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Comprehensive Electrical and Computer PE Power Exam Coverage Reference Manual
Power Reference Manual for the Electrical and Computer PE Exam Second Edition
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Power
Reference Manual for the Electrical and Computer PE Exam Second Edition
John A. Camara, PE
%
m The Power to Pass®
I www.ppi2pass.com
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POWER REFERENCE MANUAL F O R THE ELECTRICAL AND COMPUTER PE E X A M Second E dition Current printing of this edition:
1
P rin tin g H istory date Dec 2012 Mar 2014 Mar 2016
edition number
printing number
1 1
2 3
2
1
update Minor corrections. Minor corrections. New edition. Code updates. Additional content. Copyright update.
© 2016 Professional Publications, Inc. All rights reserved. All content is copyrighted by Professional Publications, Inc. (PPI). No part, either text or image, may be used for any purpose other than personal use. Reproduction, modification, storage in a retrieval system or retransmission, in any form or by any means, electronic, mechanical, or otherwise, for reasons other than personal use, without prior written permission from the publisher is strictly prohibited. For written permission, contact PPI at [email protected] pass.com. Printed in the United States of America. PPI 1250 Fifth Avenue Belmont, CA 94002 (650) 5939119 ppi2 pass.com ISBN: 9781591265023 Library of Congress Control Number: 2015954486 FEDCBA
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Topics Topic !:
Mathematics
Topic II:
Basic Theory
Topic 111:
Field Theory
Topic IV:
Circuit Theory
Topic V:
Generation
Topic VI:
Distribution
Topic VII:
System Analysis
Topic VIII:
Protection and Safety
Topic IX:
Machinery and Devices
Topic X:
Electronics
Topic XI:
Special Applications
Topic XII:
Measurement and Instrumentation
Topic XIII:
Electrical Materials
Topic XIV:
Codes and Standards
Topic XV:
Professional Practice
Topic XV!:
Support Material
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Where do I find practice problems to test what I’ve learned in this Reference Manual? The Power Reference Manual provides a knowledge base that will prepare you for the Electrical and Computer PE Power exam. But there’s no better way to exercise your skills than to practice solving problems. To simplify your preparation, please consider Power Practice Problems for the Electrical and Computer PE Exam. This publication provides you with more than 600 practice problems, each with a complete, stepbystep solution.
Power Practice Problems for the Electrical and Computer PE Exam may be obtained from P P I at p p i2 p a s s .c o m or from your favorite retailer.
Table of Contents Appendices Table of C ontents ................................ vii
Topic V: Generation
Preface ............................................................................ ix
Generation Systems......................................................331 ThreePhase Electricity and Power............................ 341 Batteries, Fuel Cells, and Power Supplies................ 351
Acknowledgments .........................................................xi Codes and References ............................................. xiii Introduction ................................................................... xv Topic I: Mathematics Systems of U n its............................................................11 Energy, Work, and P o w e r ........................................... 21 Engineering Drawing P ra ctice .................................... 31 Algebra............................................................................41 Linear Algebra................................................................ 51 V ectors............................................................................61 Trigonom etry................................................................ 71 Analytic G eom etry.......................................................81 Differential C alculus.................................................... 91 Integral Calculus..........................................................101 Differential Equations................................................ 111 Probability and Statistical Analysis of D a t a ...........121 Computer Mathematics.............................................. 131 Numerical A nalysis..................................................... 141 Advanced Engineering Mathematics.........................151
Topic ¥1: Distribution Power Distribution...................................................361 Power Transformers................................................... 371 Transmission L in e s...................................................381 The Smith Chart........................................................391
Topic VII: System Analysis Power System A n alysis...............................................401 Analysis of Control Systems........................................411
Topic VIII: Protection and Safety Protection and S a fe ty ................................................. 421
Topic IX: Machinery and Devices Rotating DC Machinery...............................................431 Rotating AC Machinery. .............................................441
Topic II: Basic Theory Electromagnetic Theory..............................................161 Electronic T h e o ry ....................................................... 171 Communication T heory..............................................181 Acoustic and Transducer T h e o ry ....................... .. .191
Topic III: Field Theory Electrostatics................................................................ 201 Electrostatic F ie ld s.....................................................211 M agnetostatics............................................................221 Magnetostatic F ields.................................................. 231 Electrodynamics......................................................... 241 Maxwell’s Equations.................................................. 251
Topic X: Electronics Electronics Fundamentals............................................ 451 Junction Transistors................................................... 461 Field Effect Transistors.............................................. 471 Electrical and Electronic D ev ices.............................. 481 Digital L og ic................................................................. 491
Topic XI: Special Applications Lightning Protection and G rounding....................... 501 Illumination................................................................... 511 Power System Management....................................... 521
Topic IV: Circuit Theory DC Circuit Fundamentals.........................................261 AC Circuit Fundamentals.........................................271 Transformers................................................................ 281 Linear Circuit A nalysis............................................. 291 Transient Analysis.......................................................301 Time R esponse........................................................... 311 Frequency Response.................................................... 321
Topic XII: Measurement and Instrumentation Measurement and Instrumentation............................531
Topic XIII: Electrical Materials Electrical M aterials..................................................... 541
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Topic XIV: Codes and Standards Biomedical Electrical Engineering..............................551 National Electrical C o d e ............................................561 National Electrical Safety C o d e ................................ 571
Topic XV: Professional Practice Engineering Economic A n alysis................................ 581 Engineering L a w .......................................................... 591 Engineering Ethics........................................................601 Electrical Engineering Frontiers................................ 611 Engineering Licensing in the United States............. 621
Topic XVI: Support Material Appendices.....................................................................A l In d e x ............................................................................... 11
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Appendices Table of Contents l.A
Conversion F a c t o r s ......................................................A l
29.A
l.B
Com mon SI Unit Conversion F a c t o r s .................... A3
30.A
Resonant Circuit Form ulas................................... A32
8.A
Mensuration o f TwoDimensional A r e a s ...............A 7
38.A
Bare Aluminum Conductors, Steel Reinforced (A C S R )................................................................... A33
8.B
Mensuration o f ThreeDimensional Volumes . . . A9
10.A
Abbreviated Table o f Indefinite Integrals . . . . A 10
11. A Laplace T ran sform s................................................ A  11 12.A
Gamma Function V a lu e s ......................................A 13
15.B
Bessel Functions J0 and J i ................................... A 14
15.C
Properties o f Fourier Series for Periodic S ig n a ls ................................................................... A 15
15.D
Properties o f Fourier Transform for Aperiodic Signals................................................. A 16
15.E
Fourier Transform Pairs........................................ A 17
15.F
Properties o f Laplace Transform .........................A18
15.G
Laplace T ran sform s................................................... A 19
15.H
Properties o f DiscreteTime Fourier Series for Periodic S ig n a ls ............................................... A20
15.1
Properties o f DiscreteTime Fourier Transform for Aperiodic S ign als........................A 21
15.J
Properties o f ^Transform s.......................................A22
15.K
zTransform s................................................................ A23
17.A
39.A
Smith C h a r t ............................................................. A35
42.A
Guidance on Selection o f Protective Clothing and Other Personal Protective Equipment ( P P E ) ...............................................A36
45.A
Periodic Table o f the Elem ents............................A37
45.B
Semiconductor Symbols and Abbreviations . . . A38
48.A
Standard Zener D iodes........................................... A42
51.A
Units, Symbols, and Defining Equations for Fundamental Photometric and Radiometric Q uantities.............................................................. A44
51.B
Luminance Conversion F actors............................A46
51.C
Illuminance Conversion F a c t o r s ......................... A47
54.A
Periodic Table: Materials’ Properties Summary.................................................................A48
54.B
Table o f Relative A tom ic W e ig h t s .................... A49
56.A
W ire (and SheetMetal Gage) Diameters (Thickness) in In ch e s......................................... A50
Areas Under the Standard Normal Curve . . . . A12
15.A
Room Temperature Properites o f Silicon, Germanium, and Gallium Arsenide (at 3 0 0 K ) .................................................................A24
Tw oPort Parameter Conversions...................... A31
56.B
Conductor A m p a cities........................................... A52
56.C
Ambient Temperature Correction Factors. . . . A53
56.D
Ambient Temperature Correction Factors Conductor Properties.............................................. A55
Based on 40°C (10 4°F ).......................................A54
18.A
Electromagnetic S p e ctru m ...................................... A25
56.E
25.A
Comparison o f Electric and Magnetic E q u a tio n s.................................................................A26
56.F
Conductor A C Properties...................................... A56
56.G
Plug and Receptacle C on figu ration s..................A57
27.A
Impedance o f SeriesConnected Circuit E lem ents................................................................... A28
58.A
Standard Cash Flow F a c t o r s ...............................A60
58.B
Cash Flow Equivalent F a c t o r s ............................ A61
27.B
Impedance o f ParallelConnected Circuit E lem ents................................................................ A 29
Preface To pass the Electrical and Computer Power PE exam administered by the National Council of Examiners for Engineering and Surveying (NCEES), you need to understand both the concepts and the practical applica tions of electrical power engineering. Therefore, my goal for Power Reference Manual for the Electrical and Com puter PE Exam is to clearly explain the concepts out lined by the NCEES exam specifications, as well as present examples that show how these concepts are applied. For this second edition, I updated all material related to and dependent on the 2014 edition of the National Electrical Code (NEC) and the 2 0 1 2 edition of the National Electrical and Safety Code (NESC). I also wrote significant new content on the IEEE SO00 Stan dards Collection™ for Industrial and Commercial Power Systems, formerly known as the IEEE Color Books, as well as on workplace electrical safety, personal protective equipment (PPE), and transformer types. The existing chapter on the NEC was expanded to include new content on special occupancies, special equipment, special conditions, and communication sys tems. I’ve also included four new appendices: one on PPE, two on ambient temperature correction factors, and one on plug and receptacle configurations.
FOR WHOM THIS BOO The Power Reference Manual is written for you. It is written to be a reference at all times in your career, whether you are an undergraduate student or an experi enced engineer. If you are an exam candidate, the Power Reference Manual is an efficient resource for exploring exam topics systematically and exhaustively. If you are a practicing power engineer, it functions as a compre hensive reference, discussing all aspects of this field in a realistic manner, and incorporating the most common formulas and data. Finally, for the engineering student, it presents a thorough review of the fundamentals of electrical engineering.
The scope of this book is beyond that of the Power PE exam. When the exam is complete and you are a licensed electrical engineer, you will need a resource for electrical engineering information, either in your work or simply to satisfy the curiosity that makes us thinking humans. In both the exam and your career, I hope this book serves you well.
LOOK! NG F OR WARD TO THE FUTURE Future editions of this book will be very much shaped by what you and others want to see in it. I am braced for the influx of comments and suggestions from those read ers who ( 1 ) want more topics, ( 2 ) want more detail in existing topics, and (3) think that continuing to include some topics is just plain lame. Computer hardware and programming change quickly, as does the grand unified (atomic) theory of everything. I expect several chapters to generate some kind of debate. Should you find an error in this book, know that it is mine and that I regret it. Beyond that, I hope two things happen. First, please let me know about the error by using the error reporting form on the PPI website (ppi2pass.com/errata). Second, I hope you learn something from the error— I know I will! I appreciate suggestions for improvement, additional questions, and recommendations for expan sion so that new editions or similar texts will better meet the needs of future examinees. Read carefully, prepare well, and you will triumph over the exam! John A. Camara, PE
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Acknowledgm ents It is with enduring gratitude that I thank Michael R. Lindeburg, PE, who, with the sixth edition of the Elec trical Engineering Reference Manual, allowed me to realize my dream of authoring a significant engineering text. In taking the reins from the title’s previous author, the late Raymond Yarbrough, I was entrusted with the responsibility of shaping that text as the NCEES Elec trical and Computer PE exam and the engineering pro fession evolved. It is my hope that the Power Reference Manual for the Electrical and Computer PE Exam— which is based on the Electrical Engineering Reference Manual—will match the comprehensive and cohesive quality of the best of PPI’s family of engineering books, which I have admired for three decades. The second edition of the Power Reference Manual would not have been possible without the profession alism and meticulous attention to detail shown by Ellen Nordman, lead editor. In addition, my thanks go out to the wonderfully responsive team at PPI, including Serena Cavanaugh, associate acquisitions editor; Sam Webster, editorial manager; Nicole Evans, associate project manager; Thomas Bliss, David Chu, Sierra CirimelliLow, Tyler Hayes, Tracy Katz, Julia Lopez, Scott Marley, Heather Turbeville, and Ian A. Walker, copy editors; Tom Bergstrom, production associate and technical illustrator; Kate Hayes, production associate; Cathy Schrott, produc tion services manager, who keeps it all moving for ward; and Sarah Hubbard, director of Product Development and Implementation, whose guidance from the beginning has been most appreciated. Gregg Wagener, PE, reviewed each of the chapters of the first edition of this book, providing valuable insight, correcting numerous errors, and in general, teaching me a great deal along the way. Thank you to James Mirabile, PE, and Nanzhu Zhang, who technically reviewed the new content for the second edition. Michael R. Lindeburg, PE, provided Chapters 114 on mathematics, as well as the chapters on economic analy sis, law, ethics, and engineering licensure. He also pro vided sections of the illumination chapter. These contributions saved me considerable time, and I am grateful to use material of this quality. John Goularte, Jr., taught me firsthand the applications of biomedical engineering, and the realworld practicality of this topic is attributable to him.
Additionally, the following individuals contributed in a variety of ways, from thoughtful reviews, criticisms, and corrections, to suggestions for additional material. I am grateful for their efforts, which resulted in enhanced coverage of the material and enriched my knowledge of each topic. Thank you to Chris Iannello, PhD; Thomas P. Barrera, PhD; Robert G. Henriksen, PhD; Glenn G. Butcher, DCS; Chen J. Chang, MSE; and Aubrey Clements, BSEE. The knowledge expressed in this book represents years of training, instruction, and selfstudy in electrical, elec tronic, mechanical, nuclear, marine, space, and a variety of other types of engineering— all of which I find fasci nating. I would not understand any of it were it not for the teachers, instructors, mentors, family, and friends who have spent countless hours with me and from whom I have learned so much. A few of them include Mary Avila, Capt. C. E. Ellis, Claude Estes, Capt. Karl Hasslinger, Jerome Herbeck, Jack Hunnicutt, Ralph Loya, Harry Lynch, Harold Mackey, Michael O ’Neal, Mark Richwine, Charles Taylor, Abigail Thyarks, and Jim Triguerio. There are others who have touched my life in special ways: Jim, Marla, Lora, Todd, Tom, and Rick. There are many others, though here they will remain nameless, to whom I am indebted. Thanks. A very special thanks to my son, Jac Camara, whose annotated periodic table became the basis for the Elec trical Materials chapter. Additionally, our discussions about quantum mechanics and a variety of other topics helped stir the intellectual curiosity I thought was fading. Thanks also to my daughter, Cassiopeia, who has been a constant source of inspiration, challenge, and pride. Thanks to my lady, Shelly, for providing insight into the intricacies of the NEC and patiently missing long hours together during the update of this edition. Also, thanks to Wyatt Wade, who is teaching me the “new math.” Finally, thanks to my mother, Arlene, who made me a better person. John A. Camara, PE
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Codes and References The information that was used to write and update this book was based on the exam specifications at the time of publication. However, as with engineering practice itself, the PE exam is not always based on the most current codes or cuttingedge technology. Similarly, codes, stan dards, and regulations adopted by state and local agen cies often lag issuance by several years. It is likely that the codes that are most current, the codes that you use in practice, and the codes that are the basis of your exam will all be different. However, differences between code editions typically minimally affect the technical accuracy of this book, and the methodology presented remains valid. For more information about the variety of codes related to electrical engineering, refer to the following organizations and their websites. American National Standards Institute (ansi.org) Electronic Components Industry Association (ecianow.org) Federal Communications Commission (fcc.gov) Institute of Electrical and Electronics Engineers (ieee.org) International Organization for Standardization (iso.org) International Society of Automation (isa.org) National Electrical Manufacturers Association (nema.org) National Fire Protection Association (nfpa.org) The PPI website (p p i 2 pa ss.co m /e e fa q ) provides the dates and editions of the codes, standards, and regula tions on which NCEES has announced the PE exams are based. It is your responsibility to find out which codes are relevant to your exam. The minimum recommended library of materials to bring to the Electrical and Computer PE Power exam consists of this book, any applicable code books, a stan dard handbook of electrical engineering, and one or two textbooks that cover fundamental circuit theory (both electrical and electronic).
COOES AND STAND ARD S 47 CFR 73: Code of Federal Regulations, “Title 47— Telecommunication, Part 73— Radio Broadcast Rules,” 2014. Office of the Federal Register National Archives and Records Administration, Washington, DC. (Communications.) IEEE/ASTM SI 10: American National Standard for Metric Practice, 2010. ASTM International, West Conshohocken, PA. (Metric.)
IEEE Std 141 (IEEE Red Book): IEEE Recommended Practice for Electric Power Distribution for Industrial Plants, 1993. The Institute of Electrical and Electronics Engineers, Inc., New York, NY. IEEE Std 142 (IEEE Green Book): IEEE Recommended Practice for Grounding of Industrial and Commercial Power Systems, 2007. The Institute of Electrical and Electronics Engineers, Inc., New York, NY. IEEE Std 241 (IEEE Gray Book): IEEE Recommended Practice for Electrical Power Systems in Commercial Buildings, 1990. The Institute of Electrical and Elec tronics Engineers, Inc., New York, NY. IEEE Std 242 (IEEE Buff Book): IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems, 2001. The Institute of Electrical and Electronics Engineers, Inc., New York, NY. IEEE Std 399 (IEEE Brown Book): IEEE Recom mended Practice for Industrial and Commercial Power Systems Analysis, 1997. The Institute of Electrical and Electronics Engineers, Inc., New York, NY. IEEE Std 446 (IEEE Orange Book): IEEE Recom mended Practice for Emergency and Standby Power Systems for Industrial and Commercial Applications, 1995. The Institute of Electrical and Electronics Engi neers, Inc., New York, NY. IEEE Std 493 (IEEE Gold Book): IEEE Recommended Practice for the Design of Reliable Industrial and Com mercial Power Systems, 2007. The Institute of Electrical and Electronics Engineers, Inc., New York, NY. IEEE Std 551 (IEEE Violet Book): IEEE Recommended Practice for Calculating ShortCircuit Currents in Industrial and Commercial Power Systems, 2006. The Institute of Electrical and Electronics Engineers, Inc., New York, NY. IEEE Std 602 (IEEE White Book): IEEE Recom mended Practice for Electric Systems in Health Care Facilities, 2007. The Institute of Electrical and Elec tronics Engineers, Inc., New York, NY. IEEE Std 739 (IEEE Bronze Book): IEEE Recom mended Practice for Energy Management in Industrial and Commercial Facilities, 1995. The Institute of Elec trical and Electronics Engineers, Inc., New York, NY.
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IEEE Std 902 (IEEE Yellow Book): IEEE Guide for Maintenance, Operation, and Safety of Industrial and Commercial Power Systems, 1998. The Institute of Elec trical and Electronics Engineers, Inc., New York, NY. IEEE Std 1015 (IEEE Blue Book): IEEE Recommended Practice for Applying LowVoltage Circuit Breakers Used in Industrial and Commercial Power Systems, 2006. The Institute of Electrical and Electronics Engi neers, Inc., New York, NY. IEEE Std 1100 (IEEE Emerald Book): IEEE Recom mended Practice for Powering and Grounding Electronic Equipment, 2005. The Institute of Electrical and Elec tronics Engineers, Inc., New York, NY. NEC (NFPA 70): National Electrical Code, 2014. National Fire Protection Association, Quincy, MA. (Power.) NESC (IEEE C2): 2012 National Electrical Safety Code, 2 0 1 2 . The Institute of Electrical and Electronics Engi neers, Inc., New York, NY. (Power.)
HEFEHEHCES Anthony, Michael A. NEC Answers. New York, NY: McGrawHill. (National Electrical Code example appli cations textbook.) Bronzino, Joseph D. The Biomedical Engineering Hand book. Boca Raton, FL: CRC Press. (Electrical and elec tronics handbook.) Chemical Rubber Company. CRC Standard Mathemat ical Tables and Formulae. Boca Raton, FL: CRC Press. (General engineering reference.) Croft, Terrell and Wilford I. Summers. American Elec tricians Handbook. New York, NY: McGrawHill. (Power handbook.) Earley, Mark W. et al. National Electrical Code Hand book, 2014 ed. Quincy, MA: National Fire Protection Association. (Power handbook.) Fink, Donald G. and H. Wayne Beaty. Standard Hand book for Electrical Engineers. New York, NY: McGrawHill. (Power and electrical and electronics handbook.) Grainger, John J. and William D. Stevenson, Jr. Power System Analysis. New York, NY: McGrawHill. (Power textbook.) Horowitz, Stanley H. and Arun G. Phadke. Power Sys tem Relaying. Chichester, West Sussex: John Wiley & Sons, Ltd. (Power protection textbook.) Huray, Paul G. Maxwell’s Equations. Hoboken, NJ: John Wiley & Sons, Inc. (Power and electrical and electronics textbook.)
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Jaeger, Richard C. and Travis Blalock. Microelectronic Circuit Design. New York, NY: McGrawHill Educa tion. (Electronic fundamentals textbook.) Lee, William C.Y. Wireless and Cellular Telecommuni cations. New York, NY: McGrawHill. (Electrical and electronics handbook.) Marne, David J. National Electrical Safety Code (NESC) 2012 Handbook. New York, NY: McGrawHill Professional. (Power handbook.) McMillan, Gregory K. and Douglas Considine. Process/ Industrial Instruments and Controls Handbook. New York, NY: McGrawHill Professional. (Power and elec trical and electronics handbook.) Millman, Jacob and Arvin Grabel. Microelectronics. New York, NY: McGrawHill. (Electronic fundamentals textbook.) Mitra, Sanjit K. An Introduction to Digital and Analog Integrated Circuits and Applications. New York, NY: Harper & Row. (Digital circuit fundamentals textbook.) Parker, Sybil P., ed. McGrawHill Dictionary of Scien tific and Technical Terms. New York, NY: McGrawHill. (General engineering reference.) Plonus, Martin A. Applied Electromagnetics. New York, NY: McGrawHill. (Electromagnetic theory textbook.) Rea, Mark S., ed. The IESNA Lighting Handbook: Reference & Applications. New York, NY: Illuminating Engineering Society of North America. (Power handbook.) Shackelford, James F. and William Alexander, eds. CRC Materials Science and Engineering Handbook. Boca Raton, FL: CRC Press, Inc. (General engineering handbook.) Van Valkenburg, M.E. and B.K. Kinariwala. Linear Circuits. Englewood Cliffs, NJ: PrenticeHall. (A C /D C fundamentals textbook.) Wildi, Theodore and Perry R. McNeill. Electrical Power Technology. New York, NY: John Wiley & Sons (Power theory and application textbook.)
Introduction PART 1s HOW YOU CAM USE THIS BOOK
Q U IC K ST A R T If you are the type of reader who desires information distilled into concise bits and you are ready to begin your study, then read this section and get started! • This book takes complex topics and parses them into condensed, readable, understandable chunks. • The National Council of Examiners for Engineering and Surveying (NCEES) Electrical and Computer— Power PE exam consists of problems that require (on average) six minutes to solve. The exam problems focus on small portions of complex electrical engi neering topics. Your study should reflect this. •
Once the exam is over, this book will remain a useful reference you can use to review basic and complex power engineering concepts that you may encounter in your professional practice.
Because the chapters in this book are independent, you can focus your study on your weakest areas first. Review the table of contents to familiarize yourself with the book, then use Table 1 in this Introduction as a guide for focusing your study. This book’s index is a great starting place for searching for specific information and for planning your review of the topics covered by the NCEES specifications. The mathematics, theory, and specialty topics, while not directly listed in the NCEES requirements, supply fundamental background on topics that may be necessary for problem solving on the exam. The appendices are succinct references for common elec trical information. I have included the Professional chapters to refine your professional engineering knowl edge and round out your knowledge base. Commence study immediately. Six months is not an unreasonable time frame to prepare for the exam, though your learning curve will vary depending upon your background. Start with your weakest areas and work toward your strongest. Review every week. Work problems three to five times a week. The key to success on the exam is to work as many problems as possible. Browse PPI’s website for additional resources, such as this book’s companion, Power Practice Prob lems for the Electrical and Computer PE Exam, as well
as Power Practice Exams for the Electrical and Compu ter PE Exam, and PPI’s Exam Cafe. To minimize the time spent searching for oftenused formulas and data, prepare a onepage summary of all the important formulas and information in each subject area. You can then use these summaries during the exam instead of searching for them in this book. You may want to consider Power Quick Reference for the Electrical and Computer PE Exam, which PPI has pub lished for this type of use. This book focuses on the basics that are the building blocks of engineering knowledge. This fundamental knowledge is useful regardless of the periodic changes to the NCEES exam requirements. The topic names might change, but the knowledge necessary to solve the types of problems necessary to pass the exam remains the same. Understanding the basics allows you to solve complex problems, which can always be divided into simpler, individual concepts. A thorough review of the basics, terminology, and problems in this text and the Power Practice Problems companion book will enable you to solve the variety of problems that might appear on the PE exam or in your professional practice. Continue to study until the exam date. Enjoy the learn ing process and revel in the knowledge gained. If you are the type of reader who desires a more thor ough review and wants a deeper understanding of the PE exam before commencing your studies, read the rest of the Introduction.
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While selecting and preparing the content in this book, I followed the NCEES Electrical and Computer— Power PE exam specifications. I have endeavored throughout the text to include the exact terminology used by these specifications. While I wrote the book with the NCEES specifications in mind, the content of the Power Reference Manual is not limited to the specifications. This book also contains the fundamental building blocks you need to build a complete understanding of power engineering. In each chapter, I explore the mathematical, theoretical, and practical applications of the chapter’s topic. You can focus your study on any or all of these types of applications. The content of the Power Reference Manual and its companion book, Power Practice Problems for the
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Electrical and Computer PE Exam, are guided by these concepts. As a result, you will find relatively easy, mod erately difficult, and difficult examples in this book. The relatively easy examples and concepts are meant to remind you of the fundamentals. Studying these exam ples will help bring you closer to the level of knowledge you had readily available when college afforded you the opportunity to concentrate on these subjects daily. The moderately difficult examples can be solved in approxi mately six minutes and are meant to simulate the exam ination. The difficult examples and topics are meant to explore more complex areas of study and demonstrate how difficult problems might be broken into simpler pieces. Each chapter builds on the knowledge gained from pre vious chapters and will refer to them, or even repeat some concepts in a slightly different manner. Finally, references to standards or to specific idiosyncrasies described in the footnotes are meant to provide jumpingoff points for further study that will round off your knowledge. If you encounter these subjects on the exam, they will likely quote from the reference or standard and ask for an interpretation or calculation based on the quoted material.
•
Become intimately familiar with this book. This means knowing the order of the chapters, the approximate location of important figures and tables, what appendices are available, and so on.
• After reviewing a chapter of this book, solve each of the related problems in the companion book, Power Practice Problems for the Electrical and Computer PE Exam. Practice in problem solving is the key to success on the exam. It is a good idea to begin with problems that fall in your weakest areas and work toward your stronger areas. •
Use the answers to the practice problems to check your work. If your answer does not match the given solution, study the solution to determine why. The majority of this book uses SI units, except where customary U.S. (English) units are the standard. Some equations in the U.S. system use the term g/gc. For calculations at standard gravity, 1.00 is the numerical value of this fraction. Therefore, it is necessary to incorporate this quantity only in problems with a nonstandard gravity, or when being meticulous with units.
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Use Table 3 as your guide to focusing your exam preparation. Table 3 is found at the end of this Introduction.
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Practice by completing as many problems as possi ble. Remember that you will have an average of only six minutes per problem on the exam, so detailed design work is not possible. Rather, it is more likely that you will need to solve a portion of a design problem, complete a piece of an analysis puzzle, or demonstrate understanding of general requirements or standard practice.
FOR TH E PE CANDIDATE While you are preparing for the Power exam, the follow ing suggestions may help. •
Concentrate primarily on the fundamentals of elec trical engineering. Take time to review the principles of any problem you find difficult.
• Use the index extensively. Every significant term, law, theorem, and concept has been indexed. If you do not recognize a term, look for it in the index and read the related text for a more comprehensive over view. The index is the quickest way to find material in an exam situation, so you should familiarize your self with it before exam day. General terms or con cepts, which are covered in a variety of subject areas, may not be indexed, though the specifics of the con cepts will be listed.
If you are a practicing engineer or an engineering student and you have obtained this book as a generalreference handbook, the following suggestions and information may be of assistance.
•
•
Both the index and table of contents are extensive and are meant as the starting point for any informa tion desired. Every significant term, law, theorem, and concept has been indexed in every conceivable way— forwards and backwards.
•
Chapters are grouped into interrelated topics, each offering a broad overview of a given area of electrical engineering. If you want a quick introduction to a given topic, refer to that portion of the book.
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The mathematics chapters can be used as quick reviews when working any engineering problems.
Some subjects appear in more than one chapter. Use the index liberally and learn all there is to know about a particular subject.
• Know which subjects in this book are covered on the PE exam, and which are provided as support material. Chapters 1—25 and 58—62 are supportive and do not cover exam subjects. They do, however, provide background for the other chapters, enhance understanding of fundamentals, and are useful after the exam for quick reference. (The exam subjects known at the time of this printing are provided in Table 1 . These are subject to change; the most uptodate information can be found on P P I’s website at ppi2pass.com/eefaq.)
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IF YO U ARE A PRACTICING ENGINEER OR AM ENGINEERING STU D EN T
© The Advanced Engineering Mathematics chapter provides, in simple but accurate terms, the
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Table 1 Detailed Analysis of Tested Subjects
E L E C TR IC A L A N D C O M P U T E R — P O W E R E X A M IN A T IO N General Power Engineering (24 questions) M easurem ent and instru m entation (6 questions): instrument transformers; wattmeters; V O M metering; insulation testing; ground resistance testing S pecial applications (8 questions): lightning and surge protection; reliability; illumination engineering; demand and energy management/calculations; engineering economics Codes and standards (10 questions): N a tio n a l E le c tric a l Code; N a tio n a l E le c tric a l Safety Code; electric shock and burns Circuit Analysis (16 questions) A n a ly sis (9 questions): threephase circuit analysis; symmetrical components; per unit analysis; phasor diagrams D evices and pow er electronic circu its (7 questions): battery characteristics and ratings; power supplies; relays, switches, and PLCs; variablespeed drives Rotating Machines and Electromagnetic Devices (16 questions) R otatin g m achines (10 questions): synchronous machines; induction machines; generator/m otor applications; equivalent circuits; speedtorque characteristics; m otor starting Electrom agnetic devices (6 questions): transformers; reactors; testing Transmission and Distribution (24 questions) System analysis (10 questions): voltage drop; voltage regulation; power factor correction and voltage support; power quality; fault current analysis; grounding; transformer connections; transmission line models P o w e r system perform ance (6 questions): power flow; load sharing: parallel generators or transformers; power system stability P ro te ctio n (8 questions): overcurrent protection; protective relaying; protective devices; coordination
information necessary to understand the importance and use of various advanced mathematical concepts. This should allow you to more fully comprehend and appreciate professional journals or engineeringfocused presentations that use such concepts without resorting to a complete review of undergraduate math texts. • The Advanced Engineering Mathematics chapter is also meant as the starting point for delving more deeply into advanced mathematical concepts by pro viding the names and defining the variables of fre quently used mathematical formulas. This is done to help you through those situations where you know you need more mathematical information, but you do not know where to begin, what problems to ask, or what mathematical concept is involved. • The appendices are meant as quick references for common electrical information.
IF YOU AR E AM WSYFttJCTOH If you are teaching a review course for the PE exam without the benefit of recent, firsthand exam experience, you can use the material in this book as a guide to prepare your lectures. Your students should use the book as a textbook. You should emphasize the subjects in each chapter and avoid subjects that were omitted. Subjects that do not appear in this book have rarely, if ever, appeared on the PE exam. In an attempt to overprepare students, I suggest assign ing homework with practice problems that are more difficult and more varied than actual exam problems. It is often more efficient to cover several procedural steps in one practice problem than to ask simple “one liners” or definition problems.
