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Attacking Trigonometry Problems
4.11 - 1251 ratings - Source
This volume offers a concise, highly focused review for high school and beginning college undergraduates. Rigorously tested examples and coherent, to-the-point explanations are presented in an accessible form. 2015 edition.Remember that when we are in Quadrant III, the tangent of an angle is positive,
but the sine and cosine are negative (All Students Try Candy). We can use the ...
58 7 . Ic=aˆ' Now we can find sin 3 Ic=aˆ' and sec 58 Example 10: If sec 95 Ic = and
sin qalt;0, find cot q. How do we figure ... Now leta#39;s look at some of the typical word
problems that you will be expected to solve using Trigonometry. Now we can find
anbsp;...
Title
:
Attacking Trigonometry Problems
Author
:
David S. Kahn
Publisher
:
Courier Dover Publications - 2015-04-15
ISBN-13
:
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You wear clothes every day, but are you aware of how much math is involved in creating the outfits you put on? How Fashion Designers Use Math colorfully illustrates how designers use math to measure, create, and produce their fashions. more...
The highly-acclaimed MEI series of text books, supporting OCR's MEI Structured Mathematics specification has been updated to match the requirements of the new specifications, for first teaching in 2004. more... | 677.169 | 1 |
Head First Algebra
4.11 - 1251 ratings - Source
Using the latest research in cognitive science and learning theory to craft a multi-sensory learning experience, the book uses a visually rich format designed for the way your brain works, not a text-heavy approach that puts you to sleep.--Publisher's note.Improper fractions An improper fraction is one where the numerator is larger than
the denominator. ... Divide to make an improper fraction proper To convert an
improper fraction to a proper fraction, you need to remember that the line in theanbsp;...
Title
:
Head First Algebra
Author
:
Tracey Pilone, Dan Pilone
Publisher
:
"O'Reilly Media, Inc." - 2009-01
ISBN-13
:
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Lesson 4-12 Polynomial End Behavior(Fri. Night 2/14)
Rating:
Description:
Objective:
TEK 2A.10.A Use quotients of polynomials to describe the graphs of rational functions, predict the effects of parameter changes, describe limitations on the domains and ranges, and examine asymptotic behavior. | 677.169 | 1 |
Pre-Algebra Documents
Showing 1 to 18 of 18
Lesson 1.04: Order of Operations
The rules for the order of operations state that operations in an expression are completed in the
following order: PEMDAS
State what each letter stands for:
P _
E_
M_
D_
A_
S_
Read each CAUTION note and fill-in-the-blank:
MVE 1.10
Going to the Grocery Store
Plan your trip. Spend some time with the adult who will be going to the store with you. Decide
which grocery store you will be going to and when you will go.
Create your shopping list and find an estimate of how much Note Taking Guide
7.08: Advanced: Integers, Opposites, and Absolute Value
Use this guide to take notes as you read each lesson in module 1. On the left side of your twocolumn notes, copy all vocabulary and other information you need. On the right sLesson 1.03: Positive and Negative Exponents
Complete each rule and copy an example of each:
1) Any number raised to the first power equals that _.
EXAMPLE:
2) Any number (except zero) raised to the zero power equals _.
EXAMPLE:
3) Any number a (except zeStudy Guide for the Semester 2 Exam- PreCalculus
Start Preparing Early:
Go back and review your notes from Modules 6 - 12.
Go back and review your old exams. Rework problems that you missed and make sure
you understand the mistake you made.
Each chapter i | 677.169 | 1 |
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Creating Rich Visualizations to Explore Your Data
Brett Champion
Harness the power of Mathematica to interactively visualize your data. This Wolfram Mathematica Virtual Conference 2011 course features a series of examples that show how to create a rich interface for exploring data in depth.
Channels: Virtual Events
Roger Germundsson, director of research and development, gives an introduction and overview of geometric computation in Mathematica 10. Types of regions, their computable properties, and the integration with solvers in Mathematica ... | 677.169 | 1 |
Numerical Trigonometry
In The Man Without Qualities, Robert Musil defines a soul as "that which runs away and hides whenever someone mentions algebra". Lecture notes covering basic algebra and trigonometry. My feeling is that if you want to use this book but do not know calculus you should go back and take calculus. This course begins with a review of fundamentals of calculus, and includes infinite series, power series, paths, and differential equations of first and second order. These sorts of questions pop up all over in geosciences - from plate tectonics to maps to ocean waves, and they require you to find either an angle or a distance.
College Algebra and Trigonometry with MathXL (12-month access) (2nd Edition)
Master Math: Pre-Calculus and Geometry
Trigonometry--Slide Rule-- Mechanics
Power Lecture with ExamView Precalculus (CD-ROM), 6th Edition
Prog Math-2e Bk 15 Trigonometry
Anyone who listens to the radio, watches television, and reads books, newspapers, and magazines cannot help but be aware of statistics, which is the science of collecting, analyzing, presenting and interpreting data , source: Test Item File. Algebra and Trigonometry. A graphing approach and Precalculus. A graphing approach. Larson/Hostetler/Edwards. 3rd Edition. 2001 Edition Test Item File. Algebra and. Topics include number theory, Peano arithmetic, Turing compatibility, unsolvable problems, Gödel's incompleteness theorem, undecidable theories (e.g., the theory of groups), quantifier elimination, and decidable theories (e.g., the theory of algebraically closed fields) pdf. Trigonometric Identities: Learn about these equations that are true for all of the possible values of the variables , e.g. By Ron Larson - Algebra and download epub By Ron Larson - Algebra and. It even has a pretty handy SSA Ambiguous Clause function, where you simply enter the 3 given sides, and it tells you all of the solutions of both triangles (if there are two solutions) Tabulae Trigonometricae Ac download for free Peuerbach's work helped to pave the way for the Copernican conception of the world system; he created a new theory of the planets, made better calculations for eclipses and movements of the planets and introduced the use of the sine into his trigonometry. His early work, Tabulae Eclipsium circulated in manuscript was not published until 1514, contained tables of his eclipse calculations that were based on the Alfonsine Tables , e.g. Trigonometry Trigonometry Tabulae Trigonometricae Ac Logarithmicae download online. Sestill Director of Programming, The Learning Channel ." A math tutorial that can truly replace fear with motivation and understanding with a very special & enjoyable approach." Unlike Vedic mathematics, their works included both astronomical and mathematical contributions epub. Adrian Mole noted that one of the benefits of joining his local Good Samaritans group is that he gets to miss maths on Mondays Trigonometric Delights read for free.
The haversine is also better for small angles, when the cosine has precision difficulties. The haversine appears to have fallen into obscurity; the HP-48 does not recognize it , cited: Functions and Change: A download online Solve system of equations ti89, SQUARE ROOTS OF FRACTIONS, plotting decimals on number lines worksheets, solution to non linear differential equations, free fourth grade division worksheets, prentice hall geometry book 1998 answers Trigonometry: A Unit Circle read for free Trigonometry is relevant in many forms of everyday living and careers including architecture By PRENTICE HALL - Algebra and download pdf By PRENTICE HALL - Algebra and. For example, for 2¼ type 2 1/4. [/frame]Exponents – Type the base before the ^ symbol and the exponent in parenthesis. Remember that the exponent tells how many times the base is multiplied by itself. [/frame]Subscripts – Your variable goes outside the bracket and the subscript goes inside Trigonometry and Algebra, Custom Publication
A system of geometry and trigonometry : with a treatise on surveying; in which the principles of rectangular surveying, without plotting, are explained
College Outline for Trigonometry (Harcourt Brace Jovanovich College Outline Series)
Euclid's Elements of Geometry: From the Latin Translation of Commandine. to Which Is Added, a Treatise of the Nature and Arithmetic of Logarithms ; ... Elements of Plane and Spherical Trigonometry
Suppose , e.g. A Treatise on Higher read for free A Treatise on Higher Trigonometry.. It is perfect for the mathematical scientist who did not study physics but wants an overview. Revolutions in Twentieth Century Physics source: Spherical TrigonometryAfter download here Spherical TrigonometryAfter The Cesaro. Functions include Trigonometric, Hyperbolic, Logarithmic - Base 10, Base 2, Natural - Random Number and many more. Also includes expression editing and custom constants. Interactive College Algebra course designed to ensure engaging, self-paced, and self-controlled e-learning process and help students to excel in their classes. Java- and web-based math course includes theoretical concepts, hands-on examples featuring animated graphics and live formulas, problem-solving lessons, and customizable real time tests with solutions and evaluations download. Princeton. 1995. 0691036632 This is a reprint with corrections of an earlier work published by another publisher. An interesting book that seems in the spirit of the first book by Acton (above) is: This is a great book for projects and for reading. I would like to know however how it has done as a text. A book by a great applied mathematician that is worth having is: Hamming, R online. The first observation which we can date ex actly was made by Ptolemy on 26 March 127 while the last was made on 2 February 141. There is no evidence that Ptolemy was anywhere other than Alexandria. Heath says "it is evident that no part of the trigonometry, or of the matter preliminary to it, in Ptolemy was new Plane Trigonometry with Practical Applications read pdf.
Solutions Manual for College Algebra and Trigonometry (Basics Through Precalculus, Temple University)
Student's Solutions Manual to accompany College Algebra
Algebra & Trigonometry Custom Edition for Baton Rouge Community College
College Algebra and Trigonometry 1
Modern trigonometry
College Algebra
ALEKS Prep Access Card 6 Weeks for College Algebra
Introduction To Trigonometry & Analytic Geometr
When you place an order to purchase any downloadable Products or Services, we will send you a confirmatory email that will contain details of what you have ordered, any delivery charges, as well as details regarding how you can download them Prealgebra, Algebra, and Trigonometry (Custom Edition for University of Rhode Island) Dartmouth College, Bachelors, Double Major in Music and Theater the information described below to the designated agent listed below A treatise on plane and spherical trigonometry The radius of the inscribed circle is r = √[(s - a)(s - b)(s - c)/s] in terms of the sides , source: Calculus and analytic geometry download for free projectsforpreschoolers.com. However, I don't check for all user errors, such as entering a degree larger than 90. This program displays a menu at the bottom of the screen, with a prompt directly above the menu options online. Radial coordinate systems are the spherical and the cylindrical systems, explained shortly in more detail. You can import radial coordinate conversion functions by using the :radial tag: All angles are in radians. Cartesian coordinates are the usual rectangular (x, y, z)-coordinates. Spherical coordinates, (rho, theta, pi), are three-dimensional coordinates which define a point in three-dimensional space , cited: A Treatise On Higher Trigonometry projectsforpreschoolers.com. Now imagine pulling that same spring apart, the force that wants to bring the spring back to its original state is called tension. Randy Akiona working with McKinley students. All structures have to take into consideration these forces of compression and tension. Students volunteered to construct the bridge and after they were done, some of them where able to cross it Elements of Geometry and Conic download pdf It always stores the solution to L1 and if there is a second solution, it stores it to L2. When drawing the triangle, it adjusts the viewing rectangle to avoid distortion and shows which angle is A, B, or C (except when solving from vertices) This is Final Version if there are any bugs email doboy0@gmail.com with program and problem This program comes with no warranty whatsoever For instructions how to use visit youtube.com/doboy What does it do , e.g. Logarithmic and Trigonometric Tables To Five Places (College Outline Series) You have to be careful here as a discussion of fractals could branch out in so many ways that it could generate an entire course on its own Note Taking Guide for Larson's Algebra & Trigonometry, 9th How you solve quadratic equation by graphing, how to solve polynomials, algabra solver, Complex Partial Fraction Expansion TI-89, integer practice worksheets. Mathematics investigatory games, 9-6 Paul A Plane and spherical trigonometry Volume 1 Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. Such behavior occurs throughout nature and led to the discovery of rapidly rotating stars called pulsars in 1967. x Continue your study of the graphs of trigonometric functions by looking at the curves made by tangent, cosecant, secant, and cotangent expressions , e.g. College Trigonometry 5th ed download here. In 'elementary' right triangle trigonometry, we teach the children, er... students that trigonometry is a type of 'indirect measurement.' So you might conduct a measurement like this: A 50-foot high tower is observed having an angle of elevation of 25 degrees from the observer Elements of plane and read epub porady.site11.com. | 677.169 | 1 |
Illustrative Mathematics
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Illustrative Mathematics provides guidance to states, assessment consortia, testing companies, and curriculum developers by illustrating the range and types of mathematical work that students experience in a faithful implementation of the Common Core State Standards, and by publishing other tools that support implementation of the standards. | 677.169 | 1 |
Pre-Calculus: Speedy Study Guide
Pre-Calculus is the academic step before the incorporation of Calculus principles applied to real and imagined universal problems. A student entering a Pre-Calculus environment should be familiar with mathematics as a language and skill useful in both defined and imagined spaces. Pre-Calculus introduces mathematicians to a compendium of symbols and universal concepts including line tangentials, maximums and minimums, logarithmic line equations, the Karp Reduction, norms and lengths, symmetric differences and the occurrence of projected spaces. Pre-Calculus courses dedicate considerable amounts of time and resources to help students become familiar with the language and standard nomenclature of advanced Calculus and its | 677.169 | 1 |
Modern Syllabus Algebra
4.11 - 1251 ratings - Source mathematical concepts such as sets, relations and functions, Boolean algebra, groups, and number systems. It also illustrates linear equations, matrices, and vector spaces and then demonstrates how to solve complex numbers and combine probabilities. Mathematics teachers will find this text a suitable and convenient way of bringing themselves up to date in what is now being taught in schools.Parallel circuit Series circuit switch switch lamp switch switch lamp p d p d O 0 O
0 0 O O 1 1 0 1 0 1 0 1 1 O 0 1 1 l l 1 1 ... For instance, suppose we wish to have a
circuit which will operate when either switch p or switch q is closed but notanbsp;...
Title
:
Modern Syllabus Algebra
Author
:
D.G.H.B. Lloyd
Publisher
:
Elsevier - 2014-05-17
ISBN-13
:
Continue
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Mathematics is Story - Taschenbuch
[EAN: 9783639143225], Neubuch, [PU: Vdm Verlag Apr 2009], Education|General, This item is printed on demand - Print on Demand Titel. - 212 pp. Englisch
Gladys Sterenberg:
Mathematics Buch / Broschur, [PU: VDM Verlag Dr. Müller, Saarbrücken]
2009. 212 S. 298 Gr.Einband kartoniert. ISBN 3639143221 . Neupreis 79,00. 2009. , fast neu, ungebraucht, ungelesen. Sofortversand mit Rechnung, keine Vorauszahlung (D,A,CH)!deas through existing views of the nature of mathematics: in this... Versand D: 4,20 EUR | 677.169 | 1 |
302A with Professor Weinbergrientation to Math 302A 302B Fall 2008 Math 302A and Math 302B are designed to help prepare future elementary school teachers to teach mathematics The topics covered parallel at a more sophisticated level the mathematics curriculum in grades K 8 Some ofthese topics are Math 302A problem solving numeration whole number arithmetic meanings of operations fractions decimals percents divisors and multiples Math 302B Algebraic reasoning indirect measurement circles and angles transformational geometry perimeter area surface area volume organization of data probability The two courses model many of the recent recommendations for reform in mathematics education presented by the National Council of Teachers of Mathematics and other professional organizations Thus these courses emphasize problemsolving use of technology calculators and computers and manipulative materials cooperative learning oral and written communication of mathematical ideas and solutions and the connections between mathematics and its uses in everyday life Required Materials Bassarear T 2008 Mathematics for elementary school teachers 43911 Ed Boston Houghton Mif in Bassarear T 2008 Mathematics for elementary school teachers Explorations 43911 Ed Boston Houghton Mif in Math 302A Class Notes Calculator any kind pencils at least two colors of pen marker highlighter etc manipulative pack Activity Based Learning Both 302A and 302B are activity based courses This means that you will be given frequent opportunities to experience on a concrete level the ideas addressed in class and in the book During an activity you can feel a concrete manipulative material move it and actually use it to solve a problem or model a concept You will learn how the senses of touch and sight can be used to understand mathematics and solve problems It is very important that you do use the materials as directed by the instructor and get in the spirit of the activity The main objective is not to arrive at a solution but to re ect on the process that led you to that solution These activities also will familiarize you with some of the materials Cuisenaire rods fraction bars geoboards tangrams etc that are currently in use in elementary schools as aids to learning and teaching mathematics The class will be organized to provide a friendly intimate noncompetitive atmosphere in which you can feel free to explore discuss and learn mathematics in a small group situation We try to insist that you do work in groups not separately We want you to experience working together in a small group situation so hopefully you will later feel comfortable using this technique in your own classes The role of the instructor will be quite different from what you might have encountered in other math courses In 302MB the instructor acts primarily as a facilitator for your own exploration and learning Frequently a given question will have more than one possible answer a problem may have no ght method of solution One answer or method might be better than another We prefer that you discover this for yourself rather than just have us tell you We believe that your own learning will be more powerful and longlasting if it comes from you Homework Homework is very important in this course The assignments are involved and we encourage you to begin working on them as soon as they are assigned Get together with some of your classmates and work on the problems However you should each turn in your own personal writeup We are interested in process in how you explain your work We do not want just answers on a page In general for most problems you will have to show and explain your work Who is your audience when you are writing the homework Well in reality it will be the instructor or a grader But for the purposes of what we are looking for think of it as if you were writing for one of your peers that is an adult learner of mathematics Hence when we say that we want you to explain your work we mean for that audience not for elementary school children In general except for some computational problems you will need to explain your work Your explanations may include diagrams charts an English paragraph or any other format that conveys what you did and how you did it Help Sessions Besides your instructor s office hours you are encouraged to make use of the Math 302 Help Sessions These will be held in the Math 302MB classrooms or in the instructors offices There will be several help sessions every week to be announced at the beginning of the semester These hours are excellent occasions to get together with classmates to work on your homework and talk math During these Help Sessions a Math 302 instructor will be available for questions and guidance Attendance Class participation is very important Thus attendance and punctuality will affect your final grade More than two absences or excessive tardiness coming late or leaving early will cause your grade to be lowered or cause you to be dropped from the course Attendance will be taken at each class session Withdrawal from course A student may withdraw from the course without a grade class will not appear on transcript through Tuesday February 6 2007 A student may withdraw from the course with a grade of quotWquot if passing through Tuesday March 6 2007 Grading What follows is the grading scheme for all sections of Math 302MB It includes the different factors that will play a role in your grade InClass Exams 40 Final Exam 25 Homework 10 Quizzes Proiects Explorations Other Class Notes participation etc M 100 Scale for Final Grade A 90 100 B 80 89 C 70 79 D 60 69 E 59 and below Students with Disabilities If you anticipate the need for reasonable accommodations to meet the requirements of this course you must register with the Disability Resource Center and request that the DRC send your instructor of cial noti cation of your accommodation needs as soon as possible Please plan to meet with your instructor to discuss accommodations and how the course requirements and activities may affect your ability to fully participate Course Coordinators Suzanne Weinberg is the coordinator for Math 302A Of ce Math 511 Phone 6261699 Email sweinbergmatharizonaedu Jose Maria Menendez Gomez is the coordinator for Math 302B Of ce Math East 244 Phone 6261522 Email jmenendezmatharizonaedu Concerns about the course grades dropping the course etc should be directed to your instructor rst and then to the coordinator | 677.169 | 1 |
"Curso rapido de calculo diferencial e integral" is a program that may help college and high school students understand and learn the basic concepts of differential and integral calculus. So how to solve some useful exercises using educational videos | 677.169 | 1 |
Master the College Mathematics CLEP Test
4.11 - 1251 ratings - Source
College Mathematics, part of Peterson's Master the CLEP, is meant to fully prepare you for the CLEP College Mathematics general examination. It offers an extensive overview of the following subject areas: real number systems, sets, probability and statistics, logic, functions and their graphs, and additional topics from algebra and geometry. You will find numerous practice question sets throughout the subject review. In addition, there are a pre- and a post-test, with 50 questions each, that will help you in your preparation for this exam. All practice questions come with detailed answer explanations.Each overview will provide you with a helpful guide to the material that is
included on each exam. ... Educational Psychology, Introduction to Psychology,
Introduction to Sociology, Principles of Macroeconomics, Principles of
Microeconomics, anbsp;...
Title
:
Master the College Mathematics CLEP Test
Author
:
Peterson's
Publisher
:
Peterson's - 2012-04-30
ISBN-13
:
Continue
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A BC NOTE
Students learn in a number of ways and in a variety of settings. They learn through lectures, in informal study groups, or alone at their desks or in front of a computer terminal. Wherever the location, students learn most efficiently by solving problems, with frequent feedback from an instructor, following a worked-out problem as a model. Worked-out problems have a number of positive aspects. They can capture the essence of a key concept -often better than paragraphs of explanation. They provide methods for acquiring new knowledge and for evaluating its use. They provide a taste of real-life issues and demonstrate techniques for solving real problems. Most important, they encourage h i v e participation in learning. We created the BookWare Companion Series because we saw an unfulfilled need for computer-baaed learning tools that address the computational aspects of problem solving across the curriculum. The BC series concept was also shaped by other forces: a general agreement among instructors that students learn best when they are actively involved in their own learning, and the realization that textbooks have not kept up with or matched student learning needs. Educators and publishers are just beginning to understand that the amount of material crammed into most textbooks cannot be absorbed, let alone the knowledge to be mastered in four years of undergraduate study. Rather than attempting to teach students all the latest knowledge, colleges and universities are now striving to teach them to reason: to understand the relationships and connections between new information and existing knowledge; and to cultivate problem-solving skills, intuition, and critical thinking. The BookWare Companion Series was developed in response to this changing mission. Specifically, the BookWare Companion Series was designed for educators who wish to integrate their curriculum with computer-based learning tools, and for students who find their current textbooks overwhelming. The former will find in the BookWare Companion Series the means by which to use powerful software tools to support their course activities, without having to customize the applications themselves. The latter will find relevant problems and examples quickly and easily and have instant electronic access to them.
We hope that the BC series will become a clearinghouse for the exchange of reliable teaching ideas and a baseline series for incorporating learning advances from emerging technologies. For example, we intend to reuse the kernel of each BC volume and add electronic scripts from other software programs as desired by customers. We are pursuing the addition of AI/Expert System technology to provide an intelligent tutoring capalo bility for future iterations of BC volumes. We a s anticipate a paperless environment in which BC content can flow freely over high-speed networks to support remote learning activities. In order for these and other goals to be realized, educators, students, software developers, network administrators, and publishers will need to communicate freely and actively with each other. We encourage you to participate in these exciting developments and become involved in the BC Series today. If you have an idea for improving the effectiveness of the BC concept, an example problem, a demonstration using software or multimedia, or an opportunity to explore, contact us. Thank you one and all for your continuing support. The PWS Electrical Engineering Team: BillBarter@PWS.Com AngiehllinkoQPWS.Com Nathan_WilburQPWS.Com PamRockwell@PWS.Com MonicaBlock@PWS.Com Acquisitions Editor Assistant Editor Marketing Manager Production Editor Editorial Assistant
20 P a r k Plaza, B o s t o n , MA 0 2 1 1 6 - 4 3 2 4
Copyright @ 1997 by PWS Publishing Company, a division of International Thomson Publishing Inc. A11 rights reserved. No part of this book may he reproduced, stored in a retrieval system, or transcribed in any form or by any means -electronic, mechanical, photocopying, recording, or otherwise -without the prior written permission of PWS Publishing Company. MATLAB and PC MATLAB are registered trademarks of The Mathworks, Inc. The Mathworks, Inc. is the developer o MATLAB, the high-performance computational software introduced in this book. For further f information on MATLAB and other Mathworks products- including SIMULINKTMand MATLAB Application Toolboxes for math and analysis, control system design, system identification, and other dwiplinescontact The Mathworks at 24 Prime Park Way, Natick, MA 01760 (phone: 508-653-1415; fax: 506-653-2997; email: info@mathworks.com). can also sign up to receive the Mathworks quarterly newsletter and register You for the user group. Macintosh is a trademark of Apple Computer, Inc. MS-DOS is a trademark of Microsoft Corporation. Bookware Companion Series is a trademark of PWS Publishing Company.
Rom the beginning of the last decade we have witnessed a revolution in computer technology and an explosion in user-friendly applications. This revolution is still continuing today with low-cost personal computer systems that rival the performance of expensive workstations. This technological prowess should be brought to bear on the educational process and, in particular, on effective teaching that can result in enhanced learning. This companion book on digital signal processing (DSP) makes a small contribution toward that goal. The teaching methods in signal processing have changed over the years from the simple Yecture-only" format to a more integrated "lecturelaboratory" environment in which practical hands-on issues are taught using DSP hardware. However, for effective teaching of DSP the lecture component must also make extensive use of computer-based explanations, software examples, and exercises. For the last several years, the MATLAB developed by The Mathworks, Inc. has established itself as the de fact0 standard for numerical computation in the signal-processing community and as a platform of choice for algorithm development. There are several reasons for this development, but one most important reason is that MATLABis available on practically all computing platforms. For several years the expensive Professional Version of MATLABwas the only version available on the market. The advent of an inexpensive Student Edition has now made it possible to use it in classrooms. Recently, several textbooks in DSP have appeared which generally provide exercises that can However, for students (and for practicing engibe done using MATLAB. neers interested in DSP) there are no "how-to" references for effective use of MATLABin DSP. In this book we have made an attempt at intewith traditional topics in DSP so that it can be used grating MATLAB to explore difficult topics and solve problems to gain insight. Many problems or design algorithms in DSP require considerable computation. It is provides a convenient tool so that many scenarfor these that MATLAB ios can be tried with ease. Such an approach can enhance the learning process.
ix
SCOPE OF THE
BOOK
I
I
This book is primarily intended for use as a supplement in junior- or senior-level undergraduate courses on DSP. We assume that the student Those topics are (or user) is familiar with the fundamentals of MATLAB. not covered since several tutorial books and manuals on MATLAB are available. Similarly,this book is not written as a textbook in DSP because of the availability of excellent textbooks. What we have tried to do is to provide enough depth to the material augmented by MATLAB functions and examples so that the presentation is consistent,logical, and enjoyable. Therefore this book can also be used as a self-study guide by anyone interested in DSP. When this project got under way, version 3.5 of the Student Edition of was MATLAB available. Since the beginning of 1995 a more advanced GUI (graphical user interface) version 4.0 of the Student Edition is available. This book is compatible with the newer version.
ORGANIZATION OF THE BOOK
I
The first eight chapters of this book discuss traditional material covered in an introductory course on DSP. The last two chapters are presented as applications in DSP with emphasis on MATLAB-based projects. The following is a list of chapters and a brief description of their contents: Chapter 1, Introduction:This chapter introducesreaders to the discipline of signal processing and discusses the advantages of DSP over analog is signal processing. A brief introduction to MATLAB also provided. Chapter 2, Discrete-time Signals and Systems: This chapter provides a brief review of discrete-time signals and systems in the time domain. functions is demonstrated. Appropriate use of MATLAB Chapter 3, The Discrete-time Fourier Analysis This chapter discusses discretetime signal and system representation in the frequency domain. Sampling and reconstruction of analog signals are also presented. hs Chapter 4, The z-Ipmnsfonn. T i chapter provides signal and system description in the complex frequency domain. MATLAB techniques are introduced to analyze z-transforms and to compute inverse z-transforms. Solutions of difference equations using the z-transform and MATLAB are provided. Chapter 5, The Discrete Fourier hnsform: This chapter is devoted to the computation of the Fourier transform and to its efficient imple mentation. The discrete Fourier series is used to introduce the discrete Fourier transform, and several of its properties are demonstrated using
X
PREFACE
MATLAB. Topics such as fast convolution and fast Fourier transform are thoroughly discussed. Chapter 6, Digital Filter Structures: This chapter discusses several structures for the implementation of digital filters. Several useful MATLAB functions are developed for the determination and implementation of these structures. Lattice and ladder filters are also introduced and discussed. Chapter 7, FIR Filter Design: This chapter and the next introduce the important topic of digital filter design. Three important design techniques for FIR filters-namely, window design, frequency sampling design, and the equiripple filter design-are discussed. Several design examples are provided using MATLAB. Chapter 8, IIR Falter Design: Included in this chapter are techniques in IIR filter design. It begins with analog filter design and introduces such topics as filter transformations and filter-band transformation. Once again several design examples using MATLAB provided. are Chapter 9, Applications in Adaptive Filtering: This chapter is the first of two chapters on projects using MATLAB. Included is an introduction to the theory and implementation of adaptive FIR filters with projects in system identification, interference suppression, narrowband frequency enhancement, and adaptive equalization. Chapter 10, Applications in Communications: This chapter focuses on several projects dealing with waveform representation and coding, and with digital communications.Included is a description of pulse-code modulation (PCM), differential PCM (DPCM) and adaptive DPCM (ADPCM), delta modulation (DM) and adaptive DM (ADM), linear predictive coding (LPC), generation and detection of dual-tone multifrequency (DTMF) signals, and a description of signal detection applications in binary communicationsand spread-spectrum communications.
SOFTWARE
I I
The book is an outgrowth of our teaching of a MATLAB-based undergraduate DSP course over several years. Many MATLAB functions discussed in this book were developed in this course. These functions are available on the accompanying diskette as a pwskdsp toolbox. Create a separate directory for this toolbox and reference it in the matlabpath environment. The book also contains numerous MATLAB scripts in many examples. These scripts are also made available on the disk and are kept in individual directories created for each chapter. In addition, many figures were plots, and their scripts are available in the figures produced as MATLAB directory. Students should study these scripts to gain insight into the MATLAB procedures. We will appreciate any comments, corrections, or compact coding of these programs and scripts. Solutions to problems and
Software
xi
the associated script files will be made available to instructors in the near future. Further information about MATLAB and related publications may be obtained from The Mathworks, Inc. 24 Prime Park Way Natick, MA 01760-1500 Phone: (508) 647-7000 Fax: (508) 647-7001 Email: infoOmathworks.com WWW:
ACKNOWLEDGMENTS
We are indebted to our numerous students in our ECE1456 course at Northeastern University who provided us a forum to test teaching ideas using MATLAB and who endured our constant emphasis on MATLAB. Some efficient MATLAB functions are due to these students. We are also indebted to our reviewers, whose constructive criticism resulted in a better presentation of the material: Abeer A. H. Alwan, University of California, Los Angeles; Steven Chin, Catholic University; and Joel Trussel, North Carolina State University. We would like to thank Tom Fbbbins, former editor at PWS Pub lishing Company, for his initiative in creating the BookWare Companion Series and for his enthusiastic support of MATLAB classroom teaching, in especially in DSP. Thanks are also due to present editor Bill Barter for his support throughout the project. Finally, we would like to thank the staff at PWS Publishing Company for the final preparation of the manuscript. Vznay K. Ingle John G. Pnmkis Boston, Massachusetts
xii
PREFACE
INTRODUCTION
Over the past several decades the field of digital signal processing (DSP) has grown to be important both theoretically and technologically. A major reason for its success in industry is due to the development and use of lowcost software and hardware. New technologies and applications in various fields are now poised to take advantage of DSP algorithms. This will lead to a greater demand for electrical engineers with background in DSP. Therefore it is necessary to make DSP an integral part of any electrical engineering curriculum. a Not long ago an introductory course on DSP w s given mainly at the graduate level. It was supplemented by computer exercises on filter design, spectrum estimation, and related topics using mainframe (or mini) computers. However, considerable advances in personal computers and software over the past decade made it possible to introduce a DSP course to undergraduates. Since DSP applications are primarily algorithms that are implemented either on a DSP processor [ll]or in software, a fair amount of programming is required. Using interactive software, such as MATLAB, is now possible to place more emphasis on learning new and it difficultconcepts than on programming algorithms. Interesting practical examples can be discussed, and useful problems can be explored. With this philosophy in mind, we have developed this book as a companion book (to traditional textbooks like [16, 191) in which MATLAB is an integral part in the discussion of topics and concepts. We have chosen MATLABas the programming tool primarily because of its wide availability on computing platforms in many universities across the country. hthermore, a student edition of MATLABhas been available for several years, placing it among the least expensive software for educational puras poses. We have treated MATLAB a computational and programming toolbox containing several tools (sort of a super calculator with several keys) that can be used to exp!ore and solve problems and, thereby, enhance the learning process. T i book is written at an introductory level in order to introduce hs undergraduate students to an exciting and practical field of DSP. We emphasize that this is not a textbook in the traditional sense but a
1
companion book in which more attention is given to problem solving and hands-on experience with MATLAB. Similarly, it is not a tutorial book We and in MATLAB. assume that the student is familiar with MATLAB is currently taking a course in DSP. The book provides basic analytical tools needed to process real-world signals (a.k.a. analog signals) using digital techniques. We deal mostly with discretetime signals and systems, which are analyzed in both the time and the frequency domains. The analysis and design of processing structures called filters and spectrum analyzers is one of the most important aspects of DSP and is treated in great detail in this book. Many advanced topics in DSP (which are generally covered in a graduate course) are not treated in this book, but it is hoped that the experience gained in this book will allow students to tackle advanced topics with greater ease and understanding. In this chapter we provide a brief overview of both DSP and MATLAB.
OVERVIEW OF DIGITAL SIGNAL PROCESSING
I
I
In this modern world we are surrounded by all kinds of signals in various forms. Some of the signals are natural, but most of the signals are manmade. Some signals are necessary (speech), some are pleasant (music), while many are unwanted or unnecessary in a given situation. In an engineering context, signals are carriers of information, both useful and unwanted. Therefore extracting or enhancing the useful information from a mix of conflicting information is a simplest form of signal processing. More generally, signal processing is an operation designed for extracting, enhancing, storing, and transmitting useful information. The distinction between useful and unwanted information is often subjective as well as objective. Hence signal processing tends to be application dependent.
HOW ARE SIGNALS PROCESSED?
The signals that we encounter in practice are mostly analog signals. These signals, which vary continuously in time and amplitude, me processed using electrical networks containing active and passive circuit elements. This approach is known as analog signal processing (ASP)-for example, radio and television receivers.
Analog signal: q,(t)
---)I
Analog signal processor
I
4 ya(t)
:Analog signal
They can also be processed using digital hardware containing adders, multipliers, and logic elements or using special-purpose microprocessors. However, one needs to convert analog signals into a form suitable for digital hardware. This form of the signal is called a digital signal. It takes
2
Chapter 1
INTRODUCTION
one of the finite number of values at specific instances in time, and hence it can be represented by binary numbers, or bits. The processing of digital signals is called DSP; in block diagram form it is represented by
I.
II
Eguivalent Analog S
i Processor
Discrete System where the various block elements are discussed below.
I
I
PrF: This is a prefilter or an antialiasing filter, which conditions the analog signal to prevent aliasing. ADC: This is called an analog-to-digital converter, which produces a stream of binary numbers from analog signals. Digital signal processor: This is the heart of DSP and can represent a general-purpose computer or a special-purpose processor, or digital hardware, and so on. DAC: This is the inverse operation to the ADC, called a digital-to-analog converter, which produces a staircase waveform from a sequence of binary numbers, a first step towards producing an analog signal. PoF: This is a postfilter to smooth out staircase waveform into the desired analog signal
It appears from the above two approaches to signal processing, analog and digital, that the DSP approach is the more complicated, containing more components than the "simpler looking" ASP. Therefore one might ask a question: Why process signals digitally? The answer lies in many advantages offered by DSP.
ADVANTAGES OF DSP OVER ASP
A major drawback of ASP is its limited scope for performing complicated signal processing applications. This translates into nonflexibility in processing and complexity in system designs. All of these generally lead to expensive products. On the other hand, using a DSP approach, it is possible to convert an inexpensive personal computer into a powerful signal processor. Some important advantages of DSP are these:
1. Systems using the DSP approach can be developed using software running on a general-purpose csmputer. Therefore DSP is relatively convenient to develop and test, and the software is portable. 2. DSP operations are based solely on additions and multiplications, leading to extremely stable processing capability-for example, stability independent of temperature.
Ovetview of Digital Signal Prtxessing
3
3. DSP operations can easily be modified in real time, often by simple programming changes, or by reloading of registers. 4. DSP has lower cost due to VLSI technology, which reduces costs of memories, gates, microprocessors, and so forth.
The principal disadvantage of DSP is the speed of operations, especially at very high frequencies. Primarily due to the above advantages, DSP is now becoming a first choice in many technologies and applications, such as consumer electronics, communications,wireless telephones, and medical imaging.
TWO IMPORTANT
Most DSP operations can be categorized as being either signal analysis tasks or signal filtering tasks as shown below. Digital signal
CATEGORIES OF DSP
r - - - - - -
I
Analysis
I
L _ _ - _ _ _ 1
Measurements
Digital signal
Signal ancrlysia This task deals with the measurement of signal prop erties. It is generally a frequency-domain operation. Some of its applications are
0 0 0 0
Signal filtering This task is characterized by the "signalin-signal out" situation. The systems that perform this task are generally called filters. It is usually (but not always) a timedomain operation. Some of the a p plications are
0
In some applications, such as voice synthesis, a signal is first analyzed to study its characteristics, which are then used in digital filtering to generate a synthetic voice. In the first half of this book we will deal with the signal-analysis aspect of DSP. In Chapter 2 we will begin with basic descriptions of discretetime signals and systems. These signals and systems are analyzed in the frequency domain in Chapter 3. A generalization of the frequencydomain description, called the t-transform, is introduced in Chapter 4. The practical algorithms for computing the Fourier transform are discussed in Chapter 5 in the form of the discrete Fourier transform and the fast Fourier transform. The second half of this book is devoted to the signal-filteringaspect of DSP. In Chapter 6 we describe various implementations and structures of digital filters. In Chapter 7 we provide design techniques and algorithms for designing one type of digital filter called finitsduration impulse response (or FIR) filters, while in Chapter 8 we provide a similar treatment for another type of filter called infiniteduration impulse response (or IIR) filters. In both chapters we discuss only the simpler but practically useful techniques of filter design. More advanced techniques are not covered. Finally, the last two chapters provide some practical applications in the form of projects that can be done using material learned in the first eight chapters. In Chapter 9 concepts in adaptive filtering are introduced, and simple projects in system identification, interference suppression, adap tive line enhancement, and so forth are discussed. In Chapter 10 a brief introduction to digital communicationsis presented with projects in such s topics a PCM, DPCM, and LPC being outlined. In all these chapters the central theme is the generous use and adtools. Most of the existing MATLAB equate demonstration of MATLAB functions for DSP are described in detail, and their correct use is demonfunctions are strated in many examples. Furthermore, many new MATLAB developed to provide insights into the working of many algorithms. We believe that this "hand-holding" approach will enable students to dispel fears about DSP and will provide an enriching learning experience.
A FEW WORDS ABOUT MATLAB@
I
MATLAB an interactive, matrix-based system for scientific and engiis neering numeric computation and visualization. Its strength lies in the fact that complex numerical problems can be solved easily and in a fraction of the time required with a programming language such as Fortran or C. It is also powerful in the sense that by using its relatively simple programming capability, MATLABcan be easily extended to create new commands and functions.
A Few Words about MATLAB
5
MATLABis available on a number of computing environments: Sun/HP/VAXstation workstations, 80x86 PCs, Apple Macintosh, VAX, and several parallel machines. The basic MATLABprogram is further enhanced by the availability of numerous toolboxes (acollection of specialized functions in a specific topic) over the years. The information in this book generally applies to all these environments. The development of this book was begun under the professional version 3.5 running under DOS. A relatively inexpensive Student Edition containing limited toolboxes and functions was also available from Prentice Hall publishers. routines and other material in Therefore we decided to make MATLAB this book compatible with the Student Edition. However, at present the current major version of MATLABis version 4.2 under graphical user interface (GUI). Also a new Student Edition in GUI is available since February 1995, containing enhanced and new toolboxes. This book is certainly compatible with this edition, and every attempt is made to identify the new functions that are available and that can be used. A new toolbox available in the Student Edition is the Symbolic toolbox, which is based on Maple engine. Since digital signal processing primarily requires numerical computations, the Symbolic toolbox is neither discussed nor used in this book. go The scope and power of MATLAB far beyond the few words given in this section. It is senseless to provide a concise information or tutorial on MATLAB when excellent books and guides are available on this topic. Students should consult the MATLAB User's Guide [Z]and FkferSimilarly, students should attempt the tutorial given in ence Guide [l]. [3]. The information given in all these references, along with the online facility, usually is sufficient for students to use this book.
6
Chapter 1
W
INTRODUCTION
c
DISCRETE-TIME SIGNALS AND
SYSTEMS
We begin with the concepts of signals and systems in discrete time. A number of important types of signals and their operations are introduced. Linear and shift-invariant systems are discussed mostly because they are easier to analyze and implement. The convolutionand the dif€erenceequation representations are given special attention because of their imporThe emphasis in this tance in digital signal processing and in MATLAB. chapter is on the representations and implementation of signals and s y s tems using MATLAB.
DISCRETE-TIME SIGNALS
I
Signals are broadly classified into analog and discrete signals. An analog signal will be denoted by x , ( t ) , in which the variable t can represent any physical quantity, but we will assume that it represents time in seconds. A in discrete signal will be denoted by x (n), which the variable TI is integervalued and represents discrete instances in time. Therefore it is also called a discretetime signal, which is a number sequence and will be denoted by one of the following notations:
.(n) = {.(n)} = {...,.(-1),x(0),.(1);..}
T
where the uparrow indicates the sample at n = 0. In MATLABwe can represent a finite-durntion sequence by a TOW vector of appropriate values. However, such a vector does not have any information about sample position n. Therefore a correct representation
7
of z(n)would require two vectors, one each for z and n. For example, a sequence z(n)= {2,1,-l,O,1,4,3,7} be represented in MATLAB can by
T
>> n-C-3,-2,-1,0,1.2,3,43 ; x=C2,1,-1,0,i,4,3,71;
Generally, we will use the x-vector representation alone when the sample position information is not required or when such information is trivial (e.g. when the sequence begins at n = 0).An arbitrary infinite-duration sequence cannot be represented in MATLABdue t o the finite memory limitations.
TYPES OF SEQUENCES
We use several elementary sequences in digital signal processing for analp i s purposes. Their definitions and MATLABrepresentations are given below.
1 Unit sample sequence: . 6(n)=
1, n = O 0, n # O -
.. . ,o,o, 1,0,0, ...
In MATLAB function zeros(1,N) generates a row vector of N zeros, the which can be used to implement 6(n)over a finite interval. However, the logical relation n==Ois an elegant way of implementing 6(n). For example,
to implement
S(n - no) =
over the nl 5 tion.
{0, n #no
1, n=no
5 n2 interval, we wl use the following MATLAB il func-
"
1 .
n = Cni:nZl; x = [(n-no) == 01;
2. Unit step sequence:
In MATLAB function ones(1 ,N) generates a row vector of N ones. It the can be used to generate u(n) over a finite interval. Once again an elegant
8
Chapter 2
DISCRETE-TIME SIGNALS A N D SYSTEMS
approach is to use the logical relation n>=O.To implement
over the n1 5 no 5 n2 interval, we will use the following MATLABfunction.
where D is called an attenuation and wo is the frequency in radians. A MATLABfunction exp is used to generate exponential sequences. For ex) 10, we will need ample, to generate ~ ( n= exp [(2 j 3 ) n], 0 n script: the following MATLAB
+
< <
>> n
= [O:lO];
L
= exp((2+3j)*n);
5 . Sinusoidal sequence:
~ ( n= cos(won )
+ e), Vn
where 0 is the phase in radians. A MATLABfunction cos (or sin) is used to generate sinusoidal sequences. For example, to generate z(n) = 3cos(O.lxn+a/3)+2sin(0.5an), 0 5 n 5 10, we will need the following MATLAB script:
Discrete-time Signals
9
6. Random sequences: Many practical sequences cannot be described by mathematical expressions like those'above. These sequences are called random (or stochastic) sequences and are characterized by parameters of the associated probability density functions or their statistical moments. In MATLABtwo types of (pseude) random sequences are available. The raud(1.N) generates a length N random sequence whose elements are uniformly distributed between [0,1]. The randn(1,N) generates a length N Gaussian random sequence with mean 0 and variance 1. Other random sequences can be generated using transformations of the above functions. 7. Periodic sequence: A sequence x(n) is periodic if x(n) = x(n+ N), Vn. The smallest integer N that satisfies the above relation is called the fundamental period. We will use Z(n) t o denote a periodic sequence. To generate P periods of i(n)from one period {x(n), 0 5 n 5 N - l}, we can copy x (n)P times:
>> x t i l d e = C X , ~ . . , X I; .
But an elegant approach is to use MATLAB'S powerful indexing capabilities. First we generate a matrix containing P rows of x (n)values. Then we can concatenate P rows into a long row vector using the construct ( :). However, this construct works only on columns. Hence we will have to use the matrix transposition operator ' to provide the same effect on rows.
>> x t i l d e = x' * ones(1.P); >> x t i l d e = x t i l d e ( : ) ; >> x t i l d e = x t i l d e ' ;
% P columns of x; x is a row vector % long column vector Y. long row vector
Note that the last two lines can be combined into one for compact coding. This is shown in Example 2.1.
It is implemented in MATLAB the arithmetic operator "+". However, by the lengths of x1 (n) and 22 (n)must be the same. If sequences are of unequal lengths, or if the sample positions are different for equal-length sequences, then we cannot directly use the operator +. We have to first augment x1 (n) and 22 (n) so that they have the same position vector n (and hence the same length). This requires careful attention to MATLAB'S indexing operations. In particular, logical operation of intersection 'W',
10
Chapter 2 DISCRETE-TIME SIGNALS AND SYSTEMS
relational operations like "<=" and "==", and the find function are required to make 2 1 (n)and 2 2 (n)of equal length. The foIlowing function, called the sigadd function, demonstrates these operations.
function [y.d = sigadd(xl,nl,x2,n2)
Its use is illustrated in Example 2.2. 2. Signal multiplication: This is a sample-by-sample multiplication (or "dot" mukipljcatjon) given by
{21(4). {.2(4)
= {n(n)Q(R.))
It is implemented in MATLABby the array operator ''.*". Once again s the similar restrictions apply for the .* operator a for the + operator. Therefore we have developed the sigmult function, which is similar to
the sigadd function.
function [y,nl = sigmdt(xi,nl,x2,n2) X implements yn () x()x() ln*2n
The plot of the sequence is shown in Figure 2 . h .
'The symbol * denotes many operations in digital signal processing. Its font (roman or computer) and its position (normal or superscript) will distinguish each operation.
will generate z (n). a. z 1 (n)= 2471 - 5 ) - 31: (n 4). The first part is obtained by shifting z (n) by 5 and the second part by shifting z (n) by -4. This shifting and the addition can he easily done using the sigshift and the sigadd functions.
+
>> [xll,nll] = sigshift(r,n,5); [x12,n121 = sigshift(x,n,-4); >> [xl,nl] = sigadd(2*xli,nli,-3*x!Z,n12); >> subplot(2,1,1); stem(n1,xl); title('Sequence in Example 2.2a') >> xlabel('n'); ylabel('xl(n) '1;
The plot of z (n) is shown in Figure 2.2a. 1 b 5 2 (n)= z ( 3 - n) f ~ ( n ) s (- 2). . n (The first term can be written as I ( n - 3 ) ) .Hence it is obtained by first folding 2 (n)and then shifting the result by 3. The second part is a multiplication of z (n)and z (n - 2), both of which have the same length but different support (or sample positions). These operations can be easily done using the sigf old and the sigmult functions.
Then any arbitrary real-valued sequence z(n)can be decomposed into its even and odd components
4 = z,(n)+ z o ( 4 . )
where the even and odd parts are given by
(2.2)
z,(n) = - [ ~ ( n(-.)I) +. 2
1
1 and z,(n)= - [.(TI) - z(-n)] 2
(2.3)
respectively. We will use this decomposition in studying properties of the Fourier transform. Therefore it is a good exercise to develop a simple MATLABfunction to decompose a given sequence into its even and odd components. Using MATLABoperations discussed so far, we can obtain the following evenodd function.
function Cxe, xo, ml = evenodd(x,n) % Real signal decomposition into even and odd parts
The sequence and its support are supplied in x and n arrays, respectively. It first checks if the given sequence is real and determines the support of the even and odd components in m array. It then implements (2.3) with special attention to the MATLABindexing operation. The resulting components are stored in xe and xo arrays.
0
EXAMPLE 2.4
A similar decomposition for complex-valued sequences is explored in Problem 2.5.
The geometric series A one-sided exponential sequence of the form {a", n 2 0}, where a is an arbitrary constant, is called a geometric series. In digital signal processing, the convergence and expression for the sum of this series are used in many applications. The series converges for l1 < 1, while the sum of its components converges to a
Reclangularpulse
Even Part
0.4 0.2
-5
0 n
5
Odd Part
I
-10
-5
0
5
10
n
FIGURE 2.4
Even-odd decomposition in Ezample 2.4
Discretetime Signals
19
We will also need an expression for the sum of any finite number of terms of the series given by
nO =
These two results will be used throughout this book.
Correlations of sequences Correlation is an operation used in many applications in digital signal processing. It is a measure of the degree to which two sequences are similar. Given two real-valued sequences x(n) and y(n) of finite energy, the crosscornlation of x(n) and y(n) is a sequence rg! z() defined as
m
n=--m
The indeq ! called the shift or lag parameter. The special case of (2.6) is when y(n) = z(n) is called autocorrelation and is defined by
n=-m
It provides a measure of self-similarity between different alignments of the functions to compute auto- and crosscorrelations are sequence. MATLAB discussed later in the chapter.
DISCRETE SYSTEMS
I
Mathematically, a discretetime system (or discrete system for short) is described as an operator TI.]that takes a sequence z(n)(called excitation) and transforms it into another sequence y(n) (called response). That is,
In DSP we will say that the system processes an input signal into an output signal. Discrete systems are broadly classified into linear and nonlinear systems. We will deal mostly with linear systems.
LINEAR SYSTEMS
A discrete system T[.] a linear operator L[.]if and only if L[.] is satisfies the principle of superposition, namely,
Chapter 2 DISCRETE-TIME SIGNALS AND SYSTEMS
m
L[al.l(n)
+ azzz(n)l = alL[.l(n)l+
azL[~2(n)l,V~l,az,~l(n),~z(n) (2.8)
Using (2.1) and (2.8), the output y (n)of a linear system to an arbitrary input z (n)is given by
y(n)=L(z(n)l=L[
n=-m
c
00
z(k)6(n-k) =
1c -
z(k)L[b(n-k)]
n=-m
The response L [6 (n- k ) ] can be interpreted as the response of a linear system at time n due to a unit sample (a well-known sequence) at time k. It is called an impulse response and is denoted by h ( n , k ) .The output then is given by the superposition summation
Y (4 =
n=-m
c
m
.(k)h(n,k)
(2.9)
The computation of (2.9) requires the time-varying impulse response h (n, ) , which in practice is not very convenient. Therefore timeinvariant k systems are widely used in DSP.
Linear time-invariant (LTI) system A linear system in which an () input-output pair, z(n)and y n , is invariant to a shift n in time is called a linear timeinvariant system. For an LTI system the L[.]and the shifting operators are reversible as shown below. z(n)
-a -
~ ( nd Shift by k )
We will denote an LTI system by the operator LTI [.I. Let z(n) and y(n) be the input-output pair of an LTI system. Then the time-varyingfunction h (n, becomes a time-invariant function h (n - k), and the output from k) (2.9) is given by
- - k) *€I L I[.]
~ ( n
y(n)
d
pzq
-
y(n - k)
y(n - k)
y(n) = LTI I).([
=
k=-m
c
m
z(k)h(n - k)
(2.10)
The impulse response of an LTI system is given by h(n).The mathematical operation in (2.10) is called a linear convolution sum and is denoted b Y
y(n)
5 .(n) * h(n)
(2.11)
21
Discrete Systems
Hence an LTI system is completely characterized in the time domain by the impulse response h(n) as shown below.
z(n)--* h(n)
U
---t
y(n) = z(n)* h(n)
We will explore several properties of the convolution in Problem 2.12.
Stability This is a very important concept in linear system theory. The primary reason for considering stability is to avoid building harmful systems or to avoid burnout or saturation in the system operation. A system is said to be bounded-anputbounded-output (BZBO)stable if every bounded input produces a bounded output.
I+)I < co =+ Iy(n)l < co,vz,Y
An LTI system is BIBO stable if and only if its impulse response is absolutely summable.
m
BIBO Stability @
-m
lh(n)I < 00
(2.12)
C a w d i t y This important concept is necessary to make sure that systems can be built. A system is said to be causal if the output at index no depends only on the input up to and including the index no; that is, the output does not depend on the future values of the input. An LTI system is causal if and only if the impulse response
h(n) =o,
n<O
(2.13)
Such a sequence is termed a causal sequence. In signal processing, unless otherwise stated, we will always assume that the system is causal.
CONVOLUTION
I I
We introduced the convolution operation (2.11) to describe the response of an LTI system. In DSP it is an important operation and has many other uses that we will see throughout this book. Convolution can be evaluated in many different ways. If the sequences are mathematical functions (of finite or infinite duration), then we can analytically evaluate (2.11) for all n to obtain a functional form of y(n).
22
The s u m in 2.14 is almost a geometric series sum except that the term u(n - k) takes different values depending on n and k. There are three different conditions ( under which u n - k) can be evaluated.
CASE i
In this last case h(n) completely overlaps i ( n ) . The complete response is given by (2.15), (2.16), and (2.17).It is shown in Figure 2.6 which depicts the distortion of the input pulse. I 3 The above example can also be done using a method called graphical convolution, in which (2.11) is given a graphical interpretation. In this method h(n - k) is interpreted as a folded-and-shified version of h(k). The output y(n) L obtained as a sample sum under t h e overlap of z(k) and h(n - k). We use an example to illustrate this.
0 EXAMPLE 2.6
Given the following two sequences
~ ( n ) =3,11,7,0,-1,4,2
IT 1
,
-35n53;
h(n)=
determine the convolution g(n) = x(n)* h(n).
Sdution
In Figure 2.7 we show four plots. The topleft plot shows x(k) and h ( k ) , the original sequences. The togright plot shows s(k)and h(-k), the folded version
Thus we have obtained two values of y(n). Similar graphical calculations can be done for other remaining values of y(n).Note that the beginning point ( i s frt nonzero sample) of y(n) is given by n = -3 (-1) = -4, while the end point (the last nonzero sample) is given by n = 3 4 = 7. The complete output is given by
+ +
y(n) = 6,31,47,6, -51, -5,41,18, -22, -3,8,2
t
Students are strongly encouraged to verify the above result. Note that the resulting sequence y ( n ) has a longer length than both the z (n) and h (n) sequences.
0
{
1
MATLAB IMPLEMENTATION
If arbitrary sequences are of infinite duration, then MATLAB cannot be used directly t o compute the convolution. MATLAB does provide a built-in function called conv that computes the convolution between two finite
25
Convolution
duration sequences. The conv function assumes that the two sequences begin at n = 0 and is invoked by
>> y = conv(x,h);
to obtain the correct v(n)values. However, the conv function neither provides nor accepts any timing information if the sequences have arbitrary support. What is needed is a beginning point and an end point of y(n). Given finite duration z ( n ) and h(n),it is easy t o determine these points. Let
{ z ( n ) ;nzb
snI nze}
and
{h(n);nhb I L we} n
be two finite-duration sequences. Then referring t o Example 2.6 we observe that the beginning and end points of y ( n ) are
nub = nzb
+ nhb
and nue= nze -k nhe
respectively. A simple extension of the conv function, called convm, which performs the convolution of arbitrary support sequences can now be designed.
function Cy,nyl = conv-m(x,nx,h,nh) X Modified convolution routine for signal processing
An alternate method in MATLAB be used to perform the convocan lution. This method uses a matrix-vector multiplication approach, which we will explore in Problem 2.13.
SEQUENCE CORRELATIONS REVISITED
If we compare the convolution operation (2.11) with that of the crosscorrelation of two sequences defined in (2.6), we observe a close resemblance. The crawamlation r&) can be put in the form
rllz(4 = Y ( 4
Fkorn the construction of y(n) it follows that y(n) is "similar" to z ( n - 2) and hence their crosscorrelationwould show the strongest similarity at l = 2.To test this out using MATLAB, us compute the crosscorrelationusing two different let noise sequences.
used in applications like radar signal processing in identifying and localizing 0
It should be noted that the signal-processingtoolbox in MATLAB also provides a function called xcorr for sequence correlation computations. In its simplest form >> xcorr(x,y) computes the crosscorrelationbetween vectors x and y, while
>> xcorr(x)
computes the autocorrelation of vector x. This function is not available in It the Student Edition of MATLAB. generates results that are identical to the one obtained from the proper use of the convm function. However, the xcorr function cannot provide the timing (or lag) information (as done by the convm function), which then must be obtained by some other means. Therefore we will emphasize the use of the convm function.
DIFFERENCE EQUATIONS
I
An LTI discrete system can also be described by a linear constant coefficient difference equation of the form
If
U N # 0, then the difference equation is of order N. This equation describes a recursive approach for computing the current output, given the input values and previously computed output values. In prxtice this equation is computed forward in time, from n = -ca to n = co.Therefore another form of this equation is
This characteristic equation is important in determining the stability of systems. If the roots Zk satisfy the condition
lzkl<l, k = l , ...,N
(2.20)
then a causal system described by (2.19) is stable. The particular part of the solution, yp(n), is determined from the right-hand side of (2.18). In Chapter 4 we will discuss the analytical approach of solving difference equations using the z-transform.
MATLAB IMPLEMENTATION
A routine called filter is available to solve difference equations numerically, given the input and the difference equation coefficients. In its simplest form this routine is invoked by
J
= fi1tercb.a.x)
where
b = [bO, b i .
..., bM1; a =
CaO, ai,
. .., aN1;
are the coefficient arrays from the equation given in (2.18), and x is the input sequence array. The output y has the same length as input x. One must ensure that the coefficient a0 not be zero. We illustrate the use of this routine in the following example.
0
EXAMPLE 2.9
The plot of the unit step response is shown in Figure 2.9. c. To determine the stability of the system, we have to determine h(n) for all n. Although we have not described a method to solve the difference equation, we can use the plot of the impulse response to observe that h(n) is practically zero for n > 120. Hence the sum C Ih(n)l can be determined from MATLAB using
r
-20 0
20
40
80
loo
120
c
D
1.
0.
I
-o&
FIGURE 2.9
20
40
n
60
80
loo
I 120
I
Impulse response and step response plots in Ezample 2.9
Difference Equations
31
>> sum(abs(h))
an6 = 14.8785
w i h implies that the system is stable. An alternate approach is to we the hc stability condition (2.20) using MATLAB'Sroots function.
>>z = roots(a);
>>magz = abs(z) m a p = 0.9487
0.9487
Since the magnitudes of both roots are less than one, the system is stable. 0 In the previous section we noted that if one or both sequences in the convolution are of infinite length, then the conv function cannot be used. If one of the sequences is of infinite length, then i t is possible t o use MATLABfor numerical evaluation of the convolution. T i is done using hs the filter function as we will see in the following example.
0
EXAMPLE 2.10
If the LTI system, given by the impulse response h(n),can be described by a difference equation, then y(n) can be obtained from the filter function. From the h(n)expression
(0.9) - I) = (0.9) 0 9 " 'u(n - 1) = (0.9)"u(n I) h(n (.)or
h(n)- (0.9) - 1) = ( . ) u(n)- ( . ) u(n - 1) h(n 09" 09"
= ( . ) [ ~ ( n )~ (-nl)]= (0.9)"6(n) 09" = 6(n)
The last step follows from the fact that b(n) is nonzero only at n = 0. By definition h(n) is the output of an LTI system when the input is 6(n).Hence substituting z(n) for 6(n)and y(n) for h(n),the difference equation is y n - O.Sy(n - 1) = z(n) () Now MATLAB'S filter function can be used to compute the convolution indirectly.
The plot o the output is shown in Figure 2.10,which i exactly the same as f s that in Figure 2.6. 0 In Example 2.10 the impulse response was a one-sided exponential sequence for which we could determine a difference equation representation. This means that not all infinite-lengthimpulse responses can be converted into differenceequations. The above analysis, however, can be extended to a linear combination of one-sided exponential sequences, which results in higher-order difference equations. We will discuss this topic of conversion from one representation to another one in Chapter 4. ZERO-INPUT AND ZERO-STATE RESPONSES In digital signal processing the difference equation is generally solved forward in time from n = 0. Therefore initial conditions on z(n) and y(n) are necessary to determine the output for n 2 0. The difference equation is then given by
y(n) =
C bmz(n- rn) - C a b y ( n - k); n 2 o
mO =
M
N
(2.21)
h=l
subject to the initial conditions:
{y(n); - N 5 n 5 -1)
and
{z(n);-M 5 n 5 -1)
A solution to (2.21) can be obtained in the form
~ ( n )Vzr(n)+ yzs(n) =
Difference Eauations
33
where yzr(n) is called the zero-input solution, which is a solution due to the initial conditions alone (assuming they exist), while the zero-state solution, yzs(n), is a solution due to input z(n)alone (or assuming that another form of the function the initial conditions are zero). In MATLAB filter can be used to solve for the difference equation, given its initial conditions. We will illustrate the use of this form in Chapter 4.
DIGITAL FILTERS
Filter is a generic name that means a h e a r time-invariant system designed for a specific job of frequency selection or frequency discrimination. Hence discrete-time LTI systems are also called digital filters. There are two types of digital filters.
