Engineering - Mechatronics

8/12/2019 Engineering - Mechatronics

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Copyright 1993-2001, Hugh Jack

Engineer On a Disk

Overview: This note set is part of a larger collection of materials available at http://claymore.engi-neer.gvsu.edu. You are welcome to use the material under the license provided at http://clay-

more.engineer.gvsu.edu/eod/global/copyrght.html. As always any feedback you can providewill be welcomed.

Copyright 1993-2001, Hugh Jack

email: [email protected]: (616) 771-6755fax: (616) 336-7215


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PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 134EMBEDDED CONTROLLERS........................................................................................... 135

TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 135CONTROLLER DESIGN EXAMPLE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 135PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 135

DISCRETE SENSORS......................................................................................................... 136INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 136SENSOR WIRING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 136CONTACT DETECTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 143PROXIMITY DETECTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 143PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 156

CONTINUOUS SENSORS.................................................................................................. 159INPUT ISSUES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 159SENSOR TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 163ANGULAR POSITION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 166LINEAR POSITION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 170

VELOCITY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 173ACCELERATION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 174FORCE/MOMENT - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 176FLOW RATE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 181TEMPERATURE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 182SOUND - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 185LIGHT INTENSITY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 186PRESSURE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 187PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 187REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 188

ACTUATORS ...................................................................................................................... 189

ACTUATOR TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 189DISCRETE ACTUATORS .................................................................................................. 189INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 189TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 190PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 195

CONTINUOUS ACTUATORS ........................................................................................... 196ACTUATOR CONTROL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 196PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 202

PROGRAMMABLE LOGIC CONTROLLERS.................................................................. 203BASIC PLCs - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 203A SIMPLE EXAMPLE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 212

PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 212PLC CONNECTION ............................................................................................................ 215

SWITCHED INPUTS AND OUTPUTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - 215PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 226

PLC OPERATION ............................................................................................................... 232PLC ORGANIZATION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 232PLC STATUS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 233MEMORY TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 234


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SOFTWARE BASED PLCS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 234PROGRAMMING STANDARDS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 234PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 236

SWITCHING LOGIC........................................................................................................... 238BOOLEAN ALGEBRA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 238DISCRETE LOGIC - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 240SIMPLIFYING BOOLEAN EQUATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - 247ADDITIONAL TOPICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 250DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 253PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 254

NUMBERING ...................................................................................................................... 271INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 271DATA VALUES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 272DATA CHARACTERIZATION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 281PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 283

EVENT BASED LOGIC...................................................................................................... 286

INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 286TIMERS, COUNTERS, FLIP-FLOPS, LATCHES - - - - - - - - - - - - - - - - - - - - - 286PROGRAM DESIGN METHODS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 303DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 306PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 314

SEQUENTIAL LOGIC DESIGN......................................................................................... 321SCRIPTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 321FLOW CHARTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 325STATE BASED MODELLING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 335PARALLEL PROCESS FLOWCHARTS - - - - - - - - - - - - - - - - - - - - - - - - - - - 360SEQUENTIAL LOGIC CIRCUITS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 370

PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 372ADVANCED LADDER LOGIC FUNCTIONS .................................................................. 406ADDRESSING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 406INSTRUCTION TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 412DESIGN TECHNIQUES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 440DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 442FUNCTION REFERENCE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 445PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 471

PLC PROGRAMMING........................................................................................................ 478PROGRAMMING STANDARDS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 478PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 480

STRUCTURED TEXT PROGRAMMING.......................................................................... 481INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 481THE LANGUAGE - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 481PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 494

INSTRUCTION LIST PROGRAMMING........................................................................... 495INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 495PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 498

FUNCTION BLOCK PROGRAMMING ............................................................................ 500


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INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 500PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 501

ANALOG INPUTS AND OUTPUTS.................................................................................. 503ANALOG INPUTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 503ANALOG OUTPUTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 511DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 514PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 515

CONTINUOUS CONTROL................................................................................................. 521CONTROLLING CONTINUOUS SYSTEMS - - - - - - - - - - - - - - - - - - - - - - - - 522CONTROLLING DISCRETE SYSTEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - 523CONTROL SYSTEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 523DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 527PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 527

PLC DATA COMMUNICATION....................................................................................... 528COMPUTER COMMUNICATIONS CATEGORIES - - - - - - - - - - - - - - - - - - - 528THE HISTORY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 529

WITH PLCs - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 529SERIAL COMMUNICATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 530PARALLEL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 541NETWORKS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 541BUS TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 554DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 562PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 563

HUMAN MACHINE INTERFACES (HMI) ....................................................................... 566INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 566HMI/MMI DESIGN - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 567DESIGN CASES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 567

PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 568DESIGNING LARGE SYSTEMS ....................................................................................... 569PROGRAMMING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 569DOCUMENTATION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 570PLC PROGRAM DESIGN FORMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 571PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 578

IMPLEMENTATION........................................................................................................... 579ELECTRICAL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 579SAFETY - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 591

PROCESS MODELLING .................................................................................................... 594REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 597

PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 597SELECTING A PLC ............................................................................................................ 599

SPECIAL I/O MODULES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 603PLC PROGRAMMING LANGUAGES - - - - - - - - - - - - - - - - - - - - - - - - - - - - 607ISSUES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 607PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 607

PLC REFERENCES ............................................................................................................. 609SUPPLIERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 609


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PROFESSIONAL INTEREST GROUPS - - - - - - - - - - - - - - - - - - - - - - - - - - - 610PLC/DISCRETE CONTROL REFERENCES - - - - - - - - - - - - - - - - - - - - - - - - 610

USING THE OMRON DEMO PACKAGE......................................................................... 614OVERVIEW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 614REFERENCE GUIDE FOR OMRON PLC DEMO SOFTWARE - - - - - - - - - - - 618

INDUSTRIAL ROBOTICS.................................................................................................. 619INTRODUCTION - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 619ROBOT TYPES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 632ROBOT APPLICATIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 663END OF ARM TOOLING (EOAT) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 666ADVANCED TOPICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 675PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 676

ROBOTIC PATH PLANNING METHODS........................................................................ 690INTRODUCTION: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 690GENERAL REQUIREMENTS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 696SETUP EVALUATION CRITERIA - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 704

METHOD EVALUATION CRITERIA - - - - - - - - - - - - - - - - - - - - - - - - - - - - 709IMPLEMENTATION EVALUATION CRITERIA - - - - - - - - - - - - - - - - - - - - 716OTHER AREAS OF INTEREST - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 719COMPARISONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 720CONCLUSIONS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 722APPENDIX A - OPTIMIZATION TECHNIQUES - - - - - - - - - - - - - - - - - - - - 722APPENDIX B - SPATIAL PLANNING - - - - - - - - - - - - - - - - - - - - - - - - - - - 725APPENDIX C - TRANSFORMED SPACE - - - - - - - - - - - - - - - - - - - - - - - - - 731APPENDIX D - FIELD METHODS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 738APPENDIX E - NEW AND ADVANCED TOPICS - - - - - - - - - - - - - - - - - - - 740REFERENCES: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 741

ROBOTIC MECHANISMS ................................................................................................. 746KINEMATICS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 746MECHANISMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 749ACTUATORS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 750PATH PLANNING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 752PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 758

MOTION PLANNING AND TRAJECTORY CONTROL................................................. 763TRAJECTORY CONTROL - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 763PATH PLANNING - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 767MOTION CONTROLLERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 771SPECIAL ISSUES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 774

PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 776MICROBOT OVERVIEW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 778CRS PLUS ROBOT OVERVIEW - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 778BASIC DEMONSTRATION STEPS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 779

CNC MACHINES ................................................................................................................ 781MACHINE AXES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 781NUMERICAL CONTROL (NC) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 781EXAMPLES OF EQUIPMENT - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 785


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PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 788CNC PROGRAMMING....................................................................................................... 789

G-CODES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 790APT - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 798PROPRIETARY NC CODES - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 801GRAPHICAL PART PROGRAMMING - - - - - - - - - - - - - - - - - - - - - - - - - - - - 802NC CUTTER PATHS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 803NC CONTROLLERS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 805PRACTICE PROBLEMS - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 805


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2. OVERVIEW

Mechatronics deals with the long ignored union of machanical and electrical systems.


