcond Theinitialconditi ition on forthe derive derived d rate rate in this this model model is zero. zero. That That means means that that when when the systemstart systemstarts, s, there there could could be an error error betwee between n the estim estimate ate of the rate rate and the actual actual rate. rate. It is easy to show that this error goes to zero as time evolves, so that the derived rate approaches thevelocityof thevelocityof themass. (Try (Try to demons demonstra trate te that that this this is true. true.)) To verifythesethreestrat verifythesethreestrategi egies es for reconstructing the derivative, let us build a Simulink model that implements all of these measurements. measurements. The model will contain the three different methods for deriving rate that we described above in a masked subsystem. The model is PID_Control_Vel_Estimates in the NCS library. It is in Figure 2.28.



76

Chapter Chapter 2. Linear Linear Differe Differential ntial Equations, Equations, Matrix Matrix Algebra, Algebra, and Control Control Systems Systems Damping for Rate Reconstruction damp

Externally Applied Force

Integrator

1

1 s

1/m

2

1

Estimate of Velocity

Divide by mass

Position of Ma ss

Derived Rate

Kspring

Spring Constant for Rate Reconstruction

Figure 2.27. Getting speed from measurements of the mass position and force applied to the mass.

P Position Gain

Integral Gain

Spring Constant

I

-K-

1 s Step at 1 second

Ex t er nal F orc e

Integrator

1 s

1/m 1/ mass

Veloc it y

Integrator1

1 s

Pos it ion

Integrator2

-KDamping

[Estimate] From

Velocity (Derivative) Gain D

Position and Velocity Plot

Externally Applied Force Derived Rate Position of Mass

[Estimate] Send Rate Estimate ot SCope using Goto

Select Method for Deriving Rate for Use Use in PID Controller (Dialog al so allows testing testing of parameter variations)

Figure 2.28. Simulink model to compare three methods for deriving the velocity of the mass.

The derivation of the rate is in the subsystem at the bottom center of the model. Right click on this masked subsystem and select “Look Under Mask.” The three rate estimation approaches are in this subsystem, which is shown by Figure 2.29. We are using two new blocks in this model. The first is the Multipor Multiportt Switch. This switch uses the variable “Select” to determine which of the three inputs come out of the block.


77

Damping for Rate Reconstruction damp Externally Applied Force

Integrator

1

1 s

1/m Divide by mass Spring Constant for Rate Reconstruction

Select

Estimate3 of Velocity

Constant


Kspring 1/Samp_Time

Position of Mass

Difference

2

z-1

1/Samp_T

Derived Rate


1

z Zero-Order Hold

Multiport Switch alpha.s


s+alpha Transfer Fcn

Figure 2.29. Methods for deriving rate are selected using the

multiport switch.

“

”

If you you doub double le clic click k on the the subs subsys yste tem m bloc block, k, the the mask mask show shown n at righ rightt open openss. The The pul pulll-do dow wn menu enu at the top deter eter-mine miness the the valu valuee of SeSelect. In the the mas mask edi editor (obtained by right clicking the block block and selectselecting Edit Mask), Mask), you will seethat thepull-do thepull-down wn menu menu selectio selection n makes makes the value of Select equal to 1, 2, or 3. The The way way the the swit switch ch is set set up, up, this this numb number er-ing ing sche scheme me need needss to be reve revers rsed ed,, so in the the IniInitializ tializati ation on part part of the editor we have the code Select Select = 4-Sele 4-Select ct. All of the data used in the estimati estimation on go under the mask using using the mask dialog, which we will discuss in more detail in the next chapter. The last blocks that we use here, but have not used before in our models, are the GoTo GoTo and the From blocks. blocks. They are very convenient convenient blocks blocks to use to help keep your model looking uncluttered. As their names imply, imply, they allow a signal to propagate from one location in the model to another without the use of a line. We use these to send the signals

78


to the Scope blocks for plotting. (In general, just as with GoTo statements in code, avoid using these blocks.) Open this model, and run it. Use the mask dialog to change the subsystem values and to experiment with the different rate derivation approaches. Verify that the various derivations of the rate of the mass are reasonably accurate. Try changing the values of the various parameters (alpha, Samp_Time, and the mass, spring constant, and damping used in the derivation of the rate). These change when you type the new values right into the dialog box of the subsystem mask. Control systems engineers call this method for deriving the rate an “observer.” If you wish to explore this in more detail, and to explore observers for more complex systems, see [1]. In this chapter, we have explored linear systems and the basics of feedback controllers that will achieve a desired result. In the next chapter, we look at some real-world complexities that make systems nonlinear. The first example is chaotic in the mathematical sense, and subsequent examples extend the pendulums that we explored in Chapter 1 to rotations that are more complex. Along the way, we explore how one can incorporate nonlinear devices in our simulation using Simulink primitives (blocks in the libraries). When the desired behavior is not available as a block, we look at how to build the requisite nonlinear behavior out of the primitives.

2.8 Further Reading Dynamic systems and state-space formulation is discussed in a nice book by Scheinerman [33]. State-space models also are the backbone of modern control design, as in Bryson [6] and Anderson and Moore [1]. A pseudomeasurement of a system state can come from a mathematical description of the function that relates the state to the measurement, or from the differential equations describing the state and the measurements. Applications of both approaches are available. For example, State flow (described in Chapter 7) comes with a demo called “fuelsys” that illustrates the first approach. The Simulink blocks in this demo illustrate how the computer provides control of the fuel injectors on a car. It also shows how pseudomeasurements provide the missing data from a failed sensor, using redundant information contained in related sensors. This demo is fun to run, since it allows you to throw switches that fail the sensors and see the result. The Help file that accompanies the demo has more details.

Exercises 2.1 In the text, the differential equation in vector-matrix form or the state-space model used the output and its derivatives up to the (n − 1)st order to create the state vector. Verify that the state-space model in the text and the original differential equation are the same. How would you modify the model to add more inputs? (Hint: Make the input a vector.) How would you add more outputs? 2.2 In the text we asked that you verify that the derivative of eAt is AeAt and that the inverse of eAt is e −At . The latter result comes easily from showing that the derivative

Exercises

79

of the product eAt e−At is zero (so the product is a constant). If you do this, you need to show that AeAt = eAt A, i.e., that these matrices commute. 2.3 The calculation of the transition matrix phi and the input in fluence matrix gamma in the code c2d_ncs uses the matrix

  A

b

0

0

. Verify that this matrix creates the phi

and gamma matrices de fined in the text. (Use the hint in the text.) 2.4 Create the Laplace transform of the cosine and sine using their exponential de finitions (as suggested in the text). 2.5 The motion of the spring, mass, and damper is the solution of the differential equation in state-space form d dt

x(t) =



0 −ωn2

1 −2ς ωn



x(t) +

  0

1 M

u(t).

Solve this equation when the input is zero (with the damping ratio less than one, the initial position of x0 and the initial velocity of v0 ) to show that the motion of the mass is

 

x(t) = e−ςω n t x0 cos

v0 ωn

+ ς x0

 

1 − ς 2 ωn t +

1−

ς 2

sin





1 − ς 2 ωn t

.

Create a Simulink model that uses this solution and a simulation of the motion to compare the numerical accuracy of the Simulink solution for different solvers. 2.6 In the text, we used the Laplace transform to develop the response of a system to a np sinusoidal input as y(t) = k =1 αk H (−pk )e−pk t + |H(iω)| sin (ωt + ϕ(ω)). Verify this result. Can you think of a way that you can use Simulink to calculate the transfer function H(iω) without the use of the Control System Toolbox?



2.7 Experiment with the three different control system models (PD, PI, and PID). Try changing the gains and observing the results. Try adding nonlinearities to the models (in particular, limit the control to a maximum and minimum value) and redo the experiments. What can you say about the resulting responses?

Chapter 3

Nonlinear Differential Equations

We have already seen some nonlinear differential equations (for example, the clock and the Foucault pendulum dynamics in Chapter 1). In this chapter, we will delve into some of the more interesting aspects of nonlinear equations, including a simple example of chaos, the dynamics of a rotating body, and the modeling of a satellite in orbit. We will also explore an interesting way of modeling motions in three dimensions using a four-dimensional vector called a quaternion. This chapter illustrates some advanced Simulink modeling and a major feature of Simulink, namely the ability to create reusable models. For the beginning Simulink user these models might not be too useful, but users that need to model complex mechanical systems will find a great many uses for many of the examples presented here (and thereforecould cutand paste the models in the NCS library directly into thenew models that they might want to build). In addition, we illustrate the following: •

how to make subsystem models that can be stored in a library for later use;

•

how to annotate a model so the simulated equations are written out in the annotation;

•

how to create a subsystem and then layer a mask on top of it, the mask providing the user with a dialog that allows parameters in the model to be transferred into the model (just as a subroutine ’s parameters are transferred into the subroutine during execution).

3.1 The Lorenz Attractor One of the more interesting uses of simulation is the investigation of chaos. Chapter 7 of Cleve Moler ’s Numerical Computing with MATLAB [29] describes the Lorenz chaotic attractor problem developed in 1963 by Edward Lorenz at MIT. The differential equation is a simple nonlinear model that describes the behavior of the earth ’s atmosphere. Cleve analyzes the differential equation ’s “strange attractors ”—the values at which the differential equation tries to stop but cannot because the solution, although bounded, is neither convergent nor periodic. The model tracks three variables: a term that tracks the 81

82

Chapter 3. Nonlinear Differential Equations

convection of the atmospheric flow (we call it y1 ), the second (y2 ) related to the horizontal temperature of the atmosphere, and a last (y3 ) related to the vertical temperature of the atmosphere. The Lorenz differential equation has three parameters, σ, ρ , and β ; the most popular values for these parameters are 10, 28, and 8/3, respectively. The differential equation is three coupled first order differential equations given by

˙ = −βy + y y , ˙ = −σy + σy , ˙ = −y y + ρy − y .

y1 y2 y3

1

2 3

2

3

2 1

2

3

Alternatively, in state-space form (even though the equations are nonlinear),

 −β 0 d (t) =  0 −σ dt x

−y

2

ρ

y2 σ 1

−

 

x

(t).

We will not duplicate the analysis from Moler ’s book; suf fice it to say that the equation has two “fixed points” where, if the equation has an initial condition at t 0 equal to one of these solutions, the derivatives would all be zero and the solution would stay at these points (hence the name fixed points). However, both of these points are unstable in the sense that any small perturbation away from them will cause the solution to rapidly move away and never return. (If the solution gets close again to one of these points, it will rapidly move away again.) The two points are determined from

=

 ρ−1   ρ−1   √   √  y =  β(ρ − 1)  and y =  − β(ρ − 1)  . √ √ β(ρ

− 1)

−

β(ρ

− 1)

We created the nonlinear equation in Simulink by building each of the differential equations around the integrator block in the Continuous library, and at the same time, we use a new block to plot the resulting solution on the two axes ( y1 versus y2 and y1 versus y3 ). The model is in the NCS library, and it opens using the command Lorenz_1. The model is in Figure 3.1 along with a typical set of two-dimensional plots. This model uses a few new tricks that you should explore carefully. First, the two dimensional plots are from the Simulink Sinks library. The scales on these plots are set by double clicking on the icon. (Review the settings when the model is opened.) In order to have a new initial condition every time the model runs, we have added a Gaussian random perturbation to each of the initial conditions in the integrator blocks. The MATLAB command randn is used in the “Constant ” block, and each integrator has its initial condition set from the “external” source (so they actually appear in the diagram; this is a good practice when the models are simple because the initial conditions are visible, but when the model is very busy, this approach can make the diagram too busy). To make the initial conditions external, you need to open the integrator dialog by double clicking the integrator block, and then select external as the source of the initial conditions.

3.1. The Lorenz Attractor

83

Figure 3.1. Simulink model and results for the Lorenz attractor.

The two x -y plots show the two fixed points for the Lorenz system. To change the values of Rho, Sigma, and Beta, simply type in new values at the MATLAB command line (see [29] for some suggestions).

3.1.1 Linear Operating Points: Why the Lorenz Attractor Is Chaotic Since Release 7 of Simulink, theControl SystemToolbox link has a very robustlinearization algorithm. Two blocks appearing in the Simulink Model-Wide Utilities library can be used to get a linear model of a nonlinear system. The first of these blocks is the time-based linearization, and the second is trigger-based linearization. The first will create a complete state-space model at the times speci fied in the dialog that opens when you double click the block. Unlike the Simulink Scope block, this block does not require a connection to the model. The block causes the simulation to stop at either the times or the triggers, and then a program from the Control System Toolbox, called linmod (for a continuous time model) or dlinmod (for a discrete time model), creates the linear state-space matrices A, B , C , and D . We changed the model Lorenz_1 to create the linearizations using the trigger block. The triggers come from the Pulse Generator in the Sources library. In this new model, called Lorenz_2 in the NCS library (Figure 3.2), the diagram uses the “GoTo” and “From” blocks in the Simulink Signal Routing library to make it easier to follow. These blocks allow connections without a line, thereby simplifying the diagram. The model was also redrawn a little to emphasize the state-space form of the Lorenz differential equations. The state-space form of the equations is in the text block in the model (created using the “Enable Tex Commands” option in the Format menu). When this model runs (slowly because the linearization is taking place every 0.01 sec of simulation time), the linear models are created and stored in a structure in MATLAB. We use the structure created by linmod — called Lorenz_2_Trigger_Based_Linearization_—to investigate the stability of the

84

Chapter 3. Nonlinear Differential Equations Lorenz Attractor Element 1 of the matrix Ay y1

Beta y2 y2*y3

y3

1 xo s

Nonlinearity: Prod of y2 and y3

y1

Simulink model for the Lorenz Attractor, whose nonlinear differnetial equation is given by:

y1

dy/dt =

Rho-1+10*randn

A

where: A =

Element 2 of the matrix Ay

| -β y2

0

| 0 -σ | - y ρ 2

y3

Sigma

1 xo s

y

y2

y2

sqrt(Beta*(Rho-1))+20*randn

y 2 | σ -1

| |

. Pulse Generator

Element 3 of the matrix Ay y3

y2

Rho

1 xo s y1*y2

y1

sqrt(Beta*(Rho-1))+20*randn Nonlinearity: Prod of y1 and y2

Trigger-Based Linearization. y3

y3

Plot Subsystem (Uses Data from GOTOs)

This set of blocks creates a linear model at the times triggered by the pulse generator. The result s are stored in a structure in th e MATL AB Workspace.

Figure 3.2. Exploring the Lorenz attractor using the linearization tool from the control toolbox.

operating points and at the same time understand why the Lorenz attractor response is chaotic. A small MATLAB program, called Lorenz_eigs.m in the NCS library, plots the first 40 eigenvalues (corresponding to the first 0.4 sec of the time history). The code in this M-file is axis([-25 15 -25 25]); axis(’square’) grid hold on nlins = tstop/0.01 for i = 1:nlins/50 Lorenz(i).eig=eig(Lorenz_2_Trigger_Based_Linearization_(i).a); plot(real(Lorenz(i).eig),imag(Lorenz(i).eig),’.’) drawnow end hold off

Figure 3.3 shows the eigenvalue plot. Let us compare this plot with the solutions generated when the simulation ran. The plot below the eigenvalue figure is the simulation results for the state variable y1 plotted against the state variable y3 . The initial position in this plane was about (10, 20).

3.1. The Lorenz Attractor

85

Eigenvalues of the Lorenz System (Every 0.01 sec. of the 20 sec. simulation) 30

20

s e u l a 10 v n e g i E e h t f 0 o t r a p y r a n i g -10 a m I

t=0

t=0

t=0 -20

-30 -20

-15

-10 -5 0 Real part of the Eigenvalues

5

10

Figure 3.3. Eigenvalues of the linearized Lorenz attractor as a function of time.

From the eigenvalues plot above we can see that the initial condition was unstable; when the simulation starts there are a complex pair of eigenvalues that have a real part of 2.91. The eigenvalues are almost in the left half plane, and during the next four times they move toward the left half complex plane. From the sixth eigenvalue on, they have moved into the left half complex plane, indicating that the solution now is

86


stable. The complex pair indicates that the solutions are oscillatory, but since they are in the left half plane, it also indicates that the amplitude is decreasing, so the solution is spiraling in toward the attractor at the bottom of the time-history plot. However, as the solution approaches this attractor, the complex pair of eigenvalues again moves into the right half plane (the eigenvalue plot ends before this happens), indicating that the solution is again locally unstable, and it begins to diverge away from the attractor. This behavior is consistent with the fact that the attractors are both unstable. (As an exercise, set the value of y2 in the A matrix to the values β(ρ 1), and compute the eigenvalues of the matrix.) It is quite instructive to play with this model and compare the eigenvalues with the solutions. Some of the questions you might want to ask: Do the three eigenvalues always appear as one complex pair and one real? What is the importance of the magnitude of the real part of the complex pair? What is the importance of the imaginary part of the complex pair? This techniqueof analyzinga systemby linearization about thecurrent operating point is very useful as systems get complex. This approach is particularly useful when systems are so large that creating linear models analytically is virtually impossible.

±√ −

3.2 Differential Equation Solvers in MATLAB and Simulink Simulink uses the same solvers that are in the MATLAB ode suite (although they are Ccoded in an independent way and automatically linked by the Simulink engine to solve the differential equations modeled in the diagram). Chapter 7 of NCM describes these and how they work, so we will not go into the details here. When you create a new Simulink model, the default solver is the variable step ode45, and the default simulation time is 10 sec. In addition, the Relative tolerance is set to 1e-3 (or approximately 0.1% accuracy), and the Absolute tolerance is set to auto. The solver also allows the user to force a Max, Min, and Initial step size if desired, but these are all set initially to auto. The solver options allow the Zero crossing control to be set so that zero crossing detection occurs for all blocks that allow it (or, optionally, none of the blocks). The default option for zero crossing is Use local settings, where the solver checks to see if a zero crossing has occurred for those blocks that this option is set to on. For example, the absolute value block has the dialog shown above. If the Enable zero crossing detection check box is not checked, then in the local settings mode, this block s zero crossing is disabled. Note that if the zero crossing control is set to Enable all, this check box selection does not appear. Some simple numerical experiments were performed (in the spirit of Section 7.14 in NCM) using the Lorenz model above. To do these experiments, we execute the Simulink “

”

“

”

“

“

”

”

“

“

”

”

“

”

“

“

”

”

“

’

”

3.3. Tables, Interpolation, and Curve Fitting in Simulink Table 3.1.

87

Computation time for the Lorenz model with different solvers and tolerances.

Solver Tolerance Tolerance

Type 0.001 0.000001

Ode45 0.0370 0.3670

Ode23 0.1445 0.4680

Ode113 0.1802 0.3749

model from the MATLAB command line. Type in the following code at the MATLAB command line and then use the arrow keys to rerun the same code segment: tic;sim(’Lorenz_2’);te = toc; te

In this set of MATLAB instructions, the sim command causes MATLAB to run the Simulink model (the argument of the command tells MATLAB what model to run —it is a string). The value of the elapsed time for the simulation is te. In between each execution of the simulation using the sim command, open the Con figuration Parameters dialog in Simulink; use the Simulation pull-down menu or click on the model window and type Ctrl + e. Then change, in turn, the solver or the relative tolerances, and—this is very important—save the model and re-execute the commands above. Table 3.1 shows the result of my experiment; I used a Dell dual Pentium 3.192 GHz computer with 1 gigabyte of RAM. With the understanding that you have obtained from what we have done so far in this book, and the discussion in NCM, it should be clear that there are no hard and fast rules that apply to selecting a solver, a tolerance, or any of the other parameters in the Simulink Configuration Parameters pull-down dialog. The best way to understand the simulation accuracy for any model you will produce is to try different solvers with different tolerances. If you think that the solvers are jumping through time steps that are too big, you can set the maximum step size, and if you think that the solver is taking too long, you can try increasing the minimum step size. In addition, if you are interested in the solution of a system that has some dynamics that change very fast, but you do not really care about the details of these fast states, then the “stiff ” solvers are best (see Section 7.9 of NCM). We will see some examples in later chapters that require stiff solvers, and when we do, we will be able to explain why they are used and how to use them.

3.3 Tables, Interpolation, and Curve Fitting in Simulink There are times when a simulation requires tabulated data for a function. Examples of this aresystems that areoperating in flowing air or water (aircraft, rockets, submarines, balloons, parachutes, etc.). In many of these applications, the tabular data represents the gradient of a nonlinear function of one or more variables. The gradient measurement comes from wind tunnels, water tanks, or actual measurements of the system in its environment (a full-scale model of an airplane or automobile). How do you use such data in Simulink? Simulink implements most of these curve- fitting tools in a set of blocks that are in the Lookup Tables library. This library has a variety of tables. The simplest is the “Lookup Table”; the versions that are more complex are the 2-D and n-D (for functions that give two outputs or more). A block that does interpolation (n-D), using what is called a PreLookup of

88


the input data, simpli fies the tables. The “PreLookup Index Search ” block that goes along with this block is also in the library, but since this block works in conjunction with the PreLookup block, it is not a stand-alone block. The last block in the library is the “Lookup Table Dynamic” block. We will investigate each of these block types and illustrate when to use them in the following examples.

3.3.1 The Simple Lookup Table To illustrate the simple lookup table, let us return to the home heating system model we introduced in Section 2.4.1. The original model used an outdoor temperature variation of 50 deg F with a 24-hour sinusoidal variation of 15 deg. This might be a typical diurnal variation in the outdoor temperature, but to compute the fuel usage for a speci fic winter ’s day we want to use the actual outdoor temperature. Assume that you have made measurements of the temperature during a particular cold snap when the temperature plummets in a 48-hour period from an average of 40 deg to about 10 deg. Figure 3.4 shows the temperature over a 24-hour day (sampled every 15 minutes). The data were actually generated from MATLAB using the following code: Time = (0:.25:48)*3600; Temp = 50-0.75*Time/3600 – 5*sin(2*pi*Time/(24*3600)+pi/4)+randn(size(Time))

How do you insert this temperature variation into the model? Open the model by typing Thermo_NCS at the MATLAB command line. Find the blocks with the constant 50 and the sinusoid that models the external temperature variation along with the summation block that adds the two, and remove them from the model. Open the Simulink library browser and find the Lookup Tables library. From this library, select the block called “Lookup Table” and insert it into the model. The input to this block will be the simulation time that is the “Clock ” block in the Simulink Sources library. Get one of these blocks and insert it into the model so that it is the input to the lookup table. Then, double click on the Lookup Table block and enter the values of the data using the MATLAB commands above. The model that results is in Figure 3.5, and the Lookup Table dialog should look like Figure 3.6. The result will now show the home heating systemoperating with the outdoor temperature given by the tabular values in the lookup table. The figure also shows the dialog that opens when you double click the Lookup Table block. It allows editing of the data (using the “Edit” button). You can also enter the data as a MATLAB vector using MATLAB’s vector notation or pasting a vector of numbers into the dialog box. If, for example, the data is in an Excel spread sheet, it can be pasted into the vector of output values window and then surrounded by the square brackets “[” and “]” to make the data into a MATLAB vector. Notice also that the dialog box allows you to select the interpolation method you want to use. The default is interpolation and extrapolation. This means that the data interpolation is linear between time values, and the last two data points in the table project the data beyond the final time. The other options are interpolation with end values used past the last data point in the table, the value nearest to the input value, or the values below or above the input value. The block will also provide a value at fixed times by setting the sample time. In the simulation the value of this is set to −1, which causes the value to be computed at

3.3. Tables, Interpolation, and Curve Fitting in Simulink

89

Figure 3.4. Pseudotabulated data for the external temperature in the home heating model.

deg F to deg C 70

F2C

Blower Heat Output

blower cmd

Terr

deg C to deg F

Mdot*ha

Tset

Set Point Thermostat

Thouse

C2F

Indoor vs. Outdoor Temp.

Plots

F2C Clock

Lookup Table for Outdoor Temp.

deg F to deg C

House Heat Cost ($)

1/s

Mdot*ha Heat Output Cal/Hr.

cost Cost of Fuel $/Cal/Hr

Figure 3.5. Home heating system Simulink model with tabulated outside temperature added.

90


Figure 3.6. Dialog to add the tabulated input.

every simulation step when the input changes. (Since the input to the block is the clock, this is every integration step used by the solver.) The Help button on the dialog opens the MATLAB help and provides a very good description of how this block works. If you have not created the model yourself during the above discussion, it is available in the NCS library using the command Thermo_table. The 2-D Lookup Table block computes an approximation to z = f ( x , y ) given some x , y , and z data points. The dialog block for this lookup table has an edit box that contains the “Row index input values” (a 1 × m vector corresponding to the x data points) and an edit box with the “Column index input values” (a 1 × n vector of y data points). The values for z are in the edit box labeled “Matrix of output values” (an m × n matrix). Both the row and column vectors must be monotonically increasing. These vectors must be strictly monotonically increasing in some speci fic cases. (See the Help dialog for more details on what these cases are.) By default, the output is determined from the input values using interpolationextrapolation—that is, linear interpolation and extrapolation of the inputs past the ends. The alternatives available are “Interpolation-Use End Values, ” “Use Input Nearest,” “Input Below, ” or “Input Above,” with the same meaning as for the 1-D Lookup Table. It is important to use the proper end extrapolation method when the simulation could cause the input to the lookup table to be outside the range of the tabulated values. For example, the heating simulation is set up to run for 2 days, which is also the extent of the time data loadedinto the lookuptable dialog. The selected extrapolation was “InterpolationExtrapolation,” so running the simulation linearly extrapolates past 48 hours, using the last two data points in the table. This means that the temperature will continue to drop, going down at the rate of about 10 deg per hour (240 deg per day), which is clearly not correct. Even selecting the option of keeping the end-point values constant ( “Interpolation-Use End Values”) is not a good approximation to what actually happens, but it is the better choice. In this case the user has clear control (via the simulation time) over whether or not the lookup table data will be exceeded; in many applications the independent variable in the table is a simulation variable whose value cannot be easily determined. In this case, a plot of the simulation results and the independent variable will allow you to verify that the simulation handles the end-point properly.

3.3. Tables, Interpolation, and Curve Fitting in Simulink

91

The 2-D and n-D Lookup Tables are set up in a very similar way to the 1-D table we just reviewed. To ensure that you understand the way these are used, use the Simulink help (from the 2-D or n-D block dialog) to review the way these tables are used.

3.3.2 Interpolation: Fitting a Polynomial to the Data and Using the Result in Simulink Thereis another optionin Simulink that in some applicationsmay bemore accurateand easier to use. Thisoption is the curve fitting block, “Polynomial,” in the Math Operations Simulink library. The use of this block requires you to do some preliminary work in MATLAB to compute the coef ficients of the polynomial that will interpolate and extrapolate your data. The process of creating a curve fit to any data uses the “Basic Fitting” tool in MATLAB. To open this tool, create a plot of the data, using the MATLAB commands that we used to create the 48-hour outdoor temperature profile above or running the MATLAB M-file simulated_temp_data in the NCS library.

When theplot appears, select “Basic Fitting” underthe “Tools” menu to open theBasic Fitting dialog shown above. In this dialog, we selected the sixth order polynomial option. The tool will overplot the curve fit on the graph of the data. Try different polynomial degree options and navigate around the dialog. To view the polynomial coef ficients as shown in the figure, click the right arrow at the bottom of the dialog. To get these values into the

92

Chapter 3. Nonlinear Differential Equations Temperature Data for Norfolk MA, Jan. 1,2006 30 Temperature (Dry Bulb) 29

Temperature (Wet Bulb)

28 e r u t 27 a r ) e p F . 26 m g e e T D b l ( 25 u s B g n t i e d 24 W a e d R n a 23 y r D

22 21 20

0

5

10 15 Time (Hours from Midnight)

20

25

Figure 3.7. Realdata mustbe ina MATLABarrayforuse inthe From Workspace “

”

block.

Simulink polynomial block, click the button in the Basic Fitting tool that says “Save to workspace. ” MATLAB will save the curve fit in a MATLAB structure called “fit.” The coef ficients are in the field called coeff (i.e., the coef ficients are the MATLAB variable fit.coeff). In the Thermo_table model in the NCS library, replace the Lookup Table block with the Polynomial block, and then open this blocks dialog and type “fit.coeff ” in the area designated “Polynomial coef ficients.” Run the simulation and verify that the simulation is using the curve fit you created instead of the data. (A version of this model is in the NCS library, but you still need to create the polynomial fit as described above before you use it; the model is called Thermo_polynomial.) The reason you might want to use this approach is to smooth the data and to improve the simulation run time. In general, it is faster to use a low order fit than it is to use the table lookup with interpolation. (It is not always the case that curve fit calculations are faster, so if improving the simulation time is your goal this assumption should be veri fied.)

3.3.3 Using Real Data in the Model: From Workspace and File A web site gives access to weather data from volunteers in your local area. We used data for Norfolk, Massachusetts on January 1, 2006, to create an M- file that can then import the actual temperature data into the home heating model. This data is in the M- file weather_data.m , and Figure 3.7 is its plot. The web site that the data comes from is http://www.wunderground.com/weatherstation.

3.3. Tables, Interpolation, and Curve Fitting in Simulink Thouse

deg F to deg C 70 Set Point (F)

F2C

93

blower cmd

Terr

Mdot*ha C2F

Thermostat

Heater Blower House

Tdata(:,1:2)

F2C

Measured Data From Workspace

deg F to deg C

Indoor vs. Outdoor Temp.

Thermo Plots

deg C to deg F Measured Outdoor Temperature

Figure 3.8. Simulink model of the home heating system using measured outdoor temperatures.

Figure 3.9. From workspace dialog block that uses the measured outdoor temperature array.

Running the M-file creates the plot in Figure 3.7 and the data in the MATLAB workspace. The data is stored in a 288 × 3 array called Tdata, where the first column is the time (in hours), the second is the measured dry bulb temperature, and the last column is the wet bulb temperature (both in degrees F). To use data in MATLAB as an input to a Simulink model, the From Workspace block is used. (An equivalent block called From File gets the data from matrices that were saved using a mat file; since the two blocks work in the same way, we will illustrate only the Workspace block.) We have modified the Thermo_table model to allow input to the model to come from a data file in MATLAB. This is Thermo_real_data in the NCS library (Figure 3.8). The data we want is in the array calledTdata, where the first column is the time and the second is the outside temperature. The From Workspace block dialog in Figure 3.9 opens when you double click on the block. We have entered the values for the data into this dialog

94

Chapter 3. Nonlinear Differential Equations Indoor vs. Outdoor Temp. 74 72 70 68 66

0

1

2

3

4

5

6

7

8

9 x 10

4

Measured Temperature 30 28 26 24 22

0

1

2

3

4 5 Time (Sec.)

6

7

8

9 x 10

4

Figure 3.10. Using real data from measurements in the Simulink model.

(calling the data “Tdata”). Since the array had three columns (the last column is the wet bulb temperature, which we do not need), the dialog box Data has Tdata(:,1:2). This truncates the tabulated data down to two columns. The sample time is every 5 minutes, or every 300 sec. For completeness, we show the result of simulating the data in Figure 3.10).

3.4 Rotations in Three Dimensions: Euler Rotations, Axis-Angle Representations, Direction Cosines, and the Quaternion We saw in Chapter 1 that the Foucault pendulum has forces that are a consequence of the rotation of the coordinate system where the pendulum is mounted. When we extend these ideas to a rigid object rotating in three-dimensional space, it is not easy to keep track of the new orientation. This is because even when doing the rotations about a single axis at a time, the order they are done will determine the result. The simplest way to illustrate this is to think about what happens if you were standing, facing north, and you then turned 90 degrees to the left followed by a 90-degree rotation to lie down on your back. You would be laying face up with your head to the east. If you did this sequence in the reverse order —i.e., you first lay down on your back and then you turned left —you would be lying on your left side with your head to the north, clearly not in the same orientation as the first sequence. Mathematically, rotations do not commute. The way we keep track of the rotation of a body is with the angles of rotation about each of the three orthogonal axes. Mathematically, rotations are transformation matrices that describe what the rotation does to the x , y , and z coordinates after the rotation. These matrices have unique representations and manipulations.

3.4. Rotations in Three Dimensions

95

z θ

y New y

x New

x

Figure 3.11. Single axis rotation in three dimensions.

3.4.1 Euler angles Euler proved that any orientation requires at most three rotations about arbitrary axes on a rigid body. (The rigidness of the body ensures that the complete body is a single mass as well as inertia; we discuss what happens when the body is notrigid in Chapter 6.) The angles that define the rotation are Euler angles, and they are not unique. In most applications that use Euler angles, it is conventional for the rotations first to be about the x -axis, followed by a rotation about the new y -axis and then another rotation about the new z-axis. (This is abbreviated “123 rotation”.) This is not unique since there are 12 possible orderings of the rotation axes (namely 123, 132, 213, 232, 312, 321, 121, 131, 212, 232, 313, and 323). To compute the orientation of the new axes after such a sequence of rotations, we need to investigate what happens with each of them alone. Only coordinates in the plane perpendicular to the axis of rotation change when a single axis rotation is used. We can think of the three Euler angle rotations as a sequence of three independent planar rotations, each plane redefined by the previous rotation. Figure 3.11 shows a rotation about the z-axis that causes the entire x -y plane to rotate. For the single axis rotation of angle θ about the z-axis shown above, the coordinates (x,y) of a point in the plane has a new set of coordinates ( xNew , yNew ) given by xNe w yNe w

=

x cos(θ ) + y sin(θ), sin(θ ) + y cos(θ).

= −x

This transformation, using the vector-matrix form, is

x y

Ne w Ne w

zNe w

    =

cos(θ ) − sin(θ ) 0

sin(θ ) cos(θ ) 0

0 0 1

 x   y  z

=

x T(θ )  y  . z

96


For a given value of the angle θ , the matrix T is orthogonal. (As an exercise, prove this by showing that TTT is the identity matrix.) As an example, if we use this result to compute the transformation of the coordinates after a standard Euler angle rotation using the axes 121 (i.e., the three rotation angles are about the original x -axis, then the new y -axis, and, last, about the new x -axis) the new coordinates are given by

x  1 y  0 New New

zNew

=

0

0 cos(ψ) − sin (ψ)

0 sin(ψ) cos(ψ)

  cos(θ )  0

0 1 0

− sin(θ )

 1  0

sin(θ ) 0 cos(θ )

0

0 cos(φ) − sin(φ)

0 sin(φ) cos(φ)

 x   y  . z

Notice that the transformation is always a 2 × 2 matrix, but when the axis of rotation changes the 2 × 2 matrix appears in different rows and columns of the full 3 × 3-rotation matrix. The three transformation matrices can be multiplied together to give a single direction cosine matrix for the entire rotation, but in most applications this is not done. Note that since each of the transformations by themselves is orthogonal, the combined direction cosine matrix product is also orthogonal. (The proof of this is an exercise.) When an object is rotating, each axis of the body has its own rotational rate. It is usual to denote these rotations with the letters p,q, and r for the x -, y -, and z-axes attached to the body, respectively. When this is the case, the Euler angles are functions of time, so we need to know what happens to them as the object rotates. We saw in Section 1.4.1 of Chapter 1 that a scalar rotation around a single axis gives rise to rotations in the two coordinates that are orthogonal to the rotation axis. Generalizing this to three dimensions gives the operator equation (where a is an arbitrary vector in the Body frame that is rotating at the vector rate ω) “

”

  dt 

d a

=

Inertial

  dt 

d a

+

ω

× a.

Body

A rotating body has accelerations induced whenever the angular momentum vector changes. If we let the rotational rates around the body axes be

ω

=

p  q , r

the angular momentum of the body around its center of mass is the result of calculating the angular momentum of every mass particle in the rigid body and then summing over all of the particles. If the rigid body is homogeneous, the summation is an integral. In either case, the result is that the angular momentum is Jω, where the inertia matrix is

J=

 

J xx −J yx −J zx

−J xy

−J xz

J yy −J zy

−J yz

J zz

 .

(Since this result might be unfamiliar, see [17] for more details.) From Newton s laws applied to rotations, the torques applied to the body and the rate of change of the angular momentum balance, so (using the derivative of a vector in inertial ’


97

coordinates above) d J ω dt

= J

d ω dt

+

d J ω

dt

+

ω

× J ω = T.

When the inertia matrix is constant over time, J factors to give d dt

(ω)

= J

−1

(T − ω × (Jω)) .

This equation is the starting point for any simulation that involves the rotation of a rigid body. (We will build a simulation model in Simulink shortly.) If this equation is integrated, the result is the angular velocity at any time t . Integrating the angular velocity gives the angular position. The angular position that results from integrating the angular rate is not a record of the Euler angle history. This is because the angular rate ω is always with respect to the body, and the Euler angles are always with respect to the original orientation of the body. Using the transformation from the body axes to the Euler axes (and the Euler rotations about x -, y -, and the new x -axes) developed above, the instantaneous angular rate in the body coordinates (with respect to the Euler angle rates) is

p q r

=

   +

dϕ dt 0 0

1 0 0

    10  +

0

0 cos ϕ − sin ϕ

0 cos ϕ − sin ϕ 0 sin ϕ cos ϕ

  

0 sin ϕ cos ϕ

  cos θ  0 sin θ

0 1 0

0

  

dθ dt 0 −

  sin θ    0

cos θ

0 0 dψ dt

  . 

Multiplying the matrices in this expression gives

p 1 q 0

0 cos φ − sin φ

=

r

0

− sin θ

sin φ cos θ cos φ cos θ

 φ   dt d  θ  . ψ

The Euler rates in terms of the body rates come from inverting the matrix in this expression as follows (noting that this matrix is not orthogonal; see Exercise 3.1):

d dt

φ 1  θ   0 ψ  01  0  =

=

0

−1

 p sin φ cos θ   q  r cos φ cos θ   sin φ tan θ cos φ tan θ  p  cos φ sin φ   q .

0 cos φ − sin φ

− sin θ

−

sin φ

cos φ

cos θ

cos θ

r

Nowthere aresome interesting andpotentiallydif ficult numerical issues associated with this equation. Whenever the Euler angle θ becomes a multipleof π2 , this matrix becomes singular

98


(itsinversebecomes infinite), so keeping track of the orientation using thisrepresentation can have numeric problems whenever the Euler angles are near 90 degrees or 270 degrees. Even if there were a nice way of overcoming the numerical issues, the fact that the transformation matrix is not orthogonal would createnumerical stability issues. What can we do about this?

3.4.2

Direction Cosines

Assume that three rotations that lead to a general rotation are sequential about the axes z, y , and x (axes 321 as defined above). The rotation matrices above give the final orientation as the product of three matrices as follows (starting with a rotation of angle ψ about z, followed by a rotation of θ about y , and, last, a rotation of ϕ around x ):

1 0 0

=

0 cos φ − sin φ

0 sin ϕ cos ϕ

  cos θ  0 sin θ

 cos θ cos ψ  sin φ sin θ cos ψ cos φ sin ψ −

cos φ sin θ cos ψ

+

sin φ sin ψ

− sin θ

0 1 0

0 cos θ

 

cos ψ − sin ψ 0

sin ψ cos ψ 0

cos θ sin ψ sin φ sin θ sin ψ + cos φ cos ψ cos φ sin θ sin ψ − sin φ cos ψ

0 0 1

 

− sin θ

sin φ cos θ cos φ cos θ

 .

As we pointed out when we were talking about the Euler angles, this matrix is the “direction cosine matrix,” and it defines the cosines of the angles between the new axis (after the rotation) and the old (before the rotation). Thus the 1,1 element of the matrix is the cosine of the angle between the x -axis after rotation and the x -axis before rotation, the 1,2 element is the cosine of the angle between the x -axis after and the y -axis before, etc. If we denote this matrix by C, then when the object is rotating about its body axes with rates p, q , and r , the direction cosine matrix satis fies the differential equation d C dt

=

 

0 −r q

r 0 −p

−q

p 0

 

C=−

 

0 r −q

−r

0 p

q −p 0

 

C = −C.

This equation comes from the fact that the rate of angular rotation is the cross product of the vector rate with the body axis vector. (In Exercise 3.2 you will show that the matrix

=

 

0 r −q

−r

0 p

q −p 0

 ,

multiplying, on the left, any column ci of C, gives a result that is the same as ω × ci .) This equation canbe thecalculation used to find theorientation of an objectas itrotates, but because there are nine elements in C and there are only six independent variables, we would be calculating nine values when only six are required. For this reason, direction cosine formulations are not preferred for calculating rotations. The next section describes the preferred method. It is instructive to build a Simulink model that uses this equation to keep track of the orientation of an object and to compare the computation time required with that required using the quaternion formulation (see Exercise 3.3).


99

3.4.3 Axis-Angle Rotations Euler also proved that any three-dimensional rotation can be represented using a single rotation about some axis (Euler ’s theorem). The “axis-angle” form of a rotation uses this observation. For now let us not worry about how we determine the axis; assume that the vector n specifies it. The desired rotation is defined to be (using the right-hand rule) a rotation about this axis by some angle. The vector n is a unit vector (i.e., nT n = 1). Most of the blockset tools in Simulink (the Aerospace Block Set and SimMechanics) do not make direct use of the axis-angle form, but this representation is the starting point for deriving other forms. The axis-angle approach forms the basis of the quaternion representation that is finding more applications in many diverse fields (for example, mechanical applications, computer-aided design, and robotics). We need four pieces of information to describe the axis-angle transformation, so we  T nx ny nz θ denote it by the four-dimensional vector [ n θ ]T = . However, because nT n = 1 (n specifying direction of the axis that we are rotating about, not its length), only three of these four numbers are independent. When there is a rotation about the body axis, the rotational rate around the axis n is d n dt

= nxω = −ωxn.

3.4.4 The Quaternion Representation There is a nicecomputationally compact method for computingrotations usingthe axis-angle form. The method uses the “quaternion representation,” invented by William Hamilton in 1843. A quaternion is a modification of the angle-axis 4-vector that represents the rotation, defined as q

of q1

=



q1

s

T

=



nx sin(θ/ 2)

ny sin(θ/ 2)

nz sin(θ/ 2)

cos(θ/ 2)



.

Because of the sin (θ/ 2) multiplying the vector n in this definition, the rotational rate (first three components of the quaternion) is d q1 dt

=

1 2

q1

× ω.

By using sin and cos of the rotation angle instead of the angle itself (as was done in the axis-angle representation), the quaternion is numerically easier to manipulate. In fact,   T 2 q1 s = q1 qT the norm of q is qT q = q1 s 1 + s = 1. (As an exercise, show that this is true.) Since q must be a unit vector, we can continuously normalize it as we compute it (which ensures an accurate representation of the axis of rotation). The other attribute of the quaternion representation that makes it more robust numerically is the way the body rates determine the quaternion rates. The derivative of the

100

Chapter 3. Nonlinear Differential Equations Scalar multiplication s omega / ||q||

1

3omega 3

omega_b2i

3

s/||q||

4

Select s from q ||q||

(q_1 cross omega) / ||q||

3

Select q_1 from q

3

q_1/||q||

3

4

Cross Product

3 omega

Normalize the quaternion[q_1 s]

3

3

3 q_1 x omega + s omega

4

x||x||

omega' q_1

Form the norm of q using a subsystem

q_1/||q||

4 3

4

-1

0.5

4

qdot 4

omega3

Integrate qdot 4 4

1 s x o 4

Initial_q

This block implements the quaternion equations.

Let the quaternion q (a 4 vector) be partitioned into a 3 vector and a scalar as follows:

q = [ q 1 s ]

T

The differential equations for the partitioned components of the quaternion vector are: dq 1

/ dt = 1/2 [ q 1 X ω + s

]

and ds

/ dt = -1/2

T

q 1

To insure that the quaternion vector has a unit norm, ||q|| is calculated at each step and the derivatives are divided by this norm before they are integrated (insuring t hat the vector has unit norm at the time t, just prior to the next integration step).

Figure 3.12. Using quaternions to compute the attitude from three simultaneous body axis rotations.

quaternion, using the partition above, is d q dt

=

d dt

    q1 s

=

d dt



n sin(θ/ 2) cos(θ/ 2)

nxω sin(θ/ 2) +

=

−

1 2

1 2

   =

cos(θ/ 2)ω

sin(θ/ 2)

dθ dt

d n dt

sin(θ/ 2) + −

   =

1 2

1 2

1 2

cos(θ/ 2)n

sin(θ/ 2)

dθ

s ω + q1 xω T −ω q1

dt



dθ dt

 

.

Thus, calculating the quaternion rate is simply a matter of some algebra involving the cross product of the quaternion with the body rate (for q1 ) and an inner product with the body rate (for s ). We have created a subsystemmodel fortheseequations (see Figure3.12). It uses some Simulink blocks that we have not encountered before. The first is the block that extracts a component from a vector (the “Selector” block). We use this block twice: first to extract

1 qdot_i2b


101

and second to extract s . The other blocks that are used are the dot product and cross product blocks in the Math Operations library. We need these because of the cross product d q in dt and the calculation of ωT q1 . The cross product block in Simulink implements the definition of the cross prod1 uct using a masked subsystem. Since we In1 have not thoroughly explored the masking of Selector1 subsystems yet, let us take a little detour here to do so. The idea behind a mask is to create a newSimulink block (i.e., a user-defined 1 Selector2 block where the data needed to execute the Out1 block appear in a dialog that the user selects 2 by double clicking on the block, as we have In2 seen many times for the built in blocks in the Selector3 Simulink browser). The mask hides the internal mathematics from the user so it has the look andfeel of a Simulink built-inblock. To Selector4 build such a block, you first create a subsystem in the usual way and then invoke the mask. For the cross product, the subsystem uses two multiplication blocks and four Selector blocks (to select the appropriate components of the vector for the cross product). The subsystem is in the figure above. If you were creating this block as a masked subsystem, the next step in masking the subsystem would be to right click the subsystem block and select “Edit Mask ” from the menu. This opens the “Edit mask ” dialog that allows you to specify the mask properties. For the cross product block, no input parameters are required, so the only thing you might want to do is provide documentation for the block using the “Documentation” tab in the dialog. We will see other masked subsystems as we start to build models that are more complex. Therefore, we will defer further discussion on building a mask until we need to use them in these models. Using the equations above and the cross product block, the Simulink model that will compute the quaternions is shown below. (This model is called Quaternion_block in the NCS library.) The input to this block is the angular velocity in body coordinates of the rigid body, and the output is the quaternion rate. To complete the quaternion calculations we must integrate the quaternion rate. We will do this outside the quaternion subsystem after we show how to extract the Euler angles and direction cosine matrix from q. We use the definition of q to find the Euler angles and direction cosine matrix. Refer back to the definition of the direction cosine matrix: the Euler angles come from the appropriate angle in the matrix and the inverse trigonometric functions. For example, in the 121 representation we developed above, the 3,2 element of the direction cosine matrix is sin φ cos θ , and the 3,3 element is cos φ cos θ , so if we divide the 3,2 element by the 3,3 c element we get tan φ , so the Euler angle φ is given by φ tan 1 ( c3,2 ). In a similar fashion, 3,3 the other angles are q1

=

θ

=

sin

1

−

 

c1,3 and ψ

=

tan

1

−

−

  c1,2 c1,1

.

102


From the de finition of the quaternion, the Euler angles are determined from the following (where the components of the vector q1 are denoted q1 , q2 , and q3 ): ϕ = tan

−1



2(q2 q3 + sq 1 ) s 2 − q12 − q22 + q32



,



.

θ = sin−1 (−2(q1 q3 − sq 2 )) , ψ = tan

−1



2(q1 q2 + sq 3 ) s 2 + q12 − q22 − q32

Again, using the same reasoning, the direction cosine matrix is

 

s 2 + q12 − q22 − q32

2(q1 q2 − sq 3 )

2(q1 q2 + sq 3 )

s 2 + q22 − q12 − q32

2(q1 q3 − sq 2 )

2(q1 q3 + sq 2 ) 2(q2 q3 − sq 1 ) 2

2(q2 q3 + sq 1 )

s +

q32

−

q22

−

q12

 

.

A Simulink block to do these transformations is in Figure 3.13 (this is the model Quaternion2DCM in the NCS library). Notice that this model implements the direction cosine matrix from the following de finition:

DC M = s 2 − |q1 |2 I3x 3 + 2q1 q1T − 2s





 

0 q3 −q2

−q3

0 q1

q2 −q1 0

 

.

(In Exercise 3.5, you are to verify that this calculation gives the same direction cosine matrix as that shown above.) When this discussion started, we noted that the rotational acceleration comes from the Euler equation d (ω) = J−1 (T − ωx(Jω)) . dt We have assumed that the inertia matrix is constant; otherwise, its derivative needs to be included since the left-hand side of this equation is really the derivative of the total momentum given by Jω. We combine the body-axis angular acceleration term above with the quaternion calculation shown in Figure 3.12 to give the Simulink model shown in Figure 3.14. (This model, called Quaternion_acceleration, is in the NCS library.) Also, note that the block implements a time varying inertia matrix using the input port number 2 for the derivative of the inertia; you can either remove the port if the inertia is constant or put the 3 × 3 zero matrix as the input. Once again, some new Simulink blocks are in this model. The multiplication block in the Math Operators library computes the inverse of the matrix J . To do this, drag the multiplication block into the model and then double click to open its block dialog. In the dialog, change the multiplication type to Matrix (*) in the pull-down menu, and then put the symbols “ /*” inthe “Number of inputs” box. Theicon will changeto thematrix multiply, and “Inv” will denote the top input (because of the division symbol you entered in the dialog).


T

Matrix Multiply

Transpose q

Form q q^T

u Reshape

103

Gain Get |q|^2

2

Form (s^2-q dot q) I 1

1

m

Quaternion Vector

Direction Cosine Matrix

square s

eye(3)

3x3 Identity

Multiply by q4

2

col vector zeros(3,1) of zeros

Horiz Cat

Form skew Sym cross product matrix Q

U1 -> Y U2 -> Y(E)

Y

-1

Select q3

U1 -> Y

-1

U2 -> Y(E)

Y

Select q2 U1 -> Y U2 -> Y(E)

Y

-1 Select q1

This block implements the conversion from Quaternions to the Direction Cosine Matrix. The Equation implemented is: 2

2

T

[A(q)] = ( s -- |q | ) I + 2 q q -- 2 s Q in Yellow in Cyan in Blue

| 0 --q 3 q 2 | Where Q = | q 3 0 --q 1 | | --q 2 q 1 0 |

Figure 3.13. Simulink block to convert quaternions into a direction cosine.

We have already encountered the text block with TeX when we explored the Lorenz attractor in Section 3.1. This model uses this feature, and it is a good habit to do so. Locate the text block in the Simulink model at any desired location by double clicking at the desired location. This approach should always be used to annotate the model by showing the equations that the model implements. If you open the model later, the combination of the Simulink model and the equations make it very easy to understand and reuse the model. After you double click on a location in the model for the equations, simply type the text. You have control over the font selected, the font size, the font style, the alignment of the font, and finally the ability to turn on TeX commands in the block. These are selectable (after the text is typed) using the “Format ” menu in the Simulink window. For the TeX commands, the options are the subset of TeX that MATLAB recognizes. Thus, for example, the annotation of theequationat thebottom of thequaternion model in Figure3.12 above uses thefollowing text (the text appears centered in the block because the alignment is set to “center”):

104

Chapter 3. Nonlinear Differential Equations This section of the block implements the equation: dω

/ = J -1 [ - ω X J dt

dJ

- /

dt

ω

+m

]

total

Inv

External Momentum (M) dJ/dt

U( : )

3

2

Make into a 4x1 column vector

Matrix Multiply

4

Body Angular Accel 1

omega

1 J

Matrix Multiply

Cross Product

omega cross H

J omega ( angular momentum H)

5 q x

||x||

Magnitude of q1

Cross Product

Normalize the quaternion[q s] 1

q cross omega1

0.5

2 dq/dt

-1 q (dot) omega1

This section implements the quarternion equations. d q

/ = 1/2 [ q X

1

dt

1

ω

+s

ω

]

and ds

/

dt

= -1/2

T

ω

q

1

To insure that the quaternion vector has a unit norm,||q|| is calculated at each step and the derivatives are divided by this norm before they are integrated (insuring that the vector has unit norm at the time t, just prior to the next integration step).

Figure 3.14. Combining the quaternion block with the body rotational acceleration.

{\bf{\it This block implements the quaternion equations.}} Let the quaternion {\bfq} (a 4 vector) be partitioned into a 3 vector and a scalar as follows: {\bfq} = {[\it { \bfq_1} } {s ] T}. ˆ

The differential equations for the partitioned components of the quaternion vector are {{ {d{\bf \itq_1}}}}/_{ \itdt} = 1/2 [ { \bf{\itq_1} X {\omega}} + s {\bf { \omega}} ] ˆ

and {ds}/_{dt} = -1/2 { \bf{\omega} T} { \it{\bfq_1}}.

ˆ

ˆ

To ensure that the quaternion vector has a unit norm, {\bf || ||q||} is calculated at each step and the derivatives are divided by this norm before they are integrated (ensuring that the vector has unit norm at the time t, just prior to the next integration step).