“Capacity assignment” should be the goal in a successful review course. To do all the homework for some chap ters requires approximately 15 to 20 hours of prepara tion per week. One student may be able to put in 2 0 hours, while another may only be able to put in 1 0 hours; assigning the full 2 0 hours of homework ensures that each student is working to his or her capac ity. After the PE exam, all students will honestly say that they could not have prepared any more than they did in your course. Instead of individually grading students’ homework, permit students to make use of existing solutions to check their own work and learn approaches to the prob lems in their homework set. This will allow you to address special needs or questions written on the assign ments, rather than simply marking the work as right or wrong. Solved practice problems are available in the companion Power Practice Problems. Lecture coverage of some exam subjects is necessarily brief; other subjects are not covered at all. These omis sions are intentional; they are not the result of schedul ing. Why? First, time is not on your side in a review course. Second, some subjects rarely appear on the exam. Third, there are some subjects that are not wellreceived by the students, because they are not subjects that are pertinent to all students’ objectives. Unless you have six months in which to teach your PE review course, your students’ time can be better spent covering topics that apply to the majority of examinees. Table 1 lists the number of problems on the exam that will be devoted to a given topic. Using this and the depth of coverage in this book, you can decide which topics to focus on in your review course. In organizing the review course format outlined in Table 2, I have tried to order the subjects in a logical,
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Table 2 Typical P E Review Course Format session 1 2 3 4 5 6 7 8 9 10 11 12 13 14
subject covered
chapter(s)
Introduction: PE Exam; Mathematics; Professional Practice Basic Theory and DC Circuits Field Theory and A C Circuits Linear Circuits Field Theory and Transformers ThreePhase Electricity Power Generation and Distribution Power Transformers and Transmission Lines Power System Analysis Protection and Safety R otating D C Machinery R otating A C Machinery Electronics Codes and Standards
Introduction, 1619, 26 20, 21, 27 2931 22, 23, 28 34
progressive manner, keeping my eye on “playing the highprobability subjects.” Early in the course, I cover those subjects that everyone needs to learn. I leave the more specific subjects to the end. Use the formulas, illustrations, and tables of data in the text to minimize chalkboard time. Avoid proofs and emphasize problem solving. The more problems solved, the higher the chances of exam success. The practice problems in the companion book, Power Practice Problems for the Electrical and Computer PE Exam, range from quickanswer to indepth problems. The more indepth problems can be used as homework or timed problems. One possible review course format is outlined in Table 2 . The sessions are weekly, with the course lasting three and a half months. Most sessions could be covered in three hours of lecture. All the skipped chapters and any related practice problems are presented as floating assignments to be made up in the students’ “free time.” I strongly believe in exposing students to a realistic sample exam, but since the review course usually ends only a few days before the real PE exam, I hesitate to make students sit for several hours in the late evening to take an inclass exam. Instead, I recommend distribut ing and assigning a takehome sample exam at the first meeting of the review course. Caution your students not to peek at the sample exam prior to taking it. Looking at the sample exam or other wise using it to direct their review will produce unwar ranted specialization in subjects contained in the sample exam. For the sample exam to be realistic practice, it needs to mimic as much as possible the challenges of the actual exam. There are many ways to organize a PE review course, depending on your available time, budget, intended audience, facilities, and enthusiasm. However, all good course formats have the same result: students struggle with the workload during the course, and then they breeze through the exam afterward.
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33, 37, 40, 42 43 44 45 56,
35, 36 38, 39 41
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PART 2: EVERYTHING YOU EVER WANTED TO ICIiOW ABOUT THE PE EXAM WHAT 0S THE FORMAT O F THE PE EXAM? The NCEES PE exam in electrical and computer engi neering consists of two fourhour sessions separated by a onehour lunch period. The exam is an openbook exam. All problems are multiplechoice. Upon registering for the exam, examinees choose to work one of three exam mod ules: Computer Exam, Electrical and Electronics Exam, or Power Exam. Each exam consists of 80 problems (40 in the morning and 40 in the afternoon) that test knowl edge in the areas specified. Examinees must work all 80 problems in the exam of their choice. For more information on NCEES and the exams neces sary to become a licensed electrical engineer, refer to Chap. 62, Engineering Licensing in the United States.
WHAT SUBJECTS ARE ON THE PE EXAM? NCEES has published a description of subjects on the exam. Irrespective of the published exam structure, the exact number of problems that will appear in each sub ject area cannot be predicted reliably. There is no guarantee that any single subject will occur in any quantity. One of the reasons for this is that some of the problems span several disciplines. You might consider a pump selection problem to come from the subject of fluids, while NCEES might categorize it as engineering economics.
WHAT IS TME TYPICAL PROBLEM FORMAT? Almost all of the problems are standalone— that is, they are completely independent. However, NCEES allows that some sets of problems may start with a
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statement of a “situation” that will apply to (typically) two to five following problems. Such grouped problems are increasingly rare, however. Each of the problems will have four answer options labeled “A,” “B,” “C,” and “D.” If the answer options are numerical, they will be displayed in increasing value. One of the answer options is correct (or will be “most nearly correct,” as described in the following section). The remaining answer options are incorrect and may consist of one or more “logical distractors,” the term used by NCEES to designate incorrect options that look correct. NCEES intends the problems to be unrelated. Problems are independent or start with new given data. A mistake on one of the problems should not cause you to get a subsequent problem wrong. However, considerable time may be required to repeat previous calculations with a new set of given data.
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answer option that is most nearly what you cal culated, regardless of whether it is more or less than your calculated value. However, if the prob lem asks you to select a fuse or circuit breaker to protect against a calculated current or to size a beam to carry a load, you should select an answer option that will safely carry the current or load. Typically, this requires selecting a value that is closest to but larger than the current or load. The difference is significant. Suppose you are asked to calculate “most nearly” the current drawn by a given electrical load. Suppose, also, that the result of the calculation is 11 A. If the provided answer options were (A) 5 A, (B) 10 A, (C) 15 A, and (D) 20 A, you would go with answer option (B), because this is most nearly what you calculated. If, however, you were asked to select the breaker that provides the required overcurrent protection for this load, you would select answer option (C).
HOW MUCH “LOGIC UP” IS HEQUIHE© OH TH E EXAM ? Since the problems are multiplechoice in design, all required data will appear in the situation statement. Since the exam would be unfair if it was possible to arrive at an incorrect answer after making valid assump tions or using plausible data, you will not generally be required to come up with numerical data that might affect your success on the problem. There will also be superfluous information in the majority of problems.
W HAT PO ES “M OST H E A R LY ” REALLY MEAM? One of the more disquieting aspects of these problems is that the available answer options are seldom exact. Answer options generally have only two or three sig nificant digits. Exam problems ask, “Which answer option is most nearly the correct value?” or they instruct you to complete the sentence, “The value is approximately . . . ” A lot of selfconfidence is required to move on to the next problem when you do not find an exact match for the answer you calculated, or if you have had to split the difference because no available answer option is close. NCEES has described it like this: Many of the problems on NCEES exams require calculations to arrive at a numerical answer. Depending on the method of calculation used, it is very possible that examinees working correctly will arrive at a range of answers. The phrase “most nearly” is used to accommodate answers that have been derived correctly but that may be slightly different from the correct answer option given on the exam. You should use good engi neering judgment when selecting your choice of answer. For example, if the problem asks you to calculate an electrical current or determine the load on a beam, you should literally select the
HOW MUCH M ATH EM ATICS IS NEEOED FOR T H E EXAM? Generally, only simple algebra, trigonometry, and geom etry are needed on the PE exam. You will need to use the trigonometric, logarithm, square root, and similar buttons on your calculator. There is no need to use any other method for these functions. For functions that I recommend you program into your calculator, see the section “What about Calculators?” There are no pure mathematics problems (algebra, geometry, trigonometry, etc.) on the exam. However, you will need to apply your knowledge of these subjects to the exam problems. Except for simple quadratic equations, you will probably not need to find the roots of higherorder equations. Occasionally, it will be con venient to use the equationsolving capability of an advanced calculator. However, other solution methods will always exist. There is little or no use of calculus on the exam. Rarely, you may need to take a simple derivative to find a max imum or minimum of some function. Even rarer is the need to integrate to find an average or moment of inertia. There is essentially no need to solve differential equa tions. On the exam, there are simple linear circuits and components that can be solved or explained with differ ential equations. Often these applications, and others that use such mathematics, can be solved by other means. Basic statistical analysis of observed data may be necessary. Statistical calculations are generally limited to finding means, medians, standard deviations, var iances, percentiles, and confidence limits. The only population distribution you need to be familiar with is the normal curve. Probability, reliability, hypothesis testing, and statistical quality control are not explicit exam subjects.
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The Power exam is concerned with numerical answers, not with proofs or derivations. You will not be asked to prove or derive formulas. Occasionally, a calculation may require an iterative solution method. Generally, there is no need to complete more than two iterations. You will not need to program your calculator to obtain an “exact” answer, nor will you generally need to use complex numerical methods.
HOW ABOUT ENGINEERING ECONOMECS? For most of the early years of engineering licensing, problems on engineering economics appeared frequently on the exams. This is no longer the case. However, in its outline of exam subjects, NCEES notes: “Some problems may require knowledge of engineering economics.” What this means is that engineering economics might appear in several problems on the exam, or the subject might be totally absent. While the degree of engineering econom ics knowledge may have decreased somewhat, the basic economic concepts (e.g., time value of money, present worth, nonannual compounding, comparison of alterna tives, etc.) are still valid test subjects.
WHAT ABOUT PROFESSIONALISE AND ETHICS? For many decades, NCEES has considered adding pro fessionalism and ethics problems to the PE exam. How ever, these subjects are not part of the test outline, and there has yet to be an ethics problem in the exam.
IS THE EXAM TRICKY? Other than providing superfluous data, the PE exam is not a “tricky exam.” The exam does not overtly try to get you to fail. Examinees manage to fail on a regular basis with perfectly straightforward problems. The exam prob lems are difficult in their own right. NCEES does not need to provide misleading or conflicting statements. However, you will find that commonly made mistakes are repre sented in the available answer options. Thus, the alter native answers (known as distractors) will be logical. Problems are generally practical, dealing with common and plausible situations that you might experience in your job. Circuit design is restricted to connecting elec trical components to create a desired outcome. You will not be asked to design a new IC chip or build an equiva lent vacuum tube circuit.
DOES NCEES WRITE EXAM PROBLEMS AROUND THIS BOOK? Only NCEES knows what NCEES uses to write its exam problems. However, it is irrelevant, because this book is not intended to ( 1 ) be everything you need to
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pass the exam, ( 2 ) expose exam secrets or exam prob lems, or (3) help you pass when you do not deserve to pass. NCEES knows about this book, but worrying about NCEES writing exam problems based on informa tion that is or is not in this book means you are placing too much dependency on this book. This book, for example, will teach you how to use aspects of many standards and codes. Expecting that this book will replace those standards and codes is unrealistic. This book will provide instruction in certain principles. Expecting that you will not need to learn anything else is unrealistic. This book presents many facts, defini tions, and numerical values. Expecting that you will not need to know other facts, definitions, and numerical values is unrealistic. What NCEES uses to write exam problems will not have any effect on what you need to do to prepare for the exam.
W HAT MAKES THE PROBLEMS DlFFICULT? Some problems are difficult because the pertinent theory is not obvious. There may be only one acceptable pro cedure, and it may be heuristic (or defined by a code) such that nothing else will be acceptable. Some problems are difficult because the data needed is hard to find. Some data just is not available unless you happen to have brought the right reference book. For example, building requirements concerning electricity are in the National Electrical Code; without the code and without knowledge of where given information is located within the code, certain problems will be nearly impossible to solve. Some problems are difficult because they defy the imag ination. Some problems are difficult because the compu tational time is lengthy. Some problems are difficult because the terminology is obscure, and you just do not know what the terms mean. This can happen in almost any subject.
DOES THE PE EXAM USE SI UNITS? The PE exam primarily uses the SI system. A few topics utilize the customary U.S. system where they correspond to commonly accepted industry standards. For example, problems involving the National Electrical Code were, until recently, given in customary U.S. units with SI provided in parenthesis. Even in this topic, SI is now foremost and customary U.S units are now in parentheses.
W HY DOES MCEES REUSE SOME PROBLEM S? NCEES reuses some of the more reliable problems from each exam. The percentage of repeat problems is not high— no more than 25% of the exam. NCEES repeats problems in order to equate the performance of one
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group of examinees with the performance of an earlier group. The repeated problems are known as equaters, and together, they are known as the equating subtest. Occasionally, a new problem appears on the exam that very few of the examinees do well on. Usually, the reason for this is that the subject is too obscure or the problem is too difficult. Also, there have been cases where a low percentage of the examinees get the answer correct because the problem was inadvertently stated in a poor or confusing manner. Problems that everyone gets correct are also considered defective. NCEES tracks the usage and “success” of each of the exam problems. “Rogue” problems are not repeated with out modification. This is one of the reasons historical analysis of problem types should not be used as the basis of your review.
P O E S N C EES USE TH E EXAH T© PHE*TEST
FtJTtJHE PROBLEM S? NCEES does not use the PE exam to “pretest” or qualify future problems. (It does use this procedure on the FE exam, however.) All of the problems you work will con tribute toward your final score.
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can use. Personal notes in a threering binder and other semipermanent covers can usually be used. Some states use a “shake test” to eliminate loose papers from binders. Make sure that nothing escapes from your binders when they are inverted and shaken. The references you bring into the exam room in the morning do not have to be the same as the references you use in the afternoon. However, you cannot share books with other examinees during the exam. A few states do not permit collections of solved prob lems such as Schaum’s Outlines series, sample exams, and solutions manuals. A few states maintain a formal list of banned books. Strictly speaking, loose paper and scratch pads are not permitted in the examination room. Certain types of preprinted graphs and logarithmically scaled graph papers (which are almost never needed) should be threehole punched and brought in a threering binder. An exception to this restriction may be made for lami nated and oversize charts, graphs, and tables that are commonly needed for particular types of problems. It is a good idea to check with your state board to learn what is and is not permitted in the examination room. PPI provides information about each state board on the PPI website (ppi2pass.com/faqs/stateboards).
ARE THE E X A M P L E PRO BLEM S IN THIS BO O K R EP R ESEN TATIVE OF TH E EXAM ?
HOW M ANY BO O KS SH OULD YO U BRING? The example problems in this book are intended to be instructional and informative. They were written to illus trate how their respective concepts can be implemented. Example problems are not intended to represent exam problems or provide guidance on what you should study.
ABE THE PRACTICE PROBLEM S
R EPR ESEN TA TIVE OF TH E EXAM ? The practice problems in the companion book, Power Practice Problems for the Computer and Electrical PE Exam, cover the most likely exam subjects. Some of the practice problems are multiplechoice, and some require freeformat solutions. However, they are generally more comprehensive and complex than actual exam problems, regardless of their formats. Some of the practice problems are marked “ Time limit: one hour!' Compared to the sixminute problems on the PE exam, such onehour prob lems are considerably more timeconsuming. All of the practice problems are original. Since NCEES does not release old exams, none of the practice prob lems are actual exam problems.
W H AT H EP IH EN C E M ATERIAL IS PERMITTEE! IN TH E E X A M ? The PE exam is an openbook exam. Most states do not have any limits on the numbers and types of books you
Except for codes and standards, you should not need many books in the examination room. The trouble is, you cannot know in advance which ones you will need. That is the reason why many examinees show up with boxes and boxes of books. Since this book is not a substitute for your own experience and knowledge, with out a doubt, there are things that you will need that are not in this book. But there are not so many that you need to bring your company’s entire library. The exam is very fastpaced. You will not have time to use books with which you are not thoroughly familiar. The exam does not require you to know obscure solution methods or to use difficulttofind data. You will not need articles printed in an industry magazine; you will not need doctoral theses or industry proceedings; and you will not need to know about recent industry events. So, it really is unnecessary to bring a large quantity of books with you. Essential books are identified in the Codes and References section of this book, and you should be able to decide which support you need for the areas in which you intend to work. This book and five to 1 0 other references of your choice should be sufficient for most of the problems you answer. M A Y T A B S BE PLACED ON P A G E S ? It is common to tab pages in your books in an effort to reduce the time required to locate useful sections. Inas much as some states consider Postit® notes to be “loose
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paper,” your tabs should be of the more permanent variety. Although you can purchase tabs with gummed attachment points, it is also possible simply to use transparent tape to secure the Postits you have already placed in your books.
For maximum speed and utility, your calculator should also have or be programmed for the following functions. • converting between polar (phasor) and rectangular vectors • performing simultaneous linear equations
CAM YOU WRITE AMD M A R K IM Y O U R BOOKS? During your preparation, you may write anything you want, anywhere in your books, including this one. You can use pencil, pen, or highlighter in order to further your understanding of the content. However, during the exam, you must avoid the appearance of taking notes about the exam. This means that you should write only on the scratch paper that is provided. During the exam, other than drawing a line across a wide table of num bers, or using your pencil to follow a line on a graph, you should not write in your books.
W HAT ABOUT CALCULATORS? The exam requires the use of a scientific calculator. However, it may not be obvious that you should bring a spare calculator with you to the exam. It is always unfortunate when an examinee is not able to finish because his or her calculator was dropped or stolen or stopped working for some unknown reason. To protect the integrity of its exams, NCEES has banned communicating and textediting calculators from the exam site. NCEES provides a list of calculator models acceptable for use during the exam. Calculators not included in the list are not permitted. Check the current list of permissible devices at the PPI website (ppi2pass.com/calculators). Contact your state board to determine if nomographs and specialty slide rules are permitted. The exam has not been optimized for any particular brand or type of calculator. In fact, for most calculations, a $15 scientific calculator will produce results as satisfac tory as those from a $200 calculator. There are definite benefits to having builtin statistical functions, graphing, unitconversion, and equationsolving capabilities. How ever, these benefits are not so great as to give anyone an unfair advantage. It is essential that a calculator used for the PE exam have the following functions. •
trigonometric and inverse trigonometric functions
•
hyperbolic and inverse hyperbolic functions
•
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y/x and x 2
•
both common and natural logarithms
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• performing rms calculations • simplifying Smith chart calculations • calculating factors for economic analysis problems • calculating determinants of matrices • linear regression • calculating factors for economic analysis problems You may not share calculators with other examinees. Be sure to take your calculator with you whenever you leave the exam room for any length of time. Laptop and tablet computers, and electronic readers are not permitted in the examination room. Their use has been considered, but no states actually permit them. However, considering the nature of the exam problems, it is very unlikely that these devices would provide any advantage.
ARE MOBILE DEVICES PERMITTED? You may not possess or use a cell phone, laptop or tablet computer, electronic reader, or other communications or textmessaging device during the exam, regardless of whether it is on. You will not be frisked upon entrance to the exam, but should a proctor discover that you are in possession of a communications device, you should expect to be politely excluded from the remainder of the exam.
HOW YOU SHOULD GUESS There is no deduction for incorrect answers, so guessing is encouraged. However, since NCEES produces defen sible licensing exams, there is no pattern to the place ment of correct responses. Since the quantitative responses are sequenced according to increasing values, the placement of a correct answer among other numer ical distractors is a function of the distractors, not of some statistical normalizing routine. Therefore, it is irrelevant whether you choose all “A,” “B,” “C,” or “D” when you get into guessing mode during the last minute or two of the exam period. The proper way to guess is as an engineer. You should use your knowledge of the subject to eliminate illogical answer options. Illogical answer options are those that violate good engineering principles, that are outside normal oper ating ranges, or that require extraordinary assumptions. Of course, this requires you to have some basic under standing of the subject in the first place. Otherwise, you
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go back to random guessing. That is the reason that the minimum passing score is higher than 25%. You will not get any points using the “testtaking skills” that helped you in college. You will not be able to eliminate any [verb] answer options from “Which [noun] . . . ” problems. You will not find problems with options of the “more than 50” and “less than 50” variety. You will not find one answer option among the four that has a different number of significant digits, or has a verb in a different tense, or has some singular/plural discrepancy with the stem. The distractors will always match the stem, and they will be logical.
HOW IS TH E EXAM GRADED AN EDSCORED? The maximum number of points you can earn on the Power PE exam is 80. The minimum number of points for passing (referred to by NCEES as the cut score) varies from exam to exam. The cut score is determined through a rational procedure, without the benefit of knowing examinees’ performance on the exam. That is, the exam is not graded on a curve. The cut score is selected based on what you are expected to know, not based on passing a certain percentage of engineers. Each of the problems is worth one point. Grading is straightforward, since a computer grades your score sheet. Either you get the problem right or you do not. If you mark two or more answers for the same problem, no credit is given for the problem. You will receive the results of your exam from your state board (not NCEES) online through your MyNCEES account or by mail, depending on your state. Eight to ten weeks will pass before NCEES releases the results to the state boards. However, the state boards take varying amounts of additional time before notifying examinees. You should allow three to four months for notification. Your score may or may not be revealed to you, depend ing on your state’s procedure. Even if the score is reported to you, it may have been scaled or normal ized to 100%. It may be difficult to determine whether the reported score is out of 80 points or is out of 1 0 0 %.
HOW IS THE CUT SCORE ESTABLISHED? The raw cut score may be established by NCEES before or after the exam is administered. Final adjustments may be made following the exam date. NCEES uses a process known as the Modified Angoff procedure to establish the cut score. This procedure starts with a small group (the cut score panel) of profes sional engineers and educators selected by NCEES. Each individual in the group reviews each problem and makes an estimate of its difficulty. Specifically, each individual
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estimates the number of minimally qualified engineers out of a hundred examinees who should know the correct answer to the problem. (This is equivalent to predicting the percentage of minimally qualified engineers who will answer correctly.) Next, the panel assembles, and the estimates for each problem are openly compared and discussed. Even tually, a consensus value is obtained for each. When the panel has established a consensus value for every problem, the values are summed and divided by 1 0 0 to establish the cut score. Various minor adjustments can be made to account for examinee population (as characterized by the average performance on any equater problems) and any flawed problems. Rarely, security breaches result in compro mised problems or exams. How equater problems, exam flaws, and security issues affect examinee performance is not released by NCEES to the public.
CHEATING .AMD EXAM SUBVERSIOM............. There are not very many ways to cheat on an openbook test. The proctors are well trained in spotting the few ways that do exist. It goes without saying that you should not talk to other examinees in the room, nor should you pass notes back and forth. You should not write anything into your books or take notes on the content of the exam. You should not use your cell phone. The number of people who are released to use the restroom may be limited to prevent discussions. NCEES regularly reuses problems that have appeared on previous exams. Therefore, exam integrity is a seri ous issue with NCEES, which goes to great lengths to make sure nobody copies the problems. You may not keep your exam booklet or scratch paper, enter the text of problems into your calculator, or copy problems into your own material. NCEES has become increasingly unforgiving about loss of its intellectual property. NCEES routinely prose cutes violators and seeks financial redress for loss of its exam problems, as well as invalidating any engineer ing license you may have earned by taking one of its exams while engaging in prohibited activities. Your state board may impose additional restrictions on your right to retake any exam if you are convicted of such activities. In addition to tracking down the sources of any exam problem compilations that it becomes aware of, NCEES is also aggressive in pursuing and prosecut ing examinees who disclose the contents of the exam in online forum and “chat” environments. Your constitu tional rights to free speech and expression will not pro tect you from civil prosecution for violating the nondisclosure agreement that NCEES requires you to sign before taking the exam. If you wish to participate in a dialogue about a particular exam subject, you must do
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so in such a manner that does not violate the essence of your nondisclosure agreement. This requires decoupling your discussion from the exam and reframing the prob lem to avoid any exam particulars. The proctors are concerned about exam subversion, which generally means activity that might invalidate the exam or the examination process. The most common form of exam subversion involves trying to copy exam problems for future use. However, in their zeal to enforce and protect, proctors have shown unforgiving intolerance of otherwise minor infractions such as using your own pencil, using a calculator not on the approved list, pos sessing a cell phone, or continuing to write for even an instant after “pencils down” is called. For such infrac tions, you should expect to have the results of your exam invalidated, and all of your pleas and arguments in favor of forgiveness to be ignored. Even worse, since you will summarily be considered to have cheated, your state board will most likely prohibit you from retaking the exam for a number of exam cycles. There is no mercy built into the NCEES and state board procedures.
PART 3; HOW TO PREPARE FOR AND PASS THE POWER PE EXAM
they have had extensive experience, in the hope that the exam will feature lots of problems in those subjects. This method is seldom successful. Some engineers plan on modeling their solutions from similar problems they have found in textbooks, collec tions of solutions, and old exams. These engineers often spend a lot of time compiling and indexing the example and sample problem types in all of their books. This is not a legitimate preparation method, and it is almost never successful.
e© YOU f^IEED A CLASSBOOHi REVIEW G O tJfiSE ? Approximately 60% of firsttime PE examinees take an instructorled review course of some form. Live classroom and internet courses of various types, as well as previously recorded lessons, are available for some or all of the exam topics. Live courses and instructor moderated internet courses provide several significant advantages over selfdirected study, some of which may apply to you. • A course structures and paces your review. It ensures that you keep going forward without getting bogged down in one subject. •
A course focuses you on a limited amount of mate rial. Without a course, you might not know which subjects to study.
•
A course provides you with the problems you need to solve. You will not have to spend time looking for them.
•
A course spoonfeeds you the material. You may not need to read the book!
•
The course instructor can answer your questions when you are stuck.
W HAT SH O ULD YOU STUDY? The exam covers many diverse subjects. Strictly speak ing, you do not have to study every subject on the exam in order to pass. However, the more subjects you study, the more you will improve your chances of passing. You should use Table 3 to help you plan which subjects you are going to study, and when. You should decide early in the preparation process which subjects you are going to study. The strategy you select will depend on your back ground. The following are the four most common strategies. A broad approach is the key to success for examinees who have recently completed their academic studies. This strategy is to review the fundamentals in a broad range of undergraduate subjects (which means studying all or most of the chapters in this book). The exam includes enough fundamentals problems to make this strategy worthwhile. Overall, it is the best approach. Engineers who have little time for preparation tend to concentrate on the subject areas in which they hope to find the most problems. By studying the list of exam subjects, some have been able to focus on those subjects that will give them the highest probability of finding enough problems that they can answer. This strategy works as long as the exam cooperates and has enough of the types of problems they need. Too often, though, examinees who pick and choose subjects to review can not find enough problems to complete the exam. Engineers who have been away from classroom work for a long time tend to concentrate on the subjects in which
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You probably already know if any of these advantages apply to you. A review course will be less valuable if you are thorough, selfmotivated, and highly disciplined.
HOW LOMQ SHOU LD YOU STUDY? We have all heard stories of the person who did not crack a book until the week before the exam and still passed it with flying colors. Yes, these people really exist. However, I am not one of them, and you probably are not either. In fact, I am convinced that these people are as rare as the ones who have taken the exam five times and still have not passed it. A thorough review takes approximately 300 hours. Most of this time is spent solving problems. Some of it may be spent in class; some is spent at home. Some examinees spread this time over a year. Others try to cram it all into two months. Most classroom review courses last for three or four months. The best time to start studying will depend on how much time you can spend per week.
g U T R O P y C T i O N
W HAT TH E W ELLH EELED ELECTRICAL ENGINEER SH O ULD BEGIH A C C yilU L A T S M G There are many references and resources that you should begin to assemble for review and for use in the exam. It is unlikely that you could pass the PE exam without accumulating other books and resources. There certainly is not much margin for error if you show up with only one book. There are many problems that require knowledge, data, and experience that are presented and described only in codes, standards, and references dedicated to a single subject. You would have to be truly lucky to go in “bare,” find the right mix of problems, and pass.
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• Books on nonexam subjects: Such subjects as materi als science, statics, dynamics, mechanics of materials, drafting, history, the English language, geography, and philosophy are not part of the exam. • Books on mathematical analysis, numerical analysis, or extensive mathematics tabulations. •
Obscure books and materials: Books that are in foreign languages, doctoral theses, and papers pre sented at technical societies will not be needed during the exam.
•
Old textbooks or obsolete, rare, and ancient books: NCEES exam committees are aware of which text books are in use. Material that is available only in outofprint publications and old editions will not be used.
Few examinees are able to accumulate all of the refer ences needed to support the exam’s entire body of knowledge. The accumulation process is too expensive and timeconsuming, and the sources are too diverse. Like purchasing an insurance policy, what you end up with will be more a function of your budget than of your needs. In some cases, one book will satisfy several needs.
• Handbooks in other disciplines: You probably will not need a civil, mechanical, or industrial engineer ing handbook.
The books and other items listed in Codes and Refer ences are regularly cited by examinees as being partic ularly useful to them. This listing only includes the major “named” books that have become standard refer ences in the industry. These books are in addition to any textbooks or resources that you might choose to bring.
• Manufacturer’s literature and catalogs: No part of the exam requires you to be familiar with products that are proprietary to any manufacturer.
ADDITIONAL REW1EW M ATERIAL In addition to this book and its companion, Power Practice Problems for the Electrical and Computer Engineering PE Exam, PPI can provide you with many targeted references and study aids, some of which are listed here. All of these resources have stood the test of time, which means that examinees continually report their usefulness and that PPI keeps them up to date. • •
Power Practice Exams for the Electrical and Com puter PE Exam. Engineering Unit Conversions
• Electrical PE Exam Cafe For a complete listing of available sample problems, other books, calculator software, videos, and other examrelated products, please go to PPI’s website at ppi2pass.com/electrical.
WHAT YOU WILL HOT NEED) Generally, people bring too many things to the exam. One general rule is that you should not bring books that you have not looked at during your review. If you did not need a book while doing the example problems in this book, you will not need it during the exam.
@ Crafts and tradesoriented books: The exam does not expect you to have detailed knowledge of trades or manufacturing operations.
• U.S. government publications: With the exceptions of the publications mentioned and referenced in this book, no government publications are required for the PE exam. • Your state’s laws: The PE exam is a national exam. Nothing unique to your state will appear on it. (However, federal legislation affecting engi neers, particularly in environmental areas, is fair game.)
SHOULD YOU LOOK FOR OLD EXAMS? The traditional approach to preparing for standardized tests includes working sample tests. However, NCEES does not release old tests or problems after they are used. Therefore, there are no official problems or tests available from legitimate sources. NCEES publishes booklets of sample problems and solutions to illustrate the format of the exam. However, these problems have been compiled from various previous exams, and the resulting publication is not a true “old exam.” Further more, NCEES sometimes constructs its sample prob lems books from problems that have been pulled from active use for various reasons, including poor perfor mance. Such marginal problems, while accurately reflecting the format of the exam, are not always repre sentative of actual exam subjects.
There are some other things that you will not need.
WHAT SHOULD YOU MEMORIZE?
•
You get lucky here, because it is not necessary to mem orize anything. The exam is openbook, so you can look up any procedure, formula, or piece of information that
Books on basic and introductory subjects: You will not need books that cover trigonometry, geometry, or calculus.
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you need. You can speed up your problemsolving response time significantly if you do not have to look up the conversion from W to ftlbf/sec, or the definition of the sine of an angle, but you do not even have to memorize these kinds of things. As you work practice problems in the companion book, Power Practice Prob lems for the Electrical and Computer PE Exam, you will automatically memorize the things that you come across more than a few times.
DO YOU..NEED A REVIEW S C H E D U LE ? It is important that you develop and adhere to a review outline and schedule. Once you have decided which subjects you are going to study, you can allocate the available time to those subjects in a manner that makes sense to you. If you are not taking a classroom review course (where the order of preparation is deter mined by the lectures), you should make an outline of subjects for selfstudy to use for scheduling your prep aration. A fillinthedates schedule is provided in Table 3 at the end of this introduction.
There are several things you can do to make your review more representative. For example, if you gather most of your review resources (i.e., books) in advance and try to use them exclusively during your review, you will become more familiar with them. (Of course, you can also add to or change your references if you find inadequacies.) Learning to use your time wisely is one of the most important lessons you can learn during your review. You will undoubtedly encounter problems that end up taking much longer than you expected. In some instances, you will cause your own delays by spending too much time looking through books for things you need (or just by looking for the books themselves!). Other times, the problems will entail too much work. Learn to recognize these situations so that you can make an intelligent deci sion about skipping such problems in the exam.