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3. PROCESS CONTROL

To put it simply - we figure out how the process behaves naturally, we determine how we want itto behave, and we insert a controller to make it do what we want.

3.1 INTRODUCTION

CONTROL - Using artificial means to manipulate the world - with a particular goal.

Continuous - the values to be controlled change smoothly. e.g. the speed of a car

Discrete - The value to be controlled are easily described as on-off. e.g. the car motor is on-off(like basic pneumatics). NOTE: all systems are continuous (Except for Heisenbergs electrons)but they can be treated as discrete for simplicity.

Linear - Can be described with a simple differential equation (not a very accurate explanation).e.g. a car can be driving around a track and can pass same the same spot at a constant

velocity. But, the longer the car runs, the mass decreases, and it travels faster, but

requires less gas, .......... etc. Basically, the math gets tougher, and the problembecomes non-linear.

- This is the preferred starting point for simplicity, and a common approximation for realworld problems.

Non-Linear - Not Linear. This is how the world works, but is very complicated. Especially whentrying to do mathematical approximations. (Note: if the coefficients in a differential equationchange it is non-linear)

CONTROL

CONTINUOUS DISCRETE

LINEAR NON_LINEAR CONDITIONAL SEQUENTIAL

e.g. PIDe.g. MRAC

e.g. FUZZY LOGIC

BOOLEANTEMPORAL

e.g. TIMERS

e.g. COUNTERS

EVENT BASED

EXPERT SYSTEMS


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Temporal/sequential - the controller must not just keep track of things that changes, but it mustknow the time, or how long since something happened.

e.g.non-linear - as rocket approaches sun, gravity increases, so control must change.linear - We are driving the perfect car with no friction, with no drag, and can predict how it

will work perfectly.discrete - When I do this, that always happens! For example, when the power is turned

on, the press closes!

How to tell the difference? An elevator is the perfect example.

Discrete1. The elevator must move towards a floor when a button is pushed.

2. The elevator must open a door when it is at a floor.3. It must have the door closed before it moves.etc.

Linear (our approximation commonly used in industry) NOTE: each floor will have a cer-tain motor position, and we know what that position is.

1. If the desired position changes to a new value, accelerate quicklytowards the new position.

2. As the elevator approaches the correct position, slow down.Non-linear (not yet common in industry) e.g. Fuzzy logic, Neural Networks, Adaptive

Control1 Accelerate slowly to start.

2. Decelerate as you approach the final position.3. Allow faster motion while moving.4. Compensate for cable stretch, and changing spring constant, etc.

3.2 CONTROL SYSTEM CHARACTERISTICS

Control systems are present in most systems we see. Example - elevator height.

Some systems are naturally unstable and tend to self destruct. Example - balanced broom.

Other systems are self regulating, and tend to some stable state - these are good candidates foropen loop control. Example - city water tank

Other systems need some sort of regulation mechanisms added, these are called closed loop sys-tems. Example - car cruise control


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If a system is simple we will say that it is linear. Example - perfect elevator

A system can be subject to complicating factors that make it difficult/impossible to model math-ematically. Example - elevator in tall building with stretchy cable

If a system has switched states it can be described as discrete. Example - buttons to call elevator,open/close doors

Temporal/sequential systems can change over time, this requires sequential control. Example -washing machine

3.3 CONTROLLER TYPES

- PLCs

- embedded controls- PCs- hardwired

3.4 PROCESS DIAGRAMS AND SYMBOLS

3.5 PRACTICE QUESTIONS

1. Consider a heater in your house, how can different forms of control be applied to this problem?

2. Why is discrete control so popular when continuous control allows more precision?


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4. DISCRETE CONTROLLER DESIGN

In controller design we try to meet some objective. Typical objectives include,- positioning - for moving from position to position, where the setpoint is changed sud-

denlydeadbeat controlfirst order

- tracking - for following paths where setpoints are constantly changingfirst order

- disturbance resistant - unexpected variations can be compensated forstep response

- multi-stage - for complex systems that need small errorsfeed forward disturbance - can reduce effects of disturbancesfeed forward command - to follow complex motionscascade - a controller that examines intermediate steps of a process

typical constraints to be developed are,- stability - does not diverge over time- stability - sampling period is short enough (no-aliasing)- realizable - does not refer to future values

4.1 POSITIONING CONTROLLERS

In these systems we will focus on how the output relates to the input.

These techniques work best with simple systems that require a jump from one setpoint toanother.

The general procedure to design a controller is,1. Model the process (i.e., find Gp)2. Determine the control objective (i.e. select the desired system response Gdr)3. Substitute these into the equation below and solve for the controller transfer function,

4. Check the equation for practicality and stability5. Develop the control equation for the computer

GdrB( )

cn

rn-----

Gc B( )Gp B( )1 Gc B( )Gp B( )+----------------------------------------= =

Gc B( ) cn

rn-----

1

Gp B( )---------------

GdrB( )1 GdrB( )--------------------------

= =


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6. Write the computer program

Try to derive the expression given above,

4.1.1 Dead Beat Control

It is possible (for a first order process) to match the output function to the input function in onestep.

The desired function is,

Given the process model below, design a deadbeat controller,

GdrB( )

cn

rn-----

Gc B( )Gp B( )1 Gc B( )Gp B( )+----------------------------------------= =

cn rn 1=

GdrB( ) cn

rn----- B= = T

desired posn

actual posn


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Trying to eliminate the error in one step can require an extremely high (power) gain. When thisgain is excessive we may use a less powerful controller.

4.1.2 Programming Examples

The following programs are examples of methods for programming using a few common lan-guages.

These examples are for the Computer Boards DAS08-AOM board (The boards we used in thelab with Labview).

Please note that these programs have not been debugged.

Gp B( ) B

1 0.5B--------------------=


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4.1.2.1 - BASIC

A simple example of the deadbeat controller is given below. This can be converted to many ofthe modern version of basic that have appeared. Comments have been added to help clarifythe operation

4.1.2.2 - C

The example program below should implement the deadbeat controller example given in thissection,

10 REM a control loop for the deadbeat controller example for 12 bit input and output15 REM Written by H.Jack, July 17, 1997 - Not debugged20 BASE = 330H; REM the base address for the card20 T = 0.1; REM control system time step40 M_LAST = 0; REM the current output is zero50 E_LAST = 0; REM the current error60 INPUT ENTER SETPOINT [0 - 4095] , R; REM get a setpoint70 DO ; REM start the control loop80 OUT BASE+2, 0; REM start the a to d conversion

90 WAIT T; REM an operating system specific timing function100 C_HIGH = INP BASE+0; REM read current position most significant 8 bits110 C_LOW = INP BASE+1; REM read current position least significant 4 bits120 C_NOW = C_LOW/16 + C_HIGH*16130 E_NOW = R - C_NOW; REM calculate the error140 M_NOW = M_LAST + E_NOW - 0.5*E_LAST; REM the controller equation150 IF (M_NOW < 0) THEN M_NOW = 0; REM prevent out of range case160 IF (M_NOW > 4095) THEN M_NOW = 4095; REM prevent out of range case170 M_HIGH = INT(M_NOW/256) ; REM find the 4 high bits of the 12 bit word180 M_LOW = M_NOW - 256*M_HIGH; REM isolate the low 8 bits190 OUT BASE+4, (M_LOW); REM set 8 low bits