3.5. Modeling the Motion of a Satellite in Orbit

105

The command \bf{ } causes the text inside inside the braces to appear in boldface. boldface. The equation that defines the quaternion vector is {\bfq} = {[\it { \bfq_1} } {s ]ˆT}, where the \it invokes italics, and the ˆ means superscript. Since the purpose of Simulink is to model the equations in a signal flow form, it is always good practice to include annotation for the actual equation. In fact, this practice will ensure that the next user of the block understands both the block flow and the equations that the block is implementing. To familiarize you with the TeX notation, open the quaternion model and review the two text blocks.

3.5 Mod Modeli eling ng th thee Motion Motion of of a Satel Satellit litee in Orbi Orbitt With the blocks we created in Section 3.4, we can now build a simulation for the rotational dynamics dynamics of a satellit satellitee in orbit. orbit. This really really is the first Simulink model that we will build that is close to a real problem. As such, you can use it for models that you might build in the future. We start by listing the specifications for the model and for what we hope to achieve. We assume we are trying to build a control system for a satellite that will orbit the earth. We also assume that the satellite’s control system will consist of a set of reaction wheels — devices devices that are essentia essentially lly large wheels wheels attached attached to a motor motor. The motor accelerate acceleratess the reaction wheel and in the process creates a torque on the vehicle through Newton ’s third law. (The spacecraft reacts to the accelerating inertia because the torque produced is equal and the opposi opposite te of the torque torque on the wheel. wheel.)) We assume assume that that there there are three three wheels wheels mounte mounted d on the spacecraft along the three orthogonal body axes. (For the exercise we will assume that the wheels are along the axes of the body, and it is symmetric so the inertia matrix is diagonal.) The firstpartof themodel themodel uses uses thequatern thequaternionand ionand accele accelerat rationmode ionmodell from from Secti Section on 3.4. 3.4. For the second part of the model, we will need to create models for the electric motors that drive each of the wheels and the torque that they create. Once we have the model, we can create create a feedba feedback ck contro controlle llerr for the reacti reaction on wheels wheels that that will will cause cause them them to move move the satell satellit itee from one orientatio orientation n to another another.. In the process of creating creating these models, we will end up with a simulation simulation for the rotationa rotationall motion of a satellite satellite in orbit. orbit. We also will will confront a problem that always exists with reaction wheels, namely that they can provide a torque only as long as the wheels wheels can be accelerated. accelerated. Because Because a motor motor cannot accelerate accelerate once it reached its maximum speed, a method is required to allow the motor to decelerate back to zero speed. This uses reaction jets to “unload” the wheel (the jets, by firing in the opposite direction from the torque created by the deceleration, allow the reduction of the wheel ’s speed without affecting the spacecraft). We will be modeling only the reaction wheel in the following, not the unloading of the wheel, and the unloading is left as an exercise. First, First, we build build a model model for the reaction reaction wheels. A DC motor converts converts an electric electric current flowin owing g thro throug ugh h an iron iron core core coil coil into into a magn magnet etic ic field that that intera interacts cts with with a statio stationar nary y magnetic field to create a torque that causes the coil to spin. Current flowing through a coil of wire creates a north pole on one side of the coil and a south pole on the opposite side. The interaction of the magnet with the stationary magnetic field will cause the coil to spin as long as the stationary poles are opposite to the spinning coil (i.e., as long as a north pole of the spinning coil is near a north pole of the stationary coil). As soon as the spin of the coil cause the north pole of the coil to align with the south pole of the stationary magnet,

106


i

Armature Inductance

Mechanical Side of the Motor:

Armature Resistance

R

J dω /dt= K i i where K i is the Torque Constant

Motor Voltage

Back emf

Back emf = K b ω where K b is the Back emf Gain Figure 3.15. Electrical circuit of a motor with the mechanical equations.

the coil will see a torque torque that stops it from rotating, rotating, and it will stop. To overcome this, this, the current in the coil must change direction at least every half revolution. The device that switches the direction of the current at the half revolution point, if it is mechanical, is a commutator. Most modern motors achieve this switching using electronics where, as the rotor turns, a device on the shaft lets the driving electronics know where the rotor is relative to the fixed magnet (the stator). The torque applied to the motor is proportional to the current flowing in the rotating coil, and as long as the gap between the rotating coil and the fixed magnet is small, the torque is nearly linear. Thus, for the first part of the model, the torque is directly proportional to the magnetic field induced in the rotor by the rotor current; since the magnetic field is also nearly linear, the torque is proportional to the current flowing in the rotor. rotor. Thus, the torque torque is T m = Ki iR . The rotor electrical circuit circuit consists of the rotor resistance RR , the rotor inductance LR , the voltage applied to the rotor V in in , and the voltage induced across the rotor because of its motion in the magnetic field of the stator, V b . (This voltage is the back electromotive electromotive force [emf ].) A simple circuit diagram (see (see Figure 3.15) can represent the electromechanical equations for the motor. In this diagram, the motor torque T m is the controlled source Ki i , and the back emf, V b , is the controlled voltage source whose voltage is Kb ωM . It is a simple matter to use the fact that the sum of the voltages around the loop must be zero to get the equation for the current in the rotor ( iR ): LR

di R dt

= −RR iR +

V in in

−

V b .

The back emf is a nonlinear nonlinear function function of how fast the motor is turning. turning. For well-desig well-designed ned motors, however, the voltage V b is proportional to the motor angular velocity ( ωM ). Thus, the motor electrical circuit is the differential equation di R dt

= −

RR LR

iR

−

Kb LR

ωM +

1 LR

V in in .


107

Because electrical energy, when converted into mechanical energy, is conserved, the constants Km and Kb are dependent. Thus the electrical power (the product of the back emf eb eb i R and the rotor current) is P Electical E lectical = 746 (where in the English unit system the 746 converts iR watts watts into into horsep horsepowe owerr [hp]. [hp]. Using Using the linear linear back back emf voltag voltagee this this term term become becomess Kb ωM 746 . In a similar way, the mechanical power is the torque produced times the motor angular iR ωM velocity: T m ωM = Ki550 (where the 550 in the English unit system converts foot-pounds Ki = 1.3564Ki (in the English unit system). into hp). Equating these gives Kb = 746 550

 

3.5.1 Creating Creating an Attitude Attitude Error Error When When Using Using Direction Direction Cosines Cosines We need need to descri describe be howone create createss an “attitude attitude error error ” forthe 3-axisrotat 3-axisrotationof ionof a spacec spacecraf raft. t. None of the descripti descriptions ons (Euler, (Euler, direction direction cosine, cosine, or quaternio quaternion) n) lends lends themselv themselves es to forming forming the difference between a desired and an actual value. We need a way to create an error that goes to zero at the desired attitude. The direction cosine matrix difference C − Cdesired , or some similar error for either the quaternion or the Euler angle representation, is not usable. (The property that causes the rotation to the desired command is that the error is zero when the command is achieved.) This is because the necessary rotational commands about the three body axes are a nonlinear function of the angles (in any of the representations). Let us perform some algebra with Maple to see what we can do to get a representation for the commands in terms of the direction cosine matrix and the quaternion approach. We will work with the direction cosine matrix first. The direction direction cosine cosine matrix we worked with earlier is

CBI

=

 cos θ cos ψ  sin φ sin θ cos ψ cos φ sin ψ

cos θ sin ψ sin φ sin θ sin ψ + cos φ cos ψ cos φ sin θ sin ψ − sin φ cos ψ

−

cos φ sin θ cos ψ

+

sin φ sin ψ

 sin φ cos θ  , − sin θ

cos φ cos θ

where the rotations are about the z-, y -, and x -axes, are in the sequence φ , θ , ψ , and we have explicitly indicated with the subscript BI that the rotations are with respect to some inertial (i.e., fixed) axis to the principal axes of the body that is rotating. When the body is rotating with angular velocity

ω =

p q r

about these principal axes, the matrix C satis fies the differential equation d C dt

=

 

r 0 −p

0 −r q

= −



c1

−q

p 0

c2

The value of  is explicitly de fined by this equation.

c3

  .

C

108


We can also write this matrix differential equation in another form by creating a 9 × 1 vector that consists of the columns of C concatenated one on top of the other. That is, we let the vector c be c

=

c

11

c21

c31

c12

c22

c23

c31

c32

c33



T

.

We also define the cross product matrix for C as (where the subscript i denotes the i th column of the matrix C)

Cxi

=

 

0

c3i

−

0

c3i c2i

c1i

−

c2i c1i

−

0

  .

In terms of this cross product matrix, the differential equation for the columns of C is d ci d t

=

Cxi ω.

Applying this equation to the vector c, we get a vector matrix differential equation for the 9vector c in term termss of the the thre threee vect vector orss ω as (wher (wheree CCross is the9 × 3 matr matrix ix defined explicitl explicitly y by this equation) Cx 1 d Cx 2 c= ω = CCross ω. d t Cx 3

 

 

You can deduce deduce someinterestingattribut someinterestingattributes es of CCross using using Maple. Maple. Developa Developa Maple Maple program program (from MATLAB) to show, remarkably, the two products below: CT Cross CCross

=

2I3

3

×

and CT Cross c

=

09

1.

×

Maple facilitates the calculations (all of the algebra and trigonometric identities involved in manipulating these matrices; they are quite extensive and are done automatically) and saves an immense amount of work. If youhave notbuiltthe Maple Maple progra program m yourse yourself, lf, youcan open open theMATLA theMATLAB B progra program m Maple_CCt_identity in the NCS library to do the computations. computations. This program sets up T the direction cosine matrix in Maple and computes CT Cross CCross and CCross c. (It computes the vector q and the matrix Q used in the following discussion.) From these identities, we can create a nice error signal for a rotation. d Multiply the equation dt c = CCross ω by CT Cross on both sides to give CT Cross

d dt

c

=

CT Cross CCross ω

=

2ω.

d So ω = 12 CT Cross dt c, or ω is the rotational rate needed to make the direction cosine matrix C. Therefore, Therefore, if we want a particular particular direction direction cosine matrix matrix whose elements elements are the 9vector c, all we need to do is calculate the value of ω from this equation and use it to drive


109

the three axes of the spacecraft. From the second identity we proved in the Maple program, when the direction cosine matrix is at the desired value, the command becomes zero, as we require. Notice how neatly this error signal works.

3.5.2 Creating Creating an Attitude Attitude Error Error Using Quatern Quaternion ion Represent Representation ationss There is a complete analogy to this result for the quaternion representation. We will summarize it here and ask you to reproduce the results in Exercise 3.7 at the end of this chapter. The first part of the quaternion result requires the formation of the 4 × 3 matrix Q as follows: s q2 −q3 q3 s −q1 Q= . q1 s −q2 −q1 −q2 −q3

  

  

Then, following similar steps used in the direction cosine matrix derivation, the following two identities are true:

QT Q

=

I3

3

×

and QT

q  q q

1 2 3

s

  

=

04 .

Finall Finally y, the differ different ential ial equati equation on for the quater quaternio nion n vector vector that that we develo developed ped above above (in terms terms of the matrix Q) is q1 q2 T d ω = 2Q . q3 dt s

  

  

This last equation becomes the de fining equation for the quaternion command (again following the steps we used for the direction cosine); the commanded rate to achieve a desired quaternion value is T ηC = 2Q qDesired . We use this in the spacecraft attitude-control model. All of the pieces we need are now in place, so we can create the Simulink model for the simulation. You might want to try to do this yourself before you open the model Spacecraft_Attitude_Control in the NCS library.

3.5.3 The Complete Complete Spacecraf Spacecraftt Model Model The model of the complete complete rotationa rotationall dynamics of the spacecraf spacecraftt is in Figure 3.16. It has four subsystems to compute the following: •

the spacecraftrotation spacecraftrotational al motion motion —this this is a variat variationof ionof thequaterni thequaternion on block block we create created d in Section 3.4.4;

•

the Reaction Wheel dynamics that we created above;

110

Chapter 3. Nonlinear Differential Equations Spacecraft Inertia Matrix

Wheel Momentum

[3x3]

PID Controller Altitude Error

[3x1]

Command Rate Error

3

J

[3x1]

Wheel Inertia

Convert (rad) to Volts [3x1] [3x1]

1

[3x1] [3x1]

V_in

Wheel Speeds

[3x1] [3x1]

omega_wheel (r/s )

[3x3]

Inertia J dq/dt

[3x1]

Jw

Quaternion and Spacecraft Rotational Dynamics

4

Input External Momentum (M)

Momentum of Wheels

Reaction Wheels

3

4

omega Body Angular Accel

3

q

Integrate omega dot 3 omega

2E' q_desired

3

omega

1 s xo 3

Initial ones(3,1)*0.01 omega

4

Quaternions Integrate qdot

Normalize the quaternion[q s] 1 [3x4] [3x1]

2

Matrix [3x1] Multiply

4

4

E' q_desired [4x1]

Mult by 2 [4x1]

Desired q_desired/norm(q_desired) altitude (q_desired)

[3x4]

Matrix E'

Quaternions

4

4 4

Form quaternion propagation matrix

||x||

x

4

1 s xo

4

Magnitude of q 4

[0.5 0.5 0.5 sqrt(3)/3 ]'/1.048 [3x4]

In1

Calculate and display E'*E (Should be I_3)

Figure 3.16. Controlling the rotation of a spacecraft using reaction wheels.

•

a block that has a PID controller in it;

•

a block that forms the matrix Q from the quaternion vector.

In addition, there are blocks that compute the command ( ηC = 2QT q Desired ) and a block that checks the identity QT Q = I3 3 . (This is used to verify the accuracy of the computations and that the model is accurate.) It is a good idea to familiarize yourself with each of the subsystems and the overall simulation before you run it. The data is all contained in the model; the inertia of the wheels is 50 slug-ft 2 and the vehicle inertia matrix is diagonal, with the diagonal values of 1000, 700, and 500. The reaction-wheel-dynamics model uses the same parameters as we used in the model above. The model uses as initial conditions for q the value [0.4771 0.4771 0.4771 0.5509], and the desired final value is set to q_desired/norm(q_desired) in the block that is labeled Desired Attitude. The value for q_ desired is [1 2 3 4]. You can change this value at the MATLAB command line to any value you would like and see the result. The PID controller is set up with only PD. (The Integral gain is set to zero; look inside the PID block to see the values.) The Proportional gain is set to 10, and the Derivative gains (on the three angular velocities p, q , and r ) are 70. We have not attempted to make these values optimum in any way; they simply make the response reasonably fast with a minimum overshoot in the quaternion values at the end of the command. When you run the model using these parameter values, you should see the quaternion values plotted in the figure at the right below. Note that these quaternion values change smoothly over the time interval, and they do indeed go to the desired final values from the initial value we used. ×

Initial q


111

Minus q1

-u

-q1

s q3 -q2 -q1

Horiz Cat

[1x4]

Minus1 q2

-u

Quaternions 1

4

-q2

-q3 s q1 -q2

Horiz Cat

[1x4]

[1x4] [1x4]

q3

Minus2 -u

q2 -q1 s -q3

Horiz Cat

[1x4]

Vert Cat

[3x4]

Matrix Q' 1

Matrix Concatenation

s

Figure 3.17. Matrix concatenation blocks create the matrix Q used to achieve a desired value for q .

One aspect of this model is new, namely the way we create the matrix QT . We use two steps to form the elements of the matrix (which has 3 rows and4 columns). First, we compute the 3 rows(each 1 × 4) one at a time, and then we concatenate them vertically to create the matrix. The block that creates the rows and columns is the Matrix Concatenation block in the Math Operations library in Simulink. This block takes scalars or one-dimensional vectors and places them into the columns (called horizontal concatenation) or the rows (called vertical concatenation) of a matrix. The initial construction of the rows places the various scalar values from the quaternion vector q into the appropriate locations to build the three rows of the matrix (using the horizontal version of the block), and then the rows are stored into a matrix using the vertical version of the block. Figure 3.17

112


shows how we used the block “Horizontal Concatenation” in the subsystem called “Form quaternion propagation matrix.” The quaternion block is applicable to more complex dynamics than a spacecraft. Examples are the rotational dynamics of an automobile or any other vehicle in motion, a robot, a linkage, or an inertial platform. Since the quaternion formulation can model these, it is a good idea to keep this model as part of a user library with the other models in the Simulink browser.

3.6 Further Reading Reference [40] (Chapter 4 in particular) has a good description of the Euler and quaternion representations of the rotational motion of a spacecraft. In his book Unknown Quantity: A Real and Imaginary History of Algebra [7], John Derbyshire describes the development of the quaternion representation as the logical leap into the fourth dimension after the two-dimensional representation of complex numbers. He notes that in 1827 Hamilton began investigating complex numbers in a purely algebraic way. He found it extremely dif ficult to come up with a scheme that would make the algebra distributive and would maintain the property of complex numbers that the modulus of the product of two numbers is the product of their respective moduli. Hamilton’s insight into the fact that you could not satisfy this modulus rule with triplets butcould do so with quadruplets led to the invention of quaternion algebra. Hamilton was so pleased with this result that he inscribed it with a knife on a stone of Brougham Bridge near Dublin. Derbyshire’s book is about the algebraic properties of the quaternion, which are not a major consideration for portraying rotations. Despite this, his book is fun to read for its mathematical discussions and its look at the personalities behind the math. There is an excellent discussion of the computational ef ficiency of the quaternion representation in [11], and you should consult this after you complete Exercise 3.3.

Exercises 3.1 Show that the matrix product dϕ dt

p   1 0  0  q   0   0 cos ϕ sin ϕ   r ϕ  10 0 0 0 sinϕ cos cos θ 0  0 cos ϕ sin ϕ   0 1 =

0 dθ dt

+

0

−

− sin θ

+

0

− sin ϕ

cos ϕ

sin θ

 

0

0 cos θ

gives

p 1 q 0 =

r

0

0 cos φ − sin φ

 φ d  θ  . sin φ cos θ  dt − sin θ

cos φ cos θ

ψ

 

0 0 dψ dt

 

Exercises

113

3.2 Show that the direction cosine matrix satis fies the differential equation

 0 d C =  −r dt

r 0 −p

q

= −

c

1

c2

 C

−q p

0 c3

.

3.3 Start with the Simulink model for the quaternion representation of rotations. Next, create a model that does the same thing using direction cosine matrices and, last, a model that creates a rotation with Euler angles. Compare thecomputational ef ficiency of the three models. In particular, see what happens when the Euler angles approach the points where the transformations become in finite. Reference [24] has a good discussion of why the quaternion approach is better. 3.4 Show that the norm of the quaternion vector is one. 3.5 Verify that the direction cosine matrix from the quaternion vector is

 DCM = s

2

2

 − |q | I 1

3×3

+ 2q1 q1T

 − 2s 

0 q3 −q2

−q3

0 q1

q2 −q1 0

 .

3.6 Try adding a simple controller to the reaction wheel model that will cause the reaction wheel speed to decrease whenever it gets to 1000 rpm. Use an external torque applied by a reaction jet. (The jet will create a force, so you need to decide on where it is relative to the center of gravity to determine the torque.) If you apply a torque using only a single jet with the force not through the center of gravity, then rotations will occur about multiple axes. How do you prevent this? How many jets will you need to unload the reaction wheels in all three axes? 3.7 Verify the quaternion will move toward the desired quaternion q Desired if the rate body rate is ηC = 2QT q Desired . In this equation, the matrix

 s  q Q=  −q 3

2

−q1

−q3 s q1 −q2

q2 −q1 s −q3

  . 

Also, verify that the matrix concatenation blocks in Figure 3.17 create this matrix.

Chapter 4

Digital Signal Processing in Simulink

We sawin theprevious chapters howto build modelsof continuous time systems in Simulink. This chapter provides insight into how to use Simulink to create, analyze, simulate, and code digital systems and digital filters for various applications. We begin with a simple example of a discrete system; one discussed in Cleve Moler ’s Numerical Computing with MATLAB [29]. This example, the Fibonacci sequence, is not a digital filter but is an example of a difference equation. In creating the Simulink model for this sequence, we illustrate the fact that the independent variable in Simulink does not have to be time. In this case, the independent variable is the index in the sequence. We set the “sample time” in the digital block to one, and then we interpret the time steps as the number index for the element in the sequence. Digital filters use an analysis technique related to the Laplace transform, called the z-transform. We introduce the mathematics of this transform along with several methods for calculating digital filter transfer functions. A digital signal typically consists of samples from an analog signal at fixed times. Since it is usually necessary to convert these digital signals back into an analog form so that they can be used (to hear the audio from a CD or a cell phone, for example), it is natural to start by asking how to go about doing this. The answer is the “sampling theorem” that shows that an analog signal with a bounded Fourier transform (i.e., F(ω) = 0 for |ω| > ωM ) can be sampled at the times ωπ or faster, and these samples can then be used to reconstruct M the analog signal. The method for doing this reconstruction is via a filter that allows only the frequencies below ωM to pass through (and for this reason, we call the filter a low pass filter). Therefore, the second section of this chapter shows how to develop low pass filters and ways to adapt their properties to make them do other useful signal processing functions such as high pass and band pass. In all cases, we will use Simulink to simulate the filters and explore their properties. The last part of this chapter will deal with implementation issues. We will look at how Simulink allows us to evaluate different implementations of the digital filter. We will also begin to look at the effect of limited precision arithmetic on the digital filter ’s performance. We will then look at a unique combined analog and digital device called a phase lock loop. Simulink allows us to build a simulation of this device and do numerical experiments 115

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Chapter 4. Digital Signal Processing in Simulink

that demonstrate its properties. In fact, Simulink is unique in its ability to do some very detailed analyses of a phase-locked loop (see [11]). This is particularly true when it comes to analyzing the effect of noise on the loop, a topic that we will take up in Chapter 5.

4.1 Difference Equations, Fibonacci Numbers, and z -Transforms One of the more interesting difference equations is the Fibonacci sequence. Fibonacci, the sequence he developed, and its rather remarkable properties and history are all described in detail, along with several MATLAB programs developed to illustrate the sequence, in Chapter 1 of Cleve Moler s Numerical Computing with MATLAB [29]; also see [26]. Let us revisit this sequence using Simulink. The Fibonacci sequence is ’

f n+2 = f n+1 + f n with f 1 = 1 and f 2 = 2.

As a reminder, the sequence describes the growth in a population of animals that are constrained to give birth once per generation, where the index is the current generation. One possible Simulink model to generate this sequence uses the Discrete library from the Simulink browser. To understand how Simulink works, we need to describe how one would go about writing a program that generates the Fibonacci numbers. In the NCM library (the programs that accompany the book Numerical Computing with MATLAB), there is a MATLAB program that computes and saves the entire Fibonacci sequence from 1 to n. If, instead, we want only to compute the values as the program runs without saving the entire set of values, the MATLAB code (called Fibonacci and located in the NCS library) would look like “

”

function f = fibonacci(n) % FIBONACCI Fibonacci sequence % f = FIBONACCI(n) sequentially finds the first n % Fibonacci numbers and displays them on the command line. f1 = 1; f2 = 2; while i <= n f = f1 + f2 f2 = f1; f1 = f; i = i+1 end

Even though we do not need to save the entire sequence in this example, we still need to save thecurrent andthe previous value in order to calculatethe next value of thesequence. This fact means that this sequence requires two states. In the theory of systems, a state is the minimum information needed to calculate the values of the difference equation. In the snippet of MATLAB code above, we save the values in f1 and f2. When we build a Simulink model to simulate a difference equation, the state needs a place to be stored for use in the solution. Simulink uses the name 1z to denote this block. To understand where “

”

4.1. Difference Equations, Fibonacci Numbers, and z-Transforms

1

117

1

z

z

Unit Delay

Unit Delay1

Scope

Figure 4.1. Simulating the Fibonacci sequence in Simulink.

this comes from, we need to show the method for solving difference equations using the discrete version of the Laplace transform. We will do that in a moment, but first let us build the Simulink model for the Fibonacci sequence. To create the model, from MATLAB open the Simulink Library Browser as we have done previously and open a new untitled model window. Select the “Discrete ” library, and then select the 1z icon (the “Unit Delay” block) and drag one into the open model window. Right click and drag on the Unit Delay block in the model window to make a copy of the Unit Delay. Connect the output of the first delay block to the input of the second. This will send the output of the first delay to the second delay block. Now we need to create the left-hand side of the Fibonacci equation. To do this we need to add the outputs of the two Unit Delay blocks. Therefore, open the “Math Operations” library and drag the summation block into the model window. Then connect the outputs of each of the Unit Delay blocks to sum the inputs one at a time. The model should look like Figure 4.1. In order to start the process, we need the correct initial conditions. To set them, double click on the Unit Delay block and set the initial condition to 2 and 1 (from left to right in the diagram). This will set the initial value of f1 to 1 and the initial value of f2 to 2, as required. Notice that the 1z block has a default “sample time” of 1 sec, which is exactly what we want for simulating the sequence, as we discussed in the introduction above. To view the output, drag a Scope block (in the Sinks library) into the model and connect it to the last Unit Delay block. Click the start button on the model to start the simulation of the sequence. Double click on the Scope block and click on the binoculars icon to see the result. The simulation makes ten steps and plots the values that the Fibonacci sequence generated as it runs. The graph should look like Figure 4.2. We created this figure using a built-in MATLAB routine called “simplot.” This M-file uses a MATLAB structure generated by the output of the Scope block. The Scope block generates this MATLAB data structure during the simulation; the plot comes from the MATLAB command simplot(ScopeData). The plot is Fibonacci Sequence.fig , and it is available in the NCS library from the Figures directory. If youwant to compute thegolden ratio “phi” as wasdone in NCM [29], thecalculation requires that you divide the value of f2 by f1. This uses the Math Operations library Product block. In this library, drag the product block into the model window and double click on it. The dialog that opens allows you to change the operations. In the dialog box that asks “Number of Inputs” (which is set to 2 by default), type the symbols * and /. This will cause one of the inputs to be the numerator and the other the denominator in a division (denoted by × and ÷ in the block). Connect the × sign on the Product block icon to the line after

118

Chapter 4. Digital Signal Processing in Simulink Fibonacci Sequence (Values for the first 20 Numbers)

15000

10000

5000

Fibonacci Sequence (Detail from 0 to 10) 100

0 0

5

10 Index

15

20

80 60 40 20 0 0

2

4

6

8

10

Index

Figure 4.2. Fibonacci sequence graph generated by the Simulink model.

1 f(n-2)

z Unit Delay

1 f(n-1)

f(n)

z Scope

Unit Delay1 f(n)

1.6180339901756

f(n-1)

Product

Display

Figure 4.3. Simulink model for computing the golden ratio from the Fibonacci sequence.

the first delay block (the “Unit Delay” block in Figure 4.1) and connect the to the “Unit Delay1” block. Connect the output of the Product block to a Display block that you can get from the “Sinks” library. This block displays the numeric value of a signal in Simulink. Figure 4.3 shows Display with the result of the division after 10 iterations. To make the simulation run longer, open the Con figuration Parameters menu under the “Simulation ” menu at the top of the Fibonacci model window. The dialog that opens when you do this allows you to change the Stop time. Change it to some large number (from the default of 10), and run the simulation. You should see the display go to 1.618. To see the full precision of this number, double click on the Display block and in the Format pull down, select “long.” This corresponds to the MATLAB long format. The result should be 1.6180339901756, the limit of this ratio is the “golden ratio,” phi, and it has the value √ 1+ 5 phi = 2 . (MATLAB returns 1.61803398874989 when calculating this, and as can be seen after 20 iterations, Simulink has come very close.) The discussions in [29] and [26] describe phi and its history in detail. This discrete sequence is only one of many sequences that you might want to find the solution of in Simulink. In a more practical vein, we often want to process a digital signal

÷

4.1. Difference Equations, Fibonacci Numbers, and z-Transforms

119

(a process called digital signal processing). Toward this end, digital filters are part of the Simulink discrete library. They appear in the Digital library as z-transforms. What is this all about?

4.1.1

The

-Transform

z

The z-transform, F(z), of an infinite sequence {f k } , k = 0, 1, . . . , n , . . . , is F(z) =

∞ 

f k zk .

k =0

There are many technical details that need to be invoked to ensure that this sequence always converges to a finite value, but suf fice it to say that because the variable z is complex, the sequence always is finite (even for sequences that diverge). For example, let us compute the z-transform for the sequence that is 1 for all values of k . (This is called the discrete step function.) Thus, we need to compute the sum F(z) =

∞ 

zk = 1 + z1 + z2 + z3 + · · · .

k =0

If we multiply the value of F(z) by z, the sum on the right side becomes zF(z) = z1 + z2 + z3 + · · · .

Now, by subtracting the second series from the first, all of the powers of z subtract (all the way to infinity), and the only term that remains on the right is the 1, so F(z) − zF(z) = 1

or F(z) =

1 1−z

.

As a second example, consider the sequence αk , k = 0, 1, 2, . . . . This sequence is the discrete version of the exponential function eat since the values of this at the times kt generate the sequence (eat )k = α k (where α = eat ) . Following the same steps as we ∞ used above, the z-transform of this sequence is F(z) = k =0 α k zk = 1−1αz . It is a simple matter to work with the de finition to create a table of z-transforms. This table will allow you to solve any linear difference equation. For example, the discrete sine iωkt −e−iωkt can be generated using sin (ωkt) = e and the above transform of α k . 2i You can use the MATLAB connection to Maple to get some z-transforms. Try some of these: syms k n w z simplify(ztrans(2ˆn))

This gives z/(z-2) as the result. ztrans(sym(‘f(n+1)’))

This gives z*ztrans(f(n),n,z)-f(0)*z) as the result.

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Chapter 4. Digital Signal Processing in Simulink ztrans(sym(‘f(n+1)’))

This gives z*sin(k)/(zˆ2-2*z*cos(k)+1) as the result. Solutions of linear difference equations using z-transforms are very similar to the techniques for solving differential equations using Laplace transforms. Just as the derivative hasa Laplace transformthat converts the differentialequationinto an algebraicequation, the 0, 1, . . . , n , . . . , converts the difference equation into an algebraic z-transform of f k+1 , k equation. To see that this is so, assume that the z-transform of f k , k 0, 1, . . . , n , . . . , is 0, 1, . . . , n , . . . , is F(z). Then the z-transform of f k +1 , k

=

=

∞ 

k 0

=

=

f k +1 zk

= f 1 + f 2 z1 + f 3z2 + · · · = z−1F(z) − z−1 f 0.

Notice that this is the same answer as we got when using Maple above. From this, we can see why Simulink uses 1 /z as the notation for the “Unit Delay.”

4.1.2 Fibonacci (Again) Using z -Transforms Let us use the z-transform to solve the Fibonacci difference equation. We use the unit delay z-transform above twice. The first time gives the z-transform of f k +2 , and the second gives the transform of f k +1 . The z transform of the Fibonacci equation is therefore z−2 F(z)

− z−2f 0 − z−1 f 1 = z−1F(z) − z−1 f 0 + F(z).

Solving for F(z) in this expression gives (z−2

− z−1 − 1)F(z) = z−1f 1 + z−2f 0 − z−1f 0;

therefore, we have the final algebraic equation for F(z): F(z)

−1 −2 −1 −1 −1 = −(zz −2+−zz−1+−21z) = zz−2 −(z z−1+−1)1 .

Now we can factor the denominator of the function F(z) and write the right-hand side of the above √ as a partial fraction √ expansion. The roots of the denominator polynomial are φ1

= − 1+2 5 and φ2 = − 1−2 5 (note that φ2 = 1 − φ1 ), so the partial fraction expansion is F(z)

= z−1 A+ φ + z−1 B+ φ 1

. 2

A and B are determined by using Heaviside ’s method, wherein the value of A is obtained by multiplying the left and right side of the above expression by z−1 φ 1 and then setting φ 1 , and similarly for B . The inverse z-transform for each of these the value of z−1 terms comes from the transforms above. There is a lot of algebra involved in this, so let us just look at the answer (see Example 4.1):

+

=−

f n

= 2φ 1− 1 (φn1+1 − (1 − φ 1)n+1). 1

This is the same result demonstrated in NCM [29].

4.2. Digital Sequences, Digital Filters, and Signal Processing

121

There is a connection between z-transforms and Laplace transforms that we will develop later, but first let us look at practical applications of difference equations. Modern technology such as cell phones, digital audio, digital TV, high-de finition TV, and so on, depends on takingan analogsignal, processing it to make it digital, andthen doing something to the signal to make it easier to send and receive. Digital filtering permeates all of this technology.

4.2 Digital Sequences, Digital Filters, and Signal Processing The advent of digital technology for both telephone and audio applications has made digital signal processing one of the mostpervasive mathematicaltechniques in use today. The methods used are particularly easy to simulate with the digital blocks in Simulink ’s “Discrete ” library. To understand what these blocks do to a digital sequence, we need to understand the mathematics that underlies discrete time signal processing and digital control.

4.2.1 Digital Filters, Using z -Transforms, and Discrete Transfer Functions To start, we will work with the exponential digital sequence we created above (see Section 4.1.1). Assume that we are going to process a digital sequence f k using the sequence α k . The digital filter is then yk +1

=

αy k

+

(1 − α)f k .

In this difference equation, the sequence f k is the signal to be processed (where f k is the value of the signal f(t) at times kt as k , an integer, increases from 0), and the sequence y k is the processed result. The simplest way to solve this equation is to use induction. Starting at k = 0, with the initial condition y0 , we get the value of y1 as y1

=

αy0 + (1 − α)f 0 .

Now, with y1 in hand, we can compute y2 by setting k the filter. The result is as follows: y2

=

αy1

+

(1 − α)f 1

=

α(αy0

+

=

1 in the difference equation for

(1 − α)f 0 ) + (1 − α)f 1 .

Thus, y2

=

α 2 y0

+

α( 1 − α)f 0 + (1 − α)f 1 .

If we continue iterating the equation like this, a pattern rapidly emerges and can be used to write the solution to the equation for any k . (Verify this assertion by continuing to do the iteration.) This solution is k 1  −

yk

=

k

α y0 + (1 − α)

j =0

α k−1−j f j .

122


Notice that α k , the sequence we wanted, multiplies both the summation and the initial condition. We now use induction to show that this is indeed the solution. Remember that a proof by induction follows these steps: •

Verify that the assertion is true for k

•

Assume that the assertion is true for k , and show that it is then true for k + 1.

=

0.

Because of the way that we generated the solution, it is clear that it is true for k = 0. So next, assume that the solution above (for the index k ) is true, and let us show that it is true for k + 1. From the difference equation yk +1 = αyk + (1 − α)f k , we substitute the postulated solution for yk to get yk+1

=

αy k

=

α(α k y0

+

(1 − α)f k



k −1

+

(1 − α)

α k −1−j f j ) + (1 − α)f k

j =0

=

α k +1 y0

+

(1 − α)

k 

α k −j f j .

j =0

This is exactly the solution that we postulated with the index at k + 1. Thus, by induction, this is the solution to the difference equation. Notice that one way of thinking about discrete equations in Simulink is that it implements the induction algorithm. It uses the de finitions of the discrete process and starts at k = 0, iterating until it reaches the nth sample. If we take the z-transform of the difference equation yk +1 = αyk + (1 − α)f , we get z−1 Y(z) − z−1 y 0

=

αY(z) + (1 − α)F(z).

Solving this for Y(z) gives Y(z)

=

=

z−1 z

−1

−

α

1 1 − αz

y0 +

y0 +

(1 − α) z−1

−

α

(1 − α)z

1 − αz

F(z)

F(z).

By comparing the z-transform above with the  solution, we can conclude that the inverse k z-transform of 1−zαz F(z) is the convolution sum j =0 αk −j f j . Note that this says that when a sequence whose z-transform is F(z) isused asan input toa digital filter whose z-transform is H(z) (called the discrete transfer function, and for this filter its value is H(z) = 1−zαz ), the product H(z)F(z) has an inverse transform that is the convolution sum. This result, called the convolution theorem, provides the rationale for the Simulink notation of using the z-transform of the filter inside a block. The notation implies that the output of the block is the transfer function times the input to the block —even though the output came from the difference equation and the result is a convolution sum (when the system is linear). This


123

Gain1 1-alpha

1

simout

z Sine Wave

Unit Delay

To Workspace

alpha Gain

Figure 4.4. Generating the data needed to compute a digital filter transfer function.

slight abuse of notation allows for clarity in following the flow of “signals” in the Simulink model because it maintains the operator notation even when the block contains a transfer function. One of the major uses of digital filters is to alter the tonal content of a sound. Since music consists of many tones mixed together in a harmonic way, it is useful to see what a digital filter does to a single tone. Therefore, we use Simulink to build a model that will generate the output sequence yk when the input is a single sinusoid at frequency ω. Thus, we assume that f k = A sin(ωkt), where kt is the sample time. The process of converting an analog signal to a digital number is “sampling.” All digital signal processing uses some form of sampling device that does this analog to digital conversion. The discrete elements in the Simulink library handle the sampling process automatically. Try building a Simulink model that filters an analog sine wave signal with the first order digital filter yk +1 = αyk + (1 − α)f k before you open the model in the NCS library. To run the model in the NCS library, type Digital_Filter at the MATLAB command line. Figure 4.4 shows the model. In this model, thesampling of thesine wave occursat theinputto theUnit Delay block. The sample time is set in a “Block Parameters ” dialog box (opened by double clicking on the Unit Delay block). In this dialog, the sample time was set to delta_t (an input from MATLAB that is set when the model opens). This illustrates an important attribute that Simulink uses. After sampling the signal, all further operations connected to the block that does the sampling treat the signal as sampled (discrete). Thus, the Gain block operates on the sampled output from the Unit Delay, and the addition occurs only at the sample times. The dialog also allows setting the initial condition for the output. (We assume that the initial condition is zero, the default value in the dialog box.)

4.2.2 Simulink Experiments: Filtering a Sinusoidal Signal and Aliasing The digital filter model in Section 4.2.1 is set up to run 50 sinusoidal signals (each at a different frequency) simultaneously, and as it runs, it sends the results of all 50 simulations into MATLAB in the MATLAB structure “simout.” The values for the various parameters in the model are in the MATLAB workspace and have the following values:

124


>> delta_t delta_t = 1.0000e-003

%(Sample time of 1 ms or sample frequency of 1 kHz)

>> alpha alpha = 9.0000e-001

>> omega omega = Columns 1 through 25 1.0000e-001 2.8786e-001 8.2864e-001 2.3853e+000 6.8665e+000

1.2355e-001 3.5565e-001 1.0238e+000 2.9471e+000 8.4834e+000

1.5264e-001 4.3940e-001 1.2649e+000 3.6410e+000 1.0481e+001

1.8859e-001 5.4287e-001 1.5627e+000 4.4984e+000 1.2949e+001

2.3300e-001 6.7070e-001 1.9307e+000 5.5577e+000 1.5999e+001

3.0171e+001 8.6851e+001 2.5001e+002 7.1969e+002 2.0717e+003

3.7276e+001 1.0730e+002 3.0888e+002 8.8916e+002 2.5595e+003

4.6054e+001 1.3257e+002 3.8162e+002 1.0985e+003 3.1623e+003:

Columns 26 through 50 1.9766e+001 5.6899e+001 1.6379e+002 4.7149e+002 1.3572e+003

2.4421e+001 7.0297e+001 2.0236e+002 5.8251e+002 1.6768e+003

The values for the 50 frequencies, the sample time for the filter, and the value of alpha are stored as part of the model through a callback set by the Model Properties dialog. The ability to cause calculations in MATLAB to run when the model opens or when other Simulink actions occur is a feature of Simulink that you should understand. After you have opened the model, go to the File menu and select “Model Properties. ” A Model Properties window will open, allowing you to enter and/or view information about the model. It also allows you to select actions that occur at various events during the model execution. There are four tabs across the top of this window, denoted “Main,” “Callbacks,” “History,” and “Description.” Figure 4.5 shows two of the tabs in the dialog. The window opens, showing the contents of the Main tab. The Main tab is the top level of the window. It shows the model creation date and the date we last saved it. It also shows a version number (every time the model is changed or updated in any way, this number changes) and whether or not the model has been modi fied. The second tab in the window is the Callbacks tab. In this section, the user can specify MATLAB commands to execute whenever the indicated action occurs. The possible actions are as follows.


125

Figure 4.5. Addingcallbacks to a Simulink modeluses Simulink ’s model properties dialog, an option under the Simulink window’s file menu.

•

Model preload function: These commands run immediately before the model opens for the first time. Note that it is here that the values of omega, the sample time delta_t , and alpha are set. The values for omega are provided by the MATLAB function “logspace,” which creates a set of equally spaced values of the log (base 10) of the output (in this case, omega), where the two arguments are the lowest value (here it is 10 1 ) and the highest value (3 × 103 here). The value set for the sample time is 0.001 sec (corresponding to 1 kHz). −

•

Model postload function: These commands run immediately after the model loads the first time.

•

Model initialization function: These commands run when the model creates its initial conditions before starting.

•

Simulation start function: These commands run prior to the actual start (i.e., immediately after the start arrow is clicked).

•

Simulation stop function: These commands run when the simulation stops. In this case, we have two MATLAB commands to firstcalculate the maximum value of all 50 signals. (Note that the maximum values are over the structure in MATLAB generated by the “To Workspace” block.)

•

Model presave function: These commands run prior to the saving the model.

•

Model close function: These commands run prior to closing the model.

126


The History and Description tabs allow the user to save information about the number of times the model opens and is changed (the History tab) and for the user to describe the model. The StopFcn callback, executed when the simulation stops, is simmax = max(simout.signals.values); semilogx(omega,simmax); xlabel(’Frequency "omega" in rad/sec’) ylabel(’Amplitude of Output’); grid

The plot at the right comesfromthiscode. Itisa semilog plot that shows what the digital filter does to the amplitudes of a sinusoid for different frequencies. Notice that all of the low frequencies —tones up to about 6 radians/sec (about 1 Hz)—are unaffected by the filter, whereas the frequencies above that are attenuated to the point where a tone at about 3000 radians/sec (about 500 Hz) is reduced in amplitude by 95%. For this reason, this filter is a “low pass” filter. This means that low frequencies pass through the filter unchanged, and it attenuates higher frequencies. Let us see what happens if we input frequencies beyond 500 Hz. To see this, change the values in the omega array by typing at the MATLAB command line omega = logspace(2,3.8);. The second plot shown at the right is the result of rerunning the model with these 50 values of omega. Instead of continuing to re-

1 0.9 0.8 t u p 0.7 t u O d 0.6 e r e t l i F 0.5 f o e d 0.4 u t i l p m0.3 A

0.2 0.1 0 -1 10

0

10

1

2

10 10 Frequency "omega" in rad/sec

3

10

4

10

1 0.9 0.8 t u p 0.7 t u O d 0.6 e r e t l i F 0.5 f o e d 0.4 u t i l p m0.3 A

0.2 0.1 0 2 10

3

10 Frequency "omega" in rad/sec

4

10


127

Sampled Sine Wave -- Illustrating Aliasing 1.1

0.75 Sampled only at the times 0.0025 and 0.0125 sec., the sine seems to be a constant for all times.

0.5 e d 0.25 u t i l p m 0 A l a n g -0.25 i S

When the samples are every 0.002 seconds, the signal looks like a sinusoid.

-0.5 -0.75

-1.1

0

0.002

0.004

0.006

0.008 0.01 0.012 Time (sec.)

0.014

0.016

0.018

0.02 Continuous Signal Sampled at 0.0025 and 0.0125 Sampled Every 0.002 sec.

Figure 4.6. Sampling a sine wave at two different rates, illustrating the effect of aliasing.

duce the amplitude of the input, the filter ’s output amplitude starts to climb back up until, 1 at the frequency t (1 kHz), there is no reduction in the amplitude of the sinusoid. Why does the amplitude not continue to decrease? It is because this is a sampled-data signal. A close examination of what happens when we sample an analog signal to convert it into a sequence of numeric values (at equally spaced sample times) reveals why. Look carefully at the effect of sampling a 100 Hz sinusoid as illustrated in Figure 4.6. When the sample time is 0.002 seconds (the * in the figure), the values are tracking the sinusoid as it oscillates up and down in amplitude. However, when the sample time exactly matches the frequency of the sinusoid (the black squares), the oscillatory behavior of the sinusoid is lost. For the precisely sampled values shown, the amplitude after sampling is always exactly 1. Thus, as far as the digital filter is concerned, the frequency of the input is 0 (it is not oscillating at all). This is what causes the plot of the amplitude of the filtered output to turn around starting at half of the sample frequency. We call this effect “aliasing.” It is exactly because of this that the digital standard for audio CDs is to sample at 44.1 kHz (a sample time of 22.676 microseconds). With this sample frequency, the audio frequencies up to 22.05 kHz are unaffected by the digital conversion. Since the human ear does not really hear sounds that are above this frequency, the aliasing effect is not perceived. There is a caveat, though: since aliasing takes the high frequencies above 22.05 kHz and reduces them, these frequencies need to be removed (with

128


an analog filter) before the sampling takes place. The filter that does this is an antialiasing filter . In Section 4.5.2, we will investigate how one must sample an analog signal to ensure that aliasing is not a problem, but first we explore some of Simulink ’s digital system tools.

4.2.3 The Simulink Digital Library Before we leave the subject of digital filters, let us look at some other methods that Simulink has for digital filtering. You may have noticed as we built the digital models above that there were other blocks in the Discrete library for digital filters. They are •

Discrete Filter,

•

Discrete Transfer Function,

•

Discrete Zero-Pole,

•

Weighted Moving Average,

•

Transfer Function First Order,

•

Transfer Function Lead or Lag,

•

Transfer Function Real Zero,

•

Discrete State Space.

Each of these uses a different, but related, method for simulating the digital filter. The preponderance of these models uses the z-transform of the filter to create the simulation of the filter. For example, the discrete filter model uses the z-transform in terms of powers of 1/ z. We use this form of the digital filter because in some de finitions of the z-transform the infinite series is in terms of 1/ z, not z. To change the numerator and denominator of the filter transfer function, use the block parameters dialog box that opens when you double click on the Discrete Filter block. The filter default is 1+01.5z 1 . The numerator is 1, and the denominator is entered using the MATLAB notation [1 0.5], which, as the help at the top of the dialog box shows, is for ascending powers of 1/ z (see Figure 4.7). Notice in the figure that when you change the numerator and denominator, the icon in the Simulink model changes to show the new transfer function. To try this block, let us simulate the Fibonacci sequence with it. Set up a new model and drag the Discrete Filter block into it. Open the dialog by double clicking and enter the vector [1 −1 −1] as the Denominator. The vector sets the powers of z−1 from the lowest to the highest (the MATLAB convention for polynomials). Notice that the icon changes for the Discrete Filter to display the denominator polynomials. Leave the numerator at 1. Transfer functions do not have initial conditions, so we need to find a way to specify that the Fibonacci sequence start with initial values of 1 and 2. To do this we use a block that causes a signal to have a value at the start of the simulation. The block we need is the IC block, which is in the library called Signal Attributes. Grab an IC block and drag it into the model. The IC block has an input, but we do not need to use it. You can leave −


129

Figure 4.7. Changing the numerator and denominator in the Digital Filter block changes the filter icon.

the input unconnected, but every time you run the model, you will receive the annoying message Warning: Input port 1 of ’untitled/IC’ is not connected.

To eliminate this message, there is a connection in theSources library called “Ground.” All it does is provides a dummy connection forthe block andthereby eliminates themessage. The last thing is to connect a Scope to the output. The IC block has a default value of 1, which is acceptable because starting the Fibonacci sequence with initial vales of 0 and 1 will still generate the sequence. Figure 4.8 shows the model (called Fibonacci2 in the NCS library) and the simulation results from the Scope block. Comparing this with the result generated in the earlier version of the Fibonacci model shows that the results are the same (except for the initial conditions). The first seven versions of the discrete filter in the list above are all variations on this block. To understand the subtleties of the differences, spend some time modeling the Fibonacci sequence using each of these.

130


Figure 4.8. Using the Digital Filter block to simulate the Fibonacci sequence.

4.3 Matrix Algebra and Discrete Systems We looked at state-space models for continuous time linear systems in Chapter 2. There is an equivalent model for discrete systems. Let us return to the Fibonacci sequence and use Simulink in a different way to show some attributes of the sequence. Before we do though, let us look at the state-space version of the Fibonacci sequence. When we talked about the digital filters in Simulink, we had a list of eight different ways, the last of which was a state-space model. We did not go into the details at the time because we had not developed the state-space model. We saw how to convert a continuous time state-space model into an equivalent discrete model in Section 2.1.1. We can convert the Fibonacci sequence difference equation into a discrete state-space model directly. The steps are as follows. Use the values on the right-hand side of the Fibonacci sequence ( f k and f k +1 ) as the components of a vector as follows: xk =



f k f k +1



.

From the definition of the sequence f n+2 = f n+1 + f n , with f 1 = 1 and f 2 = 2, we get the state-space vector-matrix form from the fact that the first component of xk +1 is the second component of xk and the second component of xk +1 is the left-hand side of the difference equation. Thus, f k +1 0 1 xk +1 = xk , = f k +2 1 1



x0 =

   

f k = [ 1

1 1



,

0 ]xk .

4.3. Matrix Algebra and Discrete Systems

131

y(n)=Cx(n)+Du(n) x(n+1)=Ax(n)+Bu(n)

144

Ground

Display Discrete State -Space

a) Simulating the Fibonacci Sequence with the Discrete Library State-Space Block

Matrix Constant

0

1

1

1

[2x2]

Matrix Multiply

[2x2] 2

2 2

Multiply

1 2

144

[1 0]* u Gain

Display

2

z

Unit Delay

b) Simulating the Fibonacci Sequence Using Simulink’s Automatic Vectorization Figure 4.9. State-space models for discrete time simulations in Simulink.

Notice that the initial conditions are not the values we used previously, but since we start the iteration in the state-space model at k = 0, we have made the initial value of f 0 = 1, which is consistent with the initial values we used previously (since f 2 = f 1 + f 0 = 2). We use two methods to create this model. The first model for this state-space description uses the Discrete State Space block in the Simulink Discrete library. The model is shown in Figure 4.9(a). (It is called Fib_State_Space1 in the NCS library.) When this model runs, the Display block shows the values for the sequence. The model has been set up to do 11 iterations. To see other values, highlight the number 11 in the window at the right of the start arrow in the model window, and change its value to any number (but be careful: the sequence grows without bound). To build a Simulink model for this equation that uses Simulink ’s vector-matrix capabilities, we use the blocks for addition, multiplication, and gains from the Math Operations library (as we did when we created the state-space model for continuous time systems). Figure 4.9(b) shows this model (it is Fib_State_Space2 in the NCS library). Note that the Constant block from the Sources library provides, as an input, the matrix on the right side of the state-space equation above. The Multiply and the Gain blocks from the Math Operations library are used to do the matrix multiply and the calculation of the output. Thedialogs from thetwo Math Operations blocksare shown in thefollowing figures.

132


The dialog on the left is for the Matrix Constant block, and as can be seen, the value for it has been set to the MATLAB variable [1 0; 1 1]. The dialog on the right is for the Multiply block. It specifies that there are two inputs through the ** (two products), which may be increased to as many values as desired (including the use of the “ / “to denote matrix inverse). The multiplication type comes fromthe pull-downmenu next to the Multiplication: annotation. The menu has only two options: Matrix(*) and Element-wise (.*), where the notation in parentheses indicates the operation as if it were MATLAB notation. The initial conditions for the iteration are set using the dialog that opens when you double click the Unit Delay block. Double clicking this block opens the dialog that allows the desired initial values to be set. As above, the initial values are 01 . The iteration we are doing creates the Fibonacci sequence in a vector form that will allow us to show some interesting facts about the sequence. So let us do some exploring. f 1 0 The matrix 01 11 can be thought of as f , since we assume that the initial f f









1

2



value of f 0 is 0 and the values of f 1 and f 2 are both 1. Therefore, after the first iteration of the difference equation we have x2 =



f k +2 f k +3

   =

0 f 1

=

1 f 2

  x1 =

0 1

f 1

f 2

f 0 + f 1

f 1 + f 2

=



f 1 f 2

f 2 f 3



1 1





f 0 f 1

f 1 f 2



x0

x0

x0 .

Notice that after this iteration the matrix multiplying x0 is in exactly the same form as when we started, except that the subscripts are all one greater than the initial matrix. If the iteration continues, the matrix form is the same (i.e., the value of xk after k iterations is f k−1 f k x0 ; in Exercise 4.2 we ask you to use induction to prove this). f f



k

k +1



If we take the determinant of this matrix, we see that it is det



f k −1 f k

f k f k +1



2

= f k +1 f k −1 − f k .

Let us use our model to show that this determinant is f k +1 f k −1 − f k2 = ±1.

4.3. Matrix Algebra and Discrete Systems

133

The Simulink library does not contain an explicit block to compute the determinant of a matrix, so we will use some of the “User Defined Functions” blocks. There are five different ways that the user may de fine a function in Simulink. In this library are blocks that •

create embedded code directly from MATLAB instructions,

•

call any MATLAB function (this block does not compile the MATLAB code),

•

create a C-code function.