FIR filter If the unit impulse response of an LTI system is of finite duration, then the system is called a finite-dumtion impulse response (or FIR) filter. Hence for an FIR filter h(n) = 0 for n < nl and for n > 712. The following part of the difference equation (2.18) describes a cawalFIR filter:
(2.22) Fbrthermore, h(0) = 4, h(1) = 61, ...,h ( M ) = 6 ~while all other h(n)'s , are 0. FIR filters are also called nontecursive or moving average (MA) filters. In MATLAB FIR filters are represented either as impulse response values (h(n)} or as difference equation coefficients (6,) and (a0 = 1). Therefore to implement FIR filters, we can use either the conv(x,h) function (and its modifications that we discussed) or the filter(b,l ,x) function. There is a difference in the outputs of these two implementations that should be noted. The output sequence from the conv(x,h) function has a longer length than both the z(n) and h(n) sequences. On the other hand, the output sequence from the filter(b,i.x) function has exactly the same length as the input z(n) sequence. In practice (and especially for processing signals) the use of the filter function is encouraged.
IIR filter
If the impulse response of an LTI system is of infinite duration, then the system is called an infinite-dumtion impulse response (or IIR) filter. The following part of the difference equation (2.18):
N
(2.23) describes a recursive filter in which the output y(n) is recursively computed from its previously computed values and is called an autoregressive (AR) filter. The impulse response of such filter is of infinite duration and
34
Chapter 2 m DISCRETE-TIME SIGNALS AND SYSTEMS
hence it represents an IIR filter. The general equation (2.18) also describes an IIR filter. It has two parts: an AR part and an MA part. Such an IIR filter is called an autoregressive moving avemge, or an ARMA, filter. In MATLAB filters are described by the difference equation coe5cients IIR {b,} and {ah} and are implemented by the f i l t e r ( b , a , x ) function.
b. Generate and plot cos(0.3lrn), -20 5 n 5 20. Is this sequence periodic? If it is, what is its fundamental period? From the examination of the plot what interpretation can you give to the integers K and N above? c. Generate and plot cos(0.3n), -20 5 n 5 20. Is this sequence periodic? What do you conclude from the plot? If necessary examine the value of the sequence in MATLAB to arrive at your answer. P . Decompose the sequences given in Problem 2.2 into their even and odd components. Plot 24 these components using the stem function. is P . A complex-valued sequence ze(n) called conjugate-symmetric if 25
respectively. a. M d f the evenodd function discussed in the text so that it accepts an arbitrary oiy sequence and decomposes it into its symmetric and antisymmetric components by implementing (2.24). b. Decompose the following sequence:
+)
=
o I I 10
into its conjugatesymmetric and conjugate-antisymmetric components. Plot their real and imaginary parts to verify the decomposition. (Use the subplot function.)
P. 26
The operation of signal dilation (or decimation or down-sampling) is defined by
= z(nW
in which the sequence z(n)is down-sampled by an integer factor M.For example, if
z (n)= {. . . ,-2,4,3, -6,5, -1,8,. . .}
t
then the down-sampled sequences by a factor 2 are given by y(n) = { ..., -2,3,5,8 ,...}
t
a. Develop a MATLAB function &sample that has the form
function
J
= &sample(s,M)
to implement the above operation. Use the indexing mechanism of MATLAB with careful attention to the origin of the time axis n = 0. b. Generate z(n)= sin(0.125ma), -50 5 n p 50. Decimate z(n) by a factor of 4 to ) generate y(n). Plot both ~ ( nand y(n) using subplot and comment on the results. c. Repeat the above using z(n) = sin(0.5m), -50 5 n p 50. Qualitatively discuss the effect of down-sampling on signals. Determine the autocorrelation sequence rvr(t)and the crosscorrelation sequence rrv(l)for the following sequences.
P2.7
~ ( n= (0.9)", )
What is your observation?
05n
5 20;
y(n) = (0.8)-",
-20
5n 50
36
Chapter 2
DISCRETE-TIME SIGNALS AND SYSTEMS
P2.8
In a certain concert hall, echoes of the original audio signal z(n) are generated due t o the al reflections at the wls and ceiling. The audio signal experienced by the listener y(n) is a combination of z(n) and its echoes. Let
y(n)= z(n) az(n - k)
where k is the amount of delay in samples and a is its relative strength. We want to estimate the delay using the correlation analysis. a. Determine analytically the autocorrelation ryv(l) terms of the autocorrelation rzr(t). in b. Let z(n)= cos(0.Zxn) +0.5cos(0.6xn), a = 0.1, and k = 50. Generate 200 samples of y(n) and determine its autocorrelation. Can you obtain a and k by observing rvv(t)? Three systems are given below.
+
P. 29
T [z(n)] 2=("); I =
T [+)I z
=3 4 4
+ 4;
T [z(n)]= z(n) 3
+ 2z(n - 1) - z(n - 2)
a. Use (2.8)to determine analytically whether the above systems are linear. b. Let q ( n ) be a uniformly distributed random sequence between [0,1] over 0 6 n 5 100, and let zz(n) be a Gaussian random sequence with mean 0 and variance 10 over 0 5 n 2 100. Using these sequences, test the linearity of the above systems. Choose any values for constants al and a2 in (2.8).You should use several realizations of the above sequences to arrive at your answers. P . 0 Three systems are given below. 21
n
n+lO
Ti [~(n)] =
0
4k);
T [~(n)] z =
n-10
4k);
T3 [=(.)I
= z(-n)
a. Use (2.9) to determine analytically whether the above systems are time-invariant. b. Let z(n) be a Gaussian random sequence with mean 0 and variance 10 over 0 5 n 5 100. Using this sequence, test the time invariance of the above systems. Choose any values for sample shift k in (2.9).You should use several realizations of the above sequence to arrive at your answers. P . 1 For the systems given in Problems 2 9 and 2.10 determine analytically their stability and 21 . causality. P2.12 The linear convolution defined in (2.11) has several properties:
zi(n)*z,(n)= m ( n ) *zz(n)
: Commutation : Association
[zi(n)* zz(n)] * 23(n) = SI(~)[52(n) d n ) ] * *
: z~(n)[zz(n) z3(n)] z~(n)22(n)+ zi(n)* ~ ( n ) Distribution * + = *
(2.25)
z(n) * s(n - no) = z(n - no)
a. Analytically prove these properties.
: Identity
b. Using the following three sequences, verify the above properties.
z1(n) = n [u(n 10) u(n
+
-
- ZO)]
n(n) = cos (O.l*n)[u(n)- u(n- 3 ) 0j
23(n)
= (1.2)" [u(n
+ 5) - u(n - lo)]
37
Use the conv_pIfunction.
Problems
P2.13 When the sequences x ( n ) and h ( n ) are of finite duration N, and N h , respectively, then their linear convolution (2.10) can also be implemented using mat*-vector multiplimtim If elements of y(n) and z(n) are arranged in column vectors x and y respectively, then from (2.10) we obtain
y=HX
where linear shifts in h (n- k) for n = 0,. .. ,Nh - 1 are arranged as rows in the matrix H. This matrix has an interesting structure and is called a Toeplitz matrix. To investigate this matrix, consider the sequences
z(n)={:,2,3,4}
and
h(n)={?,2,1}
a. Determine the linear convolution y (n) = h (n) I (n). *
b. Express I (n)as a 4 x 1 column vector x and y (n)as a 6 x 1 column vector y. Now determine the 6 x 4 matrix H so that y = Ex. c. Characterize the matrix € . this characterization can you give a definition of a From I Toeplitz matrix? How does this definition compare with that of time invariance? d. What can you say about the first column and the first row of H? P2.14 MATLAB provides a function called t o e p l i t z t o generate a Toeplitz matrix, given the first row and the first column. a. Using this function and your answer to Problem 2.13 part d, develop an alternate MATLAB function to implement linear convolution. The format of the function should be function [y ,HI -conv-tp(h ,x) % Linear Convolution using Toeplitz Matrix
THEDISCRETETIME FOURIERANALYSIS
We have seen how a linear and time-invariant system can be represented using its response to the unit sample sequence. This response, called the unit impulse response k(n),allows us to compute the system response to any arbitrary input z(n)using the linear convolution as shown below.
z(n) +
--+
y(n) = k(n) * z(n)
T i convolution representation is based on the fact that any signal can hs be represented by a linear combination of scaled and delayed unit samples. Similarly, we can also represent any arbitrary discrete signal as a linear combination of basis signals introduced in Chapter 2. Each b e sis signal set provides a new signal representation. Each representation has some advantages and some disadvantages depending upon the type of system under considerstion. However, when the system is linear and timeinvariant, only one representation stands out a the most useful. It s is based on the complex exponential signal set {ej""} and is called the Discrete-time Fourier lhnsform.
T H E DISCRETE-TIME FOURIER TRANSFORM (DTFT)
I m
If z (n) is absolutely summable, that is, discrete-time Fourier transform is given by
cymIs(.)[
<
00,
then its
The inverse discretetime Fourier transform (IDTFT) of
X(&")is given
bY
) . ( 2
= A
F1
[X(&")] = 2*
'i
-77
X(e'w)e'"vu
(3.2)
The operator 3[.I transforms a discrete signal z(n)into a complex-valued continuous function X(eJ") of real variable w, called a digital frequency, which is measured in radians.
0
Determine the discrete-time Fourier transform of the following finite-duration sequence:
= {1,2,3,4,5)
f
--QI
Sne X(e'") is a complex-valued function, we will have to plot its magic nitude and its angle (or the real and the imaginary part) with respect to w separately to visually describe X(dw). Now w is a real variable between -cu and 00, which would mean that we can plot only a part of the X(dw) function using MATLAB. Using two important properties of the discretetime Fourier transform, we can reduce this domain to the [O,m] interval for real-valued sequences. We will discuss other useful properties of in the next section.
x($-)
0
TWO IMPORTANT PRoPERT'ES
We will state the following two properties without proof.
1. Periodicity: The discretetime Fourier transform X(ej") is periodic in w with period 27r.
Implication:To plot X(eJ"), we now need to consider only a half period of X(ej"). Generally, in practice this period is chosen to be w E lo,.].
MATLAB IMPLEMENTATION
If z(n) is of infinite duration, then MATLAB cannot be used directly to compute X(ej") from z(n).However, we can use it to evaluate the expression X(e3") over [0,. frequencies and then plot its magnitude and ] angle (or real and imaginary parts).
Evaluate X ( 2 " ) in Example 3.1 at 501 equispaced points between [0,7r] and plot its magnitude, angle, real, and imaginary parts.
The resulting plots are shown i Figure 3.1. Note that we divided thew array by n pi before plotting so that the frequency axes are in the units of 7r and therefore 0 easier to read. T i practice is strongly recommended. hs
42
Chapter 3 fi THE DISCRETE-TIME FOURIER ANALYSIS
Magnitude Part
~ b aPart l
frequency in pi u n b
Ande Part
!m i
24.8
I
-0.8 0 0.5 freqwwinpiunits
1
lmaginaly Part
-0.6
0
0.5 hequency in pi units
FIGURE 3.1
Plots in Emmple 3.3
If x(n) is of finite duration, then MATLAB be used to compute can X(ej") numerically at any frequency w. The approach is to implement (3.1) directly. If, in addition, we evaluate X(@) at equispaced frequencies between [O, A], then (3.1) can be implemented as a matrix-vector mdtiplacation operation. To understand this, let us m u m e that the sequence x(n) has N samples between nl 5 IZ 5 n N (i.e., not necessarily between [0,N - 4) and that we want to evaluate X(&") at
wk
The frequency-domain plots are shown in Figure 3.2. Note that the angle plot is depicted as a discontinuous function between - x and x. This is because the 0 angle function in MATLAB computes the principal angle. The procedure of the above example can be compiled into a MATLAB function, say a dtft function, for ease .of implementation. This is explored in Problem 3.1. This numerical computation is based on definition (3.1).
44
Chapter 3
THE DISCRETE-TIME FOURIER ANALYSIS
Maanitude Part
? .
Real Part
0
0
0.5
1
-..
0
0.5
1
frequency in pi units Angle Part
frequency in pi units Imaginary Part
E -10l -?l !5 !& !O
I
0 0.5
1
0
frequency in pi units
05 frequency in pi units
FIGURE 3.2 Plots in Ezample 3.4
It is not the most elegant way of numerically computing the discrete-time Fourier transform of a finite-duration sequence. Furthermore, it creates an N x (M 1)matrix in (3.4) that may exceed the size limit in the Student Edition of MATLABfor large M and N . In Chapter 5 we will discuss in detail the topic of a computable transform called the discrete Fourier transform (DFT) and its efficient computation called the fast Fourier transform (FFT). Also there is an alternate approach based on the ztransform using the MATLAB function f reqz for finite-duration sequences, which we will discuss in Chapter 4. In this chapter we will continue to use the approaches discussed so far for calculation as well as for investigation purposes. In the next two examples we investigate the periodicity and symmetry properties using complex-valued and real-valued sequences.
+
0
EXAMPLE 3.5
Let z(n) = (0.9 exp ( j ~ / 3 ) , ~ 0 5 n 5 10. Determine X(eJ') and investigate ) its periodicity. Since z(n) is complex-valued,it satisfies only the periodicity property. Therefore it is uniquely defined over one period of 27r. However, w will evaluate and plot it e at 401 frequenciesover two periods between [-an,2 4 l o observe its periodicity.
Solution
The Discrete-time Fourier Transfwm (DTFT)
45
From the plots in Figure 3.3 we observe that X ( e i W ) periodic in w but is not is 0 conjugatesymmetric. Let z(n) = 2", -10 n 5 10.Investigate the conjugatesymmetry property of its discretetime Fourier transform. Once again we will compute and plot X(ej") over two periods to study its symmetry property.
From the plots in Figure 3.4 we observe that X(t?) is not only periodic in (J but is also conjugate-symmetric.Therefore for real sequences we will plot their Fourier transform magnitude and angle responses from0 to T . 0
THE PROPERTIES OF THE DTFT
I I
In the previous section we discussed two important properties that we needed for plotting purposes. We now discuss the remaining useful properThe Properties of the DTFT
47
ties, which are given below without proof. Let X(.'") be the discretetime Fourier transform of ~(n).
1. Lmearity: The discretetime Fourier transform is a linear transformation; that is, F[(Iz1(n) /3m(n)l = a3[z1(41+ 87 Izz(n)l
+
(3.5)
for every (I,
2.
p, z (n), xz (n). 1 and Time shifting: A shift in the time domain corresponds to the phase
F [z(n- k)] = X(ejw)e--jwk
(3.6)
shifting.
3. Fkequency shifting: Multiplication by a complex exponential corresponds to a shift in the frequency domain.
4. Coqjugation: Conjugation in the time domain corresponds to the folding and conjugation in the frequency domain.
~ [ z * ( n ) X*(e-jw) =]
(3.8)
5. Folding Folding in the time domain corresponds to the folding in the frequency domain.
F [ x ( - ~ )= X(e-jw) ]
(3.9)
6. Symmetries in real sequences: We have already studied the conjugate symmetry of real sequences. These real sequences can be decomposed into their even and odd parts a we discussed in Chapter 2. s
z(n)= z e ( 4
Then
+ XO(4
(3.10)
F[z,(n)]= Re [ ~ ( e j " ) ]
F[zO(n)]j I m [ ~ ( e j " ) ] =
Implication: If the sequence z(n) is real and even, then X ( e j w )is also real and even. Hence only one plot over [0,n] is necessary for its complete representation. A similar property for complex-valued sequences is explored in Problem 3.7. 7. Convolution: This is one of the most useful properties that m k s ae system analysis convenient in the frequency domain.
(3.12)
The convolution-like operation above is called a periodic convolution and hence denoted by @ . It is discussed (in its discrete form) in Chapter 5. 9. Energy: The energy of the sequence x(n) can be written as
(3.13)
dw (for real sequences using even symmetry)
0
This is a s known as Parseval's Theorem. From (3.13)the energy density lo spectrum of x(n) is defined as
(3.14)
Then the energy of z(n)in the [ w ~ , w zband is given by ]
J @,(w)dw,
W1
W2
0 I w1 < W2
I
In the next several examples we will verify some of these properties using finite-duration sequences. We will follow our numerical procedure to compute discretetime Fourier transforms in each case. Although this does not analytically prove the validity of each property, it provides us with an experimental tool in practice.
0
EXAMPLE 3.7
Since the maximum absolute error between the two Fourier transform arrays iO-", the two arrays are identical within the limited numerical precision of MATLAB. 0 Let z(n) be a random sequence uniformly distributed between [O, 1) over 0 5 n 5 10 and let y(n) = z(n - 2. Then we can verify the sample shift property ) (3.6) 85 follows.
Then using the evenodd function developed in Chapter 2, we can compute the even and odd parts of s(n) and then evaluate their discretetime Fourier transforms. We will provide the numerical as well as graphical verification.
From the plots in Figure 3.6 we observe that the real part of X ( e j w ) (or the imaginary part of X(e'")) is equal to the discretetime Fourier transform of z J n ) (or ~ ( n ) ) . 0
T H E FREQUENCY DOMAIN REPRESENTATION OF LTI SYSTEMS
I I
We earlier stated that the Fourier transform representation i the most s useful signal representation for LTI systems. It is due to the following result.
The Frequency Domain Representation of LTI Systems
53
RESPONSE TO A COMPLEX EXPONENTIAL
@'won
Let z(n) = eJwonbe the input to an LTI system represented by the impulse response h(n).
eJwon -+
0
h(n) -+
m
h(n)* gWon
Then
(3.15)
rn
DEFINITION 1
m e n c y Response The discrete-time Fourier transform of an impulse response is called the Frequency h p o n s e (or Transfer Function) of an LTZ system and is denoted by
H ( P ~ )5
C h(n)e-jwn
-00
00
(3.16)
Then from (3.15) we can represent the system by
-+
y(n) = ~
( 2 "x)$won
(3.17)
Hence the output seqoence is the input exponential sequence modified by the response of the system at frequency wo. This justifies the definition of If(&") as a frequency response because it is what the complex exponential is multiplied by to obtain the output y(n). This powerful result can be extended to a linear combination of complex exponentials using the linearity of LTI systems.
In general, the frequency response H ( d w )is a complex function of w . The f is magnitude IH(dw)(o H(eJW) called the magnitude (or gain) response function, and the angle LH(ejw)is called the phase response function as we shall see below. RESPONSE TO SINUSOIDAL SEQUENCES
~
Let z(n) = Acos(w0n 6'0) be an input to an LTI system h(n). Then from (3.17) we can show that the response y(n) is another sinusoid of the same frequency wo, with amplitude gained by IH(d"0)l and phase sh@ed
+
5 4
Chapter 3
THE DISCRETE-TIME FOURIER ANALYSIS
by LH(ei"o), that is,
y(n) = A ~
~ ( e i w cos (won o ) ~
+ e, + LH(PO))
(3.18)
This response is called the steady-state response denoted by ysd(n).It can be extended to a linear combination of sinusoidal sequences.
RESPONSE TO ARBITRARY SEQUENCES
Finally, (3.17) can be generalized to arbitrary absolutely surnmable sequences. Let X(ej") = F[z(n)] and Y(d")= F[y(n)]; then using the convolution property (3.11), we have
Y ( P )= H ( e q X(e3")
(3.19)
Therefore an LTI system can be represented in the frequency domain by
X(.'")
-p+q -
Y(.i") = If(&") X(eJY)
The output y(n) is then computed from Y(.'") using the inverse discretetime Fourier transform (3.2). This requires an integral operation, which is not a convenient operation in MATLAB. we shall see i Chapter 4, there As n is an alternate approach to the computation of output to arbitrary inputs using the r-transform and partial fraction expansion. In this chapter we will concentrate on computing the steady-state response.
0
EXAMPLE 3.13
Let an input to the system in Example 3.13 be O.lu(n). Determine the steadystate response yss(n). Since the input is not absolutely summable, the discretetime Fourier transform is not particularly useful in computing the complete response. However, it can be used to compute the steady-state response. In the steady state (i.e., n cm)
Magnilude Response
w o n
I
0
01
0.2
03 .
0.4 0.5 0.6 hequency in pi units
07
08
0.9
1
FIGURE 3.7
Fkquency response plots a Ezample 3.13 n
56
Chapter 3 I THE DISCRETE-TIME FOURIER ANALYSIS
the input is a constant sequence (or a sinusoid with wo = 00 = 0). Then the output is
y.,(n)
= 0 1 x H(20) 0 1 x 10 = I . = .
where the gain of the system at w = 0 (also called the DC gain) is H(e'O) = 10, 0 which is obtained from Figure 3.7. When an LTI system is represented by the difference equation
Ftom the plots in Figure 3.8 we note that the amplitude o U...(n) is approxif mately 4. To determine the shift in the output sinusoid, we can compare zero
0
10
20
3 0
40
50
60
7p
80
90
100
n output sequence
-_I
0
10
20
30
40
50
60
70
80
90
100
n
FIGURE 3.8 Plots in Ezample 3.15
5a
Chapter 3
THE DISCRETE-TIME FOURIER ANALYSIS
crossings of the input and the output. This is shown in Figure 3.8,from w i h hc the shift is approximately 3 5 samples. . 0 In Example 3.15 the system was characterized by a first-order difference equation. It is fairly straightforward to implement (3.22) in MATLAB as we did in Example 3.13.In practice the difference equations are of large order and hence we need a compact procedure to implement the general expression (3.21).This can be done using a simple matrix-vector multiplication. If we evaluate I€(@") at Ic = 0,1,.. ,K equispaced frequencies . over [O,T], then
M
respectively. Now the array H(ej"*) in (3.23)can be computed using a ./ operation. This procedure can be implemented in a MATLAB function to determine the frequency response function, given {b,} and { a t } arrays. We will explore this in Example 3.16 and in Problem 3.15.
0
EXAMPLE 3.16
A 3rd-order lowpass filter is described by the difference equation
~ ( n= 0.0181~(n) 0.0543~(n-1) )
SAMPLING AND RECONSTRUCTION OF ANALOG SIGNALS
In many applications-for example, in digital communications-realworld analog signals are converted into discrete signals using sampling and quantization operations (collectively called analog-tedigital conversion or ADC). These discrete signals are processed by digital signal processors, and the processed signals are converted into analog signals
60
Chapter 3
H
THE DISCRETE-TIME FOURIER ANALYSIS
using a reconstruction operation (called digitd-to-analog conversion or DAC). Using Fourier analysis, we can describe the sampling operation from the frequency-domain viewpoint, analyze its effects, and then address the reconstruction operation. We will also assume that the number of quantization levels is sufficiently large that the effect of quantization on discrete signals is negligible. SAMPLING Let za(t) be an analog (absolutely integrable) signal. Its continuow-time Fourier transform (CTFT) is given by
Xa(jR) i 2
J z,(t)e-j"fdt
-m
m
(3.24)
where R is an analog frequency in radians/sec. The inverse continuowtime Fourier transform is given by
za(t) = 27r .
Let X(eJ")be the discrete-time Fourier transform of ~ ( n ) . it can be Then shown (191 that X(d") i s a countable sum of amplitudescaled, frequencyscaled, and translated versions of the Fourier transform Xa(jR).
(3.26)
The above relation is known as the aZiasirzg formula. The analog and ' digital frequencies are related through T
w = RT,
(3.27)
while the sampling frequency
F, is given by
(3.28)
F. = 2, SLUII/S~C Ta
The graphical illustration of (3.26) is shown i Figure 3.10, from which n we observe that, in general, the discrete signal is an aliased version of the corresponding analog signal because higher frequencies are aliased into
Sampling and Reconstruction of Analog Signals
61
s.
t
i
Sample
I
eq. (3.27)
x(n1
I
tO
-2n -n
FIGURE 3.10
0
u
2.a
Sampling opemtion in the time and frequency domains
lower frequencies if there i an overlap. However, it i possible to recover s s the Fourier transform X , ( j n ) from X ( e J w )(or equivalently, the analog signal za(t) from its samples z ( n ) ) if the infinite ''replicas" of Xa(jf2)do not overlap with each other to form X ( e j w ) .This is true for band-limited analog signals.
m
MFlNlTlON 2
Band-limited Signal A signal is band-limited if there exists a finite radian frequency Ro such that X a ( j R ) is zerw for (01> Ro. The frequency Fo,R0/27r is called the signal bandwidth a Hz. n
Otherwise aliasing would result an x ( n ) . The sampling rate of 2Fo for an analog band-limited signal is called the Nyquist mte.
It should be noted that after s,(t) is sampled, the highest analog hs frequency that x ( n ) represents is F,/2 Hz (or w = 7r). T i agrees with the implication stated in Property 2 of the discretetime Fourier transform in the first section of this chapter.
MATLAB IMPLEMENTATION
In a strict sense it is not possible to analyze analog signals using MATLAB unless we use the Symbolic toolbox. However, if we sample x a ( t )on a fine grid that has a sufficiently small time increment to yield a smooth plot and a large enough maximum time to show all the modes, then we can appraximate its analysis. Let A t be the grid interval such that A t << T,. Then
xG(m) = z a ( m A t )
h
(3.30)
can be used as an array to simulate an analog signal. The sampling interval
T,should not be confused with the grid interval At, which is used strictly to represent an analog signal in MATLAB. Similarly,the Fourier transform relation (3.24) should also be approximated in light of (3.30) as follows
x,bn) %
m
XG(m)e-JnmAtAt= A t x Z G ( m ) e - j n m A t
m
(3.31)
Now if s a ( t ) (and hence Z G ( ~ )is of finite duration, then (3.31) is similar ) to the discretetime Fourier transform relation (3.3) and hence can be implemented in MATLAB a similar fashion to analyze the sampling in phenomenon. 63
(3.32)
which is a real-valued function since z.(t) is a real and even signal. To evaluate X ( R numerically, we have to first approximate za(t) by a finite-duration ,j) grid sequence zc(m). Using the approximation e-' x 0, we note that z,(t) can be approximated by a finite-duration signal over -0.005 5 t 5 0.005 (or equivalently, over [-5,5] msec). Similarly from (3.32), X ( Q % 0 for R 1 ,j) 2n (ZOOO). Hence c h m i n g
Figure 3.11 shows the plots of zo(t)and X ( Q . Note that to reduce the number .j) of computations, we computed X ( R over [O,4000rj radians/sec (or equiva,j) lently, over [0,2] KHz) and then duplicated it over [-400Chr,O] for plotting 0 purposes. The displayed plot of X.(jR) agrees with (3.32).
0
EXAMPLE 3.18
To study the effect of sampling on the frequency-domain quantities, we will sample z.(t) in Example 3.17 at two different sampling frequencies.
a. Sample z,(t) at
In the top plot in Figure 3.12 we have superimposed the discrete signal %(n) over x.(t) to emphasize the sampling. The plot of X Z ( ~ " ) that it i a shows s scaled version (scaled by F. I 5000) of X,(jQ). Clearly there is no aliasing.
Sampling and Reconstruction of Analog Signals
65
Discretesiid
tinmsec.
Discrete-time Fourier Transform
;
l
:
2 01
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Frequency in pi units
FIGURE 3.12 Plots in Ezample 3.18a
b Here F. = loo0 < 4000. Hence there will be a considerable amount of . aliasing. T i is evident from Figure 3.13,in which the shape of X(ejW)i diths s ferent from that of X , ( j n ) and can be seen to be a result of adding overlapping replicas of X,(jn). 0
RECONSTRUCTI ON
From the sampling theorem and the above examplea it is clear that if we sample band-limited z,(t) above its Nyquist rate, then we can reconstruct z,(t) from its samples z(n).T i reconstruction can be thought of as a hs t m s t e p process:
0
This twc+step procedure can be described mathematically using an interpolating formula [19]
where sinc(z) = is an interpolating function. The physical interpretation of the above reconstruction (3.33) is given in Figure 3.14, from which we observe that this ideal interpolation is not practically feasible because the entiie system is noncausal and hence not realizable.
Pructical D/d converters In practice we need a different approach than (3.33). The two-step procedure is still feasible, but now we replace the ideal lowpass filter by a practical analog lowpass filter. Another interpretation of (3.33) is that it is an infinite-order interpolation. We want finite-order (and in fact low-order) interpolations. There are several a p proaches to do this.
Sampling and Reconstruction of Analog Signals
67
I
t
+
Sample at t = nTs
t
x(3) sincfF,(t
- 3TJ1
t -
-T,
0
T,
ZT, 3Ts
Sampling FIGURE 3.14
Reconstruction
Rmnshrction of band-limited signal from its samples
0 Zero-order-hold (ZOH) interpolation: In this interpolation a given sample value is held for the sample interval until the next sample is r e ceived.
i.&)
= s(n),
nT, I n < (n-I1)T.
which can be obtained by filtering the impulse train through an interpolating filter of the form
68
Chapter 3
fi
THE DISCRETE-TIME FOURIER ANALYSIS
which is a rectangular pulse. The resulting signal is a piecewiseconstant (staircase) waveform which requires an appropriately designed analog post-filter for accurate waveform reconstruction.
0 First-order-hold (FOH) interpolation: In this case the adjacent samples are joined by straight lines. This can be obtained by filtering the impulse train through
0,
otherwise
Once again an appropriately designed analog postfilter is required for accurate reconstruction. These interpolations can be extended t o higher is orders. One particularly useful interpolation employed by MATLAB the following. 0 Cubic spline interpolation: This approach uses spline interpolants for a smoother, but not necessarily more accurate, estimate of the analog signals between samples. Hence this interpolation does not require an analog postfilter. The smoother reconstruction is obtained by using a set of piecewise continuous third-order polynomials called cubic splines, given by 1 1 5
2,
( t )= 0 0 (n)
+
01
(4(t - nT,) + 0 2 (n)( t - nTJ2
(3.34)
+ 0 3 (n)(t - ~LT,)~, I < (n+ 1)T. nT, n
where (ai(n),0 I i 5 3) are the polynomial coefficients, which are determined by using least-squares analysis on the sample values. (Strictly speaking, this is not a causal operation but is a convenient one in MATLAB.)
MATLAB IMPLEMENTATION
For interpolation between samples MATLAB provides several approaches. The function sinc(x), which generates the ( s i n m ) / K X function, can be used t o implement (3.33), given a finite number of samples. If { (n) n1 5 n 5 n2) is given, and if we want to interpolate z ( t ) on z , , a very fine grid with the grid interval At, then from (3.33)
z (mat) ,
=
nz
~ ( n ) [F,(mAt - nT,)] tl sinc ,
n=n,
I mat I t z
(3.35)
Sampling and Reconstructionof Analog Signals
69
which can be implemented as a matrix-vector multiplication operation as shown below.
>> n = ni:n2; t = t1:tZ; Fs = UTs; nTs = n*Ts; X Ts is the sampling interval >> xa = x * sinc(Fs*(ones(lsngth(n) ,l)*t-nTs'*ones(l,length(t))));
Note that it is not possible to obtain an exact analog z,(t) in light of the fact that we have assumed a finite number of samples. We now demonstrate the use of the sinc function in the following two examples and also study the aliasing problem in the time domain.
0
EXAMPLE 3.19
Fkom the samples zl(n)in Example 3.18a, reconstruct z.(t) and comment on the results.
Note that z,(n) was obtained by sampling z.(t) at Z = l/Fa = 0.0002 8ec. We will use the grid spacing of O.ooOo5 sec over -0.005 5 t 5 0.005, which gives z(n) over -25 5 n 5 25.
The maximum error between the reconstructed and the actual analog signal is 0.0363, which is due to the fact that s,(t) is not strictly band-limited (and also we have a finite number of samples). Elom Figure 3.15 we note that visually the reconstruction is excellent. 0
The maximum error between the reconstructed and the actual analog signal is 0.1852,w i h is significant and cannot be attributed to the nonband-limitedness hc of z,(t) alone. From Figure 3.16 observe that the reconstructed signal differs from the actual one in many places over the interpolated regions. T i is the hs visual demonstration of aliasing in the time domain. 0
The second MATLAB approach for signal reconstruction is a plotting approach. The stairs function plots a staircase (ZOH) rendition of the analog signal, given its samples, while the plot function depicts a linear (FOH) interpolation between samples.
0
EXAMPLE 3.21
Plot the reconstructed signal from the samples xi (n) in Example 3.18 using the ZOH and the FOH interpolations. Comment on the plots. Note that in this reconstruction we do not compute x,, (t) but merely plot it using its samples.
R e o o N e d Signal horn e ( n ) using sinc fun&
The plots are shown in Figure 3.17, born which we observe that the ZOH r e construction i a crude one and that the further processing of analog signal is s neexmy. The FOH reconstruction appears to be a good one, but a careful observation near t = 0 reveals that the peak of the signal is not correctly repre
R e c o n m e d Signal fmm xl(n) using zem-oder-hOld
-
5
-
4
4
-
2
-
1
0
1
2
3
4
5
linmsec
tin msec.
FIGURE 3.17
Signal rt%on3truction in Ezample 3.21
72
Chapter 3
THE DISCRETE-TIMEFOURIER ANALYSIS
duced. In general, if the sampling frequency is much higher than the Nyquist rate, then the FOH interpolation provides an acceptable reconstruction. 0
The third approach of reconstruction in MATLABinvolves the use of cubic spline functions. The s p l i n e function implements interpolation between sample points. It is invoked by xa = s p l i n e ( n T s , x , t ) , in which x and nTs are arrays containing samples ~ ( n ) nT, instances, respectively, at and t array contains a fine grid at which za(t) values are desired. Note once again that it is not possible to obtain an =act analog ~ ( t ) .
0
EXAMPLE 3.22
From the samples zl(n) and z (n)in Example 3.18, reconstruct r.(t) using 2 the spline function. Comment on the results. This example is similar to Examples 3.19 and 3.20.Hence sampling parameters are the same as before.
The maximum error between the reconstructed and the actual analog signal is 0.0317,which is due to the nonideal interpolation and the fact that z.(t) is nonband-limited. Comparing this error with that from the sinc (or ideal) interpolation, we note that this error is lower. The ideal interpolation generally suffers more from time-limitedness (or from a finite number of samples). Ftom the top plot in Figure 3.18 we observe that visually the reconstruction is excellent.
error = 0.1679 The maximum error in this case is 0.1679,which is significant and cannot be attributed to the nonideal interpolation or nonband-limitednarsof z,(t). From the bottom plot in Figure 3.18 observe that the reconstructed signal again differs 0 from the actual one in many places over the interpolated repions.
-
Sampling and Reconstructionof Analog Signals
73
ReconstructedSignal from xl(n)usingc bc s l n function u i pie
tinmsec.
FIGURE 3.18
Reconstructed signal in Ezample 3.22
From these examples i t is clear that for practical purposes the spline interpolation provides the best results.
PROBLEMS
I
I
P . Write a MATLABfunction to compute the DTFT of a fmite-duration sequence. The format 31 of the function should be
2) Determine the DTFT for N = 5, 15, 25, 100. Scale the DTFT so that X ( ' = 1. Plot the normalized DTFT over [-*,?TI. Study these plots and comment on their behavior a a s function of N. Repeat Problem 3.4 for a symmetric triangular pulse that is given by
P. 36
Repeat Problem 3.4 for a symmetric raised cosine pulse that is given by
+ 0.5 cos
P. 37
3 [ z e(n)] XR (2") =
(?)I
RN (n)
A complex-valued sequence z (n)can be decomposed into a conjugate symmetric part ze(n)and a conjugatsantisymmetric part zo(n)as discussed in Chapter 2. Show that
and
3 [zo = j X r (n)]
(2")
where X R (2") XI(2") the real and imaginary parts of the DTFT X (g"), and are respectively. Verify this property on
z (n)= 2°.1*n(n)- u (n - 20)] [u
PS.8
using the MATLABfunctions developed in Chapter 2. A complex-valued DTFT X (ej") can also be decomposed into its conjugate-symmetric . and conjugate-antispmetric part X,,(2"); is, that part X (2")
for N = 41, a = 20, and we = 0.51~. c. Determine and plot the frequency response function H (2") compare it with the and ideal lowpass filter response H d (ej"). Comment on your observations. P . 4 An ideal highpass filter is described in the frequency domain by J1
where we is called the cutoff frequency and a is called the phase delay. a. Determine the ideal impulse response hd (n) using the IDTFT relation (3.2). b. Determine and plot the truncated impulse response h(n)=
{:y
, OInIN-I (n) otherwise
for N = 31, a = 15, and we = 0.51~. c. Determine and plot the frequency response function H (2") compare it with the and ideal highpass filter response H d (ej"). Comment on your observations. P . 5 For a linear time-invariant system described by the diEerence equation 31
d. Find two other analog signals zD(t), with different analog frequencies, that will give the same steady-state output ya (t) when I, (t) = COB (2Chrt) is applied.
. to the e. To prevent aliasing, a prefilter would be required to process I (t) before it p~ A/D converter. What type of filter should be used, and what should be the largest cutoff frequency that would work for the given configuration?
c. Reconstruct the analog signal ga (t) from the samples z (n) using the cubic spline interpolation and determine the frequency in g (t) from your plot. (Ignore the end effects.) . d. Comment on your results. P . 1 Consider the analog signal I ,(t) = sin ( 2 h t n/4) , 0 5 t 5 1. It is sampled at T, = 0.05 32 , sec inter& to obtain s(n). a. Plot za( t ) and superimpose z (n) on it using the plot(n.x, ' 0 ' ) function. b. Reconstruct the analog signal go (t) &om the samples I (n)using the sinc interpolation (use At = 0.001) and superimpose z(n) on it. c. Reconstruct the analog signal ga (t) from the samples z(n)using the cubic spline interpolation and superimpose I (n) on it. d. You should observe that the resultant reconstruction in each case has the correct frequency but a different amplitude. Explain this observation. Comment on the role of phase of s (1) on sampling and reconstruction of signals. .
+
Problems
79
a
80
THEz- TRANSFORM
In Chapter 3 we studied the discretetime Fourier transform approach for representing discrete signals using complex exponential sequences. T i hs representation clearly has advantages for LTI systems because it describes systems in the frequency domain using the frequency response function H(.i"). The computation of the sinusoidalsteady-state response is greatly facilitated by the use of El(&"). Furthermore, response to any arbitrary absolutely summable sequence z(n) can easily be computed in the fre quency domain by multiplying the transform X(ej") and the frequency response El(&"). However, there are two shortcomings to the Fourier transform approach. First, there are many useful signals in p r a c t i c e such as u(n) and nu(n)-for which the discretetime Fourier transform does not exist. Second, the transient response of a system due to initial conditions or due to changing inputs cannot be computed using the discretetime Fourier transform approach. Therefore we now consider an extension of the discrete-time Fourier transform to address the above two problems. T i extension is called the hs z-transform. Its bilateral (or two-sided) version provides another domain in which a larger class of sequences and systems can be analyzed, while its unilateral (or one-sided) version can be used to obtain system responses with initial conditions or changing inputs.
THE BILATERAL Z-TRANSFORM
I
I
The z-transform of a sequence z(n) is given by
X ( z ) 5 2 [z(n)] =
n=-m
c
m
2(n)Z-"
(44
where z is a complex variable. The set of z values for which X ( z ) exists
is called the region of convergence (ROC) and is given by
&- < ( z ( < Rz+
for some positive numbers Rz- and &+. The inverse z-transform of a complex function X ( z ) is given by
(44
where C is a counterclockwise contour encircling the origin and lying in the ROC.
Comments:
1. The complex variable z is called the wmplez fquency given by z = IzI ej", where 1zI is the attenuation and w is the real frequency. 2. Since the ROC (4.2) is defined in terms of the magnitude lzl, the
shape of the ROC is an open ring as shown in Figure 4.1. Note that &may be equal to zero and/or &+ could possibly be 00. 3. If Rz+ < R=-, then the ROC is a null space and the z-transform does not exist. 4. The function Jzl= 1 (or z = ej") is a circle of unit radius in the z-plane and is called the unit circle. If the ROC contains the unit circle, then we can evaluate X ( z ) on the unit circle.
~ ( z ) l , , , ,= ~ ( e i "= ~ )
C
n=-m
m
z(n)e-jw = ~ [ z ( n ) l
Therefore the discretetime Fourier transform X(ej") may be viewed as a special case of the z-transform X ( z ) .
0
EXAMPLE 4.1
Note: XI(Z) this example is a rational function; that is, in
A B(2) 2 X,(z) = -= __
A(z)
z-a
where B ( z ) = z is the numemtorpolynomiol and A(z) = z-a is the denominator polynomaaL The roots of B(z) are called the terns of X ( z ) , while the roots of A(z) are called the poles of X ( z ) . In this example XI(Z) a zero at the origin has z = 0 and a pole at z = a. Hence 51( n ) can also be represented by a pole-sen, diagmm in the z-plane in which zeros are denoted by 'a' and poles by ' x ' as shown in Figure 4.2. 0
0
EXAMPLE 4.2
k t zz(n)= -b"u(-nsequence.) Then
&(z)
l), 0 < Ibl
-1
< 00.
(This sequence is called a negative-tame
c = - 2(2)" = - c (i)" - (t)" 1
=b"z-"
--m m
-m m
=
1
The ROC, and the pole-zero plot for this zz(n) are shown in Figure 4.3.
FIGURE 4.3
The ROC a Ezample 4.2 n
82
Chapter 4
THE %-TRANSFORM
Note: If b = a i this example, then Xz(z) XI(Z) n = except for their respective ROCs; that is, ROC1 # ROCz. This implies that the ROC is a distinguishing feature that guarantees the uniqueness of the r-transform. Hence it plays a very
important role in system analysis.
0
EXAMPLE 4.3
< Jal,the Roc3 is a null space and &(2) does not exist. If la1 < lbl, then the Roc3 is la < 1 1 < Ibl and X ~ ( Zexists in this region as shown in 1 2 ) Figure 4 4 .. 0
If Ibl
PROPERTIES
OF THE ROC
From the observation of the ROCs in the above three examples, we state the following properties.
1. The ROC is always bounded by a circle since the convergence condition is on the magnitude 121. 2. The sequence zl(n) = a"u(n) in Example 4.1 is a special case of a right-sided sequence, defined as a sequence z(n) that is zero for some n < no. From Example 4 1 the ROC for right-sided sequences is always . outside of a circle of radius &-. If no 2 0, then the right-sided sequence is also called a causal sequence. 3. The sequence ~ ( n= -b"u(-n - 1) in Example 4.2 is a special ) case of a left-sided sequence, defined 89 a sequence z(n)that is zero for some n > no. If 5 0, the resulting sequence is called an anticausal sequence. From Example 4.2 the ROC for left-sided sequences is always inside of a circle of radius
a+.
FIGURE 4.4
The ROC in Ezample 4.3
The Bilateral z-Transform
03
4. The sequence ~ ( nin Example 4.3 is a two-sided sequence. The ) ROC for two-sided sequences is always an open ring &- <1.1 < &+ if it exists. 5. The sequences that are zero for n < nl and n > n2 are called finite-duration sequences. The ROC for such sequences is the entire zplane. If nl < 0, then z = oo is not in the ROC. If n2 > 0, then z = 0 is not in the ROC. 6. The ROC cannot include a pole since X ( z ) converges uniformly in there. 7. There is at least one pole on the boundary of a ROC of a rational
X(4.
8. The ROC is one contiguous region; that is, the ROC does not come in pieces.
In digital signal processing, signals are assumed to be causal since almost every digital data is acquired in real time. Therefore the only ROC of interest to us is the one given in 2 above.
IMPORTANT PROPERTIES OF THE Z-TRANSFORM
I I
The properties of the z-transform are generalizations of the properties of the discrete-time Fourier transform that we studied in Chapter 3. We state the following important properties of the z-transform without proof.
This last property transforms the timedomain convolution operation into a multiplication between two functions. It is a significant property and are in many ways. First, if X~(Z) X ~ ( Z ) two polynomials, then their product can be implemented using the conv function in MATLAB.
0
U(AMPlE4.4
SdUtiOn
From the definition of the z-transform we observe that
zl(n) = {2,3,4} I
and
zz(n)= {3,4,5,6}
t
Then the convolution of the above two sequences will give the coefficients of the required polynomial product.
>> xl = [2,3,41; 12 = [3,4,5,61; >> x3 conv(x1,xZ) x3 = 6 IT 34 43
-
38
24
Hence
&(z) = 6
+ 17z-'+
3 4 ~ - 4 3 ~ - 3&-' ~ ~
+
+
+2 4 ~ - ~
Using the convat function developed in Chapter 2, we can also multiply 0 two z-domain polynomials corresponding to n o n c a d sequences.
0
U(AMPlE4.5
In passing we note that to divide one polynomial by another one, we would require an inverse operation called deconuolution [19, Chapter 61. In MATLAB[p,rl = deconv(b.a) computes the result of dividing b by a in a polynomial part p and a remainder r. For example, if we divide the polynomial X&) in Example 4.4 by X , ( z ) ,
>> 13
= 16,17,34,43,38,241; X l = C2,3,41; >> [x2,rl = deconv(x3,xl) I2 = 3 4 5 6 r =
0
0
0
0
0
0
then we obtain the coefficients of the polynomial X ~ ( Zas expected. To ) obtain the sample index, we wl have to modify the deconv function il as we did in the conv_m function. This is explored in Problem 4.8. This operation is useful in obtaining a pmper rational part from an improper rational function. The second important use of the convolution property is in system hs output computations as we shall see in a later section. T i interpretation is particularly useful for verifyii the z-transform expression X ( z ) using MATLAB. Note that since MATLAB a numerical processor (unless is the Symbolic toolbox is used), it cannot be used for direct z-transform calculations. We will now elaborate on this. Let z(n)be a sequence with a rational transform
86
Chapter 4 w THE &TRANSFORM
where B ( z ) and A(z) are polynomials in z-l. If we use the coefficients of B(z) and A(z) as the b and a arrays in the filter routine and excite this filter by the impulse sequence 6(n), then from (4.11) and using 2 [S(n)] = 1, the output of the filter will be ~ ( n (.h sis a numerical approach of )T i computing the inverse z-transform; we will discuss the analytical approach in the next section.) We can compare this output with the given ~ ( nto ) verify that X ( z ) is indeed the transform of ~ ( n This is illustrated in ). Example 4.6.
SOME COMMON Z-TRANSFORM PA1RS
0
EXAMPLE4.6
Using the definition of z-transform and its properties, one can determine z-transforms of common sequences. A list of some of these sequences is given in Table 4.1.
Using z-transform properties and the z-transform table, determine the ztransform of
s(n) = (n
INVERSION OF THE Z-TRANSFORM
From definition (4.3) the inverse z-transform computation requires an evaluation of a complex contour integral that, in general, is a complicated procedure. The most practical approach is to use the partial fraction expansion method. It makes use of the a-transform Table 4.1 (or similar tables available in many textbooks.) The z-transform, however, must be a rational function. This requirement is generally satisfied in digital signal processing.
Centml Idea: When X(z) is a rational function of t-', it can be expressed as a sum of simple (first-order) factors using the partial fraction expansion. The individual sequences corresponding to these factors can then be written down using the t-transform table. The inverse z-transform procedure can be summarized as follows: Method:
Given
0
express it as
Proper rational part
polynomial part
If
M>N
where the fist term on the right-hand side is the proper rational part hs and the second term is the polynomial (finitelength) part. T i can be obtained by performing polynomial division if M 2 N using the deconv function. 0 perform a partial fraction expansion on the proper rational part of X ( z ) to obtain
lnvesion of the %-Transform
89
where p k is the kth pole of x(z) and R k is the residue at p k . It is assumed that the poles are distinct for which the residues are given by
pk
For repeated poles the expansion (4.13)has a more general form. If a pole has multiplicity T , then its expansion is given by
(4.14)
where the residues R k , ( are computed using a more general formula, which is available in [19]. 0 write z(n)as
MIN
finally, use the relation from Table 4.1
2-1
[I] = PiU(n)
z-Pk -p:U(-n-1)
{
(Zkl IZkI
5 42 &+
(4.15)
to complete ~ ( n ) .
0
EXAMPLE 4.7
Find the inverse z-transform of z(z) =
Write
32'
- 4~ + 1'
32
;,-1+
1 -1
sduti
X(2) =
3(z2 - $ z +
4) - 1 -
-
1,2
or
NOW, ( z ) has two poles: 21 = 1 and zs = 4; and since the ROC is not specified, X there are three possible ROCs as shown in Figure 4.5.
90
Chapter 4
which is a left-sided sequence. c. ROCs: f <1.1 < 1 Here pole ZI is on the exterior side of the Roc3. that is, Izl( R,+ = 1-while pole zz is on the interior side-that is, (22J5 f . 2 Hence from (4.15)
23(n)= --u(-n 2
1
- 1 ) - - (-) 2 3
1 1 "
u(n)
0
which is a tw-sided sequence.
MATLAB IMPLEMENTATION
A MATLABfunction residue2 is available to compute the residue part and the direct (or polynomial) terms of a rational function in 2-l. Let
... b M K M B (2) X ( 2 ) = 4 biz-' uo u ~ z - ' .. aNzPN = A (2)
+ +
+ + +. +
be a rational function in which the numerator and the denominator polynomials are in meending powers of z-'. Then [R,p,C]=residuez(b,a)
Inversion of the &Transform
91
finds the residues, poles, and direct terms of X ( z ) in which two polynomials B ( z ) and A ( z ) are given in two vectors b and a, respectively. The returned column vector R contains the residues, column vector p contains the pole locations, and row vector C contains the direct terms. If p(k)=. . .=p(k+r-1) is a pole of multiplicity r, then the expansion includes the term of the form
which is different from (4.14). Similarly, [b ,a] =residuez(R,p ,C), with three input arguments and two output arguments, converts the partial fraction expansion back to polynomials with coefficients in row vectors b and a.
0
UAMPLE 4.8
To check our residue functions,let us consider the rational function
X(2)=
3Z2
- 42 + 1
*
given in Example 4.7.
First rearrange X ( z ) so that it is a function in ascending powers of z-'.
We can evaluate the denominator polynomial as well a the residues using MATs
LAB.
>> b
a =
-
I; a = ply(~0.9,0.9,-0.91)
-0.9000 -0.8100 0.7290
1.0000
>> [R .p, Cl =residuez(b, a) R;.
0.2500 0.5000
0.2500
P '
0.9000 0.9000 -0.9000
c -
c1
Note that the denominator polynomial is computed using MATLAB'S polynomial function p l y , which computes the polynomial coefficients, given its roots. We could have used the conv function, but the use of the poly function is more convenient for this purpose. From the residue calculations and using the order of residues given in (4.16), we have
X(%) =
0.25
~
+
0.25
Inversion of the z-Transform
93
Hence from Table 4.1 and using the z-transform property of time-shift,
Similar to the frequency response function H(ei"), we can define the z-domain function, H ( z ) , called the system function. However, unlike H(ej"), H ( z ) exists for systems that may not be BIB0 stable.
DEFlNlTlON 1
The System h n c t i o n The system junction H ( z ) is given by
-m
Using the convolutionproperty (4.11) of the z-transform, the output transform Y ( z )is given by
Y ( z )= H ( z ) X ( Z )
System Repesentation in the &Domain
:
ROC, = ROCh n ROC,
(4.18)
95
provided ROC, overlaps with Roc),. Therefore a linear and time invariant system can be represented in the z-domain by
the system function H ( z ) can easily be computed. Taking the z-transform of both sides, and using properties of the z-transform,
k=l
e=o
or
(4.20)
After factorization, we obtain
(4.21)
k=l
where Ze'S are the system zeros and Pk's are the system poles. Thus H ( z ) (and hence an LTI system) can a s be represented in the z-domain using lo a pole-zero plot. This fact is useful in designing simple filters by proper placement of poles and zeros. To determine zeros and poles of a rational H(z), we can use the MATLABfunction roots on both the numerator and the denominator polynomials. (Its inverse function poly determines polynomial coefficients from its roots as we discussed in the previous section.) It is also possible to to use MATLAB plot these roots for a visual display of a pole-zero plot. The function zplane (b,a) plots poles and zeros, given the numerator mw vector b and the denominator mw vector a. As before, the symbol "o"
96
Chapter 4
THE %.TRANSFORM
represents a zero and the symbol "x" represents a pole. The plot includes the unit circle for reference. Similarly, zplane(z,p) plots the zeros in column vector z and the poles in column vector p. Note very carefully the form of the input arguments for the proper use of this function. TRANSFER FUNCTION REPRESENTATION
If the ROC of H ( z ) includes a unit circle ( z = ej"), then we.can evaluate H ( z ) on the unit circle, resulting in a frequency response function or transfer function H(ej"). Then from (4.21)
M
H(g.U) I g ( N - M ) U 4
n(ej"
1
- zi)
(4.22)
n ( e j w- p k )
1
N
The factor (dw 21) can be interpreted as a vector in the complex z-plane r from a zero z to the unit circle at z = ej", while the factor (ej" - p k ) can be interpreted as a vector from a pole p k to the unit circle at z = d". This is shown in Figure 4.6. Hence the magnitude response function
(4.23)
can be interpreted as a product of the lengths of vectors from zeros to the unit circle divided by the lengths of vectors from poles to the unit circle and scaled by (41.Similarly, the phase response function
~ ~ ( e i ~ ) = [ ~[o N - M ) ~ ] (r ~ ]
--1 Constant linear
+
+ C L ( ~ -~ ~ (Z ~~ p) k ) C - g 1
I
M
N
nonlinear
(4.24)
Re(d
circle
FIGURE 4.6
Pole and zero vectors
97
System Representation in the z-Domain
can be interpreted as a sum of a constant factor, a linear-phase factor, and a nonlinear-phase factor (angles from the "zero vectors" minus the s u m of angles from the ''pole vectors").
MATLAB IMPLEMENTATION
In Chapter 3 we plotted magnitude and phase responses in MATLAB by directly implementing their functional forms. MATLABalso provides a function called f reqz for this computation, which use8 the interpretation given above. In its simplest form this function is invoked by
CH,wl = freqz(b,a,N)
which returns the N-point frequency vector u and the N-point complex frequency response vector H of the system, given its numerator and denominator coefficients in vectors b and a. The frequency response is evaluated at N points equally spaced around the upper half of the unit circle. Note that the b and a vectors are the same vectors we use in the filter func41) tion or derived from the difference equation representation ( . 9 . The second form
CH,vl
-
freqz(b,a,N,'whole')
uses N points around the whole unit circle for computation. In yet another form
H = freqz(b,a,w)
it returns the frequency response at frequencies designated in vector u, normally between 0 and 7c.
0
EXAMPLE 4.11
The difference equation can be put in the form
y(n) - 0.9y(n - 1) = z(n) a Fkom (4.21) .
H(z)= '
1 1 - 0,gz-1
14 .
I
'0.9
since the system is causal. There is one pole at 0.9 and one zero at the origin. W will use MATLAB illustrate the use of the zplane function. e to
98
Chapter 4 0 THE &TRANSFORM
>> b 11, 03; a = Cf, -0.91; >> zplaue(b,a)
Note that we specified b=[l,O] instead of b=l because the zplane function assumes that scalars are zerw or poles. The resulting pole-zero plot is shown in Figure 4.7. b. Using (4.23) and (4.24), we can determine the magnitude and phase of H ( 2 " ) . Once again we will use MATLAB illustrate the use of the freqz to function. Using its first form, we will take 100 points along the upper half of the unit circle.
-
>> [H,vl = freqz(b.a.100); >> magH = abs(8); phaH = angle(H); >> subplot(2,i,l);plot(w/pi,mag8);grid >> xlabel('frequency in pi units'); ylabel('Magnitude'); >> title(*bgnitude Response') >> subplot(2,1,2);plot(v/pi,ph~/pi) ;grid >> xlabel('frequency in pi units'); ylabel('Phase in pi units'); >> title('Phase Response')
The response plots are shown in Figure 4.8. If you study these plots carefully, you will observe that the plots are computed between 0 5 w 5 0.9% and fall short at w = ?r. T i is due to the fact that in MATLAB lower half of the hs the
Pole-Zero Plot
0.8
0.4
0
0
0.9
.-
-1
-0.5
0
0.5
1
RBal part
FIGURE 4.7
Pole-zero plot of Ezample 4.11a
Srjtem Representation in the 2-Domain
99
MagmtucbResponse
3
lo
1 15 '0
c
01
0.2
03
frequency5n p unlts 04 5I 0 0.6
0.7
0.8
0.9
5 1
f -0.3
-0.4
fi
0
0.1
02
03 .
04 0.5 0.6 frequency in pi units
07
0.8
09
1
FIGURE 4.8
Frequency response plots tn Ezample 4.11
unit circle begins at w = K. To overcome this problem, we will use the second form of the freqz function as follows.
Now the lOlst element of the array H will correspond to w = ?r. A similar result can be obtained using the third form of the freqz function.
In the future we will use any one of these forms, depending on our convenience. Also note that in the plots we divided the Y and phaH arrays by pi so that the plot axes are in the units of ?r and easier to read. T i practice is strongly hs recommended. c. From the z-transform Table 4.1
The poles of the system function are at z = O.9L rt r / 3 . Hence the ROC of the above causal system is1.1 > 0.9. Therefore the unit circle is in the ROC,and the discret&ime Fourier transform H(e'") exists.
RELATIONSHIPS In this and the previous two chapters we developed several system repre BETWEEN sentstions. Figure 4.9 depicts the relationships between these representaSYSTEM tions in a graphical form. REPRESENTATIONS STABILITY AND CAUSALITY
For LTI systems the BIB0 stability is equivalent to Crm Jh(k)l < M. From the existence of the discrete-time Fourier transform this stability implies that H(ejw)exists, which further implies that the unit circle1.1 = 1 must be in the ROC of H(z). This rasult is called the z-domain stability
Express H(z)in z-'
FIGURE 4.9 System representations an pictorial fown
102
Chapter 4
THE z-TRANSFORM
theorem; therefore the dashed paths in Figure 4.9 exist only if the system is stable.
THEOREM 2
z-Domain LTI Stability An LTI system is stable i f and only if the unit circfe is in the ROC of H G ) .
For LTI causality we require that h(n) = 0, for n < 0 (i.e., a rightsided sequence). T i implies that the ROC of H ( z ) must be outside of hs some circle of radius Rh-. T i is not a sufficient condition since any hs right-sided sequence has a similar ROC. However, when the system is stable, then its causality is easy to check.
THEOREM 3
z-Domain Causal LTI Stability A causal LTI system is stable i f and only if the system function H ( z ) has all its poles inside the unit circle.
A causal LTI system is described by the following difference equation:
y(n) = 0.81y(n - 2)
Since the system is causal,the ROC will be outside of a circle with radius equal to the largest pole magnitude.
a. %king the r-transform of both sides of the difference equation and then solving for Y(z)/X(r) or using (4.20), we obtain
SOLUTIONS OF THE DIFFERENCE EQUATIONS
In Chapter 2 we mentioned two forms for the solution of linear constant coefficient difference equations. One form involved finding the particular and the homogeneous solutions, while the other form involved finding the zero-input (initial condition) and the zero-state responses. Using ztransforms, we now provide a method for obtaining these forms. In addition, we w l also discuss the transient and the steady-state responses. il In digital signal processing difference equations generally evolve in the il positive n direction. Therefore our time frame for these solutions w l be n 2 0. For this purpose we define a version of the bilateral z-transform called the one-sided 2-tmnsfonn.
W
DEFINITION4
The One-sided x 3)rmsfofin The one-sided z-transform of a sequence x ( n ) is given by
2+ [2(n)] 2 [x(n)u(n)] x+[%I = 6 b
Cz(n)P
n=O
W
(4.25)
SdutMns of the Difference Equations
105
Then the sample shifting property is given by
z+[s(n - k)] = 2 [s(n- k)u(n)]
m
m
Or
Z+(~(71- k)] = 4-1)~'-~++2(-2)~~-~+...+z(-k)+z-~X+(z) (4.26)
This result can now be used to solve difference equations with nonzero initial conditions or with changing inputs. We want to solve the difference
equation
N
k=1
M
m=O
l+Caky(n-k)= XbmZ(71-m), n 2 0
subject to these initial conditions:
{y(i),i = -1,. . . ,-N} and
{ ~ ( i ) , i -1,. =
.., -M}.
We now demonstrate this solution using an example.
y(n) - y ( n
3
1 - 1) + p ( n - 2) = ~ ( n )n, 2 0
where
subject to y(-1) = 4 and y(-2) = 10.
Solution
Taking the onesided z-transform of both sides of the difference equation, we obtain
Y+(%)-5 [y(-1)
3
1 + o-'Y+(z)]+z[Y(-2) + z-lY(-I)
1 +z-2Y+(z)]= i=p
Substituting the initial conditions and rearranging,
106
Chapter 4
U
THE %-TRANSFORM
or
Finally,
Y() +.
=
2(1
+ i2-2
(I - 2-1) (1 - $2-1)
- 42-1)
Using the partial fraction expansion, we obtain
After inverse transformation the solution is
(4.29)
Fornrs of the Solutions The above solution is the complete response of the Uerence equation. It can be exprgsed in several forms.
0
Homogeneous and particular parts:
homogeneous part
particular part
The homogeneous part is due to the system poles and the particular part is due to the input poles.
rn Dansient and steady-state responses:
. ( . , = [ ~ ( ~ ) " + ( ~ ) " ] . +. , (
.
,w 2
w
Y
,
transient reaponse
steady-state response
The transient response is due to poles that are inside the unit circle, while the steady-state response is due to poles that are on the unit circle. Note that when the poles are outside the unit circle, the response is termed an unbounded response. 0 Zero-input (or initial condition) and zero-state responses: In equation (4.27) Y+(z) has two parts. The first part can be interpreted as
YZS(Z)= H(z)X(z)
Solutions of the Difference Equations
107
while the second part as
Yzr(z)= H ( z ) X r c ( z )
where X r c ( z ) can be thought of as an equivalent initial-condition input that generates the same output Yzr as generated by the initial conditions. In this example zrc(n) is
zrc(n) = (1, -2)
t
Now taking the inverse z-transform of each part of (4.27), we write the complete response as y(n)
=[; (;)n
\
-2
(f)n + ;]
u(n)
+ [3
\
(i)
- 21 4) .
,
zerc-state response
zeroinput response
From this example it is clear that each part of the complete solution is, in general, a different function and emphasizes a different aspect of system analysis.