200 OUT BASE+5, (M_HIGH); REM set 4 high bits210 M_LAST = M_NOW; E_LAST = E_NOW; REM current values become last values220 LOOP WHILE(ABS(E) < 0.001); REM continue looping until the error is small


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4.1.2.4 - 6811 Assembler

{ A deadbeat controller for a Computer Boards DAS08-AOM card using 16 bit IO }{ Written by H.Jack, July 17, 1997 - Not debugged }

program deadbeat(input, output);

varbase = 0x330, { Base address for card }r, c_now, { Set up variables for equations }e_now, e_last=0, { error values }m_now, m_last=0: integer; { control values }

T = 0.1: float; { set the system time step }begin

writeln(Enter the Setpoint [0 - 4095]);readln(r); { Input a setpoint }

while (abs(e_last) < 1) do{ loop until the error is small }begin

outp(base+2, 0); { start the a to d conversion process }---------sleep(T); { an operating system specific timing function }---------c_now := inp(base+0)/16 + inp(base+1)*16; {calculate the current value }-

---e_now := r - c_now; { calculate current error }m_now := m_last + e_now - 0.5 * e_last; { the control equation }if (m_now < 0) then

m_now := 0; { keep the output in 0-4095 range }

if (m_now > 4095) thenm_now := 4095;

outp(base+4, m_now & 255); { output 8 LSB } -------outp(base+5, m_now>>8); { output 4 MSB }---------m_last := m_now; e_last := e_now; { make current values last }

end;end.


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The example program below should implement the deadbeat controller example given in thissection for a 6811 single chip microcontroller.

As an exercise, implement the controller on a Basic Stamp chip.


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4.1.3 First Order Response

The deadbeat controller can call for extremely high gains, but this requires a high level of powernot commonly found in engineered systems.

When we want to slow down the system response so that it occurs over a number of controllersteps, we can replace the deadbeat (one time step) function with a gradual first order reduction(exponential decay) over a number of time steps.

This gradual response will use a time constant to produce the rate of response (recall that the dif-ference between input and output is reduced 63% for each duration of the time constant).

(note: this program will be similar to all others)


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The general form of this relationship is seen below,

Consider the example for the deadbeat controller,

Gdr

cn

rn-----

B 1 e

T

---

1 Be

T

---

--------------------------= =

d

dt-----

c c+ r=

First we relate the output c to the input r (Note the first order differential equation).

Next, we convert this to a discrete equation using the methods (source - table) seen before,

(from transform table)

desired

actual

T


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First we write the process model,

Gc B( ) cn

rn-----

1

Gp B( )---------------

GdrB( )1 GdrB( )--------------------------

1 0.5BB

--------------------

B 1 e

T

---

1 Be

T

---

--------------------------

1

B 1 e

T

---

1 Be

T

---

--------------------------

-----------------------------------------

= = =

Assuming the process model given below, find the general form for the control equation,

Gp B( ) B

1 0.5B--------------------=

Next, we use this and the desired response to find the controller equation,

Gc B( )

1 e

T

---

1 0.5B( )

1 Be

T

---

B 1 e

T

---

----------------------------------------------------------------

1 e

T

---

0.5 1 e

T

---

B

1[ ] e

T

---

1 e

T

---

+ B+

-------------------------------------------------------------------------= =

After some work.....this should lead the control difference equation,

mn mn 1 +=


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Assume the response time constant should be 1.0 seconds, and the sampling time is 0.2 seconds,develop the final form of the controller.

4.2 TRACKING

In tracking controllers we assume that the setpoint will be moving. As a result we will focus ona desired error function, instead of an overall transfer function.

To do this we must first develop a function that relates the elements of the control loop, to thedesired error response,

1.0=

T 0.2=


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Specific controllers can be developed using the relationships used for the positioning controllers.

As an example tracking control would be needed in your hand for curved letters, positioningwould result in block letters.

4.2.1 Minimum Error

The method behind this controller is very similar to the deadbeat controller, we try to eliminatethe error in a single step,

Gde

en

rn-----

1

1 GcGp+-----------------------= =

en rn cn= cn GcGpen=

First we look at the basic relationships for a feedback control system,

Next we relate in the desired error response, and rearrange to separate the controller,

e n rn GcGpen=

e n 1 GcGp+( ) rn=

1 GcGp+( )1

Gde---------=

GcGp 1Gde--------- 1=

Gc1

Gp------

1 Gde

Gde------------------

=

Note: The error is now treated as a position,and we want to control the error instead.A small error will correspond to a small

difference between the changing inputand the resulting output.


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Find a controller for the transfer function given below, assuming a ramped input with a samplingtime of 0.2 seconds. Use a maximum error of 0.02.

The desired error function should have a finite value at only one point in time,

en Sn 0B0

ATB 0B2

0B3

0B4 + + + + + +( )=

en SnATB=

Gc1

Gp------

1 Gde

Gde------------------

1Gp------

rn ATBSn

ATBSn----------------------------

= =

This has to be manipulated into the desired transfer function,

Gde

en

rn-----

SnATB

rn-----------------= =

At this point we would be given or assume an input forcing function, and use it also.

(A is the largest error value)

Gp B( ) B

1 0.5B--------------------= r t( ) 0.2t=


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4.3 DISTURBANCE RESISTANT

In real systems we expect that certain events will occur that are not part of our system model.

In this case we assume that the system control is happening as expected, and we add in a new

disturbance input.

The block diagram below shows one of these systems, with a disturbance injected between thecontroller and the process.

Notice that in the above form we are reducing the problem by finding differences (basically apartial differential solution) which will be a good approximation when the disturbance is nottoo fast or large.

We can develop an equation for the controller, based on the desired system response to the dis-

turbance,

enrn

cn

mn

dn

+

+

+

-

Gp

Gc

If we neglect the effects of the setpoint, we can rearrange the loop as shown below,

Gc

Gpcn+

-

dnmn

mn

Note: we can do thisbecause the sys-

tem is linear.


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The closed form expression can be calculated by replacing the desired transfer function,

These systems are often called regulators, and can be used when a system is subject to unex-pected noise. Examples of possible applications would include plumbing systems, electricalpower supplies, etc.

4.3.1 Disturbance Minimization

We can use an approach similar to the deadbeat controller, but we still need to know the type ofdisturbance expected to develop a controller.