There are also variations on these blocks where a function uses a rather arcane but useful form and a version of the C-code function that uses MATLAB syntax for those who refuse to learn C. We will use only two of these blocks: the MATLAB function and the Embedded MATLAB blocks. The model that was created is shown below. (It is called Fib_determinant in the NCS library.) In this model, the MATLAB Function block uses the single function “det” directly from MATLAB to compute the determinant (which is set using the dialog that opens when you double click on the block), and the “Embedded MATLAB Function” block has the following simple code to compute the determinant of the 2 × 2 matrix input u. Since the result of using the MATLAB function and the Embedded MATLAB function are the same, you might legitimately wonder why there are two blocks. The reason has to do with calling MATLAB from Simulink. Stand-alone code does not have access to MATLAB, so the MATLAB Function block will not work. The Embedded MATLAB block, on the other hand, creates exportable C code, so when the code compiles it works as a stand-alone application. function d = det2(u) % An embeddable subset of the MATLAB language is supported. % This function computes the determinant of the 2x2 matrix u. d=det(u);

The model is in Figure 4.10. The first time this model runs, the embedded code compiles into a dll file that executes each time the model runs. When the model runs, 13 iterations result, and the determinant plot appears in the Scope block (see Figure 4.11). We will use this model to illustrate the computational aspects of finite precision arithmetic. If the number of iterations is set to 39, the determinant from the MATLAB function shows a value of 2, and the determinant from the embedded MATLAB function shows a value of 0. If we continue past 39 iterations, say 70, we still get zero from the embedded function, but we get 2.18e+013 for the determinant from the MATLAB function. What is happening here? The problem is that we are at the limit of the precision of the computations. The values for the Fibonacci sequence are at 8e+014 so that the products of the values in the equation f k +1 f k −1 − f k2 are on the order of 1029 and the difference is therefore less than the least signi ficant bit in the calculation. The built-in MATLAB function starts to fail as soon as the terms in this function get to about 1018 , whereas the embedded MATLAB block protects the calculation from underflow by making the value 0. This still works only for so long; eventually even this strategy fails. (To see this, try 80 iterations; at this number of iterations none of the determinant values work.)

134


Figure 4.10. A numerical experiment in Simulink: Does f k +1 f k −1 − f k2 = ±1?

Figure 4.11. 13 iterations of the Fibonacci sequence show that f k +1 f k −1 − f k2 = ±1 (so far).

This exercise is an example of how you could use Simulink to check that some mathematical result is true. If you wanted to show that f k +1 f k −1 − f k2 = ±1, and you did not have any idea if it were true or not, you could build the model and try it. It would be immediately obvious that the values are ±1 for the number of iterations you used.

4.4. The Bode Plot for Discrete Time Systems

135

You then could try to prove the result. The ability to use the pictorial representation of the premise quickly to create numerical results can almost immediately tell you if the premise is true. Now that we know f k +1 f k −1 − f k2 = ±1 is true, Exercise 4.3 asks that you use induction to prove it. (Use the fact that det



0 1

1 1

 is −1.)

4.4 The Bode Plot for Discrete Time Systems In Section 4.2.2, we created a frequency response plot, but it used a numerical experiment on the Simulink model, which is ponderous at best. There is a simpler way, based on the z-transforms that we explore now. In order to compute the Bode plot for a discrete system we need to understand the mapping from the continuous Laplace variable to the discrete z-transform variable. To see this we need to investigate the Laplace transform of a sampled signal f(t) (i.e., a signal that exists only at the sample times kt) when we shift it in time by t . If we assume that the Laplace transform of f(t) is F(s) , then ∞



∞

e

−st

f (t + t)dt =

e−s(t +t) f(τ)dτ

t

0

=e



−st

F(s) − e

−st

f (0).

The last step used the fact that the Laplace transform is from t = 0 to infinity, and in the first line of the equation the integral starts at t . Since the function f(t) is discrete, f (0) is the only value not in the integral when it starts at t . Comparing this to the z-transform derived in Section 4.1.1, we see that z−1 = e−st or z = est . Withthisinformation, we would like to find thediscrete (z-transform)transfer function for the discrete time state-space model. We have seen two ways for developing the discrete state-space model. When we developed the solution for the continuous time state-space model in Section 2.1.1, we showed that the result of making the system discrete in time was the model xk +1 =

(t)xk +  (t)uk ,

yk = Cxk + Duk .

In developing the discrete state-space model of the Fibonacci sequence above, we went directly from the difference equation to the discrete state-space model. (In this case the matrices were not determined from the continuous system, and they are not functions of t .) In either case, the form of the equations is the same. Taking the z-transform of this gives zX(z) − zx0 =

(t)X(z) +  (t)U(z).

136


Thus the discrete transfer function of the system ( H(z)) is the z-transform with the initial conditions set to zero, so we have H(z) =

Z {yk } Z {uk }

= C(zI − )−1 B + D.

The Bode plot of the discrete system from the state-space form of the model comes from setting z = eiωt in the above derivation. That is we need to compute H (eiωt ) = C(eiωt I − )−1 B + D,

which has exactly the same form as the continuous transfer function except iω is replaced by eiωt . (Remember that t is the sample time of the discrete process and is therefore a constant.) The manipulations of the state-space model for both continuous and discrete systems in MATLAB are the same, so the connection from Simulink to MATLAB for the Bode plot calculation is identical. We can go back now to the digital filter example in Section 4.2.2 above and use the Control System Toolbox to calculate its Bode plot. Open the digital filter model by typing “Digital_Filter1” at the MATLAB command line; the model that opens is the same as the model we created in Section 4.2.2, but the sinusoid input has been deleted since this is not needed to create the Bode plot. As we did above, under the “Tools” menu select “Control Design” and the submenu “Linear Analysis.” The “Control and Estimation Tools Manager ” will open. In the model, right click on the line coming from the Input block and select the “Input Point” sub menu under the “Linearization Points” menu item. Similarly, select “Output Point” (under the same menu) for the line going to the Output block. Selecting these input and output points causes a small I/O icon to appear on the input and output lines in the model. In the “Control and Estimation Tools Manager” GUI, select the Bode response plot for the plot linearization results and then click the Linearize Model button. The LTI Viewer starts, and right click to select the Bode plot under “Plot Types.” The plot of the amplitude and phase of the discrete filter as created by the Viewer (Figure 4.12) stops at the frequency 3.1416 radians/sec because this is the frequency at which the Bode plot for the discrete system starts to turn around and repeat. π (This is called the “half sample frequency,” and it is equal to t .) Compare thisplot withthe amplitude plotcreated by the Simulinkmodel Digital_Filter in Section 4.2.2. You will see that it is the same. The important difference is that the computation of the Bode plot using this method is far more accurate (and faster) than using a large number of sinusoidal inputs as we did there.

4.5 Digital Filter Design: Sampling Analog Signals, the Sampling Theorem, and Filters In 1949, C. E. Shannon published a paper in the Proceedings of the Institute of Radio Engineers (the IRE was an organization that became part of the Institute of Electrical and Electronics Engineers, or IEEE) called “Communication in the Presence of Noise. ” This landmark paper introduced a wide range of technicalideasthat form thebackbone of modern communications. The most interesting of these concepts is the sampling theorem. It gives conditions under which an analog signal can be 100% accurately reconstructed from its

4.5. Digital Filter Design

137

Figure 4.12. TheBode plot forthe discrete filter modeldeveloped using the Control System Toolbox interface with Simulink.

discrete samples. Because it is such an important idea, and because it is so fundamental, it is very instructive to work through the theorem to understand why, and how, it works.

4.5.1 Sampling and Reconstructing Analog Signals When we introduced discrete signals above, they were simply a sequence of numbers. In addition, their z-transform was the sum of the sequence values multiplied by powers of z. In Section 4.4, we saw that the frequency response results when eiωt is substituted for z (thereby giving H (eiωt )). When the digital signal is the result of sampling an analog signal, we need a way of representing this fact. The method must maintain the connection with the analog process. Equating a sampled analog signal to a sequence of its values at the sample times loses the analog nature of the process (and besides is really only one of the mathematical ways of representing the process). An alternative is to represent the sampled signal as an analog signal that is a sequence of impulses (analog functions) multiplying the sample values. Doing this gives an alternate representation of the sampled sequence, {sk } as the analog signal s ∗ (t): ∗

s (t) =

∞  k=0

sk δ(t − kt).

138


With this definition, we can take the Laplace transform of s ∗ (t) as ∞

S ∗ (s) =



 

e−st s ∗ (t)dt =

0



∞

∞

e−st

0

sk δ(t − kt) dt .

k =0

The integral and the sum commute in the last term above, so ∞

∗

S (s) =

k =0

∞

∞

 

e

−st

 

sk δ(t − kt)dt =

0

sk e−skt .

k =0

Now to understand the sampling theorem, assume that we have been sampling the signal for a very long time so that the signal is in steady state. In that case, the Fourier transform describes the frequency content of the signal. The difference between the Laplace and Fourier transform is in the assumption that for the Fourier transform the time signal began at −∞. The steps that created the sampled Laplace transform above are the same, except that, since the time signal exists for all time, the sum and integral are double sided: ∞

∗

S (ω) =

k =−∞

∞

∞

 

e

−iωt

 

sk δ(t − kt)dt =

−∞

sk e−skt .

k =−∞

Now, the sampling theorem is as follows: If a continuous time signal s(t) has the property that its Fourier transform is zero for all frequencies above ωm or below −ωm (the Fourier transform of s(t),S (ω) = 0 for |ω| ≥ ωm ), then we can perfectly reconstruct the signal from its sample values at the sample times t = ωπ (or if the signal is sampled faster) using the in finite sum m

∞

s(t) =



sk

sin(ωm t − kπ ) (ωm t − kπ )

k =−∞

.

This theorem is critical to all applications that use digital processing since it assures that after the signal processing is complete the signal may be perfectly reconstructed as long as the sampling was originally done at a rate at least twice as fast as the highest frequency in the signal. As an aside, Claude Shannon proved this result, and it was published in the Proceedings of the I.R.E. in 1949 [39]. The proof of this assertion is straightforward. Refer to Figure 4.13 as we proceed through the steps in the proof. The first step is to take the function S(ω) and expand it into an infinite series (using the Fourier series to do so). The result is ∞

S expanded (ω) =



S k e

ik 2πω 2ωm

−

k =−∞

The value of S k comes from the Fourier expansion S k =

1 2ωm

ωm



−ωm

S(ω)e

ikπω ωm

dω .

.


139

s0 s1 s2 s3 s4 s5 … t

−ω m a) Signal and its Sampled Values

ω m

b) Fourier Transform of the Signal

c) Result of replicating S (ω ) an infinite number of times.

Figure 4.13. Illustrating the steps in the proof of the sampling theorem.

Since the inverse Fourier transform of S(ω) is the signal s(t) = 21π S(ω)e iωt dω , we get that the value of S k is ωπ sk . Therefore, S (ω) is the same as S expanded (ω) as shown in m Figure 4.13. That is, ∞



−∞

∗

∞ ∗

S (ω)



=

k =−∞

π ωm

sk e

−

ikπω ωm

.

The last step in the proof is now simple. In order to recover the signal we simply need to multiply the Fourier transform of the sampled signal by the function p(ω) defined by p(ω)

=



1, 0,

ωm < ω < ωm , elsewhere. −

Figure 4.13 shows the rectangular function drawn on top of the transform of the sampled signal. Clearly, the product is the transform of the original analog signal. The inverse Fourier transform of p(ω) is the “sinc” function given by ωm sin(ωm t − kπ ) π

(ωm t − kπ )

.

The proof of the result follows immediately because the inverse Fourier transform of the product of S (ω) and p(ω) is the convolution of this function and the inverse transform of S (ω) which is just the sum of impulses de fining s (t) that we started with. ∗

∗

∗

140

Chapter 4. Digital Signal Processing in Simulink View Signals

Sum of Sines Input Generate 15 Sine Waves at Frequencies given by freqs. (set by a call back when the model loads)

D/A Filter Output Digital Signal 2*pi*5000

15 15

s+2*pi*5000 A/D Converter (Sampled at 0.1 msec)

Transfer Fcn of D/A Filter

Error D/A Error

Figure 4.14. Simulation illustrating the sampling theorem. The input signal is 15 sinusoidal signals with frequencies less than 500 Hz.

The function p(ω) eliminates all frequencies in the sampled signal above the frequency ωm , and we have already seen that such a filter is a low pass filter. This is the main reason we need to create good low pass analog filters. For various technical reasons, it is impossible to build a filter that has a frequency response that is exactly p(ω). This means that we are always searching for a good approximation. The next section shows that there are numerous ways to come up with an approximation and introduces the design of analog filters. These filters are prototypes of digital filters, so this section is important for both the implementation of the sampling theorem and for digital filter design, but firstlet us do some simplesimulations to illustrate these concepts. Because of the sharp discontinuity in the filter function, the ideal low pass filter represented by the function p(ω) above isnot the resultof using a finite dimensional system (i.e., a system represented by a finite order differential equation, having a transfer function whose magnitude is |H(iω)|). Therefore, it is necessary to figure out how best to create a good approximation. As you should guess, the approximation must not have the sharp corner, so the transition from the region where the gain of the filter is 1 to the region where the gain is 0 must be smooth. In Figure 4.14, we show a Simulink model that uses the sine block (from the Sources library) to create 15 continuous time sinusoidal signals. The signals are then summed together and sampled at 0.1 msec, using a sample and hold operation (the “zero order hold ” block in the Discrete library). Once again, you should try to create this model from scratch using the Simulink library rather than loading it from the NCS library using the MATLAB command Sampling_Theorem . The approximation we use for the low pass filter is the simple first order digital filter from Section 4.2.2. The frequency for the filter has been set at 5 kHz. The sinusoidal frequencies in the simulation come from the MATLAB vector freqs, whose values are: 65, 72.8, 74.7, 82, 89.3, 91, 99.5, 103.2, 125, 180.2, 202.1, 223.3, 230.3, 310.3, and 405.2 Hz. Because the sample frequency of 10 kHz is 20 times the highest frequency in the signal, the digital to analog (D/A) reconstruction of the signal using the simple first order filter is not too bad. (The error is about 8%, as can be seen from the D/A Error Scope.)


141

Table 4.1. Using low pass filters. Simulating various filters with various sample times. Sample Time

Filter Type

Filter Freq.

0.001 0.001 0.0001 0.0001

First Order Second Order First Order Second Order

1000 Hz 1000 Hz 1000 Hz 1000 Hz

Standard Dev. Of Error Observed 2.093 1.902 0.872 0.632

This simulation can investigate changes in the sample frequency relative to the frequencies in the signal. As a first numerical experiment, try changing the sampling frequency to 500 Hz. (Make the sample time in the A/D Converter block 0.002.) Make note of the magnitude of the error. Next, change the filter to 1 kHz by changing the parametervalues in theTransfer Function block to 2*pi*1000 in both the numerator and denominator. Note the error for this simulation. Remember that this filter is not a good approximation to the ideal low pass filter. (If you go back to Section 4.2.2 and look at the frequency response plot for this filter, youcan seethatfor this filter the amplitude is reduced 50% at 1000 Hz and is reduced only 90% at 10 kHz.) A better filter is one that has a more rapid reduction in amplitude. Higher order filters will do this. For example, double click the filter block and change the filter parameters to the values in the figure above. This makes the transfer function for the filter H(s) = (2π 1000)2 . As we will see next, this is an example of a Butterworth filter. s 2 +1.414(2π 1000)+(2π 1000)2 Run the simulation using 1 kHz sampling and record this error. You should see very little difference between the two filters when the sample time and the filter functions are set to what the sampling theorem says are the appropriate values. Again, this is because we are not filtering the signal with the ideal filter required by the theorem. To see that the filters work well when the sampling is done at a higher rate than the minimum value of the sampling theorem, change the sample time back to 0.1 msec, as we started with, and rerun the simulation. Table 4.1 summarizes our results from these simulations.

142


Figure 4.15. Speci fication of a unity gain low pass filter requires four pieces of information.

4.5.2 Analog Prototypes of Digital Filters: The Butterworth Filter We did some simple analog low pass filter design in the previous section to illustrate the conversion of a digital to an analog signal. We also need to have analog filters as follows. •

Since the Fourier transform of a signal that we want to sample must have no frequencies above half the sample frequency, a low pass antialiasing filter used prior to sampling ensures that this is true.

•

We can design a digital filter using the analog filter as the starting point and then converting the result using some type of transformation,.

So, with these reasons as motivation, let us explore some analog filters.  Remember that thetransfer function of a filter is H(s) |s =iω = |H(iω)| e i H(iω). Three “bands” on the amplitude plot specify this filter, as illustrated in Figure 4.15. The first of these bands is the “pass band,” which represents the regionwhere the signal is not attenuated (the region whose frequencies are below ωm ). The second region is the “transition band, ” where the amplitude gradually reduces to an acceptable minimum. (Because the frequency response of a finite dimensional system is the ratio of polynomials, it is impossible for the amplitude to be exactly zero except at an in finite frequency.) The last region is the “stop band,” where the filter is below the acceptable value. In the figure, there are three additional parts to the speci fication: the acceptable gain change over the pass band, the acceptable amplitude of the signal in the stop band, and the frequency at the transition band limit.


143

These parameters are ε1 , ε2 , and ωmacceptable , respectively, and they specify bounds on the transfer function as follows: In the pass band, 1

− ε ≤ |H(iω)| ≤ 1 for |ω| ≤ ω

m.

1

In the stop band,

|H(iω)| ≤ ε

2

for ω

| |≥ω

macceptable .

The analog low pass filters that are most frequently used are the Butterworth, Chebyshev, and elliptic filters. The first of these, the Butterworth filter, comes from making the filter transfer function as smooth as possible in each of the regions. The filter has as many of the derivatives of its transfer function equal to 0 at frequencies 0 and in finity. Since themagnitude of thetransfer function is H(iω) H(iω)H( iω) , itisusual to use the square of the magnitude in specifying the filter transfer function (eliminating the square root). Therefore, the Butterworth filter is

| = √

|

|H

Butter

2

| =

1

+

1  ω ωm

2n

−

.

This function has the property that its derivatives (up to the (2n 1)st are zero at ω 0 and at ω . Exercise 4.3 asks that you show that this assertion is true. The Laplace transform for the filter can be determined by using the fact that

−

=∞

2

|H(iω)| =

H(s)H( s)

− |=

s iω

=

.

Therefore, filter transfer function is the result of factoring H(s)H( s)

− =

1

+

1  s iω m

2n

.

The poles of the transfer function are the roots of the denominator, given by the equation 1, or equivalently, the 2n roots of this polynomial are the 2 n complex roots of ( iωs )2n m 1 given by

−

=−

sk

= (−1)

1 2n

(iω m ) .

These poles are equally spaced around a circlewith radius ωm . The values with negative real parts define H(s), and those with positive real parts become H ( s) , so the transfer function for the desired Butterworth filter is stable. It is easy to create this filter in MATLAB. We have included in the NCS library an M- file called butterworthncs that uses the above to determine the Butterworth filter poles and gain for any filter order. The code performs the

−

144


calculations shown below: function [denpoles, gain] = butterworthncs(n, freq) % The Butterworth Filter transfer function % Use zero-pole-gain block in Simulink with the kth pole given by the % formula: % (1/2n) % Poles with real part <0 among p = i(-1) 2*pi*freq % k % Where: % freq = the Butterworth filter design frequency in Hz. % n = order of the desired filter % i = sqrt(-1) denroots = (i*roots([1 zeros(1,2*n-1) 1]))*2*pi*freq; %BW poles = roots denpoles = denroots(find(real(denroots)<=0)); % of -1 in left plane % % %

Zero out the imaginary part of the real pole (there is only one real pole, and then only when n is odd. Because of the precision in computing roots, the imaginary part will not be zero):

index = find(abs(imag(denpoles))<1e-6); % imag. part is 0 denpoles(index) = real(denpoles(index)); % when < eps. % To insure the steady state gain is 1, the Butterworth filter gain % must be n times the magnitude of the Butterworth circle squared: gain = (2*pi*freq)ˆn

To exercise this code, let us design an eighth order Butterworth filter for the problem we investigated above. The poles and gain result from typing the following command in MATLAB: [BWpoles, BWGain] =butterworthncs(8, 10000)

Now let us filter the same signal we filtered above, but this time we build a Simulink model using these poles and gain. A new model similar to the Sampling_Theorem model above is in Figure 4.16. (This model is Butterworth in the NCS library.) The only difference in this model is that instead of thetransferfunction block, the filter comes from the zero-pole-gain block in the Simulink continuous library. The dialog values for this filter are set using the output of the M- file butterworthncs. (The output of this M-file is the gain, BWgain, and the poles, BWpoles.) We designed the filter for a sample frequency of 10000 Hz. Since the maximum frequency contained in the signal is only about 500 Hz, this filter should do a good job of reconstructing the sampled signal. It does, because the root mean square (rms) error in the reconstruction is only 0.0761 (signi ficantly smaller than the rms value for the simple second order filter we used above). Notice that the model above contains a delay block (called transport delay in the model) that delays the input signal by 1.1 msec to account for the Butterworth filter ’s phase shift that delays

4.6. The Signal Processing Blockset

145 Filtered Signal

Generate 15 Sine Waves at Frequencies given by freqs. (set by a call back when the model loads ) BWgain BWpoles(s)

A/D Converter Butterworth Filter of Order n (Sampled at 0.1 msec) Compute the coefficients using the m -filebutterworthncs

Φ

Transport Delay

Error

Analog Input This model uses the data created by a callback to MATLAB when the model opens as follows: Signal The source is a sum of sine waves at the following frequencies freqs = [65 72.8 74.7 82 89.3 91 99.5 103.2 125 180.2 202.1 223.3 230.3 310.3 405.2] Hz. The pha se of each sine wave is: = [0 0.0011 -0.0014 -0.0013 0.0006 0.0002 0.0037 -0.0006 0.0044 0.0030 -0.0039 -0.0034 -0.0011 -0.0004 0 ] The ampl itude of all of the sine waves and the sample times that they are generated are: amplitudes = 1; sample time = 10 -5 sec.

Figure 4.16. Butterworth filter for D/A conversion using the results of the M- file “butterworthncs.”

the output signal. By delaying the input before the error is computed, this filter lag is accommodated. (Remember that in signal processing, time lags in the reconstruction of the data are acceptable because there is usually a large spatial difference between the source and the site of the reconstruction. (Think of transmission via the internet or via a radio link.) We do not have to go through these machinations every time we want to design a filter. Simulink easily allows the adding of new tools. These tools are blocksets, and the first of them that we will look into is the Signal Processing blockset, which contains built-in blocks that allow anyanalog (ordigital) filters to be created. In thenext section, we experiment with analog and digital filter blocks using the Signal Processing Blockset to design Butterworth and other filters.

4.6 The Signal Processing Blockset Analog filters, digital filters and many other signal-processing components are available in the Simulink add-on tool called the Signal Processing Blockset. This tool has many unique features that allow analog and digital filters to be designed, modeled, and then coded. (Signals can be analog, digital, or mixed analog-digital; digital signals can have multiple sample times.) Among the features of the tool are the ability to capture a segment of a time series (in a buffer) for subsequent “batch” processing. The tool also permits processing temporal data using “frames,” where the computations wait until a frame of data is collected, and then the processor operates on the entire frame in a parallel operation. The tool allows digital signal processing models and components using computations that have limited precision and only fixed-point calculations. This last feature couples with a coding method that allows the generation of C-code and allows HDL code output for special purpose signal processing applications on a chip. (The code comes directly from the Simulink model.) We will not describe howwe do this in this chapter,but we will touch on some of thefeatures in Chapter 9.

146


4.6.1 Fundamentals of the Signal Processing Blockset: Analog Filters To understand the capabilities and become familiar with the Signal Processing Blockset, open the library. The figure at the right, a snapshot of the Library browser, shows nine different categories of model elements. They are Estimation, Filtering, Math Functions, Quantizers, Signal Management, Signal Operations, Sinks, Sources, Statistics, and Transforms. Explaining the details of many of these blocks would take us beyond the scope of this book, so we will leave out the Estimation library and some of the blocks in the Filtering and Math Functions libraries. We start our discussion by navigating the Filtering library. Let us use the model we created above, but this time we add filtering blocks from the Signal Processing library. Open the model butterworth_sp (shown in Figure 4.17) that now includes a block from the Filtering library that automatically designs an analog filter of any type. The list of possible filters goes way beyond the Butterworth that we have been exploringso far. When youopen themodel, it will contain the Butterworth filter we designed using the butterworthncs M- file along with the Filter block from the Signal Processing Blockset. This block contains the design specifications for the same Butterworth filter. This model illustrates another powerful feature of Simulink. In the process of making a change to the model (perhaps a change that does nothing but simplify it, as we are doing here), we need to be worried that the change might introduce an error. The simple expedient of comparing the two calculations using the summation block (to compute their differences) creates an immediate and unequivocal test for the accuracy of the calculations. The difference, displayed on a Scope, should be about the numeric precision. The result, displayed in a Scope block, contains both the Signal Processing Blockset results and the difference between the design using the NCS library design and the Signal Processing Blockset design of the Butterworth filter. The Scope we call “Compare SP block with NCS block ” gives the plot shown in Figure 4.17(b). As can be seen in this figure, the difference between the


147 Analog Filter Design from SP Blockset

Compare SP block with NCS block

butter SP Blockset Result

Difference between NCS and SP Library Butterworth filters

Generate 15 Sine Waves at Frequencies given by freqs. (set by a call back when the model loads) BWgain

15

D/A Filter Output

BWpoles(s) A/D Converter (Sampled at 0.1 msec)

Butterworth Filter of Order n Computed using the NCS m-file butterworthncs

Transport Delay

D/A Error

Sum ofSines Input

This model uses the data created by a callback to MATLAB when the modle opens as follows: The source is a sum of sine waves at the following frequencies freqs = [65 72.8 74.7 82 89.3 91 99.5 103.2 125 180.2 202.1 223.3 230.3 310.3 405.2] Hz. The phase of each sine wave is: Φ = [0 0.0011 -0.0014 -0.0013 0.0006 0.0002 0.0037 -0.0006 0.0044 0.0030 -0.0039 -0.0034 -0.0011 -0.0004 0 ] The amplitude of all of the sine waves and the sample times that they are generated are: amplitudes = 1; sample time = 10

-5

Analog Input Signal

sec.

a)

Simulink Model with Butterworth Filter from the Signal Processing Blockset and the Filter Designed using butterworthtncs. SP Blockset Result 15 10 5 0 -5 -10 0

4

x 10

0.05

0.1

0.15

0.2

0.25

0.2

0.25

Difference between NCS and SP Library Butterworth filters

-14

2 0 -2 -4

0

0.05

0.1

0.15 Time

b)

Reconstruction of an Analog Signal using a Filter Designed with the Signal Processing Blockset.

Figure 4.17. Using the Signal Processing Blockset to design an analog filter.

two implementations of the filter is mostly less than 2 × 10 14 , which is well within the numerical tolerances of the calculations. We can now experiment with the entire range of filters that the Signal Processing Blockset offers. Note that as the filter changes, the icon for the filter shows a plot of the −

148


Table 4.2. Comparison of the error in converting a digital signal to an analog signal using different low pass filters. 10th Order Filter Type

Time Delay milliseconds

A/D Error

Butterworth Chebyshev I Chebyshev II Elliptic Bessel

1.10 1.47 0.58 0.49 0.132

0.0761 0.4576 0.1859 0.3393 to 0.5108 0.274

frequency response of the transfer function, along with the name of the filter type. This provides a direct visual cue to the type of filter so that when we review it in the future, we know precisely the original intent (and if for any reason the picture and/or the type do not match the original intention or the speci fications, it is readily apparent). Five different filters are available from the analog filter design block. Try each of these filters in turn and record the error in the reconstruction of the original analog signal. We did this experiment, with the results tabulated in Table 4.2. (Each filter has a different time lag, so modify the delay time in the Transport Lag block, as shown, to account for this.) The elliptic filter stop band ripples range over 0.1 –2 db, hence the error ranges.

4.6.2 Creating Digital Filters from Analog Filters We have seen how one can use the state-space model to create a digital system that has the same response as the analog system when the input to the system is a step. In Section 2.1.1, the discrete time solution of a continuous state variable model was determined to be xk+1

= (t)xk + (t)uk ,

where (t) =

 

t

eAτ dτ ,

0

(t) =

t

eAτ Bdτ ,

0

and the MATLAB file c2d_ncs computes these matrices. We can use these methods of forming a discrete time system with a response that is equivalent to the analog system to make a digital filter from the analog filter. In this instance the responses are equivalent in the sense that both the analog and digital filters will have the same step response. (In Exercise 4.5 you will show that this is true.) Digital filters are usually equivalent (using this approach) to the analog Butterworth, Chebyshev, Bessel, or elliptic filters. The other equivalence that one can have is “impulse equivalence, ” where the impulse responses of the analog and digital filter are the same (but not the step responses). A third approach for the creation of an equivalent digital filter uses an approximation of the mapping of the Laplace and z-transform variables. The approximation is the bilinear 2 1−z 1 transformation given by s = t . This mapping in the complex plane is one-to-one, 1+z 1 −

  −


149

so the inverse mapping is unique and is (noting that this form is also bilinear) z

=



t s 2 t − s 2

1+ 1

The denominator in this mapping is of the form

z

=



=

1 + ts

1+

t

2

s



+

1+

t 2

2



.

1 = 1−x

t

2

s

+

1 + x + x 2 + · · · , so the value of z is

   t

2

2

s

+···

s2 + · · · .

This approximation is very close to the Taylor series for est , the actual mapping for the Laplace variable to the z variable. However, because the bilinear mapping is one-to-one, there is no ambiguity when this transformation is used. The powerful feature of this transformation is that one frequency exists where the phases of the transfer functions for both the analog and digital filters are the same. You can select this frequency using a technique called prewarping. Any of these approaches are selectable in the filter design block from the Signal Processing Blockset. In the next section, we look at one of these.

4.6.3 Digital Signal Processing One of the consequences of using a linear system described by a differential equation to filter data is the time lag that we encountered in the low pass filters we used in the previous section. In signal processing, these are “infinite impulse response,” or iir , filters. If it is necessary to eliminate this time lag, then the filter transfer function needs to be real at all frequencies. (If it is complex, then the imaginary part of the filter transfer function corresponds to a phase shift that results in the time lag.) For an iir filter, the only way that the transfer function can be real is if the poles are symmetric with respect to the imaginary axis (i.e., for any pole with negative real part, there is an equivalent pole with positive real part). Since poles in the right half plane (with positive real parts) are unstable, it is clearly impossible to build an iir filter that processes the signal sequentially and has no phase shift. The next best attribute we can ask for an iir filter is that its phase be linear. This is possible, and many filter designs impose this criterion in the development of the filter specification. There is an alternative to an iir filter; it is called “finite impulse response,” or fir , filter. The fir filter does not require the solution of a differential equation but instead relies on the convolution of the input with the filter response to create the output. The most important attribute of a fir filter is that it always has linear phase. Let us explore this attribute with some Simulink models. The first model we create uses one of the simplest and, for many applications, mostuseful of all fir filters. This filter is themoving average filterthat computes the mean of some number of past samples of a signal. The filter is yk

=

1 n

n−1

 i =0

uk −i .

150


Band-Limited White Noise 1

[zeros(n-2,1);1]

ones(1,n-1)/n* u

z Unit Delay

b Vector

Scope

c

diag(ones(1,n-2),1)* u

A Matrix 1/n d

a) The FIR Filter Simulink Model 2 0 0 P o i n t M o v in g A v e r a g e Simulink Simulation Using State Sace Model 1 0.8 0.6 0.4

0.2

0 -0.2 -0.4 -0.6 -0.8

-1

0

50

100

1 50

200

2 50 Time

3 00

350

400

4 50

50 0

b) Simulation Results

Figure 4.18. Moving average FIR filter simulation.

The z-transform of this filter is Y(z)

=

1 n



1 + z−1 + · · · + z−(n−1)



=

1 nzn−1



zn−1

+

zn−2

+···+z+



1 .

Thus, this filter has n − 1 poles at the origin. This means only that you have to wait for n samples before there is an output. The process for computing the output yk is to accumulate (in the sense of adding to the previous values) n1 times the current and past values of the input. The Simulink model that generates the fir n-point moving average filter (Figure 4.18) is in the NCS library. It is Moving_Avg_FIR. Exercise 4.6asks that youverify that this model is indeedthe state-space model of the filter. The simulation computes the average of the input created by the Band Limited White Noise block, which we investigate in more detail in the next chapter. The important information about this block is that the output is a sequence of Gaussian random variables with zero mean and unit variance (so the moving average should be zero). When the model opens, the callback sets n to 200 so the average is over the previous 200 samples. The result is in Figure 4.18(b).


151

You can change thenumber of points in the moving average by changing thevariable n in MATLAB. Beware, however, that thisimplementation of the filter is extremely inef ficient, so as n gets larger, the time it takes to compute the moving average increases dramatically. This last point illustrates a very important aspect of filter design. That is, the design approach makes a dramatic impact on the time it takes to perform the computations and the accuracy of the result. Let us explore this with the Digital Filter design block in the Signal Processing Blockset. The model Moving_Avg_FIR_sp in the NCS library uses the signal processing Digital filter block to create an fir filter that is identical to the statespace model above. Double click on the “Digital Filter” block in the model, and the window on the right will appear. This dialog allows you to select the filter type (in the “Main” tab); in this case, we selected FIR (all zeros). Next, we select the filter structure, the feature of the block that allows a more robust implementation. This is not critical in the simulation of the filter (although using a bad implementation as we did above can cause a long wait for the simulation to be completed), but it is absolutely critical in implementing a filter in a real-time application. Using the smallest number of multiplies, adds, and storage elements can greatly simplify the filter and make its execution much more rapid. In this implementation, we have used the “Direct Form” of the filter with the numerator coef ficients equal to 1/n*ones(1,n), where in the simulation, n 200. One last feature of the Filter block is the ability to view the filter transferfunction as a frequency plot (and in other views). Clicking the “View Filter Response” button creates theplot shown at the right. The plot opens showing the magnitude of the filter transfer function plotted versus the normalized frequency. The viewer allows you also to look at =

•

the phase of the filter,

•

the amplitude and phase plotted together,

152

Chapter 4. Digital Signal Processing in Simulink Pole/Zero Plot

1

0.8

0.6

0.4

t r 0.2 a P y r 0 a n i g a m-0.2 I

199

-0.4

-0.6

-0.8

-1

-1

- 0.5

0

0.5

1

Real Part

a) Pole Zero Plot

b) Filter Properties

Figure 4.19. Pole-zero plot and filter properties for the moving average filter.

•

the phase delay,

•

the impulse or the step response,

•

the poles and zeros of the filter,

•

the coef ficients of the filter (both numerator and denominator polynomials),

•

the data about the filter ’s form,

•

issues about the filter implementation by viewing the filter magnitude response with limited precision implementations.

We select these using the buttons along the top of the figure (circled in the figure). Two of the more interesting outputs from this tool are the pole-zero plot and the filter properties. Figure 4.19 shows these for the moving average filter. Notice that the filter properties gives a calculation of the number of multiplies, adds, states, and the number of multiplies and adds per input sample. (In this case the numbers are the same since the fir filter requires only multiplications and adds for each of the zeros.) We will explore some of the filter structures and their computation counts in Exercise 4.6. For now, we can really appreciate the difference the implementation makes by using the sim command in MATLAB to simulate both of the moving average Simulink models. The MATLAB code for doing this is tic;sim(’Moving_Avg_FIR’);t1=toc; tic;sim(’Moving_Avg_FIRsp’);t2=toc; tcomp = [t1 t2]

The results from this code (on my computer) are tcomp = 18.6036

0.5747


153

The difference is so dramatic because the first implementation (the state-space model) has a state matrix a that is 199 × 199. At every iteration, this matrix multiplies the previous state. Then there is a vector addition, followed by a vector multiply and an addition (for the feed-through of the input). Use this as a guide for determining the calculation count in Exercise 4.6. In general, all fir filters have the form of a sum of present and past values of the input multiplied by coef ficients that are the desired ( finite) impulse response. Thus, the general fir filter and its z-transform are n 1  −

yk

=

hi uk −i and H(z)

i =0

=

Y(z) U(z)

n 1 

=



n−1

−

hi z

−i

=

i =0

(z−1

+

zeroi ).

i =0

Notice that this filter has n zeroes (zeroi ) that are determined by factoring the polynomial H(z). Because the highest power in this transfer function is z−n , the filter has n poles at the origin (which corresponds to the fact that the filter output is only complete after n samples; there is an n-sample lag before the correct output appears). This appears in the response of the moving average filter from the Simulink model above. The discussion above was for a filter that computes the mean of n samples. Because of the study of this type of filter in statistics, all zero filters are “moving average” (or MA filters). The statisticians also developed iir filters that were all poles, and they dubbed these autoregressive (or AR- filters). Last, if a filter has both poles and zeros, it is an ARMA- filter . The filter design block allows you to create all three types of filters. The dialog also allows the user to specify where the filter coef ficients are set in the model. The option we have used is to set them through the filter mask dialog. (They are computed from the poles and zeros set in the dialog.) You also can select an option where the filter coef ficients are an input (they can then be computed in some other part of the Simulink model for use in this block, thereby changing the filter coef ficients “on the fly”), or the coef ficients can be created in MATLAB using an object called DFILT. Finally, the dialog allows the user to force the filter to use fixed-point arithmetic. This leads us to the discussion in the next section.

4.6.4 Implementing Digital Filters: Structures and Limited Precision Very few analytic methods allow one to visualize the effects of limited precision arithmetic on a filter. The use of simulation in this case is mandatory. Therefore, let us explore some of the consequences of designing a filter for use in a small, inexpensive computer that, say, has only 16 bits available. Input Signal (Sum of sinusoids) 15 10 5 0 -5 -10

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

154


Table 4.3. Speci fication for the band-pass filter and the actual values achieved in the design. Specification

Spec. Values

Sample Freq. Pass Band #1 Edge Stop Band #1 Edge Pass Band #2 Edge Stop Band #2 Edge Stop Band 1 Gain Transition Width Pass Band Ripple

10 kHz 100 Hz 110 Hz 120 Hz 130 Hz −60 dB 10 Hz 1 dB

Generate 15 Sin e Waves at Frequencies given by freqs. (set by a call back when the model loads ) double

double

Actual Values (−3 dB point) 10 kHz 100 Hz 109.7705Hz 120.2508 Hz 130 Hz −60 dB 10 Hz 1 dB

Digital Bandpass Filter Designer Bandpass

double

Floating Pt. Bandpass Filter

View Signals

Figure 4.20. The Simulink model for the band-pass filter with no computational limitations.

The first question we need to ask is, what ’s the problem? We have created a very simple example of a digital signal-processing task that will allow us to explore some of the limited precision arithmetic features built into the signal processing blockset. The model, called Precision_testsp1 or 2 in the NCS library, is a simulation of a device that one might design to try to find a single tone in a time series that consists of a multitude of tones. We might use it, for example, in a frequency analyzer or in a device to tune a wind instrument during its manufacture. It consists of a band-pass filter with a narrow frequency range (10 Hz in this case) and a sample time of 10 kHz. The input to the filter is the same sum of sinusoids that we have used previously, namely sine waves at the frequencies 65, 72.8, 74.7, 82, 89.3, 91, 99.5, 103.2, 125, 180.2, 202.1, 223.3, 230.3, 310.30, and 405.2 Hz, in the figure above. The band-pass filter specification is in Table 4.3, and the Simulink model Precision_testsp1 that we use to test the design is in Figure 4.20. If you open this model, you will see that it contains the Bandpass Filter block from the “Filter Design Toolbox” Simulink library. This block converted the specification in the table into filter coef ficients used by the Digital Filter block. The digital filter block allows you to take the floating point design and convert it to an equivalent fixed point design. Figure 4.21 shows the amplitude-versus-frequency plot for the filter designed by the “Band Pass Filter Design ” block. (Open the model in the NCS library and double click on this block to see the speci fication and create this plot.) The solid lines in the figure are the filter amplitude, as designed, and the dotted lines are the speci fication from the table


155 Magnitude Response (dB)

0

-10

-20

) B d (

-30

e d u t i n g a M -40

-50

-60

-70 0

0.02

0.04

0.06

0.08 0.1 Frequency (kHz)

0.12

0.14

0.16

0.18

Figure 4.21. The band-pass filter design created by the Signal Processing Blockset digital filter block.

above. When this filter is used, the digital filter response is almost indistinguishable from the response of an analog filter. We have deliberately designed this filter for a frequency that is not contained in the 15 sine waves that make up the input. (The pass band is from 110 to 120 Hz; none of the sinusoids have this frequency range.) Thus, the output of the filter should be very small. This is in fact the case as can be seen in the response plot (Figure 4.22(a)). The amplitude of the output (after the initial transient dies down) is about 0.01 (about 1/1000 the amplitude of the input signal), which is very good for detecting that the tone is not present. The next part of the design is to place the filter pass band in the area where we know there is a tone. Thus, let us try to pick out the tone at 103.2 Hz by specifying that the pass band corners will be 100 and 110 Hz (so the values of Fstop1, Fpass1, Fstop2, and Fpass2 are 90, 100, 110, and 120, respectively). This is so easy to do in the design that it amounts to a trivial change (in contrast to designing this filter by hand). You should try this, if you have not already. If you have not, the Simulink model Precision_testsp2 in the NCS library has the changes in it. Open this model and double click the Bandpass Filter block to view the changes that were made. (The dialog that opens has the new design for the pass band.) When you run this model, the response should look like the plot shown in Figure 4.22(b). The maximum output is now about 1.5 (compared to the 0.15 above), showing that there is a tone in the 10 Hz frequency range of 100 to 110 Hz. The filter design method that we use for both of the band-pass filters is the “Second Order Section” (or Direct Form Type II). When a digital filter is implemented using floatingpoint operations, the structure of the filter does not usually matter, but when the filter isfor a fixed-point computer, as we intend to do next, the way that the filter is structured can make a huge difference. For this reason, the Signal Processing Blockset includes design methods for a wide variety of filter structures.

156

Chapter 4. Digital Signal Processing in Simulink Output of Bandpass Filter 0.2

0.1

0

-0.1

-0.2

0

0.2

0.4

0.6

0.8

1 Time

1.2

1.4

1.6

1.8

2

a) Output of the Bandpass Filter when no Signal is in the Pass Band

1.5 1 0.5 0 -0.5 -1 -1.5 0.2

0.4

0.6

0.8

1 Time

1.2

1.4

1.6

1.8

2

b) Bandpass Filter Output when One of the Sinusoidal Signal is in the Pass Band Figure 4.22. Band-pass filter simulation results (with and without signal in the pass band).

Why is the implementation method an issue for fixed-point calculations? In Section 4.4, we showed that the transfer function for a discrete linear system comes from the Z{y } state-space model as H(z) = Z {ukk } = C(zI − )−1 B + D. The poles of the discrete system and the eigenvalues of the system are given by the det (zI − (t)), so using the denominator polynomial to create a digital filter will result in coef ficients that range from the sum of the eigenvalues (or poles) to their product. All of the eigenvalues (poles) of the z-transform transfer function are less than one because the matrix (t) is λ1 t

(t)

=e

At

= T

−1

e  0  ... 0

0 e λ2 t .. . 0

··· ··· ··· ···

0 0 .. . eλn t

   T ,

where T is the matrix that diagonalizes the matrix A (and (t)). The denominator of n λj t ). the z-transform is the determinant of this matrix and is the polynomial j =1 (z − e (Exercise 4.7 asks that you show that this is true.)




157

If this polynomial de fines the filter denominator, the precision needed to store the coef ficients would range over many orders of magnitude. Consider, for example, the fourth order filter with poles at 0.1, 0.2, 0.3, and 0.4; its denominator polynomial is z4 + z3 + 0.35z2 + 0.05z + .0024, so the coef ficients span almost three orders of magnitude. Each coef ficient would therefore require at least 10 bits just to get one bit of accuracy. The simplest way to avoid this is to break the filter up into cascaded second order systems. Second order sections can eliminate complex arithmetic. (The denominator polynomial of ∗ ∗ 2 a second order system is always z2 + (eλi t + eλi t )z + eλi λi t , which is always real.) The cascaded second order sections that were designed by the filter designer for this model can be seen by looking under the mask (by right clicking on the Bandpass Filter block, and selecting “look under mask ” form the menu that opens). The block that appears is the “Generated Filter block, ” and we open it by double clicking. When this block opens, you will see the cascaded second order sections (unconnected) that make up the filter (as shown in the figure below). 1 Input.

CheckSigna Attributes

-K-

[Sec t1]

s(1)

[Sect1]

-K-

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a(3)(4)

b(3)(4)

The coef ficients are in gain blocks, and we can inspect them by double clicking on the block. It is very useful to become familiar with navigating around the Signal Processing Blockset generated models this way. Now that a filter design exists, we impose the requirement that the calculations must use a precision of 16 bits. Designing a filter with limited precision arithmetic is simple if the filter design block creates the design. Once it is complete, MATLAB uses the filter design block to design the implementable digital filter. Before we do this, we need to think a little about what we mean by limited precision arithmetic and digital computing without floating-point calculations. Whenever a calculation uses floating-point arithmetic, the computer automatically normalizes the result (using the scaling of the floating-point number) to give the maximum precision. (Among the many references on the use of floating point, [31] describes how to use a block floating-point realization of digital filters.) The normalized version of any variable in the computer is x = ±(1 + f ) · 2e , f is the fraction (or mantissa) and e is the exponent.

The fraction f is always positive and less than one and is a binary number with at most 52 bits. The exponent (in 64-bit IEEE format) is always in the interval −1022 ≤ e ≤ 1023. Because of the limited size of f , there is a limit to the precision of the number that can be represented. (In MATLAB, this is captured by the variable eps = 2−52 , the value for any computer using the IEEE floating-point standard for 64-bit words.) The exponent has a similar effect, except that its maximum and minimum values determine the smallest and the largest numbers we can represent. It would pay to read Section 1.7 of NCM [29] to make sure that you understand the concept of limited precision.

1 Output

158


Inexpensive computers do not have floating-pointcapability. (Theyhave “fixed-point” architectures, where the user places the binary point at a fixed location in the digital word.) Programming these computers requires that the programmer ensure that the calculation uses enough bits. When we say enough bits, we are immediately in the realm of speculation because the number of bits needed for any calculation depends on the range of values that the data will have. For example, if we are interested in building a digital filter that will act as the tone control for an audio ampli fier, the ampli fier electronics immediately before the analog-to-digital conversion determines the maximum value that the signal will ever have. The amplifier might limit the signal to a maximum of 1 volt. The result of the conversion of the analog signal will be a number that ranges from −1 to 1. It is then easy to scale the fixed-point number so its magnitude is always less than one by simply putting the binary point after the first bit. (The first bit is then the sign bit, and the remainder of the bits are available to store the value.) This scaling can be kept for all of the intermediate calculations, but depending on the calculation (add, subtract, multiply, or divide), this may not be best, so scaling needs to be reconsidered at every step in the computation. When the amplitude of the signal being filtered is not known—as, for example, when filtering a signal that is being received over a radio link —the designer needs to figure out what the extremes for the signal will be and ensure that the digital word accommodates them. When the extremes are exceeded, the conversion to a digital word reaches a limit (which is called saturation and results in the most signi ficant bit in the conversion to try to alter the sign bit). When the digital word has a sign, the computer recognizes this attempt and creates an error. When using unsigned integers, the bit overflows the register. (The computer sees a carry bit that has no place to go, resulting in an over flow error.) To handle the limited precision, the designer needs to figure out what the effect of the limited precision will be both in terms of the accuracy that is required and in terms of the artifacts introduced by the quantization. As we will see, quantization is a nonlinear effect that can introduce many different noises into a signal that may, at a minimum, be distracting and, in the worst case, cause the filter to do strange things (like oscillate). Based on the discussion above, it should be clear that fixed-point numbers accommodate a much smaller dynamic range than floating-point numbers. The goal in the scaling is to ensure that this range is as large as possible. The scaling usually used is to have a scale factor that goes along with the digital representation and an additive constant (or bias) that determines what value the digital representation has when all of the bits are zero. The scaling acts like a slope and the additive constant acts like the intercept in the equation of a straight line, i.e., V r epresentation = S · w + b. As was the case for the floating-point numbers, the constant S = (1 + f ) · 2e , where the magnitude of f is less than one and b is the bias. The difference between the floating-point and fixed-point representation is that the value for e is always the same in the fixed-point calculations. The programmer must keep track of the slope and the bias. The computer uses only w during the calculations. Simulink fixed-point tools have many different rules for altering the scaling for each of the calculations that occur ensuring the best possible precision (with constraints). Simulations insure that the input signals truly represent the full range that is expected. The simulation also must ensure that the calculations at every step use values that have ranges that are consistent with the data.


159

Figure 4.23. Fixed-point implementation of a band-pass filter using the Signal Processing Blockset.

With this brief discussion, we can begin to investigate digital signal processing on a fixed-point machine. Open the models Precision_testsp2 (see Figure 4.23(a)) and double click the Digital Filter block. The dialog in Figure 4.23(b) will appear. The infor-

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mation in the Main dialog consists of the data created by the filter design in the “Digital Band-pass Filter Designer ” block in the model. The Transfer function type is iir , and the filter structure is the biquadratic direct form that we speci fied. The dialog has a tab that forces the filter design to use a fixed-point architecture. Click this tab and look at the options. Since the filter consists of four cascaded second order sections, the first issue we need to address is how to scale each of the outputs. As a starting guess we specify that each second order filter will use the full precision of the 16-bit word, so we put the binary point to the right of the sign bit (i.e., between bit 15 and 16, so that the fraction length —the length of w in the scaling equation —is 15 bits). The coef ficients are also critical, so they need as much precision as possible. We have speci fied overall accuracy of 32 bits, with 15 bits for both the numerator and denominator coef ficients in each of the second order sections. Each multiplication in our computer has 32 bits of precision, so we specify that the word length is 32 bits, and we again allow the most precision of 31 bits for the “Fraction Length. ” The output will use a 32-bit D/A conversion, so the output is scaled the same as the accumulator. These guesses for the scaling are now tested. The first step in the process is to run the model (with the guesses) to see what the maximum and minimum values of the various signals are. To do this, just click thestart button. The simulation will run, andthe maximum and minimum values appear in MATLAB in a 45-element cell array called FixPtSimRanges. The last five values in this cell array show the simulation results for the fixed-point calculations. For example, the 45th entry, obtained by typing FixPtSimRanges{45} at the MATLAB command line, contains

Path: [1x54 char] SignalName: ’Section output’ DataType: ’FixPt’MantBits: -16 FixExp: -15 MinValue: -1.0000 MaxValue: 0.9998. The nextstepis toallowthe “Autoscale” tool to adjustthe scaling to give themaximum amount of precision to the implementation. To do this, select “Fixed Point Settings ” under the Tools menu. This selection opens the dialog in Figure 4.24. To automatically scale the calculations (and in the process compute all of the scale factors S associated with each calculation), simply click the “Autoscale Blocks” button at the bottom of the dialog. The results appear by opening the Digital Filter block and then looking under the Fixed-Point Tab. The Autoscale will have scaled all of the calculations, but you should find that the values we selected are good. Now that the filter is complete, you can experiment with different word sizes and precisions, and see what the effect is on the output of our filter. However, if you look at the “View Signals” Scope block, you will notice that even with the design optimized for precision, the fixed-point and floating-point computations are not the same.