MATLAB IMPLEMENTATION
In Chapter 2 we used the f i l t e r function to solve the difference equation, hs given its coefficients and an input. T i function can also be used to find the complete response when initial conditions are given. In this form the f i l t e r function is invoked by
J
which agrees with the response given in (4.29). In Example 4.14 we computed .rc(n) analytically. However, in practice, and especially for large order difference equations, it is tedious t o determine xrc(n) analytically. MATLAB provides a function called f i l t i c , which is available only in the Signal Processing toolbox (version 2.0b or later). It is invoked by
xic = filtic(b,a,Y,X)
in which b and a are the filter coefficient arrays and Y and X are the initialrespectively, condition arrays from the initial conditions on y(n) and ~ ( n ) , in the form y = [Y(-l), Y ( 4 , '., Id"
x = [.(-l),
If z(n)= 0,
2(-2),
..., .(-M)j
n 5 -1 then X need not be specified in the f i l t i c function. In Example 4.i4 we could have used
> Y . [4, 101; >> xic = filtic(b,a.Y) Y =
1
-2
The first two terms of y(n) correspond to the steady-state response, as well as to the particular response, while the last two terms are the transient response (and homogeneous response) terms. To solve this example using MATLAB, will need the filtic function, we ) which we have already used to determine the Z J C ( ~sequence. The solution will be a numerical one. Let us determine the first 8 samples o y n . f ()
a. Prove the above result by substituting the definition of convolution in the left-hand side. b. Prove the above result using the convolution property. by z c. Verify the above result using MATLAB choosing any two sequences 11 (n) and z (n).
P . 4 The out,put sequence y (n)in Problem 4.13 is the total response. For each of the systems 41 given in that problem, separate y (n)into (i) the homogeneous part, (ii) the particular part, (iii) the transient response, and (iv) the steady-state response. P . 5 A stable system has the following pole-zero locations: 41
Z1
=j ,
z2
= -j,
1 1 = -- +i-2'
P 2 = - 51 3 2 .1 -
It is also known that the frequency response function H (2") evaluated at w = 0 is equal t o 0.8; that is,
A digital filter is described by the difference equation
y(n) = z(n)+z(n - 1)
+ 0.9y(n - 1) - 0.81y(n - 2)
+
a. Using the freqz function, plot the magnitude and phase of the frequency response of the above filter. Note the magnitude and phase at w = ?r /3 and at w = ?r. b. Generate 200 samples of the signal z (n)= sin (7rn/3) 5 cos (m)and process through the filter. Compare the steady-state portion of the output to z (n).How are the amplitudes and phases of two sinusoids affected by the filter? P4.17 Solve the following difference equation for y (n)using the one-sided z-transform approach.
P . 9 A causal, linear, and timeinvariant system is given by the following difference equation: 41
y(n) = y(n- 1) +y(n
- 2) +z(n-
1)
a. Find the system function H ( r ) for this system. b. Plot the poles and zerm o H ( z ) and indicate the region of convergence (ROC). f c. Find the unit sample response h(n) of this system. d. Is this system stable? If the answer is yes,justify it. If the answer is no, find a stable unit sample response that satisfies the difference equation. P4.40 Determine the zero-state response of the system
y(n) = $ y ( n - 1) + % ( T I ) +31(n - 1),
n 2 0 ; y(-1) = 2
to the input
I (n)= ,ixn''u
(n)
What is the steady-state response of the system?
Problems
115
t
116
THEDISCRETE
FOURIER
TRANSFORM
In Chapters 3 and 4 we studied transform-domain representations of discrete signals. The discretetime Fourier transform provided the frequency-domain (w) representation for absolutely summable sequences. The r-transform provided a generalized frequency-domain ( z ) representation for arbitrary sequences. These transforms have two features in common. First, the transforms are defined for infinite-length sequences. Second, and the most important, they are functions of continuous variables (w or z ) . From the numerical computation viewpoint (or from MATLAB'S viewpoint), these two features are troublesome because one has to evaluate infinite sums at uncountably infinite frequencies. To use MATLAB, have to truncate sequences and then evaluate the expressions we at finitely many points. This is what we did in many examples in the two previous chapters. The evaluations were obviously approximations to the exact calculations. In other words, the discretetime Fourier transform and the z-transform are not numerically computable transforms. Therefore we turn our attention to a numerically computable transform. It is obtained by sampling the discretetime Fourier transform in the frequency domain (or the z-transform on the unit circle). We develop this transform by first analyzing periodic sequences. From Fourier analysis we know that a periodic function (or sequence) can always be represented by a linear combination of harmonically related complex exponentials (which is a form of sampling). This gives us the Discrete Fourier Series (or DFS) representation. Since the sampling is in the frequency domain, we study the effects of sampling in the time domain and the issue of reconstruction in the z-domain. We then extend the DFS to finite-duration sequences, which leads to a new transform, called the Discrete Fourier %nsform (or DFT). The DFT avoids the two problems mentioned above and is
a numerically computable transform that is suitable for computer implementation. We study its properties and its use in system analysis in detail. The numerical computation of the DFT for long sequences is prohibitively time consuming. Therefore several algorithms have been developed to efficiently compute the DFT. These are collectively called fast Fourier trans form (or FFT) algorithms. We will study two such algorithms in detail.
THE DISCRETE FOURIER SERIES
I
I
In Chapter 2 we defined the periodic sequence by Z(n), satisfying the condition
%(n)= Z(n
+kN),
Vn, k
(5.1)
where N is the fundamental period of the sequence. From Fourier analysis we know that the periodic functions c n be synthesized as a linear coma bination of complex exponentials whose frequencies are multiples (or harmonics) of the fundamental frequency (which in our case is 27r/N). From the frequencydomain periodicity of the discrete-time Fourier transform, we conclude that there are a b i t e number of harmonics; the frequencies are { $k, k = 0,1,. ..,N - 1). Therefore a periodic sequence %(n) can be expressed a s
(5.2)
k=O
. where { X ( k ) , k = 0,fl,..,} are called the discrete Fourier series c e efficients, w i h are given by hc
N-1
X(k)=
n=O
c
%(n)e-j@nk, k = O , f l ,
...,
(5.3)
Note that X ( k ) is itself a (complex-valued) periodic sequence with fundamental period equal to N, that is,
*(k
+ N)= X ( k )
(5.4)
The pair of equations (5.3) and (5.2) taken together is called the discrete Fourier series representation of periodic sequences. Using W ~f e-j@ to J
The Discrete Fourier Series
A careful look at (5.5) reveals that the DFS is a numerically computable represent.tion. It can be implemented in many ways. To compute each sample X ( k ) , we can implement the summation BS a f o r . . .end loop. To compute all DFS coefficients would require another f o r . . .end loop. T i will result in a nested two for. . .end loop implementation. T i hs hs is clearly inefficient in MATLAB. efficient implementation in MATLAB An
118
Chapter 5 m THE DISCRETE FOURIER TRANSFORM
would be to use a matrix-vector multiplication for each of the relations in (5.5). We have used this approach earlier in implementing a numerical and X denote approximation to the discretetime Fourier transform. Let i column vectors corresponding to the primary periods of sequences z(n) and X ( k ) , respectively. Then (5.5) is given by
Caution: The above functions are efficient approaches of implementing (5.5) in MATLAB.They are not computationally efficient, especially for large N . We will deal with this problem later in this chapter.
0
The plots for the above and a l other cases are shown in Figure 5.2. Note that l since R ( k ) is periodic, the plots are shown from - N / 2 t o N / 2 . c. Several interesting observations can be made from plots in Figure 5.2. The envelopes of the DFS coefficients of square waves look like "sine" functions. s The ampIitude at k = 0 i equal to L , while the zeros of the functions are at multiples of N I L , which is the reciprocal of the duty cycle. We will study t h e e n functions later in this chapter.
RELATION
Let x(n) be a b i t e d u r a t i o n sequence of duration N such that
TO THE ZTRANSFORM
x(n) = {NOT,
O<nlN-l
elsewhere
(5.8)
The Discrete Fourier Series
121
DFS of SQ. wave: L S , N=20
I
DFS of SQ. wave: L=5.N=40
I
I
-10
-5
0
k
5
10
-20
-10
0
k
10
20
DFS of SQ. wave: L S . N=60
DFS of SQ. wave: L=7, NaO
-20
0
20
-20
k
FIGURE 5.2
0 k
20
The DFS plots of a periodic square wave for various L and N .
Then we can find its z-transform:
N-1
X(z)=
c
n=O
x(71)t-n
(5.9)
Now we construct a periodic sequence Z(n)by periodically repeating x(n) with period N, that is,
Since ~ ( nin (5.8) is of finite duration of length N, i t is also absolutely ) summable. Hence its DTFT exists and is given by
Comparing (5.13) with (5.11), we have
(5.14)
then the DFS X ( k ) = X ( d w * )= X(eJkwl), which means that the DFS is obtained by evenly sampling the DTFT at w1 = % intervals. From (5.12) and (5.14) we observe that the DFS representation gives us a sampling mechanism in the frequency domain which, in principle, is similar t o sampling in the time domain. The interval w1 = is the sampling interval in the frequency domain. It is also called the frequency resolution because it tells us how close are the frequency samples (or measurements).
The sequence z(n) is not periodic but is of finite duration.
a. The discretetime Fourier transform is given by
x(ej-1 =
C z(n)e-jwn =
m n=-m
,-jw
+ 2e-j2-
+ 3,-j3w
The Discrete Fourier Series
123
SAMPLING A N D RECONSTRUCTION IN T H E Z-DOMAIN
I
I
Let z(n) be an arbitrary absolutely summable sequence, which may be of infinite duration. Its 2-transform is given by
X(z)=
2
m=-m
z(m)t-m
and we m u m e that the ROC of X (2) includes the unit circle. We sample X ( z ) on the unit circle at equispaced points separated in angle by w = i 2 r / N and call it a DFS sequence,
X ( k ) 22 x ( z ) l l , , , ~ L , k = O , i l , * 2 , . ..
=
(5.15)
m=-m
2
z(m)e-3+Pm=
2
m=-m
z(m)w$m
which is periodic with period N. Finally, we compute the IDFS of z ( k ) ,
Z(n)= IDFS [ X ( k ) ]
which is also periodic with period N.Clearly, there must be a relationship hs between the arbitrary z(n)and the periodic Z(n). T i is an important issue. In order to compute the inverse DTFT or the inverse z-transform numerically, we must deal with a finite number of samples of X ( z ) around the unit circle. Therefore we must know the effect of such sampling on hs the timedomain sequence. T i relationship is easy to obtain. (from (5.2))
k=O
N-1
which means that when we sample X(z) the unit circle, we obtain a on periodic sequence in the time domain. This sequence is a linear combination of the original z ( n ) and its infinite replicas, each shifted by multiples of fN.This is illustrated in Example 5.5. From (5.16) we observe that if z ( n ) = 0 for n < 0 and n 2 N , then there will be no overlap or aliasing in the time domain. Hence we should be able to recognize and recover z(n) from Z ( n ) , that is,
z ( n ) = Z(n) for 0 5 n
5 ( N - 1)
or
z ( . )
= Z(la)'RN(.) = z(n)
1, O S n 5 N - l 0 , else
where 7?,~(n) called a rectangular window of length N . Therefore we is have the following theorem.
and the inverse DFS computation to determine the corresponding time-domain script for N = 5 is shown below. sequence. The MATLAB
n
N=40
T-----l
0
10
20
30
40
n
n
FIGURE 5.3
Plots in Ezample 5.5
126
Chapter 5 8 THE DISCRETE FOURIER TRANSFORM
is sufficiently small to result in any appreciable amount of aliasing in practice. Such information is USeN in effectively truncating an infiniteduration sequence prior to taking its transform. 0
RECONSTRUCTION FORMULA
0N Let t(n) be time-limited to 1 , - 1).Then from Theorem-1 we should be able to recover the z-transform X ( z ) using its samples X ( k ) . This is given by
x(z)=(2 [ I). .
=z[z(n)xN(n)]
= Z[IDFS{
v
x(k)
}RN(~)]
samples of X ( z )
The above approach results in the z-domain reconstruction formula.
N-1
X ( 2 )=
c
0
N-1
5(n)t--"=
c
0
i(n)z-n
Since wikN 1, we have =
(5.17)
THE DTFT INTERPOLATION FORMULA
The reconstruction formula (5.17) can be specialized for the discretetime Fourier transform by evaluating it on the unit circle z = e j w . Then
Sampling and Reconstruction in the &Domain
127
Consider
Let
@(w) = 2 e
A
sin(&)
N sin(%)
_ . N-1 3 w (: an ) - r interpolating polynomial
(5.18)
Then
N-1
X ( e i W )=
k=O
X(k)@ w - 2) ; k
(
(5.19)
This is the DTFT interpolation formula to reconstruct X ( e y ) from its samples X (k). Since @(O) = 1, we have that X(eJ*nk/N)= X(k), which means that the interpolation is exact at sampling points. Recall the time domain interpolation formula (3.33) for analog signals:
z,(t) =
C
n=-m
m
z(n)sinc [F,(t - n ~ ~ ) ]
(5.20)
The DTFT interpolating formula (5.19) looks similar. However, there are some dfierences. First, the timedomain formula (5.20) reconstructs an arbitrary nonperiodic analog signal, while the frequency-domain formula (5.19) gives us a periodic waveform. Second, in (5.19) we use a interpolation function instead of our more familiar (sinc) function. Therefore the @(w) function is sometimes called a digital sinc function, which itself is periodic. This is the function we observed in Example 5.2.
%
MATLAB IMPLEMENTATION
The interpolation formula (5.19) suffers the same fate as that of (5.20) while trying to implement it in practice. One has to generate several interpolating polynomials (5.18) and perform their linear combinations from its computed to obtainJhe discretetime Fourier transform X ( d w ) we samples X ( k ) . Furthermore, in MATLAB have to evaluate (5.19) on a finer grid over 0 5 w 5 27r. This is clearly an inefficient approach. Another approach is to use the cubic spline interpolation function as an efficient approximation to (5.19). This is what we did to implement (5.20) in Chapter 3. However, there is an alternate and efficient approach based on the DFT, which we will study in the next section.
128
Chanter 5
THE DISCRETE FOURIER TRANSFORM
THE DISCRETE FOURIER TRANSFORM
The discrete Fourier series provided us a mechanism for numerically computing the discretetime Fourier transform. It also alerted us to a potential problem of aliasing in the time domain. Mathematics dictates that the sampling of the discretetime Fourier transform result in a periodic sequence 2(n).But most of the signals in practice are not periodic. They are likely to be of finite duration. How can we develop a numerically computable Fourier representation for such signals? Theoretically, we can take care of this problem by defining a periodic signal whose primary shape is that of the finiteduration signal and then using the DFS on this periodic signal. Practically, we define a new transform called the Discrete Fourier rrcmSfom (DFT), which is the primary period of the DFS. This DFT is the ultimate numerically computable Fourier transform for arbitrary finiteduration sequences. First we define a finite-duration sequence z(n) that has N samples over 0 5 n 5 N - 1as an N-point sequence. Let 2(n) be a periodic signal of period N, created using the N-point sequence z(n); that is, from (5.19)
00
Z(n)=
r=--00
z(n - TN)
This is a somewhat cumbersome representation. Using the modulo-N o p eration on the argument we can simplify it to
2(n)= z(n mod N)
(5.21)
A simple way to interpret this operation is the following: if the argument n is between 0 and N - 1, then leave it as it is; otherwise add or subtract multiples of N from n until the result is between 0 and N - 1. Note carefully that (5.21) is valid only if the length of z(n)is N or less. Furthermore, we use the following convenient notation to denote the modulo-N operation.
In this function n can be any integer array, and the array m contains the corresponding modulo-N values. From the frequency sampling theorem we conclude that N equispxed samples of the discrete-time Fourier transform X(eJ") of the N-point sequence z(n) can uniquely reconstruct X ( e J w ) These N samples around . t_he unit circle are called the discrete Fourier transform coefficients. Let X ( k ) = DFSZ(n), which is a periodic (and hence of infinite duration) sequence. Its primary interval then is the discrete Fourier transform, which is of finite duration. These notions are made clear in the following definitions. The Discrete Fourier TYansform of an N-point sequence is given b Y
It is clear from the discussions at the top of this section that the DFS is practically equivalent to the DFT when 0 5 n 5 N - 1. Therefore the and implementation of the DFT can be done in a similar fashion. If ~ ( n ) X ( k ) are arranged as column vectors x and X,respectively, then from (5.24)and (5.25) we have
(5.26)
where W N is the matrix defined in (5.7)and will now be called a DFT matriz. Hence the earlier d f s and idfs MATLAB functions can be renamed as the d f t and i d f t functions to implement the discrete Fourier transform computations.
function [ k = dft(xn.N) X] X Computes Discrete Fourier Transform
Note that when the magnitude sample is zero, the corresponding angle is not zero. T i is due to a particular algorithm used by MATLAB compute the hs to
132
Chapter 5 0 THE DISCRETE FOURIER TRANSFORM
Magnitude d the DTFT
; ' " .
-1
'0
02
0.4
0.6
0.8
1
12
14
1.6
1.8
2
treguency in pi units
~ O a l b D m
r - _
-100
-2"o0
0.2
0.4
0.6
0.8
1
12
1.4
1.6
1.8
2
traquency in pi undts
FIGURE 5.4
The DTFT plots m Ezample 5.6
angle part. Generally these angles should be ignored. The plot of DFT values is shown in Figure 5.5. The plot of X(erW)is also shown as a dashed line for comparison. From the plot in Figure 5.5 we observe that Xa correctly gives 4 samples of X(e'"), but it has only one nonzero sample. Is this surprising? By looking at the bpoint z(n), which contains all ls one must conclude that its ', periodic extension is
Z(n) = 1, Vn
which is a constant (or a DC) signal. T i is what is predicted by the DFT hs Xd(k), which has a nonzero sample at k = 0 (or w = 0) and has no values at 0 other frequencies.
0
EXAMPLE 5.7
How can we obtain other samples of the DTFT
Xd) ('?
W i
It is clear that we should sample at dense (or finer) frequencies; that is, we should increase N. Suppose we take twice the number of points, or N = 8 instead of 4. T i we can achieve by treating z(n) as an &point sequence by hs appending 4 zeros.
44 = {L1,1,1,0,0,0,01
r
The Discrete Fourier Transform
133
Magnitude of the OFT: N=4
'\
,'- - _
-. ,
2 k
r
-_
25 .
.
',
3
35 . 4
-1
0
0.5
1
1.5
Angle of the DFT: N=4
c
0
I
-.
1.5
-100
-200 0
FIGURE 5.5
05 .
1
2
k
25 .
3
35 .
4
The DFT plots of Eznmple 5.6
This is a very important operation called a zem-padding opemtaon. T i operhs s n ation i necessary i practice to obtain a deme spectrum of signals as we shall see. Let XS(k)be an &point DFT,then
7
Comments: Based on the last two examples there are several comments that we can make.
1. Zero-padding is an operation in which more zeros are appended t o the original sequence. The resulting longer DFT provides closely spaced samples of the discretetime Fourier transform of the original sequence. zerepadding is implemented using the zeros function. In MATLAB 2. In Example 5.6 all we needed to accurately plot the discretetime Fourier transform X(ej") of z(n) was X4 (k),the Cpoint DFT. This is because z(n) had only 4 nonzero samples, so we could have used the interpolation formula (5.19) on X4 (k) t o obtain X(eJ").However, in practice, it is easier t o obtain Xs (k) and XI6 (k), and so on, t o fill in the values of X(ej") rather than using the interpolation formula. This approach can
The Discrete Fourier Transform
135
r
MasMude of the DFT: N=16
0
2
4
6
B
10
12
14
16
k
Angle d the DFT: N-16
200
1
-200 0
FIGURE 5.7
2
4
6
8
10
12
14
1 6
J
k
The DFT plots of Emmple 5.7: N = 16
be made even more efficient using fast Fourier transform algorithms to compute the DFT. 3. The zero-padding gives us a high-density spectrum and provides a better displayed version for plotting. But it does not give us a hzghmolution spectrum because no new information is added to the sign&, only additional zeros are added in the data. 4. To get a high-resolution spectrum, one has to obtain more data &om the experiment or observations (see Example 5.8 below). There are also other advanced methods that use additional side information or nonlinear techniques.
0 EXAMPLE 5.8
To illustrate the differencebetween the high-density spectrum and the high resolution spectrum, consider the sequence
z(n) = cos (0.48~71) cos (0.52~71)
W want to determine its spectrum based on the finite number of samples. e
n5
PI
Now the plot in Figure 5.9 shows that the sequence has a dominant frequency at w = 0 . 5 ~This fact i not supported by the original sequence, w i h has two . s hc frequencies. The zero-padding provided a smoother version of the spectrum in Figure 5.8. h. To get better spectral information, we will take the first 100 samples of z(n) and determine its discrete-time Fourier transform.
Now the discrete-time Fourier transform plot in Figure 5.10 clearly shows two frequencies, which are very close t o each other. This is the high-resolution spec. trum of ~ ( n )Note that padding more zeros to the 100-point sequence will result in a smoother rendition of the spectrum in Figure 5.10 but will not reveal any 0 new information. Students are encouraged to verify this.
138
Chapter 5 U THE DISCRETE FOURIER TRANSFORM
signal x(n), 0 <= n <= 99
0
10
20
30
40
50
60
70
80
901W
n
DTFT Magnitude
FIGURE 5.10
Signal and its spectrum in Ezample 5.86: N = 100
PROPERTIES OF THE DISCRETE FOURIER TRANSFORM
I I
The DFT properties are derived from those of the DFS because mathematically DFS is the valid representation. We discuss several useful prop erties, which are given without proof. These properties also apply to the DFS with necessary changes. Let X ( k ) be an N-point DFT of the sequence z(n).Unless otherwise stated, the N-point DFTs will be used in these properties.
1. Linearity: The DFT is a linear transform
D F T [ u z ~ (f ) ~ b~z(n)] u D F T [ z I ( ~ ) ] bDFT[zz(n)] =
+
(5.27)
Note: If z1(n) and ~ z ( nhave different durations-that is, they are ) Nl-point and Nz-point sequences, respectively-then choose N3 = max(N1, N z ) and proceed by taking &-point DFTs. 2. Circular folding: If an N-point sequence is folded, then the result z(-n) would not be an N-point sequence, and it would not be possible
139
Properties of the Dixrete Fourier Transform
t o compute its DFT. Therefore we use the modulo-N operation on the argument (-n) and define folding by
T i is called a circular folding. To visualii it, imagine that the hs
se
quence z(n) is wrapped around a circle in the counterclockwise direction so that indices n = 0 and n = N overlap. Then z((-n))N can be viewed as a clockwise wrapping of z ( n ) around the circle; hence the the name circular folding. In MATLAB circular folding can be achieved by x=x(mod(-n,N)+I). Note that the arguments in MATLAB begin with 1. Then its DFT is given by
Comments: 1. Observe the magnitudes and angles of the various DFTs in Examples 5.6 and 5.7. They do satisfy the above circular symmetries. These symmetries are different than the usual even and odd symmetries. To visualize this, imagine that the DFT samples are arranged around a circle so that the indices k = 0 and k = N overlap; then the samples will be symmetric with respect to k = 0, which justifies the name circular symmetry. 2. The corresponding symmetry for the DFS coefficients is called the periodic conjugate symmetry. 3. Since these DFTs have symmetry, one needs to compute X ( k ) only for
k=0,1,
or for
which means that the DFT coefficient at k = 0 must be a real number. But k = 0 means that the frequency wk = kwl = 0, which is the DC
142
Chapter 5
THE DISCRETE FOURIER TRANSFORM
frequency. Hence the DC coefficient for a real-valued z(n)must be a real number. In addition, if N is even, then N/2 is also an integer. Then from (5.32)
X(f)
=x*((-;))N=x*(;)
which means that even the k = N/2 component is also real-valued. This component is called the Nyquist component since k = N/2 means that = the frequency W N I ~ (N/2)(2?r/N) = ?T, which is the digital Nyquist frequency. The real-valued signals can ako be decomposed into their even and odd components, ze(n) and Z, ( n ) , respectively, as discussed in Chapter 2. However, these components are not N-point sequences and therefore we cannot take their N-point DFTs. Hence we define a new set of components using the circular folding discussed above. These are called circular-even and circular-odd components defined by
(5.33)
Then
Implication: If z(n) is real and circular-even, then its DFT is also red and circular-even. Hence only the first 0 5 n 5 N/2 coefficients are necessary for complete representation. Using (5.33),it is easy to develop a function to decompose an N-point sequence into its circular-even and circular-odd components. The following circevod function uses the mod function given earlier to implement the nMOD N operation.
function Cxec, roc1 = circevod(x) 1 signal decomposition into circular-even and circular-odd parts
The plots in Figure 5.13 show the circularly symmetric components o z(n). f
10- 0
6-
-
F 6-
I 42-
0
0-
p
1 2
3
r 11:
7
8 9
10
- 4 8 0
4
5 n
6
FIGURE 5.13
&cular-ewen and circular-oddcomponents ofthe sequence in Ez-
ample 5.1Oa
144
Chapter 5 ITHE DISCRETE FOURIER TRANSFORM
From the plots in Figure 5.14 we observe that the DFT of s.,(n) i the same as s the real part of X ( k ) and that the DFT of z,(n) i the same 85 the imaginary s part of X ( k ) . 0
A similar property for complex-valued sequences is explored i Exern
cise 5.10.
5. Circular shift of a sequence: If an N-point sequence is shifted in either direction, then the result is no longer between 0 5 n 5 N - 1. e and then Therefore w first convert z(n) into its periodic extension Z(n), shift it by m samples to obtain
: i L
20
10
Z(n - m) = z((n- m))N
InwWWn)l)
(5.35)
0
0
5
10
-20
0
5
10
50-
40.
30
20
10
0
1
0
5
10
k
FIGURE 5.14
5 k
I 10
Plots of DFT symmetry properties in Example 5.10b
Properties of the Discrete Fourier Transform
145
This is called a periodic shijl of i ( n ) .The periodic shift is then converted
into an N-point sequence. The resulting sequence
Z(n - m)RN(n)= z ( ( n - m))N z N ( n ) 7
(5.36)
is called the circular shij? of z(n). Once again to visualize this, imagine that the sequence z(n) is wrapped around a circle. Now rotate the circle by k samples and unwrap the sequence from 0 5 n 5 N - 1. Its DFT is given by DFT [Z ((n- m))NR,(n)] = WkmX(k)
0
the left. b. Sketch z ((n - 3))15&(n), that is, a circular shift by 3 samples toward the right, where z(n)is assumed to be a 15-point sequence.
We will use a stepby-step graphical approach to illustrate the circular shifting of operation. This approach shows the periodic extension f(n) = z ( ( n ) ) N z(n), followed by a linear shift in i(n)to obtain f (n- m) = I ((n- m ) ) N ,and finally truncating f (n- m) to obtain the circular shift.
a. Figure 5.15 shows four sequences. The topleft shows z(n), the bottomleft shows i ( n ) ,the topright shows 5 (n 4), and finally the bottom-right shows s((n+4)),, Rll(n). Note carefully that as samples move out of the [O, N - I] window in one direction, they reappear from the opposite direction. This is the meaning of the circular shift, and it is different from the linear shift. b In this case the sequence z(n) is treated as a 15-point sequence by . padding 4 zeros. Now the circular shift will be different than when N = 11. This is shown in Figure 5.16. In fact the circular shift z ((n- 3))15 looks like a 0 linear shift z(n - 3).
+
To implement a circular shift, we do not have t o go through the s periodic shift a shown in Example 5.11.It can be implemented directly in two ways. In the first approach, the moduleN operation can be used on the argument (n- m) in the time domain. This is shown below in the c i r s h f t t function.
function y = cirshftt(x,m,N) X Circular shift of m samples u r t size N in sequence
I:
6. Circular shift i the Bequency domain: This property is a dual of n the above property given by
DFT [Wi'"~(n)] X ((k - e)), R,(k) =
(5.38)
7. Cicular convolution: A linear convolution between two N-point sequences will result in a longer sequence. Once again we have to restrict our interval to 0 5 n 5 N - 1. Therefore instead of linear shift, we should consider the circular shift. A convolution operation that contains a circular shift is called the circular convolution and is given by
N-1
%i(n)
@ zz(n) =
mO =
zi(m)z2 ((n- m ) ) N , 0 5 n 5 N
-1
(5.39)
Note that the circular convolution is also an N-point sequence. It has a structure similar to that of a linear convolution. The differences are in the summation limits and in the N-point circular shift. Hence it depends on N and is also called an N-point Circular convolution. Therefore the
146
Chapter 5
THE DISCRETE FOURIER TRANSFORM
10-
0 0
0
-
8642-
0
0-
T PTt? p
9
0 0
s
c
3
;:
10Ei 8 -
t
'; j
64-
0
2-
use of the notation convolution is
@ is appropriate. The DFT property for the circular
An alternate interpretation of this property is that when we multiply two
N-point DFTs in the frequency domain, we get the circular convolution (and not the usual linear convolution) in the time domain.
0
U(AMPLE5.13
Let zI(n) = {1,2,2) and m(n) = {1,2,3,4). Compute the 4-point circular convolution zl(n) @ zz(n). Note that zl(n) is a %point sequence, hence we will have to pad one zero to make it a Cpoint sequence before we perform the circular convolution. We will solve this problem in the time domain 85 well as in the frequency domain. In the time domain we will use the mechanism of circular convolution, while in the frequency domain we will use the DFTs.
0 Tame-domain approach:
Similar to the circular shift implementation, we can implement the circular convolution in a number of different ways. The simplest approach would be to implement (5.39)literally by using the c i r s h f t t function and requiring two nested for. .end loops. Obviously, this is not efficient. Another ~ approach is to generate a sequence x ((n- T I Z ) )for each n in [O, N - 11 as rows of a matrix and then implement (5.39)as a matrix-vector multiplication similar to our d f t function. This would require one for. . .end loop. The following circonvt fundion incorporates these steps.
The third approach would be to implement t h e frequency-domain operation (5.40) using the dft function. This is explored in Exercise 5.15.
0 EXAMPLE 5.14
Let us use MATLAB perform the circular convolution in Example 5.13. to The sequences are zl(n)= (1,2,2} and z2(n) = {1,2,3,4}.
solution
>> xi = [1,2,21; x2 = C1,2,3,41; >> y = circonvt(x1, x2, 4)
Y '
15 Hence 12
9 14
zl(n) @ zz(n) = (15, 12, 9, 14)
as before.
0
0
EXAMPLE 5.15
In this example we will study the effect of N on the circular convolution. Obviously, N 2 4; otherwise there will be a time-domain aliasing for zz(n). We will use. the same two sequences from Example 5.13.
a. Compute zl(n) 5 ~ ( n ) . b. Compute zl(n) 6 z2(n). c. Comment on the results. The sequences are zl(n)= {1,2,2} and n(n) = {l, 2,3,4}. Even though the sequences are the same BS in Example 5.14,we should expect different results for different values of N.This is not the case with the linear convolution, which is unique, given two sequences.
a. %point circular convolution:
c. A careful observation of 4, 5-, and 6point circular convolutions from this and the previous example indicates some unique features. Clearly, an Npoint circular convolution is an N-point sequence. However, some samples in these convolutions have the same values, while other values can be obtained as a sum of samples in other convolutions. For example, the first sample in f the 5-point convolution is a sum of the first and the last sample o the 6point convolution. The linear convolution between zl(n) and zz(n)is given by
~i(n)~ z ( n )= {l, 4, 9, 14, 14,8 * )
which is equivalent to the &point circular convolution. These and other issues tl are explored in the next section.
8. Mdtiplication: This is the dual of the circular convolution p r o p erty. It is given by
(5.41)
in which the circular convolution is performed in the frequency domain. The MATLABfunctions developed for circular convolution can also be used here since X1 (k) and X z (k)are a s N-point sequences. lo 9. Parsed's relation: This relation computes the energy in the frequency domain.
(5.42)
The quantity the power spectrum.
is called the energy spectrum of finiteduration se-
quences. Similarly, for periodic sequences, the quantity
Properties of the Dixrete Fourier Transform
153
LINEAR CONVOLUTION USING THE DFT
I
One of the most important operations in linear systems is the linear conve lution. In fact FIR filters are generally implemented in practice using this linear convolution. On the other hand, the DFT is a practical approach for implementing linear system operations in the frequency domain. As we lo shall see later, it is a s an efficient operation in t e r n of computations. However, there is one problem. The DFT operations result in a circular convolution (something that we do not desire), not in a linear convolution that we want. Now we shall see how to use the DFT to perform a linear convolution (or equivalently, how t o make a circular convolution identical to the linear convolution). We alluded to this problem in Example 5.15. be Let q ( n ) be an N1-point sequence and let ~ ( n ) an Nz-point by that sequence. Define the linear convolution of i l ( n ) and ~ ( n )53(n), is,
z3(n)= z1(n)* .z(n) (5.43)
k=-m
0
Then z3(n) is a (N1 Nz - 1)-point sequence. If we choose N = max(N1,Nz) and compute an N-point circular convolution q ( n ) @ zz(n), then we get an N-point sequence, which obviously is different from q(n). This observation also gives us a clue. Why not choose N = N1 Nz - 1 and perform an (N1 + Nz - 1)-point circular convolution? Then at least both of these convolutions will have an equal number of samples. Therefore let N = N I NZ - 1 and let us treat q ( n ) and ~ ( nas ) N-point sequences. Define the A'-point circular convolution by z4(n).
+
+
+
5 4 ( 4
=2 1 ( 4
N-1
04.)
z()Z( lkz(n
(5.44)
- k))N
1
RN(n)
r
1
zs(n-rN)
I
154
r=-w m=O
Chapter 5
THE DISCRETE FOURIER TRANSFORM
using (5.43) This analysis shows that, in general, the circular convolution is an aliased version of the linear convolution. We observed this fact in Example 5.15. Now since ~ ( nis )an N = ( N 1 + Nz - 1)-point sequence, we have
~ 4 ( n= 23(n); )
0 5 n 5 ( N - 1)
which means that there is no aliasing in the time domain.
N Conclusion: If we make both q ( n ) and ~ ( n )= N1+ Nz - 1 point sequences by padding an appropriate number of zeros, then the circular convolution is identical to the linear convolution.
0
EXAMPLE 5.16
In order to use the DFT for linear convolution, we must choose N properly. However, in practice it may not be possible to do SO, especially when N is very large and there is a limit on memory. Then an error will be introduced when N is chosen less than the required value to perform the circular convolution. We want to compute this error, which is useful in
155
T i is a simple yet important relation. It implies that when max(N1,N2) hs 5 N < (N1 + N2 - l), the error sample at n is the same as the linear
convolution N samples away. Now the linear convolution will be zero after ( N l + Nz - 1)samples. This means that the first few samples of the circular convolution are in error, while the remaining ones are the correct l n a convolution values. ier
0
EXAMPLE 5.17
The last case of N = 4 also provides the useful observation given below.
Observation: When N = max(N1,Nz)is chosen for circular convolution, then the first (M- 1)samples are in error (i.e., different from the linear hs convolution),where M = min(N1, Nz). T i result is useful in implementing long convolutions in the form of block processipg.
BLOCK CONVOLUTIONS
When we want to filter an input sequence that is being received continuously, such as a speech signd from a microphone, then for practical purposes we can think of this sequence as an infinitelength sequence. I f we want to implement this filtering operation as an FIR filter in which the linear convolution is computed using the DFT, then we experience some practical problems. We will have to compute a large DFT, which is generally impractical. Furthermore, output samples are not available until all input samples are processed. This introduces an unacceptably large amount of delay. Therefore we have to segment the infinitelength input sequence into smaller sections (or blocks), process each section using the DFT, and finally assemble the output sequence from the outputs of each
157
Linear Convdution using the DFT
section. T i procedure is called a block convolutron (or block processing) hs operation. Let us assume that the sequence z(n)is sectioned into N-point sequences and that the impulse response of the filter is an M-point sequence, where M < N. Then from the above observation we note that the N-point circdar convolution between the input block and the impulse response will yield a block output sequence in which the first (M - 1) samples are not ) the correct output values.If we simply partition ~ ( ninto nonoverlapping sections, then the resulting output sequence will have intervals of incorrect samples. To correct this problem, we can partition z(n)into sections, each overlapping with the previous one by exactly (M - 1) samples, save the last (N- M 1) output samples, and finally concatenate these outputs into a sequence. To correct for the 6rst (M - 1) samples in the first output block, we set the first (A4- 1) samples in the first input block to zero. This procedure is called an overlap-save method of block convolutions. Clearly, when N >> M, this method is more efficient. We illustrate it using a simple example.
Since M = 3, we will have to overlap each section with the previous one by two samples. Now z(n)is a lCLpoint sequence, and we will need (M - 1) = 2 zeros in the beginning. Since N = 6, we will need 3 sections. Let the sections be
a(n) = {0,0,1,2,3,4}
zz(n) = {3,4,5,6,7,8}
~ 3 ( n= )
{7,8,9,10,0,0)
Note that we have to pad za(n) by two zeros since z(n) runs out of values at n = 9. Now we wl compute the &point circular convolution of each section il with h(n).
yl = a ( n )
@ h(n) = I-3,
@ h(n) = {-4,
-4,1,2,2,2} -4,2,2,2,2}
yz = z2(n)
y3=z3(n) @h(n)
= {7,8,2,2,-9,-10)
Noting that the first two samples are to be discarded, we assemble the output ?An)
y(n)={1,2,2,2,2,2,2,2,2,2,-9,-10}
T
158
Chapter 5 W THE DISCRETE FOURIER TRANSFORM
The linear convolution is given by
s(n)* h(n) = {1,2,2,2,2,2,2,2,2,2,-9, -10)
f
which agrees with the overlapsave method.
0
MATLAB IMPLEMENTATION
Using the above example as a guide, we can develop a MATLAB function to implement the overlapsave method for a very long input sequence z(n). The key step in this function is to obtain a proper indexing for the segmentation. Given z(n)for n 2 0, we have to set the f r t (M- 1)samples is to zero to begin the block processing. Let this augmented sequence be
?(n)={O 0
b
(M-1) zerm
... O,z(n)}, n 1 0 u
0 5 n 5 N - 1, is
1
and let L = N - M given by
+ 1,then the kth block zk(n),
m5kLtN
~ r ~ (=?(m); I L n) C
- 1, k 2 0, 0 5 TI 5 N -
The total number of blocks is given by
where N, is the length of z(n)and 1.1 is the truncation operation. Now each block can be circularly convolved with h(n) using the circonvt function developed earlier to obtain
It should be noted that the ovrlpsav function a developed here is not s the most efficient approach. We will come back t o this issue when we discuss the fast Fourier transform.
0
EXAMPLE 5.19
To verify the operation of the ovrlpsav function, let us consider the sequences given in Example 5.18.
Solution
>> n >> y Y'
1
MATLAB Script
ovrlpsav(x.h,N)
2 2
= 0:9; x = n+l; h
2
[l,O,-l]; N = 6;
2
2
2
2
2
2 - 9 - 1 0
Ti is the correct linear convolution as expected. hs
0
There is an alternate method called an owerlap-add method of block convolutions. In this method the input sequence z(n)is partitioned into nonoverlapping blocks and convolved with the impulse response. The resulting output blocks are overlapped with the subsequent sections and added to form the overall output. T i is explored in Exercise 5.20. hs
T H E FAST FOURIER TRANSFORM
I
I
The DFT (5.24) introduced earlier is the only transform that is discrete in both the time and the frequency domains, and is defined for finite-duration sequences. Although it is a computable transform, the straightforward implementation of (5.24) is very inefficient, especially when the sequence 4 1 length N is large. In 1965 Cooley and Tukey 1 showed a procedure t o substantially reduce the amount of computations involved in the DFT. This led to the explosion of applications of the DFT, including in the
160
Chapter 5
THE DISCRETE FOURIER TRANSFORM
digital signal processing area. Furthermore, it also led to the development of other efficient algorithms. All these efficient algorithms are collectively known as fast Fourier transform (FFT) algorithms. Consider an N-point sequence z(n). Its N-point DFT is given by (5.24) and reproduced here
N-1
X(k)=
n=O
z(n)WEk, 0 5 k 5 N - 1
(5.46)
where WN = e-j2*IN.To obtain one sample of X ( k ) , we need N complex multiplicationsand (N-1) complex additions. Hence to obtain a complete set of DFT coefficients, we need N2 complex multiplications and N(N 1) N N a complex additions. Also one has to store N 2 complex coefficients {WEk} (or generate internally at an extra cost). Clearly, the number of DFT computations for an N-point sequence depends quadratically on N, which will be denoted by the notation
For large N , o (N2) is unacceptable in practice. Generally, the processing time for one addition is much less than that for one multiplication. Hence from now on we will concentrate on the number of complex multiplications, which itself requires 4 real multiplications and 2 real additions.
Goal of an Eficient Computation In an efficiently designed algorithm the number of computations should be constant per data sample, and therefore the total number of computations should be linear with respect to N. The quadratic dependenceon N can be reduced by realizing that most of the computations (which are done again and again) can be eliminated using the periodicity property
and the symmetry property
of the factor {WEk}. One algorithm that considers only the periodicity of WEk is the Goertzel algorithm. This algorithm still requires CN = o(Nz) multiplications, but it has certain advantages. This algorithm is described in C h a p ter 10. We first begin with an example to illustrate the advantages of the symmetry and periodicity properties in reducing the number of computations. We then describe and analyze two specific FFT algorithms that
161
Then each element of the resultant matrix is multiplied by where p is the row index and q is the column index; that is, the following dot-product is
perfOlXled.
{wq},
FIGURE 5.18
Signdflowgmph in Ezample 5.20
The Fan Fourier Transform
163
Finally, two more smaller %point DFTs are taken of m w vectors.
Although this interpretation seems to have more multiplications than the efficient algorithm, it does suggest a systematic approach of computing a larger 0 DFT based on smaller DFTs.
DIVIDE-ANDCOMBINE APPROACH
To reduce the DFT computation'squadratic dependence on N,one must
choose a composite number N = LM since
L2+ M2 << N 2for large N
Now divide the sequence into M smaller sequences of length L, take M smaller Lpoint DFTs, and then combine these into a larger DFT using L smaller M-point DFTs. T i is the essence of the divide-and-combine hs approach. Let N = LM, then the indices n and k in (5.46)can be written
for each of the columns m = 0,. ..,M - 1. 2. Second, we modify F p m) to obtain another may. C, (5.51) The factor Wgm is called a twiddle factor. 3. Finally, we compute the M-point DFTs
M-1
X@,q)=
m=O
G(p,m)Wzq 0 5 q 5 M
-1
(5.52)
for each of the rows p = 0,. . . ,L - 1. The total number of complex multiplications for this approach can now be given by
cN = M L + N + L M ~ o ( N ~ ) ~ <
(5.53)
This procedure can be further repeated if M or L are composite numbers. Clearly, the most efficient algorithm is obtained when N is a highly composite number, that is, N = R". Such algorithms are called mdiz-R FFT algorithms. When N = R T q ..., then such decompositions are called mized-mdix FFT algorithms. The one most popular and easily programmable algorithm is the radix-2 FFT algorithm.
RADIX-2 FFT ALGORITHM
Let N = 2"; then we choose M = 2 and L = N/2 and divide x(n) into two Nla-point sequences according to (5.48) as
This is called a merging formula, which combines two N/%-point DFTs into one N-point DFT. The total number of complex multiplications reduces to
T i procedure can be repeated again and again. At each stage the sehs quences are decimated and the smaller DFTs combined. This decimation ends after Y stages when we have N one-point sequences, which are also one-point DFTs. The resulting procedure is called the decimation-in-time FFT (DIT-FFT) algorithm, for which the total number of complex multiplications is
CN = NY= NlogZN
Clearly, if N is large, then CN is approximately linear in N, which was the goal of our efficient algorithm. Using additional symmetries, CN can be reduced to log, N. The signal flowgraph for this algorithm is shown in Figure 5.19 for N = 8.
8
FIGURE 5.19
Decimation-in-tameFFT structure for
N =8
166
Chapter 5
THE DISCRETE FOURIER TRANSFORM
In an alternate approach we choose L = 2, M = N/2 and follow the steps in (5.49). Note that the initial DFTs are %-pointDFTs, which contain no complex multiplications. From (5.50)
F(0,m) =s(O,m)+x(l,m)k@
= z(n) z(n
+
+ N/2), 0 5 n 5 N/2
F(1,m) =x(O,m)+z(l,m)Wz'
= x(n)
- x(n + N/2), 0 5 n 5 N/2
and from (5.51)
G(0, m) = F(0,m )Wi
= x(n)
+ x(n + N/2),
0 5 n 5 N/2
G(1,m) = F(l,m)W,"
=) ( . I
(5.55)
- ~ (+n N/2)] WE, 0 5 n 5 N/2
Let G(0,m) = &(n) and G(1,m) = &(n) for 0 5 n 5 N/2 - 1 (since they can be considered as time-domain sequences); then from (5.52) we have
(5.56)
This implies that the DFT values X(k) are computed in a decimated fashion. Therefore this approach is called a decimation-in-freqzlency FFT (DIF-FFT) algorithm. Its signal flowgraph is a transposed structure of the DIT-FFT structure, and its computational complexity is also equal to f logz N.
MATLAB IMPLEMENTATION
MATLAB provides a function called fft to compute the DFT of a vector x. It is invoked by X = fft(x,N), which computes the N-point DFT. If the length of I(. is less than N, then x is padded with zeros. If the argument N is omitted, then the length of the DFT is the length of x. If x is a matrix, then fft (x,N) computes the N-point DFT of each column of x. This fft function is written in machine language and not using MATLAB commands (i.e., it is not available as a .m file). Therefore it executes very fast. It is written as a mixed-radix algorithm. If N is a power of two, then a high-speed radii-2 FFT algorithm is employed. If N is not a power of two, then N is decomposed into prime factors and a slower mixed-radix FFT algorithm is used.Finally, if N is a prime number, then the fft function is reduced to the raw DFT algorithm.
167
The Fast Fourier Transform
The inverse DFT is computed using the ifft function, which has the same characteristics as ff t.
0
EXAMPLE 5.21
In this example we will study the execution time of the fft function for 1 5 N 5 2048. This win reveal the dividP-and-combinestrategy for various values of N . To determine the execution time, MATLAB provides two functions. The clock function provides the instantaneous clock reading, while the etime(t1, t2) function computesthe elapsed time between two time marks t 1 and t2. To determine the execution time, we will generate random vectors from length 1through 2048, compute their FFTs, and save the computation time in an array. Finally, we will plot this execution time versus N .
MATLAB Script Nmax = 2048; fft-time=zeros(l ,Nmar); for n=i:l:Nmax x-rand(i,n); t-clock;fft (x) ;fft_time(n)=etime(clock, t) ;
The plot of the execution times is shown in Figure 5.20. This plot is very informative. The points in the plot do not show one clear function but appear to group themselves into various trends. The uppermost group depicts a o(N*) dependence on N , which means that these values must be prime numbers b e tween 1 and 2048 for which the FFT algorithm defaults to the DFT algorithm. Similarly, therearegroups corresponding to theo(N2/2), o ( N 2 / 3 ) ,o ( N 2 / 4 ) , and so on, dependencies for which the number N has fewer decompositions. The last group shows the (almost linear) o ( N log N ) dependence, which is for N = 2",0 5 v 5 11. For these values of N , the radii-2 FFT algorithm is used. For all other values, a mixed-radixFFT algorithm is employed. This shows that the divideand-combinestrategy is very effective when N is highly composite. For example, the execution time is 0.16 second for N = 2048, 2.48 seconds for 0 N = 2047, and 46.96 seconds for N = 2039. The MATLAB functions developed previously in this chapter should now be modified by substituting the fft function in place of the dft function. From the above example care must be taken to use a highly composite N. A good practice is to choose N = 2" unless a specific situation demands otherwise.
168
Chapter 5
THE DISCRETE FOURIER TRANSFORM
FFF exewtion times
50
454035-
N
FIGURE 5.20
FFT ezecution times for 1 <= ' <= 2048 h
FAST CONVOLUTIONS
The conv function in MATLAB implemented using the filter function is (which is written in C) and is very efficient for smaller values of N (< 50). For larger values of N it is possible to speed up the convolution using the FFT algorithm. T i approach uses the circular convolution to implement hs the linear convolution, and the FFT to implement the circular convolution. The resulting algorithm is called a fast convolution algorithm. In addition, if we choose N = 2" and implement the radix-2 FFT, then the algorithm is called a high-speed convolution. Let x1 (n) be a N1-point sequence and 5 2 (n)be a Nz-point sequence; then for high-speed convolution N is chosen to be (5.57) where 1r is the smallest integer greater than x (also called a ceiling . function). The linear convolution x1 (n) * xz (n) can now be implemented by two N-point FFTs, one N-point IFFT, and one N p i tdot-product. -on x1 (n) * x~ (n) = IFFT [FFT[q(n)] FFT [ z( n ) ] ] . z (5.58)
For large values of N , (5.58) is faster than the timedomain convolution as we see in the following example.
The Fast Fourier Transform
169
0
EXAMPLE 5.22
To demonstrate the effectiveness of the high-speed convolution, let us compare
the execution times of two approaches. Let 51(n) be an Lpoint uniformly distributed random number between [O,1 ,and let zz(n) be an Lpoint Gaussian 1 random sequence with mean 0 and variance 1. We will determine the average execution times for 1 5 L 5 150, in which the average is computed over the 100 realizations of random sequences.
Figure 5.21 shows the linear convolution and the high-speed convolution times for 25 5 L 5 150. It should be noted that these times are afTected by the script. The plot in Figure computing platform used to execute the MATLAB 5.21 was obtained on a 33-MHz 486 computer. It shows that for low d u e s of L the linear convolution is faster. The crossover point appears to be L = 50, beyond which the linear convolution time increases exponentially, while the high-speed convolution time incresses fairly linearly. Note that since N = 2", 0 the high-speed convolution time is constant over a range on L.
HIGH-SPEED
BLOCK CONVOLUTIONS
Earlier we discussed a block convolution algorithm called the overlapandsave method (and its companion the overlapand-add method), which is used to convolve a very large sequence with a relatively smaller sequence. The MATLAB function ovrlpsav developed in that section uses the DFT to implement the linear convolution. We can now replace the DFT by the radix-2 FFT algorithm to obtain a high-speed overlapand-save algoChapter 5 W THE DISCRETE FOURIER TRANSFORM
170
Comparison of OOnvdutiOn times
0.35
0.3 aMvolutiMl
0.25
! O2 I
-
I
I
50
E
0.15 -
0.1 0.05 -
OL
0
100
s e q m length N
150
FIGURE 5.21 Comparison of linear and high-speed convolution times
rithm. To further reduce the computations,the FFT of the shorter (fixed) sequence can be computed only once. The following hsolpsav function shows this algorithm.
function [yl hsolpsav(x,h,N) X High-speed Overlap-Save method of block convolutions using FFT
and let &(n) be periodic with fundamental period N = 100, where one period is given by
These two periodic sequences differ in their periodicity but otherwise have equal nonzero samples. a. Find the DFS X I ( k ) of i l ( n ) and plot samples (using the stem function) of its magnitude and angle versus k. b. Find the DFS & (k)of &(n) and plot samples of its magnitude and angle versus k.
c. What is the difference between the above two DFS plots?
Clearly, Zs(n) is different from &(n) of Problem 3 even though both of them are periodic with period 100. a. Find the DFS x (k)of &(n) and plot samples of its magnitude and angle versus k. 3 b. What effect does the periodicity doubling have on the DFS? c. Generalize the above result to M-fold periodicity. In particular, show that if
EA.f(n) = Z1(n),. ..,i1(n)
PERIODIC
z(n)= {I,2,3,4,5,6,6,5,4,3,2,1)
Plot (using the stem function) its magnitude and a. Determine the DFT X (k) of z (n). phase. b. Plot the magnitude and phase of the DTFT X(e'") of
I
(n)using MATLAB.
c. Verify that the above DFT is the sampled version of X(e'"). It might he helpful to combine the above two plots in one graph using the hold function.
d. Is it possible to reconstruct the DTFT X ( g W ) from the DFT X (k)? If posible, give the necessary interpolation formula for reconstruction. If not possible, state why this reconstruction cannot be done.
Problems
173
P5.7 Plot the DTFT magnitudes of the following sequences using the DFT as a computation
tool. Make an educated guess about the length N so that your plots are meaningful. a.
51
b. Modify the circevod function developed in the chapter so that it can be used for complex-valued sequences. c. Verify the above symmetry property and your MATLAB function on the following sequence.
z(n)= (0.9d"/3)"[u(n) -u(n-20)]
P5.11 The first five values of the &point DFT of a real-valued sequence z (n) are given by
{0.25,0.125 - j0.3,0,0.125 - j0.06,0.5}
Determine the DFT of each of the following sequences using properties. a. z1 (n) = z ((2 -a)),
a. Prove the above property analytically. b. This property can be used t o compute the DFTs of two real-valued N-point sequences z using one N-point DFT operation. Specifically, let z1 ( n ) and z (n) be two N-point sequences. Then we can form a complex-valued sequence
X If m is a scalar then y is a sequence (rov vector) X If m is a vector then y is a matrix, each rov is a circular s h i f t
% in x corresponding t o entries i n vecor n x M and x should not be matrices
Verify your function on the following sequence
X I (n) = 11
- n,
0 5 n 5 10
with rn = 10 and N = 15.
P . 4 Using the analysis and synthesis equations of the DFT, show that 51
176
Chapter 5
m THE DISCRETE FOURIER TRANSFORM
This is commonly referred to as a Parsed's relation for the DFT. Verify this relation by using MATLAB the sequence in Problem 5.9. on
Clearly, ym (n) is a (2L - 1)-point sequence. In this method we have to save the intermediate convolution results and then properly overlap these before adding to form the final result y (n).To use DFT for this operation, we have to choose N 2 (2L - 1).
a Develop a MATLAB function to implement the overlapadd method using the circular convolution operation. The format should be
function Cyl
c. Using results of calculations, determine the minimum value of N necessary so that linear and circular convolutions are the same on the N-point interval. d. Without performing the actual convolutions, explain how you could have obtained the result of part c. P . 2 Let 52
where t' is an integer. Notice that 1 (n)contains ezoctly t' periods (or cycles) of the m i n e waveform in N samples. This is a windowed cosine sequence containing no leakage. a. Show that the DFT X (k)is a real sequence given by
b. The above result implies that the original frequency wo of the cosine waveform has lenked into other frequencies that form the harmonics of the timelimited sequence, and hence it is
Problems
179
called the leakage property of cosines. It is a natural result due to the fact that band-limited periodic cosines are sampled over noninteger periods. Explain this result using the periodic extension Z(n)of 2 (n)and the result in Problem 5.22 part a. c. Verify the leakage property using z (n)= co9 ( 5 4 9 9 ) RZOO Plot the real and the (n). imaginary parts of X (k) using the stem function.
P . 4 Let 52
x(n) =
Asin(2rrPn/N),
05 n 5 N -1 = A sin (2nt'n/N) E N (n) elsewhere
where e is an integer. Notice that 2 (n)contains ezactly e periods (or cycles) of the sine waveform in N samples. This is a windowed sine sequence containing no leakage. a Show that the DFT X (k) is a purely imaginary sequence given by AN X ( k )= -6(k 2j
b. Show that if
b. The above result is the leakage property of sines. Explain it using the periodic extension 5(n) of I (n)and the result in Problem 5.24 part a. c. Verify the leakage property using I (n) = sin (57499) 7 b o o (n).Plot the real and the imaginary parts of X (k) using the stem function. P . 6 An analog signal x.(t) = 2sin (4d) ~ C O S 8 d ) is sampled at t = 0.01n for 52 ( n = 0,1,.. .,N - 1 to obtain an N-point sequence I (n). An N-point DFT is used to obtain an estimate of the magnitude spectrum of z,,(t). a. From the following values of N, &owe the one that will provide the accurate estimate of the spectrum of z,(t). Plot the real and imaginary parts of the DFT spectrum (X (k)l. (i) N = 40, (ii)N = 60. (ii) N = 50, b. Fkom the followinn values of N.choose the one that will Drovide the least amount of leakage in the spectrum of s,(t). Plot the real and imaginary parts of the DFT spectrum
In earlier chapters we studied the theory of discrete systems in both the time and frequency domains. We will now use this theory for the processing of digital signals. To process signals, we have to design and implement systems called filters (or spectruni analyzers in some contexts). The filter design issue is influenced by such factors as the type of the filter (i.e., IIR or FIR) or the form of its implemehtation (structures). Hence before we discuss the design issue, we first concern ourselves with how these filters can be implemented in practice. This is an important concern because different filter structures dictate different design strategies. As we discussed earlier, IIR filters are characterized by infiniteduration impulse responses. Some of these impulse responses can be modeled by rational system functions or, equivalently,by differenceequations. Such filters are termed as auto-regressive moving average (ARMA) or, more generally, as recursive filters. Those IIR filters that cannot be so modeled are called nonmursive filters. In DSP,IIR filters generally imply recursive ones because these can be implemented efficiently. Therefore we will always use the term IIR to imply recursive filtersdFurthermore, ARMA filters include moving average filters that are FIR filters. However, we will treat FIR filters separately from IIR filters for both design and implementation purposes. We begin with a description of basic building blocks that are used to describe filter structures. In the remaining sections we briefly describe IIR, FIR, and lattice filter structures, respectively, and provide MATLAB functions to implement these structures.
BASIC ELEMENTS
I
Since our filters are LTI systems, we need the following three elements to describe digital filter structures. These elements are shown in Figure 6.1.
0 Adder: This element has two inputs and one output and is shown in Figure 6.l(a). Note that the addition of three or more signals is impla mented by successive two-input adders. 0 Multiplier (gain): T i is a single-input, single-output element and hs is shown in Figure 6.l(b). Note that the multiplication by 1 is understood and hence not explicitly shown. 0 Delay element (shifter or memory): This element delays the signal passing through it by one sample as shown in Figure 6.l(c). It is implemented by using a shift register.
Using these basic elements, we can now describe various structures of both IIR and FIR filters. MATLAB a convenient tool in the development is of these structures that require operations on polynomials.
IIR FILTER STRUCTURES
I
The system function of an IIR filter is given by
, where b, and a are the coefficients of the filter. We have assumed without loss of generality that = 1. The order of such an IIR filter is called N if
Three different structures can be used to implement an IIR filter:
0 Direct form: In this form the difference equation (6.2) is implemented directly as given. There are two parts to this filter, namely the moving average part and the recursive part (or equivalently, the numerator and denominator parts). Therefore this implementation leads to two versions: direct form I and direct form I1 structures. 0 Cascade form In this form the system function H ( z ) in equation (6.1) is factored into smaller second-order sections, called biquads. The system function is then represented as a grvduct of these biquads. Each biquad is implemented in a direct form, and the entire system function is implemented as a cascade of biquad sections. Parallel form: This is similar to the cascade form, but after factorization, a partial &action expansion is used to represent H ( z ) as a sum of smaller second-order sections. Each section is again implemented in a direct form, and the entire system hnction is implemented as a parallel network of sections.
We will briefly discuss these forms in this section. IIR filters are generally described using the rational form version (or the direct form structure) of the system function. Hence we will provide MATLAB functions for converting direct form structures to cascade and parallel form structures.
DIRECT FORM
As the name suggests, the difference equation (6.2) is implemented as given using delays, multipliers, and adders. For the purpose of illustration, let M = N = 4. Then the difference equation is
y(n) = ~ T J Z ( ~ ) blz(n - 1)
+
+ b ~ (-n2) + b32(n - 3) + b4z(n - 4)
- aly(Tl-
1) - azY(n- 2) - a3y(n - 3) - Ql/(n - 4)
which can be implemented as shown in Figure 6.2. This block diagram is called direct f o m I structure. The direct form I structure implementseach part of the rational function H ( z ) separately with a cascade connection between them. The numerator part is a tapped delay line followed by the denominator part, which is a feedback tapped delay line. Thus there are two separate delay lines in this structure, and hence it requires eight delay elements. We can reduce this delay element count or eliminate one delay line by interchanging the order in which the two parts are connected in the cascade.
184
Chapter 6
DIGITAL FILTER STRUCTURES
FIGURE 6.2
Direct form I structure
Now the two delay lines are close to each other, connected by a unity gain branch. Therefore one delay line can be removed, and this reduction leads to a canonical structure called direct form ZI structure, shown in Figure 6.3.It should be noted that both direct forms are equivalent &om the input-output point of view. Internally, however, they have different signals. MATLAB IMPLEMENTATION In MATLABthe direct form structure is described by two row vectors; b containing the {b,} coefficients and a containing the {a,} coefficients. The structure is implemented by the filter function, which is discussed in Chapter 2. In this form the system function H ( z ) is written as a product of secondorder sections with real coefficients. This is done by factoring the numerator and denominator polynomials into their respective roots and then combining either a complex conjugate root pair or any two red roots into second-order polynomials. In the remainder of this chapter we assume
CASCADE FORM
yk(n) =xk+lln)
Yk+l(")
FIGURE 6.4
Biquad section structure
that N is an even integer. Then
-
-
k=l
n
K
1
+
1 A*,lz-l+
+
Bk,iZ-l
+B ~ , z Z - ~
Ak,2z-2
where K is equal to $, and & , I , &,2, Ak,l, and Ak,2 are real numbers representing the coefficients of second-order sections. The second-order section
with
yl(z)
bOX(z);
YK+l(Z)= y ( z )
is called the kth biquad section. The input to the kth biquad section is
the output from the (k - 1)th biquad section, while the output from the kth biquad is the input to the (k 1)th biquad. Now each biquad section H k ( Z ) can be implemented in direct form I1 as shown in Figure 6.4. The entire filter is then implemented a a cascade of biquads. z As an example, consider N = 4. Figure 6.5 shows a cascade form structure for this fourth-order IIR filter.