Consider a case where the disturbance is a step function

Gddcndn

---------Gp

1 GcGp+-----------------------= =

Gc G

p

Gdd

GpGdd----------------------=

Gddcndn

---------=

Gc

Gp Gdd

GpGdd

----------------------

Gpcndn

---------

Gpcn

dn---------

----------------------dnGp cn

Gpcn---------------------------= = =

cn Sn 0B0

AB1

0B2

0B3

0B4 + + + + +( ) ABSn= =


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We can examine the previous controller for stability as well,

Gp B( ) B

1 0.5B--------------------=

Given the example transfer function and input disturbance,

dn DB0

DB1

DB2

DB3 + + + +( )Sn

1

1 B------------

Sn= =

Next we assume

Gc

dnGp cn

Gpcn---------------------------

1

1 B------------

SnB

1 0.5B--------------------

5BSn

B

1 0.5B

-------------------- 5BSn

---------------------------------------------------------------------1 5 1 0.5B( ) 1 B( )

5B 1 B( )-------------------------------------------------------= = =

cn Sn 0B0

5B1

0B2

0B3

0B4 + + + + +( ) 5BSn= =

Gc

dnGp cn

Gpcn---------------------------

1

1 B------------

SnB

1 0.5B--------------------

5BSn

B

1 0.5B

-------------------- 5BSn

---------------------------------------------------------------------1 5 1 0.5B( ) 1 B( )

5B 1 B( )-------------------------------------------------------= = =

Gc1 5 1 0.5B( ) 1 B( )

5B 1 B( )-------------------------------------------------------

1 5 7.5B 2.5B2

+

5B 5B2

-------------------------------------------------= =

mn

en------

1 5 7.5B 2.5B2

+

5B 5B2

-------------------------------------------------=

Now we put this in discrete equation form,

5mn 1 5mn 2 4en 7.5en 1 2.5en 2+=

5mn 1 5mn 2 4en 7.5en 1 2.5en 2+=

mn mn 1 0.8en 1+ 1.5en 0.5en 1+=

Note: this controller has a reference to an n+1 error term. This term refers to anerror value that is in the future. So.... unless you have a time machine, this con-troller is not REALIZABLE.

And these are used to calculate the controller equation


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If we have a first order system we only need to have a sample time that is shorter than the systemtime constant.

4.4 MULTI-CONTROLLER SYSTEMS

It can sometimes be helpful to have more than one controller in a system.

In feedforward controllers we have one controller to deal with an input value, and a second tocontrol error.

In cascade control we break the control into segments, that take advantage of readings within the

process.

4.4.1 Disturbance Feedforward

When a system has a disturbance that can be measured we can add a controller that specificallycompensates.

The figure below shows a feedforward disturbance controller,

First we get the characteristic equation from the denominator of the controller transfer function,

5B 5B2

0=

Next we find the roots of this equation,

B 1=

B 0=

In this case one root is less than 1, guaranteeing that the system will be unstable. The otherroot with a value of one would lead to a system that is critically stable.


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In this case F(B) is a function of how the disturbance actually effects the system. But Gf isthe controller to compensate. First, we need a model of F(B). Next the controller Gf is,

The error controller is designed as if the disturbance is not present.

This system is well suited to systems with large measurable disturbances. For example we couldmeasure cutting force, and adapt an axis controller on a CNC machine.

4.4.2 Command Feedforward

By separating out error control, and process modeling functions we can make a system that ismore accurate, with reduced steady state error.

This technique may be used when we have a process model, and the input will not be changing

suddenly (The feedback error controller will tend to overcompensate).

We do this using a system of the form below,

rn

en mn

dn

fn

m'n cnGp

Gf

F B( )

Gc+

-

+

+ +

+

In Computer

Gf F B( )=


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The relevant transfer functions are,

The feedback error controller can be selected to reduce the errors. The typical design techniqueis, to design for minimal error or disturbance.

Develop the system transfer function (given before) for a command feed forward controller

rn

en

fn

mn mn cn

Gf

Gc Gp+

+

+-

G B( ) cn

rn----- 1

Gf

Gc------+

GcGp1 GcGp+-----------------------

= =

The system transfer function is,

The feedforward controller is best defined as, a perfect match when there is no error,

Gf Gp( )1

=

en mn 0= =

cn rnGfGp=

and the input and output are identical,

cn rn=


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4.4.3 Cascade

By adding extra feedback loops we can add levels of robustness to a control system.

In this case the control loops appear to be nested.

Consider a cascade control system for disturbances,

cn

rn----- =

+

-

+

-

+

+

dn

rnGp

rnGc2Gc1


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Now consider a cascade controller for a multistage process, such as a sewage treatment plantwith sequential tanks with varying flow rates. Here the disturbance would be the sewage flow-ing into the first tank. the two process blocks are the treatment tanks. The output is waterreturned to rivers The controllers vary the flow rates by looking at parameters for both tanks.

To do design/analysis for these controllers we do an analysis of the innermost feedback loop.This then becomes a process, and we do design/analysis at the next higher loop.

Select controllers for the system pictured below,

+

-

+

-

++

rn

dn

Gc1Gc2

Gp1Gp2

rn

Hf


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+

-

+

-

++

dn

rn

5B

1 B------------

rnGc2Gc1

1. Design for Gc1 as if the disturbance and Gc2 are not present2. Design for disturbance minimization with Gc2


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4.5 SAMPLE TIME

In deadbeat type controllers we do not typically see mentions of the sample period, but in mostpractical systems we see a mention of the sample time T.

We can use the system transfer function to calculate the acceptable range of controller time. Inall cases the lower sample period is 0, but for practical reasons we may want to make thislarger.

A shorter sample time requires a faster computer, and steals cycle time from other processes.

The general procedure is,1. Find the characteristic equation of the overall system transfer function.2. Determine the roots (symbolically) of the equation.3. Select values of T that ensure that all of the roots have a value of 1 or more.4. Select a sample time that is less than the longest time.

Consider the example below,

G B( ) 4TB 2

B2

TB 2T2

+-----------------------------------=

Find the sample time T required for stability,

B2

TB 2T2

+ 0=

First, we find the characteristic equation, and solve for the roots, when the roots are

B T( ) T( )2 4 1( ) 2T2( )

2 1( )--------------------------------------------------------------------

T

2--- 1 1 8+( ) T

2--- 1 3( ) 1= = = =

T 1 0.5,=

Based on these results we can now determine the stable region.


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4.6 SUMMARY

The following is a map of techniques that show typical uses of the mathematical technique cov-ered so far.

realsystem

differentialequations

differenceequations

estimatedvalues

realizability

transferfunction

controllerequation

blockdiagram

desiredbehavior

designtechniques

system transferfunction

pick sampletime

stabilityanalysis

inputfunction

outputfunction

input functionsfrom data

input functionequation output

value table

error functions

steady state

b-shifttable

b-shift

algebra

differenceequation

B-shift

table

partial fractionsor long division

transferfunction

feedbackequation

transferfunction

finalvaluetheory

characteristicequation

transferfunction

tablecalculations


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4.7 PRACTICE PROBLEMS

1. a) We are developing a freight elevator control system. The first major task is developing aheight positioning controller. The elevator is moved to a new floor by issuing a step functionfor the new floor height. The transfer function below relates the voltage supplied to the eleva-

tor motor to the height of the elevator. Develop a controller transfer function that has a firstorder response, and draw a complete block diagram.

b) Develop the discrete equation for the controller if the time constant for the first order responseis 5 seconds and the sampling time is 1 second.

2. c) Write a computer program (in the computer language of your choice) that implements thecontroller in question 1. Assume the input and output values are set using the two functionsbelow.

3. Redraw the following system and add a feedforward controller. Develop the function used forthe feed forward controller.

4. The control system below will be used for positioning the height of an elevator.

GpB 1+

1 0.5B--------------------=

INPUT() - This function will return the elevator height in feet (floating point)OUTPUT(value) - This function will set the output voltage at value

+

_

10 B( )

B 1 B( )2------------------------B 5+

B 1-------------

B 2+


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a) Find a controller transfer function that gives a first order response for a time step of T=1 and atime constant of 3.

b) Develop the discrete equation and determine if the controller is realizable.

c) Develop a transfer function for the final system and determine if the system is stable.

4. We want to design a control system to minimize the effects of disturbance. Given a step input of

magnitude 1, we are willing to accept a maximum error of 0.5 for one time step.

a) Find a controller transfer function.

b) Develop the discrete equation for the system and determine if the controller is realizable?