W H A T to d o
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There are a few things you should do a week or so before the exam. You should arrange for childcare and trans portation. Since the exam does not always start or end at the designated time, make sure that your childcare and transportation arrangements are flexible.
A SIM PLE PLAMMMQ SU G G E ST IO N Designate some location (a drawer, a corner, a card board box, or even a paper shopping bag left on the floor) as your “exam catchall.” Use your catchall during the months before the exam when you have revelations about things you should bring with you. For example, you might realize that the plastic ruler marked off in tenths of an inch that is normally kept in the kitchen junk drawer can help you. Or, you might decide that a certain book is particularly valuable, that it would be nice to have dental floss after lunch, or that large rubber bands and clips are useful for holding books open. It is not actually necessary to put these treasured items in the catchall during your preparation. You can, of course, if it is convenient. But if these items will have other functions during the time before the exam, at least write yourself a note and put the note into the catchall. When you go to pack your exam kit a few days before the exam, you can transfer some items immediately, and the notes will be your reminders for the other items that are back in the kitchen drawer.
Check PPI’s website for lastminute updates and errata to any PPI books you might have and are bringing to the exam. If you have not already done so, read the “Advice from Examinees” section of PPI’s website. If it is convenient, visit the exam location in order to find the building, parking areas, exam room, and restrooms. If it is not convenient, you may find driving directions and/or site maps online. Take the battery cover off your calculator and check to make sure you are bringing the correct size replacement batteries. Some calculators require a different kind of battery for their “permanent” memories. Put the cover back on and secure it with a piece of masking tape. Write your name on the tape to identify your calculator. If your spare calculator is not the same as your primary calculator, spend a few minutes familiarizing yourself with how it works. In particular, you should verify that your spare calculator is functional.
PREPARE YO UR CAR
HOW YO U CAM M AKE YO UR REVIEW REALISTIC In the exam, you must be able to recall solution proce dures, formulas, and important data quickly. You must remain sharp for eight hours or more. When you played a sport back in school, your coach tried to put you in gamerelated situations. Preparing for the PE exam is not much different from preparing for a big game. Some part of your preparation should be realistic and repre sentative of the exam environment.
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: ] Gather snow chains, a shovel, and tarp to kneel on while installing chains. [ ] Check tire pressure. [ ] Check your spare tire. ] Check for tire installation tools. ] Verify that you have the vehicle manual. ] Check fluid levels (oil, gas, water, brake fluid, transmission fluid, windowwashing solution). [ ] Fill up with gas.
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 ] Check battery and charge if necessary. j ] Know something about your fuse system (where fuses are, how to replace them, etc.). [ ] Assemble all required maps. [ ] Fix anything that might slow you down (missing wiper blades, etc.).  ] Check your taillights. j ] Affix the current DMV vehicle registration tag.  ] Fix anything that might get you pulled over on the way to the exam (burnedout taillight or headlight, broken lenses, bald tires, missing license plate, noisy muffler). ! ] Treat the inside windows with antifog solution. [ ] Put a roll of paper towels in the back seat. [ ] Gather exact change for any bridge tolls or toll roads. [ ] Find and install your electronic toll tag, if applicable. [ ] Put $20 in your glove box. [ ] Check for current vehicle registration and proof of insurance. [ ] Locate a spare door and ignition key.  ] Find your AAA or other roadsideassistance cards and phone numbers. [ ] Plan alternate routes.
PREPARE YOUR EXAM IC1TS Second in importance to your scholastic preparation is the preparation of your two exam kits. The first kit consists of a bag, box (plastic milk crates hold up better than cardboard in the rain), or wheeled travel suitcase containing items to be brought with you into the exam room. [ ] your exam authorization notice [ ] current, signed governmentissued photographic identification (e.g., driver’s license, not a student ID card) [ ] this book [ ] other textbooks and reference books [ ] regular dictionary [ ] review course notes in a threering binder [ ] cardboard boxes or plastic milk crates to use as bookcases  ] primary calculator [ ] spare calculator  ] instruction booklets for your calculators I ] extra calculator batteries j ] two straightedges (e.g., ruler, scale, triangle, protractor)  ] protractor I ] scissors  ] stapler I ] transparent tape  ] magnifying glass
jo n r ii
j ] small (jeweler’s) screwdriver for fixing your glasses or for removing batteries from your calculator j ] unobtrusive (quiet) snacks or candies, already unwrapped ! ] two small plastic bottles of water [ ] travel pack of tissue (keep in your pocket) [ ] handkerchief I ] headache remedy [ ] personal medication [ ] $5.00 in assorted coinage  ] spare contact lenses and multipurpose contact lens cleaning solution [ ] backup reading glasses (no case) [ ] eye drops [ ] light, comfortable sweater j ] loose shoes or slippers [ ] cushion for your chair [ ] earplugs [ ] wristwatch with alarm [ ] several large trash bags (“raincoats” for your boxes of books)  ] roll of paper towels [ ] wire coat hanger (to hang up your jacket or to get back into your car in an emergency)  ] extra set of car keys on a string around your neck The second kit consists of the following items and should be left in a separate bag or box in your car in case it is needed. ] ] [ ] [ ] ] I] ] [ ] [ ] [] [ ]
copy of your exam authorization notice light lunch beverage in thermos or can sunglasses extra pair of prescription glasses raincoat, boots, gloves, hat, and umbrella street map of the exam area parking permit batterypowered desk lamp your cell phone length of rope
The following items cannot be used during the exam and should be left at home. [ ] personal pencils and erasers (NCEES distributes mechanical pencils at the exam.) [ ] fountain pens  ] radio, CD player, MP3 player, or other media player ! ] battery charger [ ] extension cords [ ] scratch paper [ ] notepads [ ] drafting compass [ ] circular (“wheel”) slide rules
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Since the limit is less than 1 , the infinite series converges. E xam ple 4.14 Does the infinite series A converge or diverge?
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Two tests can be used independently or after Eq. 4.93 has proven inconclusive: the ratio and comparison tests.
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In most cases, the expression for the general term an will be known, but there will be no simple expression for the sum Sn. Therefore, Eq. 4.92 cannot be used to determine convergence. It is helpful, but not conclu sive, to look at the limit of the general term. If the limit, as defined in Eq. 4.93, is nonzero, the series diverges. If the limit equals zero, the series may either converge or diverge. Additional testing is needed in that case. j = 0 lim an{ n>°° I ^ o
nm n+oo
By observation, the general term is _
1
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The general term can be expanded by partial fractions to
However, 1/n is the harmonic series. Since the harmonic series is divergent and this series is larger than the harmonic series (by the term 1 /n 2), this series also diverges.
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Alternatively, the ratio test can be used with the abso lute value of the ratio. The same criteria apply.
1
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25. SERIES OF A LTER N A TIN G SIGN S14 Some series contain both positive and negative terms. The ratio and comparison tests can both be used to determine if a series with alternating signs converges. If a series containing all positive terms converges, then the same series with some negative terms also con verges. Therefore, the allpositive series should be tested for convergence. If the allpositive series con verges, the original series is said to be absolutely con vergent (If the allpositive series diverges and the original series converges, the original series is said to be conditionally convergent.)
14This terminology is commonly used even though it is not necessary that the signs strictly alternate.
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Mathematics
A L G E B R
Mathematics
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Linear Algebra
1. Matrices........................................................... 51 Special Types of Matrices............................. 51 3. Row Equivalent Matrices.............................. 52 4. Minors and Cofactors.................................... 52 5. Determinants.................................................. 53 6 . Matrix Algebra............................................... 54 7. Matrix Addition and Subtraction................ 54 8 . Matrix Multiplication.................................... 54 9. Transpose........................................................ 55 10. Singularity and R ank.................................... 55 1 1 . Classical A djoint............................................ 55 1 2 . Inverse............................................................. 56 13. Writing Simultaneous Linear Equations in Matrix F orm ............................................... 56 14. Solving Simultaneous LinearEquations . . . . 57 15. Eigenvalues and Eigenvectors...................... 58 2.
after the second row and second column have been removed is an _ &31
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An augmented matrix results when the original matrix is extended by repeating one or more of its rows or col umns or by adding rows and columns from another matrix. For example, for the matrix A, the augmented matrix created by repeating the first and second col umns is an
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2* SPECIAL TY P ES OF M ATRICES Certain types of matrices are given special designations.
1. H ATftlCES A matrix is an ordered set of entries ( elements) arranged rectangularly and set off by brackets.1 The entries can be variables or numbers. A matrix by itself has no particular value— it is merely a convenient method of representing a set of numbers. The size of a matrix is given by the number of rows and columns, and the nomenclature m x n is used for a matrix with m rows and n columns. For a square matrix, the number of rows and columns will be the same, a quantity known as the order of the matrix. Bold uppercase letters are used to represent matrices, while lowercase letters represent the entries. For example, a23 would be the entry in the second row and third column of matrix A.
A =
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&32
&33
® cofactor matrix: the matrix formed when every entry is replaced by the cofactor (see Sec. 5.4) of that entry • column matrix: a matrix with only one column ® complex matrix: a matrix with complex number entries • diagonal matrix: a square matrix with all zero entries except for the a# for which i — j • echelon matrix: a matrix in which thenumber of zeros preceding the first nonzero entry of arow increases row by row until only zero rows remain. A rowreduced echelon matrix is an echelon matrix in which the first nonzero entry in each row is a 1 and all other entries in the columns are zero. • identity matrix: a diagonal (square) matrix with all nonzero entries equal to 1 , usually designated as I, having the property that AX = IA = A • null matrix: the same as a zero matrix • row matrix: a matrix with only one row
A submatrix is the matrix that remains when selected rows or columns are removed from the original matrix.2 For example, for matrix A, the submatrix remaining
• scalar matrix:3 a diagonal (square) matrix with all diagonal entries equal to some scalar k
lrThe term array is synonymous with matrix, although the former is more likely to be used in computer applications. 2By definition, a matrix is a submatrix of itself.
3Although the term complex matrix means a matrix with complex entries, the term scalar matrix means more than a matrix with scalar entries.
PPE
® w w w np p g 2 p a s s „ c @ m
52
P O W E R
R E F E R E N C E
M A N U A L
Mathematics
singular matrix: a matrix whose determinant is zero (see Sec. 5.10) skew symmetric matrix: a square matrix whose transpose (see Sec. 5.9) is equal to the negative of itself (i.e., A = —A*) square matrix: a matrix with the same number of rows and columns (i.e., m = n) symmetric (al) matrix: a square matrix whose trans pose is equal to itself (i.e., A* = A ), which occurs only when a^ = a^
4, MINOitS AMB C O F^CTO ftS Minors and cofactors are determinants of submatrices associated with particular entries in the original square matrix. The minor of entry is the determinant of a submatrix resulting from the elimination of the single row i and the single column j. For example, the minor corresponding to entry a12 in a 3 x 3 matrix A is the determinant of the matrix created by eliminating row 1 and column 2 .
minor of a\2 =
triangular matrix: a square matrix with zeros in all positions above or below the diagonal •
unit matrix: the same as the identity matrix
•
zero matrix: a matrix with all zero entries
Figure 5.1 Examples of Special Matrices 0
0
"2
18
2
18'
"1
9
0
O’
0
0
1
9
0
0
1
0
0
0
1 0
0
0
0
9
0
0
0
1
0
0
0
0
0
0
0
0
5.
(a) diagonal
.0
(b) echelon
0
0
o
0
0
3
0
0
1
0
0
0
3
0
0
1.
.0
0
0
3.
0
0
0
0
1
0
0
0
.0
0
(d) identity
"
(e) scalar
(c) rowreduced echelon "2
‘ +1
1
+1
1
+1
1
+1
1
+1
"
0
0
0
7
6
0
0
9
1
1
0
_8
0
4
5.
(f) triangular
cofactor of a\2 = —
• interchanging the zth and jth rows • multiplying the zth row by a nonzero scalar
GaussJordan elimination is the process of using these elementary row operations to rowreduce a matrix to echelon or rowreduced echelon forms, as illustrated in Ex. 5.8. When a matrix has been converted to a rowreduced echelon matrix, it is said to be in row canonical form. The phrases rowreduced echelon form and row canonical form are synonymous.
«23
031
033
5.2
2
9 1
3
4 0
7
5 9.
Solution The minor’s submatrix is created by eliminating the row and column of the —3 entry.
• replacing the zth row by the sum of the original ith row and k times the jth row However, two matrices that are row equivalent as defined do not necessarily have the same determinants. (See Sec. 5.5.)
^21
What is the cofactor corresponding to the —3 entry in the following matrix?
A =
w w w Bp p i 2 p ^ g S o C © m
5.1
E xam ple 5.1
A matrix B is said to be row equivalent to a matrix A if it is obtained by a finite sequence of elementary row operations on A:
•
033
For example, the cofactor of entry a12 in matrix A (described in Sec. 5.4) is
3. ROW EQUIVALENT MATRICES
P P B
031
i
"3
"1
................. o
0  6 0 0
1 o
0
023
The cofactor of entry a^ is the minor of multiplied by either + 1 or —1 , depending on the position of the entry. (That is, the cofactor either exactly equals the minor or it differs only in sign.) The sign is determined according to the following positional matrix.4
Figure 5.1 shows examples of special matrices.
9
021
M:
9
1
5
9
The minor is the determinant of M . M = (9) (9) — (5)(1) = 76 The sign corresponding to the —3 position is negative. Therefore, the cofactor is —76. 4The sign of the cofactor if (i + j) is odd.
is positive if (z + j ) is even and is negative
L I N E A R
A L G E B R A
5 " 3
A determinant is a scalar calculated from a square matrix. The determinant of matrix A can be repre sented as D {A }, D et(A ), AA, or A.5 The following rules can be used to simplify the calculation of determinants. • If A has a row or column of zeros, the determinant is zero. • If A has two identical rows or columns, the determi nant is zero. • I f B is obtained from A by adding a multiple of a row (column) to another row (column) in A , then B = A. • If A is triangular, the determinant is equal to the product of the diagonal entries. • If B is obtained from A by multiplying one row or column in A by a scalar k, then B = fcA. • If B is obtained from the n x n matrix A by multi plying by the scalar matrix k: then kA = fcnA. • I f B is obtained from A by switching two rows or columns in A , then B = —A.
A =
augmented A =
A =
b
c
d
a
b
c
d
b e h
c f i
+
+
+
d 5.4
A = aei + bfg + cdh — gee — hfa — ic
5.5
The second method of calculating the determinant is somewhat slower than the first for a 3 x 3 matrix but illustrates the method that must be used to calculate determinants of 4 x 4 and larger matrices. This method is known as expansion by cofactors. One row (column) is selected as the base row (column). The selection is arbi trary, but the number of calculations required to obtain the determinant can be minimized by choosing the row (column) with the most zeros. The determinant is equal to the sum of the products of the entries in the base row (column) and their corresponding cofactors. ;
Calculation of determinants is laborious for all but the smallest or simplest of matrices. For a 2 x 2 matrix, the formula used to calculate the determinant is easy to remember. a
a d 9
Mathematics
5. DETERMINANTS
A = a
e h
/
b
c
!
e
/
Lg
h
i,
b
c
 d h
i
i
b
c
e
f
5.6
+ 9
E xam ple 5.2 = ad — be
5.3
Two methods are commonly used for calculating the determinant of 3 x 3 matrices by hand. The first uses an augmented matrix constructed from the original matrix and the first two columns (as shown in Sec. 5.1) . 6 The determinant is calculated as the sum of the products in the lefttoright downward diagonals less the sum of the products in the lefttoright upward diagonals.
Calculate the determinant of matrix A (a) by cofactor expansion, and (b) by the augmented matrix method. 2 A =
3
4 '
3
1
2
4
7
6
Solution (a) Since there are no zero entries, it does not matter which row or column is chosen as the base. Choose the first column as the base. 1
3
2
4
3 1
1
N
A — 2
7
6
3
4
+ 4 1
2
= ( 2 ) ( 6  14)  (3)(—18  28) + (4 )(6  4) = 82
5The vertical bars should not be confused with the square brackets used to set off a matrix, nor with absolute value. 6It is not actually necessary to construct the augmented matrix, but doing so helps avoid errors.
PPg
® w w w » p p § 2 p a s s . e @ m
5=4
P O W E R
S i F E R I I I G i
M A N U A L
8. MATRIX MULTIPLICATION
(b)
Mathematics
A matrix can be multiplied by a scalar, an operation known as scalar multiplication, in which case all entries of the matrix are multiplied by that scalar. For example, for the 2 x 2 matrix A ,
A;A =
IA= 72  (10)  82
kan
kayi
ka2i
ka22
60 MATRIX ALG EBRA7 Matrix algebra differs somewhat from standard algebra. •
equality: Two matrices, A and B, areequal only if they have the same numbers of rows and columns and if all corresponding entries are equal.
•
inequality: The > and < operators are not used in matrix algebra.
•
commutative law of addition:
•
5.7
E xam ple 5.3
associative law of addition: A +
•
(B + C) = (A + B) + C
Matrix division can only be accomplished by multiply ing by the inverse of the denominator matrix. There is no specific division operation in matrix algebra.
5.8
Determine the product matrix C.
associative law of multiplication: C = (A B )C = A (BC )
•
5.5
4
CO
A + B= B+ A
A matrix can be multiplied by another matrix, but only if the lefthand matrix has the same number of columns as the righthand matrix has rows. Matrix multiplica tion occurs by multiplying the elements in each lefthand matrix row by the entries in each righthand matrix column, adding the products, and placing the sum at the intersection point of the participating row and column.
2
6
■ 7
12"
11
8
. 9
10.
left distributive law: Solution A (B + C ) = A B + A C
•
right distributive law: (B + C) A = B A + C A
•
5.10
5.11
scalar multiplication: fc(AB) = (JfcA)B  A(fcB)
The lefthand matrix has three columns, and the righthand matrix has three rows. Therefore, the two matrices can be multiplied. The first row of the lefthand matrix and the first col umn of the righthand matrix are worked with first. The corresponding entries are multiplied, and the products are summed.
5.12
It is important to recognize that, except for trivial and special cases, matrix multiplication is not commutative. That is, AB^BA
cn = (1) (7) + (4)(11) + (3) (9) = 78 The intersection of the top row and left column is the entry in the upper lefthand corner of the matrix C. The remaining entries are calculated similarly. ci2 = (1) (12) + (4) ( 8 ) + (3)(10) = 74
7. MATRIX ADDITION AND SUBTRACTION Addition and subtraction of two matrices is possible only if both matrices have the same numbers of rows and columns (i.e., order). They are accomplished by adding or subtracting the corresponding entries of the two matrices.
c 21
= (5)(7) + (2)(11) + (6)(9) = 1 1 1
c22 = (5)(12) + ( 2 )( 8 ) + ( 6 ) (10) = 136 The product matrix is
C = 7Since matrices are used to simplify the presentation and solution of sets of linear equations, matrix algebra is also known as linear algebra.
PP i
•
w w w . p p i 2 p a s s . c ® m
78
74
111
136
9. TR AN SPO SE
Example 5.4
The transpose, A*, of an m x n matrix A is an n x m matrix constructed by taking the ?th row and making it the ith column. The diagonal is unchanged. For example,
What is the rank of matrix A ?
■1
1
5_
"1
2
V
6
3
1
_9
4
5.
5  5
2
3
3
2
4
i a A =
2
A =
6
CO
A —
1
A L G E B R A
Mathematics
L I N E A R
2
Solution Matrix A is singular because A = 0 . However, there is at least one 2 x 2 nonsingular submatrix: 1
2 = (1)(3)  (—3)(—2) =  3
Transpose operations have the following characteristics.
3
3
( A * )4 = A
5.13
(k.A ) ‘ = jfe(A‘ )
5.14
I s= I
5.15
(A B ) 4 = B 4A 4
5.16
Example 5.5
(A + B)* = A ‘ + B 4
5.17
A4=A
5.18
Determine the rank of matrix A by reducing it to eche lon form.
Therefore, the rank is 2.
10. SINGULARITY AND RAN K
The rank of a matrix is the maximum number of linearly independent row or column vectors.8 A matrix has rank r if it has at least one nonsingular square submatrix of order r but has no nonsingular square submatrix of order more than r. While the submatrix must be square (in order to calculate the determinant), the original matrix need not be.
2
5
3
.0
4
1 0
6
1
"
_
Solution By inspection, the matrix can be rowreduced by sub tracting two times the second row from the third row. The matrix cannot be further reduced. Since there are two nonzero rows, the rank is 2 . "7
4
0
2
5
0
0
9
r
....... ,
The determination of rank is laborious if done by hand. Either the matrix is reduced to echelon form by using elementary row operations, or exhaustive enumeration is used to create the submatrices and many determi nants are calculated. If a matrix has more rows than columns and rowreduction is used, the work required to put the matrix in echelon form can be reduced by work ing with the transpose of the original matrix.
0
9
CO o
The rank of an m x n matrix will be, at most, the smaller of m and n. The rank of a null matrix is zero. The ranks of a matrix and its transpose are the same. If a matrix is in echelon form, the rank will be equal to the number of rows containing at least one nonzero entry. For a 3 x 3 matrix, the rank can either be 3 (if it is nonsingular), 2 (if any one of its 2 x 2 submatrices is nonsingular), 1 (if it and all 2 x 2 submatrices are singular), or 0 (if it is null).
4
i o
is zero. Similarly, a nonsingular matrix is one whose determi nant is nonzero.
A =
'7
11. C LA SSIC A L ADJOINT The classical adjoint is the transpose of the cofactor matrix. The resulting matrix can be designated as A adj, a d j{A }, or A adj. Example 5.6 What is the classical adjoint of matrix A? 2 A =
3
4
0 4
2
1
5
1
The row rank and column rank are the same.
PPI
® w w w . p p i 2 p a s $ . e © m
5  6
P O WE R
R E F E R E N C E
M A N U A L
Solution
Using Eq. 5.22, the inverse is
Mathematics
The matrix of cofactors is 1 8
2
1 1
14
1 0
A 1 =
3
5
“ 2
4
4 Check.
The transpose of the matrix of cofactors is ' 1 8 A adj —

1 1
2
14
4
5
10"
3 2
A A 1 =
—!
4 
8
5" 2
.
1
0
0
1
2
"6 5 3 3
1 0
+
5 +
10"
6
[OK]
12. INVERSE The product of a matrix A and its inverse, A 1, is the identity matrix, I. Only square matrices have inverses, but not all square matrices are invertible. A matrix has an inverse if and only if it is nonsingular (i.e., its deter minant is nonzero). A A ' 1 = A " 1A = I
5.19
(A B )1 = B 1A _1
5.20
13 WRITING SIMULTANEOUS LINEAR EQUATIONS IN MATRIX FORM Matrices are used to simplify the presentation and solu tion of sets of simultaneous linear equations. For example, the following three methods of presenting simultaneous linear equations are equivalent: an xi + 0 and opens down (points up) if p < 0 . (x  ft) 2 = 4p(y  k) opens vertically
Figure 8.13 Parabola
^
^ P V Ivertex at origin
8.75 8.76
The general and vertex forms of the equations can be reconciled with Eq. 8.77 through Eq. 8.79. Whether the first or second forms of these equations are used depends on whether the parabola opens horizontally or vertically (i.e., whether A = 0 or (7 = 0 ), respectively.
h=
0 and down if AE < 0.
^
[opens vertically]
[opens horizontally] 8.79
p=
0
8.90
Figure 8.15 Hyperbola y
Ax2 + Cy2 + Dx + Ey + F = 0
8.88
m ajor axis
c
1
a = y/C + c2
y2 ' f2 J
8.87
8.81
asym ptote '/ I / b \/ 'a)x
m
asym ptote y  k = ( ! ) < x  h)
A^C Equation 8.82 gives the standard form of the equation of an ellipse centered at (ft, fc). Distances a and b are known as the semimajor distance and semiminor distance, respectively. (x  ft) 2
(y  k f
=
1
8.82
The distance between the two foci is 2 c. 2c=2Va2 
&2
transverse axis
Equation 8.91 is the standard equation of a hyperbola. Coefficients A and C have opposite signs. 8.83
Ax2 + Cy* + D x + E y + F=Q\ AC < 0
8.91
The aspect ratio of the ellipse is aspect ratio = ^
8.84
The eccentricity, 6 , of the ellipse is always less than 1 . If the eccentricity is zero, the figure is a circle (another form of a degenerative ellipse).
Equation 8.92 gives the standard form of the equation of a hyperbola centered at (ft, k) and opening to the left and right. (xh f
(y  k f
=
8.92
1 opens horizontally
8.85
PPI
•
w w w . p p i 2 p a s s Bc ® m
Mathematics
A N A L Y T I C
P O W E R
R E F E R E N C E
M A N U A L
(y ~ k ) 2
(y  k f
1
a"
V ■„
8.93
21. SPHERE Equation 8.104 is the general equation of a sphere. The coefficient A cannot be zero. A x2 + Ay2 + Az2 + Bx + Cy + Dz + E = 0
Equation 8.105 gives the standard form of the equation of a sphere centered at (ft, ft, I) with radius r. 8.94
2 c = 2\/ a2 + b2
The eccentricity, e, of the hyperbola is calculated from Eq. 8.95 and is always greater than 1. \Ja2 + b2
>
1
(x  ft)2 + (y — k)2 + (z — I)2 = r2
8.95
u
The hyperbola is asymptotic to the lines given by Eq. 8.96 and Eq. 8.97. y= ± 0 
+ k
B 2A
8.106
8.107
i^ z R
8.96
!b 2 + c 2 + d 2 b
8.97
If the asymptotes are the x and yaxes, the equation of the hyperbola is simply I U/
xy= ± 
8.98
2
8.99
h
8.100
E k= ■
8.101
A
2 C
a=
V —C V —A
[opens horizontally] [opens vertically]
f VA
[opens horizontally]
[ VC
[opens vertically]
b=
PPI
•
8.102
8.103
w w w . p p i 2 p a s s . c e i
8.109
22. HELIX A helix is a curve generated by a point moving on, around, and along a cylinder such that the distance the point moves parallel to the cylindrical axis is pro portional to the angle of rotation about that axis. (See Fig. 8.16.) For a cylinder of radius r, Eq. 8.110 through Eq. 8.112 define the threedimensional positions of points along the helix. The quantity 2nk is the pitch of the helix.
The general and center forms of the equations of a hyperbola can be reconciled by using Eq. 8.99 through Eq. 8.103. Whether the hyperbola opens left and right or up and down depends on whether M/A or M/C is positive, respectively, where M is defined by Eq. 8.99. M= ^ + ^  F 4A 4C —D
e
4A
opens vertically
For a rectangular (equilateral) hyperbola, the asymptotes are perpendicular, a = 6 , c = \/2 a, and the eccentricity is e = y/2. If the hyperbola is centered at the origin (i.e., h = k = 0 ), then the equations are 3? — y2 = a (opens horizontally) and y2 — 3? = a2 (opens vertically).
8.108
2A
opens horizontally
± 7 ( 2 ;  ft) + A;
8.105
The general and center forms of the equations of a sphere can be reconciled by using Eq. 8.106 through Eq. 8.109.
II
The distance between the two foci is 2 c.
y=
8.104
opens vertically
??
Mathematics
Equation 8.93 gives the standard form of the equation of a hyperbola that is centered at (ft, k) and is opening up and down.
tol I
8  1 2
Figure 8.16 Helix
x = r cos t
8.110
y = rsin£
8.111
z  k6
8.112
Mathematics
Differential C alcu lu s 1. Derivative of a Function............................... 91 Elementary Derivative Operations.............. 91 3. Critical Points................................................ 92 4. Derivatives of Parametric Equations.......... 93 5. Partial Differentiation................................... 94 6 . Implicit Differentiation................................. 94 7. Tangent Plane Function............................... 95 8 . Gradient V ector............................................. 95 9. Directional Derivative................................... 96 10. Normal Line V ector....................................... 96 11. Divergence of a Vector Field........................ 97 12. Curl of a Vector F ield................................... 97 13. Taylor’s Formula........................................... 98 14. Maclaurin Power Approximations...............98
N ew to n ’s notation (e.g., ra and x) is generally only used with functions of time.
2.
A regu lar ( analytic or h olom orp h ic) fu n c tio n possesses a derivative. A point at which a function’s derivative is undefined is called a singular p o in t , as Fig. 9.1 illustrates. Figure 9. 1 Derivatives and Singular Points
1. DERI VATI VE OF A FUNCTION In most cases, it is possible to transform a continuous function, f ( x i, # 2 , £3 ,. • one or more independent variables into a derivative function. 1 In simple cases, the derivative can be interpreted as the slope (tangent or rate of change) of the curve described by the original function. Since the slope of the curve depends on x, the derivative function will also depend on x. The deriva tive, f \ x ), of a function / ( x) is defined mathematically by Eq. 9.1. However, limit theory is seldom needed to actually calculate derivatives.
2. ELEMENTARY DERIVATIVE OPERATIONS Equation 9.2 through Eq. 9.5 summarize the elemen tary derivative operations on polynomials and expo nentials. Equation 9.2 and Eq. 9.3 are particularly useful, (a, n, and k represent constants. f ( x ) and g(x) are functions of x.)
A/(z)
9.1
Ax
The derivative of a function f ( x ) , also known as the first derivative , is written in various ways, including
/ '( * ) .
, D / ( 4 »D i / ( 4 , / O ) >s f(s )
A secon d derivative may exist if the derivative operation is performed on the first derivative— that is, a derivative is taken of a derivative function. This is written as ^2/( z )
rf2/
dx2 ’ dx2
, D 2/ ( x ) , D 2/ ( x ) , / ( x ) , 52/ ( 5)
1A function, f (x ), of one independent variable, x, is used in this section to simplify the discussion. Although the derivative is taken with respect to x, the independent variable can be anything.
II
Az >0
0
p
f \ x ) = lim
9.2
B x n = n x n_1
9.3
D In x = —
9.4
X
D e ax = a eax
9.5
Equation 9.6 through Eq. 9.17 summarize the elemen tary derivative operations on transcendental (trigono metric) functions. D sin x = cos x
9.6
D cos x — —sin x
9.7
D tan x = sec2 x
9.8
D cot x = —csc2 x
9.9
D sec x = sec x tan x
9.10
D csc x = —csc x cot x
9.11
P P §
•
w w w . p p i 2 p a s s . c © m
92
P O W E R
R E F E R E N C E
Mathematics
D arcsin x =
M A N U A L
,
g 12
E xam ple 9.2 What are the derivatives of the following functions?
D arccos x = —D arcsin x
9.13
D arctan x — — — 77 1 4  xz
9.14
D arccot x = —D arctan x
9.15
(a) f (x) = 5 v ^
m D>arcsec x = — —1 .... xv x2 — 1 D a rccs c x = —D a rcsec x
■/£ 9.7 7
(b) f( x) = sin x cos2 x (c) f( x) = ln(cos ex) Solution (a) Using Eq. 9.3 and Eq. 9.24,
Equation 9.18 through Eq. 9.23 summarize the ele mentary derivative operations on hyperbolic transcen dental functions. Derivatives of hyperbolic functions are not completely analogous to those of the regular transcendental functions. D sinh x — cosh x
9.18
D cosh x = sinh x
9.19
f '( x) =
5D
^ / ? = 5D(a^ ) 1/3
= (5 )(i)(^ 5 )  2/ 3 D 2;5 =
( 5 ) ( 5) (:e 5) 
2/ 3( 5 ) ( i e 4 )
= 25x2/3/3 (b) Using Eq. 9.26,
D tanh x = sech2 x
9.20
D coth x = —csch2 x
9.21
= (sin x) (2 cos x) (D cos x) + cos2 x cos x
D sech x = —sech x tanh x
9.22
= (sin x) (2 cos x) (—sin x) f cos2 x cos x
D csch x = —csch x coth x
9.23
=
Equation 9.24 through Eq. 9.29 summarize the elemen tary derivative operations on functions and combinations of functions. Dkf(x) — kDf(x)
9.24
f\ x ) = sin xT> cos2 x + cos2xD sin x
—2
sin2 x cos x + cos3 x
(c) Using Eq. 9.29, f \ x) = ( — ^ D c o s e * Vcos exJ
D (/(a;) + g(x)) = D f (x ) ± Dg(x)
9.25
= (— !— V s in e ^ D e *
V(f(x)g(x)) = f{x)T>g(x) + g(x)T>f(x)
9.26
s in ex\ &x cos ex J = —ex tan ex
f { x ) \ _ g(x)Df(x)  f (x)Dg(x)
Vcos exJ
(9(X) ) 2 D (/(^ ))" =
r i(f{x ))n'~1'Df(x)
D / ( ^ ( ^ ) ) = V ;!f { g ) T ) xg{x)
9.28
3. CRITICAL POINTS
9.29
Derivatives are used to locate the local critical points of functions of one variable— that is, extreme points (also known as maximum and minimum points) as well as the inflection points {points of contraflexure). The plurals extrema, maxima, and minima are used without the word “points.” These points are illustrated in Fig. 9.2. There is usually an inflection point between two adja cent local extrema.