+
-4,1
~
2-1
4
7
- 4 1
"2-1
B2,l
186
Chapter 6
DIGITAL FILTER STRUCTURES
MATLAB
IMPLEMENTATION
Given the coefficients (, and {a,} of the direct form filter, we have to 6) obtain the coefficients b ~ {Elk,;}, and {Ak,,}. T i i done by the function , hs s dir2cas given below.
function [bO,B.Al = dirZcas(b,a); Y. DIRECT-form to CASCADE-form conversion (cplxpair version)
The above function converts the b and a vectors into K x 3 B and A matrices. It begins by computing 4, which i equal to s (assuming a0 # 1). It then makes the vectors b and a of equal length by zeroIIR Filter Structures
187
padding the shorter vector. This ensures that each biquad has a nonzero numerator and denominator. Next it computes the roots of the B ( z ) and A ( z )polynomials. Using the cplxpair function, these roots are ordered in complex conjugate pairs. Now every pair is converted back into a secondorder numerator or denominator polynomial using the poly function. The cascade form is implemented using a casf iltr function, which is described below. It employs the f i l t e r function in a loop using the coefficients of each biquad stored in B and A matrices. The input is scaled by bO, and the output of each filter operation is used as an input to the next filter operation. The output of the final filter operation is the overall output.
function y = casfiltr(bO.B,A,x); % CASCADE form realization of IIR and FIR filters
y = bO*w(K+l,:);
The following MATLAB function, cas2dir, converts a cascade form hs to a direct form. T i is a simple operation that involves multiplication of function several second-order polynomials. For this purpose the MATLAB conv is used in a loop over K factors.
function [b,a] = cas2dir(bO,B,A); % CASCADE-to-DIRECT form conversion
In this form the system function H(z) is written as a sum of second order sections using partial fraction expansion (PFE).
only
ii M > N
where K is equal to and &,o, & , Ir & , I r and A k . 2 are red numbers representing the coefficients of second-order sections. The second-order section
g,
with
is the kth proper rational biquad section. The filter input is available to all biquad sections as well as to the polynomial section if M 2 N (which is an FIR part). The output from these sections is summed to form the filter output. Now each biquad section H k ( z ) can be implemented in direct form 1 . Due to the summation of subsections, a parallel structure can be 1 built to realize H ( z ) . As an example, consider M = N = 4. Figure 6.7 shows a parallel form structure for this fourth-order IIR filter.
190
The dir2cas function first computes the t-domain partial fraction expansion using the residuez function. We need to arrange pole-and-residue pairs into complex conjugate pole-and-residue pairs followed by real poleand-residue pairs. To do this, the cplxpair function from MATLABcan be used; this sorts a complex array into complex conjugate pairs. Howal ever, two consecutive cls to this function, one each for pole and residue arrays, will not guarantee that poles and residues will correspond to each other. Therefore a new cplxcomp function is developed, which compares two shufRed complex arrays and returns the index of one array, which can be used to rearrange another array.
function I = cplxcomp(pl,p2) % I = cplxcomp(pl,p2) X Compares two complex pairs which contain the same scalar elements % but (possibly) at differrent indices. This routine should be 1 used after CPLXPAIR routine for rearranging pole vector and its % corresponding residue vector. % p2 = cplxpair(p1) % I[ ; =] for j=1:1:length(p2) for i=l:l:length(pl) if ( b ( l i - 2 j ) asp()p() < 0.0001) I=[I,il; end end end I=I';
After collecting these pole-and-residue pairs, the dir2cas function computes the numerator and denominator of the biquads by employing the residuez function in the reverse fashion. These parallel form coefficients are then used in the function parf iltr, which implements the parallel form. The parf iltr function uses the filter function in a loop using the coefficients of each biquad stored in the B and A matrices. The input is first filtered through the FIR part C and stored in the first row of a Y matrix. Then the outputs of all
To obtain a direct form from a parallel form, the function par2dir can be used. It computes poles and residues of each proper biquad and combines these into system poles and residues. Another call of the residuez function in reverse order computes the numerator and denominator polynomials.
function Cb,a] = parZdir(C,B.A); % PARALLEL-to-DIRECT form conversion
What would be the overall direct, cascade, or parallel form if a structure contains a combination of these forms? Consider the block diagram shown in Figure 6.9. This structure contains a cascade of two parallel sections. The first parallel section contains two biquads, while the second one contains three biquads. We will have to convert each parallel section into a direct form using the par2dir function, giving us a cascade of two direct forms. The overall direct form can be computed by convolving the corresponding numerator and denominator polynomials. The overall c d e and parallel forms can now be derived from the direct form.
which is a linear convolution of finite support. The order of the filter is M - 1, while the length of the filter (which is equal to the number of coe5cients) is M . The FIR filter structures are always stable, and they ate relatively simple compared to IIR structures. Furthermore, FIR filters can be designed to have a linear-phase response, which is desirable in some applications. We will consider the following four structures:
0 Direct form: In this form the difference equation (6.7) is implemented directly as given. 0 Cascade form: In this form the system function H ( z ) in (6.5) is factored into second-order factors, which are then implemented in a cascade connection. 0 Linear-phase f o m When an FIR filter has a linear phase response, its impulse response exhibits certain symmetry conditions. In this form we exploit these symmetry relations to reduce multiplications by about half. 0 Frequency sampling form: This structure is based on the DFT of the impulse response h(n) and leads to a parallel structure. It is also suitable for a design technique based on the sampling of fiequency response H ( e3") .
We w l briefly describe the above four forms along with some exami l ples. The MATLAB function d i r l c a s developed in the previous section is also applicable for the cascade form.
197
FIR Filter Structures
FIGURE 6.10
Direct form FIR structure
DIRECT FORM
The difference equation (6.7) is implemented as a tapped delay line since there are no feedback paths. Let M = 5 (i.e., a fourth-order FIR filter); then
y(n) = boz(n)
+ bl+
- 1)
+ 4 z ( n - 2) + 4 z ( n - 3) + b4s(n - 4 )
The direct form structure is given in Figure 6.10. Note that since the denominator is equal to unity, there is only one direct form structure.
MATLAB IMPLEMENTATION
In MATLAB direct form FIR structure is described by the row vector the
b containing the {b,} coefficients. The structure is implemented by the filter function, in which the vector a is set to the scalar value 1 as
discussed in Chapter 2.
CASCADE FORM
T i form is similar to that of the IIR form. The system function H ( t ) hs
is converted into products of second-order sections with real coefficients. These sections are implemented in direct form and the entire filter a a s d e of second-order sections. Elom (6.5)
H(t)= b o + b l ~ - ' + . . . + b ~ - l ~ - ~ + '
(6.8)
= bo
k=l
n
K
(1
+B k , l Z - l +
Bk,2Y2)
where K is equal to and &,I and B k , 2 are real numbers representing the coefficients of second-order sections. For M = 7 the cascade form is shown in Figure 6.11.
[%I,
FIGURE 6.11
Cascade form
FIR structure
Chapter 6
W
198
DIGITAL FILTER STRUCTURES
MATLAB IMPLEMENTATION
Although it is possible to develop a new MATLABfunction for the FIR cascade form, we will use our dir2cas function by setting the denominator vector a equal to 1. Similarly, cas2dir can be used to obtain the direct form from the cascade form.
LINEAR-PHASE FORM
For frequency-selectivefilters (e.g., lowpass filters) it is generally desirable to have a phase response that is a linear function of frequency; that is, we want
LH(e3") = /3
- aw,
-7r
< w . 7r I
(6.9)
where p = 0 or &s/2 and a is a constant. For a causal FIR filter with impulse response over [O, M - 1 interval, the linear-phase condition (6.9) 1 imposes the following symmetry conditions on the impulse response h(n): h(n)= h(M - 1- n); h(n)= -h(M - 1- 71);
p = 0, 0 5 n 5 M - 1 p = &7r/2, 0 5 n 5 M - 1
(6.10) (6.11)
An impulse response that satisfies (6.10) is called a symmetric impulse response, while that in (6.11) is called an antisymmetric impulse response. These symmetry conditions can now be exploited in a structure called the linear-phase form. Consider the difference equation given in (6.7) with a symmetric impulse response in (6.10). We have
Clearly, this structure requires 50%fewer multiplications than the direct form. A similar structure can be derived for an antisymmetricimpulse response.
MATLAB
IMPLEMENTATION
0
EXAMPLE 6.4
The linear-phase structure is essentially a direct form drawn differently to save on multiplications. Hence in a MATLAB implementation the linears phase structure i equivalent to the direct form.
An FIR filter is given by the system function
H ( t ) = 1 + 1 6 1 ~ + .~ - ~ - z
16
Determine and draw the direct, linear-phase, and cascade form StNCtures.
a. D m t f o m The difference equation is given by
y(n) = z(n)
+ 16.0625z(n - 4)+ z(n - 8) + i ( n - 8)] + 16.0625z(n - 4)
and the direct form structure is shown in Figure 6.13(a). b. Liiearphase f o m The difference equation can be written in the form
y(n) = ( ) .I [ T
For the filter in Example 6.4 what would be the structure if we desire a cascade form containing linear-phase components with real coefficients? We are interested in cascade sections that have symmetry and real coefficients. From the properties of linear-phase FIR filters (see Chapter 7 , if such a filter ) has an arbitrary m o at z = rL0, then there must be three other zeros at (l/r)LO, rL -0, and (l/r)L -0 to have real filter coefficients. We can now make use of this property. First we will determine the zero locations o the given f eighth-order polynomial. Then we will group four zeros that satisfy the above property to obtain one (fourth-order) linear-phase section. There are two such sections, which we will connect in cascade.
In this form we use the fact that the system function H ( 2 ) of an FIR filter can be reconstructed from its samples on the unit circle. From our discussions on the DFT in Chapter 5 we recall that these samples are i n fact the M-point DFT values { H ( k ), 0 5 k 5 M - 1) of the 114-point impulse response h (n).Therefore we have
Using this procedure, we obtain [see (5.17) on page 1271
This shows that the DFT H ( k ) , rather than the impulse response h (n) (or the difference equation), is used in this structure. It is also interesting to note that the FIR filter described by (6.12) has a recursive form similar to an IIR filter because (6.12) contains both poles and zeros. The resulting filter is an FIR filter since the poles at WGk are canceled by the roots of 1 - 2 -M = 0 The system function in (6.12) leads to a parallel structure as shown in Figure 6.15 for M = 4. One problem with the structure in Figure 6.15 is that it requires a complex arithmetic implementation. Since an FIR filter is almost always a real-valued filter, it is possible to obtain an alternate realization in which only real arithmetic is used. This realization is derived using the symmetry properties of the DFT and the W i kfactor. Then (6.12) can be expressed
Note that the DFT samples H (0) and H (M/2) are real-valued and that is the third term on the right-hand side of (6.13) absent if M is odd. Using (6.13)and (6.14), show a frequency sampling structure in Figure 6.16 we for M = 4 containing real coefficients.
c
:
Z-1
.
FIGURE 6.16
FI-equency sampling structure for M = 4 with real weficients
FIR Filter Structures
203
MATLAB
IMPLEMENTATION
Given the impulse response h ( n ) or the DF'T H ( k ) , we have to deterfunction, mine the coefficients in (6.13) and (6.14). The following MATLAB dir2f s, converts a direct form ( h (n) values) to the frequency sampling form by directly implementing (6.13) and (6.14).
function [C,B,Al = dir2fs(h) % Direct form to Frequency Sampling form conversion
In the above function the impulse response values are supplied through the h array. After conversion, the C array contains the gain values for each parallel section. The gain values for the second-order parallel sections are given first, followed by H (0) and H ( M / 2 ) (if M is even). The B matrix contains the numerator coefficients, which are arranged in length2 row vectors for each second-order section. The A matrix contains the denominator coefficients, which are arrangd in length-3 row vectors for the second-order sections corresponding to those in B, followed by the coefficients for the first-order sections.
204
Chapter 6 I DIGITAL FILTER STRUCTURES
A practical problem with the structure in Figure 6.16 is that it has poles on the unit circle, which makes this filter critically unstable. If the filter is not excited by one of the pole frequencies, then the output is z bounded. We can avoid this problem by sampling H ( z ) on a circle II = 7, where the radius r is very close to one but is less than one (e.g., r = 0.99), which results in
Since M = 5 is odd, there is only one first-order section. Hence
H(z)=
1-0.809 + 0.8092-' -0'5818 1 - 0.6182-1 + 5
2-5
2-2
+ 0.0848 1 + 1.6182.-1 + 2-2 +
0.309 - 0.309z-'
The frequency sampling form is shown in Figure 6.17.
0
FIR Filter Structures
205
-0.809 0.618 0.809
I
0.5818
xfn)
0.2
I
- . ...
V
t 4 &
-
-1
>
I
t2-l
0.309 -1.618
J
0.0848'.
:* y ( n )
2-1
-0.309
-
FIGURE 6.17 m e n c y sampling structure in Ezample 6.6
0
EXAMPLE 6.7
The frequency samples of a 32-point linear-phase FIR filter are given by
( H ( k ) (= 0.5, 0,
{
1,
k=0,1,2 k=3 k = 4 , 5 ,..., 15
Determine its frequency sampling form, and compare its computational complexity with the linear-phase form.
Sokrtiotl
In this example since the samples o the DFT H ( k ) are given, we could use f (6.13) and (6.14) directly to determine the structure. However, we will use the dir2fs function for which we will have to determine the impulse response h (n). Using the symmetry property and the linear-phase constraint, we assemble the DFT H ( k ) as
To determine the computationalcomplexity, note that since H (0) = 1, the limb order section requires no multiplication, while the three second-order sections require three multiplications each for a total of nine multiplications per output sample. The total number of additions is 13. To implement the linear-phase structure would require 16 multiplications and 31 additions per output sample. Therefore the frequency sampling structure of this FIR filter is more efficient 0 than the linear-phase structure.
LATTICE FILTER STRUCTURES
I
I
The lattice filter is extensively used in digital speech processing and in the implementation of adaptive filters. It is a preferred form of realization over other FIR or IIR filter structures because in speech analysis and in speech synthesis the small number of coefficients allows a large number of formants to be modeled in real time. The all-zero lattice is the FIR filter representation of the lattice filter, while the lattice ladder is the IIR filter representation.
ALL-ZERO LATTICE FILTERS
An FIR filter of length M (or order M - 1) has a lattice structure with M - 1 stages as shown in Figure 6.18. E c stage of the filter has an input ah and output that are related by the order-recursive equations [19]: fm(n)= fm-l(n) Kmgm-l(n- l), m = 1,2,. ..,M
g,(n)=Kmfm-i(n)+gm-l(n-l),
+
-1
m = l , 2 ,... , M - 1
(6.16)
where the parameters K,, rn = 1,2,. ..,M - 1, called the mfiectzon mficient s, are the lattice filter coefficients. If the initial values of fm(n) and g,(m) are both the scaled value (scaled by KO) the filter input of z(n), then the output of the (M - 1) stage lattice filter corresponds to
208
The equations (6.20) can also be used to determine the direct form coefficients {b,} from the lattice filter coefficients {K,} using a recursive procedure [19]:
Ao(2) = J ~ ( z= 1 )
A,(z)=Am-l(z)+Kmz-lJ,-~(z), Jm(z) = z-,A, (z-') ,
bm = KOCXM-I (m),
m = l , 2 ,...,M - 1 m = 1,2,. ..,M - 1 m = 0, 1, . ..,A 4
(6.21)
-1
The MATLAB function latcldir given below implements (6.21). Note ( z ) is obtained by convolving the two corthat the product K,z-'J,-l responding arrays, while the polynomial Jm(z) is obtained by using a fliplr operation on the Am(z)polynomial.
function [bl = latc2dir(K) X All-Zero Lattice form to FIR Direct form Conversion
A lattice structure for an IIR filter is restricted to an all-pole system function. It can be developed from an FIR lattice structure. Let an allpole system function be given by
1
(6.22)
m=l
which from (6.19) is equal to H ( z ) = &. Clearly, it is an inverse system to the FIR lattice of Figure 6.18 (except for factor 4).T i IIR hs filter of order N has a lattice structure with N stages as shown in Figure 6.20. Each stage o the filter has an input and output that are related by f
212
Chapter 6
DIGITAL FILTER STRUCTURES
the order-recursive equations [19]:
fN(n) =.. )(
fm-l(n) = fm(n) - K,g,-l(n gm(n) = Kmfm-l(n)
= f o ( 4 =go(4
+ grn-l(n - l),
- l),
m =N, N m =N, N
- 1,.. . ,1 (6.23) - 1,. .., l
where the parameters K,, m = 1 , 2 , ... ,M - 1, are the reflection coefficients of the all-pole lattice and are obtained from (6.20) except for KO, which is equal to 1.
MATLAB IMPLEMENTATION
Since the IIR lattice coefficients are derived from the same (6.20) procedure used for an FIR lattice filter, we can use the d i r l l a t c function in MATLAB. Care must be taken to ignore the KOcoefficient in the K array. Similarly, the latcldir function can be used to convert the lattice {Km} coefficients into the direct form {aN(m)}provided that KO= 1 is used as the first element of the K array. The implementation of an IIR lattice is given by (6.23), and we will discuss it in the next section. Consider an all-pole IIR filter given by
1 H ( z ) = 1 + & - Pz-2 + l z - 3 &+ '
The direct form and the lattice form structures of this IIR filter are shown in Figure 6.21. 0
LATTICE-
LADDER
FILTERS
A general IIR filter containing both poles and zeros can be realized as a lattice-type structure by using an all-pole lattice as the basic building block. Consider an I R filter with system function I
where, without loss of generality, we assume that N 2 M. A lattice type structure can be constructed by first realizing an all-pole lattice with coefficients K,,,, 1 5 m 5 N for the denominator of (6.24), and then adding a ladder part by taking the output as a weighted linear combination of {g,(n)} as shown in Figure 6.22 for M = N. The result
is a pole-zero IIR filter that has the lattice-ladder structure. Its output is given by
(6.25) where {Cm}are called the ladder coeficients that determine the zeros of the system function H ( z ) . It can be shown [19] that {Cm)are given by
M
(6.26) where Jm(z) is the polynomial in (6.20). Rom (6.26) one can obtain a recursive relation
Bm(z) = Bm-i(z)
. + CmJm(z); m = 0,2,..,M
or equivalently,
M
Cm = bm+
C,ai(i-m); m = M , M - 1,...,0
i=m+l
(6.27)
from the definitions of B,(z) and Am(z). To obtain a latticsladder structure for a general rational IIR filter, we can first obtain the lattice coefficients {Km} from A N ( z )using the recur62) sion ( . 0 .Then we can solve (6.27) recursively for the ladder coefficients {Cm} to realize the numerator BM(z).This is done in the MATLAB function dirlladr given below. It can a s be used to determine the all-pole lo lattice parameters when the array b is set to b;- C11.
function CK,Cl = dir2ladr(b,a) % IIR Direct form to pole-zero LatticdLadder form Conversion
Note: To use this function, N 2 M. If Ad > N , then the numerator A N ( z )should be divided into the denominator B M ( z )using the deconv function to obtain a proper rational part and a polynomial part. The proper rational part can be implemented using a latticeladder structure, while the polynomial part is implemented using a direct structure. To convert a latticeladder form into a direct form, we first use the recursive procedure in (6.21) on {K,,,} coefficients to determine { a ~ ( k ) } and then solve (6.27) recursively to obtain { b ~ ( k ) }This is done in the . MATLAB function ladr2dir given below.
function Cb.4 = ladrZdir(K,C) % Lattice/Ladder form to IIR Direct form Conversion
The latticeladder filter is implemented using (6.23) and (6.25). T i hs is done in the MATLAB function ladrf ilt,which is given below. It should be noted that due to the recursive nature of this implementation along
216
Chapter 6
DIGITAL FILTER STRUCTURES
with the feedback loops, this MATLAB function is neither an elegant nor s an efficient method of implementation. It i not possible to exploit MATLAB'S inherent parallel processing capabilitiesin implementing this latticeladder structure.
function Cyl = ladrfilt(K.C,x) X LATTICE/LADDER form realization of IIR filters
The resulting direct form and the lattice-ladder form structures are shown in Figure 6.23. To check that our lattice-ladder structure i correct, let us compute s the first 8 samples of its impulse response using both forms.
is to be implemented using a flowgraph of the form shown in Figure 6.26.Fill in all the coefficients in the diagram.
P. 67
.. Consider the linear time-invariant system given in Problem 6 5
FIGURE 6.25
Structure for Pmblem 6.5
220
Chapter 6
DIGITAL FILTER STRUCTURES
FIGURE 6.26
Strueture for Pwblem 6.6
P. 68
It i to be implemented using a flowgraph of the form shown in Figure 6.27. s a. Fill in all the mfficients i the diagram. n b. Is your solution unique? Explain. An FIR filter is described by the difference equation
Determine and draw the block diagrams of the following structures. a Direct form b. Linear-phase form
-
L
-
5-
FIGURE 6.27 Structure for Problem
6.7
Problems
221
c. C d e form d. Frequency sampling form P . A linear timeinvariant system is given by the system function 69
10
H(z)= k=O
(2z)-"
Determine and draw the block diagrams of the following structures. a. Direct form b. Cascade form c. Lattice form d. Frequency sampling form P . 0 Using the conjugate symmetry property of the DFT 61
and M is even. b. Modify the MATLAB function dir2fs (which was developed in this chapter) to implement the above frequency sampling form. The format of this function should be
[C,B .A,rM] = dirlfs(h,r) % Direct form to Frequency Sampling form conversion
1: h = impulse response vector of an FIR filter X r = radius of the circle over which samples are taken (rcl) 1:
c. Determine the frequency sampling structure for the impulse response given in Example 6.6 using the above function. P6.12 Determine the impulse response of an FIR filter with lattice parameters
FIR FILTERDESIGN
We now turn our attention to the inverse problem of designing systems from the given specifications.It is an important as well as a difficult problem. In digital signal processing there are two important types of systems. The first type of systems perform signal filtering in the time domain and hence are called dzgital filters. The second type of systems provide signal representation in the frequency domain and are called spectrum analyzers. In Chapter 5 we described signal representations using the DFT. In this and the next chapter we will study several basic design algorithms for both FIR and IIR filters. These designs are mostly of the frequency selective type; that is, we will design primarily multiband lowpass, highpass, bandpass, and bandstop filters. In FIR filter design we will also consider systems like differentiators or Hilbert transformers, which, although not frequency-selective filters, nevertheless follow the design techniques being considered. More sophisticated filter designs are based on arbitrary frequency-domain specifications and require tools that are beyond the scope of this book. We first begin with some preliminary issues related to design philos ophy and design specifications. These issues are applicable to both FIR and IIR filter designs. We will then study FIR filter design algorithms in the rest of this chapter. In Chapter 8 we will provide a similar treatment for IIR filters.
PRELIMINARIES
I
I
The design of a digital filter is carried out in three steps:
0 SpeciRcations: Before we can design a filter, we must have some specifications. These specificationsare determined by the applications. Approximations: Once the specificationsare defined, we use various concepts and mathematics that we studied so far to come up with a filter description that approximates the given set of specifications. This step i s the topic of filter design.
224
0 Implementation: The product of the above step is a filter descrig tion in the form of either a difference equation, or a system function H ( z ) , or an impulse response h(n). Rom this description we implement the filter in hardware or through software on a computer as we discussed in Chapter 6 .
In this and the next chapter we will discuss in detail only the second step, which is the conversion of specifications into a filter description. In many applications like speech or audio signal processing, digital filters are used to implement frequency-selective operations. Therefore, specifications are required in the frequency-domain in terms of the desired magnitude and phase response of the filter. Generally a linear phase response in the passband is desirable. In the case of FIR filters, it is possible to have exact linear phase as we have seen in Chapter 6. In the case of IIR filters a linear phase in the passband is not achievable. Hence we will consider magnitude-only specifications. The magnitude specifications are given in one of two ways. The first approach is called absolute specifications, which provide a set of requirements on the magnitude response function IH(ej")l. These specifications are generally used for FIR filters. IIR filters are specified in a somewhat different way, which we will discuss in Chapter 8. The second approach is called relative specifications, which provide requirements in decibels (dB), given by
This approach is the most popular one in practice and is used for both FIR and IIR filters. To illustrate these specifications, we will consider a lowpass filter design as an example. ABSOLUTE SPECIFICA-
A typical absolute specification of a lowpass filter is shown in Figure 7.la,
in which
0 band [0,wp]is called the p s b a n d , and 61 is the tolerance (or ripple) that we are willing to accept in the ideal passband response, 0 band [wS,n] is called the stopband, and 62 is the corresponding tolerance (or ripple), and 0 band [wp,ws] called the transition band, and there are no restricis tions on the magnitude response in this band.
TIONS
RELATIVE (DB) SPECIFICATlONS
A typical absolute specification of a lowpass filter is shown in Figure 7.lb,
in which
0 0
Rp
is the passband ripple in dB, and
A, is the stopband attenuation in dB.
225
Preliminaries
I
I
I I
I
i
Transition band
4
I
Stopband ripple
0
RP
I
0
I
I
FIGURE 7.1 FIR filter specificutions: (a) Absolute (a) Relative
The parameters given in the above two specifications are obviously related. Since IH(ej")l, in absolute specifications is equal t o (I 61), we have
+
and
0
EXAMPLE 7.1
In a certain filter's specifications the passband ripple is 0.25 dB, and the stop band attenuation is 50 dB. Determine 61 and 62.
Using (7.1), we obtain
The above specifications were given for a lowpass filter. Similar specifications can also be given for other types of frequency-selectivefilters, such as highpass or bandpass. However, the most important design parameters are frequency-band tolernnces (or ripples) and band-edge frequencies. Whether the given band is a passband or a stopband is a relatively minor issue. Therefore in describing design techniques, we will concentrate on a lowpass filter. In the next chapter we will discuss how to transform a lowpass filter into other types of frequency-selectivefilters. Hence it makes more sense to develop techniques for a lowpass filter so that we can compare these techniques. However, we will also provide examples of other types of filters. In light of this discussion our design goal is the following.
Problem Statement Design a lowpass filter (i.e., obtain its system function H ( z ) or its difference equation) that has a passband [O,w,] with tolerance 6 (or Rp in dB) and a stopband [w.,, n] with tolerance 62 (or 1 A, in dB). In this chapter we turn our attention to the design and approximation of FIR digital filters. These filters have several design and implementational advantages:
0
0
The phsse response can be exactly linear. They are relatively easy to design since there are no stability probThey are efficient to implement. The DFT can be used in their implementation.
lems.
0
0
As we discussed in Chapter 6, we are generally interested in lineare phase frequency-selective FIR filters. Advantages of a linear-phase r sponse are:
0
design problem contains only real arithmetic and not complex arith-
metic;
0 linear-phase filters provide no delay distortion and only a 6x4 amount of delay; 0 for the filter of length M (or order M - 1) the number of operations are of the order of M / 2 as we discussed in the linear-phase filter implementation.
We first begin with a discussion of the properties of the linear-phase
FIR filters, which are required in design algorithms. Then we will discuss
Preliminaries
In this section we discuss shapes of impulse and frequency responses and locations of system function zeros of linear-phase FIR filters. Let h(n), 0 5 n 5 M - 1 be the impulse response of length (or duration) M. Then the system function is
M-1 M-1
H(z)=
n=O
h(n)z-" = z - ( ~ - ' )
n=O
h(n)~~-'-~
which has (M - 1) poles at the origin z = 0 (trivial poles) and (M - 1) zeros located anywhere in the z-plane. The frequency response function i s
n=O
Now we will discuss specific requirements on the forms of h(n) and H(&") as well as requirements on the specific locations of (M - 1) zeros that the linear-phase constraint imposes.
IMPULSE RESPONSE h(n)
We impose a linear-phase constraint
f H ( d W ) -aw, =
-T
< w <_ -?F
where a is a constant phase delay. Then we know from Chapter 6 that h(n)must be symmetric, that is,
h ( n )= h ( M - 1 - a ) ,
0 _< n 5 ( M - 1) with a =
M-1 2
(7.3)
Hence h(n) is symmetric about a,which is the index of symmetry. There are two possible types of symmetry:
0 M odd: In this case a = (M - 1)/2 is an integer. The impulse response is as shown below.
228
Chapter 7
FIR FILTER DESIGN
0 M even: In this case a = ( M - 1)/2 is not an integer. The impulse response is as shown below.
Q
Q
Q
0
E
C
0-
P
d q
LH(e+) = p - aw
d f H(ejw)
We also have a second type of "linear-phase" FIR filter if we require that the phase response LH(ejW) satisfy the condition
which is a straight line but not through the origin. In this case a is not a constant phase delay, but
___ = -ff
dw
is constant, which is the group delay. Therefore a is called a constant group delay. In this case, 8s a group, frequencies are delayed at a constant rate. But some frequenciesmay get delayed more and others delayed less. For this type of linear phase one can show that
This means that the impulse response h(n) is antisymmetric. The index of symmetry is still a = (M - 1)/2. Once again we have two possible types, one for M odd and one for M even.
0 M odd: In this case a = (M - 1)/2 is an integer and the impulse response is as shown below.
1
g o
-
P
b
i ;
P ?
Note that the sample h(0) at a = (M - 1)/2 must necessarily be equal to zero, i.e., h((M - 1)/2) = 0. 0 M even: In this case a = (M - 1)/2 is not an integer and the impulse response is as shown below.
0
fO
--
?
d
0
I
P
P
FREQUENCY RESPONSE H(ejw)
When the cases of symmetry and antisymmetry are combined with odd and even M, we obtain four types of linear-phase FIR filters. Requency response functions for each of these types have some peculiar expressions and shapes. To study these responses,we write H ( P ) as
230
Chapter 7
FIR FILTER DESIGN
where Hp(w) is an amplitude response function and not a magnitude response function. The amplitude response i a real function, but unlike s the magnitude response, which is always positive, the amplitude response may be both positive and negative. The phase response associated with the magnitude response is a discontinuow function, while that associated with the amplitude response is a continuous linear function. To illustrate the difference between these two types of responses, consider the following example.
0
EXAMPLE 7.3
since c a w can be both p i t i v e and negative. In this c s the phase response ae is piecewise linear. On the other hand, the amplitude and the corresponding phase responses are
H,(w) = 1
LH
+SCOSW,
-7r<W<K
(2") -w, =
In this case the phase response is truly linear. These responses are shown in Figure 7.2. From this example the difference between the magnitude and the amplitude (or between the piecewise linear and the linear-phase) responses should be clear. 0
Note: At w = 0 and w = 'IT we have H,(w) = 0, regardless of c(n) or h(n). Furthermore, ej*I2 = j , which means that jH,(w) is purely imaginary. Hence this type of filter is not suitable for designing a lowpass filter or a highpass filter. However, this behavior is suitable for approximating ideal digital Hilbert transformers and differentiators. An ideal Hilbert transformer [19] is an all-pass filter that imparts a 90"phase shift on the input signal. It is frequently used in communication systems for modulation purposes. Differentiators are used in many analog and digital systems to take the derivative of a signal. 233
Properties of Linear-phase FIR Filters
m4 linear-phase FIR filter: Antisynmetric impulse response, M even This case is similar to Type-2. We have (seeProblem
7.4)
The MATLAB routine freqz computes the frequency response but we. cannot determine the amplitude response from it because there is no function in MATLAB comparable to the abs function that can find amplitude. However, it easy to write simple routines to compute amplitude responses for each of the four types. We provide four functions to do this.
1. Hr-typel:
function ~Hr,w,a,Ll = Hr-TypelCh);
These four functions can be combined into one function. T i function hs can be written to determine the type of the linear-phase filter and to implement the appropriate amplitude response expression. This is explored in Problem 7.5. The use of these functions is dmribed in Examples 7.4 through 7.7.
ZERO
LOCATIONS
Recall that for an FIR filter there me (M - 1) (trivial) poles at the origin and (M - 1 ) zeros located somewhere in the z-plane. For linear-phase FIR filters, these zeros possess certain symmetries that are due to the symmetry constraints on h(n). It can be shown (see I191 and Problem 7.6) that if H ( z ) has a zero at
Z
= 2 = T€?' 1
then for linear phase there must be a zero at
= - = Ae-jO 1
z i
T
For a real-valued filter we also know that if z1 is complex,then there must be a conjugate zero at z;l = re-j', which implies that there must be a zero at I/%: = ( l / ~eje. Thus a general zew constellation is a quadruplet )
as shown in Figure 7.3. Clearly, if r = 1, then I/T 1, and hence the =
zeros are on the unit circle and occur in pairs
e3'
and
e-je
If B = 0 or 0 = R, then the zeros are on the real line and occur in pairs
T
and
-
1
Finally, if r = 1 and B = 0 or 0 = K , the zeros are either at z = 1 or z = -1. These symmetries can be used to implement cascade forms with linear-phase sections.
236
Chapter 7
FIR FILTER DESIGN
PoleZero Plot
I
z-plane
I
4.5}
I
,
-1.5
-1
0
1
FIGURE 7.3
I
-0.5
0
0.5
1
1.5
real axis
A g e n d aem wnstelh%taon
In the following examples we illustrate the above described properties of linear-phase FIR filters.
The plots and the zero locations are shown in Figure 7.4. From these plots we observe that there are no restrictions on H, ( w ) either at w = 0 or at w = R . There is one zeroquadruplet constellation and three zero pairs. 0
0
EXAMPLE 7.5
The plots and the zero locations are shown in Figure 7.6. Rom these plots we observe that H , (w) 0 at w = 0 and at w = T. There is one zer-quadruplet = constellation, two zero pairs, and zeros at w = 0 and w = ?r BQ expected. 0
0
The plots and the zero loeations are shown in Figure 7.7. Fmm these plots we observe that H , ( w ) i zero at w = 0. There is one zerc-quadruplet constellation, s three zero pairs, and one zero at w = 0 ax expected. 0
A
gl:r
I .
I
0
5
10
n
frequency in pi units
an) cceffkients
Pde-rnrn
lo
I
.
0
-5
0
5
10
-1
0
1
n
real axis
FIGURE 7.7
Plots a Ezample 7.7 n
242
Chapter 7
FIR FILTER DESIGN
WINDOW DESIGN TECHNIQUES
I 1
The basic idea behind the window design is to choose a proper ideal frequency-selective filter (which always has a noncausal, idnite-duration impulse response) and then truncate (or window) its impulse response to obtain a linear-phase and causal FIR filter. Therefore the emphasis in this method is on selecting an appropriate windowing function and an appropriate ideal filter. We will denote an ideal frequency-selective 6lter by &(dw), which has a unity magnitude gain and linear-phase characteristics over its passband, and zero response over its stopband. An ideal LPF of bandwidth wc < n is given by
(7.18)
where w, is also called the cut08 frequency, and Q is called the sample delay (note that from the DTFT properties, e-jw implies shift in the positive n direction or delay). The impulse response of this filter is of infinite duration and is given by
hd(n) = F-' [Hd(&")] =
1
/
7 7
ffd('?)dw"dw
(7.19)
=
L
2n
7
-us
--*
1. e-jawejwn&
- sin [w,(n - a)] *(n - a)
Note that hd(n) is symmetric with respect to a,a fact useful for linearphase FIR filters. To obtain an FIR filter from hd(n),one has to truncate hd(n)on both sides. To obtain a causal and linear-phase FIR filter h(n)of length M , we must have h(n)=
hd(n), 0 5 7 l 5 &f - 1 0, elsewhere
and
a=-
M-1
2
(7.20)
This operation is called "windowing." In general, h(n)can be thought of as being formed by the product of hd(n) and a window function w(n) as follows:
h(n)= hd(n)w(n)
Window Design Techniques
(7.21)
243
where some symmetric function with respect to w(n) = a over 0 5 n 5 M - 1 0, otherwise
{
Depending on how we define w(n) above, we obtain daerent window designs. For example, in (7.20) above
w(n) =
1, O < n < M - 1 = RM (4 0, othenvise
which is the rectangular window defined earlier. In the frequency domain the causal FIR filter response H(e3") is given by the periodic convolution of Hd(ej") and the window response W ( 2 " ) ; that is,
(7.22)
T i is shown pictorially in Figure 7.8 for a typical window response, from hs which we have the following observations:
1. Since the window w ( n ) has a finite length equal to M , its response has a p k y main lobe whose width is proportional to 1/M, and has side lobes of smaller heights.
0
-n
-oc
0
Max side-lobe
height
w
Minimum stopband attenuation
FIGURE 7.8
Wandowing operation in the frequency domain
244
Chapter 7
FIR FILTER DESIGN
2. The periodic convolution (7.22) produces a smeared version of the ideal response Hd(GW). 3. The main lobe produces a transition band in H(eJw) whose width is responsible for the transition width. This width is then proportional to 1/M. The wider the main lobe, the wider will be the transition width. 4. The side lobes produce ripples that have similar shapes in both the passband and stopband.
Basic Window Design Idea For the given filter specificationschoose the filter length M and a window function w(n) for the narrowest main lobe width and the smallest side lobe attenuation possible. 1 Rom observation 4 above we note that the passbandfolerance 6 and the stopband tolerance & cannot be specified independently. We generally , , , take care of & alone, which results in S = 61. We now briefly describe various well-known window functions. We will use the rectangular window a an example to study their performances in the frequency domain. s
RECTANGULAR WINDOW
This is the simplest window function but provides the worst performance from the viewpoint of stopband attenuation. It was defined earlier by
w(n) =
1, O l n l M - I 0, otherwise
(7.23)
Its frequency response function is
which is the amplitude response. From (7.22) the actual amplitude response H, ( w ) is given by
W,(X)dX=--*
1 T s i n ( q ) dX, 2a sin(%)
~
M >> 1
(7.24)
-*
This implies that the running integral of the window amplitude response (or accumulated amplitude response) is necessary in the accurate analysis of the transition bandwidth and the stopband attenuation. Figure 7.9 shows the rectangular window function w (n),its amplitude response W ( w ) , the amplitude response in dB, and the accumulated amplitude response (7.24) in dB. From the observation of plots in Figure 7.9 we can make several observations.
Window Design Techniques
245
45
I
0
z
0
3 3
21
FIGURE 7.9
Rectangular window: M = 45
1. The amplitude response W, (w) has the first zero at w = w1, where
Hence the width of the main lobe is 2wl = 4 x / M . Therefore the approzimate transition bandwidth is 4 n l M . 2. The magnitude of the first side lobe (which is also the peak side lobe magnitude) is approximately at w = 3 r / M and is given by
Comparing this with the main lobe amplitude, which is q u a l to M, the peak side lobe magnitude is
2 - = 21.22% z 13 dB
3r
of the main lobe amplitude. 3. The accumulated amplitude response has the fist side lobe magnitude a 21 dB. This results in the minimum stopband attenuation of 21 t dB irrespective of the window length M .
246
Chapter 7
FIR FILTER DESIGN
4. Using the minimum stopband attenuation, the transition bandwidth can be accurately computed. It is shown in the accumulated hs amplitude response plot in Figure 7.9. Ti computed a a c t tmnseteon bandwidth is
w. -up = -
1.8~
M
which is about half the approximate bandwidth of 4alM. Clearly, this is a simple window operation in the time domain and an easy function to analyze in the frequency domain. However, there are two main problems. First, the minimum stopband attenuation of 21 dB is insufficient in practical applications. Second, the rectangular windowing it being a direct truncation of the infinite length h d (n), suffers from the Gibbs phenomenon. If we increase M, the width of each side Iobe will decrease, but the area under each lobe will remain constant. Therefore the relative amplitudes of side lobes will remain constant, and the minimum stopband attenuation will remain at 21 dB. This implies that al ripples l will bunch up near the band edges. It is shown in Figure 7.10. Since the rectangular window is impractical in many applications, we consider other window functions, many of which bear the names of the people who fist proposed them. Although these window functions can
M=21
~~
<
P I
=
go
-1
1 -1 50 0 frequency In pi M=Sl
1
!m
4
E"
frequencyIn pi units 0
1
M = 101
P
~~
3-1 go
I
heguencyinpiunits 0
1
I K 1
i E go
-1
B
frequency in pi units 0
FIGURE 7.10
Gabbs phenomenon
Window Design Techniques
247
also be analyzed similar to the rectangular window, we present only their results.
BARTLETT WINDOW
Since the Gibbs phenomenon results from the fact that the rectangular window has a sudden transition from 0 to 1 (or 1 to 0), Bartlett suggested a more gradual transition in the f o e of a triangular window, which is given by
otherwise
T i window and its frequency-domain responses are shown in Figure hs
7.11.
TriangularWindow : M 4 5
1
Amplitude Response in dB
I
szri
0 0
3
22 n
45
frequency In pi units
Amprtude Response
0
-1
0 frequency in pi units
1
60 -1
frequency in pi unlts
1
FIGURE 7.11 Bartlett (triangdar) window:
M = 45
Chapter 7
248
FIR FILTER DESIGN
Hanning Window :M + -
I
0
I
22
45
n
AmplitudeResponse
-1
0
1
frequencyinpiunits
frequency in pi units
FIGURE 7.12
Hanning window: M = 45
HANNING WINDOW
This is a raised cosine window function given by w(n) = {ll;[l-cm(*)],
OlnlM-1
otherwise
(7.26)
This window and its frequency-domain responses are shown in Figure 7.12.
HAMMING WINDOW
This window is similar to the Hanning window except that it has a small amount of discontinuity and is given by,
w(n) =
0.54-0.46cos(*),
O<n<M-l otherwise
(7.27)
This window and its frequency-domain responses are shown in Figure 7.13.
Window Design Techniques
249
Hamming Window :M=45
1
Amplitude Response in dB
0
* -
5
0
6 n
43 60
-1
-22
0
22
n Amplitude Response
0 1 frequency in pi units AccumulatedAmplitude Response
r
1
!
frequency in pi units
FIGURE 7.13
Hamming window: M = 45
BLACKMAN
WINDOW
T i window is also similar to the previous two but contains a second hs harmonic term and is given by
w(n) =
otherwise
(7.28)
This window and its frequency-domain responses are shown in Figure 7.14. In Table 7.1 we provide a summary of window function characteristics in terms of their transition widths (asa function of M) and their minimum stopband attenuations in dB. Both the approximate as well as the exact transition bandwidths are given. Note that the transition widths and the stopband attenuations increase as we go down the table. The Hamming window appears t o be the best choice for many applications.
KAISER WINDOW
This is one of the most useful and optimum windows. It is optimum in the sense of providing a large main lobe width for the given stopband attenuation, which implies the sharpest transition width. The window
where I0 [.] is the modified zero-order Bessel function, and p is a parame ter that depends on M and that can be chosen to yield various transition hs widths and near-optimum stopband attenuation. T i window can provide different transition widths for the same M, w i h is something other hc windows lack. For example,
0 if P = 5.658, then the transition width is equal to 7.8?r/M,and the hs minimum stopband attenuation is equal to 60 dB.T i is shown in Figure 7.15. 0 if = 4.538, then the transition width is equal to 5.8?r/M, and the minimum stopband attenuation is equal to 50 dB.
H n e the performance of this window is comparable to that of the H m ec
ming window. In additioa, the Kaiser window provides flexible transition bandwidths. Due to the complexity involved in the Beme1 functions, the design equations for this window are not easy to derive. Fortunately,
Kaiser Window :M 4 5
a
-1
0
1
-1
frequency in pi units
1
FIGURE 7.15
Kaiser window:
M
= 45,
= 5.658
252
Chapter 7
FIR FILTER DESIGN
Kaiser has developed empirical design equations, which we provide below without proof.
Using these routines, we can use MATLAB design FIR filters based to on the window technique, which also requires an ideal lowpass impulse response hd(n). Therefore it is convenient to have a simple routine that creates hd(n) as shown below.
function hd ideal_lp(vc,M); % I d e d LouPass filter computation
In the Signal Processing toolbox MATLAB provides a routine called f irl, which designs FIR filters using windows. However, this routine is not available in the Student Edition. To display the frequencydomain plots of digital filters, MATLABprovides the freqz routine, which we used in oiid earlier chapters. Using this routine, we have developed a m d f e version, called freqzsl, which returns the magnitude response in absolute as well as in relative dB scale, the phase response, and the group delay response. We will need the group delay response in the next chapter.
function [db.mag,pha,grd,ul = freqz-m(b.a); X Modified version of freqz subroutine
We now provide several examples of FIR filter design using window techniques and MATLAB routines. Design a digital FIR lowpass filter with the following specifications:
w, = 0.2?r, w, = 0.3?r,
I, = 0.25 dB ?,
A, = 50 dB
Choose an appropriate window function from Table 7 1 Determine the impulse ..
response and provide a plot of the frequency response of the designed filter. SdUtiOn
Both the Hamming and Blackman windows can provide attenuation of more than 50 dB. Let us choose the Hamming window, which provides the smaller
254
Chapter 7
W
FIR FILTER DESIGN
transition band and hence has the smaller order. Although we do not use the passband ripple value of Rp = 0.25 dB in the design, we will have to check the actual ripple from the design and verify that it is indeed within the given tolerance. The design steps are given in the following MATLAB script.
There are two transition bands, namely, Awl = wlp -wa and A m = w2. -ap. These two bandwidths must be the same in the window design; that is, there is no independent control over Awl and Am. Hence Awl = AWZ Aw. For this = design we can use either the K i e window or the Blackman window. Let us asr use the Blackman Window. We will also need the ideal bandpass filter impulse response hd (n). Note that this impulse response can be obtained from two ideal lowpass magnitude responses, provided they have the same phase response. Ti is shown in Figure 7 1 . Therefore the MATLAB hs .9 routine ideal-lp(vc,M) is sufficient to determine the impulse response of an ideal bandpass filter. The design steps are given in the following MATLABscript.
Note that the Blackman window length is M = 61 and that the actual stopband attenuation is 75 dB. The time and the frequencydomain plots are shown in Figure 7.20. 0
0
EXAMPLE 7.11
The frequency response of an idea! bandstop filter is given by
He(2") =
{
1,
0,
0 5 IWI < A/3 r / 3 5 IwlS 2 ~ / 3
1, 2 ~ / < IwI 3
5A
Using a Kaiser window, design a bandstop filter of length 45 with stopband attenuation of 60 dB. Note that in these design speofieations, the transition bandwidth is not given. It w l be determined by the length M = 45 and the parameter p of the K i e i l asr
n Magnitude Response in dB
-
i 5
0
FIGURE 7.20
2
0
4 n
0
6
0
0
0.2 0.35 0.65 0.8 frequency in pi units
1
Bandpass flteter plots in Example 7.10
Window Design Techniques
259
window. From the design equations (7.30) we can determine fl from A.; that is,
f l = 0.1102 X (A. - 8.7)
The ideal bandstop impulse response can also be determined from the ideal lowpass impulse response using a method similar to Figure 7.19. We can now implement the Kaiser window design and check for the minimum stopband attenuation. This is shown in the following MATLABscript.
The time- and the frequency-domain plots are shown in Figure 7.22, i which n 0 the designed filter satisfies the neceasBIy requirements.
0.8
1
-
0.6 0.4
z
0 1 0 2 0
0.8
Q0.6 0.4 0.2
0
02
0
-. 02
n
3
0
4
0
0
1
0
2
0 3 n
0
4
0
Actual Impulse Response
Magnitude Rebponse in dB
n
FIGURE 7.22
Bandstop filter plots in Emmple 7.11: p = 5.9533
261
Window Design Techniques
0
EXAMPLE 7.12
The frequency response of an ideal digital differentiator is given by
(7.31)
Using a Hamming window of length 21, design a digital FIR differentiator. Plot the time- and the frequency-domain responses.
solution
The ideal impulse response of a digital differentiator with linear phase is given b Y
n=a
The above impulse response can be implemented in MATLAB along with the Hamming window to design the required differentiator. Note that if M is an even number, then a = (M- 1) /2 is not an integer and hs (n)will be zero for all n. Hence M must be an odd number, and this will be a Type-3 linearphase FIR filter. However, the filter will not be a full-band differentiator since H, ( T ) = 0 for Type-3 filters.
In this design approach we use the fact that the system function H (2) can be obtained from the samples H ( k ) of the frequency response H(ej").
Furthermore, this design technique fits nicely with the frequency sampling structure that we discussed in Chapter 6. Let h(n)be the impulse response of an M-point FIR filter, H ( k ) be its M-point DFT, and H ( z ) be its
264
Chapter 7
FIRFILTER DESIGN
system function. Then from (6.12) we have
and
with
For a linear-phase FIR filter we have
h ( ~= f h ( M )
- 1 - n),
n = O , l , . ..,M - 1
where the positive sign is for the Type-1 and Type2 linear-phase filters, while the negative sign is for the Type-3 and Type-4 linear-phase filters. Then H ( k ) is given by
Basic Idea Given the ideal lowpass filter Hd(ej"), choose the filter length M and then sample Hd(ejw)at M equispaced frequenciesbetween 0 and 2n. The actual response H(ej") is the interpolation of the samples H ( k ) given by (7.34). This is shown in Figure 7.25. The impulse response is given by (7.39). Similar steps apply to other frequency-selective filters. Furthermore, this idea can also be extended for approximating arbitrary frequency-domainspecifications. From Figure 7.25 we observe the following:
1. The approximation error-that is, the difference between the ideal and the actual response-is zero at the sampled frequencies. 2. The approximation error at all other frequencies depends on the shape of the ideal response; that is, the sharper the ideal response, the larger the approximation error. 3. The error is larger near the band edges and smaller within the band.
There are two design approaches. In the first approach we use the basic idea literally and provide no constraints on the approximation error; that is, we accept whatever error we get from the design. This approach is called a naive design method. In the second approach we try to minimize error in the stopband by varying values of the transition band samples. It results in a much better design called an optimum design method.
The time and the frequency-domain plots are shown in Figure 7.26. O s r e bev that the minimum stopband attenuation is about 16 dB, which is clearly unacceptable. If we increase M, then there wl be samples in the transition hand, il for which we 40 not precisely know the frequency response. Therefore the naive 0 design method is seldom used in practice.
OPTIMUM DESIGN METHOD
To obtain more attenuation, we will have to increase M and make the transition band samples free samples-that is, we vary their values to
obtain the largest attenuation for the given M and the transition width. Ti problem is known as an optimization problem, and it is solved using hs linear programming techniques. We demonstrate the &ect of transition band sample variation on the design using the following example.
Freauency Samdes: M=20
0.2
I
5 0.1
0
0
0.20.3
frequency in pi units
1
-0.1' .
0
5
1 0
15
I 20
n Magllitltee Respollse
0
02. .03
1
frequencyin pi wits FIGURE 7.26
Naive frequency sampling design method
268
Chapter 7 B FIR FILTER
DESIGN
0
EXAMPLE 7.15
Using the optimum design method, design a better lowpass filter of Example 7.14.
solution
Let us choose M = 40 so that w have one sample in the transition band e A 0 . 2 ~ w < 0.3~. < Since W I = 27r/40,the transition band samples are at k = 5 and at k = 40 5 = 35. Let us denote the value of these samples by T I , 0 < TI< 1; then the sampled amplitude response is
-
H,(k) [l,l,l,l,l,Ti,O O,T1,1,1,1,1] = ,...,
* as
WIOB
Since a =
= 19.5,the samples of the phase response are
LH (k) =
{
-19.5gk = -0.975?rk, 0 5 k 5 19 +0.975~ - k) , (40
20 5 k
2 39
Now we can vary TI to get the best minimum stopband attenuation. T i will hs result in the widening of the transition width. We first see what happens when TI = 0.5.
From the plots of this design in, Figure 7.27 we observe that the minimum stopband attenuation is now 30 dB, which is better than the naive design attenuation but is still not at the acceptable level of 50 dB. The best value for TI 'a obtained hy varyibg it manually (although more efficient linear programws ming techniques are available, these were not used in this case), and the near optimum solution was found at TI = 0.39.
From the plots in Figure 7.28 we observe that the optimum stopband attenuation is 43 dB. It is obvious that to further increase the attenuation, we will have to vary more than one sample in the transition band. 0
Frequency Sampling Design Techniques
269
Frequency Samples: M-40,T1=0.5
Impulse Response
4
frequency in pi unb Amplitude Response
0.2
0.1
0 -0.1
0 1 0 3 0 4 n MagnitudeResponse 0 2 0 0.20.3 frequency in pi units
%"
2 n
0
0.20.3 freguency in pi units
1
0
FIGURE 7.27
Optimum frequency design method: TI = 0.5
lmpulsa Response
Frequency Samples:M=40,T1=0.39
p
0 0.20.3 frequency in pi units Amplitude Response
1
0.2 0.1
0
-0.1' 0 1 0 3 0 4 n Magnitude Response 0 2
I
0
0
0.20.3 frequency in pi units
1
0
0.20.3 frequency in pi units
n
1
FIGURE 7.28
Optimum frequency design method: TI = 0.39
270
Chapter 7
FIR FILTER DESIGN
Clearly, this method is superior in that by varying one sample we can get a much better design. In practice the transition bandwidth is generally small, containing either one or two samples. Hence we need t o optimize at most two samples t o obtain the largest minimum stopband attenuation. This is also equivalent to minimizing the maximum side lobe magnitudes in the absolute sense. Hence this optimization problem is also called a minimax problem. This problem is solved by Rabiner et al. [20], and the solution is available in the form of tables of transition values. A selected number of tables are also available in 119, Appendix B]. This but it would require the use of problem can also be solved in MATLAB, the Optimization toolbox. We will consider a more general version of this problem in the next section. We now illustrate the use of these tables in the following examples.
0
EXAMPLE 7.16
Let us revisit our lowpass filter design in Example 7.14. We will solve it using two sampla in the transition band so that we can get a better stopband attenuation. Let us choose M = 60 so that there are two samples in the transition band. Let the values of these transition band samples be TI and Tz. Then H , ( w ) is given by
Solution
H (w) = 1,. . . , 1 , Ti,Tz,O,. .. ,O ,Tz,Ti, 1,. . .,1]
L
7
ones
a
zeros
6 ones
&om tables in [19, Appendix B] TI = 0.5925 and Tz = 0.1099. Using these values, we use MATLAB compute h (n). to
Recall that for a highpass filter M must be odd (or Type1 filter). H n e we ec will choose M = 33 to get two samples in the transition band. With this choice
of M it is not possible to have frequencysamples at w, and up. The samples of the amplitude response are
H, ( k ) = U , T i , T 2 , 1 , ... ,1,Tz,Ti,O,. ..,O]
v
8
w
10
11
while the phase response samples are
LH ( k ) = [ - r z k = - $ k ,
+-T
OIk116 17 1 k
(33 - k) ,
5 32
Bandpass:M=40,T1=0.5941.T2=0.108
-o f
0 0 2 0.35 0.65 0.8
1 -4 . 4
0.2
-0.2
0
1
0
2
0
3
0
4
0
hequency In pl units
AmpHhldeResponse
n
0
0.2 0.35 0.65 0.6 frequency in pi units
1
FIGURE 7.u)
Bandpass filter design plots in Ezomple 7.17
Frequency Sampling Design Techniques
273
The optimum values of transition samples are TI = 0.1095 and Tz = 0.598. design i s Using these values, the MATLAB
The time- and the frequency-domain plots of the design are shown in F i e 0 7.31.
0
EXAMPLE 7.19
Design a 33-point digital differentiator basedon the ideal differentiator of (7.31) given in Example 7.12.
-4 . 4
0
1
0
frequency in pi units ArnpURasponse
2 n
0
3
0
Magnitk.de Response
0
.6 .8 frequency in pi units
1
0
.6 .8 frequency in pi units
FIGURE 7.31
Highpass filter design plots in Ezample 7.18
214
Chapter 7
FIR FILTER DESIGN
SOlUtiOn
From (7.31) the samples of the (imaginarj-valued)amplitude response are given b Y
k = O,..
j H , (k) =
k=
and for linear phase the phase samples are
LJ v
k = 0, ...,
k=
.,
171
M-1
+l,...,M-I
L H (k) =
{
- Z z k = -+-a(M
M-12a
M-1
ak,
171
M-1 +l,...,M-l
h (n) = IDFT [H(k)]
M
-k),
Therefore
H (12) = jH,(k) d L H ( k 0 ,5 k 5 M - 1 )
and
The t i e and the frequency-domain plots are shown in Figure 7.32. We observe 0 that the differentiator is not a full-band differentiator.
0
EXAMPLE 7.20
Design a 51-point digital Hilbert transformer based on the ideal Hilbert transformer of (7.32).
solution
horn (7.32) the samples of the (imaginary-valued)amplitude response are given b Y
p,
k=1,
k =0
...,LTj- l M
jH,(k)= { 0,
Since this is a Type3 linear-phase filter, the amplitude response will be zero at w = a. Hence to reduce the ripples, we should choose the two samples (in transition bands) near w = 7~ optimally between 0 and j . Using our previous experience,we could select this value as 0.39j. The samplesof the phase response are selected similar to those in Example 7.19.
The plots in Figure 7.33 show the effect of the transition band samples.
0
The type of frequency sampling filter that we considered i called a s Type-A filter, in which the sampled frequencies are
2r ~k=-k, M
OsklM-1
There is a second set of uniformly spaced samples given by wk =
2r (k
+ f) ,
OlkSM-1
This is called a Type-Bfilter, for which a frequency sampling structure is also available. The expressionsfor the magnitude response H(ej") and the
276
Chapter 7 m FIR FILTER DESIGN
H i l h l Transformer, frequency sampling design : M = 51
0
0.2
0.4 0.6 frequency in pi u n L
0.8
lmoulse res~onse
1 ,
0.5
P
n
FIGURE 7.33
Digital Hilbert transformer design plots a Ezample 7.20 n
impulse response h(n) are somewhat more complicated and are available in [19].Their design can also be done in MATLAB using the approach discussed in this &ion.
OPTIMAL EQUlRlPPLE DESIGN TECHklQUE
I
I
The last two techniques-namely, the window design and the frequency sampling design-were easy to understand and implement. However, they have some disadvantages. First, we cannot specify the band frequencies w p and wd precisely in the design; that is, we have to accept whatever values we obtain after the design. Second, we cannot specify both 6 and 1 6 ripple factors simultaneously. Either we have 61 = 6 in the window 2 2 design method, or we can optimize only 62 in the frequency sampling method. Finally, the approximation error-that is, the difference between the ideal response and the actual responseis not uniformly distributed over the band intervals. It is higher near the band edges and smaller in the regions away from band edges. By distributing the error uniformly, we can obtain a lower-order filter satisfying the same specifications. Fortunately, a technique exists that can eliminate the above three problems.
Optimal Equiripple Design Technique
277
This technique is somewhat difficult to understand and requires a computer for its implementation. For linear-phase FIR filters it is possible to derive a set of conditions for w i h it can be proved that the design Solution is optimal in the sense hc of minirnizzng the mazirnum approximation e m r (sometimes called the manimwor the Chebysheu error). Filters that have this property me called qpliripple filters because the approximation error is uniformly distributed in both the passband and the stopband. This results in lower-order filters. In the following we first formulate a minimax optimal FIR design problem and discuss the total number of maxima and minima ( d m tively called eztrema) that one can obtain in the amplitude response of a linear-phase FIR filter. Using this, we then discum a general equiripple FIR filter design algorithm, which uses polynomial interpolation for its solution. T i algorithm is known as the ParbMcCleifan algorithm, and hs it incorporates the Remez exchange routine for polynomial solution. T i hs algorithm is available as a subroutine on many computing platforms. In this section we will use MATLABto design equiripple FIR filters.
DEVELOPMENT Earlier in this chapter we showed that the frequency response of the four OF THE cases of linear-phase FIR filters can be written in the form MINIMAX PROBLEM H(@) = & P e - j V W H , ( w )
where the d u e s for p and the expressions for H,(w) are given in Table 7.2.
TABLE 7.2 Amplitude response and @-valuesfor lineor-phase FIR filters
Linear-phase FIR Filter Qpe
Typel: A4 odd, symmetric h(n)
P
0
MI2
HC(ei")
(M--l)/2
C
0
a(n)coswn
Type-2: M even, symmetric h(n)
0
C b(n)cos [w(n - 1/2)] 1
(M-1)/2
Type-3:
M odd, antisymmetric h(n)
?r
2
c(n)sinwn
1
Type-4:M even, antisymmetric h(n)
2
?r
MI2
1
d(n) sin [w(n - 1/2)]
278
Chapter 7
FIR FILTER DESIGN
TABLE 7.3
Q(w),
L, and P(w)for linear-phase FIR filters
1
me2
W cos-
2
-2
1
Cb(n)coswn
0
L -
Type-3
sin w
W sin -
M-3
CE(n)coswn
0
=
0
Type-4
2
-- I
2
kci(n)ooswn
Using simple trigonometric identities, each expressionfor H J w ) above can be written as a product of a fixed fundion of w (call this Q(w)) and a function that is a sum of cosines (call this P(w)).For details see [19]and Problems 7.1-7.4. Thus
where P(w) is of the form (7.41)
and &(w), L,P(w) for the four cases are given in Table 7.3. The purpose of this analysis is to have a common form for H,(w) across all four cases. It makes the problem formulation much easier. To formulate our problem as a Chebyshev approximation problem, we have to define the desired amplitude response &(w) and a weighting function W(w),both defined over passbands and stopbands. The weighting function is necessary so that we can have an independent control over 61and 62. The weighted error is defined as
These concepts are made clear in the following set of figures. It shows a typical equiripple filter response along with its ideal response.