1 B

B------------Gc

rn cn+

-

ANS.

a) Gc0.283B

2

1 2B B2

+----------------------------=

b) realizable

c) G0.283B

2

1 0.717B--------------------------= stable

Gp10B

1 B------------=

ANS.

a) Gc10 0.5 1 B( )2

5B 1 B( )---------------------------------------=

b) not realizable mn mn 19.5

5-------en 1+

1

5---en

0.5

5-------en 1+ +=


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5. DISCRETE SYSTEMS

When dealing with computers we will sample data values from the real world. These sampledvalues can then be used to estimate how a system will behave.

The term discrete refers to the use of single sampled values, instead of a continuous functions.You will see that the difference is between weighted sums and integrals.

5.1 DISCRETE SYSTEM MODELLING WITH EQUATIONS

We can write a differential equation, and then manipulate it to put in terms of time steps oflength T

In review consider how we can approximate derivatives using two/three points on a line.

T - Sample Period

sampledvalues


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Consider the example,

T T

yi 1

yi 1+

ti

yi

ti 1 ti 1+

y t( )

y ti( )ti 1

ti

yi yi 1+2--------------------- ti ti 1( )

T

2--- yi yi 1+( )=

d

dt-----y ti( )

yi yi 1

ti ti 1---------------------

yi 1+ yi

ti 1+ ti---------------------

1T--- yi yi 1( )

1

T--- yi 1+ yi( )= = =

d

dt-----

2y ti( )

1

T--- yi 1+ yi( )

1

T--- yi yi 1( )

T-----------------------------------------------------------------

yi 1+ yi 1 2yi+

T2

-----------------------------------------=

If we read the flow rate of oil in a pipeline once every hour for three hours. The read-ings in order are 1003, 1007, 1010. What is the integral and first and second deriva-tives for the values?


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5.1.1 Getting a Discrete Equation

First, consider the example of a simple differential equation,

We can continue the example and use this equation to simulate the system. Here the x valuesare given, and the first y value is assumed to be 0. Assume T=0.5 and K=0.2.

d

dt-----

y t( ) Kx t( )=

y T( ) Kx t( ) td

T

Kx t( ) td

0

Kx t( ) td0

T

+ y0 Kx t( ) td0

T

+= = =

We can solve the differential equation at time T, assuming that the slope is constant and known,

The equation above is developed for a time T after zero. We can manipulate it so that itwill be valid for any point in time,

y T( ) y0 K x0( )T+=

yn yn 1 K xn 1( )T+=

yn yn 1 K xn 1( )T=

y T( ) y0 K x0( )T+=

Now we can assume that the velocity x(t) is constant,

ASIDE: The approximations above canbe shown using a slope approxi-mated with two points on a curve.

for this particular differentialy x0t=equation we can assume

say d

dt-----A V=


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Find the discrete equation for the differential equation below. Then find values over time.

We can expand this model by also including a disturbance. This can be used to model random

i

01

23456789101112

13.....

xi

00

11111100000

0.....

yi

00

0.10.20.30.40.50.60.70.70.70.70.7

0.7.....

0.2(xi)0.5

00

0.10.10.10.10.10.100000

0.....

d

dt-----v

1

M-----F=


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noise, or changes in system loading, found in all systems.

5.1.2 First Order System Example

The water tank below has a small outlet, and left alone the fluid level in the tank will drop untilempty. There is a valve controlled flow of fluids into the tank to raise the height of the fluid.

d

dt-----

y t( ) Kx t( ) d t( )+=

Go back to the original differential equation, and add a disturbance term d that

y T( ) Kx t( ) td

T

d t( ) td

T

+ Kx t( ) td

0

Kx t( ) td0

T

d t( ) td0

T

+ += =

yn yn 1 K xn 1( )T dn 1 T+ +=

yn yn 1 K xn 1( )T dn 1 T+=

y T( ) y0 K x0( )T d0T+ +=

appears for a single sample,

d t( ) td

0

0=

assume

h

water flow out

water flow in

water tank

valve

qi

qoA = Area of tank (i.e., fluid surface)


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page 43

As long as the fluid levels in the tank are normal, the inlet and outlet flow rates are independent,We can model them both with simple differential equations,

The water flow rate out of the tank is approximately a function of the hydrostaticpressure caused by the water level in the tank,

qo Koh=

The inlet valve can be (very roughly) approximated with a simple equation,

qi Ki=

We can now relate these components to find the fluid height in the tank,

V qi qo Ad

dt-----h Ki Koh= = =

Ad

dt-----h Ki Koh=

h Ce

t

A

Ko------

-----------

Ki

Ko------

+ Ce

tKo

A-------- Ki

Ko------

0+= =

The general form of solution for this differential equation for h is, (we will review later)

h0 Ce0 Ki

Ko------

0+=

We can find the value of coefficient C by setting time to zero,

C h0Ki

Ko------

0=

h h0

Ki

Ko------

0 e

tKo

A------

KiKo------

0+ h0 et

Ko

A------

0Ki

Ko------

et

Ko

A------

KiKo------

+

+= =

The final form of this equation can be constructed, by separating out height and valve angle,

hT h0 eT

Ko

A------

0

Ki

Ko------

eT

Ko

A------ KiKo------

+ +=

Put in terms of a time step, to prepare to make discrete,

hn hn 1 eT

Ko

A------

n 1Ki

Ko------

eT

Ko

A------

Ki

Ko------

+

+=

Then to a discrete form, as a difference equation,

transient + steady state

A

d

dt-----h Ki Koh=


44/805

page 44

This difference equation can then be used to predict fluid height. If we change the valve posi-tion, this will also be reflected in the calculated values.

egr450.0.mcd

Now try varying the input valve angle,

Use the parameters given to calculate the height of the water in the tank over time.

initial height =1 m

surface area = 2 m2

Ki = 1 liter/min/degreeKo = 2 liter/min/mvalve angle = 20 degreesT = 30 sec.


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page 45

5.1.3 Second Order System Example

Consider the second order example below,

Using all of the conditions outlined before, find out what happens when the valveangle is changed to 10 degrees after 10 seconds.


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page 46

try the following problem.

d

dt-----

2y Ax=

First, the basic equation,

Second we begin to integrate, starting at time zero, to time T to get difference equations

d

dt-----

yT Ax0T d

dt-----

y0+=

d

dt-----

2yT Ax0=

yT Ax0T

2

2-----

d

dt-----

y0T y0+ +=

d

dt-----

ynd

dt-----

yn 1 Axn 1 T=

yn yn 1 Axn 1T

2

2-----

d

dt-----

yn 1 T+=

These equations can be solved in two steps, or we can recombine them to get a singledifferential equation. We do this by rearranging the last difference equation, andthen we substitute this into the previous difference equation.

T d

dt-----

yn 1 yn yn 1 Axn 1T

2

2-----=

d

dt-----

yn1

T---y

n 1+

1

T---y

n Axn

T

2---=

d

dt-----

yn

d

dt-----

yn 1 Axn 1 T

1

T---yn 1+

1

T---yn A xnT

2---

1

T---yn

1

T---yn 1 Axn 1

T

2---+ += =

2yn yn 1 yn 1++ + AxnT

2

2----- Axn 1 T

2 Axn 1

T2

2-----+=

This can then be shifted back in time one step to make it more useful.

2yn 1 yn 2 yn+ + Axn 1T

2

2----- Axn 2

T2

2-----=

yn 2yn 1 yn 2 Axn 1T

2

2----- Axn 2

T2

2-----+=

(1)

(2)

(2)

(1)


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page 47

5.1.4 Example of Dead (Delay) Time

In a real system there is a distance between an actuator, and a sensor. This physical distanceresults to a lag between when we actuate something, and when the sensor sees a result.