E xam ple 9.1 What is the slope at x = 3 of the curve f (x) = x3 — 2x1 Solution The derivative function found from Eq. 9.3 determines the slope. f { x ) = 3x2 — 2 The slope at x = 3 is /'( 3 ) = (3)(3)2  2 = 25
PPO
® w w w . p p i 2 p a s g ne ® m
The first derivative is calculated to determine the loca tions of possible critical points. The second derivative is calculated to determine whether a particular point is a local maximum, minimum, or inflection point, according to the following conditions. With this method, no dis tinction is made between local and global extrema. Therefore, the extrema should be compared with the function values at the endpoints of the interval, as
D I F F E R E N T I A L
\ 1
!
1
1
1
1 1
1
i
l
x cl
a
x c2
f " ( x 2) = ( 6 ) (  l ) + 2 =  4
local m axim um \
Therefore, X\ is a local minimum point (because f " ( x 1) is positive), and ^ is a local maximum point (because f " { x 2) is negative). The inflection point between these two extrema is found by setting f " ( x ) equal to zero.
\ \ \
\

 global  \ / m inim um I j> k^ _singular j  point  /
x c3
//
xs
f " ( x ) = 6x + 2 = 0 or x = —
b
illustrated in Ex. 9.3.2 Generally, f \ x ) ^ 0 at an inflec tion point.
Since the question asked for the global extreme points, it is necessary to compare the values of f(x) at the local extrema with the values at the endpoints. / (  2) =  l
f \ x c) = 0 at any extreme point, xc
9.30
f " ( x c) < 0 at a maximum point
9.31
f " ( x c) > 0 at a minimum point
9.32
f " ( x c) = 0 at an inflection point
9.33
/ (  1) =
2
/ ( i)  22/27 m
= 11
Therefore, the actual global extrema are the endpoints.
E xam ple 9.3 Find the global extrema of the function f(x) on the interval [—2 , + 2 ]. f(x)
= X3 +
x2 — X +
1
Solution The first derivative is f\ x ) = 3x2 j 2x — 1
4„ DERIVATIVES ©F PARAMETRIC EQ IM YIO HS The derivative of a function f ( x 1 , 0:2 , . . . , xn) can be calculated from the derivatives of the parametric equa tions / 1 (s) , f 2(s) , . . . , f n(s). The derivative will be expressed in terms of the parameter, 5 , unless the deriv atives of the parametric equations can be expressed explicitly in terms of the independent variables. E xam ple 9.4 A circle is expressed parametrically by the equations
Since the first derivative is zero at extreme points, set f'(x) equal to zero and solve for the roots of the quad ratic equation.
x = 5 cos 0
3x2 + 2x — 1 = (3x — l )( x + 1) = 0
Express the derivative dy/dx (a) as a function of the parameter 9 and (b) as a function of x and y.
The roots are x\ = V3 , x2 = —1. These are the locations of the two local extrema. The second derivative is f " ( x ) = 6x +
y = 5 sin 9
Solution (a) Taking the derivative of each parametric equation with respect to 0,
2
2It is also necessary to check the values of the function at singular points (i.e., points where the derivative does not exist).
P P B
•
w w w , p p i 2 p a s s . e Q m
Mathematics
/ " ( * 1 ) = (6 ) (  ) + 2 = 4
1 1
local] ! m inim um 1
9  3
Substituting x\ and x^ into f " ( x ),
Figure 9.2 Extreme and Inflection Points I global m axim um j\at endpoint l\ 1\ 1\ 1 \ 1 \ 1 \ / 1 \ inflection / i \ p o in t / \ /I 1 \ / i
C A L C U L U S
9  4
P O W E R
R E F E R E N C E
M A N U A L
Then,
E xam ple 9.6 dy dy_= dO _ 5 cos0 dx dx —5 sin t dQ
—cot
A surface has the equation x2 + y2 2? — 9 = 0. What is the slope of a line that lies in a plane of constant y and is tangent to the surface at (x,y,z) = ( 1 , 2 , 2 ) ? 3 Solution
(b) The derivatives of the parametric equations are closely related to the original parametric equations.
Solve for the dependent variable. Then, consider vari able y to be a constant.
dx = —5 sin# : dO dy _
:= y j 9
5 cos 6 = x
dy d y = d6 dx dx dO
— x2 — y2
d z = d ( 9  a : 2  y 2) 1/2
dx
dx
—x
y yj9
— x2 — y2
5. PARTIAL Dl FFERENTI ATI ON..................... Derivatives can be taken with respect to only one inde pendent variable at a time. For example, f \ x ) is the derivative of f(x) and is taken with respect to the inde pendent variable x. If a function, f(x\, x2, £3 ,. . .) , has more than one independent variable, a partial derivative can be found, but only with respect to one of the inde pendent variables. All other variables are treated as constants. Symbols for a partial derivative of / taken with respect to variable x are df/dx and fx(x, y).
At the point ( 1 , 2 , 2 ), x = 1 and y =
dz dx
2.
1 (1,2,2)
v/ 9  ( 1 ) 2  ( 2
)2
The geometric interpretation of a partial derivative df/dx is the slope of a line tangent to the surface (a sphere, ellipsoid, etc.) described by the function when all variables except x are held constant. In threedimensional space with a function described by z = f(x, y), the partial derivative df/dx (equivalent to dz/dx) is the slope of the line tangent to the surface in a plane of constant y. Similarly, the partial derivative df/dy (equivalent to dz/dy) is the slope of the line tangent to the surface in a plane of constant x. E xam ple 9.5
z = 3x2 — 6y2 + xy + by — 9
When a relationship between n variables cannot be manipulated to yield an explicit function of n — 1 inde pendent variables, that relationship implicitly defines the nth variable. Finding the derivative of the implicit variable with respect to any other independent variable is known as implicit differentiation.
The partial derivative with respect to x is found by considering all variables other than x to be constants.
An implicit derivative is the quotient of two partial derivatives. The two partial derivatives are chosen so that dividing one by the other eliminates a common differential. For example, if z cannot be explicitly
What is the partial derivative dz/dx of the following function?
Solution
^ = Q x  0 + y + 0  0 = 6xOX
PP1
•
w w w . p p i 2 p a s S o C @ m
3Although only implied, it is required that the point actually be on the surface (i.e., it must satisfy the equation f(x, y, z) = 0).
D I F F E R E N T I A L
dz _ dx~
dx df dz
_dl dzL_ __dy_ d y ~ df_ dz
9.34
9" 5
7. TANGENT PLANE FUNCTION..................... Partial derivatives can be used to find the equation of a plane tangent to a threedimensional surface defined by /(# , y,z) = 0 at some point, P0.
T(xo, y0, z0) = (x  xq)
df(x, y, z) dx Po df(x, y, z)
9.35
+ { y ~ Vo) + { z — Zq)
Example 9.7
dy
df(x, y, z) dz 9.36
Find the derivative dy/ dx of / ( x, y) = x2 + xy + y3 Solution
The coefficients of x, y, and z are the same as the coefficients of i, j, and k of the normal vector at point P0. (See Sec. 9.10.)
Implicit differentiation is required because x cannot be extracted from /(# , y).
Example 9.9
f * = 2x+y
What is the equation of the plane that is tangent to the surface defined by /(# , y, z) = 4x2 + y2 — 1 6 2 : = 0 at the point (2,4, 2 )? Solution
dy
dy dx
Calculate the partial derivatives and substitute the coordinates of the point.
d x ....  Q + y) df_ x\ 3 y2 dy
Example 9.8
df{x, y, z) dx
= 8*1(2,4,2) = (8)(2) = 16
df(x, y, z) dy
— 22/1(2,4,2) — (2 ) (4 ) — 8
Solve Ex. 9.6 using implicit differentiation. Solution f{x, y, z) = x2 + y2 + z2 — 9 = 0 d£_ 2x dx d l= 2 z dz
dfjx, y, z) dz
Po
T (2,4,2) = (16)(x  2) + (8)(y  4)  (16)(^  2) = I6x + 8y — 16z — 32 Substitute into Eq. 9.36, and divide both sides by
8.
2x\ y — 2z — 4 = 0 dz _ dx
df dx _ —2x _ _ x df 2z z dz
At the point ( 1 , 2 , 2 ), dz _ dx
1 2
8. GRADIENT VECTOR...................................... The slope of a function is the change in one variable with respect to a distance in a chosen direction. Usually, the direction is parallel to a coordinate axis. However, the maximum slope at a point on a surface may not be in a direction parallel to one of the coordinate axes.
P P 8 ® w w w . p p S 2 p a s s ac © m
Mathematics
extracted from f(x, y,z) = 0 , the partial derivatives dz/dx and dz/dy can still be found as follows.
C A L C U L U S
9  6
P O W E R
R E F E R E N C E
M A N U A L
Mathematics
The gradient vector function V /(x , y, z) (pronounced “del f ”) gives the maximum rate of change of the func tio n /(x , y, z). V f ( x , v, , ) = V
^
I
+ W
does not need to be, as only the direction cosines are calculated from it. ry // \ y, z) Vuf(x, y , z ) = (   l cos a dx
^ 3
, 9f(x,y±z)_, dz
+ 9.37
df(x, y, z) dy
fdf{x,y,z)\
+ (V~ ^ ^ J C0S7 E xam ple 9.10
V = U Xi + Uyj + U zk
A twodimensional function is defined by
cos a 
Ux 'U
9.39
Ux
f(x, y) = 2x2  y 2 + 3 x  y
yf
9.40
ui+ui+uj UV
(a) What is the gradient vector for this function? (b) What is the direction of the line passing through the point ( 1 , 2 ) that has a maximum slope? (c) What is the maximum slope at the point (1, —2)?
9.38
9.41
cosp =W \ 9.42
cos^ = W \
Solution
E xam ple 9.11
(a) It is necessary to calculate two partial derivatives in order to use Eq. 9.37.
What is the rate of change of /(# , y) = 3x2 + xy — 2y2 at the point (1 ,2 ) in the direction 4i + 3j?
d / 0 , y) = 4x + 3 dx
Solution The direction cosines are given by Eq. 9.40 and Eq. 9.41. Ux
dfjx, y) dy
^ ( 4 ) 2 + (3)
=  2 y  l
^ = k r !
V / 0 , y) = (4x + 3)i + (  2 y  l)j (b) Find the direction of the line passing through (1, —2) with maximum slope by inserting x = 1 and y = —2 into the gradient vector function.
The partial derivatives are 9f(x, y) = 6x+ y dx
V = ((4)(1) +3)i + ((—2)(—2) —l)j
dfjx, y) dy
— 7i + 3j (c) The magnitude of the slope is IV = y/(7)2 + (3)2 = 7.62
2
4 5
= x — Ay
The directional derivative is given by Eq. 9.38. V „/(z , y) = 0
(6z + y ) + 0
(x  4y)
Substituting the given values of x = 1 and y = —2, V « /( l, 2 ) = () ((6)(1)  2) + ( ) ( l  (4 )(—2))
9. DIRECTIONAL DERIVATIVE Unlike the gradient vector (covered in Sec. 9.8), which calculates the maximum rate of change of a function, the directional derivative, indicated by X7uf ( x , y , z ) , Duf ( x ,y , z ), or f u(x,y,z), gives the rate of change in the direction of a given vector, u or U. The subscript u implies that the direction vector is a unit vector, but it
P>p 0
•
w w w . p p i 2 p a s s . c @ m
=f (86) 10, NORMAL LINE VECTOR............................. Partial derivatives can be used to find the vector normal to a threedimensional surface defined by /(# , y, z) = 0
D I F F E R E N T I A L
0 " 7
Solution
Mathematics
at some point P0. The coefficients of i, j, and k are the same as the coefficients of x, y, and z calculated for the equation of the tangent plane at point P0. (See Sec. 9.7.)
C A L C U L U S
From Eq. 9.45, r\
9f(x, y, z) N = dx
div F =
df(x, y, z) i+ dy Po
df{x, y, z) + dz
d_ d_ xz + exy + — Ix^y = z + ex + 0 dz dx' dy
= zh ex 9.43
Po
12. CURL OF A VECTOR FIELD.......................
E xam ple 9.12 What is the vector normal to the surface f(x, y, z) = 4x2 + y1 — 16z = 0 at the point (2,4,2)?
of
Solution The equation of the tangent plane at this point was calculated in Ex. 9.9 to be
The curl, curl F, of a vector field F(e, y, z) is a vector field defined by Eq. 9.49 and Eq. 9.50. The curl F can be interpreted as the vorticity per unit area of flux (i.e., a flowing substance) in a small region (i.e., at a point). One of the uses of the curl is to determine whether flow (represented in direction and magnitude by F) is rota tional. Flow is irrotational if curl F = 0. F = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k
T(2,4,2) = 2x + y  2z  4 = 0
rdR
A vector that is normal to the tangent plane through this point is
dx
9.49
dy)
It may be easier to calculate the curl from Eq. 9.50. (The vector del operator, V, was defined in Eq. 9.47.)
11. DIVERGENCE,OF A VECTOR FIELD The divergence, div F, of a vector field F (x,y,z) is a scalar function defined by Eq. 9.45 and Eq. 9.46.4 The divergence of F can be interpreted as the accumulation of flux (i.e., a flowing substance) in a small region (i.e., at a point). One of the uses of the divergence is to determine whether flow (represented in direction and magnitude by F) is compressible. Flow is incompressible if divF = 0, since the substance is not accumulating.
A. « dP dQ dR divF = — + —^ + — ox dy ou oz
(dP _ dR\ . ^dz dxr
vQy
N=2i+j2k
F = P(x, y, z)i + Q(x, y, z)j + R(x, y, z)k
dQ\.
9.48
curl F = V x F i d dx
d_
d_ dy
x, y, z)
dz
9.50
R(x, y, z)
Q(x, y, z)
If the velocity vector is V , then the vorticity is
9.44 9.45
k
j
ijj — V x V = ujxi +
+ (jJzk
9.51
The circulation is the line integral of the velocity, V , along a closed curve.
It may be easier to calculate divergence from Eq. 9.46. div F = V •F The vector del operator, V, is defined as Y?
__
d • d x 1'
uj dA
r = (pV •ds ■
9.46
9.52
E xam ple 9.14 9.47
If there is no divergence, then the dot product calculated in Eq. 9.46 is zero. E xam ple 9.13
Calculate the curl of the following vector function. F(:r, y, z) = 3x2i + 7exyj Solution Using Eq. 9.50,
Calculate the divergence of the following vector function. F(x, y, z) = xzi + exyj + 7x3yk
i curl F =
j
d_ d_ dx dy 3a;2 7 exy
4A bold letter, F, is used to indicate that the vector is a function of x, y, and z.
PP g
•
k d_ dz 0
w w w , p p g 2 p a s § Be © m
9  8
P O W E R
R E F E R E N C E
M A N U A L
14. MACLAURIN POWER APPROXIMATIONS
Expand the determinant across the top row.
Mathematics
£


If a = 0 in the Taylor series, Eq. 9.53 is known as the Maclaurin series. The Maclaurin series can be used to approximate functions at some value of x between 0 and 1. The following common approximations may be referred to as Maclaurin series, Taylor series, or power series approximations.
J( s < ° >  1 M )
+ k ( l (7e’ 9 ,“ l i (3j:!)) = i(0 — 0) —j(0 — 0) + k (7exy  0)
✓v.3
= 7exyk
13. TAYLOR’S FORMULA
. T
O
n!
( b — a)n + Rn(b)
9.53
In Eq. 9.53, the expression f n designates the nth deriv ative of the function f(x). To be a useful approximation, two requirements must be met: (1) point a must be relatively close to point 6, and (2) the function and its derivatives must be known or easy to calculate. The last term, Rn( b), is the uncalculated remainder after n deriv atives. It is the difference between the exact and approx imate values. By using enough terms, the remainder can be made arbitrarily small. That is, Rn(b) approaches zero as n approaches infinity. It can be shown that the remainder term can be calcu lated from Eq. 9.54, where c is some number in the interval [a, b\. With certain functions, the constant c can be completely determined. In most cases, however, it is possible only to calculate an upper bound on the remainder from Eq. 9.55. Mn is the maximum (positive) value of f n+1(x) on the interval [a, 6].
n ( ) _ (n + l ) ! (
}
(6 — a )n+1
J*n(6) < M,
•
(n + 1)!
w w w . p p i 2 p a s s nc ® m
9 .5 4
9.55
rJ
.
tij
x^_ 4!
. /
\ Ti
_
J2n+l
_
9.56
,2n
n
X
(2 n)!
6!
^,2n+l . 1 . X . X . X . . smh x k x + — + — + — H h (2n+l)! u
_
t
. XT . X *
C O S h ^ l +  + 
^
. X”
+ 
,
+
l + Z+ t +  + "
9.57
9.58
^.2n
+
9.59
(2 n)\
x'° n\
9.60
n 1
/W = /(a )+ ^ (6 a )+ ^ (6 a )2 2! “+
x?_ 2!
cos xt
Taylor’s formula (series) can be used to expand a func tion around a point (i.e., approximate the function at one point based on the function’s value at another point). The approximation consists of a series, each term composed of a derivative of the original function and a polynomial. Using Taylor’s formula requires that the original function be continuous in the interval [a, b} and have the required number of derivatives. To expand a function, /(x ), around a point, a, in order to obtain /(&), Taylor’s formula is
P P> g
~5
• vLf , smISSX —_ +
1 + x + x2 + x3 +  V xn
9.61
9.62
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.
0
Mathematics
1
Integral Calculus
Integration...................................................... Elementary Operations................................. Integration by P arts...................................... Separation of Term s...................................... Double and HigherOrder Integrals............. Initial Values.................................................. Definite Integrals............................................ Average V alue................................................ A rea................................................................. Arc Length...................................................... Pappus’ Theorems......................................... Surface of Revolution.................................... Volume of Revolution.................................... Moments of a Function................................. Fourier Series.................................................. Fast Fourier Transforms............................... Integral Functions..........................................
101 101 102 103 103 104 104 104 104 105 105 105 106 106 106 108 108
Equation 10.2 and Eq. 10.3 are particularly useful. (C and k represent constants. f(x) and g(x) are functions of x.)
I II
kdx:=kx + C
xm dx =
,
X‘771+1
771+ 1
+
10.1
While most of a function, /(#), can be “recovered” through integration of its derivative, / ( # ) , a constant term will be lost. This is because the derivative of a constant term vanishes (i.e., is zero), leaving nothing to recover from. A constant of integration, C, is added to the integral to recognize the possibility of such a constant term.
2. ELEM EN TA R Y OPERATIONS Equation 10.2 through Eq. 10.8 summarize the elemen tary integration operations on polynomials and exponentials.2 lrThe difference between an indefinite and definite integral (covered in Sec. 10.7) is simple: An indefinite integral is a function, while a definite integral is a number. 2More extensive listings, known as tables of integrals, are widely avail able. (See App. 10.A.)
10.3
10.4
J ekxdx = ^ + C
10.5
xekxdx
ekx(kx  1)
C
10.6
/■
■+C a Ink
In x dx = x In x — x + C
Integration is the inverse operation of differentiation. For that reason, indefinite integrals are sometimes referred to as antiderivatives} Although expressions can be functions of several variables, integrals can only be taken with respect to one variable at a time. The differential term (dx in Eq. 10.1) indicates that variable. In Eq. 10.1, the function f ( x ) is the integrand, and x is the variable of integration.
J f ( x ) d x = f(x) + C
[m 7^  1 ]
J — d x = In x + C
kaxd x = 
1. !INTEGRATION
C
10.2
10.7
10.8
Equation 10.9 through Eq. 10.20 summarize the elemen tary integration operations on transcendental functions
/ /
sin xdx = —cos x + C
10.9
cos x dx = sin x + C
10.10
/
tan x d x = In sec x\ + C = —In cos rr + C
J
cot x dx = In Isin x\ + C
10.11 10.12
J sec x d x = lnsec x + tan x\ + C = In
I
+ C
10.13
J csc x d x = lncsc x —cot x\ + C , x = In t a n  + C
dx k2 + x2
P P
10.14
10.15
k
a
® w w w , p p i 2 p a s s no ® m
102
P O W E R
Mathematics
/ /
 
R E F E R E N C E
= arcsin y + C
dx _______ = \ arcsec y + C x V x2  k2 k *
M A N U A L
[k2 > x2]
10.16
[x2 > k2'
10.17
J sm2x d x = \ x  \ s m 2 x + C
10.18
J cos2a: d x = \ x —\ sin 2 x + C
10.19
J
tan x dx = tan x — x + C
10.20
Equation 10.21 through Eq. 10.26 summarize the ele mentary integration operations on hyperbolic transcen dental functions. Integrals of hyperbolic functions are not completely analogous to those of the regular trans cendental functions.
Solution This is a polynomial function, and Eq. 10.3 can be applied to each of the three terms.
J (3x2 + \x — 7) dx = x3 + \x2 — 7 x + C
3. INTEGRATION BY PARTS Equation 10.30, repeated here, is known as integration by parts. f(x) and g{x) are functions. The use of this method is illustrated by Ex. 10.2.
J f(x)dg(x) = f(x)g(x)  J g(x)df(x) + C
10.31
Example 10.2
//
sinh x dx = cosh x + C
Find the following integral.
cosh x dx = sinh x + C
;2exdx
70.22
/■
J tanh a; dx = In coshx + C
70.23
J coth x dx = In sinh
10.24
!
+ C
Solution a?ex is factored into two parts so that integration by parts can be used. f (x) = x2
J sech x d x = arctan (sinh z) + C
10.25
csch x dx = In tanh 
J (f(x) + g ( x ) ) d x = J f (x )d x + J g{x)dx
J J(x)dX=ln +C
df(x) = 2x dx
10.26
Equation 10.27 through Eq. 10.30 summarize the ele mentary integration operations on functions and combi nations of functions
J k f { x ) d x = k J f(x)dx
dg{x) = exdx
10.27
10.28
10.29
J f(x)dg(x) = f ( x ) J dg{x) — J g(x)df(x) + C
g(x) = J
dg(x) = J exdx = ex
From Eq. 10.31, disregarding the constant of integration (which cannot be evaluated),
I
J f(x)dg(x) = f(x)g(x)  J g(x)df{x)
x2exd x = x2ex
J ex(2x)di
The second term is also factored into two parts, and integration by parts is used again. This time,
= f(x)g(x)  J g(x)df(x) + C f( x) = x 10.30
dg(x) = exdx df(x) = dx
Example 10.1 Find the integral with respect to x of 3a;2 + \x — 7
IP P 8 •
w w w ap p i 2 p a s s . c ® m
g(x)  J
dg(x) = J
exdx 
From Eq. 10.31,
3x + 2 dx 3x — 2
/
J 2xexd x = 2 J xexdx = 2^xex — J exdx
C A L C U L U S
1+
3x2)
Idx+I
= 2(xex  ex)
10~ 3
Idx
4 dx 3x2
xhin(3a; — 2) + C
Then, the complete integral is
J x2exd x = x 2ex  2(xex  ex) + C = ex(x2 — 2x + 2) + C
4„ SEPARATION OF TER M S Equation 10.28 shows that the integral of a sum of terms is equal to a sum of integrals. This technique is known as separation of terms. In many cases, terms are easily separated. In other cases, the technique of partial frac tions can be used to obtain individual terms. These techniques are illustrated by Ex. 10.3 and Ex. 10.4. Example 10.3 Find the following integral. (2x2 + 3)2
I
B* DOUBLE AMB HIGHERORDER INTEGRALS A function can be successively integrated. (This is anal ogous to successive differentiation.) A function that is integrated twice is known as a double integral; if inte grated three times, it is a triple integral, and so on. Double and triple integrals are used to calculate areas and volumes, respectively. The successive integrations do not need to be with respect to the same variable. Variables not included in the integration are treated as constants. There are several notations used for a multiple integral, particularly when the product of length differentials represents a differential area or volume. A double inte gral (i.e., two successive integrations) can be repre sented by one of the following notations.
J J f( x,y) dxdy,
dx
J f( x,y) dxdy,
« / / / ( * , y)dA
Solution
A triple integral can be represented by one of the follow ing notations.
J J J f(x, y, z) dx dy dz,
= J y ix + I2x +  j dx = xA+ 6x2 + 9 In \x\ + C
Example 10.4
L f i x , y, z) dx dy dz,
or■ J J J f ( x , y , z ) d V
Example 10.5
Find the following integral. 3x 2 dx 3x — 2
/ Solution
Find the following double integral.
J
Solution
The integrand is larger than 1, so use long division to simplify it.
J x2 (
+
y3x)dx
=

x3 + \y3x2 + C\
(3^+\yix2 +^0dy=\yx*+1yix2 + C\y+ c 2
3x — 2 13x + 2 3z  2 4 remainder
So,
J J (x2 + y3x)dxdy = lyx3 + l y 4x2 + C \y+ C2
Mathematics
I N T E G R A L
10 4
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6. INITIAL VALUES Mathematics
The constant of integration, £7, can be found only if the value of the function f(x) is known for some value of Xq. The value f(x o) is known as an initial value or initial condition. To completely define a function, as many initial values, f(x o), f ( x i), f"(x 2 ), and so on, as there are integrations are needed, xq, x\, x2, and so on, can be, but do not have to be, the same.
A common use of a definite integral is the calculation of work performed by a force, F, that moves an object from position Xi to x^. The force can be constant or a function of x. F dx
W
10.33
Example 10.7 Example 10.6
Evaluate the following definite integral.
It is known that f(x) = 4 when x = 2 (i.e., the initial value is /(2 ) = 4). Find the original function.
rn/3 n/tr
Jn/A
/
(■3x3 — 7x)dx
sin x dx
Solution
Solution
From Eq. 10.32,
The function is / (
x) = J (3a
f n/S r /3 7i f n\ / sin x dx = —cos x ; = —c o s  — —c o s Jn /4 '* / 4 3 V A)
7x) dx
= 0 .5  (0.707)
= \x4  \x2 + C
= 0.207
Substituting the initial value determines C. 4 = (f)(2 )4  (I)(2)2 + ^
8. AVERAGE VALUE
4 = 12 — 1 4 + C The average value of a function f (x) that is integrable over the interval [a, b\ is
axis. What is the surface of revolution?
1]
10.51
0/
jNtt
2
tt/'
2
bn =
\3tj 4.77
f(t)sm ntdt
[n>l\
10.52
While Eq. 10.51 and Eq. 10.52 are always valid, the work of integrating and finding an and bn can be greatly simplified if the waveform is recognized as being sym metrical. Table 10.1 summarizes the simplifications.7
0 < t < n
0
n < t < 2n
1^11
f(f+7T) =  f ( t )
o/TVr '
2 7T
1
0
f
an =
0
[even n]
bn =
0
[all n]
a0 = 0 an =
0
[all n)
bn =
0
[even n]
rlT ?r an = l p ( f) cos n td t bn =  l fit) sin n td t Jq [odd n] ^0 [odd n]
IA,I
lAotall =2 1 ^ 1
*same as rotational sym m etry
m
quarterwave sym m etry fit + IT)   f { t )
fit)
0 / 1 )2 X77
2
tt
\2j2j
t
a0 =
0
a0 =
0
an =
0
[even n]
an =
0
bn =
0
[all n]
bn = 0 [even n]
a o =h
\A2\ = l/4 1
2tr
1f(t) cos n td t
an = §
'Aotal1 = ^IAi 1
[all n\
TT
n
77
[all n]
halfwave sym m etry*
\A2\ =
1
0
*any repeating w aveform
Example 10.11 Find the first four terms of a Fourier series that approx imates the repetitive step function illustrated.
a0 — 0
[all n]
r2n f 2 tt f(t) cos nt dt bn = 1 f(t) sin n td t J0 [a" n] J0 [a" n]
an = l
1^21 = 1^11 ^total^ —
0
an =
W
f(t) sin nt dt
2_ f T Jo
4
fit + 2 tt) = fit)
I J0 7 _ 1 bn —
t
fullw ave sym m etry*
i r ZK an = 
1 0 7
Mathematics
The coefficients an and bn are found from the following relationships.6
C A L C U L U S
[odd n]
^0
bn =^~ I f(f) sin nt dt J() [odd n]
Solution
From Eq. 10.50,
i l
This value of V2 corresponds to the average value of
.1 mdt+h
L {a)dt
2
6Several valid forms of the Fourier series are used. The one shown, Eq. 10.50, is consistent with that used by NCEES. Other forms remove the fraction V2 embedded in the Oq term in Eq. 10.50 and instead show it specifically as 1/2 a0 in an equation such as Eq. 10.47. Additionally, other forms show the interval of integration as —n to +tt. Because of the periodicity of the integrands, the interval of integration may be replaced by any other interval of length 271, or, if the period is other than 271, by the interval for the period T. rThe Fourier series of an even function is called a Fourier cosine series. The Fourier series of an odd function is called a Fourier sine series. If both the cosine and sine terms are present in a given function, the Fourier series may be written with coefficients labeled as cn that are combinations of the an and bn coefficients.
f ( t ) . It could have been found by observation.
ai = ± f ( l ) c o S^
+ ± f ( 0 > cos t dt =  sin t
n Jo
0
71
n Jn
In general, an —
1 r bi = ^ J0
1 sin n t n
n
1 rzn(0)sin t dt
(l)sm tdt + —
^ Jn
©
0]
12.34
A is both the mean and the variance of the Poisson distribution. Example 12.9
p{1} = C ) r r _ x = ( i J (005)1(095)6 = (7)(0.05)(0.95)6
The number of customers arriving at a hamburger stand in the next period is a Poisson distribution having a mean of eight. What is the probability that exactly six customers will arrive in the next period? Solution
p {x > 2} = 1 — ((0.95)7 + (7)(0.05)(0.95)6^
A = 8 and x = 6. From Eq. 12.34,
= 1  (0.6983 + 0.2573) = 0.0444
10. HYPERGEOMETRIC DISTRIBUTION Probabilities associated with sampling from a finite population without replacement are calculated from the hypergeometric distribution. If a population of finite size M contains K items with a given characteristic (e.g., red color, defective construction), then the probability
p{6} =
e~AA* x\
6!
=
0.122
13. CONTINUOUS DISTRIBUTION FUNCTIONS Most numerical events are continuously distributed and are not constrained to discrete or integer values. For example, the resistance of a 10% 1 O resistor may be any value between 0.9 and 1.1 fi. The probability of
P P § ® w w w . p p i 2 p a s s . c o m
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an exact numerical event is zero for continuously dis tributed variables. That is, there is no chance that a numerical event will be exactly x 2 It is possible to determine only the probability that a numerical event will be less than x, greater than x, or between the values of X\ and 0%, but not exactly equal to x. While an expression, f(x), for a probability density func tion can be written, it is used to derive the continuous distribution function (CDF), F ( xq), which gives the probability of numerical event Xq or less occurring, as illustrated in Fig. 12.3. fXo p { X < x0} = F(xo) = / }{x )d i Jo /o dFjx)
f (x) = '
dx
bellshaped curve, which represents the distribution of outcomes of many experiments, processes, and phenom ena. (See Fig. 12.4.) The probability density and con tinuous distribution functions for the normal distribution with mean fi and variance a2 are i (xny rf \
/(* ) = '
e 2U
i
12.41
[— OO < X < + o o ]
(t \ / 2 tu
p{fi < X < x 0} = F ( xq) x°
I
2
e“ 2V a ) dx l— ( tV^ k Jo
12.42
12.35 Figure 12.4 Normal Distribution 12.36
Figure 12.3 Continuous Distribution Function
14. EXPONENTIAL DISTRIBUTION The continuous exponential distribution is given by its probability density and continuous distribution functions.