279
Thus the maximum error in both the passband and stopband is 62. Therefore, if we succeed in minimizing the maximum weighted error to &, we automatically also satisfy the specification in the passband to 61. Substituting Hr(w)from (7.40) into (7.42), we obtain
E (w) = W (w) I&r ( w ) - Q (w) P (w)I
If we define
then we obtain
E(w) = @ ( W )
[fidr(W)
-P(W)],
wE
s
(7.44)
Thus we have a common form of E ( w ) for d four cam. l
Problem Statement The Chebyshev approximation problem can now be defined a : s
Determine the set of coefficients a(n)or 6(n)or E(n)or d(n)[or equivalently a(.) or b(n) or c(n)or d(n)]to minimize the maximum absolute value of E (w) over the passband and stopband, i.e.,
Optimal Equirippk Design Technique
281
Now we have succeeded in specifying the exact wp, w., 6 ,and 6 . In 1 2 addition the error can now be distributed uniformly in both the passband and stopband.
CONSTRAINT ON THE NUMBER OF EXTREMA
Before we give the solution to the above problem, we will first discuss the issue: how many local maxima and minima exist in the error function E(w) for a given M-point filter? This information is used by the ParksMcClellan algorithm t o obtain the polynomial interpolation. The answer is in the expression P(w). From (7.41) P (w) is a trigonometric function in w. Using trigonometric identities of the form
cos (2w) = 2 cos2 (w) 1
cos (3w) = 4 C O S ~ - 3 cos (w) (w)
. _ .
P (w) can be converted t o a trigonometric polynomial in cm (w), which s we can write (7.41) a
has only one minimum at w = n / 2 . However, it has three extrema in the closed , interval 0 5 w 5 A (i.e., a maximum at w = 0, a minimum at w = ~ / 2 and R). Now if we include the end points w = 0 and w = R, a maximum a%= then P ( w ) has at most (L 1) local extrema in the closed interval 0 5 w 5 r. Finally, we would like the filter specificationsto be met exactly at band edges w, and w.. Then the specificationscan be met at no more than (L 3) extremal frequencies in the 0 5 w 5 R interval.
+
+
Concluswn
0
EXAMPLE 7.22
The error fiction E(w) has at most (L
+ 3) extrema in S.
Let us plot the amplitude responseof the filter given in Example 7.21 and count the total number of extrema in the corresponding error function. The impulse response is h(n)=-[1,2,3,4,3,2,1), 15 and a(n)= & [ 4 , 6 , 4 , 2 ] and p(n) = [O,O,
1
Let us now turn our attention to the problem statement and equation (7.45). It is a well-known problem in approximation theory, and the solution is given by the following important theorem.
H
THEOREM1
Alternation Thc?oft?m Let S be any closed subset of the closed interval [0,r].In order that P ( w ) be the unique minimax approximation to Hdr(W) on S , it is necessary and suficient that the error function E ( w ) ezhibit at least ( L 2) "alternations" or extremal frequencies in s; that is, there must exist ( L + 2) frequencies wi in S such that
+
E (w,) = -E (wi-1) = 2~max IE (w)l
S
(7.47)
= *c5, v w o
A
< w1 < . . . < W L + l E s
Combining this theorem with OUT earlier conclusion, we infer that the optimal equiripple filter has either (L 2) or (L 3) alternations
+
+
Optimal Equiripple Design Technique
203
1.0;
1.(
Amplitude Response
0.9: Error Function
L+3=6
0.07
0
r '
extrema
'
-
-0.07
-
0.1
0.03 0.1
-0.0.
0.1
0.4
1
FIGURE 7.34
Amplitude response and ehe envrfinction in Example 7.22
in its error function over S.Most of the equiripple filters have (L 2) alternations. However, for some combinations of w, and w., we can get filters with ( L t 3 ) alternations. These filters have one extra ripple in their response and hence are called Eztm-ripple filters.
PARKSMcCLELLAN ALGORITHM
+
The alternation theorem ensures that the solution to our minimax a p proximation problem exists and is unique, but it does not tell us how to obtain this solution. We know neither the order M (or equivalently, L), nor the extremal frequencies w,, nor the parameters {a(n)}, the nor maximum error 6. Parks and McClellan [17] provided an iterative solution using the Remez exchange algorithm. It assumes that the filter length M (or L ) and the ratio 62/61 are known. If we choose the weighting function as in (7.43),and if we choose the order M correctly, then 6 = 6 when 2 the solution is obtained. Clearly, 6 and M are related; the larger the M , the smaller the 6. In the filter specifications61, 62, wp, and w, are given. 4 Therefore A has to be assumed. Fortunately, a simple formula, due to Kaiser, exists for approximating M. It is given by
284
Chapter 7
FIR FILTER DESIGN
The ParksMcClellan algorithm begins by guessing ( L 2 ) extremal frequencies {w,} and estimating the maximum error 6 at these frequencies. It then fits an Lth-order polynomial (7.46) through points given in (7.47). Local maximum errors are determined over a finer grid, and the extremal frequencies {w,} are adjusted at these new extremal values. A new Lthorder polynomial is fit through these new frequencies, and the procedure is repeated. This iteration continues until the optimum set {w,} and the gIobaI maximum error 6 are found. The iterative procedure is guaranteed to converge, yielding the polynomial P(w). Fkom (7.46) coefficients a(.) are determined. Finally, the coefficients a(n) as well as the impulse res p o b h ( n ) are computed. This algorithm is available in MATLABs the a remez function, which i described below. s Since we approximated M , the maximum error 6 may not be equal to 6 .If this is the case, then we have to increase M (if 6 > 62) or decrease 2 M (if 6 < &) and use the remez algorithm again to determine a new 6. We repeat this procedure until 6 5 62. The optimal equiripple FIR filter, which satisfies all the three requirements discussed earlier is now determined.
+
MATLAB IMPLEMENTATION
The ParbMcClellan algorithm is available in MATLABas a function called remez, the most general syntax of which is
[hl = remez(N,f ,m,veights,ftype)
There are several versions of this syntax:'
0 [h] remez(N,f ,m) designs an Nth-order (note that the length of the filter is M = N 1) FIR.digita1 filter whose frequency response is specified by the arrays f and m. The filter coefficients (or the impulse response) are returned in array h of length M . The array f contains band1.0. These frequencies edge frequencies in units of R, that is, 0.0 f .. must be in increasing order, starting with 0.0 and ending with 1 0 The array m contains the desired magnitude response at frequencies specified in f. The lengths off and m arrays must be same and must be an even number. The weighting function used in each band is equal to unity, which means that the tolerances (4's) in every band are the same. 0 [h] = remez(N,f +,weights) is similar to the above case except that the array weights specifies the weighting function in each band. 0 [hl = remez (N , .m ,ftype) is similar to the first case except when f ftype is the string 'differentiator'or 'hilbert', designs digital difit
-
+
< <
'It should be noted that the remez function underwent a s a l change from the old ml Student Edition to the new Student Edition of MATLAB from the Signal Processing (or Toolbox version 2.0b to version 3.0). The description given here applies to the new version.
Optimal Equiripple Design Technique
285
ferentiators or digital Hilbert transformers, respectively. For the digital Hilbert transformer the lowest frequency in the f array should not be 0, and the highest frequency should not be 1. For the digital differentiator, the m vector does not specify the desired slope in each band but the desired magnitude. 0 [] = remez(N,f ,m,weights,ftype) i similar to the above case h s except that the array weights specifies the weighting function in each band.
As explained during the description of the Parks-McClellan algorithm, we have to first guess the order of the filter using (7.48) to use the routine remez. After we obtain the filter coefficients in array h,we have to check the minimum stopband attenuation and compare it with the given A, and then increase (or decrease) the filter order. We have to repeat this procedure until we obtain the desired A,. We illustrate this procedure in examples. the following several MATLAB
0
D(AMPLE7.23
Let us design the lowpass filter described in Example 7.8 using the Parks McClellan algorithm. The design parameters are
W, = 0
W .
Note that we stopped the above iterative procedure when the computed s t o p band attenuation exceeded the given stopband attenuation As, and the optimal value of M was found to be 47. This d u e is considerably lower than the window design techniques (M = 61 for a Kaiser window) or the frequency sampling technique (M = 60). In Figure 7.35 we show the time- and the frequency-domain plots of the designed filter along with the error function in both the passband and the stopband to illustrate the equiripple behavior.
0
Actual impulse Response . 3 7
1
-0.1 0
46
0
n
Amplitude Response
0.0144 -
02. .03 frequency in pi units Error Response
1
0.0032
I
0
0.20.3 frequency in PI units
1
frequency in pl unlts
FIGURE 7.35
Plots for equrnpple lowpass FIR firter in Ezample 7.23
207
Optimal Equiripple Design Technique
0
EXAMPLE 7.24
Let us design the bandpass filter described in Example 7.10 using the ParksMcCleIlan algorithm. The design parameters are:
Wla
Note a h that we increased the value of M to maintain its odd value. The optimum M was found to be 37. The time- and the frequency-domain plots of the designed filter are shown in Figure 7.37. 0
0
EXAMPLE 7.26
In this example we will design a "staircase" filter, which has three bands with different ideal responses and different tolerances in each band. The design specifications are
Band-1: Band-2:
05w 0.4%5 w
Finally, we design a Hilbert transformer over the band 0 . 0 5 ~ w 5 0.95~. 5 Since this is a wideband Hilbert transformer, we will choose an odd length for o w filter (i.e., a Type3 filter). Let us c h o w A = 51. 4
The subroutine should first determine the type o the linear-phase FIR filter and then use f the appropriate Hr-Type# function discussed in the chapter. It should also check if the given h (n)is of a linear-phase type. Check your subroutine on sequences given in Examples 7.4 through 7 7 .. If H ( z ) has zeros at
Compare the filter length of this design with that of Example 7.26. Provide a plot of the magnitude response in dB. P . 5 Consider an ideal lowpass filter with the cutoff frequency we = 0.3%. We want to 71 approximate this filter using a frequency sampling design in which we choose 40 samples. a. Choose the sample at wc equal to 0.5 and use the naive design method t o compute h (n). Determine the minimum stopband attenuation. b. Now vary the sample at we and determine the optimum value t o obtain the largest minimum stopband attenuation. c. Plot the magnitude responses in dB of the above two designs in one plot and comment on t h e results. P . 4 Design the bandstop filter of Problem 7.7 using the frequency sampling method. Choose the 71 order of the filter appropriately so that there is one sample in the transition band. Use optimum value for this sample.
P . 5 D s g the bandpass filter of Problem 7.8 using the frequency sampling method. Choose the 71 ein order of the filter appropriately so that there are two samples in the transition band. Use
optimum values for these samples.
P . 6 Design the highpass filter of Problem 7.9 using the frequency sampling method. Choose the 71 s order of the filter appropriately so that there are two samples in the transition band. U e
optimum values.
P . 7 W e want t o design a narrow bandpass filter to pass the center frequency at wo = 0.5%.The 71
bandwidth should be no more than 0.1%.
a. Use the frequency sampling technique and choase M so that there is one sample in the
transition band. Use the optimum value for transition band samples and draw the frequency sampling structure. s b. Use the Kaiser window technique so that the stopband attenuation is the same a that of t h e above frequency sampling design. Determine the impulse response h (n)and draw the linear-phase structure. c. Compare the above two filter designs in terms of their implementation and their filtering effectiveness.
298
Chapter 7 W FIR FILTER DESIGN
P . 8 The frequency response of an ideal bandpass filter is given by 71
Hd
(2") =
{
0 I 5 */3 I4 1, r / 3 5 I w ~ 5 2*/3 0, 2 ~ / 5 [wl 5 1~ 3
0,
a. Determine the coefficients of a 25-tap filter based on the Parks-McClellan algorithm with stopband attenuation of 50 dB. The designed filter should have the smallest possible transition width. b. Plot the amplitude response of the filter using the function developed in Problem 7.5. P7.19 Consider the bandstop filter given in Problem 7.7. a. Design a linear-phase bandstop FIR filter using the ParbMcClellan algorithm. Note that the length of the filter must be odd. Provide a plot of the impulse response and the magnitude response in dB of the designed filter. b. Plot the amplitude response of the designed filter and count the total number of extrema in the stopband and passbands. Verify this number with the theer'etical estimate of the total number of extrema. c. Compare the order of this filter with those of the filters in Problems 7.7and 7.14. d. Verify the operation o the designed filter on the fdlowjng signal. f
z(n)= 5
- 5cos ( y ) ;
0 I n 5 300
P . 0 Using the Parks-McClellan algorithm, design a 25-tap FIR differentiator with slope equal to 72 1 sample/cycle. ~ Plot the impulse response a. Choose the frequency band of interest between 0 . 1 and 0.91~. and the amplitude response. b. Generate 100 samples of the sinusoid
z(n)= 3sin(0.25xn),
n = 0,. . ,100
.
and process through the above FIR differentiator. Compare the result with the theoretical "derivative"of z (n).Note: Don't forget to take the 12-sampledelay of the FIR filter into account.
P7.21 Design a lowest-order equiripple linear-phase FIR filter to satisfy the specificationsgiven in Figure 7.41.Provide a plot of the amplitude response and a plot of the impulse response. P7.22 A digital signal z (n) contains a sinusoid of frequency 7r/2 and a Gaussian noise w (n) of
zero mean and unit variance; that is,
2
nn (n) = 2cos -
2
+ w (n)
We want to filter out the noise component using a 50th-order causal and linear-phase FIR filter. a. Using the Parks-McClellan algorithm, design a narrow bandpass filter with passband and width of no more than 0.021~ stopband attenuation of at least 30 dB. Note that no other parameters are given, and you have to choose the remaining parameters for the remez
I I R FILTERDESIGN
IIR filters have infiniteduration impulse responses, hence they can be matched t o analog filters, all of which generally have infinitely long impulse responses. Therefore the basic technique of IIR filter design transforms well-known analog filters into digital filters using wmplez-valued mappings. The advantage of this technique lies in the fact that both analog filter design (AFD) tables and the mappings are available extenhs sively in the literature. Ti basic technique is called the A/D (analogto-digital) filter transformation. However, the AFD tables are available only for lowpass filters. We also want to design other frequency-selective filters (highpass, bandpass, bandstop, etc.). To do this, we need to apply frequency-band transformations to lowpass filters. These transformations are also complex-valued mappings, and they are also available in the literature. There are two approaches to this basic technique of IIR filter design:
Approach 1:
I
I
Approach 2:
I
The first approach is used in MATLAB design IIR filters. A straightto functions does not provide any insight into forward use of these MATLAB the design methodology. Therefore we will study the second approach b e cause it involves the frequency-bandtransformation in the digital domain. Hence in this IIR filter design technique we will follow the following steps:
The main problem with these approaches is that we have no control over the phase characteristics of the IIR filter. Hence IIR filter designs will be treated as magnitude-only designs. More sophisticated techniques, which can simultaneously approximate both the magnitude and the phase responses, require advanced optimization tools and hence will not be covered in this book. We begin with a discussion on the analog filter specifications and the properties of the magnitud*squared response used in specifying analog filters. This will lead us into the characteristics of three widely used analog filters, namely, Butterworth, Chebyshev, and Elliptic filters. We will then study transformations to convert these prototype analog filters into different frequency-selectivedigital filters. Finally, we will conclude this chapter with a discussion on the merits and comparisons of FIR and IIR digital filters.
SOME PRELIMINARIES
We discuss two preliminary issues in this section. First, we consider the magnitudesquared response specifications,which are more typical of analog (and hence of IIR) filters. These specificationsare given on the relative linear scale. Second, we study the properties of the magnitude-squared response.
RELATIVE LINEAR SCALE
Let H a ( j n )be the frequency response of an analog filter. Then the lowpass filter specifications on the magnitudesquared response are given by
where E is a passband ripple parameter, Sl, is the passband cutoff frequency in radjsec, A is a stopband attenuation parameter, and R, is the stopband cutoff in radjsec. These specifications are shown in Figure 8.1,
302
Chapter 8
IIR FILTER DESIGN
from which we observe that [Ha(jR)I2 must satisfy IH0WP)l2 = ( H , ( ~ Q )=~ I at R = R,
A2
1
at R = Q ,
The parameters E and A are related to parameters Rp and A s , respectively, of the dB scale. These relations are given by
Rp
and
= -1Olog,o
-aE=JM 1 (8.3) 1+€2
1 A, = -1Olog1, - 3 A = 104 / 2 0 A2
(8.4)
1 The ripples, 6 and 62, of the absolute scale are related to E and A by
and
Some Preliminaries
303
PROPERTIES OF IH,(jn)('
Analog filter specifications (8.1),which are given in terms of the magnitudesquared response, contain no phase information. Now to evaluate the s-domain system function Ha (s), consider a ( j f l ) = f&(s)18=jn Then we have
or
Therefore the poles and zeros of the magnitudesquared function are distributed in a mirror-image symmetry with respect to the jfl axis. Also for real filters, poles and zeros occur in complex conjugate pairs (or mirrorimage symmetry with respect to the real axis). A typical polezero pattern of Ha(s)H4(-s)shown in Figure 8.2. From this pattern we can is construct Ha(s), which is the system function of our analog filter. We () want Ha(s) to represent a cawal and stable filter. Then all poles of & s must lie within the left half-plane. Thus we assign a31 left-half poles of H,(s)H,(-s) to Ha(s).However, zeros of Ha(s) can lie anywhere in the s-plane. Therefore they are not uniquely determined unless they all are xs lying on the js2 a i . We will choose the zeros of Ha(s)Ha(-s) inside or on the j n axis as the zeros of Ha(s). The resulting filter is then called a minimum-phase filter.
FIGURE 8.2
m a pole-zero pattern of H,(s)H.(-s) i 1 c
304
Chapter 8
IIR FILTER DESIGN
CHARACTERISTICS OF PROTOTYPE ANALOG FILTERS
I I
IIR filter design techniques rely on existing analog filters to obtain digital filters. We designate these analog filters BS prototype filters. Three prototypes are widely used in practice. In this section we briefly summarize the characteristics of the lowpass versions of these prototypes: Butterworth lowpass, Chebyshev lowpass (Type I and 11),and Elliptic lowpass. functions to design these filters, it is necesAlthough we will use MATLAB sary to learn the characteristics of these filters so that we can use proper parameters in MATLAB functions to obtain correct results.
BUTTERWORTH LOWPASS FILTERS
T i filter is characterized by the property that its magnitude response is hs
flat in both passband and stopband. The magnitude-squared response of an Nth-order lowpass filter is given by
where N is the order of the filter and R, is the cutoff frequency in rad/sec. The plot of the magnitude-squared response is shown below.
from this plot we can observe the following properties:
at 0 = 0 I H ~ ( ~ = I / ~ a~ N. , o ) for at R = R,, (H4(jRc)IZ = for all N,which implies a 3 dB attenu, ation at 0 . 0 IH4(jR)12is a monotonically decreasing function of 0. 0 IH4(jn)lZ approaches an ideal lowpass filter a N + 00. s 0 IH,,(jR)(' is rnmirnally flat at R = 0 since derivatives of all orders exist and are equal to zero.
0
0
Characteristicsof Prototype Analog Filters
305
To determine the system function H,(s), put (8.6) in the form of (8.5) we to obtain
An interpretation of (8.8) is that 0 there are 2N poles of Ha(s)Ha(-s), which are equally distributed on a circle of radius R, with angular spacing of z / N radians, 0 for N odd the poles are given by p k = RedknJN, = 0,1,. . . , k 2N - 1, 0 for N even the poles are given by P k = fl,$(*+%), k = 0 , l ) ...) 2N-1, 0 the poles are symmetrically located with respect to the jfl axis,
and
0 a pole never falls on the imaginary auiS, and falls on the real axis only if N is odd.
As an example, poles of third- and fourth-order Butterworth filters are shown in Figure 8.3. A stable and causal filter H,(s)can now be specified by selecting m poles in the left half-plane, and H,(s) be written in the form
'k=O
\
\
I
I
N=3
FIGURE 8.3
N=4
Pole plots for Butternorth fllters
306
Chapter 8
IIR FILTER DESIGN
0
EXAMPLE 8.1
Given that IH&n)la = -determine the analog filter system function
1+64W'
Ha(S).
Sdution
From the given magnitudesquared response,
Comparing this with expression (8.6), we obtain N = 3 and 0,= 0.5.The poles o Ha(s)H,(-s) as shown in Figure 8.4. f are
N=3
FIGURE 8.4
Pole plot for Ezample 8.1
H&n)
=
(S
+ 0.25 - j0.433)(5 + 0.5)(s + 0.25 + j0.433)
1/8
MATLAB IMPLEMENTATION
(s
+ 0.5)(sz + 0.5s + 0.25)
0.125
0
MATLAB provides a function d e d [z,p.k]=buttap(N) to design a normalized (i.e., 0, = 1) Butterworth analog prototype filter of order N , w i h returns zeros in z array, poles in p may, and the gain value k. hc However, we need an unnormaliied Butterworth filter with arbitrary 0 . , From Example 8.1 we observe that there are no zeros and that the poles of the unnormalized filter are on a circle with radius R instead of on a , hs unit circle. T i means that we have to scale the m a y p of the normalized filter by R, and the gain k by 0 In the following function, called : . U-buttap(N,Omegac), we design the unnormalized Butterworth analog prototype filter.
307
The analog lowpass filter is specified by the parameters Rp, Rp, R,, and A,. Therefore the essence of the design in the case of Butterworth filter is to obtain the order N and the cutoff frequency R,, given these specifications. We want
0
at R = R,, -lOlogl, I H ~ ( ~ R= ~ ' or )
and
0
at R = R,, -lOloglo IHa(jR)lZ A, or =
Solving these two equations for N and R,, we have
N=
log,, [(lORJ'O
- 1) / (10AJ'O
- l)]
2hlOWPlQS)
In general, the above N will not be an integer. Since we want N to be an integer, we must choose
where the operation 1x1 means "choose the smallest integer larger than 45 $'-for example, r . 1 = 5. Since the actual N chosen is larger than required, specificationscan be either met or exceeded either at Rp or at R,. To satisfy the specifications exactly at R,,
X As = Stopband attenuation in +dB; (As > 0) X
if Up <= 0 error('Passband edge must be larger than 0') end if Us <= Up error('Stopband edge must be larger than Passband edge') end if (Rp <= 0) I (As < 0) error('PB ripple and/or SB attenuation ust be larger than 0') end
To display the frequency-domainplots of analog filters, we provide a function called f reqssl, which is a modified version of a function f reqs provided by MATLAB.This function computes the magnitude response hs in absolute as well as in relative dB scale and the phase response. Ti function is similar to the f r e q z n function discussed earlier. One main difference between them is that in the f reqs_m function the responses are computed up to a maximum frequency amax.
This H9 (s) is slightly different from the one in Example 8 3 because in that . example w used R, = 0.5, while in the &&butt function Rcis chosen to satisfy e 0 the specifications at flp. The filter plots are shown in Figure 8.5.
CHEBYSHEV LOWPASS FILTERS
There are two types of Chebyshev filters. The Chebyshev-I filters have equiripple msponse in the passband, while the Chebyshev-I1 filters have equiripple response in the stopband. Butterworth filters have monotonic response in bath bands. Recall our discussions regarding equiripple FIR filters. We noted that by choosing a filter that has an equiripple rather than a monotonic behavior, we can obtain a lower-order filter. Therefore Chebyshev filters provide lower order than Butterworth filters for the same specifications. The magnitude-squared response of a Chebyshev-I filter is
(8.13)
where N is the order of the filter, E is the passband ripple factor, which is related to Rp, and TN (x) is the Nth-order Chebyshev polynomial given b Y cos (Ncos-'(z)) cosh (cosh-'(x))
The equiripple response of the Chebyshev filters is due to this polynomial TN(z). key properties are (a) for 0 < z < 1, TN(z) Its oscillates between -1 and 1, and ( b ) for 1 < x < 00, TN(z) increases monotonically to cu. There are two possible shapes of IHo(jfl)l2,one for N odd and one for N even as shown below. Note that x = n/fl, is the normalized frequency.
314
Chapter 8
IIR FILTER DESIGN
From the above two response plots we observe the following properties:
0
At x = 0 (or R = 0);
IH,(jO)12 = 1
for N odd.
IHJ~O)~~ = for N even. 1 + €2
0
2
At x = 1 (or R = 0,); IH,(jl))' = 1+€2 For 0 5 x
for al N. l
0
1 (or 0 5 R 5 R) IHa(jz)lz oscillates between 1 and ,,
1 1 + €2.
0
For x > 1 (or R > Rc),IHa(js)lzdecreases monotonically to 0.
1 At x = R, IH,(js)l' = ,p. ,
To determine a causal and stable H,(s), must 6nd the poles of we H,(s)Ho(-s) select the left half-plane poles for H,(s). poles of and The H,(s)H,(-s) obtained by finding the roots of are
The solution of this equation is tedious if not difficult to obtain. It can be shown that if Pk = O k j&, k = 0,. ..,N - 1 are the (left half-plane) roots of the above polynomial, then
+
where
a='@2
N & ) ,
b=2t&+
1
Nfi), and
a=!+-
(8.15)
These roots fall on xs ellipse with major axis bR, and minor a i an,. Now the system function is given by
(8.16)
Characteristicsof Prototype Analog Filters
315
where K is a normalizing factor chosen to make
[
1,
N odd
(8.17)
MATLAB
MATLAB provides a function called Cz,p,k~=c~eblap(N,Rp) to
desjgn
IMPLEMENTATION
a normalized Chebyshev-I analog prototype filter of order N and passband ripple Rp and that returns zeros in z array, poles in p array, and the gain value k. We need an unnormalized Chebyshev-I filter with arbitrary R,. T i is achieved by d i g the array p of the normalized filter hs by R,. Similar to the Butterworth prototype, this filter has no zeros. The new gain k is determined using (8.17),which is achieved by scaling the old k by the ratio of the unnordized to the normalized denominator polynomials evaluated at s = 0. In the following function, called U-chblap(N,Rp ,Omegac), we design an unnormdized Chebyshev-I analog prototype filter that returns Ha (s) in the direct form.
function [b,d = u_chblap(N,Rp,~egac);
Using the U-chblap function, we provide a function called afd-chbl to design an analog Chebyshev-I1lowpm filter, given its specifications. This i shown below and uses the procedure described i Example 8.5. s n
function [b.d afd-chbl(Wp,Ws.Rp.As); X Analog Loupass Filter Design: Chebyshev-I
The specifications are satisfied by a 4th-order Chebyshev-I filter whose system function is
Ha
= (sz
+ 4233s + 0.1103) (s2 + 0.1753s+ 0.3895)
D
0.0383
The filter plots are shown in Figure 8.6.
A Chebyshev-I1 filter is related to the Chebyshev-I filter through a simple transformation. It has a monotone passband and an equiripple stopband, which implies that this filter has both poles and zeros in the 8plane. Therefore the group delay characteristicsare better (and the phase response more linear) in the passband than the Chebyshev-I prototype. If
Magnitude Response
0.8913
x
0.1585
lr'\Ir d
Magnitude in dB
3
16
OO
02 0.3 0.5 Analog trequenn, in pi unHs
30
0
1
phase Response
-0.5
-1
;I 1
=
0.05
0.2 0.3 0.5 Analog freguency in pi units
0
0.2 0.3 0.5 Analog frequency in pi unlts
0 1 0 2 0 3 0 4 0
time in 8BconQ
FIGURE 8.6
Chebyshev-I analog filter in Example 8.6
characteristics of Prototype Analog Filters
319
we replace the term e2T;(Ci/Q,) in (8.13) by its reciprocal and also the argument G = n/O, by its reciprocal, we obtain the magnitudesquared response of Chebyshev-I1 as
(8.21)
One approach to designing a Chebyshev-I1 filter is to design the corresponding Chebyshev-I first and then apply the above transformations. We will not discuss the details of this filter but will use a function from MATLAB design a Chebyshev-I1 filter. to
MATLAB IMPLEM ENTATION
provides a function called [z, ,k]=cheb2ap (N ,As) to design a p normalized Chebyshev-I1 analog prototype filter of order N and passband ripple As and that returns zeros in z array, poles in p array, and the gain value k. We need an unnormaliied Chebyshev-I filter with arbitrary a,. This is achieved by scaling the array p of the normalized filter by Q,. Since this filter has zeros, we also have to scale the array z by Q,. The new gain k is determined using (8.17), which is achieved by scaling the old k by the ratio of the unnormalized to the normalized rational functions evaluated at s = 0. In the following function, c d e d U_chb2ap(N,As,Omegac), we design an unnormalized Chebyshev-11 analog prototype filter that returns Ha ( s ) in the direct form.
M.4TLAB
The design equations for the Chebyshev-I1 prototype are similar t o those of the Chebyshev-I except that 0,= Q, since the ripples are in the stopband. Therefore we can develop a MATLABfunction similar t o the afd-chbl function for the Chebyshev-I1 prototype.
function b,al = afd,chb2(Wp,Ws,Rp,As); % Analog Lowpass Filter Design: Chebyshev-2
These filters exhibit equiripple behavior in the passband as well as in the stopband. They are similar in magnitude response characteristics to the FIR equiripple filters. Therefore elliptic filters are optimum filters in that they achieve the minimum order N for the given specifications (or alternately, achieve the sharpest transition band for the given order N). These filters, for obvious reasons, are very di5cdt to analyze and, therefore, to design. It is not possible to design them using simple tools, and often programs or tables are needed to design them. The magnitude-squared response of elliptic filters is given by (8.22)
where N is the order, e is the passband ripple (which is related to h), and UN (.) is the Nth order Jacobian elliptic function. The analysis of this function, even on a superficial level, is beyond the scope o this book. f Note the similarity between the above response (8.22) and that of the Chebyshev filters given by (8.13). Typical responses for odd and even N are shown below.
1
1 1 + <z
1 A)
0
COMPUTATION OF FILTER ORDER N
Even though the analysis of (8.22) is difficult, the order calculation formula is very compact and is available in many textbooks (16, 19, 20). It is given by
(8.23)
where
Characteristics of Prototype Analog Filters
323
and
is the complete elliptic integral of the first kind. MATLABprovides the function ellipke to numerically compute the above integral, which we will use to compute N and to design elliptic filters.
Elliptic filters provide optimal performance in the magnitude-squared response but have highly nonlinear phase response in the passband (which is undesirable in many applications). Even though we decided not to worry about phase response in our designs, phase is still an important issue in the overall system. At the other end of the performance scale are the Butterworth filters, which have maximally flat magnitude response and require a higher-order N (more poles) to achieve the same stopband specification. However, they exhibit a fairly linear phase response in their passband. The Chebyshev filters have phase characteristics that lie somewhere in between. Therefore in practical applications we do consider Butterworth as well as Chebyshev filters, in addition to elliptic filters. The choice depends on both the filter order (which influences processing speed and implementation complexity) and the phase characteristics (which control the distortion).
ANALOG-TO-DIGITAL FILTER TRANSFORMATIONS
I
After discussing different approaches to the design of analog filters, we are now ready to transform them into digital filters. These transformations are complex-valued mappings that are extensively studied in the literature. These transformations are derived by preserving different aspects of analog and digital filters. If we want to preserve the shape of the impulse response from analog to digital filter, then we obtain a technique called impulse invariance transformation. If we want to convert a differential equation representation into a corresponding difference equation representation, then we obtain a finite diflerence approximation technique. Numerous other techniques are also possible. One technique, cdled step invariance, preserves the shape of the step response; this is explored in Problem 9. The most popular technique used in practice is called a Bilinear transformation, which preserves the system function representation from analog to digital domain. In this section we will study in detail impulse invariance and bilinear transformations, both of which can be easily implemented in MATLAB. In this design method we want the digital filter impulse response to look "similar" to that of a frequency-selective analog filter. Hence we sample h,(t) at some sampling interval T to obtain h(n);that is, h(n)= h,(nT) The parameter T is chosen so that the shape of h,(t) is "captured" by the samples. Since this is a sampling operation, the analog and digital
Analog-twDigital Filter Transformations
IMPULSE INVARIANCE TRANSFORMATION
327
frequencies are related by
u = RT or e'" = dnT
Since z = eJw on the unit circle and s = j Q on the imaginary axis, we have the following transformation from the s-plane to the z-plane:
2
5
esT
(8.24)
The system functions H ( z ) and Ha(s) are related through the frequencydomain aliasing formula (3.27):
H(z)=
1 T
5
k=-a,
Ha (s - j $ l c )
The complex plane transformation under the mapping (8.24) is shown in Figure 8.9, from which we have the following observations: 1. Using u = Re(s), we note that
u
3. Since the entire left h l of the s-plane maps into the unit circle, a af causal and stable analog filter maps into a causal and stable digital filter. 4. If H,(jR) = H,(jw/T) = 0 for IRI 2 n/T, then
s-plane
FIGURE 8.9
I
z-plane
Complesplane mapping in ampuke invariance tmnsformation
320
ChaDter 8
IIR FILTER DESIGN
and there wl be no aliasing. However, no analog filter of finite order can il il be exactly band-limited. Therefore some aliasing error wl occur in this design procedure, and hence the sampling interval T plays a minor role in this design method. Given the digital lowpass filter specificationswp, w,, Rp, and A,, we want to determine H (z)by first designing an equivalent analog filter and then mapping it into the desired digital filter. The steps required for this procedure are
1. Choose T and determine the analog frequencies
DESIGN
PROCEDURE
2. Design an analog filter Ha(s) using the specificationsfip, a,, Rp, and A,. This can be done using any one of the three (Butterwotth, Chebyshev, or elliptic) prototypes of the previous section. 3. Using partial fraction expansion, expand Ha(s) into
into a digital filter H ( 2 ) using the impulse invariance technique in which T = 0.1.
solution
W first expand Ha (8) using partial fraction expansion: e
The poles are at pi = -3 and pz = -2. Then f o (8.25) and using T = 0.1, rm we obtain
Analog-to-Digital Filter Transformations
329
It is easy to develop a MATLABfunction to implement the impulse invariance mapping. Given a rational function description of Ha(s), we can use the residue function to obtain its pole-zero description. Then each analog pole is mapped into a digital pole using (8.24). Finally, the residue2 function can be used to convert H ( 2 ) into rational function form. This procedure is given in the function imp-invr.
expected. In Figure 8.10 we show the impulse responses and the magnitude responses (plotted up to the sampling frequency 1/T) of the analog and the resulting digital filter. Clearly, the aliasing in the frequency domain is evident. 0 In the next several examples we illustrate the impulse invariance de-
Design a lowpass digital filter using a Chebyshev-I1prototype to satisfy
w, = 0.27r,
Itp = 1 dB
w = 0 . 3 ~ , A. = 15 dB .
solution
Recall that the Chebyshev-I1 filter is equiripple in the stopband. It means that this analog filter has a response that does not go to zero at high frequencies in the stopband. Therefore after impulse inmiance transformation, the aliasing i l effect w l be significant; this can degrade the passband response. The MATLAB script is shown:
From the frequency response plots in Figure 8.13 we clearly observe the pass band as well as stopband degradation. Hence the impulse invariance design 0 technique has failed to produce a desired digital filter.
334
Chapter 8
*
IIR FILTER DESIGN
Magnitude Response 0.8913
-
0.1770
n
il:,
"0 0.20.3 hequencv in.Di units . . Magnitude in dB
1
Group Delay
zl :m
$ 5 0
1
0
0
0.20.3 frequency in pi units
1
0.20.3
frequency in pi units
FIGURE 8.13
S i p
Digital Chebyshev-II lowpass filter using impulse anvariance de-
0
EXAMPLE 8.14
Design a lowpass digital filter using an elliptic prototype to satisfy
up= 0.2?r, Rp = 1 dB
us= 0.3x,
Sdution
A, = 15 dB
The elliptic filter is equiripple in both bands. Hence this situation i similar to s that of the Chebyshev-I1 filter, and we should not expect a good digital filter. script i shown: s The MATLAB
Fkom the frequency response plots in Figure 8.14 we clearly observe that once 0 again the impulse invariance design technique has failed.
The advantages of the impulse invariance mapping are that it is a stable design and that the kequencies R and w are linearly related. But the disadvantage is that we should expect some aliasing of the analog frequency response, and in some cases this aliasing is intolerable. Come quently, this design method is useful only when the analog filter is essentially band-limited to a lowpas or bandpaw filter in w i h there are no hc oscillations in the stopband.
BILINEAR TRANSFORMATION
This mapping is the best transformation method; it involves a well-known function given by (8.26)
336
Chapter 8 w IIR FILTER DESIGN
where T is a parameter. Another name for this transformation is the linear fmctional transformation because when cleared of fractions, we obtain
-sz+ -s - z + 1 = 0
2 2
T
T
which is linear in each variable if the other is fixed, or bilinear in s and z. The complex plane mapping under (8.26) is shown in Figure 8.15, from which we have the following observations:
1. Using s = D
z=
+j f l in (8.26), we obtain
l+-++j- flT)/(l-!?T-j!E) aT 2 2 2
(
(8.27)
Hence
<1 =1 >1
2. The entire left half-plane maps into the inside of the unit circle. Hence this is a stable transformation. 3. The imaginary a i maps onto the unit circle in a oneto-one fashxs ion. Hence there is no aliasing in the frequency domain.
s-plane FIGURE 8.15
I
.+plane
Complez-plane mapping in bilinear tmnsforrnataon
Analog-tc+Digital Filter Transformations
331
Substituting u = 0 in (8.27), we obtain
since the magnitude is I. Solving for w as a function of R, we obtain
w = 2 tan-'
(F)
or
2 a = -tan T
(i)
(8.28)
T i shows that R is nonlinearly related to (or warped into) w but that hs there is no aliasing. Hence in (8.28) we will say that w is prewarped into
?tansform H,(s)= ___ into a digital filter using the bilinear transfors2 + 5 9 + 6 mation. Choose T = 1.
+
0
EXAMPLE 8.15
'
Using (8.26), we obtain
1
-2
1- 2-1 +1 1 +%-I
+ 2-1
0
Simplifying,
H(z) = 3
+ 22-' - z-' - 0.15 + 0.lz-l - 0.05~-~ 1 + 0.21-1 20 + 42-1
MATLAB provides a function called bilinear to implement this m a p ping. Its invocation is similar to the imp-invr function, but it also takes several forms for different input-output quantities. The Student Edition manual should be consulted for more details. Its use is shown in the following example.
0
EXAMPLE 8.16
Given digital filter specificationsup,us, and A,, we want to determine Rp, H(z). The design steps in this procedure are the following:
1. Choose a value for T.This is arbitrary, and we may set T = 1. 2. Prewarp the cutoff frequencies up and w,; that is, calculate R, and R, using (8.28):
R - - t2 n ( ? ) , a P-T
n.=Ttan($) 2
3. Design an analog filter Ha(s) meet the specifications a,,, to R,, Rp, and A.. We have already described how to do this in the previous
section. 4. Finally, set
The desired filter is once again a Gth-order filter and has 6 zeros. Since the 6th-order zero of H, ( 5 ) at '8 = -m is mapped t o z = -1, these zeros should be at z = -1. Due t o the finite precision of MATLAB these zeros are not exactly a t z = -1. Hence the system function should be
The frequency response plots are given in Figure 8.19.Note that the bilinear 0 transformation has again properly designed the elliptic digital filter. The advantages of this mapping are that (a) it is a stable design, (b) there is no aliasing, and (c) there is no restriction on the type of filter that can be transformed. Therefore this method is used exclusively in computer programs including MATLABas we shall see next.
LOWPASS FILTER DESIGN USING MATLAB
In this section we will demonstrate the use of MATLAB'S filter design routines to design digital lowpass filters. These functions use the bilinear transformation because of its desirable advantages as discussed in the previous section. These functions are a follows: s
1. Cb,a] =butter(N,wn) This function designs an Nth-order lowpass digital Butterworth filter and returns the filter coefficients in length N+ 1 vectors b and a. The filter order is given by (&lo), and the cutoff frequency M is determined by the all prewarping formula (8.29). However, in MATLAB digital frequenciesare f o Hence wn is computed by using the following relation: given in units .
w, = 2 tan-'
n
(T) 02
The use of this function is given in Example 8.21. 2. Cb, al =chebyl (N,Rp.wn) This function designs an Nth-order lowpass digital Chebyshev-I filter with Rp decibels of ripple in the passband. It returns the filter coefficients in length N + 1 vectors b and a. The filter order is given by (8.20), and the cutoff frequency wn is the digital passband frequency in units of n; that is,
w, = wp/n
The use of this function is given in Example 8.22. 3. Cb,al=chebyP(N,As,wn) This function designs an Nth-order lowpass digital Chebyshev-I1filter with the stopband attenuation As decibels. It returns the filter coefficients in length N 1 vectors b and a. The filter order is given by (8.20), and the cutoff frequency vn is the digital stopband frequency in units of n; that is,
+
w, = w,. /
The use of this function is given in Example 8.23. 4. Cb, a] =ellip(N,Rp, As ,wn) This function designs an Nth-order lowpass digital elliptic filter with the passband ripple of Rp decibets and a stopband attenuation of As decibels. It returns the filter coefficients in length N + 1 vectors b and a. The filter order is given by (8.23), and the cutoff frequency wn is the digital
345
Lowpass Filter Design Using MATLAB
passband frequency in units of
T;
that is,
w, = wpjn
The use of this function is given in Example 8.24.
All these above functions can also be used to design other frequencyselective filters, such as highpass and bandpass. We will discuss their additional capabilities in Section 8.5. There is also another set of filter functions, namely the buttord, cheblord, chebzord,and ellipord functions, which can provide filter order N and filter cutoff frequency w,, given the specifications. These functions are available in the Signal Processing toolbox but not in the Student Edition, and hence in the examples to follow we will determine these parameters using the formulas given earlier. We will discuss the filter-order functions in the next section. In the following examples we will redesign the same lowpass filters of previous examples and compare their results. The specifications of the lowpass digital filter are
wp = 0 . 2 ~ ,
which is the Same as i Example 8.20.The frequency-domain n plots were shown in Figure 8.19. 0
COMPARISON OF THREE FILTERS
In our examples we designed the same digital filter using four different prototype analog filters. Let us compare their performance. The spec$cations were w p = 0.2x,Rp = 1 dB, w, = 0.3a,and A, = 15 dB. This comparison in terms of order N and the minimum stopband attenuations i shown i Table 8.1. s n
349
Lowpass Filter Design Using MATLAB
TABLE 8.1
Compahon of three filters
Pmtotype
Onier
N
Stopband Att.
15 25 27
Butterworth Chebyshev-I Elliptic
6 4 3
Clearly, the elliptic prototype gives the best design. However, if we compare their phase responses, then the elliptic design has the most nonlinear phase response in the passband.
FREQUENCY-BAND TRANSFORMATIONS
I I
In the preceding two sections we designed digital lowpass filters from their corresponding analog filters. Certainly, we would l i e to design other types of frequency-selective filters, such as highpass, bandpass, and bandstop. This is accomplished by transforming the frequency axis (or band) of a lowpass filter so that it behaves as another frequency-selectivefilter. These transformations on the complex variable z are very similar to bilinear transformations, and the design equations are algebraic. The procedure to design a general frequency-selective filter is to first design a dzgital prototype (of fixed bandwidth, say unit bandwidth) lowpass filter and then to apply these algebraic transformations. In this section we will describe the basic philosophy behind these mappings and illustrate their mechanism through examples. MATLAB provides functions that incorporate frequency-band transformation in the s-plane. We will first demonstrate the use of the z-plane mapping and then illustrate the use of MATLAB functions. Typical specifications for mast commonly used types of frequency-selective digital filters are shown in Figure 8.20. Let H ~ p ( 2 be the given prototype lowpass digital filter, and let H ( z ) ) be the desired frequency-selective digital filter. Note that we are using two different frequency variables, 2 and z, with H L P and H, respectively. Define a mapping of the form
such that
To do this, we simply replace 2 ' everywhere in H L P by the function G(2-l). Given that H~p(2) a stable and c a d filter, we also want is H ( z ) to be stable and causal. This imposes the following requirements:
350
Chapter 8
IIR FILTER DESIGN
FIGURE 8.20
Specifications offrequency-selective filters
1. G(.) must be a rational function in z-l so that H ( z ) is implementable. 2. The unit circle of the Z-plane must map onto the unit circle of the tplane. 3. For stable filters, the inside of the unit circle of the 2-plane must also map onto the inside of the unit circle of the tplane.
The general form of the function G ( . )that satisfies the above requirements is a rational function of the all-pass type given by
where l a k l < 1 for stability and to satisfy requirement 3. Now by choosing an appropriate order n and the coefficients { a k } , we can obtain a variety of mappings. The most widely used transformations are given in Table 8.2. We will now illustrate the use of this table for designing a highpass digital filter.
TABLE 8.2
In Example 8.22 we designed a Chebyshev-I lowpass filter with specifications
= 0.21~,
w: = 0.3x,
Rp = 1 dB
A. = 15 dB
and determined its system function
1,55482-1 + 0.64932-2) Design a highpass filter with the above tolerances but with passband beginning at up = 0.61~.
HLp(z)
0.001836(1+
= (1 - 1.49962-1
+ 0.84822-2)(1-
Solution
We want to transform the given lowpass filter into a highpass filter such that the cutoff frequency u; = 0 . 2 is mapped onto the cutoff frequency up= 0 . 6 ~ . ~ From Table 8.2
Hence
-
(1
+ 0.5661r' + 0.7657r2)(1+ 1.0416r' + 0.4019z-a)
0.02426(1- z - ~ ) ~
which is the desired filter. The frequency response plots o the lowpass filter f 0 and the new highpass filter are shown in Figure 8.21. From the above example it is obvious that to obtain the rational function of a new digital filter from the prototype lowpass digital filter, we should be able to implement rational function substitutions from Table 8.2. This appears to be a difficult task, but since these are algebraic functions, we can use the conv function repetitively for this purpose. The following mapping function illustrates this approach.
function [bz,azI = pPapping(bZ,aZ,Nz,Dz) % Frequency band Transformation from 2-domain to z-domain
which is essentially identical to that in Example 8.25.
DESIGN PROCEDUR E
In Example 8.26 a lowpass prototype digital filter was available to transform into a highpass filter so that a particular band-edge frequency was properly mapped. In practice we have to first design a prototype lowpass digital filter whose specifications should be obtained from Specifications of other frequency-selective filters as given in Figure 8.20. We w l now i l show that the lowpass prototype filter specifications can be obtained from the transformation formulas given in Table 8.2. Let us use the highpass filter of Example 8.25 as an example. The passband-edge frequencies were transformed using the parameter Q = -0.38197 in (8.30). What is the stopband-edge frequency of the highpass : of filter, say w,,corresponding to the stopband edge w = 0 . 3 ~ the p r e totype lowpass filter? This can be answered by (8.30). Since a is fixed for the transformation, we set the equation
This is a transcendental equation whose solution can be obtained iteratively from an initial guess. It can be done using MATLAB, the solution and
is
w, = 0.4586~
Now in practice we will know the desired highpass frequencies w, and up,and we are required to find the prototype lowpass cutoff frequencies w and w;. We can choose the passband frequency w; with a reasonable : and value, say w; = O.~A, determine a from wp wing the formula from Table 8.2. Now w: can be determined (for our highpass filter example) from Q and
The above highpass filter design procedure csn be easily extended to other frequency-selectivefilters using the transformation functions in Table 8.2. These design procedures are explored in Problems 8.18 through 8.22. We now describe MATLAB'S filter design functions for designing arbitrary frequency-selectivefilters.
MATLAB IMPLEMENTATION
In the preceding section we discussed four MATLAB functions to design digital lowpass filters. These same functions can also be used to design highpass, bandpass, and bandstop filters. The frequency-band transformations in these functions are done in the s-plane, that is, they use Approach1 discussed on page 301. For the purpose of illustration we will use the function butter. It can be used with the following variations in its input arguments.
0 [b,a] BUTTER(N,un. 'high') designs an Nth-order highpass filter with digital 3-dB cutoff frequency m in units of A. 0 [ b . d = BUTTF.R(N,wn,)designs anorder 2 bandpass filter if unis N a two-element vector, wn= [HI w21, with 3-dB passband ul < w < u2 in units of A. 0 [b,aJ BUTTER(N,wn,'stop') is an order 2 bandstop filter if N wn- [ul u27with 3-dB stopband ul < u < u2 in u i s of n. nt
-
To design any frequency-selective Butterworth filter, we need to know the order N and the 3-dB cutoff frequency vector wn. In this chapter we
358
Chapter 8
IIR FILTER DESIGN
described how to determine these parameters for lowpass filters. However, these calculationsare more complicated for bandpass and bandstop filters. provides a function called In their Signal Processing toolbox, MATLAB buttord to compute these parameters. Given the specifications, up,w., 4,and A,, this function determines the necessary parameters. Its syntax is
CN,vnl = buttord(wp,ws ,Rp,As)
Now using the buttord function in conjunction with the butter function, we can design any Butterworth IIR filter. Similar discussions apply for chebyl,cheby2,and e l l i p functions with appropriate modifications. We illustrate the use of these functions through the following examples.
0
EXAMPLE 8.28
In this example we wl design a Chebyshev-I highpass filterwhose specifications il were given i Example 8.27. n
MATLAB Script
This is also a 10th-order filter. The frequency domain plots are shown in Figure
8.24.
0.8913
'0
-400
'a O m
Digilal frequency in pi units 0.25 0.4 0.70.8 1
0
Maenihrde in d6
0.25 0.4
0.70.8
1
D@al freqwncy in pi units
FIGURE 0 2 Digital Chebyshev-11bandstop filter in Example 8.90 .4
362
Chapter 8
IIR FILTER D% ! N
COMPARISON OF FIR VS. IIR FILTERS
So far we have seen many techniques for designing both FIR and IIR filters. In practice one wonders about which filter (FIR or IIR) should be chosen for a given application and which method should be used to design it. Because these design techniques involve different methodologies, it is difficult to compare them. However, some meaningful comparisons can be attempted if we focus on the minimax optimal (or equiripple) filters. In the case of FIR filters these optimal filters are the equiripple filters designed via the ParbMcClellan algorithm (or Fkmez Exchange Algorithm), while in the case of IIR filters these are the elliptic filters. One basis of comparison is the number of multiplications required to compute one output sample in the standard realization of these filters. For FIR filters the standard realization is the linear-phase direct form, while for elliptic filters cascade forms are widely used. Let M be the length of a linear phase FIR filter (assume M odd). Then we need
-
M+l 2
N
_
-
for large^
multiplications per output sample. Let N (assume N even) be the order of an elliptic filter with the cascade form realization. Then there are N / 2 second-order sections, each requiring 3 multiplications (in the most efficient implementation). There are an additional three multiplications in the overall structure for a total of
N 32
3N +3 N 2
(for large N )
multiplications per output sample. Now if we assume that each filter meets exactly the same specifications: (e.g., w,, w,, 61 (or passband ripple Rp), and 62 (or stopband attenuation As) for a lowpass filter), then these two filters are equivalent if
M+l 3N+3 M 1 -- ==+ - = 3 + - N 3 for large N 2 2 N N
-
This means that if the ratio M / N = 3, then two filters are roughly efficient. However, an equiripple FIR filter is more efficient if M / N < 3, or an elliptic IIR filter is more efficient if M / N > 3. It has been shown experimentally that
0
T i shows that for most applications I R elliptic filters are dqirable from hs I the computational point of view. The most favorable conditions for FIR filters are
0
0
large values of 61, small values of 62, and large transition width.
firtherrnore, if we take into account the phase equalizers (which are all-pass filters) connected in cascade with elliptic filters that are needed for linear-phase characteristics, then FIR equiripple filter designs look good because of their exact linear-phase characteristics.
PROBLEMS
I
. I
P8.1 Design an analog Butterworth lowpass filter that has a I-dB or better ripple at 30 rad/sec and at least 30 dB of attenuation at 40 rad/sec. Determine the system function in a cascade form. Plot the magnitude response, the log-magnitude respopse in dB,the phase response, and the impulse response of the filter. P . Design a lowpass analog elliptic filter with the following characteristics: 82
an acceptable passband ripple of 1 dB, passband cutoff frequency of 10 rad/sec, and stopband attenuation of 40 dB or greater beyond 15 rad/sec.
0
Determine the system function in a rational function form. Plot the magnitude response, the log-magnitude response in dB, the phase response, and the impulse response of the filter. P8.3 A signal ia(t) contains two frequencies, 100 He and 130 Hz.We want to suppress the 130-H~ component to 50-dB attenuation while passing the 100-Hz component with less than 2dB attenuation. Design a minimum-order Chebyshev-I analog filter to perform this filtering operation. Plot the log-magnitude response and verify the design. P8.4 D s g an analog Chebyshev-I1lowpass filter that has a 0.5 dB or better ripple at 250 Hz ein and at least 45 dB of attenuation at 300 H .Plot the magnitude response, the z log-magnitude response in dB, the phase response, and the impulse response of the filter. P. 8 5 Write a MATLAB function to design analog lowpass filters. The format of this function should be function [b, al =afd(type ,Fp, Fa ,Rp,As)
Use the aid-butt, afd-chbl, afd-chb2, and afd-elip functions developed in this chapter. Check your function on specifications given in Problems 8.1 through 8.4. P8.6 Design a lowpass digital filter to be used in a structure
Use the afd function developed in Problem 8.5. Check your function on specifications given in Problems 8.6 and 8.7. In this problem we will develop a technique called the step rnwariance transformation. In this technique the step response of an analog prototype filter is preserved in the resulting digital filter; that is, if 5. (t) is the step response of the prototype and if [ (n)is the step response of the digital filter, then
P . 0 Design the lowpass Butterworth digital filter of Problem 8.7 using the step invariance 81 method. Plot the log-magnitude response in dB and compare it with that in Problem 8.7. Plot the step response [ (n) and the impulse response & (t) of the analog prototype and , compare their shapes. P8.11 Consider the design of the lowpass Butterworth filter of Problem 8.7. a. Use the bilinear transformation technique outlined in this chapter and the bilinear function. Plot the log-magnitude response in dB. Compare the impulse responses of the analog prototype and the digital filter. b. U e the butter function and compare this design with the above one. s P . 2 Following the procedure used in this chapter, develop the following MATLAB 81 functions to design FIR filters via the Kaiser window technique. These functions should check for the valid band-edge frequencies and restrict the filter length to 255. a. Lowpass filter: The format should be
Ue the chebyl function and determine the system function H ( 2 ) of such a filter. Provide a s
plot containing subplots of the log-magnituderesponse in dB and the impulse response. P . 2 Write a MATLAB 82 function to determine the lowpass prototype digital filter frequencies from a bandstop digital filter's specificationsusing the functions given in Table 8.2 and the procedure outlined for highpass filters. The format of this function should be
system in which the sampling frequency is loo0 sam/sec. a. Design a minimum-order I R digital filter that will pass the 1 0 H component with I 5-z attenuation of less than 1 dB and suppress the l W H z component to a t least 40 dB. The filter should have a monotone passband and an equiripple stopband. Determine the system function in rational function form and plot the log-magnitude response. . b. Generate 300 samples (sampled at lo00 sam/sec) of the above signal z (t) and process through the designed filter to obtain the output sequence. Interpolate this sequence (using any one of the interpolating techniques discussed in Chapter 3) to obtain y (t). Plot the , input and the output signals and comment on your results. P8.24 Using the bilinear transformation method, design a tenth-order elliptic bandstop filter t o remove the digital frequency w = 0 . 4 4 ~ with bandwidth of 0.08~. Choose a reasonable value for the stopband attenuation. Plot the magnitude response. Generate 201 samples of the sequence
z (n)= sin [0.44m] ,
n = 0,. . ,200 .
and process thorough the bandstop filter. Comment on your results. P8.25 Design a digital highpass filter H ( z ) t o be used in a
a Plot magnitude response o the overall analog filter over the [O, 5Kh]interval. . f b. Plot the magnitude response of the digital lowpass prototype. c.Plot the magnitude response of the analog lowpass prototype. d.What limitations must be placed on the input signals 80 that the above structure truly acts like a highpass filter to them?
312
Chapter 8
IIR FILTER DESIGN
APPLICATIONS IN ADAPTIVE
FILTERING
In Chapters 7 and 8 we described methods for designing FIR and IIR digital filters to satisfy some desired specifications. O r goal was to determine u the coefficients of the digital filter that met the desired specifications. In contrast to the filter design techniques considered in those two chapters, there are many dietal signal processing applications in which the filter coefficients cannot be specified a priori. For example, let us consider a high-speed modem that is designed to transmit data over telephone channels. Such a modem employs a filter called a channel equalizer to compensate for the channel distortion. The modem must effectively transmit data through communication channels that have different frequency r b sponse characteristics and hence result in different distortion effects. The only way in which this is possible is if the channel equalizer has adjustable coeficients that can be optimized to minimize some measure of the distortion, on the basis of measurements performed on the characteristics of the channel. Such a filter with adjustable parameters is called an adaptive filter, in this case an adaptive equalizer. Numerous applications of adaptive filters have been described in the literature. Some of the more noteworthy applications include (1) adaptive antenna systems, in which adaptive filters are used for beam steering and for providing nulls in the beam pattern to remove undesired interference [23];(2) digital communication receivers, in which adaptive filters are used to provide equalization of intersymbol interference and for channel 11 identification 1 8 ; (3) adaptive noise canceling techniques, in which an
adaptive filter is used to estimate and eliminate a noise component in some desired signal [22, 10, 141; and (4) system modeling, in which an adaptive filter is used as a model to estimate the characteristics of an unknown system. These are just a few of the best known examples on the use of adaptive filters. Although both IIR and FIR filters have been considered for a d a p tive filtering, the FIR filter is by far the most practical and widely used. The reason for this preference is quite simple. The FIR filter has only adjustable zeros, and hence it is free of stability problems associated with adaptive IIR filters that have adjustable poles as well as zeros. We should not conclude, however, that adaptive FIR filters are always stable. On the contrary, the stability of the filter depends critically on the algorithm for adjusting its coefficients. Of the various FIR filter structures that we may use, the direct form and the lattice form are the ones often used in adaptive filtering applications. The direct form FIR filter structure with adjustable coefficients h(O),h(l), .. . ,h(N - 1) is illustrated in Figure 9.1. On the other hand, the adjustable parameters in an FIR lattice StNCtWe are the reflection coefficients K , shown in Figure 6.18. An important consideration in the use of an adaptive filter is the criterion for optimizing the adjustable filter parameters. The criterion must not only provide a meaningful measure of filter performance, but it must a s result in a practically realizable algorithm. lo One criterion that provides a good measure of performance in a d a p tive filtering applications is the least-squares criterion, and its counterpart in a statistical formulation of the problem, namely, the mean-squareerror (MSE) criterion. The least squares (and MSE) criterion results in a quadratic performance index as a function of the filter coefficients,and hence it possesses a single minimum. The resulting algorithms for adjusting the coeEcients of the filter are relatively easy to implement.
Input
Coefficient adiustment
r
Chapter 9
output
FIGURE 9.1 Dimt f o m adaptive FIR filter
374
APPLICATIONS IN ADAPTIVE FILTERING
In this chapter we describe a basic algorithm, d l e d the least-meansquare (LMS) algorithm, to adaptively adjust the coefficients of an FIR filter. The adaptive filter structure that will be implemented is the direct form FIR filter structure with adjustable coefficients h(O),h(l), ..., h(N - I), as illustrated in Figure 9.1. After we describe the LMS algorithm, we apply it to several practical systems in which adaptive filters are employed.
LMS ALGORITHM FOR COEFFICIENT ADJUSTMENT
Suppose we have an FIR filter with adjustable coefficients {h(k),O 5 k 5 N - I}. Let {z(n)}denote the input sequence to the filter, and let the corresponding output be {y(n)}, where
N-1
y ( n ) = C h ( k ) z ( n - k ) , n=O, ...,M
k=O
(9.1)
Suppose that we also have a desired sequence { d ( n ) } with which we can compare the FIR filter output. Then we can form the error sequence {e(n)}by taking the difference between d(n) and y(n). That is,
e(n)= d(n)- y(n), n = 0 , . ..,M
(9.2)
The coefficients of the FIR filter will be selected to minimize the s u m of squared errors. Thus we have
where, by definition,
n=O
M
rz2 (k)= C z ( n ) z ( n k), 0 5 k 5 N
n=O
+
-1
(9.5)
LMS Algaithm for Coefficient Adjustment
375
We call {r&(k)} the crosscorrelationbetween the desired output sequence { d ( n ) } and the input sequence {~(n)}, {rzz(k)}is the autocorrelation and sequence of { ~ ( n ) } . The sum of squared errors E is a quadratic function of the FIR filter coefficients. Consequently,the minimization of E with respect to the filter coefficients { h ( k ) } results in a set of linear equations. By differentiating E with respect to each of the filter coefficients, we obtain
and, hence
N-1
k=O
C h ( k ) r z z ( k - m) = r & ( m ) , o s m 5 N - 1
(9.7)
This is the set of linear equations that yield the optimum filter coefficients. To solve the set of linear equations directly, we must first compute the autocorrelation sequence { r z z ( k ) }of the input signal and the crosscorrelation sequence {r&(k)} between the desired sequence {d(n)} and z the input sequence ( (n)}. The LMS algorithm provides an alternative computational method for determining the optimum filter coefficients { h ( k ) }without explicitly computing the correlation sequences { r z z ( k ) }and ( r & ( k ) } . The algorithm is basically a recursive gradient (steepest-descent) method that iinds the minimum of E and thus yields the set of optimum filter coefficients. We begin with any arbitrary choice for the initial values of {h(k)}, say {ho(k)}. For example, we may begin with h ( k ) = 0, 0 5 k 5 N -1. Then after each new input sample {z (n)}enters the adaptive FIR filter, form the error signal we compute the corresponding output, say {y (n)}, e(n) = d(n) - y(n), and update the filter coefficients according to the equation
hn ( k ) = hn-l ( k )
+ A . e ( n ) . ~ ( - k), 0 < k _< N - 1, n
n = O , l , . ..
where A is called the step size parameter, z(n- k) is the sample of the input signal located at the kth tap of the filter at time n, and e (n) z(n - k) is an approximation (estimate) of the negative of the gradient for the kth filter coefficient. This is the LMS recursive algorithm for adjusting the filter coefficients adaptively so as to minimize the sum of squared errors E. The step size parameter A controls the rate of convergence of the algorithm to the optimum solution. A large value of A leads to large step size adjustments and thus to rapid convergence, while a small value of A results in slower convergence. However, if A is made too large the algorithm becomes unstable. To ensure stability, A must be chosen 118)
376
Chapter 9
APPLICATIONS IN ADAPTIVE FILTERING
to be in the range
O<A<-
1 10NP,
(9.9)
where N is the length of the adaptive FIR filter and P is the power in , the input signal, w i h can be approximated by hc
(9.10)
The mathematical justification of equations (9.9) a d (9.10) and the proof that the LMS algorithm leads to the solution for the optimum filter coefficients is given in more advanced treatments of adaptive filters. The interested reader may refer to the books by Haykin [9] and Pro& [HI.