Assume we have a simple process where after a change in x there is a delay of m time stepsbefore the proportional change occurs in y. We can write the equation as,


Calculate the position of a 5 Kg mass if a force of 2N is applied for 5 seconds and thenremoved. Try the calculations using a 1 second and a 2 second period.

yn Axn m=


48/805


49/805

page 49

We have already dealt with deriving an equation for the process. In this case it was the valvetank combination discussed before. By itself the tank is an open loop system, we set the valveangle and hope for a liquid level.

Next, we need to find an equation for the controller. This equation can be highly dependent uponthe control method to be used. If we are to use a computer it is best to have a simple equation,

as shown below, (NOTE: the form of the equation, and the values of the coefficients changethe nature of the control problem).

The controllers (equations) that follow will be put in the above form. These controllers can alsobe used individually, or combined to get more complex properties.

Keep in mind that the typical objective of a control system is to minimize the error between theinput and output. Another common goal is to do this as quickly, or efficiently as possible. Oneconstraint we must observe is that the system should not become unstable.

Controller

Feedback

Process hactualhdesired e

To make notations more generic we will associate stock variables for desired and actualpositions, as well as error, and controller output. We will also add the discretetimestep subscripts used before,

rn hdesired=

cn hactual=

+

-

en rn cn=

mn =

(valve/tank)

mn mn 1 K0en K1en 1 K2en 2 K3en 3 K4en 4 + + + + + +=

mn mn 1 K0en K1en 1 K2en 2 K3en 3 K4en 4 + + + + +=The basic difference equation is,

The output oriented form is shown below,


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page 50

5.2.1 A Proportional Controller

One of the simplest controllers is the proportional control,

The magnitude of K will determine how fast the system responds. If the value is too large thesystem will oscillate and/or become unstable (i.e. flood or go empty). If too small the systemerror will be very large. (ie, the tank will never reach the right height.)

This type of controller will always have a small error between the actual and desired values.

For the water tank (from before) add a controller, and try varying K values.

m Ke=

mn Ken=

This can be put in discrete form easily,

The basic proportional equation is,

mn mn 1 Ken Ken 1=

Finally this can be put into the general form seen before,

m n mn 1 Ken Ken 1+=

Note: This form will bemore useful whendealing with morecomplex controllers.

hn hn 1 e

TKo

A

------

n 1

Ki

Ko------ e

TKo

A

------ K

iKo------ +

+=


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page 51

We could implement this controller using the Basic stamp chip. (Note: not a full implementa-tion)

-

+

GND +5V

23

21

20

19

10K

1uF10K

+5V

220

10K

10K

10Kmotor

potentiometer

+5V

e_last=0 set previous error to zerok=2 set multiplierr=100 desired water heightm_last=0 last control output valueloop:high 15 discharge the capacitorpause 1 wait for 1msrctime 15,1,c measure charge timee_now=r-c calculate the errorm_now=m_last+k*e_now-k*e_last the controller equation

e_last=e_nowm_last=m_nowpwm 14,m,100 set the output voltagegoto loop


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page 52

5.2.2 Integral Control

Integral controllers tend to respond slowly at first, but over a long period of time they tend to

eliminate errors.

The integral controller is based on a simple integration.

If the constant K is small, the longer term error will slowly drop off. If K is large the long termerror will be reduced quickly. Too large a K value will result in a signal that grows out of con-trol.

Try controlling the water tank with the I controller,

First we write the general equation for the integrator,

m K e t d0

t

=In discrete form this becomes,

mn K Tej

j 1=

n

=

Putting this into the difference equation yields,

mn mn 1 K Tej

j 1=

n

K Tejj 1=

n 1

KTen= =

finally,

m n mn 1 KTen+=

ASIDE

integral

sum

Note: now the value of the equa-tion form become obvious.


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page 53

5.2.3 Differential Control

When there is a sudden change in the system the differential controller will be able to compen-

sate. But in terms of long term effects the controller will allow huge steady state errors.

The control equation can be derived as,

hn hn 1 eT

Ko

A------

n 1Ki

Ko------

eT

Ko

A------

KiKo------

+

+=

First, the basic equation,

m K d

dt-----

e=

Next, this can be written with the backwards difference equation,

mn K

en en 1

T----------------------- =We then apply the difference equation, and put in final form,

mn mn 1 Ken en 1

T-----------------------

Ken 1 en 2

T-----------------------------

KT----en 2

K

T----en 1

K

T----en 2+= =

m n mn 1Kd

T------en 2

Kd

T------en 1

Kd

T------en 2+ +=


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page 54

This larger the value of K the faster this controller will compensate for a change in the system.

Try controlling the water tank level with the D controller,

5.2.4 Proportional, Integral, Derivative (PID) Control

The functions of the individual proportional, integral and derivative controllers are complemen-tary. When combined we get a system that responds quickly to change (derivative), generallytrack required positions (proportional), and will eventually reduce errors (integral).

To get this we combine the expressions from the three individual controllers. Subscripts will beadded to distinguish the K gain values for each controller.

hn hn 1 eT

Ko

A------

n 1Ki

Ko------

eT

Ko

A------

KiKo------

+

+=


55/805

page 55

Quite often the three constants are made the same, giving us the simpler equation below.

This controller now allows us to vary the three different gains, and as a result we will change theperformance of the system.

Consider the examples below,

Recall the equations for the three controllers

proportional

integral

derivative

mn mn 1 Kpen Kpen 1+=

mn mn 1 KiTe n+=

mn mn 1

Kd

T------en 2

Kd

T------en 1

Kd

T------en 2+ +=

The three differences are added for a new difference equation,

mn mn 1 Kpen Kpen 1( ) KiTen( ) Kd

T------en 2

Kd

T------en 1

Kd

T------en 2+

+ +=

Rearranging results in,

mn mn 1 en Kp KiT Kd

T------+ + en 1 Kp 2

KdT

------ en 2KdT

------ + + +=

mn mn 1 en K KT K

T----+ +

en 1 K 2K

T----

en 2K

T----

+ + +=

mn mn 1 K en 1 T1

T---+ +

en 1 12

T---

en 21

T---

+ + +=


56/805

page 56

egr450.1.mcd

5.3 BLOCK DIAGRAMS AND TRANSFER FUNCTIONS

Block diagrams are the primary tool for showing process and control system models.

In a block diagram each block has one input and one output.

A transfer function provides functions that are a ratio of block output to block input.

Consider the block diagram seen before, but with transfer functions (ratio of output to input)shown for the controller and process.

Given the water tank from before (with the given variables), try a variety of differentvalues for the PID gains to see how the system responds.


57/805

page 57

Both of the transfer functions are ratios of inputs to outputs,

These transfer functions (ratios) are rarely a simple multiplication, and so we need to use analternate representation called a transform.

In this case we will use a backshift transform, hence the B in the representation.

equations for the block diagram below,

Controller Process

rn cn+

-

en mn Gp B( )Gc B( )

Gc B( ) mn

en------= Gp B( )

cn

mn------=

Controller Process

rn cn+

-

en mnGp B( )Gc B( )


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page 58

Recall the techniques for block diagram manipulation.

5.3.1 The Backward-Shift B Operator

An operator is a special part of an equation - one example is the (d/dt) of calculus.

The basic definition of the backshift operator B is,

If we apply this operator to an equation seen before,

This form then allow some very useful techniques that we will see later.