4.6.5 Batch Filtering Operations, Buffers, and Frames The Signal Processing Blockset can also process signals in a “batch” mode where the signals are captured and then buffered in a register before the entire sequence is used to calculate a desired functional. Themost frequent useof this techniqueis in thecalculation of theFourier transform of a signal. Remember that theFourier transformis an integral over alltime. Since


161

Figure 4.24. Running the Simulink model to determine the minima, maxima, and scaling for all of the fi xed-point calculations in the Band-pass filter.

we can capture only a finite time sample, any calculation for a signal processing application will at best approximate the transform. In addition, the data will be digital, so the transform will be done using an approximating summation. The most frequent approximation is the fast Fourier transform (FFT). It is the subject of Chapter 8 of NCM, so we will assume that the reader is familiar with its method of calculation. In order to illustrate the concept of buffers and the FFT in the Signal Processing blockset, consider the problem of recreating an analog signal from its samples (the sampling theorem problem we investigated above). In this case let ’s not try to find an analog filter that will do this, but let ’s try to use the Fourier transform (in the form of the FFT). The model FFT_reconstruction in the NCS library (Figure 4.25) starts with the same sum of 15 sinusoids that we have been using in the previous models. The reconstruction, however, uses the transform of the signal. Let us look at the theory behind the calculations, and then we will explore the model ’s details. Remember that the sampling theorem told us that to reconstruct the signal from its samples we need to multiply the Fourier transform of the sampled signal by the function that is 1 up to the sample frequency and zero elsewhere. When we take the Fourier transform using the FFT, we get frequencies that are up to half the sample frequency. Thus, if the

162

Chapter 4. Digital Signal Processing in Simulink Delay by: samp_time*nbuffer (Accountsfor lag in buffer)

Generate 15 Sine Waves at Frequencies given by freqs. (set by a call back when the model loads) 15

Sampled Input

Pad Transform with Zeros [512x1]

Buffer (nbuffer values)

FFT

[512x1] [512x1]

F FT

Pa dd ed F FT

Unbuffer (ndesired values) 4096

IFFT

Make Ressult Rea l (Eliminate Residual Imaginary Part)

[4096x1]

Data

Inherit Ref Complexity

IFFT

FFT

Scale by: nd/nbuf -K-

Reconstructed at: nd/nbuf

View Signals

[512x1]

|u|

[512x1]

Compute |FFT|

Freq

DSP Constant

1

Short-Time Spectrum

Frequency Domain Sampled Signal Reconstruction Ideal Low Pass Filter Created by Padding the FFT

Figure 4.25. Illustrating the use of buffers. Reconstruction of a sampled signal using the FFT.

conditions for the sampling theorem (that the signal is band limited) are valid, the FFT is just S(ω), −ωM ≤ ω ≤ ωM , where ωM is the maximum frequency contained in the signal and the values of ω are just 2ωnM , where n is the number of samples of the signal that were transformed. In creating the model in Figure 4.25, the first step was to save n samples of the input signal for subsequent processing by the FFT. The block in the Signal Processing Blockset that does this is the Buffer block. It simply stores some number of values of the input in a vector before passing it on to the next block. When the model opened, three data values (called nbuffer, npad, and ndesired) appear in MATLAB. Nbuffer is the size of the buffer, and it is set to 512. (It always must be a power of two for the FFT block.) The FFT and the inverse FFT use the blocks from the Transforms library in the Signal Processing Blockset. Thus, outwardly, all we are doing in the model above is taking the transform of the input and then taking the inverse transform to create an output. The output, though, cannot be the result of the inverse transform, since this is going to be a vector of time samples that, presumably, matches the output of the Buffer block. The way to pass the vector of samples back into Simulink as a set of time samples at the appropriate simulation times is to use the “Unbuffer ” block. (Both the Buffer and Unbuffer are in the Signal Management library under Buffers.) To apply the Sampling theorem, you need to multiply the transform of the sampled signal by the pulse function and then take the inverse Fourier transform(which is continuous so this results in a continuous time signal). We are using the IFFT block that takes the inverse Fast Fourier Transform and therefore outputs a signal only at discrete values of time. Therefore, in order for the inverse to have more time values (thereby filling in or interpolating the missing samples), we need to increase the size of the FFT before we take the inverse. We do this by padding the transform with zeros to the left of −ωM and to the right of ωM . The subsystem called “Pad Transform with Zeros ” in Figure 4.26(a) does this. To see how, double click on the block to open it. There is a block called “Zero Pad” in the “Signal Operations ” library, and we use it to do the padding. The block adds zeros at either the front or the rear of a vector. However, we need to be careful about this. Remember that when the FFT is computed, the highest frequencies are in the center of the transform vector, and the lowest (zero frequencies) are at the left and right of the vector (i.e., the transform is stored from 0 to −ωM and then from


163

Zero Pad the Transform (at the lowest neg. freq.) 1 FFT

MATLAB Function

[512x1]

512

Un-Shift the Transform so Zero is at the Ends.

2304

Shift the Transform so Zero is at the Center.

MATLAB Function

4096

Zerod Pad the Transform (at the highest pos. freq.)

4096 4096 4096

|u|

1 Padded FFT View Padded Transform as a Matrix

|FFT| padded

4096

[8x4096] 4096

Matrix Viewer

Store the |FFT| in rows of a matrix

a) Simulink Subsystem that Pads an FFT with Zeros to Increase the Number of Sample Points in the Transform 300

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b) Padding an FFT with Zeros Must Account for the Fact that the 0-Frequency is Not at the Center of the FFT Figure 4.26. Using the FFT and the sampling theorem to interpolate a faster sampled version of a sampled analog signal. to 0 with a discontinuity at the center of the array). If you run the following MATLAB code, it will generate the plot in Figure 4.26(b) to illustrate this: ωM

t = 0: .01:511*.01; y = sin(t)+sin(10*t)+sin(100*t); z=fft(y); plot (abs(z))

The plot shows the fact that the transform has the zero frequency at the 0th and the 512th points computed. Because of this FFT quirk, padding the FFT with zeros during the computation in theSimulinkmodelwill notadd zeros below −ωM and above ωM . (Exercise: What does it do?) There is a built-in command in MATLAB that we will use to rotate the fft. The function is fftshift, and it converts the FFT so its zero frequency is at the center of the

164


plot (as it appears using the Fourier transform). To see the effect of using this command, replace the last plot command with plot (fftshift(abs(z))) .

There is no function in the Signal Processing library that does the equivalent of Therefore, we usethe MATLAB function block in theSimulink library to usethe MATLAB fftshift. The function needs to be invoked twice: once before we do the zero padding andonce after to putthe FFTback into the correct form forthe inverse (IFFT block). Two additional blocks used in the model come from the Signal Processing Blockset sinks library. They are the Short Time Spectrum block and the Matrix Viewer block. The first block is used to view the FFT as it is computed, and the second allows us to view the padded FFT as it is computed (in a three-dimensional plot). In this plot, time and frequency are the two axes, and the color is the amplitude of the FFT. (In the book it is in a gray scale, but in the model that you are running from the NCS library, it is in color.) Another attribute used in the model is to display the length of the vectors and the sample times on the various lines using different colors for the lines. These attributes come from the Port/Signal Displays under the Format menu in the model. Now, with this understanding of the mathematics involved, it should be clear that if we pad the transforms by some number of zero elements to make the final padded value have 2 n points, the inverse FFT (IFFT) will result in a time series with 2 n values. Thus, making 2n > nbuff er , the output will have more time samples than the input. From the sampling theorem, this output is the result of multiplying the FFT of the input by the pulse function, and the result is a signal that perfectly—to within numeric precision, of course —reconstructs the original signal at the new sample times. The result of running this model is shown in Figure 4.27 for the reconstruction of the samples signal at 8 times the input frequency ( ndesired = 4096, nbuffer = 512, and npad = 1792). The results show very good reconstruction of the sampled signal, as we would expect. Figure 4.28(a) shows the output from the Short-Time Spectrum block in the model, and Figure 4.28(b) shows the plot created by the Matrix Viewer block. Almost every one of the 15 frequencies in the input sine wave can be seen in the peaks of the spectrum (which is really the FFT), and the padding of the FFT to create the new output can be seen in the Matrix Viewer output. We have explored about 20% of the capabilities of the Signal Processing Blockset in this chapter. For example, dramatic improvements in many signal-processing applications result from the processing of a large sample of data transferred to the computer as a contiguous block. To some extent, we have seen this in the example above, where we buffer the signal data before sending it to the FFT. The same approach applies to filters and many other signal-processing operations. When we do this in Simulink, the operation uses a vector called a “frame.” Frames make most of the computationally intensive blocks in the Signal Processing Blockset run faster. There are many examples in the demos that are part of the signal-processing toolbox that illustrate this, and now that you understand how the buffer block works, you should be able to work through these examples without trouble. Furthermore, it is a good idea to look at all of the demos, and also to open all of the blocks in the blockset to see how each of themworks, and, along withthe help, determine whateachof the blocks needs in terms of data inputs and special considerations for the use of the blocks. fftshift.

4.7. The Phase-Locked Loop

165 Analog Signal Sampled at 1kHz

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Figure 4.27. The analog signal (sampled at 1 kHz, top) and the reconstructed signal (sampled at 8 kHz, bottom).

4.7 The The Phase-L Phase-Lock ocked ed Loop An interesting signal-processing device incorporates the basics of signal processing and feedback control. The device, a phase-locked loop (PLL), is an inherently nonlinear control system. system. It is extremel extremely y simple to understand, understand, and Simulink Simulink provides provides a perfect perfect way for simulating the device. In the process of creating the simulation, we will encounter encounter some new Simuli Simulink nk blocks blocks,, we will will use some some famili familiar ar blocks blocks in a new way way, and we will will encoun encounter ter some numerical issues. We begin with how the PLL operates. Imagine Imagine that we want to track a sinusoid sinusoidal al signal. signal. The classic classic example of this is the tuner in a radio, television, cell phone, or any device that must lock onto a particular frequency to operate properly. In early radios, a demodulator followed an oscillator tuned to the desired frequency. frequency. The oscillator operated in an open loop fashion, so if its frequency drifted (or the signal ’s frequency shifted slightly), the radio needed to be manually retuned. The operation of the demodulator used the fact that the product of the incoming frequency and the local oscillator created sinusoids at frequencies that were the sum and difference of the input and oscillator frequencies. Mathematically this comes from sin(ω1 t +ϕ1 ) cos(ω2 t +ϕ2 ) =

1 2

(sin ((ω1 + ω2 )t + ϕ1 + ϕ2 ) + sin ((ω1 − ω2 )t + ϕ1

−

ϕ2 )) .

The output of the demodulator was the result of extracting only the difference frequency using a circuit tuned to this “intermediate” frequency. frequency. In most applications, applications, tuned radiofrequency amplifiers increased the intermediate intermediate signal ’s amplitude. amplitude. The PLL uses the same

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Figure Figure 4.28. Results Results of interpolat interpolating ing a sampled sampled signal signal usingFFTs usingFFTs and the sampling sampling theorem.


167 num(s) den(s)

Input Signal 101.56 Hz.

Modulator

Low Pass Filter (3rd Order Butterworth at 5 Hz.)

Error

VCOout

VCO Tracking Input

VCOin

Voltage Controlled Oscillator This Phase Locked Loop Simulation will Track a sinusoidal input that is in the frequency range of 95 to 105 Hx.

Figure 4.29. Phase-locked loop (PLL). A nonlinear feedback control for tracking frequency frequency and phase of a sinusoidal signal.

concept, except the oscillator frequency uses the difference sinusoid to change the frequency of the oscillator. The feedback uses a device called a voltage controlled oscillator (abbreviated VCO). The operation of the loop needs three parts: •

the VCO,

•

the device that creates the product of the input and the VCO output,

•

a filter to remove sin ((ω1 + ω2 )t + ϕ1 + ϕ2 ) before the result of the product passes to the VCO.

The term “locked” describes when the input frequency and the VCO frequency are the same. Thus, at lock, since the product has both the sum and difference frequencies, the difference frequency frequency is zero. Therefore, the filter that we need to remove the sum frequency is our old friend the low pass filter. We will build the loop simulation in Simulink using a third order Butterworth filter. The Simulink model of phase-locked loop is Phase_Lock_Loop in the NCS library (Figure 4.29). A MATLAB MATLAB callback (as usual) creates the parameters for the simulation. The Butterworth filter is third order, and its coef ficients cients are in the Transfer Transfer function function block. block. The modulator uses the product block. The VCO is a subsystem that looks like Figure 4.30. This implementation of the VCO ensures that the generated sinusoid always has the correct frequency despite the limited numeric precision of the simulation. The big worry is the effect of roundoff. roundoff. NCM has a discussion of the effect of calculating calculating a sinusoid using incrementsin increments in t thatlook like t next Ifthevalueof t cannot cannot be represent represented ed precisel precisely y next = t +t . Ifthevalueof in binary, then eventually the iteration implied by the equation will give an incorrect result. The integrator modulo 1 in the diagram above fixes this problem. Let us look at this subsystem (Figure 4.31). The first thing to note is that the integration is creating the ω2 t part of the argument of cos(ω2 t + ϕ 2 ) that is the VCO output. The second thing to note is that the integrator resets whenever the value of its output is one.

168


Scope Integrator Modulo 1 1

Kpll

In

Voltage Controlled Oscillator Output Ampitude cos(u[1])

I nt egral mo modulo 2p 2pi

VCOin

Amp

cosine

Gain f0

Oscill ator Ba se Oscill Frequency

Phase

1 VCOout

Oscillator Base Phase

The integrator output is approximately: approximately: 0 <= ( f0 + Input(t) Kvco ) t < 2

π

Figure 4.30. Voltage controlled oscillator subsystem in the PLL model. Gain 1+eps

In 1 1 s

>= Modulus 1

rem

Modulus

Math Function

1

1

Relational Operator

[0]

Initial Integral is set to 0 at t = 0

2*pi

Integrator

1 Integral modulo 2pi

xo Set value to be between 0 and 2 pi

************************************************************************** This integrator creates creates an output that that is between 0 and 2 π . The reset on the integration uses the state port at the top (so there is no algebraic loop). Any errors in the integration at the reset are calculated from the remain der. The integration starts at a value value that is its value value before the reset - 1. **************************************************************************

Figure 4.31. Details of the integrator modulo 1 subsystem in Figure 4.30.

This reset uses two options in the integrator block that we have not used before. The state port that comes from the top of the integrator. A check box in the integrator dialog causes the display of this port. You use it when you need the result of the integration to modify the input to the integrat integrator or.. If you fed the output back to the input, input, an algebraic algebraic loop would result that is dif ficult for Simulink to resolve. The port eliminates this loop. The second new input to the integrator is the “reset” port. This port is created when you select rising from the “External reset” pull-down menu in the integrator dialog. Thus, in the model, the integrator is reset to the initial condition whenever the relational operator shows that the output is greater than 1+eps. (eps is the smallest floating-point number in the IEEE floating-point standard; see NCM for a discussion of this MATLAB MATLAB variable.) The first is the


169 Input Signal

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Time

a) Phase Lock Loop Simulation Input and Output

Output of the Modulo Integrator

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Figure Figure 4.32. Phase-locked loop simulation results. integrator steps are not necessarily going to occur at exactly the point where the output is exactly exactly one. The external external initial conditio condition n in the integrator integrator is used at the reset to set the valu valuee of the the inte integr grat ator or to the the rema remain inde derr (the (the amou amount nt the the outp output ut exce exceed edss 1) for for the the next next cycl cycle. e. The output of the integral integral is therefore therefore a sawtooth sawtooth wave that goes from zero to one. We multiply the output by 2π before using it to calculate the cosine. Run the PLL Simulink model and observe the outputs in the Scopes. There are three Scope blocks, blocks, two at the top level of the diagram. diagram. The first shows the two sinusoids (the input and the VCO output), and the second shows the feedback signal that drives the VCO. Look at these and at the Scope block in the VCO subsystem. subsystem. (Figure (Figure 4.32(a) 4.32(a) shows the

170


plots of the input and output of the simulation.) Figure 4.32(b) shows the moduloarithmetic integr integrato atorr outputthat outputthat we descri described bed above, above, andFigure andFigure 4.32(c 4.32(c)) is thetrackingerrorin thetrackingerrorin thePLL. Because the frequency of the VCO is not exactly 101.56 Hz, the VCO frequency (VCOin) must must have have a nonzer nonzero o steady steady state state value. value. This This is clearl clearly y the case. case. Furt Further hermo more, re, the PLLgot to this steady state value with a rapid response and minimal overshoot. The design of the PLL feedback uses linear control control techniques that start with a linear model model of the PLL dynamics. Many more interesting and complex mathematical processing applications are possible using the blockset, including noise removal, more complex filter designs, and the ability to create a processing algorithm for an audio application and actually hear how it sounds on the PC. There are tools that allow the export of any signal processing application to Hardware Description Language form so that the processing ends up on a chip. Tools also will export designs to Xilinx and Texas Instruments architectures. We arenow ready ready to talk talk about about stocha stochasti sticc proces processesand sesand themathemat themathematicsof icsof simula simulatio tion n when the simulated variables are processes generated by a random quantity called “noise.” This is the subject of the next chapter.

4.8 4.8 Furthe Furtherr Readi Reading ng The Fibonacci Fibonacci sequence sequence and the golden ratio is the subject subject of a book by Livio Livio [26]. It is a very readable and interesting review of the many real (and apocryphal) attributes of the sequence. In graduate school (at MIT), Claude Shannon worked on an early from of analog computer called a “differential analyzer. ” His work on the sampling theorem links to the subject subject of simulation simulation in a very fundamen fundamental tal way. way. In fact, while he was working working on the differential analyzer he created a way of analyzing relay switching circuits that were part of these early early devices. devices. He published published a paper paper on these these results in the Transactions of the American Institute of Electrical Engineers (AIEE) that won the the Alfred Alfred Noble AIEE award. award. During World War II, Shannon worked at Bell Labs on fire control systems. As the war ended, Bell Labs published a compilation of the work done on these systems. Shannon, along with Richard Blackman and Hendrik Bode, wrote an article on data smoothing that cast the control problem as one of signal processing. This work used the idea of uncertainty as a way to model information, and it was a precursor to the discovery of the sampling theorem. The Shannon sampling theorem is so fundamental to all of discrete systems that his Proceedings of papers frequently reappear in print. Two recent examples of this are in the Proceedings the IEEE [23], [23], [39]. A paper paper in the IEEE Communications Society Magazine [27] followed shortly afterward. The best source for the proof of the theorem is in Papoulis [32]. Papoulis had had a knac knack k for for finding nding elegan elegantt proofs proofs,, andhis proof proof of thesamplin thesampling g theore theorem m is no except exception ion.. There There are hundre hundreds ds of texts texts on digita digitall filter lter design design.. One One that that is reason reasonabl ablee is by Leland Leland Jackson [22]. Reference [11] describes describes the operation of phase-locked loops. It also shows how they are analyzed using linear feedback control techniques, and how easy it is to create a loop in a digital form. To learn learn moreabout the tools, tools, blocks, blocks, and technique techniquess availablewith availablewith the Signal Signal ProcessProcessing Blockset and Toolbox, see The MathWorks Users Manuals and Introduction [43], [44].

Exercises

171

Exercises 4.1 Show that the Fibonacci sequence has the solution (completing the step outlined in Section 4.1.2) 1 f n = (φ n+1 − (1 − φ 1 )n+1 ). 2φ1 − 1 1 4.2 Verify using induction the result from Section 4.3 that:

xk

=



f k−1 f k

f k f k+1



x0 .

Is it a

fact that because you use the iterative equation in the model and the result is what you expect that this must imply that the result is true by induction? If you use a simulation of a difference equation, iteratively, to produce a consistent result, will this imply that the result is always true by induction? 4.3 Show, using induction, that the Fibonacci sequence has the property that f k +1 f k −1 − f k2 = ±1. (Follow the hint in Section 4.3.) 4.4 Verify that the Butterworth filter function 1

|H Butter |2 =

1+

2n

  ω ωm

is maximally flat. (All of its derivatives from the and infinity.)

first to the

(n − 1)st are zero at zero

d 4.5 Show that the analog system dt x(t) = Ax(t) + Bu(t) and the discrete system xk +1 = (t) xk + (t)uk have the same step responses when

(t)  (t)

=

t Aτ e dτ, 0

 

=

t Aτ e Bdτ . 0

Create a Simulink model with the state-space model of a second order system in continuous time and in discrete time using the above. Create a discrete time system using the Laplace transform of the continuous system 2 1−z−1 with the mapping s = t ( 1+z−1 ) to generate the discrete system. Compare the simulations using inputs that are zero (use the same initial conditions for each of the versions), a step, and a sinusoid at different frequencies. 4.6 Show that the fir filter for an n-sample moving average has the state-space model

xk +1

=

  

0 0 .. . 0

yk =

1 0 .. . 0 1 n



... ...

0 1 .. . 0 1

1

... ...

0 0 .. . 1

1

1

   

  

0 0 .. . 1

xk

+

xk

+ n1 uk .

  

uk ,

In this state-space model the A matrix is (n − 1) × (n − 1) and the vectors are of appropriate dimensions. Verify that the Simulink model in the text does indeed use

172

Chapter 4. Digital Signal Processing in Simulink this equation. Calculate the number of adds and multiplies this version of the moving average filter requires. Investigate other non–state-space versions of this digital filter. Use the Simulink digital filter library. Work out the computation count needed to implement the filters you select, and compare them with the number of computations needed for the statespace model. Is this model an ef ficient way to do this filter? What would be the most ef ficient implementation?

n

4.7 Show that det(zI (t)) is the eigenvalues of the matrix A. −

j 1 =

(z

−

eλj t ), where (t)

=

eAt and the λj are

Chapter 5

Random Numbers, White Noise, and Stochastic Processes Simulation of dynamic systems where we assume we know the exact value for all of the model parameters does notadequately representthe real world. Most of thetimethe designer of a system wants to know what happens when one or more components are subject to uncertainty. The modeling of uncertainty covers two related subjects: random variables and stochastic processes. Random variables can model the uncertainty in experiments where a single event occurs, resulting in a numeric value for some observable. An example of this might be the value of a parameter (or multiple parameters) speci fied in the design. In the process of building the system, these can change randomly because of manufacturing or other errors. Simulink provides blocks that can be used to model random variables using the two standard probability distributions: uniform and Gaussian or normal. Stochastic processes, in contrast to random variables, are functions of time or some other independent variable(or variables). The mathematics of these processesis complicated because of the interrelation between the randomness and the independent variable(s). In this chapter, we explore the use of Simulink to model both types of uncertainty. We introduce the two types of random variables available in Simulink and show how to create other probability distributions for use in modeling phenomena that are more complex. We extend ideas of random variables to the simplest of stochastic processes where at each of the sample times in a discrete system we select a new random variable. Finally, we show discrete time processes that converge to continuous time processes in the limit where the time steps are smaller and smaller.

5.1 Modeling with Random Variables in Simulink: Monte Carlo Simulations The simplest uncertainties one might need to model are the values of parameters in a simulation. These parameter uncertainties can be errors due to manufacturing tolerances, uncertainties in physical values because of measurement errors, and errors in model parameters because of system wear. Monte Carlo simulations use parameters selected from known probability distributions. The parameters are then random variables, so each simulation has 173

5.1. Modeling with Random Variables in Simulink: Monte Carlo Simulations From M ATLAB: Sum n+1 Uniformly Distributed Random Variables with 0 mean (values from -1 to +1 ) SumofUniforms

Variance

1/sqrt(n/6)

Sign

1/sqrt(n/2) Sum n/2 Binomial RVs

Create n/2 Random Variables with: Pr( x = 1 ) = Pr( x = -1 ) = 1/2 (Me an = 0 and Var = 1)

Display

Scope2

Summation form s Binomial Distribution for X, with -n/2 < X < n/2

Random Number

1.033

Scope1

Divide by Sigma of Uniform

5000

Running Var

Sum of Uniform rvs with mean 0 and unit variance

From Workspace

5000

175

simout

Sum of Binomial rvs with mean 0 and unit variance

To Workspace Scope

Divide by Sigma of Binomial

Running Var

0.9905 Display1

Variance1

Figure 5.1. Monte Carlo simulation demonstrating the central limit theorem.

350

300

250

200

150

100

50

0 -6

-4

-2

0

2

4

6

Figure 5.2. Result of the Monte Carlo simulation of the central limit theorem.

structure.) The first column is the time points (in this case the integers from 0 to n, which are the counters for the number of random samples we are creating and not time), and the second column is the random numbers generated in MATLAB using the rand function. The result of running this model with n 10,000 is shown in Figure 5.2. Note that the distribution is almost exactly Gaussian, with zero mean and unit variance as the central limit theorem demands. Thus, we have done a numerical experiment where we have made =

176

Chapter 5. Random Numbers, White Noise, and Stochastic Processes

multiple runs with random number generators, and we have created the probability density function for the result —and have seen that the result is Gaussian as theory predicts. This is exactly what Monte Carlo analyses are supposed to do. In the situations where it is used, the mathematics required to develop the probability distribution are so complex that this method is the only way to achieve the required understanding. In the simulation above, we generate 100 million random variables to create the histogram (50 million in MATLAB before we start the model, and 50 million over all of the time steps in the model). Using tic and toc, the total time that this took on a 3.192 GHz dual processor Pentium computer was 24.2 seconds. When the simulation ends, there is a callback (using the StopFcn tab in the File menu Model Properties submenu) that executes the following code to create the plot in Figure 5.2: y = simout.signals.values; hist(y,100).

It is important to see how the parameters were set up in the dialog box for the binomial distribution. We used the “Band Limited White Noise ” which generates a Gaussian random variable at each time step, and then we extracted from this a random variable that is +1 when the sign is positive and −1 when it is negative. This uses the sgn (sign) block from the Math Library. We add the elements in the resulting vector to create a random variable that lies between −n/2 and n/2; thus it is a binomial distribution. The dialog box for the Band Limited White Noise block has the value of the random seed set to 12345:100:12345+100*n/2-1.

The seed needs to be a vector of different values so that each of the n/2 generated random variables is independent. From this discussion, youshould be able to seehow to createa simulation that contains random variables either usingthe random number generators in Simulink or using MATLAB commands to generate random numbers. However, you might think from the example that the only options for the random variables are uniform distributions or Gaussian. Well, while you were not looking, we sneaked a binomial distribution into the model above. We actually generated this distributionusing Simulinkblocks. Let us explore another example, the Rayleigh distribution, which will give some insight into how to generate more complex (and consequently more likely to occur in practice) random variables.

5.1.2 Simulating a Rayleigh Distributed Random Variable Radio waves that re flect off two different surfaces have amplitudes that are the square root of the sum of the squares of the amplitudes from each direction. Thus, if the amplitude of each wave is Gaussian, the resulting signal is Rayleigh. To make this precise, let x(t) and y(t) be two Gaussian random variables; then the random variable z(t) = x(t)2 + y(t)2 is Rayleigh distributed. The probability density function for z(t) is



 z e for z σ  0 for z < 0. −z2 /2σ 2

f Z (z) =

2

≥

0,

5.2. Stochastic Processes and White Noise

177

sqrt Random Number

Product

Rayleigh Distributed Random Variable

Math Function

Running Var

Variance Random Number1

Scope

Product1 0.4292 Display

Figure 5.3. A Simulink model that generates Rayleigh random variables.

How do we generate a Rayleigh distributed random variable for use in Simulink? The answer is to go to the de finition. The model in Figure 5.3 creates a Monte Carlo simulation that generates a random variable with a Rayleigh distribution. This model is Rayleigh_Sim in the NCS library. When this model runs, the generated noise sample appears in the Scope, and when the simulation stops, a MATLAB callback creates the histogram for the generated process (as we did in Section 5.1.1 above). The figure at the right is the histogram that results. As an aside, the variance of the Rayleigh random variable is 2 0.4292. The simu2 lation uses the “variance” block from the Signal Processing library to compute the variance as the simulation progresses, and the value is exactly the 0.4292 for the 10 sec simulated in the example above. It should now be clear how one goes about creating random variables with different distributions. The first stepis to find a mathematical relationship between the desired random variable and the Gaussian or otherrandom variable, andthen build a setof blocksthat form that function. In Example 5.1, we askthat youmodify the Simulink model above to add the calculation of the random variable that is the phase angle of the re flected sine waves. Histogram for the Monte Carlo Experiment ofthe Rayleigh Distribution

70

60

e u l a V d e 50 i f i c e p S e h t 40 r o f

π

−

=

s e l p

m a 30 S f o r e b m20 u N

10

0

5.2

0

0.5

1

1.5

2 2.5 3 Values of the Random Variable

3.5

4

4.5

5

Stochastic Processes and White Noise

A random variable is a number that results from a random experiment. The experimenter may assign the number, or it might arise naturally because of the experiment. For example, with a coin flip, the experiment is the operation of flipping the coin and observing the result. The outcome is a member of the set that contains the two possible outcomes { “Head,” “Tail”}. A random variable for this experiment is the assignment of a number to each of the

178


outcomes (the number 1 for heads and the number 0 for tails, for example). A stochastic or random process is similar to a random variable except that the experimental result is the assignment of a function (usually, but not always, a function of time) to the event. Thus, the following are stochastic processes (in order of complexity from trivial to overwhelming). •

A random process resulting from an event with only two outcomes :

– Flip a coin, and assign the function sin (αt) if the result of flip is a head and assign the function e−αt if it is a tail.

•

The random walk process: “

”

– Start with x( 0) = 0. At every discrete time step kt , flip a coin and move left if the result is a head and move right if it is a tail.

– Thus, x(kt) = x({k − 1}t)) − 1 if a head appears and x(kt) = x({k − 1}t)) + 1 if a tail appears. •

The Brownian motion (Weiner) process:

– Let the coin flip above happen infinitely often in the sense that over any interval, t no matter how small, the coin is still

flipped

an in finite number of times.

Each of these processes is progressively more complex than the previous. The first is simple because there are only two outcomes so the probability for each is simply 12 . The second process has a growing number of possible outcomes, so at the nth time step there are n possible values for the process. The last process is inconceivably complex; it is the prototype for Brownian motion —the motion of a particle in a fluid. This process plays a fundamental role in the analysis of stochastic processes. (The process is also called the Weiner process.) Most of us have heard of white noise, perhaps only as a noise source that masks background noise to allow us to sleep well. However, this ubiquitous process is an essential part of allanalysesof systems with noise, andit is, therefore, a very convenient mathematical fiction. As a process, white noise does not exist (it is easy to demonstrate that this process has infinite energy), but it is possible to give some insight into its properties and show a way that it is related to the Brownian motion process, and as a consequence it can be subjected to a rigorous mathematical treatment. As the first step in understanding systems excited by noise, let us explore the Brownian motion process in more detail. We will see how to use it as a prototype of white noise.

5.2.1

The Random Walk Process

The starting point for our discussion of Brownian motion is the random walk process described above. Let us create a Simulink model that generates this process (Figure 5.4). We need to use the 1z block from the Discrete library to store the previous value of the Random Walk, and then we need to add this to the result of “Flipping the Coin, ” which is simulated using the sign of the Uniform Random Number Generator (a sign of +1 or −1). The simulation has a seed for the random number generator that is a vector of length nine, so each run creates nine separate instantiations of the random walk. Figure 5.5 shows


179

x

k+1

= x + sign(y ) k

k

x =0 0

+ or - 1 x_k+1

Sign

Uniform Random Number

Scope

x_k

1 z Unit Delay

Figure 5.4. Simulink model that generates nine samples of a random walk. Nine Simulations of the Random Walk for 10000 Coin "Flips" 200

150

100

s 50 e u l a V k l a 0 W m o d n a R -50

-100

-150

-200 0

1000

2000

3000

4000

5000 Index "k"

6000

7000

8000

9000

10000

Figure 5.5. Nine samples of a random walk process.

the nine samples of the random walk that result with 10,000 flips. (After n iterations, the value of the random walk must be between −n and n, but it can be shown that a random √ walk grows like n, and this is indeed the case for these nine samples.)

180


Let us compute some of the statistics for this process. The mean of the process is as follows: k

E {xk } =



E {sign(y i )} = 0.

i =0

The fact that the mean of the sign of y is zero follows from the fact that the probability that the sign is +1 (from the uniform distribution) is 12 and the probability that the sign is −1 is also 12 . The mean is therefore 12 (+1) + 12 (−1) = 0. Similarly, the variance of the process is

 

2 σ RW (k)

=E

     2

k

sign(y i )

i =0

k

E sign(y i )2 = k.

=

i =0



Our goal is to convert this discrete time random walk into a continuous process by letting the numbers of coin flips in any time interval go to in finity. That means that lim xk (k,t) = B(t).

n→∞ t →0

We need to verify that the result B(t) is a mathematical function that exists and has the desired property that it represents the random motion of a particle immersed in a fluid. We have scaled the time axis in such a way that it can become continuous, but we do not know how to scale the outcome of the coin flip (in the random walk it is +1 or −1 for each flip), so we need to answer the following very simple question: If we scale the number of coin fl ips in the random walk using a scale factor of t (we are scaling the horizontal axis that will converge to the time axis), then how must we scale the position (vertical) axis so the process converges to a valid continuous time process as t approaches zero?

This question arose at the end of the 19th century because physicists were trying to model Brownian motion as a veri fication of the atomic theory of matter in which the motion was due to collisions with atoms (see [5]).

5.2.2

Brownian Motion and White Noise

Brownian motion is a process named for the English botanist Robert Brown, who stated in a paper in 1828 that the motion of a particle immersed in a fluid was not due to something alive in the particle but was probably due to the fluid. This led to attempts to find what it was in the fluid that caused the motion. The Polish physicist Marian von Smoluchowski was trying to model the Brownian motion process using a limiting argumenton therandom walk. He realized that this approach would provide a way of verifyingthe atomic theory of matter. Physicists who believed in the atomic theory thought that the observed Brownian motion was due to the very large number of collisions between the particle and the atoms in the fluid. While Smoluchowski spent a lot of effort on solving this problem using the approach described here, Albert Einstein worked with the underlying probability density instead, deriving this density from first principles.


181

Einstein’s first 1905 paper (the year in which he also published the photoelectric effect and the special theory of relativity papers) reported his work on Brownian motion. Smoluchowski published his results abouta year later. A few years later, the French physicist Perrin won the Nobel Prize by using Einstein ’s result with experimental measurements of Brownian motion to come up with an estimate for Avogadro ’s number that was in excellent agreement with (and actually more accurate than) other measurements. The velocity of a particle undergoing Brownian motion is not computable, and Perrin showedthatthiswas true when oneattempted tomeasurethe velocityof a particleundergoing Brownian motion. In fact, Brownian motion is an example of a mathematical function that is nowhere differentiable. We will show this result and use it to show why white noise does not exist, by itself, in any mathematical sense. (However, a form of the process exists whenever it appears, as it always does in practice, in an integral.) The discussion that follows is about how one goes about modeling a random process in Simulink. To do this properly we need to understand the nature of the continuous time process called white noise and its relation to Brownian motion. The way we will do this is to let the random walk above become a continuous process by letting the time between coin flips go to zero (in a very precise way). Brownian motion is the result of the convergence in the limit, as the time between flips goes to zero, to a continuous time process. Before we explore this convergence, we need to describe what we mean by convergence. If we look at the Simulink results for the random walk simulation above, it is quite clear that the random walk does not converge to a deterministic function. In fact every time we redo the experiment, the results are different, so what is converging? Remember that the stochastic process is the result of a random experiment, so it is a function of two variables: one is time, and the other is the experimental outcome (the observed coin flips and the process generated by the particular sequence of flips). The process can converge in the following ways: •

in terms of the statistics (i.e., in the mean, mean square, or variance sense, for example);

•

in terms of the probability distribution(i.e., in the sensethat the probability distribution for the random walk converges to a probability distribution for the continuous time process);

•

in the sense that for every experimental outcome, the function of time that results converges to a well-de fined function with certainty (in a sense that is made precise but is beyond the scope of this discussion).

We will use the convergence in the probability distribution sense in the following, but it is true that the random walk converges in every one of these senses to Brownian motion. Let us assume that we scale the graphs above so that along the flips axis, the variable is replaced by kt . The question is, how do we scale the vertical axis? Since we do not k know exactly what to use, let us just scale it with an unknown value, say s . In that case, the mean of the process (before we take the limit) is still 0 and the variance is ks 2 . (Exercise 5.2 asks that you verify this using the calculations for the variance of the random walk above.) It is now a simple matter to use the central limit theorem to develop the probability density function for the Brownian motion process. Let the number of flips k go to infinity

182


and let t go to 0 in such a way that the product becomes the time variable t in the Brownian motion process. That means that lim (kt)

t n

→0 →∞

= t,

t or equivalently, the number of flips k is constrained to be t for any time t . We use the central limit theorem. The variance of the scaled random walk (scaled by t s ) is ks 2 . Thus, in the limit, when we force k to be t , the variance of the random walk

√

2

becomes ts . It is now obvious that the scale factor, s , must be proportional to t . If t it is not, the limiting variance will be either zero or in finity, neither of which makes the limiting process have the desired properties. The infinite variance is a process that could have infinite jumps in any short period, and the zero variance would mean the process is not random. Brownian motion has neither of these properties. Therefore, scaling the random walk by σ t will insure convergence to the Brownian motion process. (The parameter σ is the proportionality constant and is the standard deviation of the resulting Gaussian process.) Thus, we have the following result.

√

Create a random process B(t) from the random walk as follows:

B(t)

=

1

lim

√

→0 σ k = t σ 2 t t

k 

t i =1

σ 2 t sign(yi ).

Then B(t) is a process that has a probability density function that is normal with mean 0 and variance 1. 1 The usual way of stating the limit is to remove the scale factor σ √ on the process so t

the variance of the result is σ 2 t . Both Einstein and Smoluchowski showed this result, but using the approach we are using, Smoluchowski was able to prove that the process does not have a derivative anywhere. We can see this if we try to differentiate the random walk and then take the limit as we did above. When we do, we get the following: The Brownian motion process B(t) has an in finite derivative everywhere.

The derivative is

dB(t) dt

= limt →0 B(t +t)t −B(t) √ .

the difference in the numerator is proportional to

Based on the scaling we did above,

t . Let the proportionality constant be √

B(t t) B(t) 0 t

α; then the derivative becomes lim t → limt →0 α t t The Brownian motion process has the following properties:

+ −

=

→ ∞ for all times t .

•

The process is Gaussian at every time t , with mean zero and variance σ 2 t .

•

The process is 0 at t

•

The process has increments that are independent random variables in the sense that B(t n ) B(t n−1 ) and B(t n−2 ) B(t n−3 ) are independent random variables for any nonoverlapping intervals (t n−3 t n−2 ) < (t n−1 t n ).

−

•

= 0. − ≤

≤

The process does not have a derivative anywhere.


•

183

The integral of any function of time multiplying the derivative of B(t) is evaluated using the integral f (t)dB(t), which is well de fined even though the derivative of B(t) is not.



The fifth property is the mathematical rationale for the existence of white noise. Even though w(t) = dB(t) does not exist, it does make sense to talk about integrals that contain dt products of w(t) (white noise) with other functions because these integrals are



f(t)w(t)dt =



f (t)dB(t).

The first four of these properties follow from the discussion so far. Norbert Weiner demonstrated the last result in the 1950s, but it will take us too far a field to show the result here, so we will accept it without proof. However, because of his seminal work in showing these properties, we call the process B(t) the Wiener process, and we will use this name from now on. The name white noise comes from an analogy with white light that contains every color. White noise as a stochastic process is supposed to contain every frequency. Let us see what that means. A stochastic process is called stationary if for any two times t 1 and t 2 the correlation function, defined by E {y(t 1 )y(t 2 )}, depends on only the difference between the two times t 1 and t 2 . That is, R(τ) = E {y(t 1 )y(t 2 )} = R( |t 2 − t |1 ), where τ = |t 2 − t 1 |. If we go back to the attempt to de fine the derivative of the Weiner process, the correlation function for white noise is E {w(t)w(t + τ )} =

 0, τ > 0,  , τ 0, ∞

= σ 2 δ(τ).

=

The first part of this assertion comes from the fact that the Weiner process has independent increments, so the expected value of nonoverlapping intervals is the product of the means (and since the mean is 0, the result is 0). The last part of this assertion comes from the fact that the derivative of B(t) is infinite. The σ 2 in this definition is often called the noise power. It is not the variance of the process (which is in finite); however, the integral of this process has the variance σ 2 t , since the integral is the Weiner process. We can compute the correlation function of a stationary process directly from the process, or from the way we generated the process. If, for example, the process were the resultof using white noise as an input to a linear system, theoutput would be theconvolution of the white noise with the system ’s time response to an impulse. With this approach, the white noise process only appears in the solution through the convolution integral, and from property 5 of the Weiner process, this integral always is computable. Remember, however, that the convolution integral can be computed from the product of the transforms of the two functions. (In this case, we must use the Fourier transform because a process can be stationary only if it has been in existence for an in finite time, as we will see when we attempt to develop simulations of these processes.)

184


The Fourier transform of the correlation function is the For the white noise process the spectral density is S(ω)

=



“spectral

density function. ”

∞

σ 2 δ(τ)e

iωτ

−

dτ

=

σ 2 .

−∞

Since the spectral density is constant for all frequencies, we can now see why we use the name “white noise” to describe it. If white noise does not exist, how can we simulate a system where the white noise stochastic process excites a linear system? The answer is to use the Weiner process.

5.3 Simulating a System with White Noise Inputs Using the Weiner Process From the discussion in the previous section, it is clear that to simulate a system with white noise we must use care. When simulating a continuous time systemwith a white noise input, the method used is always to select a Gaussian random variable at each of the numerical solver’s time steps and then to scale the variable so it has the same effect on the solution as if the random variable were continuous. From the results of Section 5.2, this means that the scaling has to be proportional to the square root of the time step used to generate the “white noise” sample. We will now look into how Simulink achieves this with the built in “Band Limited White Noise ” block.

5.3.1 White Noise and a Spring-Mass-Damper System Open the model White_Noise in the NCS library. We will use this model to explore Monte Carlo simulations with noise sources and to explore methods that speed up the simulations. The first part of the model uses the built in white noise block with 10 different random number seeds to create 10 separate simulations of “white noise” exciting a linear system. The model is the state-space version of the spring-mass-damper with the damping set to make the damping ratio 0.707. This Simulink model uses the integrator and gain blocks that we did in Chapter 2. The model is in Figure 5.6, and Figure 5.7 shows the 10l responses. The model uses the fixed step solver with a nominal solver step of 0.005 sec. The undamped natural frequency of the spring is 10 rad/sec (1.6 Hz). The variance and mean calculations are from the Signal Processing Blockset library. The data forthe model come from a Callback when themodelopens. The integrator step size and the various sample times in the model are all the same (called samptime in MATLAB), so when this is changed, the simulation ’s solver and all of the calculations for the white noise and the statistics are updated too. Open this model from the NCS library and run the simulation. You should see roughly the same variances as in Figure 5.6. (The variances of the position and the velocity are dif ficult to read in this figure, so make sure you run the model; the values are all around 3.53 and 353.6, respectively.) The first thing we need to look at in this model is the Band Limited White Noise block. This block is an example of a masked subsystem. To see what the model is like under the mask, right click on the block and select “Look Under Mask ” from the menu that opens. A new window will open that contains the blocks in Figure 5.8. (The block will open with the

5.3. Simulating a System with White Noise Inputs Using the Weiner Process

185

322.3

Display Mean

348.7

0.3125

341.4

0.2534

396.3

0.04678

Running Var

411.5

0.5556

325.7

-0.6937

Variance1

342.6

In

358.2 Sample Data for Statistics1

0.02838 Mean

0.0582

330.6

-0.01026

357

0.1033

Gain2

-0.1191

Display Variance1 omega^2 Band-Limited White Noise

1 s

1 s

Integrator1

Integrator

3.706

Sample Data for Statistics

3.025

Gain1

3.611 2.584

2*zeta*omega

4.045

Running Var Gain

2.493

Variance

5.135 3.439

omega^2

3.307 4.352 Display Variance

Scope

Figure 5.6. Simulink model for the spring-mass-damper system.

8

6

4

2

0

-2

-4

-6

-8

0

1

2

3

4

5

6

7

8

9

Time

Figure 5.7. Motions of 10 masses with white noise force excitations.

10

186


[sqrt(Cov)]/[sqrt(Ts)]

1

White Noi se

Figure 5.8. White noise block in the Simulink library (masked subsystem).

Gain block small; we enlarged it here to make its contents visible.) Along with the mask is a mask dialog that allows the user to select the values for the parameters used under the mask. To see this, again right click on the block, but this time select “Edit Mask. ” A dialog window with four tabs will appear. The first tab is the instructions for drawing the little wiggly line (representing noise) on the block. The second tab is the dialog that allows the user to change the parameters he wants to use in the subsystem. Options for these are an edit box, a pop-up menu, or a check box. The names of the parameters and a user prompt can also be set up. The third tab, called “Initialization,” allows the user to set up MATLAB code to set up the actual values used in the blocks under the mask. Note that the mask is a form of subroutine, so these parameter values only appear in the masked subsystem. For the Band Limited White Noise block, the important calculations are the test done in the initialization to ensure that the terms in the variable Cov are all positive, and the calculation done in the block below. This calculation should look familiar. It is the scaling that we determined above (with the value of σ denoted by Cov and the value of t denoted by Ts). Let us perform some experiments where we change the solver step size. The goal is to verify that this scaling keeps the results statistically consistent as we make the changes. The step size is the MATLAB variable samptime. (It is changed at the MATLAB command line.) The model opens with this value set to 0.005. Select some both smaller and larger time steps (say from 0.001 to 0.05). You should observe that the mean and variance of the simulated processes are about the same. (Remember in a Monte Carlo simulation the results will not be the same from simulation to simulation.) One caveat: since the seed used for the random number generator is the same in every simulation if you keep everything in the simulation the same, the results will be identical from run to run. One final caution: when you simulate the response of the system you need to be careful to select a solver step size and a value for the sample time in the white noise block that is smaller by about a factor of 10 than the response time of the system (in this case faster than about 0.05 sec). See what happens if you make the samptime equal to 0.1 sec. (You should even see some effect when it was 0.05 sec.) This effect comes from the sampling theorem. The value of the noise power in the Band Limited White Noise block is set to 1. It would seem that this should set the value of the variance of the position of the mass at 1, but it does not. Let us try to find out why.

5.3.2 Noisy Continuous and Discrete Time Systems: The Covariance Matrix To describe a stochastic process we need to specify its joint probability density function for the values at any arbitrary set of times. White noise is characterized completely by the Weiner process because the independence of all of the increments makes the joint density


187

just the product of the densities for each of the increments. When the stochastic process results from white noise excitation of a linear system, the joint density is determined from a matrix whose dimension is the highest derivative in the underlying differential equation. This matrix is the “covariance matrix.” We need to compute this matrix for a linear system. Consider the linear state-space system excited by a white noise process w(t) given by d x(t) dt

= Ax(t)+Bw(t).

The covariance matrix, P(t), of this vector stochastic process is de fined as P(t) = E x(t)x(t) T . It is relatively easy to determine a differential equation for this matrix. We simply differentiate P(t)using the fact that the derivative and the expectation commute. Thus, d P(t)





dt

           d

T

x(t) x(t)

=

E

=

E

=

E (Ax(t) +Bw(t)) x(t) T

dt

d

dt

x(t)

x(t)

= AP(t) + P(t) A

T

= AP(t) + P(t) A

T

T

+

d

+ x(t)

dt

+

x(t)

T

E x(t) (Ax(t) +Bw(t))T

E (Bw(t)) x(t)T T

+ BSB

T

+ x(t)(Bw(t))

.

 

The last step comes from the fact that the process w(t) is the vector white noise process. The covariance matrix E (Bw(t)) x(t)T in the next to last step is therefore determined using the solution of the linear state variable equation that we developed in Chapter 2. We can now use this solution, along with the fact that E w(t)wT (τ ) is the impulse function scaled by the noise power parameter (for each of the white noise processes in the vector), to develop the solution. Remember that when an impulse appears inside an integral, it selects the values of the integrand at the point the impulse occurs (in this case when (t − τ ) = 0). Therefore, we have





E (Bw(t)) x(t) T



=

E





        (Bw(t))



(t)x0 +

(t −

τ )Bw(τ)dτ

0

t

= BE

w(t) w

T

T

(τ )B

(t −

T

t

E w(t) wT (τ ) BT (t − τ )T dτ

0

t

= B

0

=

1 2



τ ) dτ

0

= B

1 2

T

BSB

T

Sδ(t − τ )B (t −

.

 T

t

τ )T dτ

188


The factor of 12 in this sequence comes from the fact that the impulse is a two-sided function and the integration limit t is right in the center of these two sides, so only 12 of the value comes from the impulse. The other 12 BSBT comes from the second term in the expectation (i.e., from the E x(t) (Bw(t))T term).





The linear matrix equation

d P(t) T T = AP(t) + P(t)A + BSB dt

has a solution given by

t

T

P(t) = (t − t 0 )P(t 0 )(t − t 0 )

+



(t −

τ )BSBT (t − τ )T dτ .

t 0

This is easy to verify by substituting back into the differential equation. Exercise 5.3 asks that you make this substitution, but remember when you do that the derivative of an integral whose upper limit depends on the independent variable comes from the identity d dt

b(t)



c(t,τ)dτ = c(t, b(t))

a

db(t) dt

b(t)

+



∂ ∂t

a

c(t,τ)dτ.

Before we investigate the use of this procedure to model a continuous time stochastic process as a discrete time equivalent, we can use the development above to compute the steady state noise variances for the spring-mass-damper Simulink model in Section 5.3.1. This system has a state variable model given by d x(t) dt

=



0 −100

1 −14.14



x(t) +

  0 10

w(t).

This state x(t) in this model has the usual components. (The first is the position of the mass, and the second is the velocity.) Therefore, the covariance matrix for the state x(t) is



2 σ pos ρσ pos σ vel

ρσ pos σ vel 2 σ vel



.

The diagonal terms are the variances of the position and velocity and the off diagonal terms show the correlations between the two. The variable ρ has a value whose magnitude is less than 1, and it measures the correlation between the position and the velocity. We can calculate the covariance matrix to verify that the Simulink model of the mass ’s position gave the correct variance. We do this by solving for the steady state covariance in the differential equation above. Steady state means that the covariance matrix does not change, so its derivative is 0. Thus, AP(t) + P(t)A

Substituting the values for



0 −100

1 −14.14



A, B,

T

T

+ BSB

=

0.

and S from the state model above gives

P(t) + P(t)



0 −100

1 −14.14

T

  +

0 0

0 10000



=

0.

This equation, called a Lyapunov equation, is solved with an M- file that is built into the control system toolbox in MATLAB. The command is lyap(A,BSBT), where A is the matrix 0 T and S = 1 for this system.) A and BSBT is the matrix BSB . (Remember that B = 100





The MATLAB commands and the result for the steady state value of P (Pss) are as follows (a version of lyap, called lyap_ncs, has been included in the NCS library so you can try doing this calculation if you do not have lyap available):


189

>> A = [0 1;-100 -14.14]; >> BSBT = [0 0;0 10000]; >> Pss = lyap(a,q) Pss = 3.5355 -0.0000

-0.0000 353.5534.

As can be seen, the variances from this calculation match the simulation results in the Simulink model we generated and evaluated in the previous section (at least statistically, since thesimulation generatesan estimate of the 10 runs). Thisthen answers thequestionwe asked at the end of Section 5.3.1. This analysis also shows that the noise power in the Band Limited White Noise block does not directly determine the variance of the output from the simulation, only the calculations above provide the correct variances for the various state variables in the model.

5.3.3 Discrete Time Equivalent of a Continuous Stochastic Process We are now in a position to develop a discrete time model for a continuous time stochastic linear system. In general, if we want to use a discrete process to create a solution that is in some sense equivalent to the original continuous time process, we need to come up with a definition for what we mean by equivalence. The simplest and most direct is covariance equivalence, de fined as follows: A continuous time system and a discrete time system (for the sample times t ) are “covariance equivalent ” if •

they have the same mean at the sample times;

•

they have the same covariance matrix at the sample times.

Remember from Section 2.1.1 that the discrete time solution of a continuous state variable model is xk+1 = (t)xk +  (t) uk . We assume that the input in this model is a vector of Gaussian random variables (each one like the random walk) whose values are independent from step to step and Gaussian. We also  assume  that random variables have the same Covariance matrix at each time step (i.e., = Sdiscrete for all values k). The covariance matrix for the discrete process is the E uk uT k solution of the difference equation, which is determined as follows: Pk+1



T = E xk+1 xk+ 1



= E (xk + uk ) (xk +  uk )T





= Pk T +  Sdiscrete T .

The state xk is a vector random variable, and it is independent of uk (because xk only depends on the past values of u, and all of the values  of uk are independent,  zero mean T T random variables). This means that E xk ( uk ) = E {xk } E ( uk ) = 0, and     T T = E { uk } E (xk ) = 0. E uk (xk )

190


The covariance matrix for the continuous process, P(t) will be exactly the same as the covariance matrix of the discrete process Pk when t = kt if the solution to the continuous time covariance matrix equation above at the times t = kt is the same as the covariance matrix for the discrete time system. That is, Pk = P(kt) where Pk comes from iterating the equation above, and P(kt) is the solution of the continuous time covariance equation at the times kt . We need to find a value for Sdiscrete that insures this. Comparing the solution for the continuous covariance matrix at time t = t when the initial covariance (at t = 0) is 0, with iterations from the difference equation for the discrete covariance matrix with the initial value 0 gives an equation for Sdiscrete as follows: Sdiscrete 

T

=



t T

T

(τ )BSB (τ )

dτ .

0

There are many ways of factoring the left-hand side of this solution to give an explicit value for Sdiscrete, and we will explore this further as we return to the spring-mass-damper example. Exercise 5.4 asks you to verify that Sdiscrete that results from this single iteration insures that the covariance matrices are the same for all k . Let us use a sample time for the discrete model that is 0.01 sec. The value of (t) is determined, as in Section 2.1.1, using the c2d_ncs program in the NCS library. Now we need to determine  (t) using the solution of the covariance matrix at the time t . When the initial covariance matrix is 0, the solution for the covariance matrix at time t is P(t) =



t

(t −

τ )BSBT (t − τ )T dτ .