Since f(x) is difficult to integrate, Eq. 12.41 is seldom used directly, and a standard normal table is used instead. (See App. 12.A.) The standard normal table is based on a normal distribution with a mean of zero and a standard deviation of 1. Since the range of values from an experi ment or phenomenon will not generally correspond to the standard normal table, a value, xq, must be converted to a standard normal value, z. In Eq. 12.43, ^ and a are the mean and standard deviation, respectively, of the distri bution from which x$ comes. For all practical purposes, all normal distributions are completely bounded by /jl ± 3cr. x0  n
f (x ) = Xe Xx p { X < x} = F(x) = 1 — e~Xx
12.37 12.38
The mean and variance of the exponential distribution are 1 A a
_1_ A2
12.39
12.40
15. NORMAL DISTRIBUTION The normal distribution (Gaussian distribution) is a symmetrical distribution commonly referred to as the It is important to understand the rationale behind this statement. Since the variable can take on any value and has an infinite number of significant digits, we can infinitely continue to increase the precision of the value. For example, the probability is zero that a resistance will be exactly 1 S7 because the resistance is really 1.03 or 1.0260008 or 1.02600080005, and so on.
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@
www. p p g 2 p a s s »c @r o
12.43
Numbers in the standard normal table, as given by App. 12.A, are the probabilities of the normalized x being between zero and z and represent the areas under the curve up to point z. When x is less than /z, z will be negative. However, the curve is symmetrical, so the table value corresponding to positive z can be used. The probability of x being greater than z is the comple ment of the table value. The curve area past point z is known as the tail of the curve. The error function, erf(x), and its complement, erfc(a:), are defined by Eq. 12.44 and Eq. 12.45. The error func tion is used to determine the probable error of a measurement. rxo 9 Cx erf(xo) =  7 = / e~x*dx
12.44
V n Jo erfc(xo) = 1 —erf(xo)
12.45
AND
Example 12.10 The mass, ra, of a particular handlaid fiberglass (Fibreglas™) part is normally distributed with a mean of 66 kg and a standard deviation of 5 kg. (a) What percent of the parts will have a mass less than 72 kg? (b) What percent of the parts will have a mass in excess of 72 kg? (c) What percent of the parts will have a mass between 61 kg and 72 kg?
S T A T I S T I C A L
A N A L Y S I S
OF
D A T A
12“7
failures from a large population of items) as a function of time. Items initially fail at a high rate, a phenomenon known as infant mortality. For the majority of the oper ating time, known as the steadystate operation, the fail ure rate is constant (i.e., is due to random causes). After a long period of time, the items begin to deteriorate and the failure rate increases. (No mathematical distribution describes all three of these phases simultaneously.) Figure 12.5 Bathtub Reliability Curve
Solution (a) The 72 kg value must be normalized, so use Eq. 12.43. The standard normal variable is z= ■
•ii _ 72 kg — 66 kg 5 kg
failure rate
infant deterioration ! m ortality period I period steadystate period
1.2
Reading from App. 12.A, the area under the normal curve is 0.3849. This represents the probability of the mass, ra, being between 66 kg and 72 kg (i.e., z being between 0 and 1.2). However, the probability of the mass being less than 66 kg is also needed. Since the curve is symmetrical, this probability is 0.5. Therefore,
The hazard function, z{t}, represents the conditional probability of failure— the probability of failure in the next time interval, given that no failure has occurred thus far.3
p{m < 72 kg} = p {z < 1.2}  0.5 + 0.3849 = 0.8849 (b) The probability of the mass exceeding 72 kg is the area under the tail past point z.
z {t} =
f( t) R(t)
dF(t) dt 1  F(t)
12.46
p{m > 72 kg} = p {z > 1.2} = 0.5  0.3849 = 0.1151 E x p o n en tia l R eliability
(c) The standard normal variable corresponding to ra = 61 kg is
Since the two masses are on opposite sides of the mean, the probability will have to be determined in two parts.
Steadystate reliability is often described by the nega tive exponential distribution. This assumption is appro priate whenever an item fails only by random causes and does not experience deterioration during its life. The parameter A is related to the mean time to failure (MTTF) of the item.4
p {6 1 < m < 72} = £>{61 < ra < 66} + p{66 < ra < 72}
R{ t } = e~xt = e*/MTTF
x  ii _ 61kg  66 kg z= ■ 5 kg
= p{ —1 < z < 0} + £>{0 < z < 1.2}
A:
12.47
1 MTTF
12.48
= 0.3413 + 0.3849 Equation 12.47 and the exponential continuous distri bution function, Eq. 12.38, are complementary.
= 0.7262 0.3413
R { t } = 1  F{t) = 1  (1  e“ At)
At
12.49
The hazard function for the negative exponential distri bution is z{ t } = A
12.50
In tro d u c tio n
Therefore, the hazard function for exponential reliabil ity is constant and does not depend on t (i.e., on the age of the item). In other words, the expected future life of an item is independent of the previous history (length of operation). This lack of memory is consistent with the
Reliability, R {t}, is the probability that an item will continue to operate satisfactorily up to time t. The bath tub distribution, Fig. 12.5, is often used to model the probability of failure of an item (or, the number of
3The symbol z{t} is traditionally used for the hazard function and is not related to the standard normal variable. 4The term “mean time between failures” is improper, However, the term mean time before failure (MTBF) is acceptable.
1
0
1.2
z
16 APPLICATION: RELIABILITY
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assumption that only random causes contribute to fail ure during steadystate operations. And since random causes are unlikely discrete events, their probability of occurrence can be represented by a Poisson distribution with mean A. That is, the probability of having £ failures in any given period is
Solution The probability of not having failed before time t is the reliability. From Eq. 12.48 and Eq. 12.49,
A=
 aaz p{x} = ■
MTTF
# { 1200 }
12.51
 e~Xt =
1 = 1000 hr
0 .0 0 1
hr  l
e (  °  001 hr1)(1200 hr) = q 3
XI
Serial System Reliability Example 12.12 In the analysis of system reliability, the binary variable Xi is defined as one if item i operates satisfactorily and zero if otherwise. Similarly, the binary variable is one only if the entire system operates satisfactorily. There fore, $ will depend on a performance function contain ing the X {.
What are the reliabilities of the following systems? (a)
A serial system is one for which all items must operate correctly for the system to operate. Each item has its own reliability, R{. For a serial system of n items, the performance function is $ = M
2X 3
   X n = m in(Xz)
12.52
The probability of a serial system operating correctly is 1} ~ ^serial system ~ R 1 R 2 R 3 ‘ ' * Rn
12.53
Parallel System Reliability A parallel system with n items will fail only if all n items fail. Such a system is said to be redundant to the nth degree. Using redundancy, a highly reliable system can be produced from components with relatively low indi vidual reliabilities. The performance function of a redundant system is
* =1 (1  *0(1  x2)(i  x3)•••(1 x n) 12.54
R = RiR2R3R4 = (0.93) (0.98) (0.91) (0.87) = 0.72 (b) This is a parallel system. From Eq. 12.55,
= 1  (1  0.76)(1  0.52)(1  0.39) = 0.93
The reliability of the parallel system is R = p {$ = 1}
17. ANALYSIS OF^ EXPERIMENTAL DATA
= 1  (1  f l i ) ( l  R2)( 1  Rs) •••(1  Rn) 12.55
E xam ple 12.11 The reliability of an item is exponentially distributed with mean time to failure (MTTF) of 1000 hr. What is the probability that the item will not have failed after 1200 hr of operation?
® w w w . p p g 2 p a s s . c o m
(a) This is a serial system. From Eq. 12.53,
R = l  ( 1  R 1) ( l  R 2) ( l  R 3)
= max(X^)
PPB
Solution
Experiments can take on many forms. An experiment might consist of measuring the mass of one cubic foot of concrete or measuring the speed of a car on a roadway. Generally, such experiments are performed more than once to increase the precision and accuracy of the results. Both systematic and random variations in the process being measured will cause the observations to vary, and the experiment would not be expected to yield the same result each time it was performed. Eventually, a
P R O B A B I L I T Y
AND
collection of experimental outcomes (observations) will be available for analysis.
S T A T I S T I C A L
OF
D A T A
12“9
(b)
The frequency distribution is a systematic method for ordering the observations from small to large, according to some convenient numerical characteristic. The step interval should be chosen so that the data are presented in a meaningful manner. If there are too many intervals, many of them will have zero frequencies; if there are too few intervals, the frequency distribution will have little value. Generally, 10 to 15 intervals are used. Once the frequency distribution is complete, it can be represented graphically as a histogram. The procedure in drawing a histogram is to mark off the interval limits (also known as class limits) on a number line and then draw contiguous bars with lengths that are proportional to the frequencies in the intervals and that are centered on the midpoints of their respective intervals. The con tinuous nature of the data can be depicted by a frequency polygon. The number or percentage of observations that occur up to and including some value can be shown in a cumulative frequency table.
A N A L Y S I S
7
6 & 5 c CD &
0
4
'CD sf
CD CD
00 CD
ID
O r
ID
c\i
LD
LD
■'tf Is "
CD Is *
cars per hour
(c)
Example 12.13 The number of cars that travel through an intersection between 12 noon and 1 p.m. is measured for 30 consecu tive working days. The results of the 30 observations are 79, 66, 72, 70, 68, 66, 68, 76, 73, 71, 74, 70, 71, 69, 67, 74, 70, 68, 69, 64, 75, 70, 68, 69, 64, 69, 62, 63, 63, 61 (a) What are the frequency and cumulative distribu tions? (Use a distribution interval of two cars per hour.) (b) Draw the histogram. (Use a cell size of two cars per hour.) (c) Draw the frequency polygon, (d) Graph the cumulative frequency distribution. Solution
cars per hour
(a)
(d) cars per hour 6061 6263 6465 6667 6869 7071 7273 7475 7677 7879
cumulative frequency
cumulative frequency
1 4
percent 3 13
6
20
9 17 23 25 28 29
30 57 77 83 93 97
30
100
cars per hour
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18. MEASURES OF CENTRAL TENDENCY Mathematics
The rootmeansquared (rms) value of a series of obser vations is defined as
It is often unnecessary to present the experimental data in their entirety, either in tabular or graphical form. In such cases, the data and distribution can be represented by various parameters. One type of parameter is a mea sure of central tendency. Mode, median, and mean are measures of central tendency. The mode is the observed value that occurs most fre quently. The mode may vary greatly between series of observations. Therefore, its main use is as a quick mea sure of the central value since little or no computation is required to find it. Beyond this, the usefulness of the mode is limited. The median is the point in the distribution that parti tions the total set of observations into two parts con taining equal numbers of observations. It is not influenced by the extremity of scores on either side of the distribution. The median is found by counting up (from either end of the frequency distribution) until half of the observations have been accounted for. For even numbers of observations, the median is esti mated as some value (i.e., the average) between the two center observations. Similar in concept to the median are percentiles (percen tile ranks), quartiles, and deciles. The median could also have been called the 50th percentile observation. Simi larly, the 80th percentile would be the observed value (e.g., the number of cars per hour) for which the cumu lative frequency was 80%. The quartile and decile points on the distribution divide the observations or distribution into segments of 25% and 10%, respectively. The arithmetic mean is the arithmetic average of the observations. The sample mean, x, can be used as an unbiased estimator of the population mean, /i. The mean may be found without ordering the data (as was necessary to find the mode and median). The mean can be found from the following formula.
12.56
+ x n) ~
The geometric mean is used occasionally when it is necessary to average ratios. The geometric mean is cal culated as geometric mean = ^/xix2x^ •••xn
\x{ > 0]
12.59
Example 12.14 Find the mode, median, and arithmetic mean of the distribution represented by the data given in Ex. 12.13. Solution First, resequence the observations in increasing order. 61, 62, 63, 63, 64, 64, 66, 66, 67, 68, 68, 68, 68, 69, 69, 69, 69, 70, 70, 70, 70, 71, 71, 72, 73, 74, 74, 75, 76, 79 The mode is the interval 6869, since this interval has the highest frequency. If 68.5 is taken as the interval center, then 68.5 would be the mode. The 15th and 16th observations are both 69, so the median is 69 The mean can be found from the raw data or from the grouped data using the interval center as the assumed observation value. Using the raw data,
I x
2069  68.97 30
19. M E ASURES OF DISPERSION..................... The simplest statistical parameter that describes the variability in observed data is the range. The range is found by subtracting the smallest value from the larg est. Since the range is influenced by extreme (low prob ability) observations, its use as a measure of variability is limited.
12.57
The standard deviation is a better estimate of variabil ity because it considers every observation. That is, in Eq. 12.60, N is the total population size, not the sample size, n.
12.58
12.60
The harmonic mean is defined as n harmonic mean = ■ — + — +•••+ — Xi
x2
xn
The standard deviation of a sample (particularly a small sample) is a biased (i.e., is not a good) estimator of the population standard deviation. An unbiased estimator
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A N D
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of the population standard deviation is the sample stan dard deviation, s.5
OF
71 =
D A T A
12~1 1
30
Mathematics
P R O B A B I L I T Y
_ 2069 „0 x= — = 68.967 30  x)2 n —1
n —1 (a)
^ — ^rnax
^min
79
61
18
If the sample standard deviation, s, is known, the stan dard deviation of the sample, crsample, can be calculated. 72— 1 ^sample —
12.62
1143,225 30
V
(b)
The variance is the square of the standard deviation. Since there are two standard deviations, there are two variances. The variance of the sample is cr2, and the sample variance is s2. The relative dispersion is defined as a measure of dis persion divided by a measure of central tendency. The coefficient of variation is a relative dispersion calculated from the sample standard deviation and the mean.
2
V r2 Z ,xi
(c)
\
12.63
\
3(x — median)
12.64
Example 12.15
4,280,761 30 29
= 4.287
Skewness is a measure of a frequency distribution’s lack of symmetry. x — mode
(& ) 1 ^
n n  1
143,225 coefficient of variation = ;
/2069V V 30 /
(d)
cr2 = 17.77
(e)
s2 = 18.38
20. CENTRAL LIMIT THEOREM.......................
For the data given in Ex. 12.13, calculate (a) the sample range, (b) the standard deviation of the sample, (c) an unbiased estimator of the population standard deviation, (d) the variance of the sample, and (e) the sample variance. Solution J^Xi = 2069
( J > ) 2 = (2069)2 = 4,280,761
= 143,225 5There is a subtle yet significant difference between standard deviation of the sample, a (obtained from Eq. 12.60 for a finite sample drawn from a larger population) and the sample standard deviation, s (obtained from Eq. 12.61). While a can be calculated, it has no significance or use as an estimator. It is true that the difference between o and s approaches zero when the sample size, n, is large, but this convergence does nothing to legitimize the use of cr as an estimator of the true standard deviation. (Some people say “large” is 30, others say 50 or 100.)
Measuring a sample of n items from a population with mean /jl and standard deviation a is the general concept of an experiment. The sample mean, x, is one of the parameters that can be derived from the experiment. This experiment can be repeated k times, yielding a set of averages (xi,x 2 The k numbers in the set themselves represent samples from distributions of averages. The average of averages, x, and sample stan dard deviation of averages, s^ (known as the standard error of the mean), can be calculated. The central limit theorem characterizes the distribution of the sample averages. The theorem can be stated in several ways, but the essential elements are the follow ing points. 1. The averages, sc*, are normally distributed variables, even if the original data from which they are calcu lated are not normally distributed. 2. The grand average, x (i.e., the average of the averages), approaches and is an unbiased estimator of (i. fit
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The standard deviation of the original distribution, /=!.
P P 1 ® w w w . p p i 2 p a s s . c o m
= cos x Incos x\ + x sin x See Chap. 11.
[II]
Combine the homogeneous (or complementary) solution I with the particular solution II to obtain the general solution.
E N G I N E E R I N G
/(0 ) = lim sF(s)
= ci cos x + C2 sin x + cos x lncos x\ + x sin x = (ci + lncosx)cos2 + (C2 + a;)sinx
15.24
The final value theorem, Eq. 15.25, can be used to determine the value of the function in the time domain as time approaches infinity.17
5. ORTHOGONALITY OF FUNCTIONS
/(o o ) = lim sF(s) s—>o
Let gm(x) and gn(x) be two realvalued functions defined on an interval a < x < b. Further, define the integral of the two as 15.20
15*5
the circuit, reacts at ^=0, use the initial value theorem, Eq. 15.24.
V(x) = yh(x) + yp{x)
(9m,9n)= f 9m(x)9n(x)dx Ja
M A T H E M A T I C S
15.25
Example 15.3 The Laplace transform of a given voltage is shown. Evaluate the voltage at t = 0.
The functions gm and gn are orthogonal on the interval a < x < b if the integral in Eq. 15.20 equals zero.14 That is,
V(s) =
4s ( 52 +
9 ) ( 52 +
l)
Solution {9m,9n)= [ 9m(x)9„{x)dx Ja = 0 [m^n]
Expand the denominator and divide the numerator and denominator by s2. 15.21
The nonnegative square root of (gm, gm) is called the norm of gm.
15.22
V(s)
4 s s2 + 1 0 + 9
As s4 + 10s2 + 9
Apply the initial value theorem, Eq. 15.24. v(0) = lim sV(s) s—>oo
®H LAPLACE TRANSFORM: INITIAL AND FINAL VALUE THEOREMS
/ = lim s s—>oo
4 s2 +
Laplace transforms simplify the solution of nonhomoge neous differential equations. Additionally, the initial conditions are automatically included in the analysis.15
10+4
j
= lim S—>00
1
—

= 0
n C + jo o
/ F(s) estds Jc—jo o
52 + 10
V
/ o t) = L  ' { F ( s ) } 15.23
Equation 15.23 can be used to obtain the inverse trans form (i.e., moves from the 5 domain back into the time domain), although the use of transform tables is more common. Further, the exact inverse transform may not be necessary. To understand how the equation, that is, 14Both Legendre polynomials and Bessel functions are orthogonal. Important orthogonal sets of functions arise as solutions to linear secondorder differential equations. 15The terms /(0) and /( 0 ) are required when transforming a secondorder differential, thereby immediately placing the initial conditions into use.
The voltage function associated with the given voltage is v(t) = sin t sin 2t The value of v(t) is indeed zero at t = 0 as sin(0) = 0, confirming the answer.
16/(0) is more technically /(0+), that is, the value at zero when approached from the positive side. This is based on the normally used Laplace transform being defined from zero to infinity. 17Technically, Eq. 15.25 should be lim f(t) = limsF(s).
>oo
t—
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Example 15.4
The coefficients of the series are given by the following.
Mathematics
Given the voltage in the frequency domain (or s domain), as shown, what is the value of the voltage at t = oo?18
0° =  y j f /(*)0
'"'1 C !  2 ? + s )
Figure 15.1 Periodic Extensions
= lim ( —  j = 1 S+0 \s2 + 2 5 + 1 / The voltage function associated with the given voltage is v(t) = 1 — e~l — te~l The value of v(t) as t —►oo can be seen to be equal to 1 from this equation, confirming the solution.19
(a) given function f(t)
7 a HALF*HANeiE EXPANSIONS In some engineering problems, a function f{t) defined over a finite interval may be more readily manipulated and a solution more easily obtained if the function is represented as a Fourier series. (See Sec. 15.8.) Given a function f(t) defined on some interval 0 < t < I or 0 < t < T/2 where T is the period,20 the even peri odic extension of f{t) of period T = 2 l is given by the Fourier cosine series.
cos
(b) even periodic extension of f(t)
f2(t)
1526
18Asking for the value of the voltage at t = 00 is the equivalent of asking for the steadystate value. 19As t approaches 00, the value of s approaches 0. In other words, as the function, or circuit, approaches a steadystate condition, the num ber of oscillations (frequency) approaches zero. 20The definition of the interval means the period is T/2 = I or T —21.
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1530
The series represented by Eq. 15.26 and Eq. 15.29 along with their associated coefficients are called the halfrange expansions of the given function f(t). Figure 15.1 shows a given function f{t) and its halfrange expansions. In the halfrange expansion in Fig. 15.1, fi(t) = f(t) on the interval 0 < t k
15.37
ut
‘
[VIII] By making this choice for Bn, Eq. VII becomes valid and Eq. VI satisfies initial condition III. One item now remains: make Eq. VI satisfy initial condition IV. Start by taking the 28This is not counting the constrained endpoints. 29The Fourier sine series results in an odd extension of f(x). In other words, the original function f(x) is extended in the coefficient B n which, when summed, leaves only f(x). See the graphs in Sec. 15.7.
the series represented in Eq. VI must converge, and the series obtained by differentiating Eq. VI twice (termwise) with respect to x and t must converge and sum to d2u/dx2 and d2u/dt2. Other solution methods are possible. One is called the D ’Alembert solution. Another is using a variation of parameters and Laplace transforms. Also, Laplace transforms can change the partial differential equation into an ordinary differential equation that can be solved using standard methods. 30A1so,
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Equation 15.37 mathematically represents what is shown in Fig. 15.2. The symbol H ( t—k) is called the Heaviside step function.31
Figure 15.3 Unit Finite Impulse Function force
1
Figure 15.2 Unit Step Function unit step
(a) step at f0
force
■■ 1 H (f  (f0 + h)) h
Unit Finite Impyise Fyncti©n I(h, t  t 0) = H(t  t0)  H (t  (t0 + h))
I ( h , t — t0) =
0
t < to and t > (to + A)
1
to < t < (to + A)
15.38
15.39
(b) step at f0 + h
force
The unit finite impulse function is shown graphically in Fig. 15.3. Of special importance in Fig. 15.3 part (c) is that the unit finite impulse function, that is, the shaded region, represents an area equal to 1 for any A > 0. This area represents a rectangle with the base equal to — (to + K) = h and the height equal to 1/A; therefore, the area is h(l/h) = l. The impulse is the integral of force (1/A) over the interval of time (A).
I [h, t — f0)
fn
fn + h (c) unit finite im pulse
Figure 15.4 Delta Function force
Haifa Function; Dirac Function; impyise
Function; Unit impyise Function
I I area = 1
Symbols: 6t; 6(t— to) If the limit as A —> 0 is taken on the unit finite impulse function, the result is the delta, or Dirac function 32 This function is sometimes called the impulse function or unit impulse function and not always differentiated from the unit finite impulse function.
lim /(A, t  to) = 6(t — to) —
h—>0
0
to 15.40
OO t = to
This is shown graphically in Fig. 15.4.33 31Oliver Heaviside was an English electrical engineer whose treatment of the derivative process (represented by D) in algebraic terms led to the development of the Laplace transform. 32Paul Dirac was an English physicist whose theory of negative energy holes predicted the existence of positrons. 33There is much more to the delta function. The mathematics shown is not strictly true. The delta function is not a proper function and is therefore shown as equivalent to (=) the limit.
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The delta function is strictly defined only in terms of integration. The delta function displays a property called sifting that is given by Eq. 15.41.34
/
foo
S 1 ( t  to)f(t)dt = f(to)
15.41
•oo
34The impulse function is a powerful analytical tool. For example, electrical signals can be represented as weighted sums of shifted impulses—essentially this can be the interpretation of Eq. 15.41. Using this interpretation, the response of a linear time invariant (LTI) system can be analyzed for any arbitrary input, including discrete inputs, that is, those inputs that are not continuous. (Recall that the term “linear” indicates that superposition can be used; “time invariant” means a time shift in the input signal causes an identical time shift in the output signal.) Using the principle of superposition, one can then obtain the output of an electrical circuit to any input. Using the unit finite impulse function, one can do the same for continuous time systems.
damma Function
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From a mathematical handbook of integrals, the result of the previous equation is
Symbols: T(q'); (a — 1)! Also known as the factorial function, the gamma func tion is defined by the following integral.35
r(2) =
a~t (i)2
 c
/•OO
T(a) = /
e~H^a~^dt
[a > 0]
15.42
 ( t + 1)
Jo
The gamma function is shown graphically in Fig. 15.5. Use L’Hopital’s rule to determine the value at infinity. Figure 15.5 Gamma Function
~{t + 1)
(0 + 1)
= 1
This result could also have been accomplished using the functional relation of Eq. 15.43 as shown. T { a + l ) = aT {a)
r(2) = r(i + i) = ir(i) = 1 Finally, the factorial symbol could have been used.36 T ( a ) = (a  1)!
r(2) = (2 —1)! = 1! = 1 A table of gamma function values is given in App. 15.A.
Error Fyneti@si and Complementary Error Function The error function, also called the probability integral, was defined in Chap. 10 as The important functional relation of the gamma func tion is r ( a + 1) = aT(a')
erf(x) = = _2_ r VnJo
15.44
15.43
2/y/n normalizes the function so that erf(oo) = l. By contrast, the complementary error function is defined as
Example 15.6 2 r° erfc(x) = 1 — erf(x) = —— I
Find T(2).
* dt
15.45
Jx
Solution
The error function is shown graphically in Fig. 15.6.
Apply the definition given by Eq. 15.42. poo
F (a )=
f /
e~tt^a ~1^dt
[ a > l]
Bessel Functions ©I Ilie First ICind A Bessel differential equation of order v is given by
Jo
r(2)= f
x2y" + xy' + (x  v )y = 0
15.46
e W 'U t
Jo
e ltdt
Jo
35Also defined in terms of the gamma function are the beta function and incomplete gamma functions.
A particular solution is given by the Bessel function of the first kind of order v. (This result is obtained using the power series method.) Jv(x) = xv Y^ ^0 2
15.47
+Pm\T(u +771+1)
6An important value of gamma is T (l/2 ) = y/n.
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The constant 7 is Euler’s constant (also given as (7#, (7, and In 7 ) which is defined as
Figure 15.6 Error Function
Mathematics
7
= lim ( 1 + i H h   In 5 )
2s J
s—>00 V
15.50
» 0.577 The variable hm is defined as hm. = 1 + —f
1_ m 15.51
The second standard solution to Eq. 15.48 for all v is given by Integer values of v are often denoted by n. The functions J0 and Ji are important in many engineering problems. These functions are shown graphically in Fig. 15.7.
Y v(x) = (  r M Vsm vnJ X ( j v (x)cOSVTl
Figure 15.7 Bessel Functions of the First Kind
— J  V(x))
Y n(x) = lim Y u(x)
15.52 15.53
Equation 15.52 and Eq. 15.53 are known as Bessel func tions of the second kind of order v. These are also called Neumann ’s functions of order v and are sometimes given as Nu{pc). They are also called Weber’s functions in some texts. These functions are shown graphically in Fig. 15.8. Figure 15.8 Bessel Functions of the Second Kind
A table of values for J0(x) and Ji(x) is given in App. 15.B.
B essel Fyneti@ris ©f the Second Kind If the value of v or n is zero in a Bessel differential equation, and dividing by the variable x, the result is xy" + y' + xy = 0
15.48
The standard particular solution is known as the Bessel function of the second kind of order zero and is given by the following equation. This is also called Neumann’s function of order zero. \
J0(a:)(ln + 7) Yo(x)
+z
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Bessel Fyncttens ®f th e Third K in d A general solution to Bessel’s equation for all values of v is given by Eq. 15.54. y{x)  ciJv{x) + c 2 Y v ( x )
15.54
Sometimes a solution needs to be complex for real values of x. In such cases, solutions take on the following forms. H ^\x) = Jv{x) + j Y v(x)
15.55
H%\x) = Jv{ x )  j Y v{x)
15.56
These linearly independent functions are called Bessel functions of the third kind of order v. They are also called the first and second Hankel functions of order v.
Hyperbolic Bessel Functions Hyperbolic Bessel functions are given by Eq. 15.57. I m{x) = j  mJm(jx)
15.57
10. CONTINUOUSTIME SYSTEMSs FOURIER S EM ES AMD TRANSFORMS Fourier series are used to analyze periodic signals. Fourier integrals, also called Fourier transforms, are used to ana lyze aperiodic signals. The power of Fourier analysis is that if the input signal can be expressed in terms of periodic complex exponentials or sinusoids, then the out put can be expressed in the same form, simplifying analy sis. Further, representing otherwise intractable equations as a periodic Fourier series allows for analysis in ordinary and partial differential equations. Example 15.7 Show that the response of a linear timeinvariant system (LTI) to a complex exponential input is the same com plex exponential with a change in amplitude.37 In other words, show38 esi
H{s)est
The term H(s) is the complex amplitude factor and in general will be a function of the complex variable s.
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Writing the output y(t) in terms of the convolution integral gives the following equation. (Recall that the convolution integral takes two functions defined in the s domain, in this case the impulse and the input signal x(t), and gives the output in the time domain t directly without determining the inverse Laplace transform.)
/ +oo h(t —r)x(r)dr = oo
p+00 / h(r)e~STest dr J—oo
Because the term est does not affect the integral, it can be moved outside the integral.
/ +oo h(r) e~STdr ■oo This can be rewritten as y(t) = estH(s) = H(s)est +00 h(r) e~STdr
/
•oo Any complex exponential is an eigenfunction of an LTI system. The constant H(s) for a specified value s is the eigenvalue of the eigenfunction est. 0 Consider an input signal of the form x(t) = a\eSlt + a2eS2t + aseSst
Solution Assume an input signal of the form x(t) = est provided to an LTI circuit with impulse response h(t).39
The response to each term separately is a\eSlt —> aiH(si)eSlt
input
output linear tim einvariant system
t
t
) —
Because, by the definition of “linear,” superposition applies, the output y(t) is y(t) = In general terms,
input (t )8(
aseS3t >■ asH(s3)eS3t
 ► h(t)
(a) LTI system with unit im pulse response h[t)
x
a2eS2t —►a2H(s2)eS2t
linear tim einvariant system
(b) LTI system response to an im pulse at tim e t with am plitude x (t )
37“Linear” indicates that superposition applies. “Time invariant” indi cates that a shift in the input signal x(t) to x(t —r) causes a shift in the output signal y(t) to y(t —r). When such a shift occurs in an equation, the equation is referred to as a difference equation and, as will be shown, represents a discrete time system. 38This is equivalent to showing the same thing for sinusoids, because from Euler’s relation e?ut = cos ut + j sin ut. 39The impulse response h(t) occurs due to an input 6(t). This is simply a notation, which can vary considerably.
+ a2H{s2)es2t + a3H {s3)e s^
input —>output Y a ke ^ k
Y a kH (sk) e ^ k
In general, 5 is complex and of the form s = a + jw. If s is restricted to be equal to jou, that is, 5 = juo, then the exponentials are of the form e?ut and Fourier trans forms are used. If s is allowed to be of the more general form s — a + j ( j , then Laplace transforms are used.
40The value H(s) is determined by the electrical circuit and represents the circuit’s response to an impulse 6(t —r). H(s), with s = jui, is the Fourier transform of the response of the circuit to an impulse signal. Generally written as H(cu), it is termed the frequency response of the system.
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11. C0WTINUOUSYI11E SYSTEMSs FOURIER SERIES REPRESENTATION OF A PERIODIC SIGNAL A continuoustime signal is one in which the indepen dent variable is continuous, that is, the signal is defined for a continuum of values. Figure 15.9 shows an example of such a signal.
amplitude factor H(s) is determined by the circuit through which the given input (i.e., the signal) is pro cessed.42 The amplitude factor H(s) is the circuit’s response, h(t—r), to an impulse, S(t—) function can be expressed as cos (toot + ) = i In the first term k = 1. Therefore, 0>k = &1 = 2
In Fig. 15.9, the values T0 and u;0 represent the funda mental period and fundamental frequency, respectively. That is, they represent the minimum positive, nonzero value for which x(t) = x(t + T) is true for a periodic function.
In the second term k = —1. Therefore, ak = o_! = \ For all other values of A;, a& = 0. The function in terms of a Fourier series can be represented as
The Fourier series representation for any continuous time periodic signal is
+00
= ^ x{t)=
X
— 2 e^u°t+^ +
—oo
+oo
ake ^
15.58
k——oo
Jl dk = 4 r [ x(t)e jkuotdt J q J to
The shift of + is the equivalent of multiplying by e^. 15.59
There are other representations of a Fourier series. (See Sec. 10.15 for an example.) This complex exponential form is very useful in electrical analysis. It is called both the complex Fourier series and the exponential Fourier series.