MATLAB IMPLEMENTATION
To formulate the problem, let us refer to Figure 9.2. We have an unknown linear system that we wish to identify. The unknown system may be an all-zero (FIR) system or a polezero (IIR) system. The unknown system wl be approximated (modeled) by an FIR filter of length N . Both the il unknown system and the FIR model are connected in parallel and are z If excited by the same ifiput sequence { (n)}. (y (n)}denotes the output of the model and { d ( n ) }denotes the output of the unknown system, the If error sequence is {e(n)= d(n)- ~(n)}. we minimii the sum of squared errors, we obtain the same set of linear equations as in (9.7). Therefore the LMS algorithm given by (9.8) may be used to adapt the coefficients of the FIX model so that its output approximates the output of the unknown system.
PROJECT 9.1: SYSTEM IDENTIFICATION
There are three basic modules that are needed to perform this project.
1. A noise signal generator that generates a sequence of random numf bers with zero mean value. For example, we may generate a sequence o uniformly distributed random numbers over the i n t e d [-a,u]. Such a sequence o uniformly distributed numbers has an average value o zero f f and a variance of a2/3. This signal sequence, call it {~(n)}, be used will as the input to the unknown system and the adaptive FIR model. In this case the input signal {z(n)}has power P, = a2/3. In MATLAB this can be implemented using the rand function. 2. An unknown system module that may be selected is an IIR filter and implemented by its difference equation. For example, we may select an IIR filter specified by the second-order difference equation
d ( n ) =ald(n- l)+a2d(n--2)+z(n)+blz(n- 1)+h3:(n22) (9.11)
Noise signal generator
x(n)
FIGURE 9.2
Block diagmm of system identification or system modeling problem
378
Chapter 9
APPLICATIONS IN ADAPTIVE FILTERING
where the parameters {al,az} determine the positions of the poles and { b l , b} determine the positions of the zeros of the filter. These parameters are input variables to the program. This can be implemented by the f i l t e r function. 3. An adaptive FIR filter module where the FIR filter has N tap coefficients that are adjusted by means of the LMS algorithm. The length N of the filter is an input variable to the program. This can be implemented using the Ims function given in the previous section. The three modules are contigured as shown in Figure 9.2. From this project we can determine how closely the imp& response of the FIR model approximates the impulse response of the unknown system after the LMS algorithm has converged. To monitor the convergence rate of the LMS algorithm, we may compute a short-term average of the squared error e2 (n) and plot it. That is, we may compute
(9.12)
where m = n / K = 1 , 2 , . .. . The averaging interval K may be selected to be (approximately) K = 10N.The effect of the choice of the step size parameter A on the convergence rate of the LMS algorithm may be observed by monitoring the ASE(m). Besides the main part of the program, you should also include, as an aside, the computation of the impulse response of the unknown system, which can be obtained by exciting the system with a unit sample sequence 6 (n).This actual impulse response can be compared with that of the FIR model after convergence of the LMS algorithm. The two impulse responses can be plotted for the purpose of comparison.
SUPPRESSION OF NARROWBAND INTERFERENCE IN A WIDEBAND SIGNAL
I
I
Let us m u m e that we have a signal sequence { z ( n ) }that comjsts of }, a desired wideband signal sequence, say { ~ ( n ) corrupted by an additive narrowband interference sequence {s (n)}.The two sequences are uncorrelated. This problem arises in digital communications and in signal detection, where the desired signal sequence {w (n)} is a spread-spectrum signal, while the narrowband interferencerepresents a signal from another user of the frequency band or some intentional interference from a jammer who is trying to disrupt the communication or detection system.
379
Suppression of Narrowband Interference in a Wideband Signal
From a filtering point of view, our objective is to design a filter that suppressesthe narrowband interference. In effect,such a filter should place a notch in the frequency band occupied by the interference. In practice, however, the frequency band of the interference might be unknown. Moreover, the frequency band of the interference may vary slowly in time. The narrowband characteristics of the interference allow us to estimate s (n)from past samples of the sequence x(n) = s(n) w(n) and to s subtract the estimate from x(n). Since the bandwidth of { (n)} is narrow compared to the bandwidth of {w (n)}, samples of {s (n)} are the highly correlated. On the other hand, the wideband sequence {w (n)}has a relatively narrow correlation. The general configuration of the interference suppression system is shown in Figure 9.3. The signal z(n)is delayed by D samples, where the delay D is chosen s&ciently large so that the wideband signal compe nents w(n) and w(n - D ) , which are contained in s(n) and z(n - D ) , respectively, are uncorrelated. The output of the adaptive FIR filter is the estimate
+
N-1
i(n) =
k 0
c
h(k)z(n - k - D )
(9.13)
e(n) = s(n) - .^(TI). The minimization of the sum of squared errors again leads to a set of linear equations for determining the optimum coefficients.
The error signal that is used in optimizing the FIR, filter coefficients is
Due to the delay D, the LMS algorithm for adjusting the coefficients recursively becomes
PROJECT 9.2: There are three basic modules required to perform this project. SUPPRESSION 1. A noise signal generator module that generates a wideband se OF quence {w (n)} of random numbers with zero mean value. In particular, SlNUSOlDAL INTERFERENCE we may generate a sequence of uniformly distributed random numbers using the rand function 8s previously described in the project on system , identification. The signal power is denoted as P. 2. A sinusoidal signaI generator module that generates a sine wave se quence s(n)= A sinwon, where 0 < wo < T and A is the signal amplitude. The power of the sinusoidal sequence is denoted as P,. 3. An adaptive FIR filter module using the Ims function, where the FIR filter has N tap coefficients that are adjusted by the LMS algorithm. The length N of the filter is an input variable to the program.
The three modules are configured as shown in Figure 9.4. In this project the delay D = 1is sufficient, since the sequence {u'(n)} is a white noise (spectrally flat or uncorrelated) sequence. The objective is to adapt the FIR filter coefficients and then to investigate the characteristics of the adaptive filter. It is interesting to select the interference signal to be much stronger than the desired signal w(n), for example, Pa = IOP,. Note that the power P, required in selecting the step size parameter in the LMS algarithm is P, = Pa + P, The frequency response characteristic H (.') of . " the adaptive FIR filter with coefficients {h(k)} should exhibit a resonant peak at the frequency of the interference. The frequency response of the interference suppression filter is Ha (2") = 1 - H ( e j " ) , which should then exhibit a notch at the frequency of the interference. and } It is interesting to plot the sequences {w (n)},{s (n)}, { ~ ( n )It . and H, (ej") is also interesting to plot the frequency responses H (d") after the LMS algorithm has converged. The short-time average squared error ASE(rn), defined by (9.12),may be used to monitor the convergence characteristics of the LMS algorithm. The effect of the length of the adaptive filter on the quality of the estimate should be investigated.
Sine wave
s(n)
generator
Delay
D=
1
FIGURE 9.4 sion
Conjigurntion of modules for ezperiment on interference suppres-
Suppression of Narrowband Interference in a Wideband Signal
381
The project may be generalized by adding a second sinusoid of a different frequency. Then H should exhibit two resonant peaks, provided the frequencies are sufficiently separated. Investigate the effect of the filter length N on the resolution of two closely spaced sinusoids.
ADAPTIVE LINE ENHANCEMENT
I I
In the preceding section we described a method for suppressing a strong narrowband interference from a wideband signal. An adaptive line enhancer (ALE) has the same configuration as the interference suppression filter in Figure 9.3, except that the objective is different. In the adaptive line enhancer, { s ( n ) }is the desired signal and { ~ ( n ) } represents a wideband noise component that masks { ~ ( n )The desired }. signal {s(n)} may be a spectral line (a pure sinusoid) or a relatively narrowband signal. Usually, the power in the wideband signal is greater than that in the narrowband signal-that is, P, > P,. It is apparent that the ALE is a self-tuning filter that has a peak in its frequency response at the frequency of the input sinusoid or in the frequency band occupied by the narrowband signal. By having a narrow bandwidth FIR filter, the noise outside the frequency band of the signal is suppressed, and thus the spectral line is enhanced in amplitude relative to the noise power in
{W(4}.
PROJECT 9.3: ADAPTIVE LINE ENHANCEMENT
This project requires the same software modules as those used in the project on interference suppression. Hence the description given in the preceding section applies directly. One change is that in the ALE, the condition is that P, > P,. Secondly, the output signal from the ALE is {s(n)}. Repeat the project described in the previous section under these conditions.
ADAPTIVE CHANNEL EQUALIZATION
I
The speed of data transmission over telephone channels is usually limited by channel distortion that causes intersymbol interference (ISI). At data rates below 2400 bits the IS1 is relatively small and is usually not a problem in the operation of a modem. However, at data rates above 2400 bits, an adaptive equalizer is employed in the modem to compensate for the channel distortion and thus to allow for highly reliable high-speed data transmission. In telephone channels, filters are used throughout the system to separate signals in different frequency bands. These filters cause
382
Chapter 9
m APPLICATIONS IN ADAPTIVE FILTERING
amplitude and phase distortion. The adaptive equalizer is basically an adaptive FIR filter with coefficients that are adjusted by means of the LMS algorithm to correct for the channel distortion. A block diagram showing the basic elements of a modem transmitting data over a channel is given in Figure 9.5. Initially, the equalizer coefficients are adjusted by transmitting a short training sequence, usually less than one second in duration. After the short training period, the transmitter begins to transmit the data sequence {a(.)}. To track the possible slow time variations in the channel, the equalizer coefficients must continue to be adjusted in an adaptive manner while receiving data. This is usually accomplished, as illustrated in Figure 9.5, by treating the decisions at the output of the decision device as correct, and using the decisions in place of the reference {d(n)}to generate the error signal. This approach works quite well when decision errors occur infrequently, such as less than one error in 100 data symbols. The occasional decision errors cause only a small misadjustment in the equalizer coefficients.
PROJECT 9.4 ADAPTIVE CHANNEL EQUALIZATION
The objective of this project is to investigate the performance of an a d a p tive equaliier for data transmission over a channel that causes intersymbol interference. The basic configuration of the system to be simulated is shown in Figure 9.6. As we observe, five basic modules are required. Note that we have avoided carrier modulation and demodulation, which is required in a telephone channel modem. This is done to simplify the simulation program. However, all processing involves complex arithmetic operations. The five modules are as follows:
1. The data generator module is used to generate a sequence of complex-valued information symbols {a(n)}. In particular, employ four equally probable symbols s j s , s j s , -s js, and -s - j s , where s is a scale factor that may be set to s = 1, or it can be an input parameter.
+
-
+
Noise
Adaptive
FIGURE 9.5
Application of adaptive filtering to adaptive channel equalization
Adaptive Channel Equalization
383
I I
I
I I I
I
I
I
e(n) Error signal
Delay FIGURE 9.6 her
____-__
1
Eqeriment f o r investigating the perfomance of an adaptive egud-
2. The channel filter module is an FIR fdter with coefficients {c(n), 0 5 n 5 K - 1) that simulates the channel distortion. For distortionless transmission, set c(0) = 1 and e(n)= 0 for 1 5 n 5 K - 1. The length K of the filter is an input parameter. 3. The noise generator module is wed t o generate additive noise that is usually present in any digital communicationsystem. If we are modeling noise that is generated by electronic devices, the noise distribution should be Gaussian with zero mean. Use the randu function. 4. The adaptive equalizer module is an FIR filter with tap coefficients {h(k), 0 < k < N - l}, which are adjusted by the LMS algorithm. However, due t o the use of complex arithmetic, the recursive equation in the LMS algorithm is slightly modified to
h,(k) = h,-l(k)
+ Ae(n)z*(n- k)
(9.15)
where the asterisk denotes the complex conjugate. 5. The decision device module takes the estimate &(n) quantizes and it to one of the four possible signal points on the basis of the following decision rule:
Re[&(n)]>O and Im[&(n)]>O
--+
l+j
1-j
Re[&(n))>O and Im[C(n)]<O Re[&(n)]<O and Im[B(n)]>O Re[b(n)]<O and Im\zl(n)]<O
-+
+
--+
-1+j
-1-j
304
Chapter 9 I APPLICATIONS IN ADAPTIVE FILTERING
The effectiveness of the equalizer in suppressing the IS1 introduced by the channel filter may be seen by plotting the following relevant sequences in a two-dimensional (real-imaginary) display. The data generator output { a ( n ) )should consist of four points with values f *j.The effect l of channel distortion and additive noise may be viewed by displaying the sequence {~(n)} the input to the equalizer. The effectiveness of at the adaptive equalizer may be assessed by plotting its output {h(n)} after convergence of its coefficients. The short-time average squared error ASE(n) may also be used to monitor the convergence characteristics o f the LMS algorithm. Note that a delay must be introduced into the output of the data generator to compensate for the delays that the signal encounters due to the channel filter and the adaptive equalizer. For example, this delay may be set to the largest integer closest to (N+ K ) / 2 . Finally, an error counter may be used to count the number of symbol errors in the received data sequence, and the ratio for the number of errors to the total number of symbols (error rate) may be displayed. The error rate may be varied by changing the level of the IS1 and the level of the additive noise. It is suggested that simulations be performed for the following three channel conditions: a. No ISI: c(0) = 1, ~ ( n = 0, 1 5 n 5 K - 1 ) b. Mild IS1 c(0) = 1, c(1) = 0.2, c(2) = -0.2, c(n) = 0, 3 5 n _< K-1 c. StrongISI: c(0) = 1, c(1) = 0.5, c(2) = 0.5, c(n) = 0 , 3 5 n 5 K-1 The measured error rate may be plotted as a function of the signaltc-noise ratio (SNR) at the input to the equalizer, where SNR is defined as PJP,, where P, is the signal power, given as P = s2, and P, is the , noise power of the sequence at the output of the noise generator.
SUMMARY
In this chapter we introduced the reader to the theory and implementation of odaptjve FIR filters with applications to system identification, interference suppression, narrowband frequency enhancement, and a d a p tive equalization. Projects were formulated involving these applications of adaptive filtering; these can be implemented using MATLAB.
Summary
385
APPLICATIONS IN COMMUNICATIONS
Today MATLABfinds widespread use in the simulation o f a variety of communication systems. In this chapter we shall focus on several applications dealing with waveform representation and coding, especially speech coding, and with digital communications. In particular, we shall describe several methods for digitizing analog waveforms, with sp'ecific application to speech coding and transmission. These methods are p b c o d e modula tion (PCM), differential PCM and adaptive differential PCM (ADPCM), delta modulation (DM) and adaptive delta modulation (ADM), and linear predictive coding (LPC). A project is formulated involving each of these waveform encoding methods for simulation using MATLAB. The last three topics treated in this chapter deal with signal-detection applications that are usually encountered in the implementation of a r& ceiver in a digital communication system, For each of these topics we describe a project that involves the implementationsvia Simulation of the detection scheme in MATLAB.
PULSE-CODE MODULATION
I
I
Pulsecode modulation is a method for quantizing an analog signal for the purpose of transmitting or storing the signal in digital form. PCM is widely used for speech transmission in telephone communications and for telemetry systems that employ radio transmission. We shall concentrate our attention on the application of PCM to speech signal processing. Speech signals transmitted over telephone channels are usually limited in bandwidth to the frequency range below 4kHz. Hence the Nyquist rate for sampling such a signal is less than 8kHz. In PCM the analog speech signal is sampled at the nominal rate of 8kHz (samples per second), and each sample is quantized to one of 2' levels, and represented digitally by
386
a sequence of b bits. Thus the bit rate required to transmit the digitized speech signal is 8000 b bits per second. s The quantization process may be modeled mathematically a
Z(n)= s(n) q(n)
+
(10.1)
where B(n) represents the quantized value of ~ ( n ) , q(n)represents the and quantization error, which we treat as an additive noise. Assuming that a uniform quantizer is used and the number of levels is s&ciently large, the quantization noise is well characterized statistically by the uniform probability density function,
where the step size of the quantizer is A = 2-b. The mean square value of the quantization error is
E (q2) = - =
12
A2
12
= -6b - 10.8 dE?
2-2b
(10.3)
Measured in decibels, the mean square value of the noise is 1Olog
(g)
= 1Olog
(g)
(10.4)
We observe that the quantization noise decreases by 6 dB/bit used in the quantizer. High-quality speech requires a minimum of 12 bits per sample and hence a bit rate of 96,000bits per second (bps). Speech signals have the characteristic that small signal amplitudes occur more frequently than large signal amplitudes. However, a uniform quantizer provides the same spacing between successive levels throughout the entire dynamic range of the signal. A better approach is to use a nonuniform quantizer, which provides more closely spaced levels at the low signal amplitudes and more widely spaced levels at the large signal amplitudes. For a nonuniform quantizer with b bits, the resulting quantization error has a mean square value that is smaller than that given by (10.4). A nonuniform quantizer characteristic is usually obtained by passing the signal through a nonlinear device that compresses the signal amplitude, followed by a uniform quantizer. For example, a logarithmic compressor employed in U.S.and Canadian telecommunications systems, called a p-law compressor, has an input-output magnitude characteristic of the form
Pulse-code Modulation
301
where s is the normalized input, y is the normalized output, sgn (.) is the sign function, and p is a parameter that is selected to give the desired compression characteristic. In the encoding of speech waveforms the value of p = 255 has been adopted a a standard in the U.S. and Canada. This value results in about s a 24dB reductioa in the quantization noise power relative to uniform quantization. Consequently, an &bit quantizer used in conjunction with a p = 255 logarithmic compressor produces the same quality speech as a 12-bit uniform quantizer with no compression.Thus the compressed PCM speech signal has a bit rate of 64,000 bps. The logarithmic compressor standard used in European telecommunication systems is called A-law and is defined as
(10.6)
where A is chosen as 87.56. Although (10.5) and (10.6) are different nonlinear functions, the two compressioncharacteristics are very similar. Figure 10.1 illustrates these two compression functions. Note their strong similarity. In the reconstruction of the signal &om the quantized dues, the decoder employs an inverse logarithmic relation to expand the signal am-
I
0
0.2
I 0.4
I 0.6
I 0.8
I
1.0
I4
FIGURE 10.1
Comparison of p-law and A-law nonlinearities
388
Chapter 10 W APPLICATIONS IN COMMUNICATIONS
plitude. For example, in p-law the inverse relation is given by
(10.7)
The combined compressor-expander pair is termed a compander.
PROJECT 10.1: PCM
The purpose of this project is to gain an understanding of PCM compres sion (linear-to-logarithmic) and PCM expansion (logarithmic-tdinear). functions for this project: Write the following three MATLAB
1. a p-law compressor function to implement (10.5) that accepts a zero-mean normalized ( s _< 1) signal and produces a compressed zeroI1 u mean signal with , as a free parameter that can be specified, 2. a quantizer function that accepts a zero-mean input and produces an integer output after bbit quantization that can be specified, and 3. a p-law expander to implement (10.7) that accepts an integer input and produces a zero-mean output for a specified p parameter.
For simulation purposes generate a large number of samples (l0,OOO or more) of the following sequences: (a) a sawtooth sequence, (b) an exponential pulse train sequence, (c) a sinusoidal sequence, and (d) a random sequence with small variance. Care must be taken to generate nonperiodic sequences by choosing their normalized frequencies as irrational numbers (i.e., sample values should not repeat). For example, a sinusoidal sequence can be generated using s (n) = 0.5 sin (n/33), 0 5 n 5 10, OO O n o m our discussions in Chapter 2 this sequence is nonperiodic, yet it has a periodic envelope. Other sequences can also be generated in a similar fashion. Process these signals through the above plaw compressor, quantizer, and expander functions as shown in Figure 10.2, and compute the
Generate random signals
(s(n))
*
L
Ar
(yy(n)}
Av
{y,(n))
#
Is,(n))
*
Plot original and reconstructed signal
Pulse-code Modulation
389
signal-tequantization noise ratio (SQNR) in dB as
For different &bit quantizers, systematically determine the value of p that maxhizes the SQNR. Also plot tlie input and output waveforms and comment on the results.
DIFFERENTIAL PCM (
I
In PCM each sample of the waveform is encoded independently of all the other samples. However, most signals, including speech, sampled at the Nyquist rate or faster exhibit significant correlation between successive samples. In other words, the average change in amplitude between successive samples is relatively small. Consequently,an encoding scheme that exploits the redundancy in the samples will result in a lower bit rate for the speech signal. A relatively simple solution is to encode the differences between successive samples rather than the samples themselves. Since differences between samples are expected to be smaller than the actual sampled amplitudes, fewer bits are required to represent the differences. A refinement o this general approach js to predict the current sample based on the f previous p samples. To be specific, let s (n)denote the current sample of defined as speech and let 1(n)denote the predicted value of ~ ( n ) ,
(10.8)
Thus 5 (n)is a weighted linear combination of the past p samples, and the a(i) are the predictor (filter) coefficients. The a(i) are selected to minimize some function of the error between s (n)and 5 (n). A mathematically and practically convenient error function is the s u m of squared errors. With this as the performance index for the predictor, we select the a (i) to minimize
(10.9)
390
Chapter 10
APPLICATIONS IN COMMUNICATIONS
where rSs (m) is the autocorrelation function of the sampled signal sequence s (n), defined as
r,, (m) =
c (i) s
i=l
N
s (i
+ m)
(10.10)
Minimization of EP with respect to the predictor coefficients {ui(n)} r e sults in the set of linear equations, called the normal equations,
P
Z a ( i ) r s s( i - j) = T S ,
i=l
,
j = 1,2,. .,p .
(10.11)
or in the matrix form,
Ra = r ==+ a = R-'r
(10.12)
where R is the autocorrelation matrix, a is the coefficient vector, and r is the autocorrelation vector. Thus the values of the predictor coefficients are established. Having described the method for determining the predictor coefficients, let us now consider the block diagram of a practical DPCM system, shown in Figure 10.3. In this configuration the predictor is implemented with the feedback loop around the quantizer. The input to the predictor is denoted as B (n), which represents the signal sample s (n) modified by the quantization process, and the output of the predictor is
is the input to the quantizer, and d (n)denotes the output. Each value of the quantized prediction error B (n) is encoded into a sequence of binary digits and transmitted over the channel to th5receiver. The quantized error 8 (n)is also added to the predicted value 3 (n)to yield 3 (n). At the receiver the same predictor that was used at the transmitting end is synthesized, and its output S (n)is added to B (n)to yield 3 (n).The signal S (n)is the desired excitation for the predictor and also the desired output sequence from which the reconstructed signal S ( t ) is obtained by filtering, as shown in Figure 10.3b. The use of feedback around the quantizer, ax described above, ensures that the error in 3 (n)is simply the quantization error q (n) = B (n) (n) -e and that there is no amumulation of previous quantization errors in the implementation of the decoder. That is, q(n)= d(n) - e ( n ) = d(n) - s ( n ) s(n)= S(n)- s (n)
A
+
(10.15)
Hence S (n)= s (n) q (n).This means that the quantized sample S (n) differs from the input s (n) by the quantization error q (n)independent of the predictor used. Therefore the quantization errors do not accumulate. In the DPCM system illustrated in Figure 10.3, the estimate or predicted value 3 (n)of the signal sample s (n) is obtained by taking a linear combination of past values S (n- k) , k = 1 , 2 , . . ., p , as indicated by (10.13). An improvement in the quality of the estimate is obtained by including linearly filtered past values of the quantized error. Specifically, the estimate of s (n) may be expressed as
A
+
P
B (n)=
i=l
a (i) 3 (n- i )
+
m
b (i) E (n- i)
i=l
(10.16)
where b (i) are the coefficients of the filter for the quantized error sequence d(n). The block diagram of the encoder at the transmitter and the decoder at the receiver are shown in Figure 10.4. The two sets of coefficients a (i) and b ( i ) are selected to minimize some function of the error e(n) = S (n)- s (n), such as the sum of squared errors. By using a logarithmic compressor and a Cbit quantizer for the error sequence e (n),DPCM results in high-quality speech at a rate of 32,000 bps, which is a factor of two lower than logarithmic PCM.
PROJECT 10.2 DPCM
The objective of this project is to gain understanding of the DPCM encoding and decoding operations.For simulation purposes, generate correlated
392
Chapter 10
APPLICATIONS IN COMMUNICATIONS
6(n)
~
1
'
To channel
Linear filter
+
(b;)
-&-t
(a)Encoder
converter
filter
filter
(b) Decoder FIGURE 10.4
DPCM modified ba, the
lanearty jiftered e m r sequence
random sequences using a pole-zero signal model of the form
s (n)= a(1) s (n- 1) f box (n) blz (n- 1)
+
(10.17)
where x ( n ) is a zero-mean unit variance Gaussian sequence. T i can hs be done using the filter function. The sequences developed in Project 10.1can also be used for simulation. Develop the following three MATLAB modules for this project: 1. a model predictor function to implement (lO.lZ), given the input signal s (n); 2. a DPCM encoder function to implement the block diagram of Figure 10.3a,which accepts a zerc-mean input sequence and produces a quantized bbit integer error sequence, where b is a free parameter; and 3. a DPCM decoder function of Figure 10.3b, which reconstructs the signal from the quantized error sequence. Experiment with several porder prediction models for a given signal and determine the optimum order. Compare this DPCM implementation with the PCM system o Project 10.1 and comment on the results. Exf tend this implementation to include an mth-order moving average filter as indicated in (10.16).
Differential PCM (DPCM)
393
ADAPTIVE PCM fADPCM1 AND DPCM
In general, the power in a speech signal varies slowly with time. PCM and DPCM encoders, however, are designed on the basis that the speech signal power is constant, and hence the quantizer is fixed. The efficiency and performance of these encoders can be improved by having them adapt to the slowly time-variant power level of the speech signal. resulting from a In both PCM and DPCM the quantization error q (n) uniform quantizer operating on a slowly varying power level input signal will have a time-variant variance (quantization noise power). One improvement that reduces the dynamic range of the quantization noise is the use of an adaptive quantizer. Adaptive quantizers can be classified as feedforward or feedback. A feedforward adaptive quantizer adjusts its step size for each signal sample, based on a measurement of the input speech signal variance (power). For example, the estimated variance, based as a sliding window estimator, is
-
ntl
(10.18)
k=n+l-M
Then the step size for the quantizer is
A (n 1) = A (n)Bn+l
+
(10.19)
In this caSe it is necessary to transmit A (n 1) to the decoder in order for it to reconstruct the signal. A feedback adaptive quantizer employs the output of the quantizer in the adjustment of the step size. In particular, we may set the step size a s
+
A (n + 1) = a (n) (n) A
(10.20)
where the scale factor a (n)depends on the previous quantizer output. For example, if the previous quantizer output is small, we may select a (n)< 1 in order to provide for finer quantization. On the other hand, if the quantizer output is large, then the step size should be increased to reduce the possibility of signal clipping. Such an algorithm has been successfully used in the encoding of speech signals. Figure 10.5 illustrates such a (bbit) quantizer in which the step size is adjusted recursively according to the relation
A (n+ 1) = A (n). M (n)
394
Chapter 10 APPLICATIONS IN COMMUNICATIONS
output
A
7812
-
111 -Previous output
--76/2
FIGURE 10.5
Ezample of a quantizer m'th an adaptive step size (l] /o)
where A (71) is a multiplication factor whose value depends on the quan4 tizer level for the sample s (n), A ( n ) is the step size of the quantizer and for processing s(n). Values of the multiplication factors optimized for speech encoding have been given by 1131. These values are displayed in Table 10.1 for 2-, 3-, and Cbit quantization for PCM and DPCM. In DPCM the predictor can also be made adaptive. Thus in ADPCM the coefficients of the predictor are changed periodically to reflect the changing signal statistics of the speech. The linear equations given by (10.11) still apply, but the short-term autocorrelation function of s (n), ras(m) changes with time.
TABLE 10.1 Multiplacataon factors for adaptive step size adjustment ([lo])
2
PCM 3
0.85
4
0.80 0.80
0.80 0.80
2
DPCM 3
0.90 0.90
1.25 1.70
4
09 .0 0.90
0.90
0.60 2.20
1.00
1.00 1.50
0.80 1.60
0.80 0.80
0.80
0.80
0.90 1.20 1.60 2.00 2.40
Adaptive P C M (ADPCM) and DPCM
395
64 kbit/s
PCM input
-
Convert to uniform PCM
signal
* output
Signal estimate
32 kbit/s
Encoder
I
Quantized difference signal Reconstructed coding
Inverse adaptive quantizer
I
32 kbitk input adaptive
FIGURE 10.6
ADPCM block
diagram
ADPCM STANDARD
Figure 10.6 illustrates, in block diagram form, a 32, OOO bps ADPCM encoder and decoder that has been adopted as an international (CCITT) standard for speech transmission over telephone channels. The ADPCM encoder is designed to accept &bit PCM compressed signal samples at 64,000 bps, and by means of adaptive prediction and adaptive Cbit quantization to reduce the bit rate over the channel to 32,000 bps. The ADPCM decoder accepts the 32,000 bps data stream and reconstructs the signal in the form of an 8-bit compressed PCM at 64,000 bps. Thus we have a configuration shown in Figure 10.7, where the ADPCM encoder/
n
Channel FIGURE 10.7
1
ADPCM
interface to PCM system
3%
Chapter 10
APPLICATIONS IN COMMUNICATIONS
decoder is embedded into a PCM system. Although the ADPCM encoder/ decoder could be used directly on the speech signal, the interface to the PCM system is necessary in practice in order to maintain compatibility with existing PCM systems that are widely used in the telephone network. The ADPCM encoder accepts the &bit PCM compressed signal and expands it to a 14bit-per-sample linear representation for processing. The predicted value is subtracted from this 14bit linear value to produce a difference signal sample that is fed to the quantizer. Adaptive quantization is performed on the difference signal to produce a Cbit output for transmission over the channel. Both the encoder and decoder update their internal variables, based only on the ADPCM values that are generated. Consequently,an ADPCM decoder including an inverse adaptive quantizer is embedded in the encoder so that all internal variables are updated, based on the same data. This ensures that the encoder and decoder operate in synchronismwithout the need to transmit any information on the values of internal variables. The adaptive predictor computes a weighted average of the last six dequantized difference values and the last two predicted values. Hence this predictor is basically a two-pole (p = 2 ) and six-zero (m = 6) filter governed by the difference equation given by (10.16). The filter coefficients are updated adaptively for every new input sample. At the receiving decoder and at the decoder that is embedded in the encoder, the Cbit transmitted ADPCM value is used to update the inverse adaptive quantizer, whose output is a dequantized version of the difference signal. This dequantized value is added to the value generated by the adaptive predictor to produce the reconstructed speech sample. This signal is the output of the decoder, which is converted to compressed PCM format at the receiver.
PROJECT 10.3: ADPCM
The objective of this project is to gain familiarity with, and understanding of, ADPCM and its interface with a PCM encoder/decoder (transcoder). As described above, the ADPCM transcoder is inserted between the PCM compressor and the PCM expander a shown in Figure 10.7. Use the s already developed MATLAB PCM and DPCM modules for this project. The input to the PCM-ADPCM transcoder system can be supplied from internally generated waveform data files, just as in the caw of the PCM project. The output of the transcoder can be plotted. Comparisons should be made between the output signal from the PCM-ADPCM transcoder with the signal from the PCM transcoder (PCM project lO.l), and with the original input signal.
397
Adaptive PCM (ADPCM) and DPCM
DELTA MODULATION (DM\
Delta modulation may be viewed as a simplified form of DPCM in which a two-level (1-bit) quantizer is used i conjunction with a fixed first-order n predictor. The block diagram of a DM encoder-decoder is shown in Figure 10.8. We note that
-
a(n) = i ( n - 1) = I ( : ?
- 1) + a(n - 1)
(10.21)
Since
it follows that
s(n)= s ( n - 1) +q(n - 1)
(10.22)
Thus the estimated (predicted) value of s (n)is really the previous sample s (n- 1) modified by the quantization noise q (n- 1). We also note that the difference equation in (10.21)represents an integrator with an s input E(n).Hence an equivalent realization of the onestep predictor i an accumulator with an input equal to the quantii error signal .Z (n).In general, the quantized error signal is scaled by some value, say Al, which is called the step size. T i equivalent realization is illustrated in Figure hs 10.9. In effect, the encoder shown in Figure 10.9 approximates a waveform s (t) by a linear staircase function. In order for the approximation to be relatively good, the waveform s (t) must change slowly relative to
E
W
I
I
Decoder
FIGURE 10.8
Block diagmm of a delta modulation system
398
Chapter 10
APPLICATIONS IN COMMUNICATIONS
"--c7.--0
'(')
I
To *channel
Accumulator
Accumulator
output
the sampling rate. This requirement implies that the sampling rate must be several (a factor of at least 5) times the Nyquist rate. A Iowpass fiIter is usually incorporated into the decoder to smooth out discontinuities in the reconstructed signal.
ADAPTIVE DELTA MODULATION (ADMI
At any given sampling rate, the performance of the DM encoder is limited by two types of distortion as shown in Figure 10.10. One is called slope overload distortion. It is due to the use of a step size A, that is too small to follow portions of the waveform that have a steep slope. The second rm type of distortion, called granular noise, results f o using a step size that is too large in parts of the waveform having a small slope. The need to minimize both of these two types of distortion results in conflicting requirements in the selection of the step size Ai. An alternative solution is to employ a variable size that adapts itself to the short-term characteristics of the source signal. That is, the step size is increased when the waveform has a steep slope and decreased when the waveform has a relatively small slope.
t
FIGURE 10.10
~"WO types
of distortion in the
ct
Small A and slopeoverload noise
DM encoder
Delta Modulation (DM)
399
A variety of methods can be used to set adaptively the step size in every iteration. The quantized error sequence 8 (n) provides a good indication of the slope characteristicsof the waveform being encoded. When the quantized error 5 (n) is changing signs between successive iterations, this is an indication that the slope of the waveform in the locality is relatively small. On the other hand, when the waveform has a steep slope, successive values of the error 8 (n) are expected to have identical signs. From these observations it is possible to devise algorithms that decrease or increase the step size, depending on successive values of 5 (n). A relatively simple rule devised by [12] is to vary adaptively the step size according to the relation
A (n)= A (n - 1) Kz(")i("-l) n = I,Z, ... ,
(10.23)
where K 2 1is a constant that is selected to minimize the total distortion. A block diagram of a DM encoder-decoder that incorporates this adaptive algorithm is illustrated in Figure 10.11.
Encodw
Decoder FIGURE 10.11
Lowpass fiker
+ Output
An example of a delta modulation system with adaptive step size
400
Chapter 10
IAPPLICATIONS
IN COMMUNICATIONS
Several other variations of adaptive DM encoding have been i n v w tigated and described in the technical literature. A particularly effectiw and popular technique first propmed by [S] is called continuously variable slope delta modulation (CVSD). In CVSD the adaptive step size parame ter may be expressed as
A (n)= CXA - 1) (71
+ kl
(10.24)
if d (n), - l), and Z(n - 2) have the same sign; otherwise Z(n
A(n) = a A ( n - 1) +kz
(10.25)
The parameters a, kl, and kz are selected such that 0 < a < 1 and kl > kz > 0. For more discussion on this and other variations of adaptive DM, the interested reader is referred to the papers by 1131 and (61 and to the extensive references contained in these papers.
PROJECT 10.4: DM AND ADM
The purpose of this project is to gain an understanding of delta modulation and adaptive delta modulation for coding of waveforms. This project involves writing MATLABfunctions for the DM encoder and decoder as shown in Figure 10.9, and for the ADM encoder and decoder shown in Figure 10.11. The lowpass Filter at the decoder can be implemented as a linear-phase FIR filter. For example, a Hanning filter that has the impulse response
may be used, where the length N may be selected in the range 5 5 N 15. The input to the DM and ADM systems can be supplied from the waveforms generated in Project 10.1 except that the sampling rate should be higher by a factor of 5 to 10. The output of the decoder can be plotted. Comparisons should be made between the output signal from the DM and ADM decoders and the original input signal.
LINEAR PREDICTIVE CODING (LPC) OF SPEECH
m
The linear predictive coding (LPC) method €or speech analysis and synthesis is based on modeling the vocal tract 85 a linear all-pole (IIR) filter having the system function
6=1
Linear Predictive Coding (LPC) of Speech
401
White noise generator
Voiced and filter signal
impulse
FIGURE 10.12 Bloek diagram model for the genemtaon of a speech signal
where p is the number of poles, G is the filter gain, and { ,(k)} are the a parameters that determine the poles. There are two mutually exclusive excitation functions to model voiced and unvoiced speech sounds. On a short-time basis, voiced speech is periodic with a fundamental frequency Fo, or a pitch period l/Fo, which depends on the speaker. Thus voiced speech is generated by exciting the all-pole filter model by a periodic impulse train with a period equal to the desired pitch period. Unvoiced speech sounds are generated by exciting the all-pole filter model by the output of a random-noise generator. This model is shown in Figure 10.12. Given a short-time segment of a speech signal, usually about 20 ms or 160 samples at an 8 W z sampling rate, the speech encoder at the transmitter must determine the proper excitation function, the pitch period for voiced speech, the gain parameter G, and the coefficients a, (k).A block diagram that illustrates the speech encoding system is given in Figure 10.13.The parameters o the model are determined adaptively from the f data and encoded into a binary sequence and transmitted to the receiver.
Sampled
: ; !
-
-zx7
t
voiced or unvoiced model and excitation
Pitch frequency
t
Encoder
- t output -
Excitation
t
Signal generator Input Decoder Filter parameters Lowpass
(b) Decoder
FIGURE 10.13
Encoder and decoder for LPC
402
Chapter 10
W
APPLICATIONS IN COMMUNICATIONS
At the receiver the speech signal is synthesized from the model and the excitation signal. The parameters of the all-pole filter model are easily determined from the speech samples by means of linear prediction. To be specific, the output of the FIR linear prediction filter is
we can determine the pole parameters (ap (k)} of the model. The result of differentiating E with respect to each of the parameters and equating the result to zero, is a set of p linear equations
k=l
2
ap(k)ras(m - k) = -ran (m) , m = 1,2,. ., p .
(10.31)
where T,, (m) is the autocorrelation of the sequence s (n) defined as
N
r8,(m) = C s ( n ) s ( n + m )
n=O
(10.32)
The linear equations (10.31) can be e x p r d in matrix form as
Raga= -rss
(10.33)
where R,, is a p x p autocorrelation matrix, r., is a p x I autocorrelation vector, and a is a p x 1 vector of model parameters. Hence
a = -R-lr 8,
86
(10.34)
These equations can also be solved recursively and most efficiently, without resorting to matrix inversion, by using the Levinson-Durbinalgorithm 1 9 .However, in MATLAB is convenientto use the matrix inversion.The 11 it all-pole filter parameters {aP(k)} can be converted to the all-pole lattice
Linear Predictive Coding (LPC) o Speech f
403
parameters { K i } (called the reflection coefficients) using the MATLAB function d i r 2 l a t c developed in Chapter 6. The gain parameter of the filter can be obtained by noting that its input-output equation is
s (n)= -
C a, (k)s (n- k) + GX (n)
k=l
P
P
(10.35)
where z (n)is the input sequence. Clearly,
Gx (n)= s (n) c a p(k)s (n- k) = e (n) +
k=1
Then (10.36)
nO = n=O
If the input excitation is normalized to unit energy by design, then
N-1
GZ=
n=0
e2 (4 =
(0)+
c
P
k=l
a,
(k)rsa (k)
(10.37)
Thus G2is set equal to the residual energy resulting from the least-squares optimization. Once the LPC coefficients are computed, we can determine whether the input speech frame is voiced, and if so, what the pitch is. This is accomplished by computing the sequence
re
(a) =
c
P
k=l
Ta
(k)rss (n - k)
(10.38)
where ra (k) is defined a s
ra
(k)= C a p (i) ap (i + k)
i=l
P
(10.39)
which is the autocorrelation sequence of the prediction coefficients. The pitch is detected by finding the peak of the normalized sequence T , (n) /re(0) in the time interval that correspunds to 3 to 15 ms in the 20-ms sampling frame. If the value of this peak is at least 0.25, the frame of sp,eech is considered voiced with a pitch period equd to the value of n = Np,where re (Np)/ T , (0) is a maximum. If the peak value is less than 0.25, the frame of speech is considered unvoiced and the pitch is zero.
404
Chapter 10 R APPLICATIONS IN COMMUNICATIONS
The values of the LPC coefficients, the pitch period, and the type of excitation are transmitted to the receiver, where the decoder synthesizes the speech signal by passing the proper excitation through the all-pole filter model of the vocal tract. Typically, the pitch period requires 6 bits, and the gain parameter may be represented by 5 bits after its dynamic range is compressed logarithmically. If the prediction coefficients were to be coded, they would require between 8 to 10 bits per coefficient for accurate representation. The reason for such high accuracy is that relatively small changes in the prediction coefficients result in a large change in the pole positions of the filter model. The accuracy requirements are lessened by transmitting the reflection coefficients { K i } ,which have a smaller dynamic range-that is, IKil < 1. These are adequately represented by 6 bits per coefficient. Thus for a 10th-order predictor the total number of bits assigned to the model parameters per frame is 72. If the model parameters are changed every 20 milliseconds, the resulting bit rate is 3,600 bps. Since the reflection coefficients are usually transmitted to the receiver, the synthesis filter at the receiver is implemented as an all-pole lattice filter, described in Chapter 6.
PROJECT 10.5:
LPC
The objective of this project is to analyze a speech signal through an LPC coder and then to synthesize it through the corresponding PLC decoder. Use several .wav sound as (sampled at 8000 sam/sec rate), e which are available in MATLABfor this purpose. Divide speech signals into short-time segments (with lengths between 120 and 150 samples) and process each segment to determine the proper excitation function (voiced or unvoiced), the pitch period for voiced speech, the coefficients {ap(k)} (p lo), and the gain G. The decoder that performs the synthesis is an all-pole lattice filter whose parameters are the reflection coeficients that can be determined from { a p (k)}. The output of this project is a synthetic speech signal that can be compared with the original speech signal. The distortion effects due to LPC analysis/synthesis may be assessed qualitatively.
DUAL-TONE MULTIFREQUENCY (DTMF) SIGNALS
I
I
DTMF is the generic name for push-button telephone signaling that is equivalent to the Touch Tone system in use within the Bell System. DTMF also finds widespread use in electronic mail systems and telephone banking systems in which the user can select options from a menu by sending DTMF signals from a telephone. In a DTMF signaling system a combination of a high-frequency tone and a low-frequency tone represent a specific digit or the characters * and #. The eight frequencies are arranged as shown in Figure 10.14, to
405
Dual-tone Multifnquency (DTMF) Signals
COll
1209Hz
Col2 1336Hz
Col3 1477Hz
Col4 1633Hz
DTMF digit = row tone + column tone
FIGURE 10.14
DTMF digits
accommodate a total of 16 characters, 12 of which are assigned a shown, s while the other four are reserved for future use. DTMF signals are easily generated in software and detected by means of digital filters, also implemented in software, that are tuned to the eight frequency tones. Usually, DTMF signals are interfaced to the analog world via a wdec (coder/decoder) chip or by linear A/D and D/A converters. Codec chips contain all the necessary A/D and D/A, sampling, and filtering circuitry for a bi-directional analog/digital interface. The DTMF tones may be generated either mathematically or from a look-up table. In a hardware implementation (e.g., in a digital signal processor), digital samples of two sine waves are generated mathematically, scaled, and added together. The sum is logarithmically compressed and sent to the codec for conversion to an analog signal. At an 8 lcHz sampling rate the hardware must output a sample every 125 ms. In this case a sine look-up table is not used because the values of the sine wave can be computed quickly without using the large amount of data memory that a table look-up would require. For simulation and investigation purposps the look-up table might be a good approach in MATLAB. At the receiving end the logarithmicallycompressed,&bit digital data words from the codec are received, logarithmically expanded to their 16bit linear format, and then the tones are detected to decide on the transmitted digit. The detection algorithm can be a DFT implementation using the FFT algorithm or a filter bank implementation. For the relatively small number of tones to be detected, the filter bank implementation is
406
Chapter 10
APPLICATIONS IN COMMUNICATIONS
more efficient. Below, we describe the use of the Goertzel algorithm to implement the eight tuned filters. Recall from the discussion in Chapter 5 that the DFT of an N-point z data sequence { (n)} is
N-1
X ( k ) = X x ( n ) W E k , k = 0 , 1 , ..., N - 1
n=O
(10.40)
If the FFT algorithm is used to perform the computation of the DFT, the number of computations (complex multiplications and additions) is N logz N. In this case we obtain all N values of the DFT at once. However, if we desire to compute only M points of the DFT, where M < log, N , then a direct computation of the DFT is more efficient. The Goertzel algorithm, which is described below, is basically a linear filtering approach t o the computation of the DFT, and provides an alternative to direct computation.
THE GOERTZEL ALGORITHM
The Goertzel algorithm exploits the periodicity o the phase factors (W i } f and allows us to express the computation of the DFT as a linear filtering operation. Since I+',$N = 1, we can multiply the DFT by this factor. Thus
N-1
X (k) = WGkNX = (k)
m=O
x (m) i k ( N - m ) W
(10.41)
We note that (10.41) is in the form of a convolution. Indeed, if we define the sequence Y k (n) BS
N- 1
Yk
(n)=
mO =
x (m) I+'ik(n-m)
(10.42)
then it is clear that y k (n) is the convolution of the finite-duration input sequence x (n) of length N with a filter that has an impulse response
h k (n)= W p u (n)
(10.43)
The output of this filter at n = N yields the value of the DFT at the frequency W k = 2*k/N. That is,
The filter with impulse response h (n) the system function k has
(10.45)
This filter has a pole on the unit circle at the frequency w = 2sk/N. k Thus the entire DFT can be computed by passing the block of input data into a parallel bank of N singlepole filters (resonators), where each filter has a pole at the corresponding frequency of the DFT. Instead of performing the computation of the DFT as in (10.42),via convolution, we can use the difference equation corresponding to the filter given by (10.45) to compute Y k (n) recursively. Thus we have
Y k (n) = w i k u k (n - 1) f
(n)7
Yk
(-1) = 0
(10.46)
The desired output is X ( k ) = y k (N). perform this computation, we To can compute once and store the phase factor W;'. The complex multiplications and additions inherent in (10.46) can be avoided by combining the pairs of resonators possessing complex conjugate poles. This leads to two-pole filters with system functions of the form
H (2)= k
11 - 2 cos(2aklN)2-1
w p
+ 2-2
(10.47)
The realization of the system illustrated in Figure 10.15 is described by the difference equations
= X(k)
FIGURE 10.15 Realamtion of two-pole resonator fur computing the DFT
408
Chapter 10 0 APPLICATIONS IN COMMUNICATIONS
with initial conditions vg (-1) = vk (-2) = 0. This is the Goertzel algorithm. The recursive relation in (10.48) iterated for n = 0,1, . ., N , but the is . equation in (10.49)is computed only once, at time n = N . Each iteration requires one real multiplication and two additions. Consequently,for a real this input sequence z(n), algorithm requires N 1 real multiplications to yield not only X ( k ) but also, due to symmetry, the value of X ( N - k ) . We can now implement the DTMF decoder by use of the Goertzel algorithm. Since there are eight possible tones to be detected, we require with each filter tuned to one of the eight filters of the type given by (10.471, eight frequencies. In the DTMF detector, there is no need to compute the complex value X (k);only the magnitude IX(k)l or the magnitudesquared value IX(k)12will suffice. Consequently,the final step in the computation of the DFT value involving the numerator term (feedforward part of the filter computation) can be simplified. In particular, we have
The objective of this project is to gain an understanding of the DTMF tone generation software and the DTMF decoding algorithm (the Goertzel algorithm). Design the following MATLAB modules:
1 a tone generation function that accepts an array containing dial. ing digits and produces a signal containing appropriate tones (from Figure 10.14) of one-half-second duration for each digit at 8 kHz sampling fe r wency, 2. a dial-tone generator generating samples of (350+440) Hz frequency at 8 kHz sampling interval for a specified amount of duration, and 3. a decoding function to implement (10.50)that accepts a DTMF signal and produces an array containing dialing digits.
Generate several dialing l i t arrays containing a mix of digits and dial tones. Experiment with the tone generation and detection modules and sound generation capabilcomment on your observations. Use MATLAB'S ities to listen to the tones and to observe the frequency components of the generated tones.
DuaCtone Multifrequency (DTMF) Signals
409
BINARY DIGITAL COMMUNICATIONS
I
Digitized speech signals that have been encoded via PCM, ADPCM, DM, and LPC are usually transmitted to the decoder by means of digital modulation. A binary digital communicationssystem employs two signal waveforms, say s l ( t ) = s ( t ) and s z ( t ) = - s ( t ) , to transmit the binary sequence representing the speech signal. The signal waveform s ( t ) , which is nonzero over the interval 0 5 t 5 T, is transmitted to the receiver if the data bit is a 1, and the signal waveform - s ( t ) , 0 t 5 T is transmitted if the data bit is a 0. The time interval T is called the signal interval, and the bit rate over the channel is R = 1/T bits per second. A typical signal waveform s ( t ) is a rectangular p u l s e t h a t is, s ( t ) = A, 0 t T-which has energy A ~ T . In practice the signal waveforms transmitted over the channel are corrupted by additive noise and other types of channel distortions that ultimately limit the performance of the communications system. As a measure of performance we normally use the average probability of error, which is often called the bit error rate.
<
< <
PROJECT 10.7: BINARY DATA COMMUNICATIONS SYSTEM
The purpose of this project is to investigate the performance of a binary data communications system on an additive noise channel by means of simulation. The basic configurationof the system to be simulated is shown in Figure 10.16. Five MATLAB functions are required.
1. A binary data generator module that generates a sequence of independent binary digits with equal probability. 2. A modulator module that maps a binary digit 1 into a sequence of M consecutive +l's, and maps a binary digit 0 into a sequence of M consecutive -1's. Thus the M consecutive +l's represent a sampled version of the rectangular pulse. 3. A noise generator that generates a sequence of uniformly distributed numbers over the interval ( - a , a ) . Each noise sample is added to a corresponding signal sample.
4. A demodulator module that sums the M successive outputs of the noise corrupted sequence +l's or -1's received from the channel. We assume that the demodulator is time synchronized so that it knows the beginning and end of each waveform. 5. A detector and error-counting module. The detector compares the output of the modulator with zero and decides in favor of 1if the output is greater than zero and in favor of 0 if the output is less than zero. If the output of the detector does not agree with the transmitted bit from the transmitter, an error is counted by the counter. The error rate depends on the ratio (called signal-to-noise ratio) of the size of M to the additive noise power, which is P = a2/3. ,
The measured error rate can be plotted for different signal-to-noise , ratios, either by changing M and keeping P fixed or vice versa.
SPREAD-SPECTRUM COMMUNICATIONS
I
I
Spread-spectrum signals are often used in the transmission of digital data over communication channels that are corrupted by interference due to intentional jamming or from other users of the channel (e.g., cellular telephones and other wireless applications). In applications other than communications, spread-spectrum signals are used to obtain accurate range (time delay) and range rate (velocity) measurements in radar and navigation. For the sake of brevity we shall limit our discussion to the use of spread spectrum for digital communications. Such signals have the characteristic that their bandwidth is much greater than the information rate in bits per second. In combatting intentional interference (jamming), it is important to the communicators that the jammer who is trying to disrupt their communication does not have prior knowledge of the signal characteristics. To accomplish this, the transmitter introduces an element of unpredictability or randomness (pseudo-randomness)in each of the possible transmitted signal waveforms, which is known to the intended receiver, but not to the jammer. As a consequence, the jammer must transmit an interfering s'gnal without knowledge of the pseudo-random characteristics of the desi ed signal. Interference from other users arises in multiple-access communications systems in which a number of users share a common communications channel. At any given time a subset of these users may transmit information simultaneously over a common channel to corresponding receivers. The transmitted signals in this common channel may be distinguished from one another by superimposing a different pseudo-random pattern, called a multiple-accesscode, in each transmitted signal. Thus a particular
i
Spread-Spectrum Communications
411
receiver can recover the transmitted data intended for it by knowing the pseudo-random pattern, that is, the key used by the corresponding transhs mitter. T i type of communicationtechnique, which allows multiple users to simultaneously use a common channel for data transmission, is called code division multiple access (CDMA). The block diagram shown in Figure 10.17 illustrates the basic e1ements of a spread-spectrum digital communications system. It differs from a conventional digital communications system by the inclusion of two identical pseudo-random pattern generators, one that interfaces with the modulator at the transmitting end, and the second that interfaces with the demodulator at the receiving end. The generators generate a pseudorandom or pseudo-noise (PN) binary-valued sequence (jzl's), which is impressed on the transmitted signal at the modulator and removed from the received signal at the demodulator. Synchronization of the PN sequence generated at the demodulator with the PN sequence contained in the incoming received signal is required in order to demodulate the received signal. Initially, prior to the transmission of data, synchronization is achieved by transmitting a short fixed PN sequence to the receiver for purposes of establishing s y n c h r e nization. After time synchronizationof the PN generators is established, the transmission of data commences.
PROJECT 10.8: BINARY SPREADSPECTRUM COMMUNICATIONS
The objective of this project is to demonstrate the effectiveness of a PN spread-spectrum signal in suppressing sinusoidal interference. Let us consider the binary communication system described in Project 10.7, and let us multiply the output of the modulator by a binary (51) sequence. PN The same binary PN sequence is used to multiply the input to the demodulator and thus to remove the effect of the PN sequence in the desired signal. The channel corrupts the transmitted signal by the addition of a wideband noise sequence { ~ ( n and a sinusoidal interference sequence )} of the form i ( n )= Asinwon, where 0 < WO < T . We may w u m e that A 1 M ,where M is the number of samples per bit from the modulator. The basic binary spread spectrum-systemis shown in Figure 10.18. can As
Input data
Modulator
Demodulator
output data
t
sequence generator FIGURE 10.17
t
sequence generator
Basic spread spectrum digital communications system
412
Chapter 10 m APPLICATIONS IN COMMUNICATIONS
Binary data generator
I
Modulator
generator
FIGURE 10.18
Block diagram of binary PN spread-spectrum system for simulation ezperiment
be observed, this is just the binary digital communication system shown in Figure 10.16, to which we have added the sinusoidal interference and the P N sequence generators. The P N sequence may be generated by using a random-number generator to generate a sequence of equally probable fl's. Execute the simulated system with and without the use of the P N sequence, and measure the error rate under the condition that A 2 M for different values of M ,such as M = 50,100,500,1000. Explain the effect of the P N sequence on the sinusoidal interference signsl. Thus explajn why the PN spread-spectrum system outperforms the conventional binary communication system in the presence of the sinusoidal jarnming signal.
SUMMARY
In this chapter we focused on applications to waveform representation and coding. In particular, we described several methods for digitizing an anal g waveform, including PCM, DPCM, ADPCM, DM, ADM, and LPC. o These methods have been widely used for speech coding and transmission. Projects involving these waveform encoding methods were formulated for implementation via simulation in MATLAB. We also described signal-detection and communication systems where MATLABmay be used to perform the signal processing tasks. Projects were also devised for these applications.
Summary | 677.169 | 1 |
Book Description Glencoe/McGraw-Hill, Blacklick, Ohio, U.S.A., 2005. Spiral Bound. Book Condition: New. 9.0" X 11.0". "MathScapeTM: Seeing and Thinking Mathematically is a standards-based program that encourages your students to learn mathematics by doing mathematics, by using and connecting mathematical ideas, and by actively increasing their understanding. MathScape is one of four middle school programs to receive a satisfactory rating from the American Association for the Advancement of Science (AAAS).". Bookseller Inventory # 000511 | 677.169 | 1 |
Introduction to Infinite Series
In this infinite series worksheet, students explore the pattern in a series of numbers. They solve a sequence of partial sums and identify the limit. This two-page worksheet contains explanations, examples and approximately five problems. | 677.169 | 1 |
The subject of this book is the successive construction and development of the basic number systems of mathematics: positive integers, integers, rational numbers, real numbers and complex numbers. This second edition expands upon the list of suggestions for further reading in Appendix III. From the Preface: 'The present book basically takes for granted the non-constructive set-theoretical foundation of mathematics, which is tacitly if not explicitly accepted by most working mathematicians but which I have since come to reject. Still, whatever one's foundational views, students must be trained in this approach in order to understand modern mathematics. Moreover, most of the material of the present book can be modified so as to be acceptable under alternative constructive and semi-constructive viewpoints, as has been demonstrated in more advanced texts and research articles'. | 677.169 | 1 |
This is part three of a three-part manual for teachers using SMSG junior high school text materials. Each chapter contains an introduction and a collection of sample test questions. Each section contains a discussion related to the topic at hand and answers to all the exercises. Chapter topics include: (1) non-metric geometry; (2) volumes and surface areas; (3) the sphere; and (4) relative error. (MP) | 677.169 | 1 |
Solve Math Problems with JavaScript
In this beginner course, you will learn JavaScr and build dynamic web pages that solve advanced high school math problems. This course is a great starting point for those new to programming or web development. If you're new to programming and want to learn JavaScr, this course is for you. In Solve Math Problems with JavaScr, you will learn core programming concepts such as variables, iteration, branching, and functions; then apply these skills to build interactive web pages that solve advanced high school level math problems. This project-based course works through nine problems of increasing difficulty, such as calculating factorials, simulating rolling dice, and estimating area under a polynomial curve. Each lesson covers the core math and programming concepts, then works through the solution step-by-step. This course is ideal for people who like math and want to learn programming or web development. By the end of this course, you will have a basic understanding of programming in Java and you'll have the skills needed to move onto learning more complex programming topics.
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Pre-Calculus Documents
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The City College of New York
Final Examination
Department of Mathematics
Math 19500
Instructions
Please read the following instructions carefully.
This exam has four parts. Answer five questions in each part, for
a total of twenty questions.
Show all yo
Welcome
Why we do algebra
Laws of algebra
Substitution
Using parentheses
Simplifying expressions
Order of operations
Exercises
Quiz review
Welcome to Math 19500 Video Lessons
Prof. Stanley Ocken
Department of Mathematics
The City College of New York
Fall
The City College of New York
Final Examination
Math 19500
Department of Mathematics
Spring 2011
Your section: M195_ Your instructors name _
Print your name: _ Sign_
Instructions
Show all work inside this booklet.
All electronic devices must be turned of
The City College of New York
Department of Mathematics
Final Examination
Math 19500
Fall 2012
Your section:_ Your instructor_
Your name: Print _ Sign_
Instructions
Show all work inside this booklet.
All electronic devices must be turned off and out of s
Radian measure
Trig functions of general angles
Finding the reference angle
Basic trig identities
Calculating trig functions
Quiz Review
Every real number is the radian measure of some angle
Every real number, viewed as a number of
radians, represents an
Laws of logarithms
Expanding log expressions
Combining log expressions
Quiz Review
Chapter 4.4: Laws of logarithms
Every law of exponents gives rise to a corresponding law of logarithms. In the following,
suppose a > 0 and a = 1, and let a be the base for
Written assignment
Solving simple trigonometric equations
Inverse trigonometric functions
Quiz Review
New feature: short written assignment
Please read the notes on Section 5.5, posted on Blackboard
At the start of class on Wednesday, November 20, please
Polynomial sign analysis
Suppose we want to sketch the graph of the polynomial y = p(x). Before we do that
must perform sign analysis to determine the intervals where p(x) is positive, zero, or
negative. We did this earlier in Chapter 1.7 on inequalities.
Introduction: graphing a polynomial function
Some degree 3 polynomials.
Parabolas: graphs of degree 2 polynomials
Sketching graphs of polynomials
We want to draw the graphs of polynomial functions y = f (x).
The degree of a polynomial in one variable x i | 677.169 | 1 |
The well-known classical geometries, Euclidean, spherical and hyperbolic, are presented here in a general context, each linked by certain geometric themes. This 2007 textbook provides for a thorough introduction to, and examples of, the more general theory of curved spaces, and more generally, abstract surfaces with Riemannian metrics.
"Sinopsis" puede pertenecer a otra edición de este libro.