Apply the B transform to the PID expression found before,

yn yn=

yn 1 Byn=

yn 2 B2yn=

yn m

Bmy

n

=

mn mn 1 K0en K1en 1 K2en 2 K3en 3 K4en 4 + + + + +=

mn Bmn K0en K1Ben K2B2en K3B

3en K4B

4en + + + + +=

mn

en------

K0 K1B K2B2

K3B3

K4B4 + + + + +

1 B--------------------------------------------------------------------------------------------- Gc B( )= =


59/805

page 59

Apply the B transform to the valve and water tank found before,

mn mn 1 en Kp KiTKd

T------+ +

en 1 Kp 2Kd

T------

en 2Kd

T------

+ + +=

hn hn 1 eT

Ko

A------

n 1Ki

Ko------

eT

Ko

A------

KiKo------

+

+=


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page 60

5.3.2 Reducing Block Diagrams

A useful method for this form is the ratio of the actual output to the desired output, (c/r)

If we are planning on applying the PID controller to the water tank example from before we getthe following block diagram, (Note: the block diagram is overkill in this application)

Controller Process

rn cn+

-

en mn Gp B( )Gc B( )

Gc B( ) mn

en------= Gp B( )

cn

mn------=

cn Gp B( )mn Gp B( )Gc B( )en Gp B( )Gc B( ) rn cn( )= = =

en rn cn=

cn rnGp B( )Gc B( ) cnGp B( )Gc B( )=

cn cnGp B( )Gc B( )+ rnGp B( )Gc B( )=

cn

rn----- Gp B( )Gc B( )

1 Gp B( )Gc B( )+---------------------------------------- G B( )= = Closed Loop Transfer Function

hactual

hdesired

Kp KiTKd

T------+ +

B Kp 2Kd

T------

B2 Kd

T------

+ +

1 B------------------------------------------------------------------------------------------------------------

BKi

Ko------

eT

Ko

A------

KiKo------

+

1 B eT Ko

A------

---------------------------------------------------------e

+-


61/805

page 61

We can find the closed loop transfer function for this process and controller,

For the problem below, use the PID values, with the valve/tank parameters used previously tofind the system response to a step input to 20 at time zero, and 10 at time 10. (hint: convert theclosed loop equation back to a difference equation)

G B( ) =


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page 62

5.3.3 Back-Shift Transform Table

The general application of the table is,1. Develop the differential equation.2. Look for the equation in the table. Sometimes the equation will have to be rearranged to

match the form in the table. If it is not in the table, derive the transfer function thelong way.

3. Select the appropriate transfer from the right column, and substitute in values and vari-ables.

Kp 0.1=

Kd 0.02=

Kp 0.2=


63/805

page 63

A few of the transforms are given below,

Time Domain Back-Shift Domain

y Kx t D( )= y B( )x B( )----------- KB

INT DT----

=

d

dt-----

y Kx= y B( )x B( )-----------

KBT

1 B------------=

ddt-----

y y+ Kx=y B( )x B( )-----------

KB 1 e

T

---

1 Be

T

---

------------------------------=

ddt-----

2y Kx= y B( )x B( )----------- KT

2B 1 B+( )2 1 B( )2

--------------------------------=

ddt-----

2y ddt-----

y+ Kx=y B( )x B( )-----------

K b1B b2B2

+( )

1 a1B a2B2

-------------------------------------=

a1 1 e

T

---

+= a2 e

T

---

=

b1 T 1 eT

---

=

b2 Te

T

---

1 eT

---

+=


64/805

page 64

Try finding the transfer function for the system below,

Time Domain Back-Shift Domain

1

n2

------

d

dt-----

2y 2

n------

ddt-----

y y+ + Kx= y B( )x B( )-----------

K b1B b2B2

+( )

1 a1B a2B2

-------------------------------------=

1>a1 e

T

1----

e

T

2----

+= a2 e

T

1----

T

2----

=

b1 1

e

T

1----

2--------- e

T

2----

1---------1

1-----

1

2-----

------------------------+= b2 e

T

1---- T

2----

e

T

2----

2--------- e

T

1----

1---------1

1-----

1

2-----

------------------------+=

AND/OR

12( ) d

dt-----

2y 1 2+( ) d

dt-----

y y+ + Kx=

Case 1:

1=a1 2e

Tn= a2 e

2Tn=

b1 1 e Tn

Tne Tn

= b2 e Tn

e Tn

Tn 1+( )=

Case 2:

1= 0 sec

vdesired(t) = 100 for t >= 0 sec

t(sec)

0

Input type Time function Laplace function

STEP f t( ) Au t( )= f s( ) As---=

RAMP f t( ) Atu t( )= f s( ) A

s2

----=

SINUSOID f t( ) A t( )u t( )sin= f s( ) A2

s2 2+-----------------=

etc......

PULSE f t( ) A u t( ) u t t1( )( )= f s( ) =


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page 102

Therefore to continue the car example, lets assume the input below,

vdesired t( ) 100= t 0sec

vdesired s( ) L vdesired t( )[ ] 100s---------= =

Next, lets use the input, and transfer function to find the output of the system.

vactual

vactual

vdesired-----------------

vdesired=

vactual

s2

Kd( ) s Kc( ) Ki+ +

s2 M10------ K+

d s Kc( ) Ki+ +

-------------------------------------------------------------

100

s

---------

=

To go further, some numbers will be selected for the values.

Kd= 10000

Kc= 10000Ki= 1000

M = 1000

vactuals

210000( ) s 10000( ) 1000+ +

s2

10100( ) s 10000( ) 1000+ +------------------------------------------------------------------------

100

s---------

=


103/805


104/805

page 104

A s10

2

1.01----------

s2

s 0.1+ +

s s 0.113+( ) s 0.877+( )---------------------------------------------------------

s 0lim

102

1.01----------

0.1

0.113( ) 0.877( )-------------------------------------

= =

A 99.9=

B10

2

1.01----------

s2

s 0.1+ +

s s 0.113+( ) s 0.877+( )---------------------------------------------------------

s 0.113+( )

s 0.113lim=

B 102

1.01----------

0.113( )2 0.113( ) 0.1+ +0.113( ) 0.113 0.877+( )

-----------------------------------------------------------------

0.264= =

C 10

2

1.01---------- s

2

s 0.1+ +s s 0.113+( ) s 0.877+( )--------------------------------------------------------- s 0.877+( )s 0.877

lim=

C 102

1.01----------

0.877( )2 0.877( ) 0.1+ +0.877( ) 0.877 0.113+( )

-----------------------------------------------------------------

1.16= =

vactual99.9

s----------

0.264

s 0.113+----------------------

1.16

s 0.877+----------------------+=


105/805

page 105

7.1.5 System Response

Next we use a list of forward/inverse transforms to replace the terms in thepartial fraction expansion.

f t( ) f s( )

A A

s---

At A

s2

----

Ae t A

s +------------

A t( )sin A

s2 2+

-----------------

e nt nt 1

2( )sin

n 1 2

s2

2ns n2

+ +---------------------------------------- fo r 1


106/805

page 106

There are two very common systems assumed - first and second order.

First order systems are very simple, as is shown below.

7.1.6 A Motor Control System Example

Condsider the example of a DC servo motor controlled by a computer. The purpose of the con-

G s( ) As B+------------= A first order system, and a typical response toa stepped input.

G s( ) A

s2

2ns n2

+ +----------------------------------------=

A second order system, and a typical response toa stepped input.

n

Natural frequency of system, approximate frequency ofcontrol system oscillations.

Damping factor of system. If < 1 underdamped, and system willoscillate. If =1 critically damped. If < 1 overdamped, and neverany oscillation (more like a first order system). As damping factorapproaches 0, the first peak becomes infinite in height.


107/805

page 107

troller is to position the motor. The system below shows a reasonable control system arrange-ment. Some elements such as power supplies and commons for voltages are omitted for clarity.