To confirm the result, use Euler’s relation to expand the Fourier series and obtain the original equation. That is, x (£ )
= I e?Vo t+) _j_ l e j(u o t+ (f> )
= 2j(cos (u)0t + 4>) + j sin (u>ot + ))
+ 5 + j sin ( — (wo< + 0 ))
This may seem rather circular. In different terms, if one represents the input as a Fourier series, that is, as Eq. 15.58, and then replaces the ak with the Fourier coefficients (also called the spectral coeffi cients) as given by Eq. 15.59, the output is obtained by multiplying by the Fourier transform of the impulse input signal (given by H(s) earlier in this section). The s is restricted to s = ju when using the Fourier transform. 43When using the Fourier transform, the s is limited to jv and H(s) is the Fourier transform of h(t—r). If s has both real and imaginary parts, that is, s = a + jo;, H(s) is the Laplace transform of h(t —r).
Applying the trigonometric identities sin (—a) = —sin a and cos (—a) = cos a gives x(t) = (cos(u;o£ + ) + jsin (a>0£+ )) + ^cos(a;oJ + 0)  jsin (wot10)) = cos(coot h )eJwt2^ d f
15.60(b)
•00
X(u)
X(o>)
+oo
r jujtdt
15.61(a)
~j2nftdt
15.61(b)
 tr
1
■oo
(Oq
 a )0
+oo
x(t)e~j2Kftdt
COS
■oo
Equation 15.60 and Eq. 15.61 are referred to as a Fourier transform pair. Equation 15.61 is the Fourier transform or Fourier integral. Equation 15.60 is the inverse Fourier transform that is the synthesis equation of the pair.47
(Oq (b)
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Figure 16.2 Magnetization Curves
E L E C T R O M A G N E T I C
Figure 16.3 Magnetic Spin Arrangements
T H E O R Y
16*5
P A R T 3s ELECTR O STATICS
(a) ferromagnetic 6. ELECTR IC C H A R G ES
\
An electrically neutral material has equal numbers of positive and negative charge carriers. Equation 16.1 describes bodies in this unelectrified state.
(b) paramagnetic
Q+ = Q ~ = 0
16.1
Only integer numbers of charge carriers exist. Therefore, the transport of an integer number n charge carriers to a body from another body means that Q+ = np+ and Q~ = ne~
(c) antiferromagnetic
v_./
(d) diamagnetic*
(e) ferrimagnetic *The circle represents the internal magnetic field.
16.2
Like charges exhibit an electric repulsion. Unlike charges exhibit an electric attraction. An attraction also exists between the masses of any two charged particles. The force associated with the gravitational attraction is very small in comparison to the force of the electrical attraction. Therefore, the gravitational force can be ignored except when the charges are accelerated. Static electricity formed by positive charges, such as that produced by rubbing silk on a glass rod, is called vitreous static electricity. When the static electricity is formed by negative charges, such as that produced by rubbing fur on a rubber, amber, or plastic rod, it is termed resinous static electricity.
Example 16.1 What is the gravitational force of attraction between two electrons located 1 cm apart? Solution Use Newton’s law of gravitation.
5 THERMOELECTRIC PHENOMENA .............
F _ Gm\ m2
There are three significant relations between electricity and heat of importance to the electrical engineer. These are the Seebeck effect, the Thomson effect, and the Peltier effect. Each will be explained in terms of its applications. For now, the important point is that heat is capable of providing the energy to release and trans port electrons.
Newton’s universal constant or gravitational constant, G, is equal to 6.672 x 1011 Nm /k g 2. The mass of the electrons (taken from Table 16.1) and the 1 cm distance between the electrons are substituted into the equation, giving the following.
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Basic Theory
As the name implies, electrostatics is the field of elec trical engineering that deals with the electrical effects of stationary charges. An important concept in this field is that all electrical phenomena are expressible in terms of the masses and charges of the elementary particles involved.
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Example 16*2
Gm\m2 _ Gm2 e
What is the magnitude of the electrostatic force between two electrons 1 cm apart in a vacuum?
6.672 x 10” n ^ p  J (9.1096 x 1(T31 kg)2 x2/ 1 m V (I cm) ( i o o W
Solution Take the magnitude of Eq. 16.5 and substitute the values for the relative permittivity and k given in this section. The relative permittivity has no units. As the term “relative” suggests, it is a comparison. This is dis cussed more fully in Sec. 16.9.
= 5.537 x 10“ 67 N Basic Theory
7. COULOMB’S LAW Coulomb’s law states that the force between charges at rest in an isotropic medium is proportional to the prod uct of the charges and inversely proportional to both the square of the distance between them and the dielectric coefficient of the medium. The force acts in a straight line between the centers of the charges.8 Equation 16.3 describes Coulomb’s law in mathematical terms.
F i2 = k
Qi Q2
ai2
err*
F12\= F = k
Ql Q[; err2
8.987 x 109 ^ 5 0 (1.6022 x 10“ 19 C) F 12 =
Qi Q2 Aneri.12
2
16.3
The vector ari 2 is a unit vector having a direction along a straight line connecting Qi and Q2. The term e repre sents the dielectric coefficient, more commonly called the permittivity. The 4tc term occurs in many electric field calculations and, for this reason, is included in this rationalized form of Coulomb’s law in order to cancel terms in integrals over spheres and make calculations easier. This is perhaps more readily seen when Cou lomb’s law is written in terms of the electric field, E. F 12 = F 2 = Q 2 F 1
16.4
Coulomb’s law is also stated in its Gaussian SI form as follows. j?
_^Qi Q2 errz
* 1  2 —  12
 2.307 x 1CT24 N
Example 16.3 What is the ratio of the electrostatic force to the grav itational force between two electrons 1 cm apart in a vacuum?
Solution As found in Ex. 16.2, the electrostatic repulsion is 2.307 x 10~24 N. As found in Ex. 16.1, the gravitational attraction is 5.537 x 1067 N. The ratio of electrostatic force to gravitational force is
irR
16.5
R The term er represents the relative permittivity of the dielectric. (er— 1.0 for a vacuum.) The constant k has an approximate value of 8.987 x 10 Nm2/C 2. In Eq. 16.3 through Eq. 16.5 the force is measured in newtons, the distance in meters, and the charge in coulombs. 8In Eq. 16.3, the unit vector, and thus the force, is from 1 to 2. The force direction is the direction charge 2 will move due to the electric field of charge 1. See Eq. 16.4. The notation often varies. The unit vector ari_2 (see Eq. 16.3) or ai_2 (see Eq. 16.5) is sometimes written ar, a12, or even a2i The a2i notation is used as a reminder that, although the vector is from 1 to 2, it is determined mathematically by subracting the position elements of 2 from 1; for example, afc — £1 , IJ2 — 2/i3and Z2 — Z\. (The unit vector is then determined by dividing the distance between the two points.) At times, the notation 12 merely indicates the connection between the points and one must determine the correct direction from other information.
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2.307 x 10~24 N 5.537 x 1067 N = 4.166 x 1042
The electrostatic force that these elementary particles experience is 42 powers of ten greater than the gravita tional force of attraction between them. Clearly, the gravitational attraction between such particles can be ignored in all but the most exacting calculations. Coulomb’s law implies that the principle of superposi tion applies for electric charges. This means that a system of charges is a linear system. The total resultant effect at any point within the system can be determined from the vector sum of the individual components.
E L E C T R O M A G N E T I C
E xam ple 16.4 Three point charges in a vacuum are arranged in a straight line. Find the force on point charge A. 3m 400 xC
4m 200
800
1 6 7
Determine the unit vectors for these forces. Let point A correspond to the origin. The unit vectors (aB_A and acA) from B and C to A are determined as
(XA 
a BA =
julC
T H E O R Y
zb)1 +
(yA 
\]{%h  XBf +
(jlC
^/b)J
(2/A  2/b)2
Because all three charges are in line, vector analysis is not needed. The force on point charge A is the sum of the individual forces from the other two point charges. Use the Gaussian form of Coulomb’s law.
Basic Theory
(0 — 0)i + (0 — 3)j
Solution
0  0)2 + (0  3)2 = J acA =
From Eq. 16.5,
(0  V40)i + (0  3)j \ /( 0  V 4 0 ) 2 + ( 0  3 ) 2 —0.904i  0.429j
^AB&C = ^B&CA The total force is
^AB + F AC = : 8.987 x 10h
F BA&BA + F CA^CA
Nm2 C2
= (7 9 .9 N )(—j) + (58.7 N )(0.904i  0.429j)
((4 0 0 x 10~6 C )(—200 x 10~6 C) \ (3 m ? +
The magnitude of the force is
(400 x 10“ 6 C)(800 x 10~6 C)
V
(7 m)
=  5 3 .li N + 54.7j N
/
FAb&c = \ j(53.1 N)2 + (54.7 N)2 = 76.2 N
= 7 9.9 N + 58.7 N The direction (counterclockwise angle with respect to the horizontal) is
 2 1.2 N Because the sign is negative, the force on A is toward B and C.
9 = 180°  arctan 5  7^ = 180°  45.9° = 134.1° 53.1 N
E xam ple 16.5 Determine the magnitude and direction of the force on point charge A caused by point charges B and C.
8. ELECTRIC FIELDS The electric field, E, with units of V /m , is the region in which an electric charge exerts a measurable force on another charge, above and beyond the gravitational force that the two charges exert on each other. It is a vector force field because a test charge placed in the field will experience a force in a given direction and with a specific magnitude. The direction of the field is along flux lines. These lines, ideally, represent the direction of the force on an infinitely small, isolated positive test charge.9 The actual direction is represented by the unit vector a. Equation 16.6 represents the electric field strength, or electric field intensity, in a substance with permittivity eat a distance r from a point charge Q.
Solution
E:
The charges and distances are the same as in Ex. 16.4. Fq a — ^AB = —79.9 N
[attractive]
F q a — F aq = —58.7 N
[repulsive]
Q Aner2
16.6
Coulomb forces obey the principle of linear superposi tion in free space and in many dielectrics. Therefore, the electric field strength or intensity acting on a charge Q 9The positive test charge is arbitrary, but universal. The charge would ideally have to be infinitely small in order to prevent its own electric field from distorting the original field.
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R E F E R E N C E
M A N U A L
located at point 1 as a result of charges Qi, •••?Q%at distances ri± from point 1 can be represented as the following sum. The vector aril is the unit vector with direction from Qi to point 1. ft
Ei = Z
curve drawn so that it has the direction of the force acting on the test charge, (i.e., it is in the direction of the electric field). A tube of force is created by drawing lines of force through the boundary of any closed curve. The lines form a tubular surface not cut by any other lines of force.
16.7
Basic Theory
Additionally, the electric field intensity at any point p produced by charges Qi, Q2> ••• ,Qn at distances r1? r2, ... ,rn from the point is the vector sum of the field strengths produced by the charges individually.
Tahie 16.6 Electric Fields and Capacitance for Various Configurations
isolated point charge
If one is dealing with charges distributed uniformly throughout space, the electric field strength at a dis tance r that is large when compared to d Fvoiume is given by Eq. 16.8.
L #
p d Vvolume
Ei = 1^volume
infinite coaxial cylinder P1
Q E=•
Q
4tt6r2
C= 16.8
0
isolated sphere
infinite line distribution inside an infinite cylinder
The unit vector a.ril is directed from dv to point 1, at which E is evaluated. The unit vector, the distance, r, and the charge density, p, are usually functions of the variable of integration. The total charge, (J, in the volume Vvoiume that has the charge density p is
_ P/
2 'rrer
91 C = 4ire/?
Q— I
pd Vvolume
16.9
” ^volume
The electric fields discussed to this point have been radial in nature. Not all fields are radial. As seen in Eq. 16.8, the electric field depends on the orientation of the surface or volume that contains the charge. A uniform electric field, commonly used in sensing instru ments and other important applications, can be created between two flat plates, as shown in Fig. 16.4.
concentric spheres
infinite sheet distribution
infinite line distribution
infinite parallel plates
Figure 16A Uniform Electric Field
 r
9
Pi tr = 2'rrer
Pi
^plates .
infinite isolated cylinder
dipole (doublet)
With a potential difference of V volts between the flat plates separated by distance r, the electric field strength is p1 •^'uniform —
16.10 E=
P/ 2tte(r + R)
( r >
0)
E=
■—^ (2cos0 ar + sin 6ae)
4776r3
The field strengths, or intensities, for several configura tions are given in Table 16.6. An electric field is an abstraction that allows one to visualize how charges in different locations within the field will experience a force. A line of electric force is a
P P 1 •
w w w . p p i 2 p a s s Bo © m
Reprinted with permission from Core Engineering Concepts for Students and Professionals, by Michael R. Lindeburg, PE, copyright 2010, by Professional Publications, Inc.
E L E C T R O M A G N E T I C
The forces between electric charges depend upon the environment in which they are located. The maximum coulombic force occurs in free space. In all other materi als, the coulombic force is reduced. Another way of viewing this phenomenon is to say that the electric flux does not pass equally through all materials.10 In fact, it cannot pass through conductive material at all and is diminished to varying degrees by various dielectrics. The permittivity of free space, that is, a vacuum, e0, may be considered to be representative of the dielectric prop erties of free space. In actuality, it is a factor that provides consistent units for Coulomb’s law when units of force, charge, and distance are arbitrary. The product of the permittivity of free space and the relative permit tivity of a medium gives the permittivity of that medium. 6 = e$er
16.11
In general, the relative permittivity is a complex num ber. Except for very high frequencies, the imaginary component, or quadrature component, can be ignored. The real component is often termed the dielectric con stant Because the value of er depends on frequency as well as other factors, the term dielectric coefficient is more appropriate. The relative permittivity can also be viewed in terms of capacitance, as shown in Eq. 16.12. _ __ ^w ith dielectric er — — —
m m lb. 12
^ vacuum
Typical relative permittivities are given in Table 16.7. The permittivity can also be expressed as follows. e = e0( l + X e)
16.13
The term x e represents the electric susceptibility of a given dielectric. It is a numeric measure of the polariza tion or displacement of electrons in the atoms or mole cules of the dielectric. The units for permittivity are C2/Nm 2 or F/m . The permittivity of free space, or vacuum, e0, in SI units is 8.854 x 10“ 12 C2/Nm2 (F/m ). E xam ple 16.6 A certain capacitor is constructed of two square parallel plates with air as the dielectric. If a teflon insert is used in place of the air, by what factor does the capacitance increase? Solution The capacitance is directly proportional to the permit tivity and is given by the following formula. c __ eA __ e0erA r r
10The concept of flux is another abstraction used to help one visualize electric interactions. It is discussed more fully in Sec 16.10.
1 6 9
Because the example asks for the factor by which the capacitance increases, not the actual value of the capac itance, one need merely compare the permittivity of teflon to that of air. Using data from Table 16.7, e r,teflon __ ^r,air
2.0 1.00059
__ ^ q
One could use the upper bound of the permittivity for the teflon, in which case the factor by which the capac itance increases would be approximately 2.2.
Table 16.7 Typical Relative Permittivities (20°C and 1 atm) material
€r
acetone air alcohol amber asbestos paper asphalt bakelite benzene carbon dioxide carbon tetrachloride castor oil diamond glass glycerine hydrogen lucite marble methanol mica
material
21.3 1.00059 1631 2.9 2.7 2.7 3.510 2.284 1.001 2.238 4.7 16.5 510 56.2 1.003 3.4 8.3 22 2.5 8
mineral oil mylar olive oil paper paper (kraft) paraffin polyethylene polystyrene porcelain quartz rock rubber shellac silicon oil slate sulfur teflon vacuum water w ood
er 2.24 2.8—3.5 3.11 2.0—2.6 3.5 1.9—2.5 2.25 2.6 5.7—6.8 5 ~ 5 2.3—5.0 2.7—3.7 2.2—2.7 6.6—7.4 3.6 4 .2 2.0—2.2 1.000 80.37 2.5—7.7
Reprinted with permission from Core Engineering Concepts fo r Stu dents and Professionals, by Michael R. Lindeburg, PE, copyright 2010, by Professional Publications, Inc.
10. ELECTRIC FLUX Flux is defined as the electric or magnetic lines of force in a region. The electric flux, *F, is a scalar field representing these imaginary lines of force. Flux lines leave or enter any surface at right angles to that surface, and the orien tations of the electric field lines and the electric flux lines always coincide. By convention, the flux lines are directed outward from the positive charge and inward toward the negative charge, or to infinity if no negative charge exists in the area, as shown in Fig. 16.5. In Fig. 16.5, only a few representative lines of flux are shown. Also, if the charge on a hollow sphere is all the same sign, the flux outside this sphere is the same as the flux that would exist from a point charge at the center. Therefore, for the purposes of drawing flux lines and calculating forces, a hollow charged sphere can be replaced by a point charge at the center.
PPI
•
w w w . p > p § 2 p a s s . c @ m
Basic Theory
9 d PERMITTIVITY AND SUSCEPTIBILITY
T H E O R Y
1 6  1 0
P O W E R
R E F E R E N C E
M A N U A L
also loosely referred to as the vector flux density. It is a vector quantity that, like the electric flux, *F, cannot be directly measured. It is given in Eq. 16.15. Unlike the electric field strength, E, the electric flux density or displacement, D, is independent of the medium.11
Figure 16.5 Flux Lines

point ♦ charge
hollow sphere Basic Theory
The flux, *F, is numerically equal to the charge. That is, by definition, one coulomb of electric charge gives rise to one coulomb of electric flux.
D — eE — 6QerE
16.15
The magnitude of D, or D, is also referred to as the flux density. It is the number of flux lines per unit area perpendicular to the flux. It is a scalar quantity with units of C/m 2. 16.16
'¥= Q
16.14
The electric flux is also called the displacement flux. This is because the flux is a quantity associated with the amount of bound charge, Q, displaced in a dielectric material that is subject to an electric field. This termi nology can be understood by referring to Fig. 16.6. Figure 16.6 Displacement Flux Experiment
When the medium is nonisotropic, the permittivity, e, is a tensor and the displacement, D, and electric field strength, E, are not necessarily in the same direction. Many terms with differing units are used to refer to “flux density,” such as pi in units of C /m and py in units of C /m 3. This confusion is avoided in texts by referring to such terms as pt and p v as charge densities.12 Care must be exercised when interpreting the phrase “flux density.”
12. GAUSS’ LAW' FOR ELECTROSTATICS......
Q
Early experimenters enclosed a fixed positive charge inside a spherical conducting shell as shown in Fig. 16.6. When the switch was closed, a charge of —Q, equal in magnitude to the enclosed positive charge, was found on the shell and was believed to be caused by the transient flow of negative charge induced by the flux from the + Q. In other words, the positive charge displaced the —Q charge and forced it onto the surface. The scalar electric flux field, XF, is not directly measurable as is the electric field, E. Instead it is inferred from such experiments.
11. ELECTRIC FLUX DENSITY........................ The electric flux density, D, is also referred to as the displacement density or simply the displacement. It is
P P I
® w w w . p p § 2 p a s s a©®m
Gauss’ law for electrostatics is a necessary consequence of Coulomb’s law. Gauss’ law states that the amount of electric flux, V F, passing through any closed surface is proportional to (and in our rationalized system of SI units, equal to) the total charge, Q, contained within the surface. The surface is called a Gaussian surface. Gauss’ law in a variety of forms is given in Eq. 16.17, Eq. 16.18, and Eq. 16.19. ds is an element of the surface area with magnitude ds and direction normal to (is, and 9 is the angle between the normal to the surface area and D.
= Q
[dA parallel to D]
16.17
¥ = J j Dds = J J D cos 8 d.s = Q
[arbitrary surface]
*  / / / pdv = Q
[arbitrary volume]
16.18
16.19
n This is because E is multiplied by the permittivity. See Eq. 16.6 by way of explanation, noting the permittivity in the denominator. 12This is consistent with the electric flux, 'P, being numerically equal to the charge. Another way of considering this is to realize that a rationalized set of units (the SI system) that makes the flux equal to the charge is being used. If the system were not rationalized, the flux would merely be proportional to the charge.
E L E C T R O M A G N E T I C
=
The surface area of a sphere is given by 4tcr2. >= D d s = DAnr2 D=
//
eEdA = Q
[dA parallel to EJ
16.20
Special Gaussian surfaces are used in highly symmetri cal charge configurations to simplify calculations with Gauss’ law. Such surfaces meet the following criteria.
16 11
Anr2
Using the same logic as in Ex. 16.7 on a uniform line charge of density pi of length I at a distance of r gives the following equation for displacement magnitude. D
1. The surface is closed.
Pi 2nr
16.21
2. At each point on the chosen surface, D, is either normal or tangential to the surface. 3. The flux density, or displacement, magnitude, D, is sectionally constant over the portion of the surface where D is normal. Requirement (1) is a general requirement of Gauss’ law. Requirement (2) completely eliminates integrals and makes the calculation simple. Requirement (3) allows D to be removed from the integral. E xam ple 16.7 Given a point charge of constant value Q, what is the flux density, D, at a distance r from the charge? Solution First, arbitrarily construct a sphere around the point charge as shown.
13. CAPACITANCE AND ELASTANCE If charges Q+ and Q~ exist on conductors separated by free space or by a dielectric, the mutual attraction “stores” the charges, holding them in place. The charges produce a stress in the dielectric measured by the elec tric field strength, E, and a potential difference exists between the plates. Capacitance is the property of a system of conductors and dielectrics that permits this storage of charges. If a potential difference is applied across two conduc tors, charges will build up, creating an electric field between the conductors. The amount of the charge, (J, is proportional to the applied voltage. The proportion ality constant is called the capacitance and has units of farads, F. This is shown in Eq. 16.22. A farad is a large capacitance and very few capacitors are constructed with a value of even 1.0 F. Most circuit elements have capacitances in the range of p F (10~6) or pF (10~12). Q = CV
c
16.22(a)
Q
16.22(b)
V
The elastance, 5, is the reciprocal of the capacitance. A sphere surrounding a point charge meets all the require ments of a special Gaussian surface. Using Eq. 16.18 gives the following. ¥
//
°  ds=
//
U „ = i_ = v = Q C Q Q
16.23
D cos 6ds = Q
Because the lines of flux emanate radially the point charge, D and ds are parallel Additionally, the magnitude of the flux constant at a given r. This results in manipulations.
outward from and cos 9 = 1 . density, Z), is the following
As shown in Eq. 16.23, the elastance can also be defined as the amount of energy, [/, required per charge, Q, to transfer a unit charge between two conductors sepa rated by a dielectric.
14. CAPACITORS Q = j ) D cos Ods = j)D (l)d s — D(p ds
A capacitor (once called a condenser) is a device that stores electric charge. It consists of conductors separated by a dielectric. A common type of capacitor consists of two parallel plates of equal area A separated by a
PPI
® w w w . p p S 2 p a s s . C Q m
Basic Theory
Because the electric field strength or intensity, E, is related to the displacement or electric flux density, D, by the permittivity, Gauss’ law as given in Eq. 16.17 can also be stated mathematically as
T H E O R Y
1 6 1 2
P O W E R
R E F E R E N C E
M A N U A L
distance r. This type of capacitor is called a parallel plate capacitor and its capacitance is determined by Eq. 16.24. C
eA r
16.24
The average energy density at any point in this electric field is given in Eq. 16.29. ^ave —
Ua ^volume
Basic Theory
Equation 16.24 ignores the effects of fringing, as do many calculations involving capacitors. An example of fringing is shown in Fig. 16.7. The formula for the capacitance of other configurations is given in Table 16.6. Assuming the plates in a capacitor begin with the same potential ( F = 0 ) , the total energy, U (in J), in the electric field of a capacitor with capacitance C charged to a potential V is given by Eq. 16.25.
u =\cv2
■1 7 0 = 1 ^ 2vw 2 q
16.25
= Ci + C2 + Cs + •••+ Cn
16.29
PART 4: ELECTROKINETICS 16 S P E E B AND MOBILITY OF CHARGE CARRIERS A positive free charge moving in an electric field through a vacuum is shown in Fig. 16.8. Such a particle would experience a force given by the following.13
For n capacitors in parallel, the total capacitance is Ctotal
= eE2 = i2 —6 2
F = QE = ma
16.30
16.26 Figure 16.8 Free Charge in the Electric Field of a Vacuum
For n capacitors in series, the total capacitance is
—
—I— —_L + JL + _L+ . ■+ _1_ Ctotal
C2
Ci
C3
Cn
16.27
V + Q «—
 —
Figure 16.7 Fringing 
The force the particle experiences is unopposed and results in the constant acceleration of the particle as long as it remains in the electric field. In metals, according to electrongas theory, the electrons instead reach an average drift velocity, v^, because they undergo collisions as they move through the electric field. These collisions are between the electrons and the atoms within the crystalline structure of the metal and are caused by the thermal vibrations of the individ ual atoms.14 The average drift velocity is = / » < « = / — dt = Q tmE m m
15 E N E R G Y DENSITY IN M i ELECTR IC FIELD
= /iE
In a rectangular dielectric solid of surface area A and thickness t with a potential difference of V applied to either surface, the average energy (in J) is given by
^ave = \ ^voltage Q
The term tmis the mean free time between collisions and m is the mass of the particle.15 The mobility of the moving charge is given by /i. Because the particles in the metal undergo an increase in vibratory motion as the temperature increases, the mobility varies with the temperature— and with the crystalline structure of the
= \(Et)(DA) = \ E D V yoiume
— 2
^volume
2
^volume
16.28
Equation 16.28 in all its forms assumes that D and E are in the same direction. If not, the dot product must be used to determine the energy.
P P i
16.31
® w w w . p p i 2 p a s s . c o m
13This is identical to Eq. 16.4 without the subscripts. Nominally, sub scripts need only be used to avoid confusion when multiple fields or charges are involved. 14The thermal vibrations can be treated in a quantum mechanical fashion as discrete particles called phonons. 15The variation in the overall field within the metal is caused by the periodic variations established by the lattice atoms. The field varia tions can result in collisions, accounted for in some formulas by an effective electron mass, m*. Only the thermal vibrations must be determined when such a mass is used.
ELECTROMAGNETIC THEORY
16*13
material. As the temperature increases, the mobility decreases, resulting in a lower drift velocity and a lower current. In circuit analysis this phenomenon is called resistivity. The resistivity is the reciprocal of the conductivity.
with an average speed of vave, the convection current will be given by Eq. 16.34.
17. CURRENT10
t
An electric current is the movement of charges past a particular reference point or through a specified sur face. The symbol I is generally used for constant cur rents while i is used for timevarying currents. Current is measured in amperes (A), commonly called amps. 1 A = 1 C/s. By convention, the current moves in the direction a positive charge would move (from the posi tive terminal to the negative terminal), that is, oppo site to the flow of electrons. Ohm’s law relates current to voltage and resistance as shown in Eq. 16.32.
In terms of thecurrent density vector, the convection current is given by Eq. 16.35 and canbevisualized as shown in Fig. 16.9.
/ = pAv,ve = (  ) 4 v ave =
AQ
V = IR
16.34
J=
pVd
16.35
Figure 16.9 Convection Current
16.32
For simple DC circuits, this relationship is adequate and the current is defined by Eq. 16.33.17
However, when the charges exist in a liquid or a gas, as in the field of electronic engineering, or when the charges are both positive and negative and have differ ent characteristics, as in semiconductors, Eq. 16.33 is inadequate and one must be more specific in defining current. As a result, the current density vector, J, with units of A /m 2 is more frequently utilized in electromag netic theory. The total current is given by the sum of the convection current, the displacement current, and the conduction current. The convection and conduction currents are con sidered true currents and the displacement current a vir tual current. Unless otherwise stated, the “current” will normally be the conduction current, but care must be used when determining what “current” is being referenced.
Convection, as the name implies, involves the diffusion of charges in a medium, as opposed to the movement of charges under the influence of an electric field. Although not of general use in electrical engineering, the concept of a convection current is sometimes useful in electro magnetic field theory.
19. DISPLACEMENT CURRENT19 When the electric field varies with time, a current can be made to flow in a dielectric material. That is, charges can be displaced within a dielectric by the electric fields of the charges moving outside the borders of the dielec tric. This is because the electric fields of the charged particles extend well beyond the dimensions of the charges themselves. The current so created is called the displacement current, i^ and is given by Eq. 16.36.
1®. CONVECTION CURRENT18 In a material containing a volume V of mobile charges with a charge density of p moving across a surface area A The displacement current in terms of the current den sity vector, J, is given by Eq. 16.37. 16Compare an electric field’s induced current to a magnetic field’s induced voltage in Sec. 16.33. 17The use of the capital I indicates that the speed of the charges referred to is constant. If the speed is not constant then the current varies with time and i would be used as the appropriate symbol. In practical terms, this means that currents in DC circuits (unidirectional current flow) in which the current magnitude varies could be labeled i. 18Compare convection current with the movement of plasma (see Sec. 17.3).
The displacement current, i^ through any given surface can be obtained by integration of the normal component 19Compare displacement current to the displacement magnetic effect in Sec. 16.35.
PP§ ® wwwappi2pass„c@m
Basic Theory
= Q
1 6 1 4
P O W E R
R E F E R E N C E
M A N U A L
of the displacement current density vector, J d, over the surface as in Eq. 16.38. —
D d A
16.38
dt J a
Basic Theory
The displacement current is equal to zero if the electric flux is constant with respect to time. That is, to cause a current to “flow” through a dielectric without the bene fit of a direct connection to a conductor (as in the case of a capacitor), the electric flux must be changing with respect to time.20 CO N DUCTION C U R R E N T 21
of motion. The transverse fields produce magnetic forces between moving charges. This force also acts between the conductors in which the charges move. It is this force that constitutes the basis for motor and generator theory in electric power engineering. Moving charges represented by the current, z, and the magnetic field established by their motion are shown in Fig. 16.10(a). Another completely valid view of the same situation is shown in Fig. 16.10(b). If the charges move with a uni form motion, the field is referred to as magneto static. If the charges move nonuniformly, the field is referred to as magnetokinetic.
Figure 16.10 Magnetic Fields of Moving Charges
A conduction current occurs in the presence of an elec tric field within a conductor. The magnitude of this current is given by the number of mobile charges trans ferred per unit time within the conductor. The charges may be electrons (in metals and semiconductors), holes (in semiconductors), or ions (in gases and electrolytes). Conduction current is given by Eq. 16.39. Ic = a (^ jV = G V
16.39
The symbol a indicates the conductivity of the material and is a property of a unit volume of the material. The area of the conductor is given by A and the potential difference by V. The length represented by I is the length over which the potential is applied. The quantity G is the conductance of the specified conductor and depends on the dimensions of the material used. The conduction current in terms of the conduction cur rent density vector, J c, is given by Eq. 16.40. Jc =
pV d
16.40
This is identical to Eq. 16.35. Applying Eq. 16.31, spe cifically v rf= / i E , puts Eq. 16.40 in the more widely recognized form shown in Eq. 16.41. J =
crE
16.41
Equation 16.41 is called the point form of Ohm’s law.
PART 5: MAGHETOSTATICS
*The symbol indicates the current flows into the page. The symbol “” indicates flow out of the page. That is, the dot indicates the head of an arrow and the “” represents the tail. Assume positive current flow. This correlates with the righthand rule that states that placing the thumb in the direction of the current flow and curling the fingers inward gives the direction of the magenetic field.
A magnetic field can exist only with two equal and opposite poles, referred to as the north pole and the south pole.22 The combination of the north and south pole linked together is referred to as a dipole. The mag netic dipole moment, m, depends on the pole strength, p, and the distance between the poles. m = pd
16.42
Mobile charges set in motion by an electric field carry their own electric fields with them. The subsequent threedimensional field can be resolved into longitudinal and transverse components, referenced to the direction
The magnetic moment is a mathematical construct. Because the poles cannot be separated for measurement, the distance, d, is not measurable. Nevertheless, the magnetic moment is an important and fundamental quantity, and it can be measured. Magnetic moments are measured in terms of Bohr magnetons. In ferromag netic material, a Bohr magneton is the moment
20The charges do not flow through the dielectric but are instead displaced within the dielectric. Also, this displacement ceases if the electric flux within the dielectric no longer changes. 21Compare conduction current with the principles of the magnetic circuit in Sec. 16.32.
22An electric charge can exist as a single charged object. Magnetic poles cannot exist as single north or south poles. (Mathematically there is no reason for the nonexistence of such a magnetic monopole, but none has yet been found.)
21. M AGNETIC PO LES
P P §
•
w w w . p p i 2 p a s s . c o m
E L E C T R O M A G N E T I C
Figure 16.10(c) represents a loop of current creating a magnetic field. As the diameter of the current loop becomes infinitely small, the field approaches that of a magnetic dipole. A magnetic dipole field is shown in Fig. 16.11.