Review:
'… the patient reader will acquire substantial techniques and methods that are part of differential geometry and along with that, much, much more. … The book is certainly a welcome addition to the literature. It is clear to the reviewer the the text is a labour of love' Mathematical Reviews
'This book paves its way through a number of geometries with a clear intent: to promote the understanding of the basic concepts in differential geometry. The classical 2-dimensional geometries such as Euclidean, spherical and hyperbolic geometry are the first few steps. The geometry of the torus, triangulations and Euler numbers are further issues. Every chapter is followed by an assortment of helpful examples. The chapters gently direct the reader [towards] Riemannian metrics and further to geodesics and abstract surfaces. The Theorema Egregium and Gauss-Bonnet Theorem are also given ample attention. What distinguishes this book from other lecture notes on elementary differential geometry? The author does not abruptly define concepts which otherwise never would have come to the reader's mind. He rather conveys the ideas, subtly initiated in the former chapters. In this sense the book is remarkably self-contained. On the other hand the author does not shun detailed proofs. All these ingredients make for a successful volume.' Johann Lang, Zentralblatt MATH
'The book is written in a nice and precise style and explicit computations and proofs make the book easy to understand. A detailed and explicit discussion of the main examples of classical geometries contributes well to a better understanding of later generalisations. A list of examples at the end of each chapter helps as well. It is a very good addition to the literature on the topic and can be very useful for teachers preparing their courses as well as for students.' EMS Newsletter
About the Author:
Pelham Wilson is Professor of Algebraic Geometry in the Department of Pure Mathematics, University of Cambridge. He has been a Fellow of Trinity College since 1981 and has held visiting positions at universities and research institutes worldwide, including Kyoto University and the Max-Planck-Institute for Mathematics in Bonn. Professor Wilson has over 30 years of extensive experience of undergraduate teaching in mathematics, and his research interests include complex algebraic varieties, Calabi-Yau threefolds, mirror symmetry, and special Lagrangian submanifolds70 Cambridge University Press. Paperback. Estado de conservación: new. BRAND NEW, Curved Spaces: From Classical Geometries to Elementary Differential Geometry, P. M. H. Wilson, B9780521713900 | 677.169 | 1 |
Buy references and guides for algebra and trigonometry at Amazon
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When the fundamentals of a subject are not clear, one tends to shy away from the subject. Having said that, the importance of these chapters cannot be undermined, as they occupy a higher weightage in the exams. Not to forget, these chapters are closely linked with a myriad of other domains as well. Hence, if you do not attain a minimum level of proficiency in Algebra and Trigonometry, then it might negatively affect your year-end scorecard. Easy to understand and illustrative books on algebra and trigonometry at Amazon will definitely clear the basic concepts in the minds of the children. Once they are clear on the basic concepts, then solving difficult problems will be like a cake walk for them.
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There are many references and guides of Algebra and trigonometry written by some of the most reputed scholars, who have substantial experience in this domain. One of the best books for under graduates is "Linear Algebra: A geometric Approach" by Kumaresan. This book deals with the basic concepts of linear algebra and is specially designed for beginners. Geometrical illustrations of linear algebra equations clarify the basic concepts in the minds of the students. This book can definitely make you friends with linear algebra, if you can persevere enough. Another book is "Algebra and trigonometry" by Dutta Ushri, Muktibodh A S and Mahgaonkar. This book not only explains the various concepts of algebra and trigonometry, but also contains problems that are generally asked in competitive exams. "Schaum's outline of linear algebra" is an amazing book that contains 612 solved problems of Algebra. Practising these problems can be very beneficial for you in your competitive and graduation exams. "Book of Abstract algebra", "Problems and solutions in mathematics", "Algebraic theories of numbers", etc. are some other books that can guide you on the path of success in mathematics.
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Glencoe Algebra (New York State Edition)
New York Algebra 1 is the first of three books in Glencoe's New York High School Mathematics Series. This series offers complete coverage of New York's Mathematics standards, strands, and performance indicators. As students learn to integrate a comprehensive array of tools and strategies, they become proficient in mastering concepts and skills, solving problems, and communicating mathematically.
This series of books helps your students identify and justify mathematical relationships; acquire and demonstrate mathematical reasoning ability when solving problems; use visualization and spatial reasoning to analyze characteristics and properties of geometric shapes; and succeed on the Regents Examinations78733162 New York material is in section at the front. Otherwise, identical to national edition. Multiple copies available. Bookseller Inventory # Z0078733162ZN | 677.169 | 1 |
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Mathematics for Physics
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An engagingly-written account of mathematical tools and ideas, this book provides a graduate-level introduction to the mathematics used in research in physics. The first half of the book focuses on the traditional mathematical methods of physics ñ differential and integral equations, Fourier series and the calculus of variations. The second half contains an introduction to more advanced subjects, including differential geometry, topology and complex variables. The authors' exposition avoids excess rigor whilst explaining subtle but important points often glossed over in more elementary texts. The topics are illustrated at every stage by carefully chosen examples, exercises and problems drawn from realistic physics settings. These make it useful both as a textbook in advanced courses and for self-study. Password-protected solutions to the exercises are available to instructors at
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Teaching Statistics
4.11 - 1251 ratings - Source
A recent survey showed that for every student taking a statistics course in a statistics department, more than three take such a course in a mathematics department, but many math departments have no statistician on staff. This instructor's manual was originally intended as a companion volume to math-teacher workshops that addressed the issue, but it may also be helpful to experienced teachers of statistics at the undergraduate or secondary level. It includes classic and original articles on various aspects of statistical education as well as descriptions of innovative projects. The text has no subject index. c. Book News Inc.9 l0. ll 12 13 14 l5 16 17 18 19 20. ... A 3-point scoring rubric is used to assign
points for write-ups on ten in-class activities, and a 2-point rubric is used for ten
chapter summaries. ... Egge, E., S. Foley, L. Haskins, and R. Johnson, Statistics
Lab Manual, 3rd edition, 1995, Carleton College, Mathematics and Computer
Scienceanbsp;...
Title
:
Teaching Statistics
Author
:
Thomas L. Moore (Ph. D.)
Publisher
:
Mathematical Assn of Amer - 2000-01-01
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Acces — The Teacher's Database
Overview
Acces is an electronic publishing system for teachers, sort of a combination
between a database, desktop publisher, and word processor. The program stores
test items and curricular material very efficiently, supports many different
page layouts, and produces beautiful, typeset-quality documents.
While some people like to call Acces a "test generator," it is really much
more than that. In fact, it can be a great help to teachers in their regular
instruction, and it can benefit students immensely. For example, teachers can
use the software to:
customize lessons for students with special needs.
supplement textbooks with interesting and challenging questions.
create a variety of classroom materials, such as overheads, flash cards, game cards, assignment schedules, and calendars.
prepare students for state assessments and standardized tests like the SAT.
produce daily assignments, review worksheets, class warm-ups, and other documents, that are closely aligned with a school district's or state's curriculum.
Although Acces addresses many subjects, it is especially well-suited for
mathematics, because it has built-in support for formulas, graphics, and special
symbols. Therefore, most of our modules are designed for math teachers.
However, do not be disappointed if you work in a different field; we also
have science and language arts modules. History will be covered
soon. If you are a curriculum supervisor, technology specialist, or testing
coordinator, then you will definitely want to read on about Acces' capabilities.
Making use of database modules
One of Acces' most impressive features is the size of its database.
Currently, the program offers more than 300,000 problems in over 40 modules.
This is no doubt the largest computer-based collection of math problems
available—and it continues to grow!
When we use the term database, we are really talking about two things:
a computerized storage and retrieval system, and various "add-on modules" or
item banks that are available for Acces. In this section, we go over some
general points about the database system and describe features that are common
to all modules. We invite you to see our Add-on
Modules page to determine which, if any, modules are appropriate for your
needs. But please keep in mind that you can also use Acces to write your own
problems and store them on the computer.
Here is some general information about the database:
Acces is a print-based system. This means that
you select problems by looking at a printed catalog and telling the computer
what you want. We adopted this method because it is much faster than scrolling
through problems on the screen. It also makes locating specific kinds of
problems very simple, because the catalogs are divided into sections (or
topics) with dozens of related problems on each page. The process is actually
similar to flipping through a teacher's edition of a textbook. But the software
has a tremendous advantage: it puts the equivalent of dozens of textbooks and
your entire filing cabinet on the computer. Plus, it handles all of the cutting and
pasting, so problem sets and exams are just a few keystrokes away.
All of the material in Acces' database is authored by real people, then
stored on the computer. The software does not "generate" items as do some
other testing programs. This means you get lots of interesting and subtle
variations of problems, rather than endless repeats. Put differently, there is
no trade-off between quality and quantity; Acces not only gives you a huge
number of problems, it provides an excellent balance between introductory and
advanced topics, or between basic and higher-order thinking skills. (Another
advantage of a real database system, as opposed to a "test generator," is in
the archiving of existing material. The module containing New York Regents
Exam questions or any of the math contest databases are good examples of this
archival capability.)
Acces never tries to outguess you. You select the items you want from the
database in the order you want them. Since the questions are not generated on
the fly, there are no surprises. Even the answers are shown in the printed
catalog, so you know exactly what you are getting. But don't get us wrong,
there are lots of ways to automate the selection process, if you wish. Acces
can be told to pick problems at random, scramble their order, or produce an
alternate version of a test or a quiz. The important point is that Acces
follows your instructions: it is designed to make a teacher's job easier; it
does not pretend to do the job better.
Acces' database can be used to store just about any kind of problem:
multiple-choice, free-response, fill-in-the-blank, true-false, column-match,
etc. The program can even handle some very unusual question types, such as
quantitative-comparisons found on the SAT I or grid-in answers found on many
modern assessments. The kind of material you write and store on the computer
is entirely your decision. EducAide's database modules tend to include either
multiple-choice or free-response questions, but also have excellent open-ended
questions. By the way, one of the options in the software is to hide the
answers, so that a multiple-choice question can do "double duty" as a
free-response question, or even be used with an answer grid.
All database modules come complete with clipart (pictures), tables, charts,
and any auxiliary files, such as reading passages, that are necessary to make
use of the items. You simply install a module into Acces and the items are
ready to go. As a side benefit, you can also make use of the included clipart
in any new problems that you write, or you can "extract" an existing problem
and make any changes to it that you like. This makes the material in the
database go even further.
Desktop publishing capabilities
In addition to managing database modules (or banks of questions), Acces is a
very powerful authoring and publishing tool. If you like to create your own
problem sets and exams on the computer, then Acces can serve as a complete
replacement for your word-processor or desktop publisher. In other words, you
can do all of your writing inside the program, or do a combination of writing,
selecting problems from the database, and modifying existing problems.
As mentioned earlier, Acces is especially suitable for mathematical and
scientific material, because it can create virtually every mathematical notation
that you can think of. In some ways, Acces works like an equation editor,
but it goes one step further: math is totally integrated with text, so problems
are fast to write and easy to modify later.
We have commented several times on Acces' documents. So as not to sound immodest,
we should explain that Acces is built around a very powerful typesetting system
called TeX (pronounced tech). The system is widely known in academic and
publishing circles for producing beautiful documents. The system has particular
advantages for Acces, because it is programmable. This has allowed us to create
numerous document types or "templates" and to support page layouts that are too
difficult even to consider doing with a word-processor—say, a two-column
document with varying amounts of "workspace" and a rectangular answer box next
to each problem.
Currently, Acces supports six document types. These are: Test/worksheet,
Standardized test, Overheads, Flash cards, Weekly calendar, and Monthly
calendar. There are numerous options for each document type, and you have
essentially complete control over font size, page headers, margins, spacing,
numbering, and many other fine-tuning features. You will likely find many
creative uses for each document type. For example, with the overhead and flash
card options, you can produce game cards, bookmarks, bulletin board items, and
other classroom materials. The calendar templates are just as interesting; you
can use them to create assignment schedules or to provide students with daily
warm-up exercises. To switch from one document type to another, you simply
indicate your preference to Acces.
There are two other advantages to using Acces as a production tool. First, it
provides exactly the same options for the problems you write as it does for the
problems you select from a database. Second, the process of writing or selecting
problems is almost completely independent of page layout decisions. This means
you can do things in almost any order: select some problems, see how they look
on a certain type of document, make some changes, try a different type of
document, and so on. Acces handles all the finer details of production and, as
you will quickly discover, it is extremely good at what it does.
Finally,
the reason Acces is so flexible is that it does something called dynamic formatting.
Problems are not formatted when they are written or stored on the computer, but
rather when they are made part of an actual document. This approach is
considered state-of-the-art in electronic publishing, and it has much in common
with HTML, the language of the World Wide Web. From the standpoint of database
development, there are many advantages, including more efficient problem writing,
where the focus is on content, not layout. Even if you are most interested in
using existing problems from a database, and not writing your own, you still
benefit from some amazing formatting capabilities.
What teachers like most about Acces
There is no question that Acces saves time, reduces the drudgery of certain
tasks, and makes teachers' jobs more enjoyable. The time-savings alone is
significant; Acces users tell us they are able to spend much more "quality time"
with students, helping with their lessons and evaluating their work. But that
is not actually what impresses teachers the most...
Acces is exceptionally easy to use. You can produce professional-looking
documents in just a few minutes, regardless of your level of experience. In
fact, Acces is a great way to get "non-computer" persons to start using
technology.
Acces is highly adaptable. You do not have to make any compromises in the
way you do things.
Acces' database modules are excellent sources of material. There are lots
of interesting problems from which to choose and most topics in the curriculum
are well covered.
Acces makes testing much simpler and more secure. You can easily produce
multiple versions of a test or quiz, and offer make-up tests without the
usual hassles.
Acces is really liked by students. No kidding! Many students are relieved
not to be given handwritten problem sets (or word processing documents with
missing symbols). But they are most happy to see materials targeted to their
own needs. Plus, they feel better prepared for tests when they can get
plenty of extra practice or review problems.
Pricing & Licensing
Price of new school site license (engine only):
Windows 7/Vista/XP
$595.00
Acces and its various database modules are designed for use by elementary and
secondary schools or college departments. Therefore, all of the list prices on
our Web site include a Site License.
Briefly, the Site License allows you to install Acces on any computer at your
school (or department office) or at a teacher's home. Database modules are
licensed in the same way, but there is an important rule governing the use of
their items: you may reprint the items freely and as often as you like, but you
may distribute them only to students and other teachers at your site.
We have prepared a sample site license agreement
for your review. Please note that this is only a copy of the Acces Site License
Agreement that has been shipped with orders prior to March 2003. Since the
Acces Site License Agreement is subject to change, the copy posted on this web
site is for review only. All use of the software is governed by the license that
is included in each shipment of Acces. | 677.169 | 1 |
ISBN 13: 9780007353835
New GCSE Maths - Grade C Booster Workbook: AQA Modular
Collins New GCSE Maths AQA Modular Grade C Booster Workbook provides extra support for students to achieve a Grade C in mathematics. The write-in format, with answers in a tear-out section, makes it easy to use and revise.
• Written by a trusted maths teacher and author, Collins New GCSE Maths AQA Modular Grade C Booster Workbook is perfect for students on the D / C borderline who want to make sure they get their Grade C. • Lots of questions at Grade C (with some Grade D and some Grade B) help students focus on the key topics required to get their grade. • Assess how well students are doing with Grade progression maps that clearly show how to move from a Grade D to C. • More in-depth questions assess students' understanding, and test problem-solving and functional maths skills • Perfect for revision, booster classes, retakes, intervention groups.
Book Description Collins Educational13 | 677.169 | 1 |
Derivative Notation
students expand their knowledge of derivatives, and to learn about finding second derivatives (and beyond).They will use the concepts learned in this lesson in future lessons where they will use both first and second order derivatives to generate graphs of | 677.169 | 1 |
Lial/Salzman/Hestwood's Basic Mathematics, 7e, gives students the necessary tools to succeed in their current math courses, preparing them for future math courses and the rest of their lives. The Lial team creates a pattern for success by emphasizing problem-solving skills, vocabulary comprehension, real-world applications, and strong exercise sets. In keeping with its proven track record, this revision includes an effective new design, many new exercises and applications, and increased Summary Exercises to enhance comprehension and challenge students' knowledge of the subject matter. | 677.169 | 1 |
Home
Welcome to Math.info
The
Math.info Web site contains
numerous features for increasing comprehension of math topics ranging
from elementary arithmetic through advanced
topics including calculus and differential equations. These
features include the following:
1) Informational Content (Including numerous worked examples)
1) Simply click a topic area listed on the
horizontal menu located at the top portion of each page.
2) After
clicking on a specific tab, associated topics will be visible on the
left side under 'Subjects' while applicable online calculators and/or
conversion utilities will be visible on the right hand side under
'Resources'.
3) Alternatively, clicking on one of the topic areas
contained within the left margin followed by selecting one of the
subtopic areas on the resulting page is also an option. | 677.169 | 1 |
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Making Your Algebra Class Interactive
Greg Hurst
With products like Wolfram|Alpha, Wolfram Demonstrations Project, and CDF, Wolfram Research brings algebra into the twenty-first century. This Wolfram Technology for STEM Education: Virtual Conference for Education talk showcases these great products and demonstrates their use in a classroom setting.
Channels: Virtual Events
Jon McLoone of Wolfram Technical Communication and Strategy delivers a fast-paced overview of some of the 500+ new functions and thousands of improvements in the latest version of the Wolfram Language ...
See how easy it is to use the Wolfram Language to solve real-world statistics and probability problems with quantity data, enhanced time series support, and over 150 distributions, including random matrices.
Explore functions and their derivatives, integrals and limits using the Wolfram Language. This session shows practice problems with the Wolfram Problem Generator and lesson plans and tips on using Wolfram|...
Learn about using Wolfram|Alpha for your physics class computations to compliment your existing curricula. Interesting blog posts and online resources are shared that help foster curiosity in areas such ... | 677.169 | 1 |
Designed as the bridge to cross the widening gap between mathematics and computer science, and planned as the mathematical base for computer science students, this maths text is written for upper-level college students who have had previous coursework involving proofs and proof techniques. The close tie between the theoretical and algorithmic aspects of graph theory, and graphs that lend themselves naturally as models in computer science, results in a need for efficient algorithims to solve any large scale problems. Each algorithm in the text includes explanatory statements that clarify individual steps, a worst-case complexity analysis, and algorithmic correctness proofs. As a result, the student will develop an understanding of the concept of an efficient algorithm. Author : G. Chartrand ISBN : 0071125752 Language : English No of Pages : 384 Edition : International edition Publication Date : 3/1/1993 Format/Binding : Paperback Book dimensions : 9.13x6.57x1.11 Book weight : 0.02 | 677.169 | 1 |
"With an emphasis on the history of mathematics, this book offers a well-written introduction to number theory and calculus and presents numerous applications throughout to illustrate the accessibility and practicality of the topic. It features numerous figures and diagrams and hundreds of worked examples and exercises--and includes six chapters that allow for a flexible format for a one-semester course or complete coverage for a two-semester course"--
Now in its fifth edition, A Mathematics Sampler presents mathematics as both science and art, focusing on the historical role of mathematics in our culture. It uses selected topics from modern mathematics including computers, perfect numbers, and four-dimensional geometry to exemplify the distinctive features of mathematics as an intellectual endeavor, a problem-solving tool, and a way of thinking about the rapidly changing world in which we live. A Mathematics Sampler also includes unique LINK sections throughout the book, each of which connects mathematical concepts with areas of interest throughout the humanities. The original course on which this text is based was cited as an innovative approach to liberal arts mathematics in Lynne Cheney's report, "50 HOURS: A Core Curriculum for College Students," published by the National Endowment for the Humanities." Schaum's Outline of Mathematics for Liberal Arts Majors helps students understand basic concepts and offer extra practice on such topics as logic, truth tables, axiom statements, consumer mathematics, probability and counting techniques, the real number system, and more. Each chapter offers clear concise explanations of topics and include hundreds of practice problems with step-by-step solutions.
Presents a clear bridge between mathematics and the liberal arts Mathematics for the Liberal Arts provides a comprehensible and precise introduction to modern mathematics intertwined with the history of mathematical discoveries. The book discusses mathematical ideas in the context of the unfolding story of human thought and highlights the application of mathematics in everyday life. Divided into two parts, Mathematics for the Liberal Arts first traces the history of mathematics from the ancient world to the Middle Ages, then moves on to the Renaissance and finishes with the development of modern mathematics. In the second part, the book explores major topics of calculus and number theory, including problem-solving techniques and real-world applications. This book emphasizes learning through doing, presents a practical approach, and features: A detailed explanation of why mathematical principles are true and how the mathematical processes work Numerous figures and diagrams as well as hundreds of worked examples and exercises, aiding readers to further visualize the presented concepts Various real-world practical applications of mathematics, including error-correcting codes and the space shuttle program Vignette biographies of renowned mathematicians Appendices with solutions to selected exercises and suggestions for further reading Mathematics for the Liberal Arts is an excellent introduction to the history and concepts of mathematics for undergraduate liberal arts students and readers in non-scientific fields wishing to gain a better understanding of mathematics and mathematical problem-solving skills.
Practical, scientific, philosophical, and artistic problems have caused men to investigate mathematics. But there is one other motive which is as strong as any of these — the search for beauty. Mathematics is an art, and as such affords the pleasures which all the arts afford." In this erudite, entertaining college-level text, Morris Kline, Professor Emeritus of Mathematics at New York University, provides the liberal arts student with a detailed treatment of mathematics in a cultural and historical context. The book can also act as a self-study vehicle for advanced high school students and laymen. Professor Kline begins with an overview, tracing the development of mathematics to the ancient Greeks, and following its evolution through the Middle Ages and the Renaissance to the present day. Subsequent chapters focus on specific subject areas, such as "Logic and Mathematics," "Number: The Fundamental Concept," "Parametric Equations and Curvilinear Motion," "The Differential Calculus," and "The Theory of Probability." Each of these sections offers a step-by-step explanation of concepts and then tests the student's understanding with exercises and problems. At the same time, these concepts are linked to pure and applied science, engineering, philosophy, the social sciences or even the arts. In one section, Professor Kline discusses non-Euclidean geometry, ranking it with evolution as one of the "two concepts which have most profoundly revolutionized our intellectual development since the nineteenth century." His lucid treatment of this difficult subject starts in the 1800s with the pioneering work of Gauss, Lobachevsky, Bolyai and Riemann, and moves forward to the theory of relativity, explaining the mathematical, scientific and philosophical aspects of this pivotal breakthrough. Mathematics for the Nonmathematician exemplifies Morris Kline's rare ability to simplify complex subjects for the nonspecialist. | 677.169 | 1 |
Basic Math and Pre-Algebra Workbook For Dummies
Basic Math and Pre-Algebra Workbook For Dummies by Mark Zegarelli
Book Description
When you have the right math teacher, learning math can be painless and even fun! Let Basic Math and Pre-Algebra Workbook For Dummies teach you how to overcome your fear of math and approach the subject correctly and directly. A lot of the topics that probably inspired fear before will seem simple when you realize that you can solve math problems, from basic addition to algebraic equations. Lots of students feel they got lost somewhere between learning to count to ten and their first day in an algebra class, but help is here! Begin with basic topics like interpreting patterns, navigating the number line, rounding numbers, and estimating answers. You will learn and review the basics of addition, subtraction, multiplication, and division. Do remainders make you nervous? You'll find an easy and painless way to understand long division. Discover how to apply the commutative, associative, and distributive properties, and finally understand basic geometry and algebra.
Find out how to:* Properly use negative numbers, units, inequalities, exponents, square roots, and absolute value* Round numbers and estimate answers* Solve problems with fractions, decimals, and percentages* Navigate basic geometry* Complete algebraic expressions and equations* Understand statistics and sets* Uncover the mystery of FOILing* Answer sample questions and check your answers Complete with lists of ten alternative numeral and number systems, ten curious types of numbers, and ten geometric solids to cut and fold, Basic Math and Pre-Algebra Workbook For Dummies will demystify math and help you start solving problems in no time!
Books By Author Mark Zegarelli
Basic Math & Pre-Algebra For Dummies, 2nd Edition (9781119293637) was previously published as Basic Math & Pre-Algebra For Dummies, 2nd Edition (9781118791981). While this version features a new Dummies cover and design, the content is the same as the prior release and should not be considered a new or updated product.
The fun and friendly guide to really understanding math U Can: Basic Math & Pre-Algebra For Dummies is the fun, friendly guide to making sense of math. It walks you through the "how" and "why" to help you master the crucial operations that underpin every math class you'll ever take.
Helsp you learn and practice your Basic Math & Pre-Algebra skills. This two-book bundle brings two popular For Dummies math guides together to offer readers essential Basic Math & Pre-Algebra instruction combined with real-world practice problems to put their knowledge to the test and help reinforce key Algebra I | 677.169 | 1 |
Most Recent Documents for Amarillo College
12.8 Lab - Length of Curves
Student: _
Calculus III
1.) Find the lenth of the entire spiral < /+) , where ) 0 and + !.
1.) _
2.) A cycloid is the path traced by a point on a rolling circle (think of a light on the rim of a moving bicycle
wheel). The cyclo
Lab 13.8 - Maximum/Minimum Problems
Student: _
Calculus III
Based on the level curves that are visible on the following graphs, identify the approximate locations of the
local maxima, local minima, and saddle points. Express the locations as ordered pairs
Lab 14.3 - Double Integrals in Polar Coords.
Student: _
Calculus III
1.) Carefully sketch the region in Quadrant I outside the circle < # and inside the lemniscate
< #-9=#). Then find the exact value of the area using a double integral in polar coordinate
12.3 Lab - Dot Products
Student: _
Calculus III
1.) Find the angle between the vectors v $ & # and
u * & " to the nearest degree.
1.) _
2.) Determine the work done if a suitcase is pulled 50 ft along a flat sidewalk
with a constant force of 30 lbs at an a
12.6 Lab - Calculus of VVFs
Student: _
Calculus III
Let the position of an object moving in three dimensional space be given by r> B> C> D > , for
> !. The velocity of the object is v> rw > Bw > Cw > D w > . The speed of the object is the Realvalued funct
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Student: _
Calculus III
1.) Find the exact area of the region bounded by the curves C ( (=38B and C ( (=38B over the
interval ! 1 using a double integral. Carefully sketch the area in the space below.
1.) _
12.2 Lab - Vectors in Three Dimensions
Student: _
Calculus III
1.) A fisherman wants to know if his fly rod will fit in a rectangular 2 ft x 3ft x 4 ft packing box. What is the
longest rod that fits in this box?
1.) _
2.) A small plane is flying horizonta
Lab 14.5 - Triples in Cylindrical, Spherical Coords
Student: _
Calculus III
1.) Use a triple integral in rectangular coordinates to find the volume of the region bounded by the parabolic
cylinder C B# and the planes D $ C and D !.
1.) _
2.) Use a triple i
Lab 13.9 - LaGrange Multipliers
Student: _
Calculus III
1.) The following figure shows the level curves of 0 and the constraint curve 1B C !. Estimate the
maximum and minimum values of 0 subject to the constraint. At each point where an extreme value occu
Lab 13.6 - Directional Derivatives, Gradient
Student: _
Calculus III
1.) You and a friend are standing on the surface of hilly terrain described by 0 B C -9=B C. Suppose
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Lab 13.1B - Quadric Surfaces
Student: _
Calculus III
The standard form of a quadric surface is given in each problem. Identify the surface as an ellipsoid, elliptic
paraboloid, hyperbolic paraboloid, elliptic cone, hyperboloid of one sheet, or hyperboloid
Lab 14.4 - Triple Integrals
Student: _
Calculus III
1.) Find the volume of the wedge of the cylinder B# #&C# #& created by the planes D % B and
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1.) _
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Student: _
Calculus III
Consider the paraboloid D B# $C# and the plane D B C %, which intersects the paraboloid in a curve
G at # " (. See the figure below. Find the equation of the tangent line to G at the point # " ( by
followi
Part 1
Comprehensive Problem 1:
The following is a comprehensive problem which encompasses all of the elements learned in previous
chapters. You can refer to the objectives for each chapter covered as a review of the concepts.
Kelly Pitney began her consu | 677.169 | 1 |
Teach Yourself: Trigonometry - A Complete Introduction
4.11 - 1251 ratings - Source Trigonometry - A Complete Introduction
Author
:
Hugh Neill
Publisher
:
- 2014-06-27
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Mathematical Models in Biology
4.11 - 1251 ratings - Source
Mathematical Models in Biology is an introductory book for readers interested in biological applications of mathematics and modeling in biology. A favorite in the mathematical biology community, it shows how relatively simple mathematics can be applied to a variety of models to draw interesting conclusions. Connections are made between diverse biological examples linked by common mathematical themes. A variety of discrete and continuous ordinary and partial differential equation models are explored. Although great advances have taken place in many of the topics covered, the simple lessons contained in this book are still important and informative. Audience: the book does not assume too much background knowledge--essentially some calculus and high-school algebra. It was originally written with third- and fourth-year undergraduate mathematical-biology majors in mind; however, it was picked up by beginning graduate students as well as researchers in math (and some in biology) who wanted to learn about this field.Although great advances have taken place in many of the topics covered, the simple lessons contained in this book are still important and informative.
Title
:
Mathematical Models in Biology
Author
:
Leah Edelstein-Keshet
Publisher
:
SIAM - 1988
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Trades Math Calculator 2.01a
Some mathematical problems require a lot of calculations in order to find out exact results, which would easily be obtained if you would use a software application. There are a lot of mathematical computers, but not all offer the same functionality and features. Trades Math Calculator is a reliable all-in-one calculator that allows you to compute certain complex mathematical functions and equations. Besides mathematical equations, you can solve certain technical problems and calculations. Reliable equation solver
The program can help you solve mathematical equations with ease, by entering the input values required, then selecting the proper instruments and measure units. The application will calculate the missing variables and display them accordingly. Furthermore, certain trigonometric problems can offer you visual solutions, not just resolved values.
In addition to mathematics, Trades Math Calculator can be used to calculate certain technical problems. You can easily compute the optimal speed for drilling a certain material, such as wood or metal, or the thickness of several threads and wires. Keyseat values can also be computed, depending on component width and height. This feature helps you compute the specifications for certain component joints and binding points. Fast unit converter
Trades Math Calculator helps you easily convert certain measurement units. This way, you can compute the correct value of a measurement unit through various sizes and unit fields, such as density, energy, flow, force, mass, power , acceleration, pressure, area, volume or time.
Additionally, you can define custom unit conversions, in order to meet your requirements. This allows you to combine the predefined unit conversions with the custom ones, in order to find out the solutions for complex mathematical and technical problems. A seamlessly powerful scientific calculator
Trades Math Calculator offers you the possibility to find out the results for various mathematical equations and functions, whilst offering optimal solutions for several technical problems. | 677.169 | 1 |
About
Overview
As in previous editions, the focus in ESSENTIAL MATHEMATICS with APPLICATIONS,Additional Product Information
Features and Benefits
The Aufmann Interactive Method (AIM) allows students to try each skill as it is presented. Each set of matched-pair examples is organized around an objective and includes a worked example and a You Try It example for students. Complete worked-out solutions are available in the appendix.
Each chapter begins with a set of learning objectives woven throughout the text in Exercises, Chapter Tests, and Cumulative Reviews. These elements connect the text to the print and multimedia ancillaries, resulting in a seamless, easy-to-navigate learning system.
To help students master and retain core concepts, exercises that follow the presentation of a new operation require students to connect verbal phrases and mathematical processes.
Problem Solving sections at the end of each chapter introduce students to various problem-solving strategies. Students are encouraged to write their own strategies and draw diagrams in order to find solutions; several open-ended problems are also included.
Applying the Concepts exercise provide further exploration of topics, requiring analysis or incorporating concepts discussed earlier in the text.
Projects and Group activities are featured at the end of each chapter. They can be used for extra-credit or cooperative learning activities.
ESSENTIAL MATHEMATICS with APPLICATIONS addresses students' broad range of study styles by offering a wide variety of tools for review:
Table of Contents
Note: Each chapter begins with a Prep Test and concludes with a Chapter Summary, a Chapter Review, and a Chapter Test. Chapters 2-12 include Cumulative Review Exercises. AIM for Success. 1. WHOLE NUMBERS. Introduction to Whole Numbers. Addition of Whole Numbers. Subtraction of Whole Numbers. Multiplication of Whole Numbers. Division of Whole Numbers. Exponential Notation and the Order of Operations Agreement. Prime Numbers and Factoring. Focus on Problem Solving: Questions to Ask. Projects and Group Activities: Order of Operations; Patterns in Mathematics; Search the World Wide Web. 2. FRACTIONS. The Least Common Multiple and Greatest Common Factor. Introduction to Fractions. Writing Equivalent Fractions. Addition of Fractions and Mixed Numbers. Subtraction of Fractions and Mixed Numbers. Multiplication of Fractions and Mixed Numbers. Division of Fractions and Mixed Numbers. Order, Exponents, and the Order of Operations Agreement. Focus on Problem Solving: Common Knowledge. Projects and Group Activities: Music; Construction; Fractions of Diagrams. 3. DECIMALS. Introduction to Decimals. Addition of Decimals. Subtraction of Decimals. Multiplication of Decimals. Division of Decimals. Comparing and Converting Fractions and Decimals. Focus on Problem Solving: Relevant Information. Projects and Group Activities: Fractions as Terminating or Repeating Decimals. 4. RATIO AND PROPORTION. Ratio. Rates. Proportions. Focus on Problem Solving: Looking for a Pattern. Projects and Group Activities: The Golden Ratio; Drawing the Floor Plans for a Building; The U.S. House of Representatives. 5. PERCENTS. Introduction to Percents. Percent Equations: Part I. Percent Equations: Part II. Percent Equations: Part III. Percent Problems: Proportion Method. Focus on Problem Solving: Using a Calculator as a Problem-Solving Tool; Using Estimation as a Problem-Solving Tool. Projects and Group Activities: Health; Consumer Price Index. 6. APPLICATIONS FOR BUSINESS AND CONSUMERS. Applications to Purchasing. Percent Increase and Percent Decrease. Interest. Real Estate Expenses. Car Expenses. Wages. Bank Statements. Focus on Problem Solving: Counterexamples Projects and Group Activities: Buying a Car. Final Exam. Appendix. Solutions to You-Try-Its. Answers to Selected Exercises. Glossary. Index.
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This online solutions manual allows instructors to create customizable solutions that they can print out to distribute or post as needed. This is a convenient and expedient way to deliver solutions to specific homework sets. Visit Cengage Learning's Mathematics CourseMate brings course concepts to life with interactive learning, study, and exam preparation tools that support the printed textbook. Access an integrated eBook, learning tools including flashcards, quizzes, and more, all designed specifically to work with ESSENTIAL MATHEMATICS WITH APPLICATIONS, 8th Edition.As in previous editions, the focus in ESSENTIAL MATHEMATICS with APPLICATIONS | 677.169 | 1 |
Shortly after the invention of differential and integral calculus, the calculus of variations was developed. The new calculus looks for functions that minimize or maximize some quantity, such as the brachistochrone problem, which was solved by Johann Bernoulli, Leibniz, Newton, Jacob Bernoulli and l'Hopital and is sometimes considered as the starting point of the calculus of variations. In Woodhouse's book, first published in 1810, he has interwoven the historical progress with the scientific development of the subject. The reader will have the opportunity to see how calculus, during its first one hundred years, developed by seemingly tiny increments to become the highly polished subject that we know today. Here, Woodhouse's interweaving of history and science gives his special point of view on the mathematics. As he states in his preface: ""Indeed the authors who write near the beginnings of science are, in general, the most instructive; they take the reader more along with them, show him the real difficulties and, which is the main point, teach him the subject, the way they themselves learned it. | 677.169 | 1 |
DO
m< OU 166593 >m
> a: 7) 73
^ CD -< co
AN INTRODUCTION TO
THE STUDY OF
NUMERICAL TRIGONOMETRY
Cambridge University Press
Fetter Lane, London
Nc e w York
Bombay, Calcutta, Madras
Toronto
Macmillan
Tokyo
Maruzen Company, Ltd
All rights reserved
AN INTRODUCTION TO
THE STUDY OF
NUMERICAL TRIGONOMETRY
by
E. A. PRICE, B.A.
Assistant Master at the Royal Naval College, Dartmouth
Formerly Assistant Master at Winchester College
Cambridge :
at the University Press
First Edition 11)12
licpnnted iDltt, 1021, l'.)24, 1930
PK1NTLJ) IN OllKAT lUUTAIN'
PREFACE
IN reprinting this little work, which formerly appeared
under the title of Examples in Numerical Trigonometry,
some slight alterations and additions have been admitted. .
The book is an attempt to introduce the elements of
Trigonometry on heuristic lines, and is intended for those
pupils who do not intend to pursue the study of Mathematics
very far. It has for some time been an accepted principle
that for such pupils a wider range of subjects treated in a
less formal fashion should take the place of that drill in the
manipulation of symbols which is essential for the specialist
in Mathematics. The author has endeavoured to carry out
this principle, and has to a large extent omitted formal
proofs and deductions the theory of the machine which it
is proposed to handle.
Even the slight acquaintance with this branch of Mathe-
matics which is aimed at here, will to a large extent miss its
true opportunity unless some notion of "functionality" or the
growth of variable quantities, is acquired, and with this end
in view a graphical treatment of the Ratios is introduced,
the same idea being applied in dealing with the ratios of the
Obtuse Angle.
Among the new features it will be found that headings
have been added to the various sections ; and the preliminary
drawing exercises have been reduced and questions substi-
tuted which are intended to call into play the "intuitive".
VI PREFACE
faculty the ability to deduce the general rule from few
concrete examples by the exercise of the imagination, with,
of course, subsequent verification. Additions have also been
made to Chapter X.
A chapter on Logarithms has been introduced before the
general treatment of triangles, but it should be noted that
their use is by no means imperative.
Thanks are due to several friends who have kindly
assisted in the working out of the answers, and especially to
the Director of Naval Education and the Headmaster for
permission to make use of questions set in the Examination
papers at Osborne.
E. A. P.
January 1918
CONTENTS
CHAP. PAOE
L Measurement of Angles. Angles of Elevation and
Depression. Subtended Angles. Bearings . . 1
II. The Tangent and its use in Problems .... 6
III. The Sine and its use in Problems .... 13
IV. The Cosine and its use in Problems . . , . 17
V. The Secant, Cosecant, and Cotangent .... 20
VI. Miscellaneous Problems involving right-angled triangles 24
VII. Areas 35
VIII. Logarithms 38
IX Solution of triangles which are not right-angled . . 48
X. Ratios of Obtuse Angles 53
XL The Sine Formula 55
XII. The Cosine Formula 58
XIII Problems 61
APPENDICES.
Definitions of the Trigonometrical Ratios and relations
between them 72
Proof of the Sine Formula 74
Proof of the Cosine Formula ..... 76
Examples of orderly arrangement of a Problem , . 76
ANSWERS , 79
CHAPTER I
MEASUREMENT OF ANGLES.
If two straight lines cut one another in such a way that all
the angles between them are equal, then each of these angles is
called a right anyle.
A right angle is divided into 90 equal degrees. (90)
A degree is divided into 60 equal minutes. (60')
A minute is divided into 60 equal seconds. (60")
[A second is such a small angle that it is used only in very accurate
calculations, e.g. in Astronomy. In this book only Degrees and Minutes
n*e used to measure angles.]
1. How many degrees are there in 2 right angles, J right
angle, ^ right angle, T ^ right angle, 0'3 right angle?
2. How many degrees arid minutes in ^ right angle, y 1 ^ right
angle, f right angle, right angle (to nearest minute) ?
3. Find the sum of each of the following pairs of angles :
(1) 22 10' (2) 29 25' (3) 38 51' (4) 79 57'
31 15' 47 35' 19 43' 59 49'
4. If two angles together make up a right angle they are
said to be complementary, and each is called the complement of
the other..
Calculate the complements of each of the angles in question 3.
5. The three angles of a triangle are together equal to two
right angles. If each pair in question 3 represents two angles of
a triangle, what is the third angle in each case 1
1
MEASUREMENT OF ANGLES
[CH.
6. If ABC is a right-angled triangle with C a right angle
and CD drawn perpendicular to AB, calculate the number of
degrees in every acute angle of the figure, and insert the results
in a small sketch when A contains (1) 30, (2) 67, (3) 43 21'.
7. Find the number of degrees in each angle of an isosceles
triangle if each of the base angles is (1) twice, (2) four times,
(3) half the vertical angle. Insert answers in a sketch.
Note. In the two annexed figures the angle ACB is said to be subtended
at C by the line (or arc) AB, and the angle APB is the angle subtended by
AB at P.
Fig. 1.
Fig. a.
8. How many degrees are there in the angle at the centre
of a regular 5-sided figure subtended by one of the sides 1
How many degrees are there in each of the interior angles,
and in each of the exterior angles of the 5-sided figure ?
9. Repeat question 8 for (1) an 8-sided, (2) a 9-sided,
(3) a 7-sided regular figure.
Note. In recording the bearing of one point from another the angle is
always measured from a meridian (i.e. a North and South line). If a ship
lies N. 31 E. of a lighthouse, a man on the lighthouse will find it by
looking first due North and then turning through an angle of 31 to the
East.
10. If A is N. 31 E. of B, what is the bearing of B from A 1
11. If P is N. 39 E. of Q, and R is S. 51 E. of Q, what
angle does PR subtend at Q)
I] MEASUREMENT OF ANGLES 3
12. If X is N. 21 W. of Y, and Z is N. 43 E. of X, and if
XY subtends an angle of 30 at Z, what is the bearing of Z
from Y 1
13. Q is N. 60 E. of R, P is S. 10 E. of a If RQ subtends
an anglo of 35 at P, what is the bearing of P from R?
Note. From a boat B a man is observed at the top of a cliff AC. The
angle ABC is said to be the angle of Elevation of the man as seen from the
boat, and the angle BAD is said to be the angle of Depression of the boat as
seen from the top of the cliff.
Tig. 8.
Remember that angles of Elevation and Depression are always measured
from the Horizontal.
14. There is a church-tower surmounted by a spire. If the
angle of elevation of the top of the spire as seen from a certain
point on the ground is 53, and the spire subtends an angle of
29 at the same point, what is the angle of elevation of the top
of the tower from the same point?
15. From a boat at sea the angle of elevation of the top of
a certain cliff is 37. What is the angle of depression of the
boat as seen from the top of the cliff?
12
4> MEASUREMENT OF ANGLES [CH. I
16. From the masthead of a ship the angles of depression of
two buoys one behind the other are found to be 19 and 24.
What angle does the line joining them subtend at the masthead?
17. What angle does one-sixth of the circumference of a
circle subtend (1) at the centre, (2) at any point on the circum-
ference]
18. What fraction of the circumference is subtended by
angles of 120, 60, 20, at points on the circumference?
19. If in Fig. 1 above C = 90% L. CPB = 73 and L. BAG = 25,
how many degrees are there in each of the angles PBC, ABC, ABP?
20. If in the same figure C = 90, L. CPB = x* and L BAG = y,
how many degrees are there in u ABPl
21. If the edges of the square pyramid in Fig. 8 are each
10 cm. long, how many degrees are there in each of the angles
ACB, ACD, BCD, CAK?
Is L AKN larger or smaller than L ACN ?
If K were moved along DC towards C, would the angle AKN
increase or decrease ?
22. From a ship sailing parallel to a straight coastline the
angle subtended by the line joining two iixed points on shore is
observed. Does this angle vary in size? (Draw a sketch.) If
so, when is it greatest ?
23. If the angle observed in the last question remained the
same for any length of time, would the ship be still moving
parallel to the shore? If not, describe carefully her course
during that time.
CHAPTER II
THE TANGENT AND ITS USE IN PROBLEMS
Instruments, squared paper, and Table of Natural Tangents
required.
Note. Express all results in decimals to three significant figures.
1. Drawing accurately with
instruments construct an angle
Q of 59*. Take a point P in four
different positions somewhere on
one arm of the angle, and draw
PN perpendicular to the other arm.
Measure PN and NQ in inches and
decimals of an inch, and calculate
PN .
the value of the fraction in deci-
NQ
mals for all four positions. Arrange
in a table thus:
Fig. 4.
1.
2.
3.
4.
PN
What do you observe with regard to these values of ]
NGl
2. Repeat question 1, making Q equal to 39.
* The angle Q in the diagram is purposely drawn incorrectly to avoid
being copied.
6
THE TANGENT AND
[cn.
3. If another angle (20 say) were drawn in the same way,
would you obtain a similar result ?
What do you observe with regard to the value of the fraction
PN
for each angle Q? Express clearly in words and remember
the result.
PN
4. The fraction (or ratio) - is called the Tangent of the
NQ
angle at Q, and from question 1 is obtained : tan 59 = 1-66.
Write down the result of question 2 in the same way.
5. By drawing accurate figures, find the values of tan 47,
tan 29-5, tan 34 24' as exactly as possible.
6. Copy the annexed figure ac-
curately, making QN = 5cm., PNQ=90
and the angles at Q each 10, the
whole angle at Q being 60.
Measure the length of PN when
the angle Q is 10, 20, 30, 40, 50,
60, and calculate the tangent of each
of these angles.
7 Turn to the Table of Natural
Tangents and search for the results
of the last question. Make a com-
parative table showing the results
obtained by drawing and from the
printed Tables thus:
Fig. 5.
Angle
Tang
by Drawing
;ent
from Tables
10
0-1763
20
0-3640
30
etc.
etc.
II] ITS USE IN PROBLEMS 7
8. Make a similar table showing the results of questions 1,
2, and 5 above.
Note that the values given in the Tables are correct to four significant
figures only. Tan 20 = 0-3639702 correct to seven figures, but this is not
absolutely exact. Four significant figures give results sufficiently accurate
for ordinary calculations.
9. Plot on squared paper the values obtained in question 7,
i.e. draw the graph of y tan x for values of x between and
60. What are the tangents of 0, 45, 90 ?
Write down the following sentence and insert the word
increases or decreases, whichever is appropriate :
"As the angle x increases from to 90, tana? "
10. Draw A ABC having C = 90, AC = 3 cm., BC - 4 cm.
Measure AB in cm. and the angles A and B in degrees and
minutes as accurately as possible.
Write down (in decimals to three significant figures) the values
of the tangents of the angles A and B as found from the figure.
11. If drawn very accurately, the angle A in the last question
should he 53 8', and tan 53 8' = 1-3335. Search for this in
the Table of Natural Tangents and write down the values of
tan 53 6' and tan 53 12'.
12. The angle B in questioi 10 should have been 36 52', and
tan 36 52' = 0*7499. Search this out and write down the values
of tan 36 48' and tan 36 54'.
13. Write down each of the following and find its value
from the Tables: tan 56, tan 56 24', tan 56 30', tan 56 27',
tan 28, tan 28 12', tan 28 18', tan 28 15', tan 16 25', tan 73 22',
tan 3 51'.
14. Is the tangent of an angle half the tangent of twice the
angle? Compare tan 40 and tan 80 and draw a freehand sketch
to illustrate your answer. (See figure of question 6.)
15. From the Tables find the angles whoso tangents are
0-4663, 3-0777, 0-3153, 1-4388, 0-4023, 3-6888, 0-1974, 2-3164,
1-1980, 0-6633.
8
THE TANGENT AND
[CH.
16. If an angle X is such that tan X - 0-7 (or ^) it may be
constructed by drawing a right angle included between two lines,
one of which is seven-tenths of the other. Draw such a triangle
having sides 7 cm. and 10 cm., and measure the angle whose
tangent is 0'7. Verify the result hy reference to the Tables.
17. Find in the same way the angles whose tangents are
(1) 0-8, (2) -2-0, (3) 3*5. Verify from the Tables.
EASY PROBLEMS
18. From a distance of 100 feet measured along level ground
from the foot of a vertical tower the angle of elevation (see p. 3)
of the top is found to be 29. Find by drawing a small freehand
sketch and using the Table of Natural Tangents what the height
of the tower must be.
19. The angle of elevation of the top of a flagstaff from a
point 40 feet from its base is 50. Calculate its height.
20. If a ladder is placed with its base 5 feet from the
bottom of a wall 12 feet high and just reaches the top, what
angles does the ladder make with the ground and the wall, and
how long must it be 1
21. A ship sails 5 miles due West and then 7 miles due
North. When it is in this position calculate the distance and
bearing of the point from which she started.
22. If the angle of elevation of a monument from a point
200 feet from its base and on the same level is 34, what is the
height of the monument ?
23. A vertical pole 12 feet high casts a shadow 20 feet long.
What must the angle of elevation of the sun be?
shadow
Fig. 6.
Il] ITS USE IN PROBLEMS 9
RIGHT-ANGLED TRIANGLES
Note. In a triangle ABC the sides opposite the angles A, B, C are usually
called a, ft, c.
All units of length in the following examples are centimetres. All lengths
should be given to four significant figures, all angles in degrees and minutes.
In the triangle ABC having C = 90
24. Find b if B = 27 and a = 10 units of length.
25. Find b if B = 27 12' and a - 10.
26. Find b if B = 27 16' and a= 10.
27. Find b if B = 27 16' and o = 2-5.
28. Find a if A = 49 28' and b = 2-3.
29. Find a if A = 84 24' and b = 12-5.
30. Find A if a = 4 and b = 5.
31. Kind A and B if a = 8 and b = 10.
32. Find A and B if a = 2-4 and 6=1-5
33. Find A and B if a = 13'6 and b = 19'9.
SOLUTION OF A TRIANGLE
Note. To solve a triangle when three of its sides and angles are given
means to calculate the remaining sides and angles.
Solve the triangle ABC if C = 90 and
34. A = 42 17' and 6 = 25 -4.
35. B = 29 11' and a =1-62.
36 A =13 53' and a = 10-4.
37. B = 72 41' and b -4-07.
38. a = 12-4 and b = 9-3.
39. a = 124-5 and 6 = 150.
PROBLEMS
40. From a point 27 feet from the wall of a building a
window sill is observed, and its angle of elevation found to bo
63 26'. Find its height above the ground to the nearest inch.
10 THE TANGENT AND [OH,
41. To find the range of a distant object C, two men A and B
take up positions BO that AC is at right angles to AB. If AB is
20 yards, and the angle ABC is found to be 85*4, find AC.
42. A thin stratum of gold-bearing rock crops out on the
level surface and is known to slope down at an angle of 7 13',
If a shaft is driven vertically down at a distance of 1J miles
from the outcrop, at what depth in feet will it meet the stratum?
43. In order to measure the width of a river a distance AB
is marked out straight along the bank; a point C is observed on
the bank exactly opposite to A and the angle CBA is measured.
If AB50 yards and CBA = 54 25', find the width of the
river.
44. The rails approaching a tunnel under a certain estuary
slope down at an angle of 5 37' and are at water level at a
distance of 1700 feet from the water's edge. At what depth (in
feet) below the surface of the water are they when exactly under
the water's edge ?
45. In a triangle ABC having O = 90, AC = 5 inches and
BC = 12 inches, BD is drawn at right angles to A B to meet AC
produced in D. Find the length of CD.
46. A chord 6 cm. long is 2 cm. from the centre ot a circle.
What angle does it subtend (see p. 2) at the centre ?
47. If a chord 10 cm. long subtends 140 at the centre of a
circle, how far must it be from the centre?
48. From a point on level ground 75 feet from the base of a
flagstaff, the angles of elevation of the top and of the point at
which a yardarm is fixed are found to be 49 37' and 38 29'.
Find the height of the top and of the yardarm above ground.
49. On a map a piece of straight road between the 100-foot
and 400-foot contours seems to be J mile long. What is its
slope and how many yards long is it really?
50. The poles for a certain Wireless Telegraphy Station con-
sist of three parts fixed vertically one above the other, each part
II] ITS USB IN PROBLEMS 11
being 50 feet long. The whole is held erect by wire stays fixed
to the top of each part and to points in the ground 75 feet away
from the base. Find the angles which each of the wire stays
makes with the ground and with the poles. Find also the
angles between the three wire stays fixed to the same point on
the ground. Draw a sketch and insert the answers,
51. From a ship sailing up a river the angle of elevation
of a point on a bridge 176 feet above the water and straight in
front is found to be 29 14'.
If the ship is moving at a rate of 5 miles an hour, in how many
seconds will it be exactly under the bridge?
52. The figure is a sketch for the following problem :
200 Art
Pig. 7.
From a boat the angle of elevation of an object on the top of
a cliff 200 feet high is found to be 29. The boat sails directly
towards the object, and after a certain interval the angle of
elevation is found to be 50. How many feet did the boat sail!
[Calculate L ACD, L BCD, AD, BD ; then AB can be found.]
53. Repeat question 52, making L A = 27 13', CD = 274 feet
and L CBD = 63 29'.
*
54. The angle of elevation of the top of a tower is observed
to be 25, and from a point 100 feet nearer to the foot of the
tower it is 43. Find the height of the tower.
12
THE TANGENT AND ITS USE IN PROBLEMS [CH. II
55. Repeat question 52, with angles 18 47' and 51' 29' and
distance 124 feet.
56. The figure represents a square pyramid each of whose
edges is 10 cm. long. It is required to find the angle (ABN)
A
Fig. 8.
between one slant-edge and the base, and also the angle (AKN)
between one face and the base.
Note that ANB, ANC, ANK, AND are right angles. Calculate BN and
AN by Pythagoras' Theorem.
57. Calculate the height of a regular tetrahedron (triangular
pyramid) whose edges are each 10 cm. long. Find also the angles
between slant-edge and base, and between two faces.
Note NK = JBK.
A
Fig. 9.
CHAPTER III
THE SINE AND ITS USE IN PROBLEMS
Instruments, squared paper , and Table of Natural Sines
required.
1. Construct an angle Q of 59 carefully with instruments.
Take a point P in four different positions somewhere on one arm
of the angle and draw PN perpendicular to the other arm.
Measure PN and PQ for each position of P and calculate the
PN
value of the fraction in each case. Tabulate the values of
PN
PN, PQ, and . [Drawings for Chap. IT may be used.]
2. Ilepeat question 1, making Q = 39.
PN
3. What do you observe with regard to the fraction -- in
each case? Express clearly in words and remember the result.
PN
4. The fraction (or ratio) is
PQ
called the Sine of the angle GL What
is the value of sin 59 t
Write down the result of ques-
tion 2 in the same way.
5. By drawing accurate figures,
find the values of sin 47, sin 2 9 -5,
sin 34 24' as exactly as possible.
6. Copy the annexed figure ac-
curately, making the radius of the
circle 10 cm. long and putting in
radii at intervals of 10.
14
THE SINE AND
[CH.
If P be the end of a radius and PN be drawn perpendicular
to NQ, then by measuring PN in all positions the sines of the
angles 10, 20, 30, 40, 50, 60, 70, 80 can be calculated.
Find these values and make a table. What is the value
of sin and of sin 90 ?
7. Plot these values on squared paper (i.e. draw the graph
of y = sin x for values of x between and 90). From the graph
find the values of sin 47 and sin 29*5, and compare with the
results of question 5. Does sin x increase or decrease as x
increases from to 90 ? Answer quite clearly.
8. Compare the results obtained in question 6 with the values
given in the Table of Natural Sines. Arrange in a table thus :
Angle
Si
by Drawing
ne
from Tables
o-oooo
10
0-1736
20
0-3420
etc.
etc.
9. In Ex. 10, Chap, n a triangle ABC having = 90,
AC = 3 cm. and BC = 4 cm. was drawn, and it was found that
AB = 5 cm. and A = 53 about. Draw a neat freehand sketch of
this triangle. If drawn very accurately the angle A should be
53 8', and sin 53 8' = 0*8000 (or |). Search for this in the Table
of Natural Sines and write down the value of sin 53 6', and of
sin 53 12'.
10. The angle B in the triangle above should be 36 52', and
sin 36 52' = 0-5999 (or nearly).
Search for this in the Table of Natural Sines, and write down
the value of sin 36 48' and of sin 36 54'.
Ill] ITS USE IN PROBLEMS 15
11. Write down each of the following and find its value
from the Tables :
sin 35 12', sin 35 18', sin 35 15', sin 35 13',
sin 72 55', sin 43 9', sin 10 53', sin 81 49'.
12. Is the sine of an angle half the sine of twice the angle ?
Compare sin 40 and sin 80 and draw a freehand sketch to
illustrate your answer. (See figure of question 6.)
13. From the Tables find the angles whose sines are
0-4226, 0-9455, 0-8141, 0-2890,
0-2898, 0-9795, 0-9701, 0-7202.
14. By drawing a right-angled triangle with one side 5 cni.
long and the hypotenuse 10 cm. long and measuring the angles,
find the angle whose sine is *5 (= T 6 ^). Verify from the Tables.
15. Find in the same way the angles whose sines are (1) 0-4,
(2) 0-75, (3) 0-43. Verify from the Tables.
RIGHT-ANGLED TRIANGLES
16. If in A ABC having C = 90 (draw freehand sketch)
(1) B = 27, c= 10 units, find b.
(2) B = 2712', c = 10, find b.
(3) B = 271G', c = 10, find b.
(4) B = 27 16', c = 2-5, find b.
(5) A = 43 37', c = 12-2, find a.
(6) A = ll22', c=l 1-22, find a.
17. In the triangle ABC, C = 90, B = 63 28' and o - 75 cm.
Find b, a (by Pythagoras) and A. Verify by drawing to scale.
18. Solve the triangle ABC if C = 90 and
(1) A = 72 14', c = 21. (Do not use Pythag.)
(2) B = 2523', c-10-4.
(3) B = 4213', c = 105-2.
(4) A=ll19', c=19-32.
16 T11E SINK AND ITS USE IN PROBLEMS [CH. Ill
PROBLEMS
19. A straight road rises 48 feet in 400 yards measured
along the road. Find its inclination to the horizontal.
20. What is the angular slope of a railway lino whose
gradient is 1 in 100? (The 100 is measured along the rails.)
21. What angular slopes are represented by the boards at
the side of a railway line which
are shown in the figure 1
22. A ladder 25 feet long
is placed against a wall and
makes an an^le of 73 22' with
the ground. To what height Fig. 12.
does it reach? Answer to nearest inch.
23. What is the height of an isosceles triangle whose equal
sides are each 2-5 inches long and whose base angles are 17 35'
each?
24. A lighthouse 13 miles away is observed from a ship to
be 19 off the course of the ship. If the ship continues on her
course, how far from the lighthouse will she be when she is
nearest to it?
25. A rod 15 inches long hangs from a nail in the wall. If
it is pulled away so that it makes an angle of 25 33' with the
wall, what is the distance of the lower end from the wall ]
26. What angle will the rod in the last question make with
the vertical when its lower end is 8 inches from the wall ?
27. A trap-door is 3 feet 3 ins. wide from hinge to opposite
edge. If it is raised through four-fifths of a right angle, how
high above the ground will the outer edge be ?
28. A framework of four rods each 1 foot long, hinged at the
ends, and with elastic diagonals, is laid upon the table. Find
the lengths of the diagonals when the angle between two adjacent
sides is (1) 45 22', (2) 145 22'.
CHAPTER IV
THE COSINE AND ITS USE IN PROBLEMS
Definition. If from any point P in one arm of an angle Q a
perpendicular PN be drawn to the
Q,N
other arm, the ratio is called
the cosine of the angle Q.
Kemember this definition.
1. Draw any acute angle ac-
curately with instruments and show
by careful measurements that the
cosine of this angle remains the same for any four different
positions of the point P. Arrange results in a table.
2. Find by accurate drawing the value of cos 59 and of
cos 39 (cos 59 means the cosine of an angle of 59).
3. Find by accurate drawing the angles whose cosines are
(1) 0-5, (2) 075, (3) 0-36.
4. Copy the annexed figure ac-
^urately making the radius 10 cm.
long and putting in radii at intervals
of 10.
By drawing PN in all positions
and measuring NQ, find the cosines
of the angles 0, 10, 20, 30, ... 90.
Tabulate the resulting values.
Does the cosine increase or de-
crease as the angle increases from
to 90 7
Fig. 14.
P. T.
13 THE COSINE AND [OH.
5. Compare the results obtained in question 4 with the
values given in the Table of Natural Cosines.
To the table of results in that question add a column showing
the values given in the printed Table.
6. From the Table of Natural Cosines find and write dowr
the values of cos 56, cos 56 24', cos 56 30'.
Now cos 56 27' lies between the last two values. Find it,
and write it down.
7. Find from the Tables the values of
cos 28 15', cos 16 25', cos 73 21',
cos 81 51', cos 53 8', cos 12 28'.
8. In a triangle ABC having C = 90 and
(1) A = 27, c=10, find*.
(2) A = 43 22', c=10, find b.
(3) B = 19 25', c= 8-8, find a.
(4) B = 85 17', 0=12-2, find a.
PROBLEMS
9. A wire 100 feet long is stretched from the top of a flagstaff
to a point in the ground. If it makes an angle of C3 29' with
the ground, how far is this point from the foot of the flagstaff?
10. -A ship has sailed North- West for 21 miles. How far
North of her starting-point is she ?
11. A straight road running uphill at an angle of 9 35' with
the horizontal is 1 mile long. How many yards long will it
appear to be on the map?
12. A ladder 25 feet long leans against a wall making an
angle of 59 43' with the ground. How far is its foot from the
foot of the wall ?
13. What is the length of the base of an isosceles triangle
whose equal sides are 2 '5 inches long and whose base-angles are
each 1734'f
IV]
ITS USE IN PROBLEMS
19
14. Gloucester is 79 miles N. 29 W. of Southampton. How
far N. of Southampton is Gloucester?
15. A rod 3 feet long hangs from the ceiling. How far is
the lower end from the ceiling when the rod makes an angle of
69 26' with the vertical?
16. What is the length of the shortest side of a set-square
with angles of 90, 60, 30, if the longest side is 15 cm. long?
17. A flagstaff whose top when vertical is 70 feet above the
ground is found to lean to one side making an angle of 82 20'
with the ground.
What is the length of its shadow when the sun is vertically
overhead ?
18. A roof is made of pieces of corrugated iron leaning
against a wall and supported at their lower onds by poles. If
the length of each piece of iron is 6 feet 6 inches, and it makes
an angle of 32 with the horizontal, what width will be kept dry
when rain falls vertically ?
19. If the latitude of New York be taken as 40 N. and the
radius of the Earth as 4000 miles,
calculate the distance of New York
from the axis NS of the Earth. Through
what distance does New York revolve
each day? (See Fig. 15.)
20. How many miles a minute is
London moving through space owing
to the spin of the Earth on its axis ?