This system can then be redrawn with a block diagram.

The block diagram can now be filled out with actual values for the components. Do this below.

+5V -5V

5K potentiometer

12Vdc motor

shafts are coupled

-+

LM675op-amp

2.2K

1K

PCI-1200dataaquisitioncard

from

NationalInstruments

Computer Running Labview

-+

desired positionvoltage Vd

gain K

X

desiredpositionvoltage +

-

potentiometer

gain K op-amp motor shaftVd


108/805

page 108

Given values:- desired potentiometer voltage- gain

For the op-amp:

For the potentiometer:

For the motor:

For the shaft:

- assume that the potentiometer has a range of 10 turns

- use the differential equation below

- it turns angular velocity into position

- use the speed curve below from rest when 10V is applied

d

dt-----

K2

JR------

+ KJR------

Vs=

1s 2s 3s

1400 RPM


109/805

page 109

Convert the block diagram into a transfer function for the entire system.


110/805

page 110

Pick a value of the gain K to give a system performance with the damping factor = 1.0.

7.1.7 System Error

We typically will be interested in system error and feedback error.


111/805

page 111

Consider a simple negative feedback system with various inputs,

re c

b

G

H

+-

System error,

Feedback error,

D r c=

D r b=

G s( ) Kp

s------= H s( ) 1=

c

r-- G

1 GH+------------------

Kp

s Kp+---------------= =

r A=

First, a step input,

Given,

c AKp

s Kp+---------------

=

D r c A AKp

s Kp+---------------

AKp AKp( ) s A( )+

s Kp+-------------------------------------------------

As

s Kp+---------------= = = =

ess s As

s Kp+---------------

s 0lim 0= =

Next, try a ramp input,


112/805


113/805


114/805

page 114

Consider the previous example, the transfer function for the whole system was found, but thenonly the denominator was used to determine stability. So in general we do not need to find thetransfer function for the whole system.

1

K1

s---

+

-

G s( ) K

s

----= H s( ) 1=

Note: This controller has adjustable gain. After this designis built we must anticipate that all values of K will beused. It is our responsibility to make sure that none of

the possible K values will lead to instability.

First, we must develop a transfer function for the entire control system.

GS s( ) G s( )1 G s( )H s( )+---------------------------------

K

s----

1 K

s----

1( )+---------------------------

K

s K+------------= = =

s K+ 0= K

0123etc..

root

Next, we use the characteristic equation of the denominator to find the roots asthe value of K varies. These can then be plotted on a complex plane. Note:the value of gain K is normally found from 0 to +infinity.

j

K K 0=

Note: because all of the roots for all values of K are real negative this system will

always be stable, and it will always tend to have a damped response. The large thevalue of K, the more stable the system becomes.


115/805

page 115


GS s( ) G s( )1 G s( )H s( )+---------------------------------=

Consider the general form for a negative feedback system.

Note: two assumptions that are not often clearlystated are that we are assuming that the control

system is a negative feedback controller, andthat when not given the feedback gain is 1.

1 G s( )H s( )+ 0=

The system response is a function of the denominator, and its roots.

It is typical, (especially in textbook problems) to be given only G(s) or G(s)H(s).

G s

( )H s

( )

K s z0+( ) s z1+( ) s zm+( )

s p0+( ) s p1+( ) s pn+( )-------------------------------------------------------------------=

The transfer function values will often be supplied in a pole zero form.


116/805

page 116

7.2.1 Approximate Plotting Techniques

The basic procedure for creating root locus plots is,1. write the characteristic equation. This includes writing the poles and zeros of the equa-

tion.

G s( ) K

s2

3s 2+ +--------------------------= H s( ) 1=

First, find the characteristic equation,. and an equation for the roots,

Given the system elements (you should assume negative feedback),

1 K

s2

3s 2+ +--------------------------

1( )+ 0=

s2

3s 2 K+ + + 0=

roots3 9 4 2 K+( )

2------------------------------------------------ 1.5

1 4K

2--------------------= =

Next, find values for the roots and plot the values,

K

0123

root

0-1-2-3

j

******CALCULATE AND PUT IN NUMBERS


117/805

page 117

2. count the number of poles and zeros. The difference (n-m) will indicate how many root

loci lines end at infinity (used later).3. plot the root loci that lie on the real axis. Points will be on a root locus line if they have

an odd number of poles and zeros to the right. Draw these lines in.4. determine the asymptotes for the loci that go to infinity using the formula below. Next,

determine where the asymptotes intersect the real axis using the second formula.Finally, draw the asymptotes on the graph.

5. the breakaway and breakin points are found next. Breakaway points exist between twopoles on the real axis. Breakin points exist between zeros. to calculate these thefollowing polynomial must be solved. The resulting roots are the breakin/breakoutpoints.

6. Find the points where the loci lines intersect the imaginary axis. To do this substitutethe fourier frequency for the laplace variable, and solve for the frequencies. Plotthe asymptotic curves to pass through the imaginary axis at this point.

Consider the example in the previous section,

1 G s( )H s( )+ 1 Ks z1+( ) s z2+( ) s zm+( )s p1+( ) s p2+( ) s pn+( )

----------------------------------------------------------------+ 0= =

k( ) 180 2k 1+( )n m

-----------------------------------= k 0 n m 1,[ ]

p

1

p2

pn

+ + +( ) z1

z2

zm

+ + +( )n m--------------------------------------------------------------------------------------------=

d

ds-----A

B A

d

ds-----B

0=

B s z1+( ) s z2+( ) s zm+( )=A s p1+( ) s p2+( ) s pn+( )=

1 Kj z1+( ) j z2+( ) j zm+( )j p1+( ) j p2+( ) j pn+( )---------------------------------------------------------------------------+ 0=


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Plot the root locus diagram for the function below,

Step 5: (find the breakout points for the roots)

B s2

3s 2+ +=

d

ds-----B 2s 3+=

A 1=

d

ds-----A 0=

A d

ds-----B

B dds-----A

0=

1 2s 3+( ) s2 3s 2+ +( ) 0( ) 0=

2s 3+ 0=

s 1.5=

j

-1-2

-1.5

Note: because the loci do not intersect the imaginary axis, we know the system will bestable, so step 6 is not necessary, but we it will be done for illustrative purposes.

Step 6: (find the imaginary intercepts)

1 G s( )H s( )+ 0=

1 K1

s2

3s 2+ +--------------------------+ 0=

s2 3s 2 K+ + + 0=

j( )2 3 j( ) 2 K+ + + 0=

2 3j 2 K+ + + 0=

3j 3j( )2

4 2 K( )2

--------------------------------------------------------------3j 9 8 4K+ +

2----------------------------------------------

3j 4K 12

-------------------------------= = =

2 3j( ) 2 K( )+ + 0=

In this case the frequency has an imaginary value. This means that there will be nofrequency that will intercept the imaginary axis.


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7.2.2 State Variable Control Systems

State variable matrices were introduced before. These can now be used to form a control system.

G s( )H s( ) K s 5+( )

s s2

4s 8+ +( )---------------------------------=


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8. Draw a detailed root locus diagram for the transfer function below. Be careful to specify anglesof departure, ranges for breakout/breakin points, and gains and frequency at stability limits.

9. Draw the root locus diagram for the system below. specify all points and values.

10. Draw the root locus diagram for the transfer function below,

11. Draw the root locus diagram for the transfer function below,

12. The block diagram below is for a motor position control system. The system has a propor-tional controller with a variable gain K.

G s

( )

2K s 0.5+( ) s2 2s 2+ +( )

s3 s 1+( ) s 2+( )-----------------------------------

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