0

Many numerical approaches will compute this integral, but let us try doing it in Simulink. We will use the ability of Simulink to solve matrix differential equations by setting up the differential equation d Pdt (t) = AP(t) + P(t)AT + BSBT . Figure 5.9 shows the model (called Covariance_Matrix_c2d in the NCS library). Running this model produces the solution for the matrix P(t) in the MATLAB workspace. To send P to MATLAB at the end of the simulation, the model uses the Triggered Subsystem block from the Ports and Subsystems library. In a triggered subsystem, the trigger causes any blocks inside the subsystem to run when the trigger signal increases in value (the default). The icon at the top of the subsystem, where the trigger signal is connected, indicates the trigger type; if the icon looks like a step discontinuity that is increasing, the trigger is “rising,” and the options are increasing, decreasing, or both. The contents of the subsystem appear when the block opened. From the annotation for this model, which shows the differential equation simulated, you can work through the diagram and verify the Simulink model. You should do this by carefully going through the model or, better still, by creating the model yourself. Before running the simulation, set up the values of A and BSBT in MATLAB. (From the Simulink model you can see that the variables are called A and BSBT respectively, and if you used the lyap_ncs code above, these should already exist in MATLAB.) After the simulation is complete, you can see what P is in MATLAB by typing P at the command line. The result should be the matrix P = 0.0030 0.4333

0.4333 86.8229.

5.3. Simulating a System with White Noise Inputs Using the Weiner Process A

A Matrix

[2x2]

Product Matrix [2x2] Multiply

191

[2x2] [2x2]

>= deltat Clock [2x2]

BSBT

[2x2]

Noise Covariance

[2x2]

[2x2]

Initial Value of P (zero matrix)

[2x2]

[2x2]

1 xo s

[2x2] [2x2]

Integrator

zeros(n)

Product1 Matrix [2x2] ultiply

Compare To Constant In1

Triggered Subsystem to Send P(deltat) to MATLAB Workspace

[2x2] [2x2] [2x2]

A Matrix (transposed)

A'

Simulink Mode l for Computing P( ∆ t) from the Differe ntial Equation: dP/ dt = AP + PA' +Q For values of A and Q in MATLAB Workspace.

Figure 5.9. Continuous linear system covariance matrix calculation in Simulink. The result is the covariance matrix that can be used for simulating with a covariance equivalent discrete system.

Armed with this value for the covariance matrix we can create the equivalent digital model for the original spring-mass-damper continuous system. Figure 5.10 shows the Simulink model. We use the discrete state-space model from the discrete library in the model, and we have set it up for three different sample times (0.01 sec as derived above, 0.05 sec, and 1 sec). The model also contains the original continuous time model as a subsystem. The resulting time histories for the four simulations are in Figure 5.11. The figures show both the position and the velocity of the mass; the lighter line is the velocity, and the smaller and darker line is the position. Notice that the simulations for the discrete time models at 0.01 and 0.05 sec match the continuous time system very nicely. (Again, remember that the match is only in the statistical sense.) For the last sample time, namely 1 sec, the match is not good. Why is this? The reason is our old friend the sampling theorem. When we create a discrete model at 1 sec, the conversion to the discrete model has errors. These arise because the system has a natural frequency of 100 radians/sec (1.592 Hz corresponding to a period of 0.628 sec). The 1-sec sample time is almost two times slower than the natural oscillation of the mass and as such violates the requirements of the sampling theorem for accurate reconstruction. (The discrete time system aliases the oscillation.) As an exercise, you should isolate each of these discrete models and see how long Simulink needs to createthe results. Remember that to do this you need to run thesimulation

192


Figure 5.10. Noise response continuous time simulation and equivalent discrete systems at three different sample times.

from MATLAB using tic and toc before and after the simulation command (in MATLAB type tic;sim(‘modelname’);toc;). If you don’t want to do this exercise yourself, we have simulated each of the models 10 times (with each of the simulations using a stop time of 1000 sec) using the M-code shown below to record the simulation times and compute their averages (see the MATLAB file Test_Noise_Models.m in the NCS Library). for j =1:10 for i = 0:3 str = [’tic; sim(’’test’ num2str(i) ’’’); t’ num2str(i) ’(’ … num2str(j) ’)=toc;’]; eval(str) end end % Averages: Avgsim0 = mean(t0) Avgsim1 = mean(t1) Avgsim2 = mean(t2) Avgsim3 = mean(t3)


193

80 Continuous Time at 0.01 sec.

60

Sample Time = 0.01 sec. 50

40 20 0

0

-20 -40 -50

-60 -80

0

10

20

30

40

50

0

10

20

Time

30

40

50

40

50

Time 80

Sample Time = 0.05 sec.

Sample Time = 1.0 sec.

60

50

40 20 0

0 -20 -40

-50

-60 0

10

20

30

40

50

-80

0

10

20

Time

30 Time

Figure 5.11. Simulation results from the four simulations with white noise inputs.

Table 5.1. Simulink computation times for the four white noise simulations. Model

Continuous Discrete at 0.01 sec. Discrete at 0.05 sec. Discrete at 1.00 sec.

Avg. Simulation Time 1.0490 sec. 0.2571 sec. 0.0572 sec. 0.0105 sec.

The results we obtained from this code are in Table 5.1. There is an improvement of about a factor of 5 in the simulation time comparing the fast sample time with the continuous solution using the ode45 solver (with a max step size of 0.01 sec), so when it comes to large and complex Monte Carlo simulations, this approach can offer a considerable computational advantage.

194


Based on the work we have done in this chapter, you should be able to simulate any system with arbitrary noise sources. If you are given the properties of a noise in the form of a power spectral density function (which implies that the noise is stationary), it is always possible to find a linear system that can simulate this noise source. In the next section, we look at one particular and important process that has a power spectral density function that is proportional to 1 /f (one over the frequency). The approach we discuss is a simpleway to createa noise that hasa desired power spectral density function.

5.3.4 Modeling a Specified Power Spectral Density: 1/f Noise In many electronic devices (such as sensors that are used to create an image from radiation in visible, infrared, or ultraviolet light) quantum mechanical effects manifest themselves as a noise that has a power spectral density that is proportional to the reciprocal of the frequency. These noise sources are not from systems that have rational transfer functions, so a method to create a typical noise sample for a Monte Carlo simulation needs to use some approximation. Let us see why this is true. The power spectral density (PSD) for a linear system excited with white noise comes from thecorrelationfunction of theprocess because, by de finition, the powerspectraldensity function is S(ω) =



∞

R(τ)e−iωτ dτ ,

−∞

where R(τ) is the correlation function R(τ) = E {x(t)x(t + τ )} .

If the process x(t) comes from a linear differential equation, then it is the result of convolving the input with the impulse response of the system. Since the correlation function involves the product of x(t) at two different times, the correlation function R(τ) requires two convolutions to compute. In fact, if a linear system with impulse response h(t) has the (real) stochastic process u(t) as its input, then the correlation function of the output is R(τ) = h(τ) ∗ Ru (τ ) ∗ h(−τ ), where Ru (τ ) is the correlation function of the input, and the * operator denotes the convolution integral. Thus, we have the following very important result. It is the reason that stochastic process models use linear systems excited by white noise: If a real, stationary, stochastic process excites a linear system with impulse response h(t) it has the correlation function R(τ) = h(τ) ∗ Ru (τ ) ∗ h(−τ ). Therefore, the power spectral density function of the output is S out (ω) = |H(iω)|2 S in (ω).

The proof of this is very straightforward and is an exercise (Exercise 5.5 at the end of this chapter).


195

Since the input is white noise S in (ω) 1, it is easy to see that the PSD of the output is simply the magnitude squared of the transfer function of the system. Since the transfer function of an nth order linear system is the ratio of polynomials of order (at most) n (in s ), the PSD is always the ratio of real polynomials with only even powers of ω, and it is of order 2n. (These properties are because of the square of the magnitude.) From this, we can see the problem with generating a spectrum that has a PSD proportional to 1/f . The linear system that generates such a spectrum has a transfer function given by H(ω) 1/ iω because the magnitude squared of this transfer function is 1/ω 1/(2πf ). The impulse response of the linear system with this transfer function comes from the inverse Laplace transform of 2π/s , which is 2/t . So how do we come up with an approximation for this spectrum? The method is to approximate the transfer function on a semilog plot (i.e., on the Bode plot). Since the spectrum is real, we do not have to worry about the phase. We can approximate a system using products of first order rational transfer functions:

=

=

=

√

√

H(s)

=

n  s + zi

= s + pi

√

.

i 1

On a Bode plot, this transfer function can be used to approximate any straight line up to the frequency of the last pole. The trick is to place the zeros and poles symmetrically with respect to the desired straight line. p +z For our approximation, we use the fact that the transfer function at the pole is 2i pi i and at the zero it is

2zi zi pi

+ , a difference of pi + zi −(pi − zi )2 . − = zi + pi 2pi 2pi (zi + pi )

2zi

We allocate this error around the desired frequency response as follows: •

First, the number of poles used to make the approximation is arbitrary, but the more that we use, the better the approximation will be.

•

The simplest way to set the number of poles is to assume that each pole is a factor of 2 in frequency away from the previous pole (i.e., they are an octave apart). In general this separation will ensure that the approximation is within 1 db of the desired spectrum (1 db equates to a maximum error of about 12% in the magnitude at the frequencies of the pole and zero; the error is zero at the midpoint between them).

•

The value of the zeros is set to 2 times the poles (this places the amplitude symmetrically around the poles and zeros).

√

There is no way that any finite representation of the 1/f noise can be valid over all frequencies (doing so requires an infinite number of terms). Thus, we assume that the approximation starts at some frequency ωLow , and it stops some number n of octaves above

196


d

Scope

Gain3

B* u Band-Limited White Noise

Gain1

1 s

1/f noise 1/f noise

C* u

Integrator

Gain2

Input Signal

Plot Spectal Estimates

Gain A* u

Simulation of a noise sam ple that has a Power Spectral Desnity Function that is approximately 1/f. The approximation is valid from:

ω

low

to 2n ω

low

In MATLAB the variable nam es are: omegalow =

ω

low

; numoctaves = n.

Figure 5.12. Simulation of the fractal noise process that has a spectrum proportional to 1/f .

this.5 The transfer function is then

H(s)

=

√ n  s + 2 2i −1 ωLow i 1

=

s

+ 2i−1ωLow

.

A Simulink model that generates a noise sample using this approximation is in Figure 5.12 (called Oneonf in the NCS library). This model generates a plot of the sample and calculations of the resulting PSD. Note that this simulation, once again, uses the vector capabilities of Simulink to simulate the state space model of the differential equation represented by the transfer function above. The differential equation thatgenerates thistime sample has an orderthat is determined by the number of octaves used. In addition, the state-space matrices A, B , C , and D are set up in a callback from the model. The number of first order systems is set to the default value of 10 when the model opens. Also, the starting pole ( ωLow ) is set to 1 radian/sec. These values are MATLAB variables using the names that are in the figure. Changing their value in MATLAB causes subsequent simulations to use the new values (again, because of an “Init ” callback from the model that executes at every simulation run).

5

In Chapter 6, we examine methods for solving partial differential equations using a finite number of ordinary differential equations. This discussion parallels the approach that we use there. It is a “lumped parameter ” or “finite element” method.


197

The value for the A and B matrices that come from the callbacks are

A

=

  z  z 

p1 p2 p3 .. . pn

−

2

−

3

−

zn

−

0 p2 z3 p3 .. . zn pn −

−

−

0 0 p3 .. . zn pn −

−

· · ·

· · ·

· · ·

· · ·

· · ·

0 0 0 .. . pn

−

   , 

B

=

z  z  z 

1

−

2

−

3

−

zn

p1 p2 p3

.. . −

pn

   . 

The code that sets up these matrices (shown below) is in the Callbacks tab of the Model Properties under the File menu of the model: n n p z A for i A end B C d

= = = = = = =

0:numoctaves-1; 2.ˆn; omegalow*n; sqrt(2)*p; diag(-p); 1:numoctaves-1 A+diag(p(i:end-1)-z(i:end-1),-i);

= (z-p)’; = ones(size(B))’; = 1;

The Simulink simulation results in the 1 /f noise sample shown in the figure at the right. In addition, the model contains a subsystem that has four different methods for computing an estimate of the Spectrum. When the model is open, if you double click on the “Plot Spectral Estimates ” subsystem you will see that these estimators are all from the Signal Processing Blockset. (Figure 5.13(a) shows the Simulink model.) The details of how each of these estimates operates arebeyond thescope of this text, but in the broad sense, all of them, except the magnitude of the FFT block, use a method that estimates a linear model that matches the statistical properties of the data and then computes the estimate of the spectrum using the model. This approach always produces an estimate of the PSD that gets better the longer the time over which the estimate is performed. (Mathematically these estimates are consistent, meaning that the variance of the estimate goes to zero as the estimation time goes to in finity.) The magnitude of the FFT is a very poor estimate of the PSD because it does not have this property. (In fact, the variance of this estimate never gets smaller.) This can be seen in 4

3

2

e s i o N 1 f / 1 f o s e 0 u l a V d e t a l u -1 m i S -2

-3

-4

1

1.1

1.2

1.3

1.4

1.5

Time

1.6

1.7

1.8

1.9

2

198


Figure 5.13. Power spectral density estimators in the signal processing blockset used to compute the sampled PSD of the 1/f noise process.

the Vector Scope plot of all of the PSD estimates in Figure 5.13(b). (The magnitude of the FFT is the top estimate with the + signs as line markers.) The three consistent estimators all have the same overall estimate, and they would indeed be identical except that each of them is given an arbitrary order for the estimate (starting at 2 for the Yule-Walker estimator “

”

Exercises

199

and going to 6 for the Burg estimator; as an exercise, change them all to order 2 and verify this result). In this chapter, we have explored the mathematics of stochastic processes and the methods for simulating them in Simulink, using some of the tools in the Signal Processing Blockset. It should be evident from what we have done so far how Simulink ’s functionality can go beyond simulating differential and difference equations. We have shown how to interchange discrete time and continuous time systems and work in the frequency domain. We consider next how Simulink can model systems described by partial differential equations.

5.4 Further Reading There are many books on probability, statistics, and random processes. Among the best is by Papoulis [33]. A recent book by Childers [7] has many MATLAB examples. Norbert Weiner ’s book Cybernetics [49] hasa wonderfuldescriptionof theimportance of being versed in many disciplines. (For more on this point, see Chapter 10.) He wrote: “Since Leibniz there has perhaps been no man who has had a full command of all the intellectual activity of his day. Since that time, science has been increasingly the task of specialists, in fields that show a tendency to grow progressively narrower. A century ago, there may have been no Leibniz, but there was a Gauss, a Faraday, and a Darwin. Today few scholars can call themselves mathematicians, physicists, or biologists without restriction. “A man may be a topologist, an acoustician, or a coleopterist. He will be filled with the jargon of his field, and will know all its literature and all its rami fications, but, more frequently than not, he will regard the next subject as something belonging to his colleague three doors down the corridor, and will consider any interest in it on his own part as an unwarrantable breach of privacy. ” For more on Weiner, visit Wikipedia at http://en.wikipedia.org/wiki/Cybernetics. Both the Weiner process and the 1 /f noise process are examples of a fractal. The fern in NCM is another example. There is an iterative method for generating fractals and a discussion of fractal dimension in Chapter 5 of Scheinerman [37]. A MATLAB code that will create Sierpinski ’s triangles is in the appendix of [33]; this example is fun to run, and you might want to try creating a Simulink model to do the same thing.

Exercises 5.1 Modify the Rayleigh distribution Simulink model in Section 5.1.2 to generate the random variable that is the phase angle of the re flected sine waves. The book by Papoulis [33] has a derivation of the probability density function for this random variable. Check to see if the histogram that you generate has the required form. 5.2 Verify the property that the random walk has zero mean and a variance of ks 2 (the properties shown in Section 5.2.2).

200


5.3 Show that the covariance matrix differential equation d P(t) dt

T

T

= AP(t) + P(t) A + BSB + 

t

has the solution P(t) (t t 0 )P(t 0 )(t t 0 )T (t τ )BSBT (t t 0 Use the hint given in Section 5.3.2 when you differentiate this solution.

=

−

−

−

T

− τ)

dτ .

5.4 We iterated the discrete covariance matrix equation one time and showed that the covariance matrix of a discrete system xk+1 (t)xk  (t) uk will be the same as the covariance matrix of the continuous system if the discrete system noise input uk has the covariance matrix E uk uT Sdiscrete for all values k , where Sdiscrete is k t

=

{

given implicitly by SdiscreteT for all possible iterations.

=

+

}=

 t 0

T

(τ )BSB

(τ)T dτ . Show that this is true

5.5 If a real, stationary, stochastic process excites a linear system with impulse response h(t) it has the correlation function R(τ)

= h(τ) ∗ R (τ ) ∗ h(−τ ). u

Use this fact to show that the power spectral density function of theoutput (the Fourier transform of R(τ)) is S out (ω) H(iω) 2 S in (ω).

=|

|

5.6 Show that the state-space √ model in Section 5.3.4 is the representation of the transfer function H(s)

==

n 2 2 i −1 ωLow s i 1 s 2i −1 ωLow .

+

+

Chapter 6

Modeling a Partial Differential Equation in Simulink As engineers design systems with more stringent requirements, it has become far more common to find that the underlying dynamics of thesystem are partial differentialequations. Examples of this permeate the engineering design literature. For example, designers of computer disk drives are always striving to store more bits. They do this in two major ways: by making the number of bits per unit area on the surface as large as they can and by increasing the rotational speed of the disk. The smaller the area on the disk that contains the ones and zeros, the more accurate the positioning of the read/write head needs to be. At the same time, the speed of rotation of the disk makes it possible to access the stored ones and zeros faster, meaning that the read/write head needs to move faster making the assembly more prone to vibrations induced by the rapid accelerations. Thus, disk drive designers have to design the read/write head positioning system with the following: • flexible

motions of the read/write heads, their mounts, and the disks themselves;

•

aerodynamic induced vibrations as the heads move across the spinning disk;

•

vibrations induced by thermal effects caused by uneven heating of the drive.

The models for these dynamics are speci fic partial differential equations, and when they all must be included, they interact with each other. In a similar vein, automatic flight control systems for advanced aircraft (particularly aircraft that deliberately have unstable rotational dynamics) impose requirements that require the designer to include the effects of flexible motions and their interactions with the aerodynamics. Designers of spacecraft control systems, particularly those with precision pointing requirements, must model the vibrations of appendages that are induced by moving parts on the vehicle and, in some cases the combined effects of these vibrations interacting with (and creating) fluid sloshing in propellant and oxidizer tanks —once again, dynamics that are described by specific partial differential equations. In Numerical Computing with MATLAB , Cleve Moler showed methods for solving pdes using finite difference methods. Since Simulink has the ability to model and solve ordinary differential equations, it is natural to ask whether or not a partial differential 201

202

Chapter 6. Modeling a Partial Differential Equation in Simulink

equation may be solved using a method that converts the partial differential equation into a set of coupled ordinary differential equations. The answer is a resounding yes.

6.1 The Heat Equation: Partial Differential Equations in Simulink In the next chapter, we will begin the design of a home heating controller that replaces a thermostat, with the following bene fits: •

better control of the temperature of the home,

•

less energy use for the same or better comfort level,

•

a significantly better feel for the residents.

The way one goes about achieving these objectives is with a thorough understanding of the dynamics of heat flow from a home, so we need to model the home in detail. With this in mind, let us build a two-room home heating model in Simulink and run it to understand the issues. Let the temperature of the house be T (x,y,z,t), where the three orthogonal spatial directions x , y , z are relative to some coordinate system and the temperature is explicitly a function of time because of the heat supplied by the heating system. The underlying equation for computing the temperature is the heat equation ρc p

∂T (x,y,z,t) ∂t

= ∇ · ( k(x,y,z,t) ∇ T (x, y,z, t) ) + Q(x,y,z,t).

The source of heat in the rooms is the convectors (radiators) that are located either along the walls or in the floor, depending on the geometry. These heat sources are the heat input Q(x,y,z,t) .

6.1.1 Finite Dimensional Models There are many simplifying assumptions that we can use to model the home heating system using this equation. First, we assume that the walls, ceilings, floors, and glass areas have constant heat conductivities (i.e., k will be a constant for each of the different types of surface). Second, we can assume that each room in our house is a separate volumetric ∂T ( (t),t) in the heat equation) “lumped” entity so that the heat stored in each room (the ρc p ∂t is independent for each of the rooms. The last assumption we can make is that the heat flow in and out of the rooms (through the walls, ceilings, etc.) will result in a linear thermal gradient in the material. We can also assume that the heat exchangers have some thermal capacity (i.e., they can store the heat for some time). Each heat exchanger will transfer their heat to the room with some thermal loss. This heat exchange occurs through boundaries of air that, while not conducting (the heat exchange is actually through convection), are modeled assuming a linear gradient in the air around the exchanger. We make the reasonable assumption that the heat flowing out of the house does not change the outside air temperature (at x

6.1. The Heat Equation: Partial Differential Equations in Simulink

203

least over the time scale at which we are the heating the house). We also assume that the combustion process is a constant source of heat, with appropriate ef ficiencies for the combustion; this means that each gallon of oil or 100,000 BTUs ( “therms ”) of gas burned produces a constant amount of heat. Under these assumptions, the heat equation becomes a set of interconnected “lumps” with the “lumps” consisting of the individual rooms and the individual heat exchangers. These assumptions—constant outside air temperatures, linear gradients in the walls and aroundthe heat exchangers, and constant heat flow from the combustion process into the medium used for heat exchange—mean that the heat flow in steady state from the warmest parts of the home to the cold outside is constant. This follows from the fact that with these assumptions the term k(x,y,z,t) ∇ T (x,y, z, t) is constant for each of the rooms. We modeled a house in Chapter 2 (the model called Thermo_NCS), where we simply provided the model as a first order differential equation for the entire home. Because the house was a single entity, the temperature of each room in the house was the same. This is clearly not the case in a typical home. Each room’s thermodynamics are interconnected but also independent in the sense that the heat loss from in filtration, conduction, and radiation depend only on the materials in the walls, the number of windows (architects use the term fenestration) and doors, and the spaces that surround the room (basements, attics, other rooms, sunspaces, etc.). In the lumped parameter approximation, the individual room temperatures are the same throughout the room’s volume. These assumptions also mean that the net heat flowing into or out of a compartment (a volume of space that has constant thermal capacity and uniform temperature) sums to zero. From the heat equation when the divergence of the gradient is zero, there is no net increase in the temperature of the compartment from this flow—and a linear gradient ensures that this is true. We will see that this assumption is equivalent to the fact that, in an electrical circuit, currents flowing into and out of any node must sum to zero. Thus, for our two-room house, the heat flow equation breaks up into four separate ordinary differential equations: two for the room temperatures and two for the heat exchanger

6.1.2 An Electrical Analogy of the Heat Equation Once we have “lumped ” therooms and heaters into single entities and made the assumptions on how the heatis flowing, the heat equation becomes analogous to an electrical circuit with resistors, capacitors, current sources and voltage sources. The analogs of the different heat-equation element values are as follows: •

Heat flow is analogous to current (and a current source is a heat source).

•

The temperature of an element is analogous to the voltage in the circuit (and therefore a voltage source is a constant temperature regardless of the heat flowing in or out).

•

Thermal conductivity of the material is analogous to a resistor (the analogy is really to the conductance, i.e., 1/R is the thermal conductance, which will be important when we need to combine heat flow paths).

204


Rwall Rfl-ceil Rconvect Rwindow

Losses

Losses

Ta

RL 3 TH1

Tp RL 2 Flame Heat “Qf ”

RL 1

Heat Source:

RHE 1 CHE 1

TR1 CR 1

TH2 RHE 2 CHE 2

Room 1

TR2

Rwall Rfl-ceil Rconvect Rwindow

Ambient Outdoor Temp. “Ta”

CR 2

Room 2

Figure 6.1. An electrical model of the thermodynamics of a house and its heating system.

•

Thermal capacitance times the volume of the element, ρc p V , is analogous to the electrical capacitance. Using this analogy, we have the following:

•

The individual rooms in our home can each be modeled as a single capacitor.

•

The heat flow between rooms and the outside are modeled as current flowing through resistors.

•

The room’s heat exchangers are also modeled as capacitors.

•

The heat losses from the heat source (the modeled as resistors.

•

The heat source (oil or gas being burned) is a current source.

•

The outside temperature (constant) is a voltage source.

flame)

to the heat exchangers can also be

Putting these together, we can model our two-room house as the electrical circuit in Figure 6.1. This model is easy to understand because it presents a (lumped) picture of the underlying dynamics. It should be obvious that the picture in this case is not the equations (in the sense that a Simulink model explicitly represents the differential equations), but we can go through each piece in this circuit (picture) to understand exactly what it implies in terms of the thermodynamics.

6.2. Converting the Finite Model into Equations for Simulation with Simulink

205

Therefore, following Figure 6.1, we have the following: •

Each room and each heat exchanger is represented by a capacitor (this is the heat capacity of the room or heat exchanger given by ρcV). For each room this is the product of the air density, the specific heat of the air, and the volume of the room, and for each heat exchanger, it is the product of the density of the heat exchange medium (water or air), the specific heat of this material, and the volume of the heat exchanger.

•

The heat flowing into or out of theroom depends on thethermal conductivity, k , which we convert into a thermal resistance using 1 /k . Thus, the four resistors in parallel, at the top of each of the room compartments, are the paths for the heat flow due to the thermal losses from the walls, the ceilings, the convection of air from the outside to the inside, and the windows. These conduct the heat (current is heat) from the room to the fixed outside temperature which is the voltage source titled “Ambient Outdoor Temp.”

•

The heat flow from the heat exchangers to the rooms involves some reduction in the temperature of the medium so the room does not get as hot as the heat exchangers (the resistors RH E 1,2 ).

•

The heat from the heat source is not completely captured. (The process of transferring the heat from the boiler to the heat exchanger involves losses that are modeled as resistors RL1,2,3 .)

•

The power of this electrical analog for the heat equation is that it is easy to modify the model to make it more complex and include more effects. (We will do this in Chapter 8.)

6.2 Converting the Finite Model into Equations for Simulation with Simulink Thecircuit picture modelsthe heat flowand the roomtemperatures, but itis not inthe formof differential equations. To convert the picture into a set of coupled differential equations we need to use Kirchhoff ’s current law that states that the sum of all of the currents entering any node of the circuit must be zero (where the signs of the currents are determined by assigning arbitrary inequalities to the various capacitor voltages). For this model, the temperatures for each of therooms andheat exchangers (the capacitorvoltages) arethe “nodes.” The external sources and sinks of heat are the voltage and current sources that represent the “constant” ambient temperatures and heat flow. (These are not constant over a long time interval, but over the time interval that the heating systemis operating they arevery reasonably constant.) The result of these manipulations on the heat equation is the equations used in the Simulink model. We have captured the salient features of the home for design of the heating system without explicitly solving the heat partial differential equation. The method we used here is only one of the ways of converting the partial differential equation into a coupled set

206


of differential equations. They are all called “finite element models.” In this version, the elements are large masses that have more or less constant temperatures (each room and each heat exchanger). For many reasons it would be nice to have a tool that would convert a picture like this into the appropriate equations for use by Simulink. It also would be nice to maintain the picture above in the model because it is a lot easier to modify the picture to change the attributes of the house. Adding more rooms, heat exchangers, and intermediate losses like that from the heat exchangers, etc., using the picture is a lot easier than working with gradually more complex equations. This is true because the picture clearly shows what interactions are taking place among the various components, and it is easy to cut and paste to create additional components with a computer. If such a tool existed, it would allow the pictorial representation of the circuit to be visually available in Simulink, just as the signal flow is available. Such a tool exists. The SimPowerSystems Blockset from The MathWorks would convert this picture into the differential equations that Simulink needs. We describe this tool later, but for now let us work through the model above and get the differential equations. (In Chapter 8, as we investigate how to use the SimPowerSystems tool, we will create a multiroom model.)

6.2.1 Using Kirchhoff’s Law to Get the Equations Kirchhoff ’s current lawat each of thecapacitancenodesin themodelgivesus thedifferential equations for the simulation. Thus we have the following. For Heat Exchanger 1:

CH E 1

dT H 1 dt

= −



1 RH E 1

+

1 RL2



T H E 1 +

1 RH E1

T R1

1

+

RL2

T p .

For Room 1: CR1

dT R1 dt

= −

1 Req 1

T R1

1

+

RH E1

1

T H 1 +

Req 2

T a ,

where 1 Req 1

=



1 Req 2

1 Rwall 1 =



+

1 Rf l −ceil1

1 Rwall 1

+

+

1 Rconvect 1

1 Rf l −ceil 1

+

+

1 Rwindow1

1 Rconvect 1

+

+

1 RH E 1

1 Rwindow1





.

For Heat Exchanger 2:

CH E 2

dT H 2 dt

= −



1 RH E2

+

1 RL3



T H 2

+

1 RL3

T R2

+

1 RL3

T p .

,


207

For Room 2: dT R2

CR2

1

= −

dt

T R2

Req 3

1

+

RH E 2

T H 2

1

+

Req 4

T a ,

where 1



=

Req 3

1

+

Rwall 2

1

=

Req 4



1 Rf l −ceil 2

1 Rwall 2

+

1

+

Rconvect 2

1 Rf l −ceil 2

+

+

1 Rwindow2

1 Rconvect 2

+

1 RH E2

1

+

Rwindow2





,

.

There is also an algebraic(nondifferentialequation) forthe temperature of the heat exchange medium, T p , given by summing the heat flow into this node at the heat source. For the heat fl ow to the heat exchangers (algebraic equation): Qf T p

=



T H 1

+

1 RL1

+

RL2 1 RL2

+

+

T H 2 RL3 1 RL3



Req 5 Qf +

=



T H 1 RL2

+

T H 2 RL3



,

where 1 Req 5

=



1 RL1

Substituting this algebraic result into the of these two equations as CH E 1

dT H 1 dt

= −



1 RH E1

+



1−

Req 5 RL2

+

1 RL2

+

1 RL3



.

first and third equations above gives the final form

  1

RL2

T H 1 +

1 RH E 1

T R1 +

Req 5 RL2 RL3

T H 2 +

Req 5 RL2

Qf

and CH E 2

dT H 2 dt

= −



1 RH E2

+



1−

Req 5 RL3

  1

RL3

T H 2 +

1 RL3

T R2 +

Req 5 RL2 RL3

T H 1 +

Req 5 RL3

Qf .

These differential equations were relatively easy to write out because there are only four differential equations (and one algebraic equation); however, if we were to try to model a house with 15 rooms, the task would be quite dif ficult. In the development of these equations, we have de fined some equivalent resistors. These equivalents are meaningful in terms of the heat losses for the rooms and the heat exchangers. The combined loss for each room is the equivalent resistor that comes from placing the resistors for the losses in parallel, so the net loss is the combined effect of the heat flowing through the walls, floors ceilings, doors, and windows and the conductive path of the heat exchangers. In practice, these resistances account for the convective losses that come from air in filtration into the rooms. Since the air in filtration depends on the external winds, these added terms include the wind speed. We have ignored this effect in our model, but it is important enough that we will revisit the model and add this effect in Chapter 8.

208


6.2.2 The State-Space Model Now that we have the equations to model the house heating system, we will create a statespace model for use in Simulink. Assume that the voltage (temperature) of each capacitor is a state in a four state model (one differential equation, or state, for each temperature). The equations above then become

 T d  T  T

  

R1

C

R2

dt

H E1

T H E 2

 T  T  T

R1

= G

R2

H E1

T H E2

  

+ B1 T a + B2 Qf .

Let the state vector be x(t) , and we can invert the matrix d dt

−1

x(t) = C

−1

Gx(t) + C

C

to give the state-space model, −1

B1 T a + C

B2 Qf .

The matrix C is diagonal so its inverse is simply the reciprocal of its elements, and therefore the final form of the state equations has the following matrices: −1

C

    

−

G = 1 CR1 Req 1

0

0 −

1 CR 1 RHE 1

0

0

1 CR2 RHE 2

1 CR2 Req 3

1 CHE 1 RHE 1

0

0

1 CHE 2 RHE 2

−



1 + CHE 1 RHE 1



1−

Req 5 RL2



1 CHE 1 RL2

Req 5 CHE 2 RL2 RL3

−1

C

B1 =

   

1 CR 1 Req 2 1 CR 2 Req 4

0 0

   , and 



Req 5 CHE 1 RL2 RL3

−

−1

C

B2 =



1 + CHE 2 RHE 2

   

0 0 Req 5 CHE 1 RL2 Req 5 CHE 2 RL3



1−

Req 5 RL3



1 CHE 2 RL3

    , 

   

This set of equations is the basis for the Simulink model of the heating system. We can run the model to learn what the temperatures do when the heating plant starts, and when the heating plant is off. To complete the model, we need numeric values for the parameters. To determine the parameters we need to de fine the house. We assume that room 1 is a rectangular solid with dimensions of 20 × 30 × 8 ft; thus the areas of the floor and ceiling are 600 square feet. Room 1 has four windows and one door, with a total area of 72 square ft. Room 2 is also a rectangular solid with a size of 20 × 40 × 8 ft. We assume that the two rooms are adjacent along the 20-foot dimension and that the heat flow through this wall is negligible (and we will ignore it). Room 2 has eight windows and two doors, with a combined area of 180 square ft. The area of the exposed walls in Room 1 is therefore the total surface area minus the glass and door areas, which is 568 square feet. Similarly, Room 2 has an exposed surface (minus the glass and doors) of 620 square ft. The floors and the ceilings of both rooms have 12 inches of fiberglass insulation. (This insulation has an equivalent R (i.e., including the in filtration loss, value of 30). The


209

thermal conductance is the reciprocal of theresistance, so an R of 30 translates into a thermal conductivity of 1/30 BTU/hour per square foot of floor and ceiling surface area per degree F temperature difference between the room and the outside. The walls of each room also have been insulated with fiberglass with the equivalent R-value of 15. (Again this represents a heat conductance of 1/15 BTU/hour per square foot of insulated wall surface area and per degree F.) The last equivalent thermal conductance value we need is that of the windows and doors. With double panes of glass, most modern windows (and doors) have equivalent R-values of 3, which gives a thermal conductance of 1/3 (same units as above). In all of these numbers we assume that the convection has been included in the equivalent R so the explicit term in the room equivalent resistances for the convection is not used (however, we will investigate the effect of added convection losses using this term). For the calculation of all of the thermal resistance values above, the values are per hour, but the simulation will be in seconds, so the data needs to be converted. The heating system useswater as the heatexchangemedium. The thermal conductance (one over the resistance) that couples the heat exchangers into the rooms is .01 BTUs per hour per linear foot of convector per degree F temperature difference between the rooms and the heat exchangers. We assume that Room 1 has 25 ft of heat exchangers, and Room 2 has 30 ft. We also assume that the losses in the heat exchange process are 0.05. The specific heat of air is 0.17 BTU per pound per degree F, and the density of air is 0.0763 pounds per cubic foot. Combining these quantities together, we get the following values for the components in the state-space model: The thermal capacitance of Room 1 is the mass of the air in the room times the specific heat of air. The room volume is 4800 cubic ft, so the mass of air is 4800 × 0.0763 = 366.31 pounds, and the capacitance is 0 .17 ∗ 366.31 = 62.27 BTU per deg F. Similarly, the capacitance of Room 2 is 83.03 BTU per deg F. For the heat exchangers, we assume that the volume of water in them is 1 cubic ft. Since the speci fic heat of water is 1, the thermal capacity of each of the heat exchangers is therefore 62.5 BTU per deg F. Thus the capacitances used in the model (in BTU/deg F) are CR1

=

62.27,

CR2

=

83.03,

CH E 1

=

CH E2

=

62.5.

The thermal resistances for the insulation in the various parts of the house are not the thermal resistance numbers in the differential equations (the dimensions of the various quantities show this). To get these numbers we need to multiply the conductance numbers for the various surfaces by their areas. Thus, we have (before the conversion from hours to seconds) RH E 1 = RH E 2 = RH E 3 = .01 · 20 = 0.2,

Rwall 1 Rwall 2

=

=

15 568

15 620

=

=

0.0264,

0.0242,

RL1

=

Rf l

ceil1

Rf l

−

ceil 2

−

RL2 =

=

=

RL3

2·

30 600

2·

30 800

= =

=

0.05, 0.1,

0.075,

Rwindow1

=

Rwindow2

=

3 72

=

3 180

0.0417,

=

0.0167.

The Simulink model we created once again uses the state-space model explicitly, and it uses the vectorizing capability of Simulink. The model resulting from this exercise is extremely simple; it is in the figure below. This model will work no matter how many rooms and heat

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Open Loop (no control) Investigation (double click on the switch): - no heat from furnace (top) - max heat from furnace (bottom) Heat Input (Furnace OFF)

Closed Loop vs. Open Loop Investigation (double click on the switch): - Open Loop (top) - Cl osed Loop (bottom)

Heat Zones (B1 is zone 1 B2 is zone 2)

Swicth Furnace On or Off

B1 from m-file

Outside Temperature (deg. F)

B1* u

10

Switch Heat Controller On or Off

0 K*u Qf Heat Input (Furnace ON) 70

Saturation (Heat command is > 0 and <1)

Temperatures

Closed Loop Set Point Temeprature

1 xo s

Heat from Furnace Set Temperature Linear Hating Controller

Matrix Multiply

Integrator Product

[50 50 120 120]'

Temperatures (R1 R2 HE1 HE2)

Plot 4 Temperatures Room 1 and 2 Heat Exchanger 1 and 2

K*u Initial Temperatures Zone 1 & 2 Heat Input Vectors (B2 is zone 1; B3 is zone 2) both vectors are from m-file

A A Matrix from m-file

Temperatures (R1 R2 HE1 HE2)

Select Room1 and 2 Temperatures

Simulation of the Heating System in a Two Room House

Figure 6.2. Simulink model for the two-room house dynamics.

exchangers we model; all you need to do is change the matrix values. The data are all in an M-file called TwoRoomHouse.m that is in the NCS library. This M-file runs when the model opens because of the callback from the “Model Properties ” dialog. At this point, you might try creating this model yourself. If you look back at the state-space model that was developed in Section 2.1.1, this Simulink model will be similar. (The state here is of dimension 4 and in the earlier model it was 2, but this does not matter since Simulink will use the matrices provided to get the dimensions.) Remember to put the initial conditions into the integrator as a 4-vector. Our version of the model is shown in Figure 6.2. (It is in the NCS library and is called TwoRoomHeatingSystem .) This model has been set up to run the M- file TwoRoomHouse when it starts. This populates all of the matrices in the model with the correct data. Two Simulink “manual switches” are in the model. Remember that these switches change state whenever you double click on the icon. We use them to investigate the open loop (no control) attributes of the house temperatures. The first switch (at the top left of the diagram) is set up when the model opens to run the simulation with the heater off. The temperatures that result from this analysis are in Figure 6.3. (The values of the four states appear on the same graph.) The second, Figure 6.4, shows what happens when the heater is turned on (double clicking the switch at the top left) and left on (the graphs are in the same order as the first figure). The simulations are for 3 hours (10,800 sec). The outside temperature is 10 deg F, and the heat from the furnace, Qf , is 22.22 BTU/sec (or 80,000 BTU per hour, which is the heat from a gallon of fuel oil when burned in one hour with about 80% ef ficiency). These numbers are very typical for a reasonably well insulated house. (Remember that this simulation uses an outside temperature of 10 deg F.) Double click on the second switch in the diagram, and rerun the simulation. The resulting temperatures are shown in Figure 6.5. We will reserve comment on these results


211

Temperatures (R1 R2 HE1 HE2) 120 Room 1 Temperature Room 2 Temperature Heat Exchanger 1 Temp.

100

Heat Exchanger 2 Temp. 80

60

40

20

0

0

1000

2000

3000

4000

5000 6000 Time

7000

8000

9000

10000

Figure 6.3. Simulation results when the heat is off.

Temperatures (R1 R2 HE1 HE2) 240 220

Room 1 Temperature

200

Room 2 Temperature Heat Exchanger 1 Temp.

180

Heat Exchanger 2 Temp.

160 140 120 100 80 60 40 0

1000

2000

3000

4000

5000 6000 Time

7000

8000

9000

10000

Figure 6.4. Simulation results when the heat is on continuously.

Temperatures (R1 R2 HE1 HE2) 150 140 130 120 110 Temp. Room 1

100

Temp. Room 2 HE 1 Temp.

90

HE 2 Temp. 80 70 60 50

0

1000

2000

3000

4000

5000 6000 Time

7000

8000

9000

10000

Figure 6.5. Using PID control to maintain constant room temperatures ( 70 deg F).

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until the next chapter. For now let us assume that a thermodynamics design team has just performed the analysis of the house heat losses and developed the heating controller. The next step in the process of designing a controller will be to put this heating system model together with a model that has the controller logic. This logic will use a tool that allows complex logic with a signal flow graph that is similar to Simulink (i.e., with a visual programming environment that is “tuned” to the needs of logic). The tool is Stateflow, which is the subject of the next chapter. We will revisit this model and create a new type of home heating controller, one that will provide heating that is more comfortable. The system will use the PID control system that we developed here.

6.3 Partial Differential Equations for Vibration The techniquesused in developingthe finite dimensional model forthe heat equation apply to other types of equations (such as the partial differential equation that describes the vibration of the read/write head in a disk drive). The idea is the same; the difference lies in the order of the time derivatives. In the case of the heat equation (which, because of the electric analog we showed, also de fines electric fields), the differential equations that result are always first order. This class of equations (called parabolic equations) also includes any type of diffusion. When the time derivative is zero (i.e., the solution in steady state is desired), they are called elliptic equations. The equations that describe vibration are more complex in that the underlying ordinary differential equations that result are second order (these equations are hyperbolic). The method for creating the finite element model follows the above method; we assume a particular “shape” for the spatial part of the vibration, and these shapes break the total motion into a large number of finite elements. In the process of developing these elements, we create masses and inertias that interact through springs (and damping elements) so the resulting equations are second order in time. The state-space models that result are then a set of coupled second order differential equations that describe the oscillations. This topic can be the subject of an entire book by itself, and if you are interested in learning more, see [6], [21], and [28]. We will also look at an example of a vibrating string in Chapter 8.

6.4 Further Reading The modeling of thermal systems using electrical analogues is pervasive. A recent special issue of the Proceedings of the IEEE describes this modeling approach for the calculation of the temperature of a chip. ([34] is one particular paper that discusses the thermal modeling of a VLSI chip.) An online PDF version of a book on heat transfer also is available [25]. The use of equivalent thermal resistance to model both the heat conduction and convection is pervasive but prone to error. The equivalent thermal resistance is good for a single convection rate. The convective losses depend on the pressure difference between the interior spaces and the exterior. These are a function of wind speed, so the greater the wind speed, the larger the heat loss due to convection. It is typical for oil companies to calculate oil consumption using a number called degree days. This number is the accumulated sum of the difference between 65 deg F and the average outdoor temperature over some period. (Generally, oil companies use the

Exercises

213

number to determine when to deliver oil, so the period is approximately a month). You can compute this number yourself, and check your oil or gas consumption per degree day. You will soon see that your consumption per degree day is not constant. However, if you use the average wind-chill–corrected temperature (which captures the convective loss), you should see a very good match. In the model we have created, you can change the thermal resistances to account for the effect of the wind by lowering the resistance by some amount. We ask you to work this out in Exercise 6.2.

Exercises 6.1 The two-room house model with the 24-hour external temperature variation can calculate the homeowner ’s heating bill. Adda setof blocksto themodel that accumulates the total heat used over the 24-hour period. Then modify the model to use the thermostat that we introduced in Chapter 2. Compare the heating costs for the two methods of heating. Is there any difference? If there is, why is this so? 6.2 Develop a model for the heating system that has a thermal loss for each room that depends on the external wind speed. Modify the resistances Req 1 , Req 2 , Req 3 , and Req 4 (remember that they each contain an equivalent resistance for convective losses) so they are of the form Req 1 (t ) = Req 1 (0) + Req 1 w(t)

where w(t) is the wind speed. Make Req 1 10% of the nominal resistance when the wind speed is 10 mph. That means that the modified thermal resistances are Req 1 (t )

=



Req 1 (0) 1 + 0.1

w(t)

10



.

Now investigate the effect of the wind on the heat loss. What can you conclude about the importance of reducing the convective losses?

Chapter 7

Stateflow: A Tool for Creating and Coding State Diagrams, Complex Logic, Event Driven Actions, and Finite State Machines Simulation of systems that have logical operations or that contain actions triggered by external asynchronous actions is dif ficult. For example, if we are simulating the action of a person throwing a switch, or if we are simulating the action that causes a computer to process an interrupt, thesimulation needs to process the event at the time it occurs. When we create such a model using Simulink, the blocks and their connections do not always provide clear indications of the data flow making the diagrams dif ficult to read and understand. The first attempt at creating a visual representation of complex logic flows was David Harel’s introduction, in the 1980s, of a visual tool to represent all of the possible states that a complex reactive system might have. He called the idea a “statechart ” [18]. The major innovation that a statechart provided was “And” states, where parts of the diagram could be acting in parallel with other parts. Until the introduction of this idea, the modeling of finite state machines relied on the ideas of Mealy and Moore from the 1950s. (Thus, in the Harel approach, states could be active at the same time —one and two are both active —whereas in the Mealy and Moore approach, all states were “Or ” states.) In 1996, The MathWorks introduced a new tool called State flow, which encompasses many modeling and programming ideas into a single and easily understood tool. The name hints that the tool allows modeling automata using the statechart approach, while also allowing the picture to represent signal flow. Designing a system with complex logic and state transitions that depend upon events (either external or internal) using the State flow paradigm is remarkably easy. It also has the major advantage that it integrates with Simulink. This last feature makes it possible not only to design the logic for an automata but also to model its interaction with the environment. The result is an insight into how well the automata design works; that is, it allows you to verify that the design meets the speci fication as it operates in a simulation of the real world. In the early stages of any design, the ability to make decisions about the structure of the automata can help clarify and tighten the requirements as the speci fication for the design also evolves. The result becomes an “executable speci fication,” which we will have more to say about in Chapter 9.

215

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Chapter 7. Stateflow

7.1 Properties of Stateflow: Building a Simple Model We used the concept of “state” extensively in this book. Stateflow expands the concept of a dynamic state to systems that do not require differential or difference equations. The basic building blocks for creating a State flow machine are the eight icons at the left. Starting at the top, we have the following: •

The first icon is used to create a state in the State flow chart.

•

The second is the icon that denotes a state that should remember its history.

•

The third icon is used to denote the initialization of the State flow chart.

•

The fourth icon is used to make signal flow connections. (Typically the connections are like resting points prior to single or multiple decisions.)

We will explore each of these in turn. The last four icons are truth tables, functions, embedded MATLAB code, and the “box.” We will consider only the truth table object and the embedded MATLAB object in this chapter. The box icon allows the user to change the shape of the State symbol to a square (box) and treat the result as a subsystem. The easiest way to get familiar with State flow is to build a simple model. Therefore, we will create a model that uses State flow to track the position of a simple on-off switch. Clearly such a switch has only two (exclusive or) states. Therefore, the diagram will be quite simple. To create the Stateflow model we need to open Simulink and create a new Simulink model using the blank paper icon in the menu. Then, in the Simulink library browser, open Stateflow by clicking on its icon, and after it opens, drag a new chart into the Simulink model. These steps should create a Simulink model that looks like the figure below. Double click the Chart icon, and the State flow window will open. This window contains the icons above on the left hand side. This window is where we construct the actual Stateflow chart. We have already established that we need two states in this chart, so let us start by dragging two copies of the state icon (at the top of the icon chain) into the diagram. After you place Chart each state icon, you will see a flashing insertion point at the top left side of the icon. If you immediately start typing, the name you assign to the state appears in the state icon. Let us call the first state “Off ” and the second “On.” If you do not type these names immediately, the icon displays with a question mark where the name should go. To add the name, simply click the question mark and type the name. With the states in place, we need to specify the transitions. If you place the mouse cursor over any of the lines that specify the boundary of the state (any of the straight lines that bound the state will do), you will see the cursor change from an arrow to a plus sign. When this happens, left click and drag the mouse to create a connection line. This line is

7.1. Properties of Stateflow: Building a Simple Model

217

actually a Bezier curve that will leave the state perpendicular to the boundary, and when you make the connection to the terminating state, it will be perpendicular to the bounding lines of that state. You need to drag two lines, one starting at the On state and terminating at the Off state and the second going the opposite direction. When these lines are in your diagram, click in turn on each of them. You will see a question mark on the line that will allow you to specify what causes the transition. We will set up an event that we will call “Switch ” that will cause the transition next, so for now all you need to do is click on each of the transitions and type the word Switch in place of the question marks. As a final step, we need to specify where in the diagram we want the state chart to start. Select the third icon in the menu to do this. When you drag the icon into the chart, a transition arrow will appear. You want to attach this to the state we have called Off. This will cause the Stateflow chart to start in the Off state. When you have completed all of these steps, the diagram will look like the picture at the right. We cannot execute the diagram yet because we have not specified what the event “Switch ” means, nor have we created anything in Simulink that will simulate the throwing of a switch. So let us specify what the event is. At the top of the chart is a menu with the traditional File, Edit, View, Help, and Simulation options that are in Simulink. In addition, there are two options called Tools and Add. Click the word “Add” in the menu and a list will appear; from this list select “Event” and in the submenu of this list we need to select “Input from Simulink.” When you do this, the dialog box shown at the left will open. Enter the name of the event (“Switch”) in the box titled “Name” (replacing the default name “event1”), and where it says “Trigger,” select “Either” from the pull-down menu. Finally, click Apply. After you complete this, look at the Simulink model that contains the Chart; you will see that the Chart icon has a Trigger input at

218

Chapter 7. Stateflow 1

0 Manual Switch

Chart

Figure 7.1. First State fl ow diagram (in the chart) uses a manual switch to create the event.

the top center of the chart and the icon inside the chart indicates that both rising and falling events will cause the event to trigger. The trigger icon is the same as that used for triggering a Simulink subsystem. This process has created not only the variable “Switch ” in Stateflow but also a data type (in the sense of a strongly typed programming language). All variables in Stateflow have to be “typed” using this process (i.e., by using the Data or related dialog). We now need to build a switch in Simulink that causes the trigger event “Switch.” In the Signal Routing library, you will find a Manual Switch icon that we can use. Drag it into the model and connect the switch arm to the trigger input of the chart. Last, from the Source library drag a Constant block into the model and duplicate it. Set one of the constants to 1 (leave the other at the 0 it came in with), and connect them as shown in Figure 7.1. The model is now complete. The model has only discrete states, so you can go into the “Configuration Parameters” menu (under Simulation or you can use Ctrl+e) and change the Solver to discrete (no continuous states), and while you are at it, change the stop time to “inf ” (the MATLAB name for infinity). If you now run the model and double click the switch icon while looking at the Chart, you will see the transitions from state to state take place each time the switch is thrown. (An active state will be highlighted in blue, and each time a transition takes place, it also will be highlighted in blue.)

7.1.1

Stateflow Semantics

The Stateflow chart that we have created so far is extremely simple. It should have given you an idea of how the chart works. (In the language of automata theory, this knowledge is called semantics.) The chart we created in the introduction does not accomplish anything. In order to do something, we need to export an action from the chart. However, before we do that, let us describe some of the semantics of State flow. Look at the diagram in Figure 7.2. It depicts most of the semantics of Stateflow. Semantics are associated with each of the State flow objects as follows. •

States:

– States can be hierarchical . A state that contains other states is called a super state (examples: SuperState_1 and SuperState_2 in the figure).


219

Figure 7.2. A State fl ow diagram that contains most of the state fl ow semantics.

– Statescancausean action whenthey are entered or exited, during their execution, or when an event occurs while the state is active. (Examples: State SS1.1 has entry action SS1.1_A1, during action SS1.1_A2, and exit action SS1.1_A3.) Note that thenamesof theactions canbe anything youchoose, andthe semantics allow you to abbreviate the names to en, du, and ex , respectively.

– States are, by default, exclusive “Or ” states. However, you can change them to parallel (“ And ”) states by highlighting the states that you want to be parallel and then right clicking on the super state that contains them and selecting the Parallel (AND) option under Decomposition in the menu that opens. Parallel states appear in the diagram with dotted boundary lines (SS2.1 and SS2.2, for example).

– States may be grouped together (as a graphical object in the diagram) using the Group state icon in the menu at the top of the chart. (It is the tenth icon.) You can also double click on the super state that contains them. (Example: SuperState_2 in the diagram is grouped, as can be seen from the wide border and the shading.) “

”

“

”

– Super states or combinations of states may be made into a subsystem (or a subchart in the semantics) by selecting this option from the Make Contents menu “

”

220

Chapter 7. Stateflow after you right click on the combination. (Example: The SuperState_3 in the diagram is a subchart. It has internal structure that appears when you double click on it, just as a subsystem opens in a new window in Simulink.)

•

Transitions, Conditions, and History:

– A transition is a directed arrow in the diagram that implies the action of leaving one state (the source) and moving to another (the destination).

– A default transition is the method Stateflow uses to determine which of the exclusive (“Or”) statesare to be activewhen there is ambiguity(for example, when the chart is started). These transitions may also occur with conditions attached or because of events. The initialization of SuperState_1 and the state SS1.1 in this super state are both defaults that occur on the first entry into the chart.