1 6 * 1 5
22. BIOTSAVART LAW The BiotSavart law is basic to magnetostatics and magnetokinetics just as Coulomb’s law is basic to elec trostatics and electrokinetics. The law, in conjunction with Ampere’s law, relates the force between two current elements as in Eq. 16.43.
Figure 16.11 Magnetic Dipole Field
F i‘ 12
JL (l2dl2) x (h d h ) x (n_2) 4tt
16.43
The term ^ is the magnetic permeability and ri_2 is a unit vector from the point represented by differential current element d\i to c?l2.24 Both current elements are in the magnetic field of the other. For parallel currents Eq. 16.43 becomes
■^12
The infinitely small loop of current can be viewed as a single electron moving in an orbit. Hence, the magnetic moments of individual atoms and the usefulness of the Bohr magneton.23 For ease of presentation, and because in magnetic mate rials many groups of atoms tend to align themselves into domains, individual electrons or loops of current are not usually shown. Instead they are more often shown as in Fig. 16.12, which illustrates the standard convention that lines of magnetic flux are directed outward from the north pole (that is, the magnetic source), and inward at the south pole (that is, the magnetic sink).
J L  /i4
(a) dielectric electric field, Ed
©
0 '© 0 © 0 ©
©©00 © © 0 © © © O © © © © © © © © © © © 0 ©
© 00© '
© © 0
00.
'
©.
'0 © , "© 0© © © © © © © © © © © © ©
© © © © © © © ©
© © © © © © ©
© © © © © ©
©.
(b) dielectric surface charge 3. ELECTRIC DISPLACEIVSENT......................... The electric flux density, D, is also called the displace ment density. It represents the displacement of charges within a dielectric. The motion of the charges within the dielectric solid is not translational, as it is for conduction current. Instead, the dipole moments rotate under the influence of the electric field (see Fig. 20.4). If the angle n/2 is chosen as the arbitrary zero energy reference, the work of rotating a dipole is W
P P I
pE
21.8
® w w w . p p i 2 p a s § B& ®m
Field Theory
Figure 21.2 Dipole Alignment About a Free Positive Charge
FIELDS
214
P O W E R
R E F E R E N C E
M A N U A L
This motion of charges is the displacement current. In terms of the current density vector, J, the displacement current is defined by Eq. 21.9.
Equation 21.9 indicates that a timevarying electric field, and subsequently a timevarying electric flux den sity, results in current flow in a dielectric material that contains no free charges. In this manner, current is made to “flow” from one side of a capacitor to another. For a poor conductor, or lossy dielectric, the ratio of the con duction current density to the displacement current density depends on the ratio of the conductivity, which indicates that the voltage is arbitrarily referenced from A to B. That is, terminal A is assumed to be the high potential (+ ). Assign polarities to the resistors based on the assumed direction.
15D Kl RC HHOFF’S VOLTAGE LAW
Circuit Theory
Kirchhoff’s voltage law (KVL) states that the algebraic sum of the voltages around any closed path within a circuit or network is zero. Some of the voltages may be sources, while others will be voltages caused by current in passive elements. Voltages through the passive ele ments are often called voltage drops. A voltage rise may also appear to occur across passive elements in the initial stages of the application of Kirchhoff’s voltage law. For resistors, this is only because the direction of the current is arbitrarily chosen. For inductors and capacitors, a voltage rise indicates the release of mag netic or electrical energy. Stated in terms of the voltage rises and drops, KVL is voltage rises = ]Tvoltage drops loop
26.24
loop
Kirchhoff’s voltage law can also be stated as follows: The sum of the voltages around a closed loop must equal zero. The reference directions for voltage rises and drops used in this book are shown in Fig. 26.6.6 Figure 26.6 Voltage Reference Directions H  lf rise
^ A /W 1 rise
± A /W ^ ^ drop
(a)
(b)
(c)
E xam ple 26.3 Consider the following circuit. Determine the expression for Fab across the load in the circuit in the figure.
6This alternative statement of Kirchhoffs voltage law and the arbi trary directions chosen are sometimes useful in simplifying the writing of the mathematical equations used in applying KVL.
P P I
•
w w w . p p § 2 p a s s . c ® m
The KVL expression for the loop, clockwise starting at terminal B, is
Vi  VRl + 72  V*  Fab = 0 Rearrange and solve for
V a b 
yAB= vx vRl +  y* 16. KIRCHHOFF’S CURRENT LAW Kirchhoffs current law (KCL) states that the algebraic sum of the currents at a node is zero. A node is a connection of two or more circuit elements. When the connection is between two elements, the node is a simple node. When the connection is between three or more elements, the node is referred to as a principal node. Stated in terms of the currents directed into and out of the node, KCL is ^ currents in = ^ currents out node
26.25
node
Kirchhoff’s current law can also be stated as follows: The sum of the currents flowing out of a node must equal zero.7 The reference directions for positive cur rents (currents “out”) and negative currents (currents “in”) used in this book are illustrated in Fig. 26.7.
7This alternative statement of Kirchhoffs current law and the arbi trary directions chosen are sometimes useful in simplifying the writing of the mathematical equations used in applying KCL.
PC
C I R C U I T
F U N D A M E N T A L S
26"7
Figure 26.9 Parallel Circuit
Figure 26.7 Current Reference Directions
I
(a) I a>l b\ negative currents or currents in (b) I ct + 0)
27.29
If the rotating portion of Eq. 27.31 is assumed to exist, the voltage can be written v{0) = Vmej9
27.32
Changing Eq. 27.32 to the phasor form yields V = VmlB
or
^rms 10 ■
Vm Z< 9= vie V2
27.33
The properties of complex numbers, which can represent voltage, current, or impedance, are summarized in Table 27.3. The designations used in the table are illus trated in Fig. 27.9.
13. RESISTORS Resistors oppose the movement of electrons. In an ideal or pure resistor, no inductance or capacitance exists. The magnitude of the impedance is the resistance, R, with units of ohms and an impedance angle of zero. Therefore, voltage and current are in phase in a purely resistive circuit. ’ZiR — jR/0 — R
® w w w  p p i 2 p a s s a©Qm
27.31
27.28
The effective value phasor notation represented by the first and second parts of Eq. 27.28 is used by NCEES. When the notation is not specified (for example, when a subscript is not used as in the third part of Eq. 27.28), assume the effective notation. Additionally, many texts do not show phasors in bold font, but instead let the angle symbol indicate that a phasor is being used. Care should be taken to determine the type of notation in use.
PP§
27.30
add com plex num bers
jO
27.34
AC
Figure 27.10 Phasor Rotation in the Complex Plane
C I R C U I T
F U N D A M E N T A L S
27~ 9
14. CAPACITORS Capacitors oppose the movement of electrons by stor ing energy in an electric field and using this energy to resist changes in voltage over time. Unlike a resistor, an ideal or p e r fe c t ca p a citor consumes no energy. Equa tion 27.35 gives the impedance of an ideal capacitor with capacitance C. The magnitude of the impedance is termed the capacitive rea cta n ce, X & with units of ohms and an impedance angle of —tt/2(—90°). Conse quently, the current leads the voltage by 90° in a purely capacitive circuit.10 Z c = X c Z —90° = 0 + j X c
27.35
x c = ii)C ~7; = 7 rb 271/(7
2736
Inductors oppose the movement of electrons by storing energy in a magnetic field and using this energy to resist changes in current over time. Unlike a resistor, an ideal or p e r fe c t in d u ctor consumes no energy. Equa tion 27.37 gives the impedance of an ideal inductor with inductance L. The magnitude of the impedance is termed the inductive r ea cta n ce, X L, withunits of ohms and an impedance angle of n/2 (90°). Conse quently, the current lags the voltage by 90° in a purely inductive circuit.11 ZL = X l 190° = 0 + j X L
(b) evMfor 0 cof = 0 © «r=  © cof=f
X L = ooL = 2 n fL
27.37 27.38
16. COMBINING IMPEDANCES Impedances in combination are like resistors: Imped ances in series are added, while the reciprocals of imped ances in parallel are added. For series circuits, the resistive and reactive parts of each impedance element are calculated separately and summed. For parallel cir cuits, the conductance and susceptance of each element are summed. The total impedance is found by a complex addition of the resistive (conductive) and reactive (sus ceptive) parts. It is convenient to perform the addition in rectangular form. Ze = ^ Z =
(c) for 0 (df = 0 © Wt= f ©
[series]
27.39
10The impedance angle for a capacitor is negative, but the current phase angle difference is positive— hence the term “leading.” This occurs mathematically because the current is obtained by dividing the voltage by the impedance: I = V /Z . n The impedance angle for an inductor is positive, but the current phase angle difference is negative— hence the term “lagging.” This occurs mathematically because the current is obtained by dividing the voltage by the impedance: I = V /Z .
PPI
•
w w w , p p S 2 p a s s aG ©m
Circuit Theory
15 im U G T O R S
2710
P O W E R
R E F E R E N C E
M A N U A L
TabBe 27.3 Properties of Complex Numbers
rectangular form
polar/exponential form Z = ZZ6>
relationship between forms
complex conjugate
addition
z = x + jy
Z = \7i\e>e = Z cos0 +jZ sin#
x = Z cos 6
Z = J x 2 + y2
y = Z sin#
6 = arctan X
V
Z* = x  jy
Z * = Ze^ = ZZ6>
ZZ* = (x 2 + y2) = \z \2
ZZ* = (Z^)(Ze^) = Z2
Zi + Z2 = (sj + X2) + j(yi + y2)
Zi + Z2 = (Zi cos # 1 + Z2j cos 02) + j(Z1sin0i + Z2sin02)
Circuit Theory
multiplication
Z iZ 2 = (X1X2  y1y2) + j{x\y2 + x2y1)
division
Zi Z2
_1_
UY 
Z iZ 2 = ZiZ2Z#i + 02
(xix2 + yiy2) +j{x2y1  xiy2)
Z2
Z22
Solution (a) From Eq. 27.39, 27.40
The impedance of various seriesconnected circuit ele ments is given in App. 27.A. The impedance of various parallelconnected circuit elements is given in App. 27.B. E xam ple 27.6
Z = JR 2+
Sly + (4 Siy = 4.47 ft
6 = arctan ^S — arctan — = 63.4° Y R 2 ft Z = 4.47 ftZ—63.4° From Eq. 27.9,
Determine the impedance and admittance of the follow ing circuits.
(a)
2
20
1 Z
1 4.47 ftZ—63.4°
0.224 SZ63.4°
(b) Because this is a parallel circuit, work with the admittances.
g = 5 = oJ £ T 2 S B c = X ^ = 0.25 n = 4 s Y = \ [g 2 +
= \J(2 S)2 + (4 S)2 = 4.47 S
(b) = arctan
v(t) = 180 sin cot
=0.25 0
= arctan = 63.4° G 2o Y = 4.47 SZ63.4° nr
PPI
»
w w w . p p i 2 p a s s . c o m
1
1
Y
4.47 S/63.4°
= 0.224 f2Z—63.4°
AC
17. ©HITS LAW Ohm’s law for AC circuits with linear circuit elements is similar to Ohm’s law for DC circuits.12 The difference is that all the terms are represented as phasors.13
C I R C U I T
F U N D A M E N T A L S
Again, the sin 2ut integrates to zero over the period, and the average power is zero. Nevertheless, power is stored during the expansion of the magnetic field and returned to the circuit during the contraction of the magnetic field. PL = 0
V = IZ V 1.9V = (
27“11
27.48
27.41
)
27.42
19.
r e a l pow er an d t h e po w er f a c t o r
In a circuit that contains all three circuit elements (resis tors, capacitors, and inductors) or the effects of all three, the average power is calculated from Eq. 27.14 as
18. POWEU The instantaneous power, p, in a purely resistive circuit is * (* M 0
1 fT PaVe = j , J o i(t)v(t)dt
= (Im&n(jt)(Vm®nUjt)
= I mVmsin2 cut
^
— 2^m ^m
2 ^171^ m
2cut
27.43
Let the generic voltage and current waveforms be repre sented by Eq. 27.50 and Eq. 27.51. The current angle (p is equal to #±.14
The second term of Eq. 27.43 integrates to zero over the period. Therefore, the average p o w er dissipated is P R = \ImVm =
— ^rms ^"rms “ I V
i(t)v(t )
= (/mcos ujt)(Vmsin ut) — I mVm sin (Jt cos u t 27.45
Because the sin 2cut term integrates to zero over the period, the average power is zero. Nevertheless, power is stored during the charging process and returned to the circuit during the discharging process. P C —0
27.46
Similarly, the instantaneous power in an inductor is Pl (*) = \ !m Vm sin 2u t
27.50
v(t) = Vmsin (cot + 0)
27.51
Substituting Eq. 27.50 and Eq. 27.51 into Eq. 27.49 gives
In a purely capacitive circuit, the current leads the volt age by 90°. This allows the current term of Eq. 27.43 to be replaced by a cosine (because the sine and cosine differ by a 90° phase). The instantaneous power in a purely capacitive circuit is
= /mVm sin 2u t
i(t) = I msin (cot + tp)
2.7.44
Equation 27.28 was used to change the maximum, or peak, values into the more usable rms, or effective, values.
Pc(t) =
27.49
27.47
12Ohm’s law can be used on nonlinear devices (NLD) if the region analyzed is restricted to be approximately linear. When this condition is applied, the analysis is termed small signal analysis. 13In general, phasors are considered to rotate with time. A secondary definition of a phasor is any quantity that is a complex number. As a result, the impedance is also considered a phasor even though it does not change with time.
Pave = ^ 2
^
^pf = I rms ^nns COS 0pf
27.52
The angle pf = 10 — ip\ is the p o w er fa c to r angle. It represents the difference between the voltage and cur rent angles. This difference is the impedance angle, ±. Because cos(—) = cos(+ 0 ), only the absolute value of the impedance angle is used. Because the absolute value is used in the equation, the terms leading (for a capac itive circuit) and lagging (for an inductive circuit) must be used when describing the power factor. The power factor of a purely resistive circuit equals one; the power factor of a purely reactive circuit equals zero. Equation 27.52 determines the magnitude of the prod uct of the current and voltage in phase with one another, that is, of a resistive nature. Consequently, Eq. 27.52 determines the power consumed by the resis tive elements of a circuit. This average power is called the real p ow er or true p o w er , and sometimes the active pow er. The quantity cospf, or cos0, is called the p o w er fa c to r or phase fa c to r and given the symbol pf. The power factor is also equal to the ratio of the real power to the apparent power (see Sec. 27.22). Often, no subscript is used on P when it represents the real power, nor are subscripts used on rms or effective values. The real power is then given by P = /F c o s 0 pf = JVpf
27.53
14Using a current angle simplifies the derivational mathematics (not shown) and clarifies the definition of the power factor angle given in this section.
PP>
•
w w w . p p i 2 p a s s ne o m
Circuit Theory
PR,(t) =
2712
P O W E R
R E F E R E N C E
M A N U A L
It is important to realize that the real power cannot be obtained by multiplying phasors. Doing so results in the addition of the voltage and current angles when the difference is required. P ^ I IxpV IB = VIlB + ip
27.54
If phasor power is used, the difficulty can be alleviated by using Eq. 27.55. P = Re (VI*)
27.55
emphasizes that the power angle, , associated with S is the same as the overall impedance angle, , whose magnitude equals the power factor angle, 0pf. The relationship between the magnitudes of these powers is Sz = P z + Qz 27.58 A drawing of the power vectors in a complex plane is shown in Fig. 27.11 for leading and lagging conditions, and the following relationships are determined from it.
Equation 27.55 is equivalent to Eq. 27.53.
P = S cos (
27.59
Q = S sin (
27.60
20. R EACTIVE POW ER The reactive power, (J, measured in units of VAR, is given by Q — I V sin (
Figure 27.11 Power Triangle
27.56
Circuit Theory
The reactive power, sometimes called the wattless power, is the product of the rms values of the current and voltage multiplied by the quadrature of the current. The quantity sin 0pf is called the reactive factor.15 The reactive power represents the energy stored in the induc tive and capacitive elements of a circuit.
21. A P P A R EN T POW ER The apparent power, 5, measured in voltamperes, is given by S = IV
pow er
27.57 (b) lagging
The apparent power is the product of the rms values of the voltage and current without regard to the angular relationship between them. The apparent power is representative of the combination of real and reactive power (see Sec. 27.22). As such, not all of the apparent power is dissipated or consumed. Nevertheless, electrical engineers must design systems to adequately handle this power as it exists in the system.
The ratio of reactance, X, to resistance, R, given in Eq. 27.62 and often used in fault analysis and trans former evaluation, can be determined from the power factor, pf, and vice versa. The power factor is the rela tionship between the current and the voltage. Since this relationship shifts significantly during a short circuit, in fault analysis the shortcircuit power factor must be used.16 pf :
220 C O M P LEX POW ER JIM© TH E POWER TRIANGLE The real, reactive, and apparent powers can be related to one another as vectors. The complex power vector, S, is the vector sum of the reactive power vector, Q, and the real power vector, P. The magnitude of S is given by Eq. 27.57. The magnitude of Q is given by Eq. 27.56. The magnitude of P is given by Eq. 27.53. In general, S = I*V where I* is the complex conjugate of the current, that is, the current with the phase difference angle reversed. This convention is arbitrary, but 15The power factor angle, pf, is used instead of the absolute value of the impedance angle (as in the power factor) because the sine function results in positive and negative values depending upon the difference between the voltage and current angles. If pf were not used, ±4> would have to be used.
P P B •
w w w . p p § 2 p a s s » e @ m
R
P S
27.61
= tan (arccos pf)
27.62
E xam ple 27.7 For the circuit shown, find the (a) apparent power, (b) real power, and (c) reactive power; and (d) draw the power triangle.
120 V (rms) 60 Hz
10 ft
6See Chap. 39 for more on the short circuit phenomenon.
AC
Solution
C I R C U I T
F U N D A M E N T A L S
27" 13
23. MAXIMUM POWER TR A N SFER
The equivalent impedance is 1 + T J _ = _j0.25 + 0.10 S
z
j4 n
10 n
Z = 3.714 n 0.25 S = arctan = 68 . 2° 0.10 S
Assuming a fixed primary impedance, the maximum power condition in an AC circuit is similar to that in a DC circuit. That is, the maximum power is trans ferred from the source to the load when the impedances match. This occurs when the circuit is in resonance. Resonance is discussed more fully in Chap. 30. The conditions for maximum power transfer, and therefore resonance, are given by
The total current is j _V _ Z
120 VZ0° = 32.31 AZ —68.2° 3.714 QZ68.2°
^load — R s
(a) The apparent power is
27.63
27.64
S = I * V = (32.31 AZ68.2°)(120 VZ0°)
24. A C CIRCUIT ANALYSIS
(The angle of apparent power is usually not reported.) (b) The real power is
All of the methods and equations presented in Chap. 26, such as Ohm’s and Kirchhoff’s laws and loopcurrent and nodevolt age methods, can be used to analyze AC circuits as long as complex arithmetic is utilized.
p = f = f iS r r = 1440 W Alternatively, the real power can be calculated from Eq. 27.59. P = S'cos 0  (3877 VA)cos68.2°
1440 W
(c) The reactive power is q
= Yl = (12Q^ ) 2 = 3600 V A R Xi 4 iZ
Alternatively, the reactive power can be calculated from Eq. 27.60. Q = S'sin 0 = (3877 VA)sin68.2° = 3600 V A R (d) The real power is represented by the vector (in rectangular form) 1440 + j 0. The reactive power is represented by the vector 0 + j3600. The apparent power is represented (in phasor form) by the vector 3877/68.2°. The power triangle is
25® COMPARISON OF AHAL0Q, PISCRETE, AND DIGITAL SIGNALS A number of different signal types are encountered in AC analysis. Each has distinctive uses and is best suited to distinctive analysis techniques. Discrete signals are used in power systems to sample analog values and convert these values into a form that can be used by control circuitry. Digital signals are on /off or one/zero signals often used in computers and digital systems that monitor and manipulate control circuitry. Each signal type is shown in Fig. 27.12, and each part of the figure shows a different way of representing the same signal. An analog signal represents a variable as a continuous measured parameter, as shown in Fig. 27.12(a).17 A dis crete signal is composed of separate and distinct parts. It represents a continuous timevarying signal at discrete time intervals known as the sampling rate, as shown in Fig. 27.12(b). The amplitude of the discrete signal is determined by the continuous signal that is sampled. A digital signal makes both the sampling rate (time) and the amplitude discrete, as shown in Fig. 27.12(c).
17An analog signal can also be defined as a signal that represents a timevarying feature (variable) of another timevarying quantity. A typical example is an AC sinusoidal voltage, which can represent a transducer output, an audio signal, or any of a multitude of sensor signals. In the case of an AC machine, the voltage does not represent another variable per se, though it may be viewed as representing the timevarying magnetic field impressed on the armature or stator. The voltage represents not a “signal” but a continuous voltage variable output directly. The distinction is not usually made, but the term “signal” is used more often when referring to sensor outputs.
P P g ® w w w  p p § 2 p a s s . c @ m
Circuit Theory
= 3877 VAZ68.20
2714
P O W E R
R E F E R E N C E
M A N U A L
Figure 27.12 Representations of a Single AC Signal
Circuit Theory 1.0
0.5
0 a tim e
CL
E
0 .5

1.0 (c) digital signal
P P i
•
w w w . p p i 2 p a s s . c o m
Transform ers Fundamentals.............................. Magnetic Coupling..................... Ideal Transformers..................... Impedance Matching.................. Real Transformers....................... Magnetic Hysteresis: BH Curves Eddy Currents............................. Core Losses..................................
Nomenclature a turns ratio B magnetic flux density C capacitance frequency f H magnetic field strength i variable current I constant or rms current k coupling coefficient L inductance or selfinductance M mutual inductance N number of items 0 origin resistance t time V variable voltage V constant or rms voltage Vs effective voltage X reactance z impedance R

T F Hz A /m A A H H 

Q s V V V
0
n
Symbols /x $ lj
permeability magnetic flux angular frequency
H /m Wb rad/s
281 281 283 284 285 286 286 286
1. FUNDAMENTALS A transformer is an electrical device consisting of two or more multiturn coils of wire in close proximity so that their magnetic fields are linked. This linkage allows the transfer of electrical energy from one or more alternating circuits to one or more other alternating circuits by the process of magnetic induction. Transformers can be used to raise or lower the values of capacitors, inductors, and resistors. They enable the efficient transmission of electrical energy at high volt ages over great distances, and then can be used to lower the voltages to safe values for industrial, commercial, and household use. The principle of the transformer is also the basic princi ple of induction motors, alternators, and synchronous motors. All these devices are based on the laws of mag netic induction.
2* MAGNETIC COUPLING Any coil of wire, when energized by a sinusoidal wave form, produces a magnetic flux, i and u;2 are the halfpower points (70% points, or 3 dB points) because at those frequencies, the power dissipated in the resistor is half of the power dissipated at the reso nant frequency.
C —r— Vr
S2 +  ! 
For frequencies below the resonant frequency, a series RLC circuit will be capacitive (leading) in nature; above the resonant frequency, the circuit will be inductive (lagging) in nature.
= 0
30.19
Zy i — y/2 R J
Wl
30.21
= _V i_ =
V I
Zai
V2R
_ _££_
30.22
\/2
The solution is
nr
5i,52 = ± j X/ — = ±j(J0
p „1 = i *r
No damping term, 0 and is called resonance.
Figure 30.9 Series Resonance (BandPass Filter)
In a real circuit, resistive components are present and will yield a damping term, cr1? in the solution to the circuit equation. The result is that an external source of energy is required to make up the loss. Resonance is then defined as a phenomenon of an AC circuit whereby relatively large currents occur near certain frequencies, and a relatively unimpeded oscillation of energy from potential to kinetic occurs. The frequency at which this occurs is called the resonant frequency. At the resonant frequency the capacitive reactance equals the inductive BW 4This type of circuit is more than academic and can be realized, or nearly so, with superconducting inductors.
P P B
®
w w w ap p i 2 p a s s ne @ m
= ( ^ ) 2r = ±p 0
30.20
30.23
T R ^ M S i E ^ T
A I A L Y S I S
3© ~ 9
11. S EM ES HESONAUCE
Figure 30.10 Parallel Resonance (BandReject Filter) high R
In a resonant series RLC circuit, impedance is minimum impedance equals resistance current and voltage are in phase current is maximum power dissipation is maximum The total impedance (in rectangular form) of a series RLC circuit is R + j(X L — X q) At the resonant frequency, uj0 = 2nf0, Xl = Xc
The frequency difference between the halfpower points is the bandwidth, BW, a measure of circuit selectivity. The smaller the bandwidth, the more selective the circuit. BW = / 2  A
[at resonance]
1
3 0.29
CJqC
30.24
1
2tt/0 =
loq =
30.28
30.30
The quality factor, Q, for a circuit is a dimensionless ratio that compares the reactive energy stored in an inductor each cycle to the resistive energy dissipated.5 The effect the quality factor has on the frequency char acteristic is illustrated by Fig. 30.9. maximum energy stored per cycle energy dissipated per cycle
1= 2k
/o /2 ~ fl
_
^0 ~
V2 p _ 1 r2 r>____ m =
i 2r
= 'Y l
30.31
R
The quality factor for a series RLC circuit is X _ u;0L
/o _ (BW)Hz (BW)ra,
.
The power dissipation in the resistor is
Q
parallel or series
R~
30.25
R
1 uoRC
Then, the energy stored in the inductor of a series RLC circuit each cycle is
R \ C
Uo
fo
(BW)rad/s
(BW)Hz
G
t,=¥=^=«(0
= GluqL = 30.26
Example 30.5
The relationships between the halfpower points and quality factor are 1
3 0.32
CJqC
1
A series RLC circuit is connected across a sinusoidal voltage with peak of 20 V. (a) What is the resonant frequency in rad/s? (b) What are the halfpower points in rad/s? (c) What is the peak current at resonance? (d) What is the peak voltage across each component at resonance? 50 ft 2^ AAA,1
3 0 .27
Various resonant circuit formulas are tabulated in App. 30.A.
H
200 pF
L
20 sin oot
5The name figure of merit refers to the quality factor calculated from the inductance and internal resistance of a coil.
PPg
® w w w „ p p i 2 p a § s ae © m
Circuit Theory
Vlc
3 0  1 0
P O W E R
R E F E R E N C E
M A N U A L
The power dissipation in the resistor is
Solution
(a) Equation 30.30 gives the resonant frequency.
wo :
1
p _ 1 T2 7? —
2m
1 =
\J(200 x 10“ 6 H)(200 x 1CT12 F)
i 2r
=Yl
5 x 106 rad/s
30.36
R
(b) The halfpower points are BW
The quality factor for a parallel RLC circuit is u0 =FA 2L
: CJ0 =F
50 s
R X
Q
n
R u>0L U>0 (BW)„
(2) (200 x 10“ b H) R \l L
= 5.125 x 106 rad/s, 4.875 x 106 rad/s (c) The total impedance is the resistance at resonance. The peak resonant current is V, z0
V2 m ~2R
Vm R
20 VZ0° so n
_u>oC __
1
G
Gu0L
/0 (BW)Hz 30.37
Example 30.6 A parallel RLC circuit containing a 10 fi resistor has a resonant frequency of 1 MHz and a bandwidth of 10 kHz. To what should the resistor be changed in order to increase the bandwidth to 20 kHz without changing the resonant frequency?
= 0.4 AZ0°
Circuit Theory
(d) The peak voltages across the components are VR = I0R = (0.4 AZ0°)(50 ft) = 20 VZ0° Vl = I qX l = loj^oL
Solution
= jf(0.4 AZ0°) (5 x 10® — ) (200 x 10"6 H)
From Eq. 30.25 and Eq. 30.37,
= 400 VZ90°
Vc = Il, X c = , I° ~
°'4AZ0‘ x 10e £ ^ (2 0 0 x 1 0 12 F)
ju0C
400 YZ —90°
/o = 2nf0RC BW 1 1 C= ■ 2nR(BW) (2ji)(10 ft)(10 x 103 Hz)
Qold '■
= 1 x lO5/ 2^ F
12. PARALLEL RESONANCE
The new quality factor is
In a resonant parallel GLC circuit,
Q
• impedance is maximum
w
= —
BW
106 Hz 20 x 103 Hz = 50
• impedance equals resistance • current and voltage are in phase • current is minimum
From Eq. 30.37, the required resistance is
• power dissipation is minimum The total admittance (in rectangular form) of a parallel GLC circuit is G + j(B c — BL). At resonance,
R=
Q
2nf0C
50 (2tt)(106 H z) (
30.34
CJ0c LUq — 2tc/o =
3 0.35
y/LC
P P §
® w w w . p p S 2 p a s s . c o m
F 271
30.33
=5Q
Time Response Fundamentals............................................... 311 FirstOrder Analysis.....................................312 FirstOrder Analysis: Switching Transients . . 313 FirstOrder Analysis: Pulse Transients . . . . 316 SecondOrder Analysis................................. 318 SecondOrder Analysis: Overdamped......... 319 SecondOrder Analysis: Critically Damped . . 3111 SecondOrder Analysis: Underdamped....... 3112 SecondOrder Analysis: Pulse Transients . . 3113 HigherOrder Circuits.................................. 3113 Complex Frequency......................................3114 Laplace Transform Analysis........................ 3115 Capacitance in the sDomain...................... 3116 Inductance in the 5Domain.........................3116 Laplace Transform Analysis: First and SecondOrder Systems..............................3116 16. PID Controllers............................................ 3117
Subscripts
0 c d
DC i L L /C n V
P s
so ss tr Th
initial or resonant capacitor, compensator, or controller derivative direct current integral inductor inductor or capacitor natural peak proportional source switch open steady state transient Thevenin
Circuit Theory
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
Nomenclature
a, A A b, B c C
constant natural response coefficient constant constant capacitance F G {s) forward transfer function — i variable current A K error constant, gain constant, or — scale factor /, I constant or rms current A L selfinductance H Q quality factor — R resistance O s, s complex frequency Hz s Laplace transform variable s 5domain variable t time s v variable voltage V V, V constant or rms voltage V Z, Z impedance Q Symbols a
f3 ( 6 k
a
r u cjq
damping oscillation frequency damping ratio phase angle steadystate constant neper frequency or napier frequency time constant angular frequency resonance frequency
Hz rad/s — rad — Np/s or Hz s rad/s or Hz rad/s
1. FUNDAMENTALS Time domain analysis is analysis that takes place with time as the governing variable. Sinusoidal analysis is analysis that occurs when the current and voltage func tions are manipulated in their trigonometric form. Fre quency domain analysis deals with Fourier or Laplace transforms of time functions, which are then functions of frequency. The Fourier transform occurs in £;space which, when k is replaced by a;, is called the frequency domain. The Laplace transform occurs in 5space, also considered the frequency domain. The variable s is called the complex frequency and is given by Eq. 31.1. (In Eq. 31.1, the variable s is shown in bold to empha size its phasor properties, which are important in polezero analysis.)
s=
a
+ jo;
31.1
The variable a is termed the neper frequency and is assigned units of Np/s or Hz. The variable a represents the ratio of two currents, voltages, or other analogous quantities. The number of nepers is the natural log of this ratio. The term “napier” is sometimes used in place of neper.1 The Laplace transform and the complex fre quency s are used to represent exponential and damped sinusoidal motion in the same manner that ordinary frequency expresses simple harmonic motion. That is, using the variable s allows constants, exponential decay, sinusoids, and damped sinusoids (sinusoids with an 1This occurs since the natural logarithm is alternatively called the Napierian or Naperian logarithm.
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exponential envelope) to be represented in the same form, Aest (see Sec. 31.11). Sinusoidal analysis is also considered any analysis using the assumption that the forcing function, the source, is sinusoidal in nature. Sinusoidal analysis in Chap. 27 through Chap. 30 assumed a single frequency. When the frequencies are allowed to vary, the circuit output is called the fre quency analysis or frequency response.
parameters of the time constant. The transient response will be of the form Ae~^T. step 4' Sum the steadystate and transient solutions,
that is, the complementary and particular solu tions, to obtain the total solution.3 This will be of the form ft + Ae~^T. step 5: Determine the initial value from the circuit
initial conditions. Use this information to calcu late the constant A.
2. FIRSTORDER ANALYSIS
step 6: Write the final solution.