– The History (the encircled H in the diagram) overrides the default. In this example, the history forces State flow to remember the substate that was active when the exit action from SuperState_1 occurs. (In the example, the transition from SuperState_1 is to SuperState_2, and it occurs regardless of which substate in SuperState_1 is active.) The subsequent return to SuperState_1 will go back to the last active substate because of the history. Otherwise, State flow uses the default transition.

– A transition may occur because of an external event (as we saw in the previous section) or because of an event that is internal to thechart. Events arenongraphicalobjects; they do nothave a pictorialrepresentation, butthey areshown in a dialog that is opened using the “Explore” option under theTools menu in thechart.

– A transition can also use a condition. Conditions can be one or more Simulink data variables. (Data variables arealso nongraphical, and you canview them using the Explorer option.) The semantics for conditional operations on Simulink data is to enclose theoperation in squarebrackets. (Example: in thechart above, the conditions are Condition_1, Condition_2, and Condition_3.) Thus, the condition [t > 10] causes a transition when “t” (data from Simulink) is greater than 10; note that this means that when the State flow model is C-coded, “t” will be input data.

– We can also have actions associated with a transition. Actions can occur in two different ways. The first is during a transition, when the condition evaluated is true (before the transition takes place). The second is when the transition actually takes place. The first action is denoted by putting the action inside of curly brackets { and }. The second is denoted using a diagonal slash / (Example: [Condition_3] {CA_3} / TA3 in the diagram above denotes that Condition_3 is being testedfor thetransition, andif it is true, theconditionalaction CA_3 will be created; the transition will then occur, and the transition action TA3 will occur.) •

Connective Junctions:

– A connective junction is a place that allows a transition to terminate while a decision is made. (It is denoted by a circle; transitions can be made to or from


221

this circle.) This device is the “flow” part of State flow. It allows pictorial representations of all kinds of programming decisions and as such greatly contributes to the visual programming approach that State flow embodies. As an example, in the chart above, there is an if, elseif, and else implied by the connective junction with the three transitions labeled 1, 2, and 3. The code that would be created from this is as follows: If Condition_1 then, if Condition_2 cause action CA_2, elseif Condition_3 cause action CA_3 and make the transition to state SS1.2 and cause transition action TA_3, else transition back to state SS1.1. Notice that in this example, the diagram has three lines with six clearly visible actions, whereas the verbal description contains 30 words, each of which must have precisely the stated order for the proper actions. (The code that it implies also is verbose.) As we develop some more complex State flow examples, the power of this visual programming environment will become even more evident. This is nota complete description of State flow’s semantics. We will see how Stateflow programs a chart when we build a model that does something.

7.1.2 Making the Simple Stateflow Chart Do Something Before we make our chart do something (have outputs), let us look at how State flow differs from the classical Mealy and Moore state machines. In the Moore paradigm, outputs from the finite state machine depend only on the machine ’s states. In the Mealy paradigm, an output can occur when there is a change in the input, so the output depends on both the states and the input. The newest version of State flow allows you to restrict the State flow chart to be one or the other of these two paradigms. If you wish to do this, you need to open the “Model Explorer ” window in the chart by selecting the “Explore” option under the “Tools” menu and then selecting the Mealy or Moore option in the “State Machine Type” pull-down menu. In addition to the method for de fining outputs, State flow also allows parallel ( “And”) states, while both the Mealy and Moore schemes are purely exclusive “Or ” machines. Finally, the ability to do signal flow using transitions and connection icons makes it possible to implement (and visualize) complex branching, including do loops and all types of if statements. Now we can returnto theexample we started and createsome outputs. We will assume that we have a simple speci fication for the automata that we want to create, and we will then build the State flow model to satisfy the speci fication. We will then look at the way State flow processes the chart by looking at the transitions and the actions both during the transitions and in the states themselves. We assume we are creating a device that will cause a neon sign to flash with a period of 0.5 sec with a duty cycle of 50%. The first requirement that we have is that the drive electronics for the neon light must be “warmed up” for at least 0.25 sec before the flashing commences. (This ensures that the voltages are at the correct values.) After the warmup is complete, we commence flashing using a timer to trigger the state chart. The final requirement is to turn on a fault detection circuit that will monitor the voltage and current going to the lamp, but this can be done only 1 sec after the start command is issued, to allow

222


Pulse Generator 12:34

start t

on trip

Digital Clock Chart

start on trip

Scope

Figure 7.3. The Simulink model for the simple timer example.

the start transient to die down. (This time is greater than 3 time constants, so we select a time of 1 sec). We assume that the earlier model was the first step in the development of this timer, so we will modify this model to incorporate the complete speci fication. Here, then, is the specification: •

A timer that sends out a pulse every 0.125 sec will replace the switch we created in the early version (and because this triggers the chart, it will cause the states and transitions to occur every 0.125 sec).

•

We will send the clocks time to the chart so we can create outputs that are “Timed.”

•

We will have three outputs. The first will be a variable called “start” that starts the light’s electronics. The second will be a variable called “on” that causes the neon light to switch on and off at the 0.25-sec rate. The final variable, “trip,” occurs after a delay of 1 sec.

Figure 7.3 shows the Simulink model with the state chart that achieves this speci fication (called SecondStateflowModel in the NCS library). Run this model several times, carefully watching the transitions and the actions that occur on entry and exit from the states. (This is easy to do because the chart in Figure 7.4 is animated.) Along with the animation, also look at the three outputs in the Scope block. The model has been set up so all of the transitions in the Chart occur every 0.6 sec. (This is done using the State flow Debugger under the Tools menu; we will look at this tool in more detail later.) The chart has the same two states that we used in the initial example. We have added entry and exit actions for the “On” state and an exit action for the “Off ” state. In addition, we have put the connective icon at the transition from the “Off ” state to the “On” state. This connection is an if-then-else statement. It executes the following logic: •

If the time t is greater than 1 sec, then trip is set to 1; otherwise (the else), go unconditionally to the “On” state. (Remember that t is data from Simulink; when the device is implemented, this is the clock time beginning when the timer is turned on.)

•

When we enter the “On” state, State flow sets the variable “on” to 1, and when the chart exits the “On” state, the variable “on” is set to 0 (causing the flashing).


223

Figure 7.4. Modi fied State fl ow chart provides the timer outputs “Start,” “On,” and “Trip.”

•

All of the transitions take place because the event Switch is triggered by the clock pulses. (Remember that either rising or falling edges of the pulse trigger the event.)

To create the input “t” for the Statechart we used the “Add” menu in the State flow window. When youselectAdd, theoptions that youcan useare “Event,” “Data,” or “Target.” We want “t” to be Data that is an input from Simulink. Therefore, select Data, and then from the submenu select “Input from Simulink. ” In a similar way, we need to add outputs called “on,” “start, ” and “trip” (as in the chart diagram above) The Simulink model contains the chart and models of the timer. (We use the Pulse Generator from the Simulink Sources library with the dialog values set to “Pulse type Sample based,” “ Period 4,” “ Pulse Width 2,” and “Sample Time 0.125.” The digital clock is also from the Sources library, and its sample time is 0.125 sec. We have now looked at the semantics for State flow, and a simple example with inputs and outputs. We can now create a model that actually simulates a device that has both complex dynamics and a State flow chart that provides the code for some complex logic.

=

=

=

=

7.1.3 Following Stateflow’s Semantics Using the Debugger When you ran the model and viewed the animation, you should have noted that there is a logical pattern to the execution of the chart. This pattern is part of the State flow semantics that you should understand. Let us look at how State flow translates the chart into actions.

224

Chapter 7. Stateflow start 1 Event 2 0.5

0 0

0.5 Event 1

1

1.5

Event 3

2

2.5

3

3.5

4

2.5

3

3.5

4

2.5

3

3.5

4

on

1

0.5

0 0

0.5

1

1.5

2

trip 1 Event 6 0.5

0 0

0.5

1

1.5

2 Time

Figure 7.5. Outputs resulting from running the simple timer example.

First, the chart does nothing until something that the chart recognizes occurs. This recognition might be an event, as we have used in the model above, or it might be a change in some data from Simulink. State flow can recognize a Simulink variable change as a “condition” for a transition. In the (anthropomorphic and cute) semantics of State flow, the chart is “asleep” when it starts. When something outside the chart causes an event or the fulfillment of a condition, the chart wakes up. When the chart wakes up for the first time, the default transition takes place and the chart initializes. Note that this takes 0.25 sec because the first event occurs then. Thus at the end of 0.25 sec the first state is active, but the chart is asleep again. The next event is at 0.5 sec. (The clock ticks occur every 0.25 sec because the Pulse Generator has a period of 4 samples, with a pulse width of 2 samples.) At the second Switch event, the chart transitions from the “Off ” state to the “On” state, and the entry action start=1 occurs. (Figure 7.5 shows the Scope that displays all of the outputs from the chart.) If you watched the transitions carefully in the State flow window, you will see that the first action that occurs is the transition. Then the connective decision occurs, with the flow stepping up to the border of the “On” state. Before the chart enters this state, the “Off ” state exit action occurs. Then and only then does the chart enter the “On” state. You can understand the details of the actions and their timing if you start the State flow debugger. This is accessible in the Tools menu at the top of the chart window or by typing Ctrl+g when the chart window is active. The debugger opens with the check box “Disable all” checked. To have the debugger stop at the various actions in the chart, deselect this check box. Run the model from the debugger by selecting Start. The debugger will stop the chart at every action, and you can use the “Step” button to move to the next change in

7.2. Using Stateflow: A Controller for Home Heating

225

the chart. As you step through the chart, note carefully the actions that State flow takes at each step. This is the best way to understand the State flow semantics. In the next section, we will use State flow to create a new concept in home heating control. This control will be an alternative to a thermostat, we will create the controller using State flow to handle all of the logic, and we will create the control system in Simulink, all the while using Simulink to model the furnace that heats the home. This example builds on the simple home heating example (using a thermostat) that we investigated in Chapter 2, and the Two Room House model that we developed in Chapter 6.

7.2 Using Stateflow: A Controller for Home Heating The simple home heating example we investigated in Chapter 2 assumed that the heat source was an electric heater. The thermostat that did the control was a bimetal device that worked because of the unequal coef ficients of expansion of the two metal pieces. This type of heating system is still in common use in most homes, but it is not very ef ficient. First, the hysteresis of the thermostat causes the temperature of the house to overshoot the value that the thermostat is set to. (This overshoot can be as much as 4 or 5 deg F for a typical thermostat.) Second, the typical house heat source is oil or gas (not electricity), and the heateritself typicallyheatseither airor water that circulates aroundthe house. (In thecase of water, or hydronic, systems, the water may be circulated as superheated water or as steam.) In either case, the heat exchanger that transfers the heat generated by the flames into the water or the air has signi ficant heat loss. (A good boiler has an ef ficiency of about 75%, and a good air heat exchange plenum has an ef ficiency of about the same.) One way to avoid these losses is to circulate water or air at a temperature that varies (and is typically only slightly above the ambient temperature of the house). The ideal method for doing this is to create a controller that varies the flow rate of the heat exchange medium. (A variable flow rate requires a variable-speed fan in the case of an air heat exchange or a variable-speed pump in the case of a hydronic system.) Let us develop an electronic controller in State flow that will provide all of the functionality that a variable flow rate system needs. The first step in the design is to create the Simulink model for the physical system in which the controller will operate. This part of the design process is critical because it provides the basis for capturing the speci fications by allowing the developer to see the effect of a change in the requirements or the speci fications on the way the system operates. It is the first step in the creation of an “executable speci fication” for the design. This step is so crucial that we will do it in some detail.

7.2.1 Creating a Model of the System and an Executable Specification We will assume that we are creating a device that will control the hydronic heating system that we investigated in Chapter 6. We will divide the system into four parts: •

the combustion process,

•

the heat exchanger that converts the heated gas or oil into hot water,

226 •

Chapter 7. Stateflow the pumps that deliver the heated water to the heat exchangers in the rooms,

• finally,

the dynamics of the heat exchange in the rooms (the flow of heat into the room from the heat exchangers and the loss of heat through conduction, in filtration of external air, and radiation).

We modeled a home that has two “rooms” in Chapter 6. This model had all of the attributes of a multiroom home but was easier to build. In the next chapter, we will revisit the model and, using another tool, build a model that has multiple rooms. This approach is typical of the way a design progresses. Initial modeling tends to be simple but expandable. Its main thrust is to capture the salient details that drive the design of the product and allow the speci fication to be re fined. When systems engineers talk about this phase, they often talk about “flowing” the requirements down. This phrase captures one of the most important attributes of the development of a speci fication. We start with a requirement for the system as a complete unit. We specify the very top part of the system; in this case we want an energy ef ficient heating system that provides control over each individual room in a multiroom home with heat flowing into the room on a continuous basis, not turned on and off with a thermostat. The next step is modeling the dynamics to investigate how the various parts of the system interact. This is where the simple model plays a signi ficant role. So-called trade-offs can now take place where a more complex subsystem in one area can make the design of other subsystems less dif ficult. Trade-offs of tighter speci fications in one area for less onerous speci fication in another require a very robust model. In fact, this is the only method that the designer can use to develop a complex speci fication. However, in most engineering the early work to create these models is lost because a written speci fication is all that the engineers see in the subsequent design phases. The transmission of the speci fication through a written document that omits the simulations used to develop it result in a loss of intellectual content and it signi ficantly slows down the design process. In order to understand the process we will create the new home heating controller using the model of the house that we developed in Chapter 6. Presumably, the model was the starting point for the design of the new heat controller, and the model showed the limits on what is possible using the new controller. When we described the model in Chapter 6, we noted that it contained a simple control system that started each time the manual switch changed. So let us open the model and look at what the speci fication developers want. Open the model TwoRoomHeatingSystem in the NCS library. Look at the responses when you run the system without the controller. (The model should open with this option —the switch on the right should be in the “up” position—and then throw the switch on the left by double clicking and see what happens when the furnace is turned on and off.) The responses you see were in Figure 6.5 of Chapter 6. Now, turn on the controller by throwing (double clicking) the switch on the right. You will see that the design team has devised a good controller. It seems to keep the house temperature right at the 70 deg F set point. (This value comes from the constant input at the lower left side of the Simulink diagram.) So letus look at howthey didthis. The Simulink subsystemthatcontainsthe controller is in Figure 7.6. The salient features of this subsystem are as follows: •

The design uses a PID controller, which makes sense because the heating system should provide the homeowner with a constant temperature under all circumstances. (The integral control forces the temperature to match the set point.)


227

Integral Control K Ts

Sample at 1 sec

Set Temperature

z-1 Discrete Integrator with Integral Gain Ki

2

Control_Gain

1 Heater Command

Proportional Control Sample at 1 sec.

Measured Temperatures

K (z-1) Ts z

1

Derivative Control

Figure 7.6. PID controller for the home-heating example developed in Chapter 6. •

The design uses derivative control, which allows the controller to anticipate changes in the outside temperatures and accommodate them.

•

The proportional control makes the heating system follow variations in the indoor temperature caused by disturbances such as the opening and closing of doors.

•

The control system is digital. (This is implicit in the speci fication since we are using a digital device for the control; the sample time of the controller is 1 sec, which is more than adequate but is long enough that it should not cause any computational issues.)

The simulation results from the executable speci fication are good, but we probably shouldsee what happens when theoutsidetemperature moves around. Back in Chapter 2, we developed a temperature pro file (from real data) for a typical day in Norfolk, Massachusetts. We can now excite the model with this temperature variation. Open the two-room house model, and delete the fixed outside temperature block and make it a From Workspace block. For the data type Tdata(:,1:2) in the “From Workspace” block. This causes the outdoor temperature in the model to follow the outside temperature pro file in the figure that opened when you started the two-room heating system model (the temperature pro file is the curve in red). Also, change the stop time for the simulation from Tend to Tend1. (This makes the simulation 24 hours long.) If you do not want to do this, you can open the model TwoRoomHeatingSystem24 in the NCS library. When you run the simulation, the four temperatures in the model are as shown in Figure 7.7. Notice that the individual room temperatures go from 50 deg F to the set point of 70 deg F with no overshoot in about 20 minutes. Furthermore, the heating system controller maintains a constant temperature of 70 deg F for both rooms despite the diurnal variations in the temperature. In addition, the circulating water temperature goes up and down as the outside temperature changes; this helps maintain a constant wall temperature and significantly improves the comfort level in the rooms. This simulation allows you verify that this controller works well. At this point, you are ready to build the State flow block that will implement this controller. However, there are features that the controller should have that you know but the control designer did not.

228

Chapter 7. Stateflow Temperatures (R1 R2 HE1 HE2)

130 120 110 100 Room 1 Temp. 90

Room 2 Temp.. Heat Exchanger 1 Temp.

80

Heat Exchanger 2 Temp.

70 60 50

0

1

2

3

4

5

6

Time

7

8 x 10

4

Figure 7.7. Response of the home heating system with the external temperature from Chapter 3 as input.

Heating Controller Temp . . Zone 1 Temp .. Zone 2

69 .7 68. 5

…

..Set Temp .Set Temp

70 .0 68.5

…

…

…

Time

Time Tem

Run

10 : 20 AM

1

2

3

4

5

6

7

8

9

0

Enter

Figure 7.8. A tentative design for the new home heating controller we want to develop.

You know that you want to raise and lower the set temperature from the device. You know that you need to display the set temperatures for each zone. You know that you need to display the status of the system for the user, including the current time and the current indoor temperature for each room (zone). Assume that a human factors engineer and a commercial artist working together have created the tentative interface in Figure 7.8 for the heating controller. From this, we can begin to build the requirements for the State flow logic. First, however, we need to talk about some of Stateflow’s Action Language constructs that we have not used so far.


229

7.2.2 Stateflow’s Action Language Types Stateflow hasvarioustypesof “Action Language” constructs, someof which we havealready seen, that will make the programming of the controller easy. For example, actions can be associated with both states and transitions. These actions have certain keywords associated with them (we have already encountered most of them), but there are some implicit actions that are very useful. These are as follows. •

Binary and Fixed-Point Operations: The rules for fixed-point and binary operations are quite extensive and beyond the scope of this introduction. You should carefully read the Stateflow manual before using any of these operations.

– In their order of precedence, the binary operations are a*b, a/b, a+b, a-b, a > b, a < b, a

= b, a <= b, a==b, a ∼= b, a != b, a <> b, a&b, a|b, a && b, a ||b.

>

– The unary operations are ∼a (unary minus), !a (logical not), a++ (increment a by 1), a – (decrement a by 1).

– Assignments: There are certain assumptions that State flow makes about the calculation results when you are using fixed point. Some of the operators can override these assumptions. The Stateflow manual describes these assumptions, so if you are using the assignments below, you need to use care.

– With this in mind the assignments are a = expression, a: = expression, a += expression, a -= expression, a *= expression, a /= expression, a /= expression, a |= expression, a &= expression. Most of these are the same as they would be for C code; however, in some cases they are bit operations and not operations on an entire fixed-point word. •

Typecast operations: These operations are data conversions and are of the form int8(v), int16(v), int32(v), single(v), and double(v).

•

C function calls: You can use a subset of the C Math Library functions. These are abs, acos, asin, atan, atan2, cell, cos, cosh, exp, fabs, floor, fmod, labs, ldexp, log, log10, pow, rand, sin, sinh, sqrt, tan, and tanh. State flow also uses macros to enable max and min. The user may call his or her own C code from State flow. The procedure for doing so is in the manuals.

•

MATLAB calls from State flow use the operator ml. For example, a = ml.sin(ml.x) will use MATLAB to compute sin(x), where x is in the MATLAB workspace.

•

Event broadcasting: Any action in State flow can create and broadcast (to other states) a new event. For example, the entry into a state can have “on: e1:e2,” which is read as “On the occurrence of the event e1, broadcast the event e2. ” An event broadcast may also occur during a transition using the code “e1/e2,” which is read “Make the transition when e1 occurs andbroadcast the event e2.” The eventscan also be directed to specific states in the chart using “send(event, state),” where event is the name of the event and state is the name of the state it will be used by. (No other state will be able to use the event.)

230 •

Chapter 7. Stateflow Temporal logic: This logic uses constructs like “Before,” “After,” “Every,” and “At.” It also allows

– conditionals and supports some special symbols like “t” for the absolute time in the simulation target;

– “$” for target code entries that will appear as written between the $ signs in the code but will be ignored by the State flow parser;

– ellipsis to ( …) denote continuation of a code segment to the next line; – the MATLAB, C, and C++ comment delimiters (%, /* – the MATLAB display symbol “;”. It supports the single-precision notation.

floating-point

…

*/, and //);

number symbol “F” and hexadecimal

7.2.3 The Heating Controller Layout We are ready to begin building thehome heating controller. We will start with thesimulation model for the heating controller TwoRoomHeatingSystem24, and modify it by adding the Stateflow chart that will handle all of the operations that the user will perform. The first step is to write the speci fication for the State flow chart that outlines all of the actions that the user can perform. The actions are as follows: •

The user may select any of the three switch settings: Run, Set Time, and Temp.

•

In the “Run” position, the controller operates to maintain the temperature set for each of the zones. The controller also displays the time of day.

•

In the “Time” position, the user can set the time. The user sets the time with the pushbutton keyboard (with the integers 1 through 0) and the Enter pushbutton. Pressing the Enter key saves the time in the display. The display cycles through the hour and minute display values every time a new number is pressed. If the Enter key is pressed before the complete time is entered, the Time entry command recycles back to the beginning—as if no entry has been made.

•

In the “Temp” position, the user can set the temperature for each zone. The display begins with zone 1. The desired temperature is set by using the number keys. After entering each digit, the option is to change it (pressing the new number causes the display to cycle through the two digits). By pressing the enter key, the number in the display is accepted. Temperatures are set to two digits, and the selection of enter saves the temperatures.

•

Zones are selected using the enter key immediately after the selected.

“Set Temp”

option is

In all cases, the system continues to run properly even if the switch is not in the “Run” position. If the users fail to put the switch to “Run,” the display will continue to accept new inputs. With this speci fication for the operation, we can begin building the State flow chart.


231

Two Zone Home Heating System and Controller

Update Time Display MATLAB Function

Clock tic

2

MATLAB Reset Time Function in GUI

next

Setting

SetClock

Setting

SetClock

PushButton Reset

Enter

Enter

User_Selections 2

SetTemps

MeasuredTemps

Command

Stateflow Chart ProcessCommands

Outdoor Temperature from Workspace

MATLAB Function MATLAB Update Set Function Temps in GUI

Clock Set Time

Time PushButton

Reset Pushbuttons in User_Selections

Time for Display

2

2 2

Command

Heat Cmd.

2

2 Zone PID Heating Controller

2 2 Heat Command Gain ( 0 < Heat cmd .
4

4

Matrix Multiply

A/D Conversion

2

1 4 xo s Integrator

A Matrix from m-file

4

4 -C-

[4x4]

Output (double) 2 Analog Inputs Digital Output (single)

Plot Room Temps (State s)

4

4

Zone 1 & 2 Heat Input Vectors

MATLAB Update Room Function Temps in GUI 2

K*u

Tdata(:,1:2)

B1 from m-file

Heater Commands

2

2

B1* u

4

4

Initial Temps

A

4 Temperature States (Room1 Room2 HE1 HE2)

Figure 7.9. The complete simulation of the home heating controller.

7.2.4 Adding the User Actions, the Digital Clock, and the Stateflow Chart to the Simulink Model of the Home Heating System The Simulink model needs a simulation of different user actions so we can test the various modes of the State flow chart. (When we test the chart, we want to make sure that we cover every possible action and that no states in the chart are unused.) The GUI will contain all of the buttons and displays that the design team speci fied (as in the sketch provided above). We also need a simulation of the digital clock that will trigger the controller. (The digital clock also drives the clock display.) The complete simulation is the model Stateflow_Heating_Controller in the NCS library (Figure 7.9). When you open this model, a Heating_Control_GUI display opens also (Figure 7.10). We made this GUI with GUIDE(GUI Development Environment) in MATLAB. The GUI simulates all of the user interactions, including •

the display of the time,

•

the display of the actual room temperatures,

•

the display of the desired (set) room temperatures,

•

the operating modes (through a pull-down menu): “Run,” Set Temp,” and “Set Time,”

•

a functional keyboard that allows the user to set the time and the two-room set temperatures.

When the user changes the settings, they immediately are available in the simulation and the GUI displays the result.

232


Figure 7.10. Graphical user interface (GUI) that implements the prototype design for the controller.

We will go through all of the details in the creation of this GUI in the next section, but first let us see how it works. The Simulink model with the controller and all of the code to drive the GUI are in the figures and code descriptions that follow. A Stateflow chart handles the logic for the feedback control (the command to the heating controller comes from the bottom right port in the chart) and all of the pushbutton and selection options. These include the options for running the system and changing the time and the desired temperature settings. Thesimulationcontains a digital clock that drivesthe display andtriggers theState flow chart. This is above the chart in Figure 7.9 and provides the trigger inputs to the chart. Five MATLAB blocks handle the changes to the GUI. These are as follows. •

Update Time Display: As its name implies, this code updates the GUI-displayed time every minute.

•

Reset Time in GUI: When the user changes the time by selecting “Set Time” in the pull-down menu, this code changes the time display in the GUI to the time entered by the user.

•

Reset PushButtons in “User_Selections”: The communication between the GUI and the Simulink model uses a Gain block whose value is changed by code in the GUI M-file. (We will spend some time describing this feature since we have not used it before.)


233

Setting_SW_Gain 1

2

1 Setting

1

-1

2

11_PB_Gain 1

PushButton

-1

3 Enter

Enter_Gain

Figure 7.11. Interactions between the GUI and State fl ow use changes in the gain values in three gain blocks in the subsystem called “User_Selections.”

•

Update Set Temps in GUI: The user inputs from the keyboard change the set temperatures, and this code displays them in the GUI.

•

Update Room Temps in GUI: The measured room temperatures are the output of a Simulink block that simulates the A/D conversion of the measured temperatures into 3 digits (in the form XX.X), which are displayed in the GUI with code in this block.

The digital controller and the simulation of the two rooms are directly from the TwoRoomHouse model from Chapter 6. We added an empty State flow chart to the model, and we used the Add menu selection (in Stateflow) to add the data inputs. The inputs are •

the settings (an integer from 1 to 3 from the GUI pull-down menu);

•

the 11 PushButton values that come from the interaction of the GUI and the User_Selection block (these values are −1 0 1 2 9); ,

•

,

,

, . . . ,

the result of pressing the Enter pushbutton (when the buttons are reset the value is −1, and it changes to +1 when the Enter button is pressed; these result from interactions of the GUI with the User_Selections block).

The inputs, from the block called User_Selections at the left of the model, result from changes in the gains in Gain blocks. These Gain blocks are in the User_Selections block shown in Figure 7.11. To see howthe interactions withthe GUIoccur,open the model and the User_Selections subsystem. You do not have to start the model to see the interactions. Start by selecting one of the three options in the pull down menu in the GUI. When you do, you should see the value in the “Setting_SW_Gain” (the top block) change. This change uses a feature of the interaction of Simulink with MATLAB that we have not used before. The command, in MATLAB, is set_param , and it allows you to change the values of the parameter settings in a block. For the Gain block, there are only two parameters, the Gain itself and the sample time. In order to use setparam, you need to get the handle to the graphics object (the block) that you want to change from the GUI. We do this in two steps. First, when the model opens, a callback (under Model Properties in the edit menu) is executed and opens the GUI

234


using the command hGUI = Heating_Control_GUI. This saves the graphics handle for the GUI in the variable named hGUI in MATLAB. We can then use this handle to access the GUI and to make the appropriate changes. The second step is to set the gain values in each of the blocks above. The handle for these blocks uses the names of the model, the subsystem that contains the gain blocks, and the names of the gain blocks. Thus, to set the gain in the first block, the following code is used: set_gain = ’1’; block_handle =… ’Stateflow_Heating_Controller/User_Selections/Setting_SW_Gain’; set_param(block_handle,’Gain’,set_gain);

The fully quali fied name for the block is the string variable block_handle. This is used in the set_param command to set the value of the Gain to the string variable set_gain (which is the string for the character 1). This set of commands will change the gain of the top block to 1. (In Figure 7.11 the value is 2.) Each time you change the value of the gain, the State flow Chart sees the result of multiplying the gain (from the set_param value) by the input 1 (i.e., it sees the values 1, 2, or 3). The “PostLoadFcn” callback also sets the value of the 11_PB_Gain to an initial value of −1 (so we can tell when no pushbuttons are selected); this changes to 0 1 2 or 9, depending on which button is pushed. The last gain, the “Enter_Gain” is changed to 1 whenever the Enter pushbutton is pressed; otherwise it is −1. In the Stateflow chart, the action “Reset” occurs after the pushbutton data is used. This action is an output from the chart. The other outputs are the Set Clock action and the time to which the clock is to be set, and the SetTemps that is a 2-vector of the desired temperatures created by the user ’s pushbutton actions. The Stateflow chart has two parallel states called Interactions (Figure 7.12) and Process_Buttons (Figure 7.13). The Interactions state computes the two commands for the controller (each command is the difference between the actual and set temperatures for each zone) and processes the changes in the state of the switch. The second parallel state processes the pushbuttons. This chart is rather easy to follow. The annotation at the top of the state has enough detail so you subsequently can see how this part of the chart processes the commands. Run the model and watch the chart. Then as you select different options with the pull-down menu, watch the chart as the paths change. In addition, notice how the events broadcast from this state to the parallel Process_Buttons state and note what happens in this part of the chart (shown in Figure 7.13). If you are sure that you understand the working of the Interactions state, focus your attentionon theProcess_Buttons chart. Theentry into this chart activatesthe Startstate. The broadcastof oneof thethree events “run,” “time,” or “temperature” causesthe chart to stay in this state (run), go to the state “SetTime ” (time), or the state “Temperature ” (temp). In each of the SetTime states, we wait for a pushbutton selection. When the selection occurs, the value of the variable “Pushbutton” is 0 1 9 depending on which button was pushed, and we exit the state and go to the state Hours2, where the first digit of the desired hour setting is set. We then reset the pushbuttons, and when the event “next” occurs, we exit the state and go to a Wait state. If the user does not continue after 25 of the next clock pulses (25 ,

,

, . . . ,

,

,

, . . . ,


235

At Every Clock tic after the Start --Test the Switch condition and act as follows: Switch Setting = 1 -- send event "run" to Process_Buttons Switch Setting = 2 -- send event "time" to Process_Buttons Switch Setting = 3 -- send event "temperature" to Process_Buttons Then calculate the heater command "Command" and Reset all of the pushbuttons. Interactions

2

Start 3

1

2

tic [Setting == 1]/send(run,Process_Buttons)

tic [Setting==3]/send(temperature,Process_Buttons)

tic [Setting==2]/send(time,Process_Buttons)

Output en: Command = SetTemps -MeasuredTemps; Reset = 0;

Figure 7.12. The first of the two parallel states in the state chart for the heating controller.

sec), this state is abandoned and we start all over again. If the user continues, we process the next digit in the same manner. When all of the digits are set, we are in the Min1 state where we wait for the user to press the Enter button. This sends the chart to the Output_Time state where we calculate the Time from the four digits and the variable SetClock is set to 1. This updates the display with the new time. First, we need to decide whether the user is setting the clock to am or pm. This happens in the state AMPM. The MATLAB code that resets

236


Process_Buttons

1

time 1

[Zone == 1] / Zone =2;Reset = 1;

run

Start en: SetClock = 0;

SetTime

3 2

temperature

1

2

temperature

run 2

1

1

next [Enter == 1]

2

3

time

tic [PushButton >= 0]

Temperature

Minutes1 en: h2 = PushButton; du: Reset = 1;

4

/Zone = 1;Reset=1;

next [PushButton == -1]

2

[Enter == -1]

Wait2

next [PushButton >=0]

after(25,next)

1

tic [PushButton >= 0] Minutes2 en: h1 = PushButton; du: Reset = 1;

Units1 en: m2 = PushButton; du: Reset = 1;

next [PushButton >= 0] next [PushButton == -1] after(25,next) 2

next [PushButton >=0]

after(25,next)

Wait3

1

next [PushButton == -1] Wait4

Tens1 en: m1 = PushButton;

next [PushButton >= 0] [Time >100]/Time = Time-100; 1

next [Enter == 1]/ Reset =1;

2

2

2

Hour1 en: m2 = PushButton; du: Reset = 1;

Wait1 1

next [PushButton == -1]

SetTemp

next [Enter==1]

/Time=Time+100;

[Zone == 1] /SetTemps[1] = 10*m2+m1;

2

1

after(25,next)

Hour2 en: m1 = PushButton; Reset = 1;

next [Enter == 1]

1 1

Output_Time en: Time=10*h2+h1+(10*m2+m1)/100; SetClock = 1;

AMPM en: Reset = 1;

[Zone == 2] /SetTemps[2] = 10*m2+m1;

2

after(4,next)/SetClock = 0;

Figure 7.13. The second of the two parallel states in the home heating controller state chart.

the clock looks at the time, and if it is greater than 100 it knows that the time is pm, and if not, it is am. Thus, in this state, every time the user hits the Enter button we alternately add or subtract 100. This sequentially sets am or pm in the time display. Try this! The path that sets the temperatures begins with the state Temperature. At the start of this, we wait to see if the user presses the Enter button. Every time the user presses Enter, the zone changes; it cycles through 1 and 2. The rest of this chart should be easy to follow, particularly if you run the chart while you select the temperatures. The only remaining parts of the model are the MATLAB code blocks that change the display. We will describe one of these in detail, and allow you to look at the remaining code blocks to see what each of them does. There are 5 different MATLAB functions in the model. One of these is the code that resets the pushbuttons, and we have seen this already. The other four all cause changes to be made to the GUI. Two of these make the changes that the user commands using the pushbuttons, while the other two change the time and the temperature as the simulation progresses. We look at the block that updates the time in the GUI. The code is “

”

“

“

“

”

”

”


237

function updatetime(Clock) persistent AMPM Clocksec Clockmin Clockhr changehr changemin changeampm if Clock == 0; return; end hGUI h Hrfield

= evalin(’base’,’hGUI’); = guidata(hGUI); = get(h.text15,’String’);

if strcmp(Hrfield,’00’) AMPM = ’am’; Clocksec = 0; Clockmin = 0; Clockhr = 12; changemin = 1; changehr = 1; changeampm= 1; end Clocksec = Clocksec + 1; if Clocksec == 60 Cm = get(h.text13,’String’); Clockmin = str2double(Cm); Clocksec = 0; Clockmin = Clockmin + 1; changemin = 1; if Clockmin == 60 Ch = get(h.text15,’String’); Clockhr = str2double(Ch); Clockhr = Clockhr + 1; Clockmin = 0; changehr = 1; if Clockhr == 13; Clockhr = 1; end if Clockhr == 12 && Clockmin == 0 changeampm = 1; if strcmp(AMPM,’am’) AMPM = ’pm’; else AMPM = ’am’; end end end end if changemin == 1; Clockmins = num2str(Clockmin); if Clockmin < 10; Clockmins = [’0’ Clockmins]; end set(h.text13,’String’,Clockmins) changemin = 0;

238


end if changehr == 1; Clockhrs = num2str(Clockhr); set(h.text15,’String’,Clockhrs) changehr = 0; end if changeampm == 1; set(h.text11,’String’,AMPM) changeampm = 0; end

This code shouldbe easy to follow. Theonly part that might be newto you(particularly if you have never used MATLAB to handle graphics before) is thecode that finds thegraphic object in theGUI.The GUI was created using GUIDE. GUIDE automatically creates a name (a handle) for each of the objects in the GUI. In this case, we want to change the attributes of the text in the GUI. At any time we want to use them, we can retrieve these names using the MATLAB command guidata. To do this we need to have the handle for the GUI. Remember that when we opened the GUI from the callback, MATLAB created a number called hGUI. This number exists in the MATLAB workspace, so to retrieve it for use in this function you need to bring it into the function’s workspace. The evalin command does this. Ifthis isthe first time you have seen this command, go to MATLABand type help evalin at the command line. The text in the GUI is always a MATLAB string variable, and its name is always “String. ” Try typing some of the “set” and “get” commands that appear in this code at the MATLAB command line. (The model does not have to be running to execute these commands in the GUI.) The controller (with the PID control) and the Stateflow chart (along with the A/D converters) are the only parts of this model that will be used in the actual controller. (The Simulink pieces can be C coded using Real Time Workshop, and the State flow chart can be coded with the Stateflow coder.) Figure 7.14 shows the final form of the Controller. The measured temperature goes to the display through a MATLAB block that computes the temperatures to three-digit accuracy (tens, units, and tenths) in the blocks to the left of the A/D converter block in Figure 7.9. When the design is complete, Real Time Workshop creates C-code for the Stateflow chart and the digital controller. Thus, the design will be complete when the Simulink diagram works as required. One of the neat aspects of this design approach is the fact that the Simulink model can be used to verify that the controller works as a customer would expect. You can show him. If you run this model, you will see it controlling the room temperatures both through the GUI and in the plots. The plots are very similar to those that we created in the simple models, which is further veri fication of the accuracy of the design (the design and the specifications match).


239

Integral Control 1 2

Samp Time 2

delt

2

2

2

z Delay 2

Integral Gain

Command 1

2

Ki

2

2

Control_Gain

2

2

1 Heater Command

2

Proportional Control

2 2

1 2

2

2

2

z

2

Kd/delt

2

Derivative Gain

Unit Delay Derivative Control

Figure 7.14. The final version of the PID control for the home heating controller.

7.2.5 Some Comments on Creating the GUI The GUI was created using GUIDE (the GUI Development Environment in MATLAB). The creation of the GUI is quite easy. The tool opens when you type guide at the command line in MATLAB. You insert each of the objects for the GUI by clicking and dragging the appropriate icon into the blank window that opens. When all of the objects (pushbuttons, pull-down menus, text, etc.) are in the window, simply click the green arrow at the top to create the GUI. The result will be an active window (which will do nothing) and an M- file that you can edit to create the actions for each of the objects. The M-file contains a subfunction that is executed for every action in the GUI. The code below, for example, is the code that runs when you double click pushbutton 1. % --- Executes on button press of pushbutton1. function pushbutton1_Callback(hObject, eventdata, handles) % hObject handle to pushbutton1 (see GCBO) % eventdata reserved - to be defined in a future version of MATLAB % handles structure with handles and user data (see GUIDATA) % This is the Keyboard Entry for the number 1. block_handle = … ’Stateflow_Heating_Controller/User_Selections/11_PB_Gain’; set_gain = ’1’; settimeortemp(hObject, eventdata, handles,set_gain,block_handle)

240


The first five lines come from GUIDE. The comment and the last three lines are the added code that processes the button. Note that this code changes the value of the gain in the User_Settings subsystem in the Simulink model (we have seen this already). The gain is changed to 1 here. (Other buttons change the gain to the values 0 1 2 9, as we have seen.) When the gain change occurs, the appropriate action in the Stateflow Chart occurs because the chart input sees the value of the output of the gain block. With this knowledge, you should be able to debug the GUI. Edit the GUI (called Heating_Control_GUI.m in the NCS library). Select some of the lines in the M- file that correspond to actions, and click the dash to the left of the lines. This will put a large red dot where the dash was, and the next time the M- file runs, MATLAB will pause at this point. As is always the case, once you understand how the GUI works, you can learn more by reading the help files. ,

,

. . . ,

7.3 Further Reading The paper by Harel [18] forms only a part of State flow. The most significant difference is the ability to do both the Mealy and Moore approaches (mixed syntactically into the same chart if you desire). In addition, State flow’s flow chart feature allows coding of if statements and do loops. This is a powerful and significant addition to the Harel formulation. However, the true genius of Stateflow is its ability to connect to Simulink. This allows integration of the software design and the hardware right from the start. We will return to this subject in Chapter 9. Stateflow semantics are subtle. The Stateflow User ’s Guide should be consulted as you work with the tool to ensure that you use the most effective techniques when you build a model.

Exercise 7.1 Use the modifications you made to the two-room house model in Exercise 6.2 to investigate the heating controller design. In particular, see what happens when the wind speed increases and decreases. Try creating a wind gust model and use it to drive the two-room house model. What do you think you could do to the controller to better accommodate the heating system response to wind gusts?

Chapter 8

Physical Modeling: SimPowerSystems and SimMechanics Almost every engineering discipline has itsown unique method fordeveloping a mathematical representation for a particular system. In electrical networks, the pictorial representation is the “schematic diagram” that shows how the components (resistors, capacitors, inductors, solid-state devices, switches, etc.) are connected. In mechanical systems, depending on their complexity, diagrams can be used to develop the equations of motion (for example, free body diagrams), and in more complex systems, the Lagrange and Hamiltonian variation methods are used. In hydraulic systems, models consist of devices such as hydraulic pumps and motors, storage reservoirs, accumulators, straight and angled connecting lines, valves, and hydraulic cylinders. The models of the devices and components used in various disciplines have unique nonlinear differential equation representations. Their inputs and outputs come from both the internal equations and the equations of the devices with which they interact. Before an engineer can use Simulink ’s signal flow paradigm, he must develop the differential equations he needs to model. However, because almost every discipline has a unique pictorial representation that helps in the derivation of these equations, an intermediate process leads to the ultimate Simulink model. Unique pictorial tools can capture the domain knowledge for different disciplines, and this domain knowledge has evolved so practitioners can easily develop the underlying differential equations for complex models. The domain models pictorially represent interconnections of components. Once the picture is complete, analysts use various methods for reducing them to the required equations. These methods become quite complex when the system has many parts, and even for simple problems, developing the equations from the picture can be a bookkeeping nightmare. Producing Simulink models for these disciplines seems to demand a method that will allow the user to draw the relevant pictures and then invoke a set of algorithms that will convert the picture into the Simulink model. The MathWorks has developed the concept of physical modeling tools that provide a way of drawing the domain knowledge pictures in Simulink. We will describe two tools that have been created for doing this. The first, used for electrical circuits and machinery, is SimPowerSystems. It allows the user to draw a picture of the circuit, and when the simulation starts, the tool converts the picture into the differential (and algebraic) equations 241

242

Chapter 8. Physical Modeling: SimPowerSystems and SimMechanics c 2

Step

Cap. Current

1

Switch Resistor 2 DC Voltage Source Capacitor

Multimeter Cap. Voltage

Continuous powergui

Figure 8.1. A simple electrical circuit model using Simulink and SimPowerSystems.

that the picture describes. The second tool is SimMechanics, which allows the user to draw pictures of complex mechanical linkages, and simulate them in Simulink. The approach used to create the Simulink equations for each of these modeling tools is very clever because the final simulation takes place in Simulink. Consequently, inputs and outputs to and from the electrical or mechanical components can be Simulink blocks (or blocks from other domains besides the electrical and mechanical systems). At the current time, tools exist for hydraulic systems (Physical Networks), aerospace vehicles (Aerospace Blockset), and automotive vehicle power plant and drive systems (SimDriveline). The final simulations are hybrids of the specific domains, and the combined tool for each of the engineering disciplines becomes the actual simulation in Simulink. We start our tour with SimPowerSystems.

8.1 SimPowerSystems An electrical network diagram shows how electrical components interact. Therefore, a circuit diagram is not the signal flow diagram used in Simulink. One of the easiest examples is thecircuitdiagram in Figure8.1. TheSimPowerSystems tool wasused to draw this circuit. In order to run this and other examples in this chapter, you will need to obtain a demo copy of this tool. (If you do not already own it, a demo copy is available from The MathWorks.) The model Simple_Circuit is in the NCS library. The circuit consists of a series connection of a 100 ohm resistor and a .001 farad capacitor with a 100 volt DC source that is switched on at t 1 sec. Because the circuit uses the standard symbols for the DC source, the resistor, and the capacitor, the pictorial representation should be easy to follow. The results of running this simulation are shown in Figure 8.2. As can be seen, the switch closes at one second, and because this circuit has only a single state (i.e., it is a first order differential equation), the current and voltage have exponential decay and growth exactly as we would expect. This is the first example of a physical model that we have encountered, and since a physical model is different from Simulink, let us spend some time learning how these =

8.1. SimPowerSystems

243

Current in Resistor & Capacitor

Voltage Across Capacitor

1

100

0.9

90

0.8

80

0.7

70

) s 0.6 p m A ( t 0.5 n e r r u 0.4 C

) s t l o V ( e g a t l o V

60 50 40

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Figure 8.2. The electrical circuit time response. Current in the resistor and voltage across the capacitor.

differences manifest themselves. First, a physical component, such as a resistor, capacitor, or inductor (or a linkage or joint in the SimMechanics tool), has two types of connections. The first type of connection creates the network (i.e., it connects components together). The icon for this is a small circle that does not have an arrow (since the current flowing in an element can flowin orout ofthenode, ortheforces ata joint act onallof the links connected to the joint). The second connection type is the small arrow that is the standard Simulink connection. In early releases of the tools, it was possible to connect the two types, but in the current release, the connection is not possible. If you use one of the earlier versions of the tools, be careful not to make incorrect connections. Two blocks in the diagram are unfamiliar and the way in which the switch changes is different, but even these blocks should be clear. To understand the way this tool is used, let us go through the steps that are required to build this model in Simulink. The blocks we need are in SimPowerSystems under the Simulink browser. Open this from the Simulink Library browser. When you double click on the SimPowerSystems icon in the browser, the Library browser should resemble Figure 8.3(a). The library for the circuit models consists of the following: •

+ Application Libraries,

•

Electrical Sources,

•

Elements,

•

+ Extras Library,

•

Machines,

•

Measurements,

•

Power Electronics.

244

Chapter 8. Physical Modeling: SimPowerSystems and SimMechanics

Figure 8.3. The SimPowerSystems library in the Simulink browser and the elements sublibrary.

The categories with the + sign in front have further subdirectories that we will discuss later. The blocksthat arerequiredto build this model are in theElectrical Sources, Elements, and Measurements libraries. To start, open a new Simulink window (as usual); then open the Electrical Sources browser and find the “DC Voltage Source.” Now drag a copy into the Simulink window. Next, open the Elements library (Figure 8.3(b)). This library contains all of the basic elements needed to create a circuit. One of the clever attributes of the Elements library is the way that the three basic elements (resistor, capacitor, and inductor) are a single icon. The options are to select a


245

series combination or a parallel combination of these elements. This minimizes the number of basic elements in thelibrary. The parallel combinationsare theninth andtenthicons in the library, and the series combinations are the fourteenth and fifteenth icons. The model we are building uses single elements for the resistor and capacitor, obtained from either the series or parallel combinations. The single elements in the diagram are the result of placing the series or parallel combinations into the diagram and then double clicking them to open the dialog box. The dialog that opens has a pull-down menu that allows the selection of a single R, L, or C; a combination of RL, RC, and LC; or the complete R, L, and C circuit element. The dialog also allows the initial voltage (for a capacitor) and the initial current (for an inductor) to be specified. Last, the dialog allows the user to select what data in the element are measured. (The options are the current in the branch, the voltage across the branch, or both.) If you need to plot one of the signals in the circuit, use the Multimeter block in the Measurements library. As can be seen in Figure 8.1, this block stands alone in the diagram. The outputs from this block are a subset (or all) of the measurements that are speci fied in the dialog. To select what is an output, double click on the Multimeter block, highlight the desired measurement in the left pane, and then click the double right arrow icon to place the selected items in the right pane. Buttons allow you to reorder the measurements or to remove a highlighted measurement from thelist. If you have not created the circuit elements yet, drag them into your diagram and set the parameter values to 100 ohms and 0.001 farads. In the capacitor icon, specify that you want to measure the current and voltage of the branch. The last part of the diagram is the switch. Two switch elements are available for use in the model. The first is the circuit breaker that is in the Elements library, and the second is the ideal switch that is in the Power Electronics library. For this model, either switch can work in the circuit. We have selected to use the circuit breaker. Drag a copy of the breaker into your diagram and double click the icon. The dialog that opens allows the selection of •

the on resistance for the breaker (note that this cannot be zero);

•

the initial state of the breaker (0 is off and 1 is on);

•

the specification of what is called a “Snubber” (a resistor and capacitor in series that are placed across the switch—this represents either actual components that are in the circuit or they model the “stray” capacitance and the open circuit resistance of the breaker when it is in the off state).

We will see why this snubber is important later. We now have all of the elements in the model, and we are ready to connect them. If you look carefully at the various icons, you should see that the electrical elements have an icon that is different from the Simulink icons that we have seen so far. First, they do not have an arrow to indicate a preferred flow direction (since there is none in a circuit), and second, the connection points are small circles rather than the familiar little arrow in Simulink. The exception is the circuit breaker, where the input on the top left is the arrow (the arrow goes to the letter “c” in the icon). This small distinction speci fies when icons in the diagram expect a Simulink (signal flow) input/output or are part of the topology of the electrical circuit. Usually these inputs and outputs are ports that give access to the internal workings of the element. (For example, the state of the switch is determined by whether or not the input “c” is greater than 0, and for the electrical machine elements, outputs provide Simulink values for the internal states of the machine, such as speed of the motor.)

246


Connect the circuit together to match Figure 8.1, and finally drag two Scope blocks and a demux block from Simulink into the diagram to show the current and voltage across the capacitor. When you run the simulation, the results should look like Figure 8.2. Once again, draw the diagram neatly so that it is easier to follow. Toward this end, if you highlight an icon, you can use the command Ctrl+R to rotate it.

8.1.1 How the SimPowerSystems Blockset Works: Modeling a Nonlinear Resistor Modeling a circuit as a schematic diagram represents its topology, and not the underlying differential equations. Kirchhoff ’s laws applied to the circuit (we encountered them in Chapter 6) give the differential equations. These laws state that in a connection of elements, the voltages around any closed loop sum to zero, and the currents entering and leaving any node sum to zero. In creating these sums, you must be careful to maintain the signs of the voltages and the direction of flow of the currents. (For example, assume current flow out of the node is negative and that a capacitor connected to a node has a current that flows out of the node.) With this convention, the currents flowing into the node from other elements are positive, and similarly for an inductor and resistor. One of the easiest numerical methods for reducing any circuit to a set of differential equations is to create a state-space model where the voltages across the capacitors and the currents in the inductors are the states. The topological representation of resistors connected in series or parallel lead to algebraic equations. SimPowerSystems creates the state-space model automatically using this approach, and it eliminates algebraic equations so the model is of the lowest possible order. When there are nonlinear elements in the circuit, SimPowerSystems isolates the nonlinear part of the model from the linear part and creates a small feedback loop in Simulink that models the linear pieces (using state-space models) with the nonlinear parts as an equation (or equations) that manipulate the currents and/or voltages. The currents and voltages are the values that come from the linear circuit. We will build a simple model to give some insight into howSimPowerSystems does this. Therefore, let us build a model for a nonlinear resistor. We will use this model in the next section, where we will create a simulation of an electric train on a track. The Electrical Sources library of the SimPowerSystems blockset contains two extremely useful blocks: the Controlled Voltage and Current Sources. Each of these blocks generates an output (a voltage or a current) that is equal to the value of the Simulink signal at its input. They are the way in which Simulink couples into SimPowerSystems and are the dual of the Voltage and Current Measurement blocks in the Measurements library. To create a device that is nonlinear, we need to use a measurement and a controlled source block. The equations that relate the voltage and current for resistors (Ohm ’s law), inductors, and capacitors are e

=

Ri,

e

=

L

i

=

C

di dt dv

,

dt

.

If the elements are functions of the current or voltage in the network (or if they are functions of some other variable that is indirectly a function of the voltages and currents), then the


247 Current Measurement

1

Resistance (R)

Current (i)

i

+

Desired R + iR

2

}

s +

-

-

Voltage Source ( V = to Simulink signal s)

Output Voltage V = iR

1

Figure 8.4. A Simulink and SimPowerSystems subsystem model for a nonlinear or time varying resistor.

circuit is nonlinear. To build a circuit element where the component values (R, L, or C) are functions of some other variable, we need to use the de finitions above to implement them. So let us build a model of a nonlinear (or time-varying) resistor. The nonlinear resistor model is shown in Figure 8.4. (It is in the NCS lbrary and is called Nonlinear_Resistor.) Try to create the model yourself before you look at the model as we created it. To create it you need the current measurement block (in the Measurements library of SimPowerSystems) and the Controlled Voltage Source in the Electrical Sources library. You also need two electrical connection ports (in the Elements library). The Current Measurement block has three ports: a pair for the electrical circuit labeled + and – and a port that is a Simulink signal output that is the measurement of the current passing through the block. The desired resistance multiplies the measured current, which is this Simulink signal. To create the product you need a Product block from the Simulink Math Operations library. Thesignalscreate a voltage outputthatsatis fies Ohm’s law, as indicatedin thediagram. It should be clear how one could use this approach to create an electrical circuit element in Simulink that has any desired nonlinear characteristics. The final step in the creation of this block is to add a mask (icon) to the block that makes the element look like a variable resistor. We do this by selecting first the entire model (use Ctrl+A to do this) and then the Create Subsystem option from the Edit menu. After creating the subsystem, right click on the Subsystem block and select the Mask Subsystem option from the menu. This will bring up the Mask dialog that allows the selection of the attributes of the mask. We do not need to specify any parameters for this block, so only the Icon, Initialization, and Documentation tabs are used. The masked icon we created looks like Figure 8.5. To draw this picture, we neededto define theplotpoints. This is under theInitialization tab, where the following MATLAB code is entered: yvals = [0.5 0.5 0.45 0.55 0.45 0.55 0.45 0.55 0.45 0.55 0.5 0.5]; xvals = [0 0.3 0.325 0.375 0.425 0.475 0.525 0.575 0.625 0.675 0.7 1]; Wyvals = [0.75 0.75 0.5 0.6 0.6 0.5]; Wxvals = [0.2 0.5 0.5 0.47 0.53 0.5];

248


i

R

--

+

Nonlinear Resistor

Figure 8.5. Masking the subsystem with an icon that portrays what the variable resistor does makes it easy to identify in an electrical circuit.

The first set of x and y plot coordinates draw the resistor, and the second set draws the wiper (the arrow). The actual drawing is under the Icon tab in the dialog, and it uses the code plot(xvals,yvals) plot(Wxvals,Wyvals).

The points that are plotted use the convention that the lower left corner of the icon is the point (0,0) and the upper right corner is the point (1,1). This will plot correctly if the Units (the last entry in the Icon options of the Icon dialog) are “Normalized. ” We also have specified in this dialog that the Rotation of the Icon is fixed (i.e., it will not rotate if the user clicks on the icon and types Ctrl+R). The last part of the mask dialog is the documentation. This tab in the dialog allows the user to name the block and create a description of the block that appears after double clicking on the block; it also provides the user with help on the block. To navigate through all of these, open the NLResistor model and select the block. Double click on it to see the mask, look under the Mask to see the model, and finally edit the Mask to see the various dialogs. If you choose, you can add this block to the SimPowerSystems Elements library so it will be available in the future. This is essentially how the Elements library was developed. SimPowerSystems is extendable to include any type of electrical device, and the library contains many such models. Take a few minutes to browse through the complete library. Note that theApplication libraries contains a set of wind turbine induction generator models, various induction and synchronous motor models, a set of DC drives that use oneor three-phaseAC in a bridge rectifier, a set of mechanical elements that model a mechanical shaft (with compliance and damping), and a speed reducer. Many of these models contain blocks from the Power Electronics devices library, where models for nonlinear electronic elements such as SCRs, IGBTs, diodes, and MOSFETS are located. These models use Simulink S-function code in the background to do the calculations that are required to implement the voltage and current characteristics of the devices. The outputs from, and input to, these S-functions use the same approach that we used to create the nonlinear resistor.

8.1.2 Using the Nonlinear Resistor Block Before we createa circuit with thenonlinear resistor, youshould be awareof a fewsubtleties.


249

Signal Builder

Nonlinear Res.

Nonlinear Res. Current

Switch On at @ 0.05

Switch c 1

2

Nonlinear Resistor1 100 V DC

RL Branch (1 Ohm & 0.1 Henry)

Multimeter

Continuous

RL Branch Voltage

1

powergui

Figure 8.6. Circuit diagram of a simulation with the nonlinear resistor block.

The nonlinear resistor block introduces an algebraic loop in Simulink. This loop automatically generates an error message that will appear in the MATLAB window every time the simulation starts. To get rid of this message, select the Con figuration Parameters option in the Simulation pull-down menu after you build the model. The Diagnostics tab in this dialog allows you to change the algebraic loop diagnostic from warning to none. Every timeyou build a new model with this block, you willneedto do this. Ifyou are going to build this model (and not use the version in the NCS library), then make this change. In addition, the SimPowerSystems tool prompts you to use the Ode23t solver for the simulation. If you do not do this, again an error message appears every time you run the model. The reason this solver is suggested is that, for technical reasons having to do with the order of the resulting differential equations, the electronic devices that have discontinuities (switches, diodes, thyristers, and related power electronics, etc.) all have a snubber circuit that is an RC circuit in series across the switched part of the device. The dynamics associated with the snubber makes the system stiff. Ode23t is a stiff solver that minimizes the number of calculations at every time step, making it run fast and accurately. Thus, if you are creating this model yourself, make sure that you select this solver. We can now try to create a circuit that uses the nonlinear resistor and demonstrate how it works. The circuit, called NL_Res_Circuit in the NCS library, is in Figure 8.6. Before we explore the model, we need to describe a new block in this diagram, the Signal Builder block. This block allows the user to create an input that consists of a piecewise continuous sequence of steps and ramps. The tool allows the user to specify

250


Figure 8.7. Using the signal builder block in Simulink to de fine the values for the time varying resistance.

when the discontinuities occur, the amplitude of the steps, and the slopes of the ramps. Figure 8.7 shows how the user interface looks. In this figure the rightmost discontinuity of the waveform has been selected (highlighted with a little circle), and consequently the time (0.5) and amplitude (3) of this edge are displayed at the bottom of the graph. By clicking and grabbing, any line can be moved (up and down if the line is horizontal, or left and right if it is vertical). If a point is selected it may be moved up and down. The duration of the waveform comes from the Axes menu (where the first choice is “Change Time Range”). This waveform drives the nonlinear resistor block so the resistance will vary from 1 ohm to 4 ohms over the first 0.1 sec. The resistor then changes from 4 to 5 ohms over the next 0.2 sec and then drop from 5 to 3 ohms over the next 0.2 sec, where the value willbe constant until the end (at 1 sec). The circuit we created in Figure 8.6 consists of the nonlinear resistor in series with the RL branch (the branch resistance is 1 ohm and the inductance is 0.1 Henry) with a 100 volt DC source that is applied at time t 0.05 seconds. Ignoring the transient due to the inductance, the current in the RL branch would be 28.5 amps at the time the switch changes state; it would be 20 amps at 0.1 sec and it would be 33 amps at the end. The simulation results are in Figure 8.8(a) and 8.8(b). The transient due to the inductance shows up quite clearly in the first graph of the voltage across the RL branch, and the nonlinear resistance is quite clear from the current =

8.2. Modeling an Electric Train Moving on a Rail

251 Ccurrent in the Nonlinear Resistor (Amps)

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Figure 8.8. Using SimPowerSystems: Results from the simple RL network simulation.

flowing

in the Nonlinear Resistor block. Note that this model is time varying and linear because the resistance does not depend on any of the internal states in the model. (It follows the time variation speci fied by the Signal Builder block.) However, it should be obvious to you that the input to the nonlinear resistor could be any variable in the Simulink model. Before we leave this discussion, the SimPowerSystems blockset manual has a description of a nonlinear resistor wherein Ohm ’s law is implemented using a controlled current source and a voltage measurement (instead of the voltage source and current measurement we used). It pays to look at the description of that in the user ’s manual (see [47]).

8.2 Modeling an Electric Train Moving on a Rail As a practical application of the SimPowerSystems tools, we model an electric train moving along a rail. We create the electrical circuits using SimPowerSystems, and we use Simulink to model the dynamics of the train ’s motion. The example illustrates how the combination of these two different modeling tools into one diagram makes the simulation far easier to understand. A train moves along a set of rails called traction rails since they provide the friction force that the wheels need to propel the vehicle. The train gets its power either from a third rail (the power rail, usually found at the side of the traction rails) or from an overhead wire. The power transfers from the rail or the overhead wire to the train using a connector that slides along the rail or wire. The power circuit completes through the traction rails (i.e., the return path for the current uses the traction rails as the conductor). As the train moves along the traction rails, the resistance between the power supplies and the train for both the traction and power rails changes. For example if a train is traveling at 60 mph (88 ft per sec) on a set of rails with a resistance of 10 −6 ohms per ft, then the resistance of the rail from the supply to the train is 0.001 ohms when the train is 1000 ft away from the source. If

252


Blocks in Light Blue Model the PI Motor Controller, the Motor Converter, and Converter Time Constant From1 [Accel]

PID Controller

Goto

i 1

2

Accel. Cmd.R Equivalent

Ac

[Speed]

Sum -KProduct Square

Air Drag Inverter Time Constant

2

-KGoto1

F 10 Motor Drive Controller

[Accel]

+

+

-

Vector of States

s+10

s

1

Friction

+

ForceConst

i -

Traction Motor Force

Integrator 1/TotalMass

Acceleration

Current Measurement

1 xo s

1 xo s

Velocity

Position

Velocity Int. Armature RL

Initial V0 Speed

Pos

Initial Position

Field RL +

Back EMF s

Blocks in Cyan Model the Dynamics of the Train

x Speed

[Speed]

-

From3 [Speed]

VConst

From 2 -[Accel]

Blocks in Green Model the Electrical Components of the Motor

Acceleration

From2

Figure 8.9. Simulink and SimPowerSystems model of a train and its dynamics.

the train is moving away from the source, the rail resistance will increase by 0.00088 ohms every second. The voltage and current flowing in the motor give the train’s speed, and the rail resistance determines them. This means that the interaction of the train propulsion with the track geometry is nonlinear. Thus, in order to model the interactions, the train and rail interactions need to make extensive use of the nonlinear resistor block we created. To start the modeling, we first need to create a DC or AC traction motor. We assume that the system uses DC, and we further assume that a control system keeps the supply voltage constant. We also assume that the traction motors are controlled and that we are modeling an urban transit system where each of the cars in the train has identical traction motors. We will lump all of these traction motors together into one equivalent motor. The model of the traction motor uses a masked subsystem where all of the parameters needed to describe the entire train drive motors are in the mask dialog. Open the model in Figure 8.10 from the NCS library, using the command Train Simulation. The train subsystem model is the green block in this diagram (labeled “Train”), and to view it you need to select “Look Under Mask ” from the menu that opens when you right click on the block. (The Train subsystem is in Figure 8.9.) This train model consists of a combination of Simulink and SimPowerSystems blocks. These blocks are in color, but here they are in black and white. The color makes it easier to understand the diagram. The color keys each part of the diagram to its function. The SimPowerSystems blocks are in green in the model, the control system that controls the acceleration of the train is in light blue, and the dynamics of the motion are in cyan. We describe the blocks, and their interconnections, beginning with the power electronics blocks (at the lower left of the figure). We will work our way clockwise around the diagram.

ux

1


253

We have already seen in Section 3.5 how to model a motor. The electrical and mechanical sides of a motor are LR

di R dt

= −RR iR +

V in

−

V b

and dω M dt

=

1 J w

(Km iR

−

Bw ωM ) .

These equations are not required in this model since SimPowerSystems will create them for us. The circuit for the motor consists of the input ports (from the Elements library) and the inductance and resistance of the train lighting, heating, ventilation, and air conditioning systems connected across the ports. This load is in parallel with the series circuit consisting of the armature inductance, armature resistance, and the back emf from the motor. In this model, the back emf is a voltage source whose voltage is the product, using the gain block, of “Vconst” andthe train speed. In building this motor model, we deduced therotational rate of the motor from the train ’s speed and the gearing that connects the motor to the traction wheels. The diagram was cleaned up using “GoTo” and “From” blocks, so the train speed is sent into the gain using the “GoTo” at the top right of the diagram. Another controlled voltage source is also in series with the armature circuit. This source models the DC voltage converter that controls the acceleration of the train. This controller is a simple PI control of the type that we have discussed before, so we will not go into details. The only important aspect of the control is that it essentially implements the nonlinear resistor block (although the block is not explicitly used). This approach emulates the way the actual converter on the train operates. We use the measurement of the current in the armature for both the train controller and the force that propels the train. The right side of the diagram implements this. The force is the result of multiplying (through the gain block) the motor parameter “ForceConst” and the current in the armature. From the armature current measurement on, the diagram consists of only Simulink blocks. These implement the acceleration from the sum of the forces on the train. These forces are •

the external force “F,” comes from the force of gravity acting when the train is moving on a grade;

•

the friction force (proportional to the speed);

•

the aerodynamic drag (proportional to the square of the speed);

•

the force from the motor (traction motor force).

The summing junctions before the gain block (with the gain 1/TotalMass) creates the total force on the train, and the gain creates the acceleration. As usual, we integrate the acceleration and speed to get the entire state for the train. The state vector uses the “Mux” block to send the state out of the subsystem. The other outputs from the subsystem are the motor current and, of course, the connections to the motor. This entire motor model is a masked subsystem. The train model connects to the simulation of the track in the diagram using the SimPowerSystems connection icons. The

254

Chapter 8. Physical Modeling: SimPowerSystems and SimMechanics Location Speed Acceleration Pos. Along Track (Feet)

Currents in Rails and Powering Train

Speed (mph)

States

I in West Traction Rail

R1Current

Accel. (mph/sec)

I in West Power Rail

R2Current

I Train

Train_Current

I in East Traction Rail

R3Current

I in East Power Rail

R4Current

Rail Resistances Acceleration Command [Accel_Cmd]

Signal 1

Track Slope Signal 1

Weight_of_train

R1

West Traction Rail Res.

R2

West Power Rail Res.

R3

East Traction Rail Res.

R4

East Power Rail Res.

[Slope_Accel]

Stop Cmd.

powergui Constant_drag

Stop Simulation when Train is at SubStation3

Continuous

West Power Rail SS2 to Train NL Resistance R2

STOP

Train to Substatoin Rail Resistances Calculated from Train Location

East Power Rail SS3 to Train NL Resistance R2Current

Ri

R4Current

+--

i

R

--

+

Power Rail 2

R4

Power Rail 1 Supply R1

Train Supply R2

[Accel_Cmd]

Ac Accel. Cmd.

State x [Slope_Accel]

Supply R3

Supply R4

SubStation3

SubStation4

States

F Ext. Forces +

Current i SubStation1

R1Current

Traction Rail 1

Location Pos1 = 42543 ft.

Location Pos 2 = 47543 ft.

Train_Current

--

SubStation2

i

R

--

+

R1

R3

Ri

R3Current

Traction Rail 2

+--

West Traction Rail SS2 to Train NL Resistance

East Traction Rail SS3 to Train NL Resistance

Train Motion___

>



Figure 8.10. Complete model of the train moving along a track.

final

model is in Figure 8.10. The model contains four DC voltage sources and their series source resistances. These four sources represent power substations along the track. We model the train moving between substations 2 and 3 (i.e., in the middle of the various substations). Four copies of the nonlinear resistor model the variable resistance between the train and the adjacent substations. We assume the train is moving to the right, so the nonlinear resistors to the left of the train (denoted “West” in the annotation) start at 0 ohms and rise to their maximum value when the train is at substation 3 (on the right). This model will not be correct when the train position exceeds the location of this substation (the topology of the model will change). Therefore, we calculate where the train is relative to this substation and stop the simulation when the train position reaches it. The block that does this also calculates the resistances for the nonlinear resistor blocks. Each of theblocks in this model areSimPowerSystems blocksexcept forthe GoTo and From blocks, the Scope blocks, and the Signal Generator blocks that create the slope over which the train operates and the acceleration commands that are sent to the train. Running the simulation creates plots of the four tracks resistances, the state (position, speed, and acceleration), the currents in the four nonlinear resistors, and the current that the train is drawing from the track. The Simulink subsystem that computes the resistances and generates the stop command is in the top right of the diagram. It has the Simulink blocks shown in Figure 8.11.


255

Train Position

States

>=

From2

Selector

Sub Sta.3

Pos2

West Traction Rail Res.

Location of Sub Station 3

East Traction Rail Res.

Rtraction

West Power Rail Res.

Rtraction Location of Train (relative to SubStation 2)

Gain1

2

Pos3

Pos3

Initial Train Position & Location of Sub-Station 2

Gain2

1

5 Stop Cmd.

Relational Operator

3

Gain4

Location of Train (relative to Sub Station 3)

East Power Rail Res.

Rpower

Rpower

4

Gain3

Figure 8.11. The Simulink blocks that compute the rail resistance as a function of the train location.

Resistance of Rail Segments Between Train and Power Substations

I in West Traction Rail

0.07

10000 5000

0.06

0

0

20

40

60

80 100 120 I in West Power Rail

140

160

180

200

0

20

40

60

80

120

140

160

180

200

0

20

40

60

80 100 120 I in East Traction Rail

140

160

180

200

20

40

60

80 100 120 I in East Power Rail

140

160

180

200

20

40

60

80

140

160

180

200

10000

0.05

5000 0

Traction Rail to SS3

0.04

Power Rail to SS3 Traction Rail to SS2

100 I Train

5000

Power Rail to SS2

0.03

0

0.02

1000 500

0.01

0

0

1000

0

500 0

-0.01

0

20

40

60

80

100 Time

120

140

160

180

a) Rail Resistances as a Function of Time

200

0

100 Time

120

b) Currents – Train, Traction & Power Rails

Figure 8.12. Simulation results. Rail resistances, currents in the rails, and current to the train. When the simulation opens, a MATLAB M- file called Train_Data runs (using the standard Model Properties callback). Edit this file and look at how it is constructed. You can increase the drag on the train by changing the variable Tunnels in MATLAB from one to two. You can also change any of the parameters. The simulation creates the graphs in Figure 8.12(a) and (b). The leftmost is the nonlinear resistor values as a function of time (and since the train is moving, as a function of

256


the location of the train). The rightmost picture is the currents in the nonlinear resistors and in the train. The five graphs, from the top to the bottom, are the current in the traction and power rails to the right (west) of the train, the train current, and the currents in the traction and power rails to the left (east) of the train. To get familiar with this model, open it and try changing some of the parameters. In addition, you can use the command generators to change the slope that the train is on and the acceleration command. We have only touched on a few of the many possible models that SimPowerSystems can create. The user ’s guide and the demos provided with the tool should be the next step if you want to become an expert with it.

8.3 SimMechanics: A Tool for Modeling Mechanical Linkages and Mechanical Systems Just as theSimPowerSystems tool allowsa picture of an electrical network that interactswith Simulink, the SimMechanics tool allows you to draw a picture of a complex linkage with multiple connected coordinate systems. The model you create will then run in Simulink. It also can use Simulink inputs, and the motions of any of its components are available for use in Simulink. It is also possible to combine SimPowerSystems and SimMechanics models. SimMechanics has a look and feel that are similar to SimPowerSystems. However, there are significant differences. When mechanical elements are connected to form a “machine,” there are physical attributes that need to be speci fied. First is the coordinate system that the entire model will use. The second is the coordinate system that applies to each of the bodies in the device. SimMechanics assigns a coordinate system to each body (usually at the center of gravity) and to every point on the body that can be connected to another part of the machine. (There is a large variety of possible connections: joints that allow rotations in one, two, or three directions and joints that are rigid, called “welds.”) Entering these coordinates means that there are a lot more data associated with the elements in SimMechanics than were the case for SimPowerSystems. One rule is still inviolate: no connections are permitted between a SimMechanics port (a connection with a small circle on it just like SimPowerSystems) and a Simulink signal. These connections must be made with measurement and controlled sources (as was the case for SimPowerSystems) that are in the Sensors and Actuators library in SimMechanics. The entire SimMechanics library looks like the figure at the left. The Bodies sublibrary contains four blocks. The first specifies a body that is the basic element of the mechanical system. The second block in the library is the Ground block. It specifies the coordinates of the base of the device. It is possible for the Ground to represent the ground (for example, when you want to model the trajec-

8.3. SimMechanics

257

tory of a projectile). Two entries are needed in this block: the coordinates of the origin and a check box that will create an output from the Ground block that you can use to connect the machine. Every SimMechanics model must have a Ground block . The third library block (Machine Environment) speci fies the environment for the device (i.e., the value of the gravity vector, the values used for the differential equation solver, etc.). Solvers for SimMechanics exist for solving the differential equations forward, as in Simulink, or backward. The backward solver uses the motions of each element to find what forces would create them. Kinematical solvers give the steady state solution without going through the differential equations. Last, a trimming solver does the calculations needed for the simulation to start with all of the elements at rest (so no motion will occur in the device until an external force or torque is applied). Every SimMechanics model must have an Environment block . When you double click on the Environment block, you will see that the dialog opens with the tab called “Parameters” selected. All of the above settings (and a few more) are accessible with this tab. You also will see that this is one of four tabs across the top. The constraints allow you to specify thesolver tolerances when a hard constraint on themotion of a body or joint occurs. Constraints are usually limits that the motions must not exceed. The third tab, “Linearization” allows you to select the size of the perturbation used to determine the linear model. (Chapter 2 contains a discussion of linearization and a description of perturbation.) The last tab, “Visualization, ” allows you to invoke the SimMechanics tool that draws and animates a picture of the device as the simulation progresses. Make sure that you check this box before you close the block; this will show the motion of the device during the simulation. The Constraints and Drivers library has a large variety of constraints that the bodies and joints in the simulation model can have. The idea behind a driver is to force a body or joint to have predetermined displacements or angles as a function of time; they provide the inputs when we compute the forces that cause a particular motion. The Force elements in the library are blocks that add a spring or a damper to a body to cause the linear or rotational motion to have stored energy or to lose energy due to friction. Without the blocks, the model would require a separate sensor and actuator connected through Simulink to provide the spring and damping forces. The meat of SimMechanics is the Joints library (which is comparable to the Elements library in SimPowerSystems). There are 22 different joints in this library. We will discuss a few of them, but if you want to use this tool in models that are more complex, you should be aware of and understand each of these blocks. The Sensors and Actuators blocks are the means used to allow Simulink to talk and listen to SimMechanics models. These blocks allow connections to and from Simulink To understand how SimMechanics works we replicate the development of the clock model in Chapter 1 using SimMechanics. We start by modeling the pendulum.

8.3.1 Modeling a Pendulum with SimMechanics One of the easiest models to create is a single joint and body configured to be a swinging pendulum. Remember that in Chapter 1 we created this model using Simulink alone, so this exercise is an interesting contrast of the differential equation development and the pictorial

258

Chapter 8. Physical Modeling: SimPowerSystems and SimMechanics Ground Coordinates Env

Pivot

Pendulum

BF

CS1

Damping at Pivot ap av

Scope Position Velocity

Joint Sensor

Figure 8.13. Modeling a pendulum using SimMechanics.

Figure 8.14. Coordinate system for the pendulum.

representation provided by SimMechanics. The pendulum model we will develop is in Figure 8.13. You can open the SimMechanics model (called SimMechanics_Pendulum in the NCS library), or you can launch SimMechanics and try building the model from scratch yourself. Since we will take you through all of the steps to create this model, it is probably better to build the model yourself. Because every SimMechanics model needs an Environment and Ground block, open an empty Simulink window and drag one of each of these blocks from the SimMechanics library into it. Then connect the Environment and Ground blocks together. Double click on the Environment and Ground blocks to view their dialogs. The Environment block opens m with the default gravity vector set to −9.81 sec 2 in the negative y direction. The default value for the Ground block coordinates (de fining the coordinate system for the entire mechanical device) is [0 0 0]. These defaults mean that the coordinates for the environment are as shown in Figure 8.14. We add the pivot for the pendulum and the pendulum body using this World coordinate system, as shown in the figure.

8.3. SimMechanics

259

The pivot for the pendulum is a revolute block from the Joints library. The revolute block allows rotations around a single axis (a single degree of freedom in the Base). The axis of revolution can be referenced to the World, to the body that the joint is connected to (called the Base), or to the body that rotates relative to the base (called the Follower). All joints connect to two bodies. (In this case, the Base or B port connects to the Ground and the pendulum; the second body connects to the follower or F port.) The coordinate system for the Pivot is the World coordinate system. (Open the Revolute dialog for the Pivot block and look at the Axes tab, where the rotation axis is set to the z-axis and the coordinates, called “Reference CS” in the dialog, are the World coordinates.) A joint can have sensors that measure the rotational angle, velocity, etc. and can have forces or torques applied. To allow this, the Joint blocks have a dialog that allows selection of a different number of sensor/actuator ports. In our model, we need a port for sensing the motion and a port to apply the damping force, so the number of ports is set to two. To make the measurements we need a Body Sensor from the Sensors and Actuators library, and to create the damping force, we need a Joint and Spring block from the Force Elements library. The joint sensor dialog for the model has the check boxes selected that provide measurements of the angle and angular velocity of the joint, using degrees as the angle units. The Joint Spring and Damper block is con figured to only provide damping. (The spring constant is set to 0, and the damping coef ficient is set to 0.00005.) This block creates equal and opposite torques for each element connected through the joint. The last block needed to describe the physical model is the Body block that models the pendulum. This block connects to the joint follower port (the F port). So grab a Body block and drag it into the model. Any number of bodies may be in the model, and SimMechanics automatically generates the equations of motion. The tool even takes care of computing the rotations of the bodies relative to each other and the base, using one of the methods we developed in Chapter 3 (Euler, angle-axis, or quaternion). For now we need only to enter the body mass properties (mass and inertia) and the location of the center of gravity and the attachment point to the pivot. Open the dialog on the body and set the mass to 1 kg and the inertia matrix to the diagonal matrix with 0.001 kg*m ˆ2 along the diagonal. The center-of-gravity (CG) coordinates in the Ground Coordinates block are set to the origin, and the CG coordinates in the Pendulumblock are set to[0 .05 −0.2485 0]. This corresponds to a slight offset in x (5 cm) for the pendulum. (This will cause the simulation to start with the pendulum moving because of the gravity force.) The value of the CG displacement of 0.2485 m along the negative y axis will make the pendulum move with a period of 1 √ sec. (Remember that the frequency of the pendulum is g/ l , which can be solved for the pendulum’s length to give 0.2485 m.) Connect all of the blocks as shown in the diagram above, making sure that you understand why these connections are all required. Also, make sure that all of the dialogs that we specified along the way are in place and that you have checked the animation box in the Visualization Tab of the Environment block. Finally, connect a scope block (with two inputs) to the joint sensor. We have set the simulation time to 100 sec, and using the Configuration Parameters dialog, we have set the maximum step size for the solver to 0.1 sec. You can start the model now. When thesimulationis complete, thependulumsimulationwill start andthe animation will appear. You should be able to watch the simulation evolve both through the animation and the graphs displayed in the Scope. This ability will be there no matter how complex the

260

Chapter 8. Physical Modeling: SimPowerSystems and SimMechanics Position 0

-5 -10 -15

-20

-25

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80

90

100

60

70

80

90

100

Velocity 100

50

0

-50

-100 0

10

20

30

40

50 Time

Figure 8.15. Simulating the pendulum motion using SimMechanics and its viewer.

mechanical system you have modeled. (A snap shot of the animation and the Scope plots is in Figure 8.15.) Remember that this simulation, unlike the one we developed in Chapter 1, did not require us to develop the equations of motion. Furthermore, it was created quite rapidly. (Of course if this model was the first time you used SimMechanics, there was a learning curve you had to go through, but you can easily appreciate how the tool could save a large amount of time when you are modeling complex linkages, machines, or dynamics.)

8.3.2 Modeling the Clock: Simulink and SimMechanics Together The simple pendulum model can redo the simulation of the clock from Chapter 1. The approachused in the SimPowerSystems tool, wherea Simulinkvalue convertsinto a physical value (a voltage or current) is also the physical modeling approach. For SimMechanics, the physical measurements can be quite numerous, as we have already seen. The SimMechanics actuatorscan apply a torque, a force, or a motionbased on anycombinationof measurements and Simulink calculation that we want. The Damping force block for the pendulum is an example of this; it is really a masked block with both an actuator and a sensor built in. The calculations that provide the desired damping and spring force are in a Simulink block. To create the clock model, we will actually build the damping, so we will not need the Damping block. We also will use the Escapement force block that was used in Chapter 1, and finally we will build a Signal using the “signal builder ” to give the pendulum an initial displacement. The initial displacement will allow us to place the center of gravity of the pendulum at the pivot location in both x and z. (Since we will be “kicking” the clock pendulum to start it, we will not need the offset of 0.05 speci fied for the pendulum CG to get the simulation started.) It should be easy for you to add the blocks needed to create the clock simulation, so try to do so. The model in Figure 8.16 is SimMechanicsClock in the NCS library. The joint sensor measures both the angle and the angular velocity of the pendulum; these are inputs to the escapement block. The angle and angular velocity determine the forces applied by the escapement and the angular velocity determines the damping force. The signal generator is a pulse of amplitude 1 with duration 0.1 sec applied at the start of the simulation.

8.3. SimMechanics

261

Ground Coordinates

Pivot

Env

B

Pendulum F

CS1

Joint Actuator Pendulum Angle

Ang. Position

Escapement Model Pendulum AngleForce

1.6

ap Escapment acceleration

av Joint Sensor

Ang. Vel.

0.0005 Damping Coefficient Signal 1

Initial Pendulum "Kick"

Figure 8.16. Adding Simulink blocks to a SimMechanics model to simulate the clock as in Chapter 1. Pendulum Position 40 ) s 20 e e r g e 0 D (

e l g n -20 A

-40 0

5

10

15

20

25

30

Time

Figure 8.17. Time history of the pendulum motion from the clock simulation.

Most of the parameters in the model remain the same, except that in Chapter 1 we used radians as the units, and here we use degrees. Thus, the escapement values Penlow and Penhigh are 5.6 and 8.6, respectively, and the gain for the escapement force increases to 1.6 (in the Gain block). The signs of the damping and the escapement forces applied to the joint are negative since they reference the follower. The result of the simulation from this model is in Figure 8.17. “

“

”

”

262


8.4 More Complex Models in SimMechanics and SimPowerSystems There are many examples of complex motions in the demos that ship with SimMechanics, and it is very useful to look at them. Before we leave this tool, though, there is one example thatis worthlookingat. In Chapter 7, we talked aboutmodeling partial differentialequations, and we said that one could model vibration using the same approach as we used to model the heat flow in the house. We also said that we could build a more complex home heating system house model. Let us illustrate both of these now. The first model is a modi fication of the vibrating string model demo that is included with SimMechanics ([2] was the source for the model and the data). A modi fied version of this model is the model SimMechanics_Vibration in the NCS library. Open this model andrun thesimulation. The model uses 20 finite linkages to model a steel string (like a guitar string) that is 1 m long. Each of the elements is a two degree of freedom representation of the string ’s tension and rotation. Each of the elements is a small mass that can rotate and translate (while transferring the forces that result from element to element). Browse through the model and locate each of the elements. Look at the dialogs that each element has to see what the data are, and then look at their values using MATLAB. The data is stored in a MATLAB structure called “fBeamElement” that has the components material: ’Mild Steel’ density: 7800 youngsM: 210000000 diameter: 5.0000e-004 length: 0.0500 cgLeng: 0.0250 beamDir: [1 0 0] Inertia: [3x3 double]: matDamping: 5.0000e-004

The inertia matrix for each of the elements is 1.0e-007 * 0.0005 0 0

0 0.2500 0

0 0 0.2500

The model (Figures 8.18 and 8.19) looks complicated, but it is simply the same block replicated 20 times. The 20 inputs allow you to pluck the string at any of the 20 lumped locations(every 5 cm)along thestring. It is interesting to do this while viewing thevibration to see how many of the various harmonics are excited as you pluck the string near the edges versus near the center. Several examples are in the output of SimMechanics shown in Figure 8.20. It is easy to modify this model to make it have more elements. We explore more attributes of this model in Exercise 8.3. The last model we will build in this chapter shows a complex electrical circuit. We have modeled a four-room house heat flow ( FourRoomHouse in the NCS library, shown in

8.4. More Complex Models in SimMechanics and SimPowerSystems

Env

Left End

Machine Environment Signal 1

Yf orce

Signal Builder -- String Pluck

263

Pluck

Pluck1

Ground1

Pluck2 Pluck3

Connect to any of the 20 Pluck Ports on the String Model to see what happens to the string motion.

StringPlucker

Pluck4 Pluck5 Pluck6 Pluck7

This is a model of a 1 meter long string made of "mild steel" that is plucked and then vibrates in a uniform gravitationla field. There are 20 elements all together, and e ach element is 5 cm long.

Pluck8 Pluck9

The elements are identical ( they consist of a spring constant and damping in each of the 2 degrees of freedom -- 1 rotation that transfers torques from element to element and 1 translation to transmit tension from element to element).

Pluck10

B

Right End

F

Pluck11 Pluck12

Each of the 5 cm segments c an be plucked by changing the connection point for the "External Pluck". The Pluck ports on the model are numbered from the left to right.

Pluck13

The mass proopserites for the elements are s eparately defined in the "Mask" for each of the 20 elements.

Pluck15

Ground2

Connection to Ground (Joint)

Pluck14

Pluck16

See Klaus-Jurgen Bathe, Finite Element Procedures, Prentice Hall Inc, 1996 for details on this model.

Pluck17 Pluck18

This model is a modification of the Vibrating String model that ships with SimMechanics.

Pluck19 Pluck20

String Model

Figure 8.18. SimMechanics model Sim_Mechanics_Vibration. LeftEnd 1

Base Tip

3 Pluck1 4

Base

Pluck

WireElement1

Tip

Base

Pluck

Tip

WireElement2

Pluck2 5

Base

Pluck

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Base

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WireElement5

Pluck4 7

Base

Pluck

WireElement6

Pluck3 6

Tip

Base

Pluck

WireElement7

Tip

Base

Pluck

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Tip

Base

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Base

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Tip Pluck

WireElement20

Pluck8 11 Pluck9 12 Pluck10 13 Pluck11 14 Pluck12 15 Pluck13 16 Pluck14 17 Pluck15 18 Pluck16 19 Pluck17 20 Pluck18 21 Pluck19 22 Pluck20

Figure 8.19. String model (the subsystem in the model above). It consists of 20 identical spring mass revolute and prismatic joint elements. “

”

Figure 8.20. String simulation. Plucked 15 cm from the left, at the center, and 15 cm from the right.

2 RightEnd

264

Chapter 8. Physical Modeling: SimPowerSystems and SimMechanics Heat Exchanger Thermal Loss1

HE Loss to Room 1

Heat Exchanger Room 1

Heat Exchanger Thermal Loss2

HE Loss to Room 2

Wall Thermal Loss Room 1

Thermal C Room 1


Thermal C Room 2


Wall to Outside Room 1

Wall C Room 1


Wall C Room 2

s + -

Heat Source

Piping/Duct Thermal Loss


HE Loss to Room 3





Thermal C Room 3

HE Loss to Room 4


Thermal C Room 4

Outside Temperature Wall to Outside Room 3

Wall C Room 3


Wall C Room 4

Figure 8.21. Using SimPowerSystems to build a model for a four-room house.

Figure 8.21). In addition to the four rooms, instead of the two that were created in Chapter 6, we have also modeled the thermal capacity of the wall. By modeling the wall capacity, we can use a measurement of the wall temperature as a way to account for the outside temperatures. The model is not complete, and in the exercises at the end of the chapter, we ask that you finish the model (Exercise 8.4). The model uses the SimPowerSystems tool. The rooms are all modeled using an identical circuit. This allows you to draw one room and then copy the circuit elements to make the next room, and so on. Look carefully at the way Simulink creates the circuit element names. Because Simulink does not allow identical names, copying an element causes Simulink to change the number at the end of the name to the next sequential number. For this reason, copying the circuit elements automatically changes the name to the next room. (Try making a fifth room to see how this works.)


265

We pointed out when we built the two-room model in Chapter 6 that the equivalent R-values for the various losses (in particular for the ceilings, floors, walls, and windows include an approximation to the heat loss from convection (i.e., from the flow of air in and out of the various spaces). Unfortunately, this approximation does not account for the effect of increased convection when the outside winds increase in speed. We can include this effect using the nonlinear resistor model we created in Chapter 6. It is simply a matter of changing the resistor at the right of each room to one of the nonlinear resistor blocks. Note that in this model, the resistor will be time varying —a function of the wind speed —and therefore the block is linear. We leave this as part of Exercise 8.4. The models created in this chapter show how Simulink includes many physical modeling concepts. We have not shown all of the uses of these tools, nor have we shown how the SimMechanics tool can use a CAD tool to build the SimMechanics model automatically from the CAD drawings. If you need to pursue this further, see [46]. Newer physical modeling tools are useful in the automotive industry and for modeling hydraulic systems. In addition, The MathWorks is developing new tools for other physical systems. They collectively will allow engineers rapidly to create simulation models. We are almost at the end of our tour of mathematical modeling with Simulink. The next chapter tries to show, with a relatively simple example, how to use Simulink in the overall designand developmentprocess. The keyto doing so is thedevelopmentof a process that makes it possible for all of the people working on the design of a system to interact using the same environment. By wrapping a process around Simulink and its various tools, a company can easily create new designs. One of the features is that the staff can maintain a clear path to the corporate legacy (intellectual property) so previous designs can easily and seamlessly be incorporated into their newer designs. The models also provide a simple way to document designs and, last but not least, have a way of generating code that will satisfy the design requirements. The next chapter takes us into the realm of product development and all of the issues associated with ensuring that the process creates a reliable and accurate device at the lowest possible cost.

8.5 Further Reading The nonlinear resistor model and the model of the train are from [15]. I developed these to investigate a method for adding high temperature superconducting wire for power at the center of the Bay Area Rapid Transit (BART) Trans-Bay Tube between Oakland and San Francisco. This report used a SimPowerSystems model for multiple trains operating over various segments of BART. It allowed a preliminary design of the power distribution system and allowed experiments that investigated alternative approaches. Once again, we are becoming aware of the importance of energy conservation in the world. The model of the four-room home is useful to answer personal questions about the cost effectiveness of adding insulation to your home (particularly when there are tax advantages for adding insulation). It is fun to create a model of you own home and play the “What if?” game with insulation and other energy ef ficiency improvements. In [8], The MathWorks’s technical staff describes a method forcreating a SimMechanics model that computes the motion of a flexible beam. The paper describing the method

266


and the models is available from The MathWorks’s web site. In the paper that accompanies the model, the authors show an alternative method that uses Simulink and the LTI modeling approach from the Control Systems Toolbox (see Chapter 2). The model they use is the state-space model. The steps that one follows to use this model provide insight into the power of Simulink ’s automatic vectorization, so it is a useful exercise. The last part of the reference develops the equations for a multi-degree-of-freedom model in some generalized coordinate system q. When a finite element modeling approach is used, the coordinates that result are usually a mix of translations and rotations for each element—anywhere from all three translations and three rotations at an element to perhaps one or two at each element. The masses (or inertias) of the elements, the damping forces, and the spring force between elements are represented by generalized mass matrices M, damping matrices C, and stiffness matrices K , so the dynamics of all of the elements are encapsulated by the differential equations

M

d 2 q dt 2

+

C

d q

+

dt

Kq

=

F.

Any finite element modeling tool will create this general form for a vibrating system. The finite element codes also will reduce this model to the form d 2 ζ

+

dt 2

2

d ζ dt

+

2 ζ = T−1 F.

This is done using the orthogonal transformation T to simultaneously diagonalize the symmetric matrices M, C, and K . The steps for this are q

=

Tζ .

Then M

d 2 Tζ dt 2

+

C

d Tζ dt

+

KTζ = F.

Multiplying this equation by the inverse of T (T−1 ) gives d 2 ζ

T−1 MT dt 2

+

d 2 ζ + dt 2

d ζ

T−1 CT dt or d ζ

2 dt

+

+

T−1 KTζ = T−1 F,

2 ζ = T−1 F.

This last step comes from the fact that the matrix T diagonalizes all of the matrices in the first equation. The algorithm that does this is the qz algorithm; it is a built-in routine in MATLAB (see the MATLAB help). Both the SimPowerSystems and the SimMechanics user’s guides [46], [47] are excellent references for creating models and for understanding the demonstrations that are part of the products. Consider them an extension of the material in this chapter.

Exercises

267

Exercises 8.1 Modify the model in Section 8.1.2 to make it nonlinear. For example, measure the voltage across the RL branch and add a nonlinear resistor that has the following value: R(t)

where R0

=

1 ohm, R

=

=

R0

+

R V RL ,

0.1 ohm, and V RL is the voltage across the RL branch.

8.2 Build a simple Simulink model of an electric car using the DC motor model in the train simulation. Set the mass of the car to 2000 pounds. Investigate the amount of power and energy used to accelerate the car from 0 to 60 mph in 10 sec. (Remember that power is the product of the current and voltage used by the motor and that the energy used is the integral of the power.) Use the air drag numbers in the train model. Try different rates of acceleration in the model. What can you conclude about the energy used as a function of the rate of acceleration? 8.3 Add more elements to the string model. You need to allocate the mass element data in the fBeamElement structure. 8.4 Add a fifth room to the model FourRoomHouse. Create data for all of the components using the values from Chapter 6. Add the time varying resistor (from Section 8.1.2) to model the convective losses from the wind. Make runs to answer some of the hypothetical questions in the Further Reading section.

Chapter 9

Putting Everything Together: Using Simulink in a System Design Process We have reached the point in our travels through the world of Simulink where we can talk about how Simulink can help in the design of complex systems. There are many ways to partition a design process. Furthermore, every person that works in research and development has a favorite approach to design. The one common aspect of all of these approaches is the use of a methodical, documented, and traceable procedure. A design process is methodical when it has a clear specification, the design has connections to the specification, the design has clear documentation, and the design validation is performed via a thorough test program. The obvious aspects of this statement are that the design meets the objectives and performs as required. It is ef ficient, reliable, and buildable within some budgeted cost. However, there are some nonobvious bene fits: the design has legacy (it is reusable), the design exploits legacy from previous intellectual property, and the ultimate audience for the system can exercise the simulation model so that their inputs have a beneficial influence on the design. (In fact, many times this feature can sell a design.) The entire design process needs careful documentation. Doing otherwise makes it impossible to revisit decisions and convince engineers, designers, and technicians working with the designs that the decisions made during the creation are valid (at least not without a large and often futile replication of the earlier work). A process that provides traceability means that the parameters in the design trace back to one or more of the specifications for the design. Traceability also provides an understanding of a how a top-level requirement flows into specifications for various subsystems. Subsystem specifications often come from a speci fication imposed by top-level system design requirements. Once the design is finished, it requires a method for its validation. This is particularly true when the design requires embedding computer code in the system. The design process also should not require the creation of a new speci fication just for this embedded code. In fact, the process should be seamless so the code generated as the speci fication evolves is always tested quality code. Simulink allows you to create a design process that will perform all of these functions in a seamless way. In fact, the Simulink model can play the role of the specification. This is an “Executable Specification.” To make Simulink the tool for system design, the R&D 269

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Chapter 9. Putting Everything Together

parts of a company, the manufacturing arm of the company, and the computer engineers need to use the same Simulink model throughout the process. Obviously, every company approaches system design with a bias that comes from their previous experience. For this reason, a process using Simulink must come from the “Systems Engineering” part of the company. Whatever way the process uses, the broad outlines of the steps are •

speci fication development and capture,

•

modeling the system to incorporate the speci fications,

•

design of the system components that meet the speci fications,

•

veri fication, and validation of the design ,

•

creation of embedded code.

In the 1960s I was part of the team that developed the original digital autopilot for the lunar module. The design of this system was dif ficult because there was no way that the control system was testable on earth. This meant that simulation was the only way to create, optimize, and verify the design. In the process of creating the control system, the first step was to understand the requirements and translate these into a mathematical model for the design. The modeling started with a digital simulation. We used it to develop candidate designs that in turn were optimized using simulation. In the 1960s this work used a mainframe computer that allowed, at best, two computer runs per day. Simulink and the PC change every aspect of the design approach we used then. Therefore, the exposition that follows should be entitled “How we would do it today. ” I wrote an article that appeared in the summer 1999 issue of The MathWorks ’s magazine News and Notes that describes the design process and the lunar module reaction-jet attitude-control system design [16]. We will use the lunar module digital autopilot design as a case study to illustrate how Simulink provides a natural tool for the design process steps above.

9.1 Specifications Development and Capture In the actual lunar module design, there were modes for the following. •

Coasting flight: the initial mode for preparation of the lunar descent, where the lunar module was powered up and checked, the inertial measurement unit was calibrated (using an optical telescope to track stars), the computer was veri fied, etc.

•

Descent: using the descent engine gimbals for control during the landing.

•

Ascent: using reaction jets but with the control actively combating the rotation caused by the engine misalignment.

•

Combined lunar module and command-service module control: where the lunar module controlled bothvehicles, the modeused for returning the astronauts after theApollo 13 explosion.

9.1. Specifications Development and Capture

271

Figure 9.1. The start of the speci fication capture process. Gather existing designs and simulations and create empty subsystems for pieces that need to be developed.

The requirements for each of these modes were traceable to the overall mission requirements. (For example, the amount of fuel needed to be carried on the lunar module was a function of how long the descent would take, which is relatively easy to calculate.) Another example is how much time a search for an alternate landing could use. This site selection was required if there was a problem with the preselected site. A gimbal on the descent engine might allow the thrust to go through the center of gravity, or we could have fixed the engine at a particular orientation (with a fuel penalty for the thrust needed to counteract the torque from the engine). The requirements optimization needed to exercise alternative designs. Calculations provided some of the answers but were time consuming. This severely limited the number of alternatives that we investigated. An easily modi fied fast simulation of multiple systems would clearly have made it possible to do many more trade-offs of options and theoretically would have resulted in a better design. To keep the model of the lunar module understandable, we will limit the discussion of the digital autopilot to the rotational motion using the reaction jets only (i.e., we will work only with the first phase of the mission, coasting flight).

9.1.1 Modeling and Analysis: Converting the Specifications into an “Executable Specification” We already have most of the modeling pieces that we need for the lunar module simulation from the discussion on the rotation dynamics of a spacecraft in Chapter 3. The ability to use a previous Simulink block is one of the major features of Simulink, allowing reuse of previous work (and the possible exploitation of corporate intellectual property). Working with the results of Chapter 3, the building of the simulation part of the model is straightforward. The model is in Figure 9.1. We placed an empty subsystem block in

272


this model to denote the control system. This allows the development of the simulation model to proceed independently of the development of the control system design. When you are working with several teams each tasked to create a different aspect of the system, this approach (sometimes called “top-down design”) enables teams to work independently but in concert. Also, note that all of the blocks in this model come from the discussion in Chapter 3, so no new dynamics are required. This is a good example of the independent design approach that we have been discussing. In this design, a team separate from the modeling team might be creating the control system. The modeling team can still proceed with the creation of the simulation using the empty subsystem. (Open the model called LM_Control_System in the NCS library and look at how this subsystem was created.) The lunar module had an inertial platform that provided Euler angle measurements for the control system. The Euler angles were a direct measurement of the angular orientation of the vehicle without the transformations needed to give the exact orientation in inertial space. Astronaut inputs were Euler angle commands that incrementally caused the vehicle to be oriented properly. The model shown here uses the quaternion representation from Chapter 3, and the measured angles for the autopilot are the direction cosines (as given by the quaternion to direction-cosine-matrix block we developed in Section 3.4.4). The computations needed to convert the Euler angles into a direction cosine matrix were well beyond the capability of the lunar module guidancecomputerat the time, so we approximated the angles using Euler angles as if they were the actual rotational angles. This is not a bad approximation since the control system was operating to maintain the orientation at the desired value every 0.1 sec. Before we leave the discussion on modeling, it might be of interest to discuss how we built the first simulation model in the 1960s. An engineer, working with FORTRAN, developed this design model (for Grumman, the lunar module designer). He spent over six months developing a model that hadonly a single axis of rotation. Thereasonsthe model was so simple had to do with the computer resources available (we used an IBM 7090 computer that allowed each user, less than 100 Kbytes) and the available time (the simulation was created using punch cards —one card for each FORTRAN instruction—and each attempt at compiling the model took one day). Simulink and the PC have dramatically changed the way we approach such designs today. So let us now change hats and make believe we are the design team that is developing the control system for the lunar module and embark on the second of the system design tasks, modeling the system to incorporate the speci fications.

9.2 Modeling the System to Incorporate the Specifications: Lunar Module Rotation Using Time Optimal Control Control of a satellite using reaction jets is possible with two different strategies for the firing of thejets. The firstis to usea modulator to emulate a linearcontrolsystem where theaverage thrust is proportional to some error signal. A modulator pulsing the jets on and off over time creates this average torque so it can be made proportional to the error signal. For several reasons this approach is not good: first, it causes the mechanical device that is pulsing the jets to wear faster than need be, and second, the typical reaction jet has an ef ficiency curve

9.2. Modeling the System to Incorporate the Specifications

273

that causes less fuel to be used if the jets are turned on for a longer time. In the early design of the lunar module, jets were turned on and off with a modulator. This analog device was a pulse ratio modulator (a combination of pulse-width and pulse-frequency modulation). For several reasons this system did not have a backup, which worried the system designers (see [16]). NASA proposed that the digital computer used for guidance and navigation might provide a backup control system. In a competition described in more detail in [16], MIT Instrumentation Labs proposed a control system that used a time optimal controller instead of a modulator. This approach greatly simpli fied theneeded computationsto reorient the vehicle, a simpli fication that made it practical to implement the autopilot in the existing computer. This is an excellent example of how independent design teams working on the same system model can optimize the design.

9.2.1 From Specification to Control Algorithm The lunar module digital autopilot design was based on a very simple conceptual approach: look at the error between the actual (measured) angular orientation and the desired orientation, and then fire the jets with either a positive or a negative acceleration (as appropriate) in such a way that the error is reduced to zero as fast as possible. Working out the details of how to do so is not dif ficult. Once the control strategy was worked out, the “control law” (i.e., the exact strategy for what to do in every situation) was modi fied to limit the amount of fuel that was used. To getan idea of what a minimum time control is like, consider thefollowing simpli fied version of the lunar module control: You are in a drag road race where you must start when a light goes green and then travel exactly 1500 ft at which point you must be stopped. You will be penalized if you either fall short or exceed the 1500-ft distance.

Before you read the following, you might ask, How do you drive the car to win this race? After a little thought you should be convinced that the way to win is to accelerate for as long as you can (to the exact point where if you did not brake you would skid past the stop line) and then stomp on the brakes and decelerate until the car stops exactly at the finish line. The car that accelerates the fastest and the driver that best finds the exact point to make the switch between accelerating and decelerating the car will determine the winner of the race. The form of this problem makes the answer easy to calculate. To perform the calculations, we will assume the following: •

the car instantaneously achieves its maximum acceleration;

•

the driver applies the brakes at exactly the correct time;

•

there is no change in the deceleration because of tire slippage, etc.

Let us put some numbers on the table. Assume that the car can accelerate at 10 feet per second per second (10 ft/sec 2 ) and the braking force allows a constant deceleration of 5 ft/sec2 . It the driver accelerates for t 1 sec, the speed will then be 10 t 1 ft/sec. During this

274

Chapter 9. Putting Everything Together t 2

acceleration, the car will travel 10 21 ft. The deceleration will now commence and the car must slow down from 10 t 1 ft/sec to zero. This will require a time of 2 t 1 sec (since the car decelerates half as fast as it accelerates). The distance traveled during the deceleration is 2 therefore 5 (2t 21 ) = 10t 12 . The total distance traveled is 15 t 12 , and this must equal 1500 ft. Thus, we have the equation that provides the critical time: 15t 12 = 1500; → t 12 = 100; → t 1 = 10. Clearly, by working through this process for any acceleration/deceleration rates, the strategy for any car can be determined. To control the lunar module, the same rules apply, except that both the acceleration and deceleration require firing the reaction jets (which produce the same force) so the acceleration and deceleration are the same. In addition, the location of the stop is different every time we apply the control, so we need a generic control that will work for any starting and stopping conditions (including the possibility that the lunar module is rotating at the start). We will develop this control system (which was only one of several on board the lunar module) and insert it into the TO BE DESIGNED block in the rotational dynamics model developed previously by the system design team. If the jet forces are a couple around the center of gravity, the linear model for a single axis rotation is simply d 2 θ e Fl u = αu, = dt 2 I where θ e is the control system error (i.e., θ measured − θ d esired ), F is the force from the jets, l is the distance from the jets to the center of gravity, I is the inertia of the vehicle around the rotational axis.