Circuit Theory
Firstorder analysis is performed upon circuits with one energy storage element—that is, one equivalent capacitor or inductor. In circuits containing numerous sources and resistors, the Thevenin or Norton equivalent is taken as seen from the terminals of the single capacitor or induc tor. This allows the methods of Chap. 30 to be applied, modified so that the Thevenin voltage, *s used in place of the source voltage and the Thevenin current, h> is used as the current through the capacitor and inductor. Figure 31.1(a) represents a generic circuit with a single capacitor or inductor. The source can be either AC or DC. Figure 31.1(b) shows the Thevenin equivalent of the circuit with the switch closed. Figure 31.1(c) shows the new Thevenin equivalent with the switch open. Part (c) of this figure is a convenient way of showing that if the source is eliminated (or changed to a DC source), the resulting currents and voltages will be exponentials (the natural response) and will share a new time constant and possibly new initial currents and voltages.
A capacitive firstorder circuit generally uses voltage as the dependent variable. Applying KVL to Fig. 31.1(b) gives ^Th = ^Th^Th + VC Figure 31.1 FirstOrder Circuit
The differential equation for firstorder circuits has the general form f(t) = b ^ + c x
31.4
D
*UC 
;Th
31.2
LorC
The solutions are of the form 31.3
The time constant r = b/c. ^ is a constant, the steadystate component of the solution. The method for deter mining the complete solution is as follows.
(b) Thevenin equivalent: DC or AC, switch closed
‘UC, s
oo.
T I ME
The Thevenin current is the current flowing through the capacitor, frh — i c — C ( d v c / d t ) . Substituting gives ^Th = ^Th C (
]+
vc
RESPONSE
3 1 ”3
f= 0 20
a
—AAA/— / +
31.5
I ion
In order to determine the constant A , the voltage con dition must be known at a certain time. Because the capacitor voltage cannot change instantaneously, vc(®~) — ^c(0+) The switch was opened three time constants into the transient determined by the condi tions in Ex. 31.4. The transient response of the circuit in Ex. 31.4 is vc (t) = Ae~^RThC = A e n ) ( w x i o ~ « f))
_ j[e t / 250xlO6 s The total solution for the circuit in Ex. 31.4, as given by Eq. 31.6, and noting that the initial voltage is zero, is Using the voltage divider concept, the Thevenin voltage is 5 V. The Thevenin resistance is calculated as seen from the terminals of the capacitor. it/Th = Rl\\R2 =
R xR2 Ri +
r2
(10 ft)(10 ft) 10 ft + 1 0 ft
5 ft The steadystate condition is determined from Eq. 31.5, noting that the voltage is constant ( dv/dt = 0). ^Th =
Rt\i
C
+
VC
Vc(t) = VCiSS + Ae~^T = 5 V + ^e  (/(250x10"6 s>
uc (0) = 5 V + ^ eo/(250xio6s) = 5 V + ^ = 0 y A=  5V tlC(f) = 5 V  5eV(250xlO6 s) y
At three time constants into the initial transient, Vc(3t) = 5 V  5e((3)(250xl0“6s))/(250xl06 s) y _ 4 75 y Use 4.75 V as the initial voltage for the network in Ex. 31.5 to determine the value of A.
= (5 ft)(50 x 10“ 6 F)(0 ) + vc
= 5V Vc,ss = 5 V
The capacitor is an open circuit for steadystate DC conditions, as expected.
PPI
® w w w np p i 2 p a s s » c @ m
vc (t) = Vc,ss + A eV T = 0 V + J4e_t^500x10’ 6s)
vc (0“ ) = vc (0+) = 4.75 V 4.75 V = 0 V + ,4e0/(500xi06s)
= A
T I M E
Therefore, the response of the network from the time the switch opens is
R E S P O N S E
31*5
Convert to the time (sinusoidal) form. 1L = 1.06 AZ—55°  1.5 A sin(377t  55°)
vc (t) = Vc ,ss + Ae~^T = 0 V + 4.75e_(/(500x10”6 s) V
= 4.75e(//'500xlir6 s) V
Example 31.6 For the network shown, Ri = 10 Cl, R2 = 10 Cl, v = 170 Vsin(377£ + 30°), and L = 0.15 H. Determine the steadystate current with the switch closed.
Effective or rms values are used with phasor quantities. Peak values are used with sinusoidal quantities.5 Radians and degrees are mixed in the equations strictly for clarity. Radians are necessary for calculations. Transient calcula tion using sinusoids follows a similar pattern, with phasor analysis being an efficient calculational method. During transients using sinusoids, the exact time of the transient must be specified. Example 31.7
0.15 H
For the network shown, R i = 10 Cl, R 2 = 10 Cl, v = 636 Vcos(2513£ + 60°), and (7 = 1 fi F. Determine the steadystate capacitance voltage with the switch closed.
Solution
^Th = lL =
= LjujiL + Rt^ l VTh
The source is given as sinusoidal. Analysis is the same for sinusoidal sources, but the mathematics is more cumbersome. Instead, use the phasor representation of the sinusoid source and Eq. 31.8. ^Th i^Th + j^L VTh _
II
VTh
z
Write the source voltage in phasor form as 120 V Z30°. Using the voltagedivider concept, the Thevenin voltage is onehalf this value. The impedances of the resistor and inductor are Z RTh = 5 ftZ0° Z L = ju>L = j( 377
Solution
Using Eq. 31.5 gives
jioL + i?Th
il 
Circuit Theory
Using Eq. 31.8 and noting that multiplying by ju is equivalent to taking the derivative, gives
(0.15 H) = 56.55 ftZ90°
Z Th = Z*Th + ZL = (5 + jO) + (0 + j 56.55) = 5 + j 56.55
^Th = ^Th C
+ vc
Represent the source voltage in phasor form as 450 VZ60°. Using the voltage divider concept, the The venin voltage is onehalf this value. Recall that, in the frequency domain, multiplying by ju is equivalent to taking the derivative. Combining the two gives ^Th = ^ThC (^~^j + vc V Th = Z*ThC juV c + V c = (juZRThC j 1) V c V Th jcjZ#ThC + l
V C= T
225 VZ60° rad j ( 2513 —^ )(5 fi)(l x 10"6 F) + : s 225 VZ60° 1.01 ftZ0.72° = 225 VZ59.280
 56.77 Cim .9° Transforming into the time domain gives Substitute to determine the steadystate current. vc,ss(t) = 318 V cos(2513£ + 59°) t
_ V Th _
z
V Th
z*Th+ z L
1.06 AZ—55°
_
60VZ300 56.77 C im .9°
The representation of phasors as rms quantities and sinusoidal vari ables as peak quantities is standard in this text. Care should be taken to determine how phasors are being represented in a given document.
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At the angular frequency given, equivalent to 400 Hz, the capacitance is essentially an open circuit and has little impact on the network. The voltage source in this example was given as the cosine function to emphasize that the choice is arbitrary and does not affect the analysis.
The unit step is defined by Eq. 31.10. The unit impulse is the derivative of the unit step function. The relation ships between various forcing functions are shown in Table 31.1. u(t — ti) = 0
[t < ti)
u(t — ti) = 1
[t > t\\
31.10
4* FIRSTORDER ANALYSIS^ P U L S E TRANSIENTS Table 31.1 Operations on Forcing Functions
A pulse is generally defined as a variation in a quantity that is normally constant. When the variation occurs at a single point in time, the pulse is called a step pulse. If the magnitude of the step pulse equals one, it is termed a unit step. A pulse has a finite duration that is normally brief compared to the time scale of interest. When the duration of the pulse is so short that it can be thought of as infinitesimal, it is called an impulse. Figure 31.2 shows a unit pulse of each type. Figure 31.2 Pulse Transients   ^
1
Circuit Theory
U (t)
t= 0 (a) unit step u(t)
u(t
function when function
differentiated
integrated
unit impulse unit step unit ramp unit parabola unit exponential unit sinusoid
unit impulse unit step unit ramp unit exponential unit sinusoid
unit step unit ramp unit parabola (third degree) unit exponential unit sinusoid
The response of a circuit to a unit step is called a step response. The source voltage magnitude, in this case the Thevenin voltage magnitude, is one for a unit step. For a firstorder linear circuit like the one shown in Fig. 31.3, the response is determined in terms of the state variables of the energy storage device, that is, inductive current or capacitive voltage.6 If other electri cal quantities are required, they can be found from any of the circuit relations. The response equations are vc (t) = u(t)(l  e t,T)
t,) t=t,
31.11
pVn
31.12
(b) unit step u(t  t,) Figure 31.3 FirstOrder Circuit
^Th
MAr
u(t)  u[t — f,) f=0 (c) pulse u(t)  u(t  t)
''Th
Lor C
1
A
The time constant, r, for the capacitive circuit is R ^ C and for the inductive circuit is L /R Th.
8(f) 0
t
The step and impulse responses for RC and RL circuits are shown in Table 31.2. Of importance, the magnitude of the voltage and current sources for a unit step response is one.7
(d) unit im pulse 8(f)
1 A
8 (f — t,) 0
t,
t
(e) unit im pulse 8 ( f  tj)
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6A state variable is a variable sufficient to specify the condition of the network and define its future behavior. 7The magnitude of one for a unit step is the reason for the apparent nonappearance of the current and voltage in the equations.
T I M E
3 1 7
R E S P O N S E
Table 31.2 Step and Impulse Responses for FirstOrder Circuits
circuit
unit step response
unit impulse response VTh{ t) =
RTh
vTh(t) = u(t)
W r *Th
O
6(t)
'C y .
C
vc(t) =
u ( t ) ( l — e _ ( / * * rhC) ^
ic (t) =
U( * ) ( ^ T ^ )
+m {±
vC
= u(t)
*Th(t) = S(t)
«Th(i) = «(*) vc (t) = u(t)RTh(l  e*/(i2xhC)^ ic(t) = u(t) = u(t)(e~t/(RTkC^
VTh ( t ) =
U(t)
VL{t) =
u(t)
iL { t ) =
u(t) (
V'Th
O
l
3 VL
= u(i) (—p ) J (l 
e
(/ ( £/ / i Th) )
*Th(0 = 4 ac. The network is overdamped.
y ( 5 ) 2  (4)(2)(2)y = 16
step 3: The solution form is x(t) = hi + AeSlt + BeS2t
Vb2 — 4ac
step 4' The initial conditions are given. The forcing
function is a constant 24 V. At steadystate, dv/dt and d2v/di9 equal zero. The steadystate voltages are11
(u(0+)  vss)
B = i(l 
V b2 —4 ac. /
L)
(0  12 )
^/(5)2  (4)(2)(2)y
24v=2(‘ili?)+5(^)+2»— 0 + 0 + 2vss 24
2
(0 0 )
(5)2 — (4) (2) (2) y
12 V
10Dividing the equation by two shows that 12 is the value of the forcing function. Further, the 12 actually represents a step change of the source (Thevenin) voltage at t = 0. After the transient ends, the final value of the voltage across the capacitor will be 12 V. n Units are not shown where they might cause confusion with the variables used, such as the voltage v and volts V. Units should be shown in all final solutions.
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step 7: Write the final solution.
v(t) = hi + AeSlt + BeS2t = 12
+ 4e2t
T I M E
7* SECONDORDER AHALYSISi CRITICALLY DAMPED
R E S P O N S E
31*11
behavior of the circuit, in terms of the capacitance voltage, for t > 0.
If the two roots of Eq. 31.14 are real and the same, equivalent to the condition b2 = 4ac, the solution is of the form x(t) =
k
+
+ Btes
31.26
The first term, ft, is a constant, that is, the steadystate component of the solution. The exponential terms repre sent the natural response of the circuit. The root of the characteristic equation is Solution 31.27
The coefficients A and B in Eq. 31.26 are found from the initial and steadystate values of x and dx/ dt. By the same reasoning used in Sec. 31.6, the initial conditions for a series RLC circuit remain unchanged and are repeated here for convenience. ^c(0+)  V0 dvc(0+) = h dt C
31.28
31.29
Similarly, the initial conditions for parallel RLC cir cuits are
u (o +) dkj 0+)
Io
M 0 +)
dt
Using KVL and writing the resulting equation in terms of the capacitance voltage, 10 V — iRxh +
+ vc
Substituting i = C(dv/dt), 10 V = c ( ^ ] R Th + L c ( ^ f ] + v c dt2 10 V = L C
d vc
dvc dt2 J 1 ^ iX1~V dt
Vc
The characteristic equation follows.
31.30
L C s2 + R C s + 1 = 0
31.31
Determine the solution form from the relationship between the coefficients of the characteristic equation.
The constants A and B are determined by substituting Eq. 31.28 through Eq. 31.31 into Eq. 31.26 and its derivative (multiplying by s is the equivalent of taking the derivative) and then solving for the specific circuit. Equation 31.28 through Eq. 31.31 can be used if net work initial conditions are given and circuit element parameters such as L and C are known. The results of such a substitution, in generic terms, are z(0+) = A + :rss
31.32
i'(0 +) = sA + B + x'ss
31.33
b2 = (RC )2 = ((80 ft)(50 x 10“ 6 F))2 = 16 x 10~6
4 a c = (4)(LC)(1) = (4)(80 x 10~3 H)(50 x 106 F )(l) = 16 x 10“ 6 b2 = 4 ac
The circuit is critically damped. The solution is of the form given by Eq. 31.26. vc (t) — k + Aest + Btest
Solving Eq. 31.32 and Eq. 31.33 simultaneously gives the values of the constants. A — x(0 ) B =
xss
x '( 0 + )  x ' s s + { £ ) A
31.34 31.35
Example 31.12 Consider the series RLC circuit shown, with Vrh = 10 V, R = 80 ft, L = 80 x 10“ 3 H, and C = 50 x l0 “ 6 F. The capacitor is initially charged to 5 V and the initial current in the circuit is 187.5 x 10~3 A. Determine the
w ‘Lc{% + R ^c{ift
The initial conditions are given. The steadystate con ditions are found from the results of the KVL analysis.
VC
= LC(0) + RThC(0) + vc VC, ss = 10
^,ss = 0
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2a
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The roots of the characteristic equation are found from Eq. 31.27 or actual calculation to be12
This can also be represented in the sinusoidal form as x(t) = hi + eat (A cos (3t + B sin f3t)
_ —b _ —Rtyi C S ~ ~2a ~ 2LC (80 ft)
~^Th 2L
The first term, ft, is a constant, that is, the steadystate component of the solution. The exponential terms repre sent the natural response of the circuit. Equation 31.36 is the ringing response, or damped oscillation, shown in Fig. 30.7(b). The sinusoidal response is enveloped by the exponential, since a is negative.
(2) (80 x 10” 3 H) = 500 Because the initial conditions and circuit parameters are known, instead of using the generic equations (see Eq. 31.32 through Eq. 31.35), use the solution form directly with the initial conditions from Eq. 31.28 and Eq. 31.29.
The variable s used in the overdamped and critically damped cases has been expanded to clarify this behavior and is represented by Eq. 31.38 and Eq. 31.39.14 si = a + j(3
31.38
jf3
31.39
S2 = OL 
vc(t) =
k
+ Aest + Btest The roots of the characteristic equation are
Vc(0+) = 10 + Aes^ + B (0) es(°)
5 = 10+ .4 A—
5
Circuit Theory
(3 =
The second initial condition results in the following value for B. vc(t) = k + Aest + Btest dvcj 0+) = k = o + Asest + B(est + tsest) dt O
187.5 x 10 3 A = Q 50 x 10 F
( _ 5)(_50 0 )/'e(500)(0)>\ V ) / g(500)(Q) (0)(—500) \
+ B
x e(—500)(0)
b 2a
31.40
VA clc — b2 2a
31.41
The coefficients A e and Be in Eq. 31.36, or A and B in Eq. 31.37, are found from the initial and steadystate values of x and dx/dt. By the same reasoning used in Sec. 31.6, the initial conditions for a series RLC circuit remain unchanged and are repeated here for convenience. vc( 0+) =
31.42
,t term
>k o A
step 3: Solve for the transformed variable, V(s) or /(s),
in terms of
step4' Use partial fraction expansion, or any other
The Laplace transform method of analysis moves the state variables, that is, the capacitive voltage and the inductive current, from the time domain into the sdomain. In the 5domain, many forms of excitation, such as the constant, the exponential, and the sinusoid, can be united and easily handled mathematically. The Laplace transform is given by20 POO
31.76
The calculation of Eq. 31.76 seldom needs to be accom plished. Numerous tables of Laplace transforms exist (see App. ll.A ). Using the data in a Laplace transform table, the properties given by Eq. 31.77 and Eq. 31.78 are used to transform a differential equation represent ing the behavior of an electrical network.
m ) = F(s )
31.77
= szF ( s )  s f ( 0 + )  f \ 0 +)
31.78
s F ( s )  f ( 0+)
31.79
Because the energy in a capacitor or inductor cannot change instantaneously, the conditions at t = 0 are those at t = 0+. Once the transformation is complete, the inverse Laplace transform can be taken using Eq. 31.80. 1
ra+joo
£ i{F (s)}= m = ± 
F(s) estdt
31.80
Jcr—700
desired method, to rearrange the terms into those recognizable in Laplace transform tables. step 5: Equate the coefficients of the powers of s. Solve
for the constants. step 6: Obtain the inverse Laplace transform.
The classical method is useful when the desired quantity is one of the state variables, capacitor voltage or induc tor current. If the desired variables are other than the state variables, the circuit transformation method is the most direct. The circuit transformation method follows. stepl: Transform each individual circuit element into
the 5domain. Initial condition sources will exist for energy storage elements. For the capacitor, an initial condition voltage source will be added in series. For the inductor, an initial condition current source will be added in parallel. step 2: If the desired quantity is one of the variables of
an energy storage device (i.e., a state variable), obtain the equivalent Thevenin or Norton form of the circuit as seen from the terminals of the capacitor or inductor. stepS: Write the differential equation, describing the
circuit behavior from known circuit relationships and laws. step 4: Solve for the desired variable in terms of 5. step 5: Use partial fraction expansion, or any other
desired method, to rearrange the terms into those recognizable in Laplace transform tables. step 6: Equate the coefficients of the powers of 5. Solve
for the constants. 20Also, taking the Laplace transform of a physical quantity introduces an additional unit of time. For example, I(s) has units of As or C. Because the inverse transform removes the extra time unit, such units will be omitted in the sdomain and I(s) is still referred to as the current. The same is done for the voltage, V(s).
step 7: Obtain the inverse Laplace transform.
Regardless of the method used, the transformed equa tions obtained in step 3 are arranged using partial
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Circuit Theory
12, LAPLACE TRANSFORM ANALYSIS
£ {/(* )} = / f(t)e~stdt J0+
5.
3 1  1 6
P O W E R
R E F E R E N C E
M A N U A L
fraction expansion in order to be inversely transformed. This is necessary for transient analysis of first and secondorder systems, because the initial conditions impact the transient. Because the initial conditions do not impact the steadystate value, in sinusoidal analysis, usually called frequency analysis or frequency response, the equations of step 3 reduce to phasor form, so steps 4 through 7 can be skipped and the response analyzed using graphical means. (See Chap. 32.)
Figure 31.9 Capacitor Impedance and Initial Voltage Source Model I c(s)
h = Zc
+ VAs)
13 CAPACITANCE INI THE^DOMAIN 14. INDUCTANCE IN THE sDOMAlN Capacitors oppose rates of change in voltage. In capac itive circuits, voltage is the state variable and is preferred in system equations. Because the capacitor voltage cannot change instantaneously, vc(0+) = vc(0). With this infor mation, the capacitive voltage, vc(t), and the capacitive current, C(dvc(t)/dt), are transformed by Eq. 31.81 and Eq. 31.82. jC{vc (t)} = Vc(s)
31.81
Inductors oppose rates of change in current. In inductive circuits, current is the state variable and it is preferred in system equations. Because the inductor current can not change instantaneously, ^(0+) = ^(0). With this information, the inductive current, z^(t), and the induc tive voltage, L(diL(t)/ dt), are transformed by Eq. 31.83 and Eq. 31.84. C{iL(t)} = I L(s)
31.83
Circuit Theory
C { i c (t)} = C\ C C{vL(t)} = cS.L = s C V c ( s )  C v c (0)
The transformed capacitance must exhibit the proper ties at its terminals given by Eq. 31.82. Such a model is shown in Fig. 31.7. Using the extended Ohm’s law on the current source sCVc(s), that is, taking the voltage across the source’s terminals, Vc(s), and dividing by the current, sCVc{s ), gives the model of Fig. 31.8. Taking the Thevenin equivalent of the model shown in Fig. 31.8 results in a third model given in Fig. 31.9 that shows the initial voltage explicitly. Any of the models is valid; Fig. 31.9 is perhaps the most intuitive. Figure 31.7 Capacitor TwoSource Model
J
sLI l (s)
 LiL{0)
31.84
The transformed inductance must exhibit the properties at its terminals given by Eq. 31.84. Such a model is shown in Fig. 31.10. Using the extended Ohm’s law on the voltage source s L Il (s ), that is, dividing by the cur rent through the source’s terminals, Il ( s), gives the model shown in Fig. 31.11. Taking the Norton equiva lent of the model in Fig. 31.11 results in a third model given in Fig. 31.12, which shows the initial current explicitly. Any of the models is valid; Fig. 31.12 is perhaps the most intuitive. Figure 31.10 Inductor TwoSource Model
Ids)
Vc(s) sCVc(s )Q
dt
31.82
I L(s)
D0
Cvc(0)
r
) sUL{s)
) LiL(0)
Figure 31.8 Capacitor Impedance and Initial Current Source Model Icis)
15. LAPLACE TRANSFORM ANALYSIS^ FIRST AND SECONDORDER SYSTEM S Either of the methods in Sec. 31.12 may be applied. If variables other than the capacitor voltage or inductor current are required, the circuit transformation method is the most direct method. The Thevenin form is preferred for capacitive circuits because the final capacitive voltage must equal the Thevenin voltage, thereby acting as a check on the solution. The Norton equivalent is preferred
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Figure 31.11 Inductor Impedance and Initial Voltage Source Model I L(s)
R E S P O N S E
3 1 1 * 7
The initial capacitor voltage can be determined from the initial charge, using C = Q/V. The constantsource voltage can be considered a unit step at t = 0. The initial capacitor voltage is a constant as well. Both transform to the sdomain using 1/s. (See App. 11.A.) Integration equates to multiplication by 1/s. Substitut ing and transforming gives V , = i(t)R + ^
M V = / (» ) ( 2 f !) + Figure 31.12 Inductor Impedance and Initial Current Source Model
+ m
+ l
i(t)dt
« x i< r Jc (75 x 1CT6 F)(s) I(s)
(75 x 1CT6 F)(s)
30 A s + 6.65 x 103
i(t) = 30 Ae_6'65xl°3*
for inductive circuits, because the final inductive current must equal the Norton equivalent current, thereby acting as a check on the solution. Any of the network analysis techniques is applicable when working in the sdomain. Example 31,13
In the circuit shown, the capacitor has an initial charge of 4.5 x 103 C. The switch closes at t — 0, applying a steady 120 V source. Determine the current using the Laplace transform method. R=
2fl
'•— V W — 120 V _zi_
i(t)
C =
1®. PID CONTROLLERS A proportionalintegralderivative controller, or PID controller, can compensate for a wide variety of electri cal system responses. A PID controller combines the functions of a proportional controller, an integral con troller, and a derivative controller. The proportional controller regulates the magnitude of the difference between the desired and actual levels of a monitored variable, the integral controller regulates the total change in the desired quantity over time, and the deriv ative controller regulates the rate of change in the mon itored variable.
7 5 jlF
Gc{s ) = K p f — + KdS Solution
The classical approach will be used. Using KVL in the direction shown gives V$  i ( t ) R  ± ;
J
i(t)dt = 0
[PID controller]
31.85
In general terms, the proportional gain controls the response time. The integral gain determines the number of oscillations over time (ringing control). The deriva tive gain controls the magnitude of the overshoot. The combination of all three will determine the overall response to any given transient.
Though not stated explicitly, the behavior of the circuit from t = 0 onward is the desired result. So, Vs = i (t )R+ VC (0+) + — J
i(t)dt
PPI
9 www.ppi2pass.eQm
Circuit Theory
Performing the inverse Laplace transform on /(s) gives
Frequency Response Fundamentals............................................... 321 Transfer Function........................................ 322 SteadyState Response................................. 323 Transient Response...................................... 324 Magnitude and Phase Plots.........................325 Bode Plot Principles: Magnitude P lot........ 327 Bode Plot Principles: Phase P lot................ 3211 Bode Plot Methods.......................................3211
Nomenclature A
factor or constant
B
constant

C
capacitance
F
D
denominator magnitude
dB
G
gain
i
instantaneous current
I, I
effective or
K
constant
L
selfinductance
H
N
numerator magnitude

P
pole
Hz
r(t)
response function, time domain
R (s )
response function, 5domain

DC current
A A 
s
complex frequency
Hz
T (s )
transfer or network function
V v, v
instantaneous voltage
X
constant
Y
constant
V V 
z
zero
Hz
Z
impedance
0
effective or
DC voltage
Symbols a
angle
rad
0 6
angle
rad
angle
rad
a
neper frequency or napier
N p /s or
frequency L0
break or corner frequency
N p /s or
angular frequency
rad/s
Subscripts d
number or denominator
G
gain
n
number or numerator
net
network
V s
pole
z
zero
source
In any network with capacitors or inductors or both, transients will occur.1 Transients can be represented by mathematical circuit models consisting of differential equations. The solutions to these equations depend on a circuit’s initial conditions, and the methods outlined in Chap. 31 can be used to obtain these solutions.2 The Laplace transform method can be used to change differ ential equation manipulation to algebraic manipulation, and the phasor method can be used to simplify sinusoi dal mathematics. Transient analysis requires the complete analysis of a circuit, including initial conditions, because both the transient and steadystate portions are relevant. The transient portion (natural response) of any output even tually settles on the steadystate (forced response) value determined by the forcing function, that is, the signal.3 Steadystate analysis of a circuit, on the other hand, is independent of the circuit’s initial condition. Solutions to problems in steadystate analysis can be determined from the circuit characteristics alone. These character istics are contained within a transfer function, T(s), as shown by Eq. 32.1.
t(s ) = £ {
m
}
The circuit’s input signal or forcing function is f(t), and the response to this function is r(t). The transfer func tion, T(s), is analyzed in the frequency (or s) domain rather than in the time (or t) domain, and this type of analysis is called frequency response analysis. Unlike phasor analysis, in which a fixed frequency is assumed, frequency response analysis involves the examination of a range of frequency values. Moreover, frequency response analysis can be used to analyze the effects of variable frequency inputs and spectral inputs, which have multiple frequencies. When the signal is sinusoidal (that is, s = jcu), sinusoi dal steadystate analysis can be used. If the time response to a sinusoidal signal is needed, the signal magnitude, \F(s)\, is multiplied by the transfer function 1In purely resistive circuits, the response is considered instantaneous. 2The arbitrary constant occurring in integration of electrical param eters represents the initial condition. 3The natural response is determined from the homogenous solution (i.e., the forcing function is equal to zero) of the differential equation. The forced response is from the particular solution whose form must be that of the forcing function.
P P I
® w w w ap p § 2 p a s s no ® m
Circuit Theory
1. 2. 3. 4. 5. 6. 7. 8.
P O W E R
3 2  2
R E F E R E N C E
M A N U A L
magnitude, T(s) , and their product is the magnitude of the output or response, i?(s). The angle is found by adding the signal angle, IF(s), to the transfer function angle, ZT(s), which gives the angle of the response, lR(s). The result is a phasor that can be changed into sinusoidal form using the known value of juj. The transfer function contains the circuit’s characteristics, so it can be used to determine the output for a variety of inputs.
is found by substituting the value of the complex fre quency s = crs\jujs for the source into Eq. 32.2. The variable a is the neper frequency and is assigned units of Np/s or Hz. (See Sec. 32.4.) The variable a represents the ratio of two currents, voltages, or other analogous quantities. In this case, as represents the ratio of two source frequencies. The properties of the circuit ele ments themselves are held in the complex zeros and poles, z or p.
2. TRANSFER FUNCTION4 The term transfer function refers to the relationship of one electrical parameter in a network to a second elec trical parameter elsewhere in the network.5 The network function, Tnet(s), refines this definition to be the ratio of the complex amplitude of an exponential output, Y(s), to the complex amplitude of an exponential input, X(s). If x(t) is Xest and y(t) is Yest, and if the linear circuit is made up of lumped elements, the network function can be derived from an inputoutput differential equation as a rational function of 5, and can be written in the following general form.6
Circuit Theory
Tnet(s) =
Example 32.1
Determine the network function of the circuit shown in terms of a current output, /(s), for a given voltage input, V(s).
5ft
Y (s)
X(s)
= A
(s 
Zj ) ( s 
Z2) ••• ( s
 zn)
{ s  p 1) ( s  p 2)   ( s  p d)
3 2 .2
The complex constants represented as Zi, Z2, and so on are the zeros of Tnet(s), and are plotted in the 5domain as circles. Zeros are values within a transfer function that result in a zero response. A zero response is the minimum a circuit can provide, which is no output. The complex constants represented as pi, p2, and so on are the poles of Tnet(s) and are plotted in the 5domain as X ’s. Poles are the values within a transfer function that result in an infinite response, which is the maximum response a circuit can provide.
The network function is Tnet{s) —
M _ 1 V(s) Z(s)
Because the calculation will be accomplished in the 5domain, transform the circuit using the properties given in Table. 31.4. The result is 1
Zc —
sC
mF\
40 F
V5(250 mF),
Zl =
sL
= 5(2.67 H) = 2.675
Equation 32.2 can be interpreted as the ratio of the response in one part of the 5domain network to the excitation in another part of the network. When 5 equals any zero, the response of the network will be zero regardless of the excitation. When 5 equals any pole, the response will be infinite regardless of the excitation. Equation 32.2 is independent of the forcing function. That is, the equation contains the characteristics of the network itself. The response to some source (input) 4In many texts, functions of s are shown with an italic s to avoid confusion with phasors, even though s is a complex number and most complex numbers are shown in bold. When s appears in equations with other complex, bolded variables, s should be treated as a complex number and not as a scalar. The zero and pole variables, which are also complex numbers, are shown as italic for consistency. 5When the function is unitless— that is, when it is the ratio of voltages, the ratio of currents, and so on— it is called the transfer ratio. 6This equation is often shown with positive signs. In this case, the zeros are the —z values. The minus signs are shown here to clarify how the s and 2 values would be combined in the 5domain (see Sec. 32.3).
PPO
»
w w w . p p i 2 p a s s . c o m
The impedance consists of a resistor in series with the parallel combination of a capacitor and an inductor.
Z(s) — 15 +
40 s
■2.67s
FREQUENCY
After manipulating mathematically, the result is
Tnet(s) = A s2 + 15 ft (s + 3)(s + 5)
= A
(/ViZai)(/V2ZQ2) •••{Nnla n) (D10 1)(D20 2)   ' ( D M N i N 2    N n\
f ai + \. The phase angle variation with fre quency is shown in Fig. 32.2(c) from (3 = —arctano;/0 or directly from 1/j, which is —j or 1Z—90°.
PPI
9
w w w . p p g 2 p a s s . c o m
32~@
P O W E R
R E F E R E N C E
M A N U A L
The frequency is allowed to vary, as illustrated in Fig. 32.3(a).12 The magnitude is plotted in Fig. 32.3(b) from T(ju)\ = yjuj1 + o\. The phase angle variation with frequency is shown in Fig. 32.3(c) from a = arctan co/an.
Figure 32.2 Single Pole at Origin yea (rad/s)
a)! < (jl)2 < o)3 < a)4
004 o>3
Figure 32.3 Single Zero on Negative Real Axis (~crn)
(02 (D1 /\
or (Np/s)
(a) pole at origin, frequency increasing
m
G)j < 0)2< W3 < CO4
1
(a) zero on negative real axis, frequency increasing
1
Circuit Theory
Kl 1 Kl (*)“!
w2 w3 w4
to (rad/s)
(b) transfer function m agnitude, frequency increasing
(3 (degrees)
0° a) (rad/s) 45°
90°
(b) transfer function m agnitude, frequency increasing
W ] co2 ot>3 0)4
(c) transfer function angle, frequency increasing
Consider a transfer function with a single zero but located on the negative real axis at —