We will let the acceleration term Fl/I be α , and u is either ±1 ( firing jets for positive or negative rotations) or 0 (no rotation). The minimum time control sets the control u to either +1 or −1 until we reach the critical point where (as was the case for the car above) the sign of the control changes. This critical point is along the unique trajectory that causes the angular position error θ e and the ˙e to be exactly zero at the same time. The theory of optimal control states angular rate error θ ˙e ) can be achieved with at most one switch in the sign that the goal (getting to zero θ e and θ of the acceleration. The plausibility of this should be clear from the discussion of the auto race above. (As an exercise, create a logical argument why this must be so.) To start the development of the control system, let us assume we are at the critical point described above (the point where we can force the position and rate to be zero at exactly the same time using a single jet firing). We use a graphical approach to visualize the ˙e when u is ±1. To develop “trajectory” that will be followed. The graph is a plot of θ e vs. θ this graph, let us solve the simple second order differential equation for the rotation above. The solution is simply ˙e (t) = αut + θ ˙e (t 0 ), θ 2 t ˙e (t 0 )t + θ e (t 0 ), θ e = αu + θ 2

9.2. Modeling the System to Incorporate the Specifications

275

where t 0 is the initial time and u is constant (either +1 or −1) over the duration of the solution. To create a plot of θ e vs. θ ˙e we need to eliminate time (t ) from these equations. Therefore, solving the first of the equations ˙ −θ ˙e (t 0 ) gives t = θ e αu . Substituting this time into the second equation gives the equation for a parabola: θ e (t) − ˙ (t) 2 −θ ˙ (t )2 θ θ e (t 0 ) = e 2αue 0 . The unique parabolas that go through the origin from an arbitrary initial condition are obtained by setting ˙e (t) = 0 θ e (t) = 0 and θ in this equation (i.e., t is the time at which the trajectory goes through 0). A plot of these parabolas in the θ e , θ ˙e plane (called the phase plane or, in higher dimensions, phase space) is in the figure above. For each value of u, one and only one parabola goes through the origin (i.e., [ θ e θ ˙e ] = [ 0 0] ). These parabolas are the locus of all of the critical points at which the sign of the jet firings must be reversed. (These curves are therefore called the switch curves.) If the initial conditions are located anywhere in the area above and to the right of the curves shown in the figure the jets are fired using −1 as the control (i.e., negative acceleration) to force the motion to approach the switch curve in the second quadrant. When the values of [θ e θ ˙e ] reach the switch curve, the sign is reversed (to +1) and the motion will track the switch curve to the origin at which point the control is set to 0. In the absence of any disturbance, the system would stay at the origin forever. Similarly, if the initial conditions are to the left of the switch curves, the jets are fired with a positive acceleration ( u = +1) until the switch curve in the fourth quadrant is intersectedand theacceleration is reversed to allow themotion to approach theorigin from above. The measurement of the values of [θ e θ ˙e ] is always off slightly, and the exact vehicle inertia and the jet forces are known only to within some error, causing the acceleration α to be in error when the switch curve is calculated. These errors will cause the control to miss the origin slightly so instead of having the control objective be to hit the origin exactly; a box around the origin is the control objective. This box is denoted using the name “dead zone,” indicating that any time the vehicle has errors inside this zone, the control is unused (i.e., dead). The resulting switch curves have the form shown in Figure 9.2. Now that we have calculated the switch curves, let us try to justify the assertion that the minimum time control is the control that gets the trajectory in the phase plane to the originwith one switch. Assume we have made one switch and areon the unique trajectory in phase space that takes the solution to the origin. Were we now to switch again, the solution

276

Chapter 9. Putting Everything Together 0.25

0.2

Coast Region -- Jets are Off 0.15

0.1 ) . c e S / . d a R (

0.05

Target Region

Turn Jets Off at this Curve

0

e t a R r a l u g -0.05 n A

-0.1

Turn Jets On at this Curve

-0.15

-0.2

-0.25 -0.8

-0.6

-0.4

-0.2 0 0.2 Angular Position (Radians)

0.4

0.6

0.8

Figure 9.2. The physics of the phase plane logic and the actions that need to be developed.

will immediately move off the only solution path that goes to the origin, and we would then have to undo that move, all of which would increase the amount of time it would take to achieve the goal of reaching the origin. Thus, the minimum time path must be that which results from using the logic described in the figure above. To implement this control using feedback in continuous time is simply a matter of tracking the location in the phase plane, performing the switch when the trajectory crosses the appropriate switch curve. However, because we are implementing the control system in a computer, there are computer usage issues we must consider. This leads to the third component of the design process: designing the system components to meet the speci fications.

9.3 Design of System Components to Meet Specifications: Modify the Design to Accommodate Computer Limitations When using a real-time computer implementation of a control algorithm, the time it takes to perform the control calculations can interfere with other computations that need to be done. (Typically a computer is too expensive to devote solely to the control system, so other functions are performed on a time sharing basis.) This is the first time we must deal with creating embedded software (and simulating its effects), so let us spend some time on this.

9.3. Design of System Components to Meet Specifications

277

The computer is totally occupied every time it looks at the location in the phase plane. The more frequently we look to see where we are, the more computer time used and the more overloaded the computer will be. The discussion above on the control system design shows we cannot wait to make a switch. Every time the control system looks at the position and rate to make a decision, other critical tasks are left waiting. (An interesting historical footnote is that when we first attempted to design a digital control system for the lunar module, we concluded that it was impossible for exactly this reason.) Thanks to George Cherry, an MIT Instrumentation Laboratory engineer at the time, the lunar module control system had a unique implementation. His idea was to make the control into a timed task. Anycomputer hasthe ability to do a timed task. The timed task interrupts thecomputer (i.e., temporarily terminating an existing task for this one) and then a program runs that starts an external timer. When the computer is finished with the task of starting the timer, it can return to performing the original interrupted task (while the timer ticks away in the background). Whenthe timer counts down, indicating thatthe taskis complete, the computer is once again used (usually again via an interrupt) to complete the timed task. This was George Cherry’s idea for the lunar module digital control system. When the control logic required jets to be on, the control algorithm computed the time needed to reach the desired location in the phase plane (either the switch curve or the small rectangle at the origin). The computer turned on the jets and sent the desired on time to an external timer. The control task ended, and the computer was free to do other tasks. When the timer counted down, the computer was interrupted again and the jets were turned off. This operation was very quick since it consisted of simply resetting bits in one of two output channels (each bit connected directly to a jet firing solenoid). We ran a large simulation of the design twice a day, so to verify that this logic worked was a time consuming process. With Simulink, we can have the same simulation running on a desktop computer. Furthermore, the simulation can use the same logic used in the actual lunar module code (including a simulation of the timers). To implement the design, we need to compute the amount of time needed to get from any initial location in the phase plane to some desired position. Let us assume that we are in the region where negative accelerations are required (i.e., to the right and above the switch curves). At the switch curve we want to intersect, the control will change from −1 to +1 (positive acceleration). From the analysis above the switch curve equation is: θ e (t ) =

˙e (t)2 θ

2α

.

Note that when we are on the switch curve, the jet on time is simply T on =

˙e (t ) θ

α

.

Until we reach the switch curve, the jets are on, providing negative acceleration, and the trajectory in the phase plane is θ e (t) − θ e (t 0 ) = −

˙e (t)2 − θ ˙e (t 0 )2 θ

2α

.

We need to compute the intersection of this curve with the switch curve. Once we know the intersection values, we can compute the time that the jets need to be on using the equation

278


for the time t we developed above (in this case with the value of u T on

=

˙ (t ) − θ ˙ (t) θ e 0 e

α

= −1):

.

One of the important ways to do an analysis like this now is to use symbolic manipulation. MATLAB has an interface to Maple that easily allows this. The MATLAB M- file with the implicit Maple calls to solve for the on time is in the code segment below. Note that the Maple variables in the code are MATLAB objects created with the “syms” command.

syms Ton t alph th thdot th0 thd0 thdot = solve('th = thdot^2/(2*alph)',thdot) thdot = thdot(2) % Select negative square root intsect = subs('th = th0 +(-thdot^2+thd0^2)/(2*alph)',thdot) thdot = solve(intsect,'thdot') Ton = (thdot(1)-thd0)/alph pretty(Ton)

The solution proceeds by first setting up the variables that are symbolic in MATLAB (Ton, t, alph, th, thdot, th0, and thd0). Next, the switch curve is defined, and the solution along thecurve as a function of θ ˙e is developed. (Realizing thatthe solution involves a square root, the negative value is selected since the intersection when the acceleration is positive will occur in the second quadrant where θ ˙e is negative.) Next, the curve followed during negative acceleration is developed. Then, we find the intersection of the two parabolas. The value of θ ˙e (t ) at the intersection determines the time T on . The MATLAB script that invokes Maple to do the algebra is in the figure above. The result from running this script (after the pretty instruction in MATLAB) is

[ 1/2 1/2 2 1/2 ] Ton = [(2 2 (th alph) alph + 2 th0 alph + thd0) - thd0] [-------------------------------------------------------] [ alph ].

This equation defines the time that the jet needs to be on in order to reach the switch curve. At every sample time (0.1 sec), the lunar module digital control system computes the total time that the jets need to fire. This equation applies for both negative and positive accelerations by using absolute values in some of the terms. The coasting region of the control system (when the jets are off and the lunar module drifts at its last angular rate) uses the approach of looking every 0.1 sec to see if the jets should be on (based on the location in the phase plane). If there were no need to fire the jets, the coast would continue for another 0.1 sec. When the motion is just about to leave the dead zone region this strategy adds an extra 0.1 sec of drift. We deemed this error was acceptable.


279

The Lunar Module Digital Autopilot Design Jet Torques1 Jet Commands [I]

Inertia

Attitude Cmd.

Yaw Jets

Jet Torques -K-

dq/dt

Integrate qdot

Two Jet

Pos. Error

2

Read data

||x||

Four Jet

Two Jet or Four Jet Yaw Couples

Write Var

2 1 Ascent Single Jets

Initial Rates

Derived_Rates

[4x1]

x

Normalize the Magnitude of q quaternian[q s] 1 Read the "News & Notes" article about this model over the web. (Double-Click Here)

Nominal Two Jet Couples NofJets

4

Derived

Initial Quaternions

1 sx o

q

Euler Angles (Yaw Roll Pitch) from Quaternions

Attitudes

e_edot

-C-

q

ψφθ

Rates Actual Actual

Input Jet Torque

Reaction Jet Control

Phase Plane Plot

1 xo s

omega

Pitch-Roll

Attitude Meas. Pitch/Roll Jets A/D Conversions (Sample time delt)

Body Angular Accel

Yaw

u

Astronaut Attitude Command to Autopilot

Integrate omega dot

Inertia J

Mux

Cmd.

Inertia Matrix Quaternion and Spacecraft Rotational Dynamics

Initialize Data Stores PitchRollJets

Write Single Jet (Ascent) PitchRoll Var or Two Jet Couples for Pitch and Roll

e_edot NofJets

PitchRollJets

Figure 9.3. Complete simulation of the lunar module digital autopilot.

9.3.1 Final Lunar Module Control System Executable Specification We are now ready to look at the Simulink model for the control system. Open the model LMdap3dof from theNCS library in MATLAB, andcarefully look at themodel(Figure 9.3). This model is the complete simulation of the control system in the three degree of freedom simulation we developed in Section 9.1.1. The control system has a number of features that areimportant forthe overall development process that we aretrying to describe. These are as follows. •

Data used in the model are saved and retrieved using Data Store blocks.

•

The control system design began with a single axis, and the result was stored in a library. Each of the three axes then used the library block.

•

All of the logic for determining where we are in the phase plane uses State flow.

•

To ensure that the blocks we developed are available for reuse, the library also has the block for the quaternion calculations, the block used to derive the digital rates for the control system from the attitudes, and the block that converts the quaternion values into Euler angles.

When you open the model, it will appear with the model browser turned on. This feature allows you rapidly to navigate aroundthe model using thesame navigation technique that allows you to navigate in Windows. At the top level, zero order hold blocks before the Reaction Jet Control subsystem convert the continuous time attitudes into measured values (at a time interval of every 0.1 sec). The zero order hold measures the value of the analog signal and then does the

280


analog to digital conversion. Because the astronauts communicated with the control system both by using a data entry key board on the computer and by turning switches on and off, we need to keep track of the switches that they used. All of the calculations above (since the acceleration from the jets determines the parabolas) use information about the available acceleration. Therefore, the control system must know the state of the switches that the astronaut could use to disable reaction jets. Use of one of these switches caused a computer update that saved the state of the switch and changed the values of the affected jet accelerations. In the Simulink model, we use blocks called Data Store Write, Data Store Read, and Data Store Memory to record and interrogate the switch states. These blocks are at the bottom of the Simulink model. They update the number of jets in the model and modify the acceleration used by the control system. Another neat feature is that the user can double click on the switch icon to change the number of jets used in the simulation. The switches change both the yaw axis jets (where the number of jets used can be set to either 2 or 4) or the pitch and roll axes jets (where control using only upward firing jets is an option). As long as the jets fired downward during ascent from the surface of the moon, all of the control forces contributed to the lifting thrust from the ascent engine. This idea reduced the amount of fuel used during the ascent. The Data Store Read, Write, and Memory blocks work by creating internal memory storage locations for the values of the variable. (The stored values are created by the Data Store Write block.) The initial values for the data are created using the Data Store Memory block (which are in the diagram but are not connected to any other blocks; they are at the bottom right side of the model). Data Store Read blocks are the source in the model of all of the volatile data required by the control system. In this model, the Read blocks are all inside the Reaction Jet Control subsystem. There is also a block that provides information about the revision number, date of revision, and so on; it appears at the bottom right of the model. Every time you save the model, Simulink updates this block. The block for the Quaternion calculations is the same block that we used in Chapter 3, but it is now stored in the library. The last new block at this level is the quaternion to Euler angle block. The blocks contain annotation of the equations used inside the block, so we will not go through them. With this introduction, you can now begin to explore each of the blocks in the final model (in Figure 9.3). A good place to start is the simulation of the counter that times the jets. This model is in Figure 9.4. (There are three of these blocks one in each of the “Reaction Jet Control ” subsystems. Browse to one of them to see the model.) The timer clock was 625 microsec, so the simulation uses two digital clocks: one at this rate and one at the sample time (0.1 sec). The jet on time counts off the 625microsec clock in the subsystem labeled “Jet On Time Counter.” This subsystem looks like Figure 9.5. Since the Sample Time clock changes at the sample time, this value is constant over the duration of the countdown of the jets. (Remember that we compute the on time anew at every sample time.) Thus at some time before the next sample the difference between the time of the last sample and the time from the counter clock will be equal to the on time. The relational operator block looks for the difference to be greater than zero, so the output goes to zero when the countdown is complete. Returning to the blocks that use the counter, we see that the logic uses this output, along with the jet on command and the counter enable that comes from a State flow chart. The logic implemented is the logical NOR of the enable and the counter output. Next, we compare the output, using a logical OR, with the counter output. This finally is multiplied by the jet on command. (This command is the number

9.3. Design of System Components to Meet Specifications 1 On Command 2 Enable Counter 3 Jet On Time

Clock at counter time tic (0.000625) clockt

Product Counter Output

Logical Operator

Logical OR1

1 Jet Command

Logic implements: ~(enable | stopjets) | stopjets

ton

12:34

Clock at tics

12:34

Clock at Sample Time

Clock at time delt (Sample time)

281

Stop jets

Counter Output

Jet On TIme Counter

Figure 9.4. Simulating a counter that ticks at 625 microsec for the lunar module Simulink model.

Enable

1 ton

2

Count down of ton ton

Time in clockt multiples

Clock at tics

3

>

Stop Command

Relational Operator

1 Stop jets

Constant 0

Time in delt multiples

Clock at Sample Time

Figure 9.5. The blocks in the jet-on-time counter subsystem in Figure 9.4.

of jets with a sign indicating the direction, so a logical AND cannot be used to provide the final jet command output.) If you start this model in the usual way, a phase plane plot for the system response appears (Figure 9.6) as the simulation runs. The plot is the control response for motion around the x (yaw) axis, which is the axis with the smallest inertia. The model also contains plots (Scopes) that show the attitude and attitude rate errors for all three axes. If you navigate through the reaction jet control subsystem, you can see the logic and the computations for the switch curves. In Section 9.1.1, we created the model and used an empty subsystem for the controller. (We called the subsystem “Control System TO BE DESIGNED”.) We have now completed the design. It is easy to see that a different group could have created the control system design. The group that created the simulation model with the quaternion block might even have been working at the same time.

282

Chapter 9. Putting Everything Together Phase Plane Plot for LM Yaw Axis Digital Control 0.1 Switch Curves for Turning Jets Off Switch Curves for

0.08

Turning Jets On 0.06

0.04 r o r r 0.02 E e t a R 0 e d u t i t t A w -0.02 a Y

-0.04

-0.06

-0.08

-0.1

-0.1

-0.08

-0.06

-0.04

-0.02 0 0.02 Yaw Attitude Error

0.04

0.06

0.08

0.1

Figure 9.6. Lunar module graph of the switch curves and the phase plane motion for the yaw axis. (The graph uses a MATLAB function block in the simulation.)

Continuing our travels through the model, we have a block that computes the three axis rotational rates from the Euler angle measurements. This block is in the library called LMdapLibrary_NCS.mdl in the NCS library. We have not encountered library blocks before, so we need to describe how to create a library and how they work. When you select “New” in the File menu, there are two options: a Simulink model or a library. To create a library you simply select the latter. A library is different from a model in that none of the Simulink menu items (like the start button) appears in the window. Furthermore, if you use any of the blocks in the library to develop a model, future changes to any of the library blocks will propagate to the Simulink models that use them. This feature makes it easy to keep track of design modifications and far easier to reuse a block in the future. When you convert a library block into embedded computer code, and the code is used multiple times (as is the case here where the same control system block applies to the yaw, pitch, and roll controllers), the code is created as a “reentrant” subroutine. The conversion of the diagram into embedded code occurs once. The Real Time Workshop conversion makes only one copy of the control system code (not three versions). Each instantiation uses the same code, minimizing the amount of storage needed for the code. While we are on the subject, libraries work best if a tool for maintaining con figuration control is also in place. Allof themodels in Simulink (and in MATLAB) allow theuser to use a configuration management tool with the model. Source control systems must comply with the Microsoft Common Source Control standard. If you have a compliant source control


283

system on your computer, the Source Control options in the MATLAB Preferences dialog will show it. Revision Control Systems (RCS) and Concurrent Versions System (CVS) are two tools that use the Microsoft Source Code Control. When a tool such as this is used, a designated configuration manager must make all changes to models and libraries. This ensures that all models that are in the current design are veri fied and validated. Individuals can work with their own version of the models, but then the con figuration manager must test them before he allows the changes to migrate to the actual design.

9.3.2 The Control System Logic: Using Stateflow Now that all of the basic analyses for the control system design are complete, we can focus on implementing the control logic in the simulation.

function [Firefct1, Coastfct1, Firefct2, Coastfct2, tcalc1, tcalc] = ... fcn(e,Njets,DB,alph,alphs)

% Time at full accel for e_dot = 0 (cross the e axis) is tcalc1. % (This time is positive in RHP and negative in the LHP, and % this is accounted for in the calculation of the total on time % in the Stateflow chart). % The intersection of the switch curve and the "on" trajectory % determines the value of xdot at the switch curve. This value % divided by the acceleration is the time needed to go from % the e axis crossing to the switch curve (tcalc). The two times % tcalc and tcalc1 are added together in the State Chart. % Note that the square plus or minus value are accounted for % in the State Chart (when in the RHP we add tcalc and tcalc1, % and in the LHP we subtract tcalc1 (which is <0) from % tcalc (>0 always). % ac

Constants used in the calculations: = 2*alphs/(alph+al phs);

% Changeable constants used in the calculation: accel = Njets*alph; accels = Njets*alphs; % Evaluate the Firefct1 = e(1) Coastfct1 = e(1) Firefct2 = e(1) Coastfct2 = e(1) %

location in the phase plane w.r.t the 4 parabolas: -DB +e(2)^2/(2*acce l); -DB -e(2)^2/(2*accels); +DB -e(2)^2/(2*acce l); +DB +e(2)^2/(2*accels);

Compute the on time for a jet firing based on phase plane locations:

x1 x2 x3 tcalc1

= = = =

tcalc

=

e(1)/accel; % Scal position for jet on time calc. e(2)/accel; % Time at full accel for e_dot = 0. D B/ acc el; % Sca le De ad Ban d for on t ime ca lc. x2; % Time to reach the e axis from edot. u = (abs(x1) + x2^2/2 - x3)*ac; sqrt(u*(u>=0)) ;

284

Chapter 9. Putting Everything Together [count==0]

/count=2; Wait_for_stable_rate

1

2

Start [ (e[1]<0 & Coastfct1<=0 & Firefct2>0) ] 4

[ (e[1]<0 & Firefct2<0) | (e[1]>0 & Coastfct2>0) ]

1

/count--; 2

3

[ (e[1]>0 & Firefct1>0) | (e[1]<0 & Coastfct1<0) ]

Fire_region_1 en: jets=-Nofjets; ton=tjcalc+tjcalc1; [ton<2*delt]{enable=1;} [ton>tmin]{count=2;} 1

2

1

[ (e[1]>0 & Coastfct2>=0 & Firefct1<0) ] [e[1]>0 & Firefct1>0] [count==0]

3

[e[1]<0 & Coastfct1<=0]

2

/ton=ton-delt;

2

{enable=0;}

[e[1]>0 & Coastfct2>0]

2

1

Skip_a_Sample_1 en:count--;

Skip_a_Sample_2 en:count--;

{enable=0;} 2

1 1

/ton=ton-delt;

[ton>tmin]{count=2;}

[count==0] Coast_region_1 en: jets=0; ton=0; enable=0;

Coast_region_2 en: jets=0; ton=0; enable=0;

[e[1]<0 & Firefct2<0]

[ton<2*delt]{enable=1;} 1

2

Fire_region_2 en: jets=Nofjets; ton=tjcalc-tjcalc1; 3

Figure 9.7. State fl ow logic that determines the location of the state in the phase plane.

The Control Law blocks from the library use an Embedded MATLAB Function block to compute the data needed to determine where the error is in the phase plane. We do this by substituting the appropriate yaw, pitch, or roll attitude and attitude rates into the equations for the four different switch curve parabolas. The functions that we generate are Firefct and Coastfct (with 1 or 2 appended to denote the quadrant). In addition, this code block calculates the jet on time using the equations from the Maple derivation above. This Embedded MATLAB code is in the window above. Note that the code is very readable because it uses standard MATLAB. As an exercise, follow the code and verify that it uses the equations above for the jet on times. Also, note that this code segment is in the LMdapLibrary_NCS.mdl as part of the Control System block, so it also is reentrant. The Embedded MATLAB block does not do the logic for determining which jets to fire and counting down the on time. This decision making uses Stateflow for the logic. (The Stateflow block decides whether to fire a jet with positive or negative acceleration and whether or not to coast at any decision point.) The logic in the State flow block is easy to follow (see Figure 9.7 for the chart). The original control design for the lunar module did not use a process like this at all; all of the logic had to be programmed in assembly language. The Stateflow chart is identical for all three axes. One of the techniques used to develop this chart was to lay it out visually so it follows the four quadrants in the phase

9.4. Verification and Validation of the Design

285

space. Skipping the top of the chart for a moment, there are four states that are at the vertices of a rectangle: •

quadrant 4 & 1 coast region is at the top right;

•

quadrant 1 & 2 jet firing (negative acceleration) is at the bottom right;

•

quadrant 2 & 3 coast region is at the bottom left;

•

quadrant 3 & 4 jet firing (positive acceleration) is at the top left.

When the state-space locations are near the switch curves, we start the counter to count down the jet on times. This is not necessary unless we are near the switch curves. (When we are not close, the on times are greater than the sample time so we turn the jets on, leave them, and reevaluate the on time at the next sample.) The State flow states that handle this logic we called “Skip_a_Sample.” There are two such states: one for the left half of the phase plane and one for the right. The Stateflow logic should be easy to follow, since every decision uses the sign of the attitude rate and the coast and fire functions evaluated in the Embedded MATLAB block. When you run the model, open the Stateflow block for the yaw control and watch the logic and the phase plane plot. The correlation between the Stateflow switching and the phase plane should make it easy to follow the logic. We are now ready for the fourth step in the design process: verification and validation of the design.

9.4 Verification and Validation of the Design If you browse through the Simulink library, you will see a group of blocks called Model Verification. Figure 9.8 shows this library. You can use these blocks to evaluate whether or not the signals in the Simulink model are doing what you expect. The block description and their icons help you to understand what they do. For example, the Check Discrete Gradient block (the second block in the library) evaluates the difference between the signals at two different times and flags you if this exceeds some number that you specify in the block. This ensures that the (digital) derivative of the signal is within bounds. You can check to see if signals are within certain bounds, that they do not exceed bounds, that they lie within a dynamic range that is possibly changing with time, and that the signal is nonzero. With these blocks and a built-in tool that gives an analysis of the coverage of the signal flow (for both Simulink and State flow), you can assure yourself that the simulation is valid (and therefore, when the time comes to create the code, that it is valid). Since this is the kind of tool that provides verification and validation, which are important only when the design process has the goal of creating a system design that may have embedded code, we will not spend more time on it. However, if the overall system design is your goal, this tool is extremely valuable. We are now at the last step in the design process, the creation of embedded code.

286


Figure 9.8. Simulink blocks that check different model attributes for model veri fication.

9.5 The Final Step: Creating Embedded Code We now come to the last step in the design process: the creation of the embedded code. Simulink offers the user a multitude of paths for achieving this goal. The simplest code generation process is to generate code that runs under Windows. The code generation tool Real Time Workshop® (RTW) will typically default to this option. When you invoke RTW, it will generate a stand-alone application that runs from the Command window. It is well beyond the scope of this book to take you through this process, but it is easy enough to do and you can generate some very interesting applications. Reference [41] discusses the need for structured development of real-time system applications; it should be clear that you could not find a more structured (since it is automatic) way than using RTW.


287

The Signal Processing Blockset comes with blocks that allow the capture of signals fromthe audio devices built intoyour computer. You can design a complex signal-processing task and try it out on your computer. The blockset comes with several interesting examples that you should try. Once you understand what the applications do, try modifying them to do something different. It is a lot of fun.

9.6 Further Reading Theprocess of creating newsystems is dif ficult to manage. Several books have been devoted to this subject (see [10], for example), and their main conclusion is that a tight integration of the research, development, and manufacturing groups is the only way to ensure that new technology rapidly migrates into products. The example in this chapter illustrates a powerful feature of Simulink. You can open a model and actually play with it. There is no better way for a research team to communicate an innovative idea to the development and manufacturing sides of a company (and, for that matter, to the company ’s managers—in fact, why not the CEO and the board of directors). Imagine the R&D team briefing the top management of a company using a simulation of the proposed new device. Imagine that the simulation has all of the bells and whistles, shows all of the manufacturability components, clearly delineates computer software required, and has a clear path to the embedded computer code. When the design team then quotes a cost and schedule, it will have far more meaning than a simple recitation of the same information. Controlling the costs of the software in a system has been extremely dif ficult. There are innumerable instances of software development cost overruns that are multiples of the initial estimates. Reference [35] describes many such horror stories. The stories generally result in the de finition of a process that will make the software development work better. In my experience, these process improvements always lack one major component: the software designers never have a way of verifying that the software will do what is required. Reference [19], for example, decries the fact that software requirements appear well before the allocation of the system requirements to the software is complete. This is justified because the software development takes so long that it must start early. Another example that I remember well was the first time I tried to put a process together. It was before Simulink but after the introduction of MATLAB. The CEO of the company I worked for asked me to fix a severe schedule slippage in the development of a complex system. The first thing I did was to force the development team to use MATLAB (instead of FORTRAN or C) to build the system components. The existing process required that we send a monthly specification document to the software group. Each of these transmittals was hundreds of pages. About two months after we adopted MATLAB for the design team, I noticed that the software group would ask for the MATLAB code. In a conversation with the software team, I found out that they were using the MATLAB code as the speci fication and checking the written speci fications only when they found con flicts or processes that were not in the MATLAB code. This was the first example of an executable speci fication that I had ever encountered. I have had the same experience many times; these occasions are what have convinced me that spending the time and effort to build a process around Simulink (and the other tools we have discussed) is cost effective and possible and will lead to faster, cheaper, and more ef ficient designs.

Chapter 10

Conclusion: Thoughts about Broad-Based Knowledge I hope that you have reached this point in the book with a much better understanding of the power of a visual programming environment. I also hope that you can see how an integrated system-engineering environment can improve your designs. In the process of working through this text, you should have become familiar with a large number of different disciplines —perhaps some that you had not previously learned. This was partly my goal. One of the major attributes of a good engineer (or mathematician or any other discipline youwould like to inserthere) is both breadth and depth of knowledge. As I was completing this book, I read a marvelous editorial in the New York Times by Thomas Friedman, entitled “Learning to Keep Learning” [12]. In the editorial, Friedman makes the case for better and broader education. He quotes Marc Tucker, who heads the National Center on Education and the Economy: “It is hard to see how, over time, we are going to be able to maintain our standard of living. ” Friedman then adds: “In a globally integrated economy, our workers will get paid a premium only if they or their firms offer a uniquely innovative product or service, which demands a skilled and creative labor force to conceive, design, market, and manufacture — and a labor force that is constantly able to keep learning. We can’t go on lagging other major economies in every math/science/reading test and every ranking of Internet penetration and think we’re going to field a work force able to command premium wages.” A little later in the editorial, Friedman again quotes Tucker as saying, “One thing we know about creativity is that it typically occurs when people who have mastered two or more quite different fields use the framework in one to think afresh about the other. ” The goal of Tucker is to make “that kind of thinking integral to every level of education. ” Tucker thinks that this requires a revamping of our educational system, designed in the 1900s for people to do routine work, into something different. The new system must teach how to “imagine things that have never been available before, and to create ingenious marketing and sales campaigns, write books, build furniture, make movies, and design software, ‘that will capture people’s imaginations and become indispensable for millions.”’ In the last part of his editorial, Friedman tempers the nationalistic sound of the statements by noting that innovation is not a zero-sum game. It can be win-win. He says, “We, China, India and Europe can all flourish. But the ones who flourish the most will be those 289

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Chapter 10. Conclusion: Thoughts about Broad-Based Knowledge

who develop the best broad-based education system, to have the most people doing and designing the most things we can’t even imagine today.” It is my hope that this book and the products and processes we described here go some smallway to making Friedman’s thoughts real for you. It is clearly possible to learn multiple disciplines, to bring them to bear on complex design problems, and to create devices using more automation. It is also possible, working with the visual programming paradigm that Simulink exemplifies, to work faster, smarter, and more accurately. I truly hope that this book helps to further that process. As a final footnote, one of the reviewers of this manuscript made the prescient observation that using Simulink or any modeling software without care can lead to disasters. This is very true. However, I am old enough to remember engineers who, after they graduated, used only handbooks for the models in their designs. The result for these engineers was often a disaster, too. Because what they selected from the handbook was not valid in the context of the design they were developing, their conclusions were wrong. Simulink does not solve this problem. In fact, no tool can. How many times have you seen someone using a wrench as a hammer? Workers always have the opportunity to abuse their tools. The main reason for this book was to show you how to use Simulink the tool, not by simply learning how to build models but by showing you how the numerical methods in the background work. Along the way, we have emphasized the design process and the fact that modern design is a team effort. If you are new to this world, or if you are a student looking forward to working as a design engineer, remember that your colleagues want to help. Use their expertise. Share the Simulink models with them —early and often. Get their feedback and use it. If you are an engineer with experience in the design of systems, work to incorporate a visual programming tool into your design process. If you do, do not forget to train the engineers that will be new to this process. Also, ensure that your design teams have an ample number of old-timers who know when something looks wrong. Finally, remember the various tricks and comments that I have made in the text: •

Annotate your models.

•

Use subsystems to keep the diagram simple.

•

Vectorize the model where you can (and let the user know that you have done so).

•

Make neat models; avoid spaghetti (very messy) Simulink models.

•

Run the models with the verification and validation tools to see if the results of the simulation make sense.

•

Check thedatayou areusing against therealworld by seeingif thesimulationproduces results that match experiments.

•

If you replace a block in a model with a new block, run the model with both the old and new block, subtracting the outputs of each to verify that the differences between them is essentially zero. In other words, be a good engineer.

Bibliography [1] Anderson, Brian D. O. and Moore, John B., Linear Optimal Control, Prentice-Hall, Englewood Cliffs, 1971. [2] Bathe, Klaus-Jurgen, Finite Element Procedures, Prentice Hall, Englewood Cliffs, 1996. [3] Bernstein, Dennis S., Feedback Control: An Invisible Thread in the History of Technology, IEEE Control System Magazine, Vol. 22, No. 2, pp 53–68, April 2002 [4] Bolles, Edmund Blair, Galileo’s Commandment , pp 415–419, W. H. Freeman, New York, 1999. This book contains the fragment from Galileo ’s 1638 book Two New Sciences, as translated by Henry Crew and Alfonso de Salvio. [5] Brush, Stephen G., A History of Random Processes, I , BrownianMovement from Brown to Perrin, Archive for History of Exact Sciences, Vol. 5, pp 1 –36, 1968. Reprinted in Studies in the History of Statistics and Probability , II, edited by M. Kendall & R. L. Plackett, Macmillan, New York, pp 347 –382, 1977. [6] Bryson Jr., Arthur E., Control of Spacecraft and Aircraft , Princeton University Press, Princeton, N.J., 1994. [7] Childers, Donald G., Probability and Random Processes — Using MATLAB with Ap plications to Continuous and Discrete Time Systems, Irwin, a McGraw-Hill Company, Chicago, 1997. [8] Chudnovsky, Victor, Kennedy, Dallas, Mukherjee, Arnav, and Wendlandt, Jeff, Modeling Flexible Bodies in SimMechanics and Simulink , MATLAB Digest, Available on The MathWorks web site http://www.mathworks.com/company/newsletters/ digest/2006/may/simmechanics.html, May 2006. [9] Derbyshire, John, Unknown Quantity — A Real and Imaginary History of Algebra, Joseph Henry Press, Washington, D.C., 2006. [10] Edosomwan, Johnson A., Integrating Innovation and Technology Management , John Wiley and Sons, New York, 1989. [11] Egan, William F., Phase-Lock Basics, John Wiley and Sons, New York, 1998. 291

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Index

1/f noise Need for, 194 Simulating, 194 Using cascaded linear systems, 195

Setting noise power of, 189 Band Limited White Noise block In moving average simulation, 150 Numerical experiments with, 184 To generate binomial distribution, 176 Using the Weiner process, 184 Bessel filter In Signal Processing Blockset, 146 Bode plots Calculating, 65 Creating, 64 Using Control Systems Toolbox, 67 Brown, Robert, 180 Brownian motion, see also Random processes; Weiner process And white noise, 178 Random walk and, 178 Brownian motion, connection White noise, 180 Buffers, 160 Using with the FFT, 161 Butterworth filter Analog, 142 Compared to other filters, 147 Definition, 143 In phase-locked loop, 167 MATLAB code for, 144 Using Signal Processing Blockset, 148

Absolute value block In clock model, 22 In Math Operations library, 23 Zero-crossing, turning on and off, 86 Zero-crossing detection, 11 Air drag, 12 Force from, 12 In Leaning Tower model, 12 In train model, 253 Leaning Tower Simulation results, 13 Analog filters As prototypes for digital filters, 142 Signal Processing Blockset, 145 Animation SimMechanics blocks, 259 Annotation Creating wide lines for vectors, 15 Of models, 16 Using TeX, 103 Attitude errors 3-axis direction cosine matrix, 107 3-axis quaternion, 109 Axis angle Rotation, 94 Axis-angle representation Euler’s theorem, 99

Callback Executing MATLAB code from, 196 Finding graphics handles after, 233

Band Limited White Noise, 184 295

296

Loading Data from, 254 Options in Model Properties, 124 PostLoadFcn, 234 PreLoad function, 39, 51, 123, 150, 167 StopFcn, 126, 176, 177 Central limit theorem, 174 Monte Carlo Simulink model, 174 Chebyshev filter In Signal Processing Blockset, 146 Clock Pendulum, 17 Simulation, 23 Using SimMechanics, 260 Clock model, see NCS library Computing Mean and Variance With Signal Processing Blockset blocks, 184 Constant block, see also Simulink blocks MATLAB input for, 4 Control System Toolbox Lyapunov equation solver, 188 Control Systems Bode plots, 64 Comparing position and velocity feedback, 60 Early development, 44 Example, thermostat, 46 Full state feedback, 72 Getting derivatives, 74 Linear differential equations and, 48 Observers, 74 PD control, 69 PID control, 71 Position feedback, 56 Simulink model, constructing derivatives, 77 Velocity feedback, 57 Control Systems Toolbox, 64 Convergence Limits for random processes, 181 Correlation function Definition, 194 Of white noise, 183 Spectral density function from, 183

Index

Covariance equivalence Definition, 189 White noise, 189 Covariance matrix, 186 Covariance equivalence, 189 Definition, 187 Differential equation for, 187 Creating an Executable Speci fication In the system design process, 271 Cross product, 98, 100 Cross product block In Simulink, masked subsystem, 101 Curve fitting with MATLAB for the polynomial block, 91 Damping ratio, 51 Data store, data read, data write Storing and retrieving data in embedded code, 279 DC motor, 105 Demux block, see also Simulink blocks Using, 246 Deriving rates For derivative feedback, 75 Difference block, see also Simulink blocks Validating model changes with, 146 Digital filters, 121 Bode plot for, 135 Discrete library, 129 In Discrete library, 121 Limited precision arithmetic and, 153 State-space models in Simulink, 129 Digital signal processing, 115 Bandpass filter design, 154 Definition, 118 Digital filter from analog using bilinear transformation, 148 fir filter from state-space model, 151 Impulse sample representation, 137

Index

Sampling and A/D conversion, 123 Uses of, 121 Using Signal Processing blockset, 145 Digital Transfer Function block, see Simulink blocks Direct Form Type II, see second order sections Direction cosines, 94 From quaternion, 101 Discrete time Comparing continuous with discrete, 39 Converting continuous system to, 38 Discrete transfer functions Calculation, 135 Dynamics Connection with stiff solvers, 87 Control system, 55 Control system sensors, 63 DC motor, 105 Forces from rotation, 24 Foucault pendulum, for, 26 House, heat flow in, 44, 48 Pendulum, 42 Reaction wheel, 105 Rotation, see also Axis angle: Direction Cosines; Quaternion Rotations, 94 Satellite rotation, 105 Einstein, Albert Brownian motion, 180 Electric train Rail resistance with nonlinear resistor block, 251 Traction motor and train dynamics, 252 Electric Train Model Model, 251 Elliptic Filter In Signal Processing Blockset, 146 Embedded code In the system design process, 286

297

Using data read, data write, data store blocks, 279 Equations of Motion For Euler angle rotations, 97 For quaternion representation of rotations, 99 Escapement, 17, 31 Description of, 19 History of, 17 Simulink subsystem for, 21 Euler angles, 95 From quaternion, 101 In lunar module model, 272 Euler’s theorem, 99 Executable Speci fication, 215, 225, 269 Analysis to create, 276 Definition and example, 225 For lunar module digital control system, 279 Stateflow and Embedded MATLAB in, 283 Fast Fourier Transform (FFT) block, 161 Fibonacci sequence, 116 Using the z-transform on, 120 Filter design, 141 fir filter, 149 White noise in, 150 Fixed-point filtering in, 159 Foucault pendulum, see also NCS library, 24 Dynamics, derivation of, 26 Model parameters, 30 Model, creating, 28 Solvers, experimenting with, 30 Vibration of moth ’s antennae, 31 Friedman, Thomas, 289 From Workspace block, 92 Gain block, 1, 12, 14, 23, 123, 184, 186 Creating a product with, 253 GUI interface to Simulink with, 232 In Signal Processing Blockset, 157 In SimMechanics, 261

298

Galileo Comparing objects of different mass, 14 Experiments by, 17 Inclined plane experiments, 31 Internet, references to, 31 Leaning Tower of Pisa, 5 Pendulum period and, 19 GoTo block, 77, 83, 253, 254 Graphical User Interface (GUI), 231 MATLAB Graphical User Interface Development Environment, 232 GUI, see Graphical User Interface GUIDE, see Graphical User Interface Hamilton Quaternion, 99 History, 112 Heat equation Electrical analog of, 203 Four-room house model, 262 Partial differential equations, 202, 213 Two-room house model, 262 Heating control With thermostat, 44 Home heating control Two-room house model, 233 Hooke, 18 Huygens, Christiaan, 18, 31 Hysteresis, 48, 225 Hysteresis block, 45 iir filter, 149

Integration Observer, 75 Integrator, see also Simulink blocks Block dialog for, 8 Block dialog options, 8 Denoted by 1/s , 7 Dialog, 8 Discrete Reset in phase-locked loop, 167 External initial conditions, 57 In integral compensation, 72

Index

In state-space model, 210 Initial condition, external, 9 Simulink Block, 7 Simulink C-coded dll, 7 Integrator modulo 1 block In the PLL model, 167 Interpolation In n dimensions, 87 In zero crossing, 11 Polynomial block with MATLAB curve fit, 91 Laplace Transforms, 39 Of state-space model, 41 Transfer function of state model, 42 Leaning Tower Simulation, two bodies, 14 Library block Quaternion, making, 279 Limited precision arithmetic, 115, 145, 152, 153 Fixed-point, 158 Floating-point, 157 Linear control systems, see control systems Linear Differential Equations Complete solution of, 35 Computing in discrete time , 37 Eigenvalues and response, 51 General form of, 33 In Simulink, state-variable form, 38 Poles and zeros, 55 State-space form of, 34 Undamped natural frequency and damping ratio, 51 Linear feedback control, see Control systems Linear models Analysis of Lorenz attractor with, 83 Creating from differential equations, 49 From Control System Toolbox, 67 Linearization Of a Nonlinear System, 49

Index

Lookup Table, 90 2-D, 90 Entering tabular data in, 88 n-D, 91 Simple, 88 Lookup Table block, see also Simulink blocks Lookup Table library, 87 Lorenz attractor Chaotic motion model, 82 Linear models as function of time, 83 Parameters and fixed points, 82 Time varying root locus for, 83 Lorenz attractor simulation, 81 Lorenz, Edward, 81 LTI object Creating and using, 67 Definition, 66 LTI Viewer For digital filters, 136 For string vibration model, 266 Lunar Module Euler angles in, 272 Quaternion block in, 272 Lunar module digital flight control System design process example, 272 Lyapunov equation solver Control System Toolbox M- file, 188 In Control System Toolbox, 188 Maple-MATLAB Interface Determining direction cosine attitude error with, 107 Using for lunar module control law, 277 Using Maple for z-transform, 119 Masked subsystem Deriving rates model, 75 Example, cross product, 101 Masked block parameters, 76 Masked subsystem, creating SimPowerSystems nonlinear resistor, 247

299

Math Operations, see Simulink blocks MATLAB Calculation of stop time, 6, 10 Commands, in Courier type, 1 Creating a discrete time model, 35 Creating a discrete time model, c2d_ncs, 37 Default Simulink data in, 10 Input for Constant block, 4 ODE Suite, compared to Simulink, 7 Opening Foucault_Pendulum, 28 Opening Leaningtower, 6 Opening Leaningtower2, 12 Opening Leaningtower3, 16 Opening Simulink, 2 Vector notation in blocks, 14 MATLAB code Updating GUI with, 236 MATLAB connection to Simulink Graphical User Interface (GUI), 232 MATLAB GUI Code for updating, 236 Creating, 238 Interface for State flow and Simulink, 230 Model Properties Tab, see Callback Modeling and Analysis In the system design process, 271 Moler, Cleve, xx, 30, 81, 115, 201 Monte Carlo simulation, 173 Mux block, see also Simulink blocks using Using, 45 NCS library 1/f noise, model, 194 Batch processing, model, 160 Butterworth filter model, 141 Butterworth filter, Signal Processing Blockset, 145 Clock_transfer_functions, 42 Control systems, model, 44 Digital filter aliasing, model, 123 Digital filter transfer function, model, 121

300

Index

Digital filter, Bode plot, 135 Fibonacci sequence Digital filter block model, 127 Model, 116 State-space model, 129 Final lunar module speci fication, model, 278 fir filter model, 149 Foucault pendulum, model, 24 Heat equation, model, 207 Heating controller Executable speci fication model, 225 MATLAB connection, 230 Model, 230 How to download, 1 Including computer limitations, lunar module model, 276 Leaning Tower, model, 9 Limited precision filter design, model, 157 Linear differential equations, model, 38 Lorenz attractor, model, 81 lunar module State flow logic, model, 282 Monte Carlo, central limit theorem model, 174 Naming conventions, 1 NCS definition, xx Nonlinear resistor, model, 246 Observer, model, 74 Partial differential equations, 207 PD control, model, 69 Phase-locked loop, model, 164 PID control, model, 71 Random walk, model, 178 Rayleigh noise, model, 176 Reaction wheels model, 105 Rotation Axis-angles, 98 Direction cosines, 98 Euler angles, 94 Quaternions, 99 Sampling theorem using FFT, model, 160

Sampling theorem, model, 140 Saving new models, 4, 5 Set MATLAB path, 1 SimMechanics clock, model, 260 SimMechanics pendulum, model, 257 SimMechanics vibrating string, model, 262 SimPowerSystems Four-room house model, 263 Simple model, 242 SimPowerSystems train, model, 251 Spacecraft with reaction wheel, model, 109 Specification capture, lunar module model, 271 Specification to design, lunar module control law model, 273 Spring-mass Model, 55 State-space model, 51 State-space, continuous and discrete, 39 State flow Chart, 215 Debugger, 223 Model input-output, 221 Simple model, 215 Systems excited by white noise, model, 189 Thermo_NCS, 46 Using data in tables, models, 87 Using filter design block, model, 153 Using the nonlinear resistor, model, 248 White noise, model, 184 New York Times

Thomas Friedman editorial, 289 Nonlinear controller Home heating thermostat, 44 Nonlinear differential equations, 79

Index

Numerical Computing with MATLAB, 30 Fibonacci sequence in, 115 Lorenz attractor in, 81 Partial differential equations in, 201 Numerical integration, see Solvers

Observers, 77 Ode113, 86 Ode23t, 249 Ode45, 193 Opening Simulink Clicking on icon, 2 Command for, 2 Partial differential equations Creating a Simulink model for, 205 Finite-dimensional models for, 202 Heat equation, 202 Electrical analog, 207 Model using SimPowerSystems, 262 State-space model, 207 Modeling in Simulink, 200 Vibrating string, 261, 264 Vibration, models for, 211, 261, 265 Pendulum, see also Clock; Foucault pendulum Clock, 17 Using SimMechanics, 257 Perrin, Jean And Brownian motion, 180 Phase-locked loop (PLL), 164 How it works, 165 Model of, 167 Simulation of, 169 Voltage controlled oscillator (VCO), 167 Physical modeling, 241 SimMechanics, 242 Poles and zeros Bode Plot calculation with, 65 Definition, 42 Digital filter implementation and, 155 For 1/f noise approximation, 195

301

From state-space model, 54 In fir filters, 150 In Signal Processing Blockset, 152 LTI object and Control Systems Toolbox, 66 Maxwell and, 44 Of Butterworth filter, 143 Phase shift in filters, 149 Using Control System Transfer function, 55, 62 Polynomial block Curve fitting in MATLAB for, 91 PostLoadFcn, see Callback Power spectral density function 1/f noise, 194 Creating with white noise, 194 Definition, 194 Preload function, see Callback Simulink model properties, 30 Quaternion, 94 Converting to direction cosines, 101 Converting to Euler angles, 101 Definition, 99 Derivative of, 99 Hamilton, discovery of, 112 In lunar module model, 272 Library block, 280 making, 279 Norm of, 99 Subsystem block for, 101 Random processes, 173 Convergence of, 181 Random walk, see also Random processes Prototype for white noise, 178 Reaction wheels Model of, 107 Operation of, 105 Relational Operator block, see Simulink blocks Reset integrator For phase-locked loop, 167

302

Root locus plot Comparing position and velocity feedback, 61 Definition, 60 For mass velocity feedback, 60 Transfer functions, numerical issues with, 63 Rotating bodies Forces on, 25 Rotations Axis-angle representation, 94 Direction cosine matrix representation, 94 Quaternion representation, 94 Sampling theorem, 115, 136 Implementing, low pass filters, 140 Numerical experiments, 140 Proof of, 138 Simulink model for, 140 Using FFT to implement, 161 Satellite in orbit Dynamics, rotational, 105 Second order sections, 157 In Filter Design, 156 Second order systems Parameters in, see also Damping ratio; Undamped natural frequency, 51 Simulink model with transfer functions, 42 Shannon, Claude, 136, 138, 170 Signal builder For input to nonlinear resistor, 249 Signal processing blocks FFT, 161 Signal Processing Blockset Butterworth filter in, 148 Signal Processing blockset, 145 Analog filters in, 146 Bessel filter in, 148 Blocks in the library, 146 Chebyshev filter in, 148 Comparing blockset and Simulink models, 146

Index

Filter design with second order sections, 157 Implementing a digital filter in fixed point, 159 Limited Precision Bandpass filter design with, 154 Using buffers for batch processing, 160 SimMechanics, 240 Environment, 258 Ground, 258 Library, 256 Physical modeling, 242 Vibrating string, 262, 264 SimMechanics Blocks Animation, 259 Body, 259 Body sensor, 259 Environment, 257 Ground, 256 Revolute joint, 259 SimPowerSystems, 240 Algebraic loops in, 248 Circuit with nonlinear resistor, 250 Connection icon, 242 Connections icon, 245 Connections to and from Simulink, 252 Library of blocks, 243, 244 Mask for nonlinear resistor block, 247 Mask for train model, 253 Modeling an electric train, 251 Nonlinear devices in, 248 Nonlinear elements in, 246 Nonlinear resistor model, 246 Simple example —RC circuit, 242 Simulink inputs, 245 SimPowerSystems, blocks DC voltage Source, 243 Resistors, inductors, capacitors, 244 Switch and circuit breaker, 245 Simulink Adding a block, 3 Automatic connections, 4 Automatic vectorization, 33

Index

Block diagram basics, 1 Block dialogs, 3 Click and drag, 3, 7 Continuous, library, 7 Creating a new model, 3 Drawing neat diagrams, 4 Library browser, 2 Logic and bit operations, library, 7 Masked subsystem, creating, 247 MATLAB, setting constants in, 9 Modeling partial differential equations, 200 Preload function in model properties, 30 Signal flow connections, 4 Simple model, 3 Sinks, library, 7 Solver Default, 10 For SimMechanics, 257 For SimPowerSystems, see also Ode23t Making changes, 30 Selecting, 86, 87 With noise, 184, 186 Sources, library, 7 Starting a simulation, 9 Viewing results, Scope block, 9 Zero crossing Blocks that detect, 11 Detection, 10 Interpolation, 10 Simulink blocks Absolute value, 22, 23 Band Limited White Noise, 184 Clock, from Sources library, 88 Comparing, using differences, 39 Constant, 88 Constant block, 4, 7 Control systems, time-based linearization, 83 Control systems, trigger-based linearization, 83 Cross product, 101 Data store, read, write and memory, 280

303

From File, 92 From Workspace, 92 Gain, 12, 23, 72 Vectorizing, 14 GoTo, 77 Hysteresis, 45 Integrator, 7, 82 Integrator notation, from Laplace operator, 41 Leaning Tower simulation, blocks for, 7 Lookup Interpolation, 87 Lookup Table, 87, 88 Dynamics, 87 Library, 87 Math Operations, 3 Matrix Concatenation, horizontal and vertical, 111 Multiplication, for matrix operations, 102 Multiport switch, 76 Mux, 45, 253 Polynomial, curve fitting, 91 Prelookup, 87 Product, 12 Relational Operator block, 7 Scope, 13 Scope block, 7 Selector, 100 Sign, 23 Signal Builder, 249 Sine Input, 21, 46 State-space Continuous, 38, 52, 58 Discrete, 39 Stop block, 7 Subsystem, 21, 190 Sum, 12, 20, 88, 253 Sum, used as difference, 72 Transfer function, 42, 63 Trigonometric functions, 4 Unit Delay, 116 Zero-Pole-Gain, 42 Simulink data In MATLAB by default, 10

304

Simulink Models, see also NCS library Annotating, 15 Wide nonscalar lines, 16 Smoluchowski, Marian von And Brownian motion, 180 Solvers, see also Ode113:Ode23t; Simulink, see also Ode45 For Numerical Integration, see also Foucault pendulum In SimMechanics, 257 ODE suite in Simulink, 86 Setting step size, 259 Simulink implements MATLAB Solvers as standalone dlls, 30 Simulink ’s use of, 7 Stiff, 105 Using Different in a simulation, 31 Using Ode23t in SimPowerSystems, 249 Spacecraft rotation Model for, 109 Spaghetti Simulink models, 290 Specification development and capture In the system design process, 270 Spectral density function Definition, 194 Using correlation function, 183 State-Space And switch curves for lunar module, 285 State-Space model, see also Simulink blocks Calculating transfer function from, 42 For 1/f noise approximation, 196 For discrete time systems, 131 For pendulum, 38 For spring-mass-damper system, 58 Full state feedback and, 74 Getting Bode plot from, 65 Getting linear model for Lorenz attractor, 83 In SimPowerSystems, 246 Of two-room house, 209 Transfer function for, 42 Discrete systems using, 135 Using lti object in MATLAB, 67

Index

State flow, 282 Action language, 228 Adding events, 221 Adding inputs and outputs, 221 Heating control speci fication, 230 Home heating controller using, 225 Semantics, 218 Simple chart, 215 Using a GUI for inputs, 230, 233 Using the debugger, 223 State flow Executable Speci fication In the system design process, 283 Stochastic processes, see Random processes Subsystems Annotating, 81 Completing top down design, 281 Converting Fahrenheit to Celsius and back, 45 Counter, for RCS jet timing (lunar module), 280 Creating, 21 Creating a library for, 81 Creating a matrix in, 111 Deriving rate, for, 75 Empty, for top-down design, 271 Heating controller, executable specifications, 226 House dynamics, heating system model, 45 Interacting with a GUI, 233 Lunar module reaction jet control, 279 Mask dialog for parameter inputs, 78 Masked, 76, 81 documentation, 247 drawing an icon on, 247 for white noise, 184 Noise simulation, discrete and continuous time, 191 Nonlinear resistor, in SimPowerSystems, 247 Padding a buffer, signal processing, 162

45283438 Numerical Computing With Simulink Volume I

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