Take London's latitude as 51 30'.
21. What is the distance round
the world in latitude 60? On Mer-
cator's Projection of the World the
width of the Atlantic Ocean in latitude 60* N. appears to be the
same as its width on the Equator.
What is the true relation between these distances ?
22
CHAPTER Y
THE SECANT, COSECANT, AND COTANGENT
Note. These ratios are shortened to sec A, cosec A, cot A.
1. Find the value of sec 36 from the Table of Natural
Secants. Find also the value of cos 36 u Multiply these
together and give the answer correct to four significant figures.
2. Repeat twice with any other angles.
What do you observe with regard to the secant and cosine
of any one of these angles 1
Note. } x f = 1, and J is called the reciprocal of } .
3. Find in a similar way some relation between the cosecant
and some other ratio known already.
4. Repeat for the cotangent and tabulate the relations found
in the first four questions. The results should be remembered.
5. If ABC is a triangle
with C = 90, express cosec A,
sec A and cos A each as a ratio
of two sides.
6. Find from the Tables
and write down the values of
sec A, cosec A and cot A when
A is
56, 56 24', 56 30', 56 27';
28, 28 12', 28 18', 28 14';
16 25', 73 21'.
CH. V] THE SECANT, COSECANT, AND COTANGENT 21
7. Write down the following and complete it by inserting
the words " increases" and "decreases" :
" As the angle A increases from to 90, sec A..., cosecA...,
cot A...." Illustrate with a freehand drawing.
PROBLEMS
Note. Use secants in questions 8 to 11.
8. A wire is stretched from the top of a vertical pole to
a point in the ground 20 feet from the base. How long must
the wire be if it makes with the ground an angle of (1) 28,
(2) 53 19', (3) 61 58'?
9. The straight line joining the tops of two vertical poles
31 feet apart is found to make an angle of 19 23' with the
horizontal. What is the distance between the tops?
10. From the top of a cliff 470 feet high the angle of
depression of a boat is found to be 23 14'. How far is the
boat from the top of the cliff? [Find the complement.]
11. Find the radius of the circle circumscribing a regular
five-sided figure whose sides are each 12 cm. long. [Calculate
the angle at each corner.]
Note. Use cotangents in questions 12 to 16.
12. A ladder makes an angle of 41 26' with a vertical wall.
If its lower end is 10 feet from the base of the wall, to what
height does the top reach?
13. One side of a rectangle is 15 inches long and the
diagonal makes an angle of 17 23' with the other side. Find
the length of the other side.
14. How far from a tower 110 feet high must a man be
standing if the angle of elevation of the top is 35 4'!
15. A is 25 miles due N. of B, arid C is due E. of B. If AB
makes an angle of 13 58' with AC, Hud the angle C and BC.
22 TUB SECANT, COSECANT, [CH.
16. P is a point on the circumference of a circle whose
diameter is XY. If the straight line from P to X is 53 inches
long and subtends an angle of 47 13' at Y, find the length of
the chord YP.
Note. Use cosecants in questions 17 to 20.
17. How long must a ladder be which just reaches a window
15 feet above the ground, if it makes an angle of 53 19' with
the ground 1
18. What is the diameter of a circle in which a chord
5 inches long subtends an angle of 25 28' at the circumference ?
[Draw the diameter through one end of the chord.]
19. If the slope of a road is 4 6', what is its gradient?
(Of. Ex. 20, Chap. III.)
20. If the line joining the tops of two vertical posts makes
an angle of 17 13' with the horizontal, and if one post is 6 feet
longer than the other, what is the distance between the tops 1
EXERCISES ON THE Six RATIOS
21. Draw a freehand sketch of a triangle ABC having C = 90,
o = 4 inches, b = 3 inches. From this figure find the six trigono-
metrical ratios of A and of B.
Find the size of the angle A by looking up sin A in the Table
of Natural Sines, and verify the remaining five ratios by searching
for this angle in the appropriate Tables. Repeat for the angle B
and arrange all results in columns.
22. Repeat question 21 for a triangle having C = 90,
c M 40 cm. and 6=9 cm.
23. Observing that the angles A and B in questions 21 and
22 are complementary, what do you note with regard to the
ratios of complementary angles)
24. If tan A = T 6 r , find (by drawing a freehand sketch) the
other five ratios of A and write them down. Verify from the
Tables as before.
25. If sin as= ^j., find the other ratios and verify.
V] AND COTANGENT 23
26. Find the sine, cosine and tangent of 35 from the
Tables. Calculate the value of ^- , and of sin 2 35 + cos 3 35.
cos 35
[sin a 35 means the square of the sine of 35.]
What do you observe with regard to these values!
27. Repeat question 26 with any angle chosen at random.
Do the same relations hold good ?
28. Prove that these relations are true for any acute angle
by drawing a sketch of a right-angled triangle and making use
of Pythagoras' Theorem.
29. Making use of your tables if necessary, draw a rapid
freehand sketch of the graph of sin A when A has values between
and 90.
30. As A increases from 80 and becomes more and more
nearly equal to 90, to what value does PN (see figure of Ex. 6,
Chap. III.) more and more nearly approach ?
What then is the ultimate value of sin 90 ?
Write down the values of cos 90, tan 90% sin 0% cos 0,
tan 0.
31. Draw rapid freehand sketches of the graphs of cos A
and tan A for values of A between and 90.
32. What can be asserted with regard to the greatest and
least values of the sine of an acute angle ?
What can be asserted with regard to the cosine and tangent
of an acute angle 1
33. By means of the Tables find a relation between tan* 35
and sec* 35.
34. Choose an angle at random and show that the relation
is still true.
35. Prove as before (question 28) that this relation is true
for any acute angle.
36. Find and prove a relation between cosec'A and cot 9 A
when. A is an acute angle.
CHAPTER VI
MISCELLANEOUS PROBLEMS INVOLVING RIGHT-ANGLED
TRIANGLES
Note. At this stage it is most essential that the pupil should be able to
recognize immediately the ratios of the angles in a right-angled triangle and
readily to express the length of one line in terms of another. Constant
practice with figures similar to those in questions 1 and 6 below is
recommended for pupils who hesitate.
Revise Ex. 6, Chap. I.
1. In the figure C = 90 and p is perpendicular to c cutting
it into two parts x and y.
C
Fig. 17.
Which other angle in the figure is equal to B, and why?
Show that the whole triangle and the two triangles in to 'which p
divides it are all equiangular (and therefore the same shape).
Now - sin B. Write down two other fractions equal to sin B.
a
CH. VI]
MISCELLANEOUS PROBLEMS
2. Write down three different expressions for cos B, tan B,
sec B, cosec B, cot B, in the figure of question 1.
3. Write down three expressions for each trigonometrical
ratio of the angle A in the figure of question 1.
4. If in the figure of question 1, a = 20 cm. and 2? = 12 cm.,
find B, A, 6, x, y and c, and verify the results of questions 1 3.
5. If in the figure of question 1,6 = 6-5 cm. and p = 6 cm., find
A, B, a, x y y and c.
T i ,. AN
6. In the figure = cos a,
.*. AN = x cos a.
90'
Pig. 18.
Write down in a similar way the values of p, y, AC, CN, in
terms of x and a.
7. In the figure of question 1 what lengths are represented
by a cos B, b sin A, c cos B, a cos A, p tan B, p cot B, p sec A,
p cosec A, b sin B ?
8. In the figure of question 6 what lengths are represented
by x sin a, x tan a, p cot a, y cos a, p tan a ? '
26 MISCELLANEOUS PROBLEMS [CH.
Without using Pythagoras' Theorem solve the triangles in
which C = 90 and
9. B = 2419', a =10-4.
10. A = 29 53', a =20-1.
11. A = 44 31', c = 2-4.
12. a = 24-7, 6=14-6.
13. c = 140, a = 77.
14. A = 72 13', 6=15-5.
15. B = 3t37' 1 c=1200.
16. 6 = 5-7, c=ll-4.
17. c=!G-75, a = 3-35.
18. c=3a.
Note. In the following questions the answers may be obtained in a
variety of ways, but by using the appropriate ratio the answer may be
arrived at by one operation.
19. From the top of a cliff 250 feet high the angle of
depression of a boat is found to be 17 43'. How far is the boat
from the top of the cliff?
20. A ship sails 17 miles due N. and then 25 miles due E.
In what direction does the starting-point lie ?
21. The angle of elevation of the top of a tree is found to be
29 11'. If the observer is 120 feet from its base, how high is
the tree ?
22. The string of a kite (supposed to be straight) is 225 feet
long and is inclined to the ground at an angle of 52 5'. Find
the height of the kito above the ground.
23. What is the slope of a railway line whose gradient is
Iin60?
24. Find the length of one of the equal sides ot an isosceles
triangle whose base is 12-4 cm. long and whose base angles are
each 41 28'.
Vl] INVOLVING RIGHT-ANGLED TRIANGLES 27
25. Find the radius of a circle in which a chord 3-4 cm. long
subtends an angle of 104 at the centre.
26. From the top of a flagstaff 108 feet high a wire 125 feefc
long is stretched to the ground. What is the angle between the
wire and the flagstaff 1
27. In a triangle PQR, PQ = 1 2'2 cm. and Q = 39 '37'. What
is the length of the straight line PN drawn from P perpendicular
toQR?
28. Find also the length of QN in the last question.
29. If in the question above, QR = 17'8 cm., what is the
length of NR? Hence find the angle at R.
30. An aeroplane is observed from a point A and its angle
of elevation is found to be 67 32'. If at the same instant it is
exactly above B (on the same level as A), and AB = 75 yards, how
high is it ifbove B 1
31. The angle A of a parallelogram ABCO is 31* 3' and the
perpendicular distance between AD and BC is 3*52 inches. Find
the length of the side AB.
32. PN is a tangent to a circle whose centre is O, N being
on the circumference. If the radius is 5-7 cm., and OP = 9*5 cm.,
what angle does PN subtend at the centre?
33. If a door 4 ft. 2 inches wide is opened through an angle
of 112 Q 42', how far (in a straight line) is the bottom corner from
its position when the door was closed ]
34. What is the radius of a circle which touches the five
sides of a regular pentagon if each side is 5 cm. long ? [Find the
angle at the centre,]
35. Find the radius of the circle which passes through the
five angular points of a regular pentagon whose sides are each
5 cm. long.
36. Repeat questions 34 and 35 for a regular hexagon
(6 sides).
28 MISCELLANEOUS PROBLEMS [CH.
37. Repeat for a regular decagon (10 sides).
38. A ship is sailing N. 11 15' E. at 9 miles an hour. At
what rate is she moving due N. ?
39. From a point 27 feet from the base of a pole and on the
same level, the angle of elevation of the top is 63 26'. Calculate
its height to the nearest inch. Find also the distance of the
point of observation from the top.
40. Find the angle of elevation of the top of the pole in
the last question from a point 20 feet from the base.
41. A rod 5 inches long is hinged at a point on the ground,
and from the other end hangs a plumb-line. What angle does
the rod make with the plumb-line when the top of the rod is
(a) 4 inches above the ground, (b) 2 inches above the ground 1
42. A pole is 126 feet high. A man 6 feet high stands at
a distance of 100 feet from the foot of the pole and observes the
angle of elevation of the top of the pole. What does he find the
angle of elevation to be I
43. A diameter AB of a circle is 5 cm. long. P is a point on
the circumference 2-34 cm. from A. Find the angle subtended by
PA at B.
44. The shadow of a vertical stick 4 feet high is found to be
5*72 feet long. Find the altitude of the sun.
45. A lies 5 miles due North of B, and C lies 6 miles due East
of B. Calculate the distance and bearing of C from A.
46. At a certain Marconi Station a vertical pole 96 feet
high is made in three equal sections and supported by wire stays
stretched from the top of each section and fixed to points in the
ground. If wires from the top of each section are fixed to a
point on the ground 25 feet from the base of the pole, Calculate
the angles which they make with the ground.
47. Find the height of an isosceles triangle whose base is
10 cm. long and whose- vertical angle is 35 28'.
VI] INVOLVING RIGHT-ANGLED TRIANGLES U9
48. In the figure, CP represents the connecting- rod and OP
the crank of a steam engine. P revolves round O, and C moves
backwards and forwards along OC.
If CP is 40 inches long and OP is 12 inches long, find the
angle at C when OP is at right angles (1) to CP and (2) to OC.
49. ABC is a triangle right-angled at C. If BC = 17*3 feet
and if ABC = 30 27', calculate the length of AB to the nearest
inch.
50. From the top of a cliff 300 feet high the angle of
depression of a ship is 13 46' and the angle of depression of a
rock between the ship and the shore is 32 10'. Calculate the
distance of the ship from the rock (to the nearest foot).
51. On one occasion while the Atlantic cable was under
repair, it was found that when the grappling iron seized the cable
on the ocean bed, 4890 fathoms of hawser had been paid out, and
that when drawn taut the hawser made an angle of 72 with the
horizontal. What was the depth of the water at this spot in
fathoms ?
52. How many degrees are there in an angle in a semicircle 1
In a semicircle of diameter 10 inches, a chord 4 inches long is
put, one, end of which coincides with one end of the diameter.
Calculate (a) the angle between the chord and the diameter, and
(b) the length of the chord joining the other ends of the diameter
and chord.
30 MISCELLANEOUS PROBLEMS [CH.
53. A diameter AB of a circle bisects a chord CD at right
angles at O. The angle OCA = 50. Find the lengths of AO
and BO, given that the chord CD is 20 inches long. What are
lengths of AC and BC ]
54. A lighthouse bears 5 miles due N. of a cruiser steaming
N. 18 -6 E. By how many miles does the cruiser clear the
lighthouse ?
55. A cruiser and a torpedo boat set out at the same time
from point C. The cruiser steams at the rate of 20 knots in
a direction N. 15 24' E., and the torpedo boat at the rate of
30 knots in a direction N. 74 36' W. Calculate the distance
and bearing of the torpedo boat from the cruiser at the end of
30 minutes.
56. A, B, and C are buoys. The beciring of B from A is
N. 47 R, and the bearing of C from A is S. 43 E. B is known
to be 5 miles from A, whilst C is due S. of B. Find the distances
of C from B and A.
57. A flagstaff on the far bank of a river is seen by a man
immediately opposite to subtend an angle of 57, whilst on
retiring 100 ft. the elevation is only 35. Find the breadth
of the river to the nearest foot.
58. The Rock of Gibraltar is 1396' in height; the angle of
elevation of its summit from a ship is 22: how far will the ship
have to move directly towards it before its angle of elevation
is 31?
59. A straight railroad rises 2-4 feet in every 100 yards of
rail : find its inclination to the horizontal.
60. A straight line from Southampton pier to Ryde pier
subtends 90 at Portsmouth pier; Southampton pier lies
15*2 miles N. 46 W. from Ryde pier, and Portsmouth pier
bears N. 28 E. from Ryde pier ; find the distance between
Portsmouth and Rvde Diers.
VI] INVOLVING RIGHT-ANGLED TRIANGLES 81
61. At Southampton on Midsummer's Day the altitude of
the sun is 62 45'. If a mast is 100 feet high, find the length of
its shadow.
Find also the length of its shadow on December 21st, when
the altitude of the sun is 15 45'.
62. From the top of a vertical cliff 200' high the angle of
depression of a boat anchored out in the bay is 4G 28 ; . Find
how far the boat is from the foot of the cliff.
63. A kite is held by a string 100 yards long, and its angle
of elevation is found to be 50 16'. What is the vertical distance
of the kite from the ground? (Neglect the sag of the string.)
64. A ship X is 10 miles S. 52 W. of a harbour at the
moment that another ship Y is leaving the harbour. If Y
steams S. 38 E. at 8 knots and the ships meet in 2 hours,
find X's course.
65. From the top of a tower A, 60 feet high, the angle of
depression of the foot of another tower B is 23 16'; whilst from
the foot of A the elevation of the top of B is 45 17'. Find how
far A is from B, and also the height of B to the nearest foot.
66. A ship on leaving port steams 10 miles N. 50 11' E.,
and then 5 miles N. 39 49' W. Find the distance she must go,
and the course she must steer, to reach port again.
67. In a circle of radius 27 inches find the length of a chord
which subtends an angle of 163 10' at the centre. Find also
how far this chord is distant from the centre of the circle.
68. From a ship sailing on a course due N. a lighthouse,
13 miles away, is observed to bear N. 19 34' E. When the
ship is nearest to the lighthouse, how far from it will she bel
What will be the bearing of the lighthouse from the ship
when they are 7 miles apart t
32 MISCELLANEOUS PROBLEMS [CH.
69. (The diagonals of a rhombus bisect each other at right
angles and also bisect the angles through which they pass.)
Make use of these properties to find the angle at B of the
rhombus ABCD, the length of each side of which is 10 inches,
the length of the diagonal AC being 5 '234 inches.
70. Observations were taken from a point 500 feet from the
foot of a tower with a flagstaff on the top, and the angles of
elevation of the top of the tower and of the top of the flagstaff
were found to be 28 15' and 30 47' respectively. Find (1) the
height of the tower and flagstaff; (2) the height of the tower
alone ; and hence (3) the height of the flagstaff alone.
71. A ship sails 4 miles due West and then 3 miles due
North ; find its distance from the starting point and its bearing
from that point.
72. A bears 10 miles due N. of C, and B bears N. 25 E.
of C, and B bears S. 65 E. of A. Find the distances of B from
A and C.
73. A lighthouse 10 miles away is seen to be 15 off a ship's
course. At what distance will the ship pass the lighthouse if
she holds her course?
74. At 10 a.m. a ship is observed from a lighthouse to bear
9 miles N. 57 32' E. and is known to be sailing S. 32 28' E.
At 11 a,m. her bearing is S. 6045'E. Find (1) rate of ship's
sailing ; (2) distance of lighthouse at second observation.
75. In a triangle ABC, the angle B 36 52', AB = 4-5 inches,
and BC = 4-68 inches.
By drawing AD perpendicular to BC, and calculating the
length of AD and BD, find the length of DC and AC, and the size
of the angle C.
Make a table of the sides and angles of the triangle ABC.
VI] INVOLVING RIGHT-ANGLED TRIANGLES 33
76. A theodolite was used to determine the height of a flag-
staff and the following data were obtained :
Height of theodolite telescope above ground 3 ft. 4 in.
Distance of theodolite from foot of flagstaff 120 ft
Angle of elevation of the top of the flagstaff 27*5.
Calculate the height of the flagstaff.
77. A and B are two points on opposite sides of a tower.
The distance AB is 140 feet. The angles of elevation of the top
of the tower from A and B are respectively 62 and 31. Cal-
culate the height of the tower.
78. O and P are points on a straight stretch of shore 1 mile
apart, and O bears N. 74 W. of P. From a ship at sea O bears
N. 15 W. and from the same ship P bears N. 75 E. Calculate
the distance of the ship from O and also its distance from the
nearest point of the shore.
79. The ropes of a swing are 25 feet long and the height of
the seat above the ground at the highest and lowest points are
14 feet and 3 feet respectively.
What is the angle through which the swing moves from side
to side 1
80. A flagstaff 90 feet high subtends an angle of 45 at a
point A on the ground due South of it. Find the angle of eleva-
tion of the top of the flagstaff at a point B, 120 feet due East
of A.
81. The Tay Bridge extends North and South a distance of
1*764 miles. A man walks due East from the North end of the
bridge until the bridge subtends an angle of 33 17' at his eye.
How far is he now in a straight line from the South end of the
bridge ?
82. A flagstaff consists of two poles one fixed above the
other. From a point 64 feet from the base the angles of elevation
of the tops of these poles are found to be 40 29' and 61 51'.
Find the length of each pole.
34 MISCELLANEOUS PROBLEMS [CH. VI
83. From the top of a cliff 320 feet above sea level the
euigles of depression of two boats in a line are found to be 11 21'
and 34* 19'. Find their distance apart.
84. From a ship sailing due S. at 12 miles an hour the
directions of two objects on shore are observed to be S. 29 14' W.
and S. 43 7' W. After sailing for 25 minutes they are observed
to be in a line due W. of the ship. Wfcat is the distance between
them and how far was the ship from each of them at the first
observation ?
85. (See figure of Ex. 56, Chap. II.) If A is the vertex of a
square pyramid whose base is BCD, and if BC = 10 cm. and
AB= 12 cm., find (1) the angle between AB and BC, (2) the angle
between AB and BD, (3) the inclination AKN of a side-face.
86. Find the height of a square pyramid each of whose edges
is 5 inches long. Find also the inclination of (1) a side-edge,
(2) a side-face to the base.
87. If the edges of a rectangular block are 2, 3 and 6 inches
respectively, find the angle between a diagonal of the block and
the diagonal of each face.
88. The angle of elevation of a tower 250 feet high is
observed from A to be 20. How far is A from its foot? If A is
due E. of the tower and due N. of a point B from which the
tower bears N. 40* W., how far is B from the foot? Find the
elevation of the top from B.
89. Repeat question 88 making height of tower a? feet and
18 the angle of elevation from A.
90. P is 1000 yards due N. of Q and on the same level. An
airship is observed from P and Q simultaneously. From P its
elevation is 23 and it bears due W. From Q it bears N. 42 W.
Find its height above the ground.
CHAPTER VII
AREAS
1. What is the area of a rectangle whose adjacent sides are
(1) 5 cm. and 10 cm., (2) 131*4 yards and 101'2 yards?
2. Draw a sketch of each of the rectangles above and draw
a diagonal.
By considering these figures find the area of a right-angled
triangle whose shorter sides are (1) 5 cm. and 10 cm., (2) 131*4
yards and 101*2 yards.
3. What is the area of each of the right-angled triangles
ANB, ANC in the figure? What is then the area of the whole
triangle ABC?
A
4cm
N
Fig. 20.
4. How is the area of a triangle found when the base BC
and the height AN are known 1 Express the answer quite clearly
in words.
86 AREAS [CH.
5. If two sides of a parallelogram are 7 cm. and 5 cm. long
respectively and one of its angles is 40, what are the perpen-
dicular distances between the opposite sides ? The parallelogram
is twice the triangle ABC. Calculate its area, (See Fig. 21.)
Fig. 21.
6. How is the area of a parallelogram found when the sides
and the perpendicular distances between them are known 1
7. How is the area of a parallelogram found when two sides
(as, y) and the included angle (A) are known ?
8. How is the area of a triangle found when two sides and
the included angle are given ?
9. Find the area of each of the following triangles :
(1) a = 5 cm., 6 = 4 cm., C = 30.
(2) 6 = 10 cm., c = 3 cm., A = 36 52'.
(3) 6-6-49 cm., c = 5*73 cm., A -65 29'.
(4) a = 21-43cm., c = 13-91cm., B = 5219'.
10. In a circle of radius 10 cm., a chord AB subtends an
angle of 35 at the centre O. Calculate the area of the triangle
OAB.
11. What is the area of a regular hexagon (six
inscribed in a circle of radius 10 cm. ?
(Find angle subtended by one side at the centre.)
12. Find the area of regular figures of 5, 7, 8, 9 sides
inscribed in a circle of radius 10 cm.
VII] AREAS 37
13. Find the area of regular figures of 5, 6, 9 sides described
with their sides touching a circle of radius 10 cm.
14. The area of a rhombus (parallelogram with equal sides)
is 648 square yards and one of its angles is 150*. Find the
length of one side.
15. The area of a rhombus is 14'58 sq. cm. and one diagonal
is twice as long as the other. Find its angles and sides.
16. The base of an isosceles triangle is 4 cm. long and the
area is 20 sq. cm. Find all the angles of the triangle.
17. Find the area of each face and the area of the whole
surface of each of the solids given in Exs. 56, 57, Chap. IL
18. A regular hexagon (6 sides) is inscribed in a circle oi
radius 7 cm. What is the area of the circle ? If two adjacent
corners A and B be joined to the centre O, what is the area of th<
sector AOB 1 Find also the area of the triangle AOB and hence
the area of the segment cut off by AB.
19. Find the area of the segment cut off by one side of a
regular pentagon (5 sides) inscribed in a circle of radius 10 cm.
20. If two tangents OA, OB be drawn to a circle of radius
4 cm., and if L AOB = 30, what is the area of the figure bounded
by the tangents and the larger arc AB
CHAPTER VIII
LOGARITHMS
L 10 3 means the product of three tens or 10 x lOx 10.
What does 10 4 mean? How many tens must be multiplied
together to be equal to 10 3 x 10M
Express 10 8 x 10 4 as a power of ten.
What is done to the 3 and 4 to obtain the new index 9
Express 10 a x 10 7 and 10 8 x 10 9 each as a power of ten
2. To assign a meaning to 10' 5 (or
[" The product of half a ten " has no intelligible meaning.]
What was done with the given indices to obtain the new
index in each of the sums in question 1 ]
Use the same rule to express 10^ x 10^ as a power of ten.
Now 4x4 = 16, and 4 is called the square root of 16.
What is then the meaning arid value of 10* (or 10 08 )?
Remember this result.
3. Express 10* x 10* x 10* as a power of ten. What is the
meaning of 10M (Compare with question 2.)
What is the meaning and value of 10* (or 10 " 25 )!
4. Express 10* x 10* as a power of ten and find its value.
SH. VIII] LOGAEITHMS 39
5. Make a table showing the values of 10 ' 28 , 10 10 075 , 10 1<0 ,
and use it to draw a graph on squared paper showing the powers
of ten which are equivalent to numbers between 1 and 10.
From this graph find the numbers which correspond to 10'*,
10' 4 , 10 08 , 10 ' 88 . Also make a table showing the powers of ten
which correspond to the numbers 2, 3, 4, 5, 6, 7, 8, 9; 10.
The index of the power of ten which is equal to 3 is called
the logarithm of 3 (to base 10). That is, log 3 =0-4771, for
3 = 10o-477i. and log 10000 = 4, for 10000 = 10 4 . This should be
remembered.
6. Turn to the printed Table of Logarithms and search for
the logarithms of the whole numbers from 2 to 10.
[The position of the decimal point is not shown in the table.]
Add a column to the table constructed in question 5, and
compare the values obtained from the graph with the values
given in the printed Tables.
7. The arrangement of the Table of Logarithms is very
similar to that of Tables of Natural Tangents, etc., already
familiar.
Now 3-5 = 10' c441 , 3-56 = 10 ' 8514 , 3-563 = 10' M18 . Search for
these in the printed Tables. Then write down each of the
following and express it as a power of ten by means of the
printed Tables :
3-7, 3-75, 3-758, 7-64, 7-649, 9-8, 9-805.
8. Find from the Tables the numbers which are equal to
100-3010 j 00'8808 JQO'4398 JQO f 4409 JQO'4404
100789C 1007903 1007901 J 00-8819 100'81.
(These may be found in the Tables of Logarithms or in those o:
Antilogarithnis.)
40 LOGARITHMS [OH.
9. From the Tables 3-563 = 10 08518 and 1 -212 = 10 ' 08 ".
Therefore 3-563 x 1-212 10 ' 8518 x 10 ' 0888 .
Now 10* x 10 8 = 10 8 by adding the indices ; thus by the same
process
3-563 x 1-212 = 10 ' 6518 x lO ' 0886 ^ 10' 63M .
Write down this line of work and find from the Tables the
number whose logarithm is 0*6353 ; this is the product of
3-563 x 1-212.
Verify by multiplying these two numbers in the ordinary way,
but remember that values given in the Tables are not quite exact ;
they are correct only to four significant figures.
10. Find the value of each of the following by means of
logarithms and verify the answer in the usual way :
(1) 2x1-427. (2) 3x2-379. (3) 2-5x3-624.
(4) 4x1-767. (5) 3*5x2-147.
11. Find by using logarithms the value of each of the
following and verify by ordinary multiplication :
(1) 4-651 x 1-843. (2) 3*794 x 2-156.
(3) 1-732x1-732. (4) (2-54) 2 .
(5) 7-342x1-112. (6) (3-152)*.
10
12. It is clear that y^ 8 = 10 a by subtracting the indices.
T .1 3 ' 563 1AO-4683
In the same way = 10 04683 .
Write down this line of work and find from the Tables the
number whose logarithm is 0*4683. This number is the value
of ,777, . Verify by long division.
*
VIIl] LOGARITHMS 41
13. Find by logarithms the value of each of the following
and verify by division :
1X 4-794 /ov 6-315 /ox 9-637
W (2) ' (3) "2^5- '
MN 3-295 7-694 5-407
< 4) T3T- < 5) ?795- < 6 > R36*
LOGARITHMS OF NUMBERS GREATER THAN TEN
14. Hitherto only numbers between 1 and 10 have been dealt
with. Now consider a number greater than 10, for instance 356*3,
356-3 = 100 x 3-563 = 10 2 x 10 M18 = 10 aw
by adding indices as before.
Write down 35-63, 3563, 35630, 356300, and express each as
a power of ten in a similar way.
15. In question 14 what do you observe with regard to the
part of each logarithm after the decimal point ?
(This part is sometimes called the mantissa, which in Latin means " a
small part added." The whole number before the decimal point is called the
characteristic.)
Explain in words why it is that no decimal points are shown
in the Table of Logarithms.
16. Write down each of the following and express it as a
power of ten. (It is advisable to express each as in the first
example of question 14.)
463-4, 79-18, 16-29, 721*4, 72-14,
3114, 76300, 204600, 73-51, 213'5.
17. Find the numbers which are equal to 10 21J51S , 10 1W8B ,
79 , 10 14300 .
LOGARITHMS
[CH.
18. Find the value of each of the following by logarithms :
(1)
2-572x14-63.
(2)
31-79 x 43-72.
(3)
727-6x1-571.
(4)
6297-x 23-67.
(5)
97-64x94-67.
<)
212-5x409-5.
(7)
21-35x37-67.
(*)
3-244 x 315-4.
(9)
61-12x43150.
(10)
73050 x 4350.
(11)
14-63
(12)
43-72
2-572 '
31-79*
(13)
727-6
lETi'
(14)
6297
23-67*
(15)
57-93
41-52*
(16)
7296
'47 -
(17)
312400
76-53 '
(18)
49-35
17-21*
(19)
13-69
(20)
200-5
"OHf"
13-61 '
LOGARITHMS OF NUMBERS LESS THAN UNITY
19. All the numbers treated above are greater than 1. In
order to find the logarithm of a number less than 1 (say 0*03563),
it is necessary to assign a meaning to expressions such as 10~ 8
(we know that 10 3 means "the product of three tens" but "the
product of minus three tens " is not intelligible).
Write down the following in a column and insert their values :
10 J = 100,000
10 4 =
10* =
10* =
Note that the indices on the left-hand side decrease by one
each time. How is each of the values on the right-hand side
obtained from the one above it? Continue the column downwards,
VlllJ LOGARITHMS 4&
decreasing the indices as before and find .values of 10, 10" 1 , 10~ 8 ,
10-, 10- 4 , 10~ 5 . From this it will be found that 10~ 8 = ^J^-T. or
~jr- 3 , and this has an intelligible meaning. Remember this result.
20. Now to find the logarithm of 0*03563 we have
0-03563 = ^j^ = JTJQ * 3*563 = 10- 2 x lO^^lO- 8 * 18 by adding
indices as before, and this is written 10* M18 to show that the
minus sign applies to the 2 only. This arrangement is adopted
in order to simplify the Tables, as will be seen hereafter.
By writing each of the following in the same way as in
question 20 above, express it as a power of ten : 0*01212, 0*1212,
0-3563, 0*0003563.
21. Write down the following in a column and express each
as a power of ten : 3563, 356*3, 35-63, 3*563, 0-3563, 0-03563,
0-003563, 0*0003563. By adopting the system of writing the
minus sign over the characteristic (e.g. 2*5518), it is found that
the same Table of Logarithms may be used for numbers greater
than or less than 1.
Discover and express in words a rule connecting the
characteristic of the logarithm with the position of the decimal
point of the number it represents.
22. Express each of the following as a power of ten :
0*0345, 0-7691, 0*00463, 0*0001105,
0*9971, 0-9065, 0-0435, 0-5432.
Note. To find the value of 0-02 x 0-007 we have
0'02 x 0-007 = 10 5>301 x 10 2-3010
3-8451
=0-00014. 4-1461
The addition is done in the usual way, but it must be remembered that
2 and 3 are simply - 2 and - 3.
44 LOGARITHMS [OH.
23. Find the value of each or wie following products by
logarithms and verify each one :
(1) 0'5x40. (2) 0-004x50. (3) 0-025x160.
(4) 0-5x0-5. (5) (0-04) 2 . (6) 0'04 x 0-05.
(7) 0-012 x 0-2. (8) 0-0008 x 40. (9) 0-0011 x 0-02.
(10) 0-25 x 0-0003.
24. Find the value of the following by logarithms :
(1) 0-02572x0-1463. (2) 0-02367x0-6297.
(3) 0-7276x0-01571. (4) 0-4372x0-003179.
(5) 946-7 x 0-09764. (6) 3-244 x 0-3154.
Note. To find the value of 0-2 H- 0-007 we have
0-2 IQI'3010
0-0007 10^ 51
The subtraction is best effected by dealing with the 3 first n
The operation 1-3010
8-8451
1-4559
is then the same as 2-3010
0-8451
1-4559
for - 3 is the same as + 3. Check by adding the last two numbers.
25. Find the value of each of the following by logarithms :
0-001463 m 0-7276 0-002367
W ^2572 ' () 0-01571' () 0-6297
/ ' 43 H ) ^ 764 (Q) 3 ' 244
W 6-003179 ' w 0-9467' ^ ; (F3T54
VIIlJ LOGARITHMS 45
26. In questions 2 and 3 of this Chapter it was found that
10* meant VlO and 10* meant ^To.
In the same way 1/50 = (50)* = (10 r8900 )4.
Now (10 4 ) s =10 4 x 10 4 xl0 4 =10 12 , the new index being ob-
tained by multiplying 4 by 3.
In the same way (lO 1 * 990 )* = lO ' 6863 by multiplying 1-6990 by
J, i.e. by dividing it by 3.
Find the number whose logarithm is 0-5663. It is <s/50.
27. In a similar manner find the values of
N/320,
28. In dealing with the logarithm of a number less than 1,
remember that the decimal part (or mantissa) must always be
kept positive, the characteristic being negative.
What is half of 3-3010 (i.e. - 3 + 0-3010) ?
To transform the answer so that the decimal part remains
positive, it is most convenient to think of 3*3010 as being equal
to 4+1-3010. Dividing this by 2, the logarithm of N/-002
may be found, and hence its value from the Tables.
29. In a similar manner tind by logarithms the value of the
following, verifying the answers :
J&QOOi.
(1) -s/004.
(4) ^O-OOS.
(2) 76 : 004.
(3)
(6)
(9)
(7) Vo-765.
{ioy -yo-oofs.
(8)> ^0-057.
LOGARITHMS
[OH.
MISCELLANEOUS EXAMPLES ON LOGARITHMS.
When the principles have been grasped it is convenient to arrange the
work as in the following Examples.
No.
A. 23-57x3-721=87-70.
372-1
2^357
= 157-9.
Find the following :
(1) 72-47 x 26-05.
23-57
3-721
log
1-37124
0-5706
No.
372-1
2-357
1-9430
log
2-5706
0-3724
2 -1982
(2) 631-4 x 0-05641.
,,s 326-4
w
3-461 *
w
435-6*
(5)
(17-49)'.
( 6 )
(31-62) 8 .
(7)
>/29 T 95.
(8)
#12.
(9)
(0-7462) 4 .
(10)
(0-06925) 8 .
(11)
N/0-007534.
(12)
#0-04691.
(13)
s/(0-1193)'.
(14)
#(0-007 2) 2 .
0-3157
/73-91V
(Ib)
(16)
0-04694 *
(17)
'/ 0-0724
(18)
7-529x0-04392
V 7-4 x 3-142'
0-5497
no\
3-142
/9m
(0-325) 8
0-247 x(0-04627)*
VIII]
LOGARITHMS
47
Note. In working out an example such as ^J5 it is convenient to
14o
make a skeleton arrangement before looking out logs from the Tables, thus :
57
sin 69
148
The logarithm of sin 69 may be found direct from the Table of Log-
arithmic Sines ; in some Tables 10 is added to the characteristic to avoid
the clumsy printing of minus signs above the numbers. The work then
stands thus :
67 x sin 69
143
=0-3722.
Find the value of :
(21) 16 sin 35*.
(23)
(25)
(27) 1341 x 7-eixcos 14* 19'.
(29) 73-56 coa 19 21'
61 sin 49
57
sin 69
1-7559
1-9702
143
1-7261
2*1553
1-5708
11-61 tan 49.
114 tan 19 24'.
243 cos 47.
63 sin 17 24'.
1 -64 cos 23 47'.
(22)
(24)
(26)
(28) gj
6 1-42 sin 25 52'.
13 sin 29 46'
(31)
(32)
(33)
(34J
(35)
95
loft sin 43 17'
12 - 6x sin^F24"
2 x 3-24 x 5-62 cos 72 31'.
2 x 7-63 x 8-5 x cos 31 37'.
? sin 23' 14'. (36)
14
19-43 sin 73 29'
10-65
CHAPTER IX
SOLUTION OF TRIANGLES WHICH ARE NOT
RIGHT-ANGLED
Note. The calculations in this and the following chapters may often be
much simplified by the use of logarithms.
A. Given two sides and the angle included.
1. Draw accurately a triangle ABC having angle A = 60,
6=10 cm. and c = 9 cm. Measure B, C and a.
2. If, in the triangle above, CN is drawn perpendicular to
AB, the right-angled triangle ACN can be solved by calculation,
and then the parts of the triangle BCN can also be calculated.
Calculate B, C, a by this means.
3. Solve the same triangle by drawing a perpendicular from
B to AC.
4. Draw a freehand sketch of the triangle ABC in which
A = 53 8', b = 7-5 cm., c = 9cm. Solve it by calculation and
verify by accurate drawing.
5. Repeat for a triangle having A = 23 35', 6 = 10 cm. and
c- 12 cm.
6. In the questions 1, 2, 3 above it was found possible to
solve the triangle by drawing a perpendicular from C or from B.
Is it possible to do so by drawing a perpendicular from A?
Express very carefully in words the line (or lines) which should
be drawn in order to solve a triangle in which two sides and the
included angle are given.
OH. IX] SOLUTION OF TRIANGLES 49
7. Solve the following triangles and verify the answers by
drawing to scale.
Note. An example of orderly arrangement will be found in the Appendix.
(1) a = 25, 6 = 24, C = 4125'.
(2) a =2-6, c=2-3, B = 6124'.
(3) 6 = 44-7, c = 46-9, A = 24 17'.
8. If A = 60, 6 = 10 cm., c = 3 cm., solve by drawing a per-
pendicular from B to AC.
9. Solve the triangle of question 8, by drawing CN perpen-
dicular to AB produced and dealing with the right-angled triangles
ACN and BCN.
10. Solve the triangle in which A = 126 52', 6 = 10 cm.,
o = 7*5 cm. by drawing the perpendicular from B to AC produced.
11. Is it possible to solve the last triangle by drawing a
perpendicular (1) from A, (2) from B]
If either of these is possible, complete the solution.
12. Solve the following triangles by drawing a perpendicular :
(1) 6 = 23-4, c = ll-6, A = 37 49',
(2) a =149, 6 = 51, C = 5823'.
(3) a = 26-4, b = 31-7, C = 135 16'.
(4) 6 = 3-49, c = 4-37, A = 98 23'.
(5) c = 0'45, a = 0-19, B = 16147'.
B. Given two angles and one side.
1. Draw as accurately as possible a triangle ABC in which
A = 53 8', B = 59' 29' and c = 10 cm. Measure C, a and 6.
2. From B drop a perpendicular BN to AC and calculate
C, AN, NB, NC, CB and CA in this order.
3. Draw a sketch of the triangle given above and write
down how it can be solved if a perpendicular is drawn from A
instead of B. Indicate the parts to be found in their proper
order as in question 2.
4
50 SOLUTION OF TRIANGLES WHICH [CH.
4. Is it possible to solve the triangle by drawing a perpen-
dicular from C? Describe in words the line (or lines) which
should be drawn to solve a triangle when two angles and the side
between them are given.
5. If two angles and the side opposite to one of them (A, B, a)
be given, from which angle should a perpendicular be drawn in
order to solve the triangle 1
Is it possible to solve it in more than one way? Indicate
how the triangle should be solved.
6. Solve by this method the triangles in which
(1)
A = 43 27',
B = 619',
a = 3-5.
(2)
A = 71 42',
B = 6519',
c = 16-9.
(3)
A = 29 17',
B = 108 53',
6 = 214.
(4)
A = 13 39',
= 41 11',
a =14-63.
(B)
= 112 41',
A = 47 25',
6=29-46.
C Given three aides.
1. Draw a sketch of the triangle ABC in which a = 5 cm.,
6 = 6 cm. and c = 4cm. Draw AN perpendicular to BC. Let
BN = a; cm. What is then the length of CN ?
Using Pythagoras' Theorem in the triangles ABN, ACN, find
two expressions for the square on AN and by equating them find
the length x. Complete the solution of the triangle ABC and
verify by drawing an accurate figure.
2. Verify the results of question 1, by drawing the perpen-
dicular from C and solving as before.
3. Draw the perpendicular from B in the same triangle and
indicate (without actual calculation but by writing down the
parts to be found in their proper order) how the triangle may be
solved.
IX] ARE NOT RIGHT-ANGLED 51
4. Solve the following triangles. Consider carefully which,
perpendicular will lead to the simplest calculation:
(1)
a =
7
cm.,
b
=
8cm.,
c
= 10cm.
(2)
a =
4,
b
=
3,
c
9
(3)
a =
2
5,
b
=
3,
c
= 1.
(4)
a =
2
5,
b
=
3-5,
c
= 4.
(5)
a =
6
34,
b
=s
7-42,
c
= 3
5.
D, Given two sides and an angle opposite one of them,
1. Draw as accurately as possible a triangle having
A = 36 52', 6 = 10 cm., a = 8 cm.
Measure B, C and c ; there are two answers to each.
2. Draw CN perpendicular to AB in the triangle above and
calculate the values of B, C and c by solving the triangles ACN
and BCN.
3. Is it possible to solve the same triangle by drawing the
perpendicular (1) from B, or (2) from A? Illustrate the answer
with a sketch.
4. Are there two possible values for c in a triangle having
A = 36 52', b = 10 cm. and a = 12 cm. 1
Solve this triangle completely.
5. In the triangle of question 1, a is less than 6, and in
question 4, a is greater than 6. Express clearly in words the
conditions under which two different solutions of a triangle may
be expected when two sides and an angle opposite one of them
are given. Illustrate with sketches.
E. Miscellaneous examples on solution by dividing into two
right-angled triangles.
1. There are six parts of a triangle three sides and three
angles. When three of these are known it is generally possible
to calculate the other three and solve the triangle.
52 SOLUTION OP TRIANGLES [CH. IX
The list below gives the possible ways in which three of
these parts may be given.
Copy each item and determine by drawing a sketch (1) whether
the triangle can be solved, and if so (2) the first step in the
solution :
(I) 3 sides (0, 6, c).
(ii) 2 sides and the included angle (a, 5, C).
(iii) 2 sides and an angle not included (a, 6, A),
(iv) 2 angles and the side included (A, B, c).
(v) 2 angles and a side not included (A, B, a),
(vi) 3 angles.
In which cases may two different solutions possibly occur ?
2. Solve the following triangles by drawing a perpendicular:
(1)
A =
54
23',
6 =
19-45, c =
23-51.
(2)
A =
117'
3 21',
B =
35
24',
C =
467.
(3)
A =
41
41',
b =
26-7
,
a =
20-63
(4)
a
4-35,
6 =
7-91
9
c =
6-41.
(5)
B =
63
26',
6 =
114,
a
91-5.
(6)
B =
24
8',
C =
79
53',
a
3-47.
(7)
c =
47
19',
a =
191,
6 =
179.
(8)
a =
13-5,
6 =
14,
c =
16-5.
(9)
c=37 a
23',
o =
91,
a =
124.
(10)
C=
71
35',
A =
52
13',
6 =
7-47.
(11)
a =
147
6 =
92,
c =
69.
(12)
A =
51*
29',
B =
79
47',
a =12-39.
Note. The problems in Chapter XIII may all be solved by the process
above, but the general formulae in Chapters X, XI, XII will be found t*
shorten the work considerably.
CHAPTER X
OBTUSE ANGLES
1. In the figure P revolves round the circle whose centre is
O and whose radius is 10 cm.
AO= 17 cm. PN is drawn per
pendicular to AOQ. / *\P
Calculate the lengths of ON
and AN when the angle POGl
has the values : 36 52', 53 8', A ; 6 FT?**
60, 120, 126 52', 143 8' and \ /
arrange the results in a table.
It will be seen that when *** *""'
POO is obtuse, AN is less than pjg 22.
17 cm.
But in the figure as drawn, AN = 17 + 10 cos POGl If this
is also true when POGl is obtuse, what do you conclude with
regard to the value of the cosine of an obtuse angle ?
2. For which pairs of the angles above is the length of ON
the same? What is the sum of each pair? Write down the
values of cos 120, cos 126 52', cos 143 8' and express in words
the relation between the cosines of supplementary angles.
3. Find the length of PN for each of the angles in question 1.
For which pairs is the length of PN the same? It is customary
to say that PN = 10 sin POQ for all values of the angle POGl acute
or obtuse.
Write down the values of sin 120, sin 126 52', sin 143 8'.
Note that PN is drawn above the line in every case, bat ON is drawn
from O in one direction for an acute angle POQ, bat in the opposite direction
for an obtuse angle POGL
54
OBTUSE ANGLES
[CH. X
4. The figure shows that 1
eosll2'37'=:- T V
Calculate the length of the
perpendicular (Pythag.), and
write down the values of the
other five ratios of this angle,
paying special attention to
sign.
5. From the Tables find
the values of the sine and co-
sine of : 140, 123, 104 13',
166 51', 152 22'. u Fig. 23.
6. Find the tangent of each of these angles. Draw a rough
sketch to determine the sign.
7. Find the cotangent, secant and cosecant of each of the
angles above.
8. Find the value of
sin 64, sin 164, cos 53 30', cos 153 30',
sin 101 25', cos 121 33', sin 111 11', cos 174 19',
tan 41 23', tanl4123', cot 19 21', cot 99 2V,
cosec3141', sec 12 17', sec 131 41', cosec 112 17'.
9. Find two angles less than 180, whose sine is 0-7660.
10. If A is an angle of a triangle, find A if
sin A = 0-9. cos A = - 0-6225, sin A = 0-2860,
cos A = 0-1225, cos A = - 0-1225, sin A = 0-3333,
tan A = 2-6051, tan A = - 2-6051, cot A = -0-2401,
cosec A = 1 -1383, sec A = 3-1085, sec A = - 3-1085.
11. In dealing with triangles any angle between and 180
may occur. What can be asserted with regard to the sine of
any angle of a triangle ? What can be asserted with regard to
the cosine of any angle of a triangle ] These results should be
carefully remembered.
CHAPTER XI
THE SINE FORMULA
1. In the triangle ABC having A = 40, B = 80 and a = 10 cm.
calculate C, 6 and c by drawing a perpendicular, and show that
__ sin B
a ~~ sin A *
n oi. i xi_ i. c sinC ,6 sin B .... . .
2. Show also that - = . and - = . in this particular
a sin A c sin C r
triangle.
3. Draw any acute-angled triangle at random (i.e. without
making the angles and sides any chosen size or length). Measure
the sides and angles and show that = -: = .
sin A sin B sin C
4. Explain clearly in words the facts verified above.
5. Consider whether the rule is true for obtuse-angled
triangles. Verify by drawing such a triangle at random and
carefully measuring its sides and angles. (Remember the fact
with regard to the sines of supplementary angles which was dealt
with in Chapter X.)
6. Making use of the sine rule above, calculate a in the
triangle having A = 62, C = 53 8', c = 8cm., and verify by
accurate drawing.
56 THE SINE FORMULA [OH.
7. Calculate the following and verify by accurate drawing.
Note. It is convenient to write the required part as the numerator of a
fraction.
(1) A = 63, C = 50, c = 6 cm., find a.
(2) B = 47, C m 73, c = 11 cm., find b.
(3) C = 30, A = 81 30', a= 6 cm., find a
(4) A = 44, B = 3r, a=4cm., find 6.
(5) B = 39 24', C = 49 12', 6=5 cm., find c.
sin A
8. If 10 sin A = 3 sin B, what is the value of -; - ?
sin B
i , .. e sin C . T 7 ,, , a c
And rf a = ETA ' show dearly thafc =
9. To prove the Sine Formula for an acute-angled triangle
draw a sketch of any such triangle ABC. Draw ON perpendicular
to AB. In the triangle CAN express CN in terms of b and A ; find
also an expression for CN in terms of a and B in the triangle
BAN. By equating these prove that -. - - - for any acute-
sin A sin o
angled triangle.
Note. An alternative proof will be found in the Appendix.
10. Bemembering that sin (180 - A) = sin A, prove the sine
formula for any obtuse-angled triangle.
11. Use the Sine Rule to find the following :
(1) If A = 36 52', B 30, a = 10 cm., find 6.
(2) If A 58 13', C = 23 35', c = 4 cm., find a.
(3) If B = 68 26', C 63 8', c = 3-5 cm., find b.
Note. Much time may be saved by Baying "a over b equals sin A over
tin B " and at the same time writing a* b x
XI] THE SINE FORMULA 57
(4) If A = 5116 / , C = 73'49', a-12cm,, findo.
(5) If B = 20 11', A = 59 51', b = 17 cm., find a.
(6) If A = 43 32', B = 61 39', c = 10 cm.; find C and a.
(7) If B = 2113', = 51 42', <* = 4cm., find 6.
(8) If A = 44 32', C = 67 21', 6 = 5 cm., find a.
(9) If A = 21 6', 6 = 10, a = 6, find two values for B.
(10) If C = 32 37', c 3-6, 6 = 5, find two values for B.
(11) If B = 37 23', 6 = 2, a = 2 -5, find A. Are there two values t
(12) If B = 37 23', 6 = 3, a = 2'5, find A. Are there two values ?
12. If in Fig. 1, p. 2 the angle A=24, ^BPC = 54' and
AP SB 10 cm., calculate L ABP, BP (by the sine formula) and hence
find BC.
13. In the same tig. if A = 41, L BPC = 61 and AP = 200 feet
find a trigonometrical expression for BC without working with
tables. By substituting the values given in the tables calculate
the length of BC.
14. Use the method indicated above to solve the following.
The angle of elevation of the top of a tower is observed to be 25
and from a point 100 feet nearer it is 43. Find the height of
the tower.
15. From a boat the angle of elevation of the top of a cliff
200 feet high is found to be 29. The boat sails directly towards
the cliff. How many feet must it sail so that the angle of eleva-
tion may be 50* 1
16. From the top of a cliff the angles of depression of two
buoys in a straight line and 124 feet apart are 18 47' and
51 29'. How high is the cliff?
Nott. Questions B 6 and several in B 2 of Chapter IX may be solved by
means of the sine formula.
CHAPTER XII
THE COSINE FORMULA
1. In the triangle ABC if b - 3 in., c - 4 in. and A = 90, show
that a must be 5 in. long.
If A is acute, will a be greater or less than 5 in. 1
If A is obtuse, will a be greater or less than 5 in. ?
To find a relation between the length of a and the size of the
angle A in this triangle, draw an accurate figure with A = 20, 40,
60, 80 in succession, and measure the length of a in each case.
Verify that 6 a + c a a 3 = 2bc cos A in each case. Arrange results
in a table.
2. By drawing a perpendicular CN from C to AB in each
case above and calculating the lengths of AN, CN, NB, a, verify
that 6 2 + c a - a* = 2bc cos A.
3. In Chapter IX, question A 2, it was found that if
6 = 10 cm., c = 9 cm. and A = 60, then a = 9-539 cm.
Verify the formula above by substituting these values.
Note. This formula is very useful and should bo remembered. A formal
proof will be found in the Appendix.
4. Verify the following by means of the Cosine Formula :
(1) If b = 7-5 cm., c = 9cm., A = 53 8', then a = 7-5 cm.
(2) If = 10cm., c= 12 cm., A = 23 35', then a = 4*903 cm.
(3) If b = 25 cm., c = 24cm., A = 41 25', then a = 17-35 cm.
CH. XIl] THE COSINE FORMULA 59
5. If in any one of the figures above the letters denoting
the sides and angles were altered so that A became B, B became
C, a became 6, etc., then what relation between the sides and
cos B would be true ?
Write down the cosine formula in a triangle ABC in three
different forms, thus :
a 2 = 6 2 + c a -..., 6 a = ............ cosB,
6. If the angle A is a right angle, what is the value of cos A,
and what does the formula become 1
Observing that cos B = - in this triangle, show that the formula
is true in each of the three forms given in question 5.
7. Consider whether the same formula is true in an obtuse-
angled triangle.
To do this calculate the length of a in a triangle ABC having
6 = 8 cm. and c = 10cm. when A is 90, 110, 130, 150, 170.
(Draw CN perpendicular to AB produced.) Arrange results in a
table and (remembering the important fact about the cosine of
an obtuse angle) verify that a* = 6 a + c a - 26c cos A.
8. Verify the cosine formula in each of the following :
(1) If 6 = 10 cm., c=7-5cm. and A = 126 52', then a =15-69 cm.
(2) If a = 19cm., c = 45cm. and 8 = 161*47', then b = 63 -33 cm.
9. Find the third side in each of the following triangles :
(1)
(2)
(3)
(4)
a = 5,
6 = 3-5,
6 = 3-5,
c
c
= 4,
= 4,
= 10
= 10
C = 32.
C = 148*.
, A = 71 36'.
, A = 108 24'.
<*)
a = 6-25,
c - 3-2, 1
3 = 21
13'.
(6)
a =17-3,
6
= 23
4, i
C = 34
25'.
(7)
a=17-3,
6
= 23
4, (
= 145
35
/
60 THE COSINE FORMULA [CH. XII
10. By substituting in the cosine formula, find the angles of
the following triangles. Verify by accurate drawing.
(1) o = 4 in., 6 = 5 in., c = 6 in.
(2) o = 5 in., 6 = 6 in,, c = 7 in.
(3) o = 4 in., b = 5 in., c = 3 in.
(4) o=7'5 cm., b = 4*4 cm., = 4-3 cm.
11. If the sides of a triangle are 3, 5 and 6 inches long,
calculate the smallest angle.
12. Find the largest angle in the triangle whose sides are
4, 5, 2 inches long.
13. Repeat question E 1, Chap. IX, showing how each of the
required parts may be found by using either the cosine formula
or the sine formula.
14. Two trains start at the same time from the same station
along straight tracks making an angle of 40. If their average
speeds are 20 and 24 miles an hour, how far apart are they in
half an hour ?
15. What angle does a straight rod 29 feet long subtend at
the eye of an observer who is 18 feet from one end and 25 feet
from the other ?
Note. Questions 4 and several of those in E 2 of Chapter IX may be
solved by means of the cosine formula.
CHAPTER XIII
PROBLEMS
Note. An example of orderly arrangement will be found in the Appendix.
1. From the ends of a breakwater a mile long and running
E. and W. the bearings of a certain buoy are found to be
S. 27 W. and S. 72 E. Find the distance of the buoy from
each end in yards, and also its distance from the nearest point
on the breakwater.
2. From the same breakwater the bearings of another buoy
are S. 63 53' W. and S. 41 19' W. Find its distance from each
end.
3. From the E. end of the same breakwater a point on shore
is found to be 1500 yds. N. 15 28' W. Find the distance of this
point from the other end.
4 If A is 6-4 miles S. 17 35' E. of P, and B is 4-3 miles
S. 31 17' W. of P, what is the distance AB ?
5. A balloon is observed from two points 1250 feet apart on
a level plain at the moment when it passes above the line joining
them. If the angles of elevation are found to be 39 51' and
61 23', find the distance of the balloon from each point and its
height above the plain,
6. A vertical wall runs horizontally across the side of a hill
which slopes at an angle of 10 25' with the horizontal From a
point down the slope 110 feet from the base of the wall the angle
of elevation of the top is found to be 15 57'. Find the height
of the wall.
62 PROBLEMS [CH.
7. The angles of elevation of the top of a certain tree from
two points on opposite sides of it are found to be 52 31' and
63 P 5'. If the points are 125 feet apart, find the height of the tree.
8. In a certain steam engine the crank OP is 18 inches long
and revolves about O. The other end C of the connecting rod
CP (which is 72 inches long) moves backwards and forwards
along a straight line passing through O, Find the angle at C
and the distance CO when the angle COP is
(1) 24, (2) 90, (3) 124.
9. Find the angle at O and the distance CO in the same
engine when the angle at C is (1) 13, (2) 10 23'. (There are
two answers to each.)
10. A tower stands on a slope which is inclined at an angle
of 17 Q 15' with the horizontal. From a point further up the
slope and 370 feet from the base of the tower the angle of
depression of the top of the tower is found to be 9 38'. Find
the height of the tower.
11. From a ship sailing N. 22J W. a lighthouse bears
N. 31 '3 E. After the ship has sailed 12 miles the lighthouse is
found to bear due E. Find its distance from the ship at each
observation.
12. P and Q are points at opposite ends of a wood. From
a point R outside the wood the angle PRO, is observed to be
79 47', and PR = 272 yds. and QR = 349 yds. Find the distance
PCt in yards.
13. In the triangle ABC having B = 49* 52', C = 61 34',
a = 17*4 cm., the bisectors of the angles meet at I. Find the
radius r of the circle which touches AB, BC and CA.
(If ID is perpendicular to BC, it is a radius. Find BD in terms of r,
CD in terms of r.)
11 In the triangle ABC, A = 43, a = 12-5 cm. What angte
does BC subtend at the centre of the circle passing through A, B,
C) Calculate the radius of the circle. If 6 = 97 cm., find the
angle which it subtends at the centre.
XIIl] PROBLEMS 68
15. A tower subtends an angle of 34 at a point P which is
80 yards due East of it. What angle will it subtend at a point
Q which is 60 yards due South of P ?
16. A ladder 20 feet long makes an angle of 30 with the
wall of a house. How much nearer to the wall must the foot of
the ladder be brought in order to reach 18 inches higher than
before, arid what angle will the ladder now make with the wall ?
17. A vessel starts from O, sails 4 miles N. 38 E. to P, and
then 6 miles S. 16 E. to Q. She then returns direct to O. How
far and in what direction has she to go, and how near will she
pass to P?
18. From the top of a wall 5 feet high the angle of elevation
of the top of a flagstaff is observed to bo 63 31', while the angle
of depression of the foot is 5 42'.
Find (a) the distance of the foot of the wall from the foot of
the flagstaff, (6) the height of the flagstaff.
19. A straight road runs N.E. for 300 yards from A to B,
From A a tower bears N. 19 27' E., and from B, N. 70 33' W.
Find the distance of the tower from A.
20. Solve the triangle in which a = 28*3 cm., c = 12*5 cm. and
B = 32,
21. In a triangle B = 40, C = 70, a = 123 yards. Find the
side c and also the area of the triangle,
22. In a triangle a = 5 in , 6 = 7 in., c = 9 in. Find the
greatest angle.
23. The length of a breakwater which lies along a meridian
is 789 yards. From a boat the Southern extremity bore
S. 67 30' W. and its distance was 426 yards.
Find (i) the distance of the boat from the breakwater, (ii) the
bearing and distance of the boat from the Northern end of the
breakwater.
64 PROBLEMS [OH.
24. Two ships leave a port at the same instant. . One steams
& 11 16' E. at 15 knots, and the other S. 22 30' W. at 19 knots.
Find their distance apart at the end of half an hour and the
bearing of one from the other.
25. A parallelogram ABCD has AB=3-64", 80 = 5-82", and
the angle 8 = 67. Calculate the length of the perpendicular
drawn from A upon BC and also the angles which the diagonal
AC makes with the sides of the parallelogram.
26. Two tangents to a circle of radius 5 cm. from an external
point are each 8 cm. long. What length of arc is included
between the tangents?
27. The sides of a parallelogram are 12" and 8", and include
an angle of 37 32'. Find the area of the parallelogram, and the
lengths of its diagonals.
28. A ship S at sea finds the bearings of a lighthouse L and
a buoy B to be N. 32 W. and N. 14 E. respectively. B is found
to be 3*5 miles N. 66 E. from L. Find the distances of the ship
from the lighthouse and the buoy.
29. A ladder reaches a window ledge 26 feet above the
ground on one side of a street, and makes 70 with the ground.
Find the length of the ladder.
On turning the ladder over without moving its foot, it is
found that when it rests against a wall on the opposite side of
the street it makes 20 with the ground. Find the width of the
street.
30. Two lighthouses are known to be 8f miles apart. A
ship observes their bearings to be S. 41* 48' W. and S. 19 T W.
respectively* After making due South for some time she observes
that they are both in a straight line due West. How far has
she sailed 1
31. A, B are two points 1200 feet apart on the straight bank
of a river flowing due East. A point C on the opposite bank
bears N. 53* 19' E. from B and N. 24J W. from A. Find the
perpendicular width of the river at (X
XIII] PBOBLEMS 66
32. In a circle of radius 3 cm. an isosceles triangle ABC is
described with base BC 3*5 cm. long. Solve the triangle ABC.
33. The diagonals of a rhombus are 13 cm. and 9 cm. long.
Calculate the length of the radius of the circle inscribed in the
rhombus.
34. For a quick survey of a bay a ship observed two points
P and Q at sea level.
Bearing of P, S. 73 W.
Q, S. 87 E.
At P angle of elevation of masthead is 3.
^ > >i
Hc3ight of masthead above water-line 120'.
Find (i) distance of P from ship, in yards ;
(ii) length of base line PQ, in yards.
35. (i) Solve a triangle ABC, given AB = 23'6 in., BC = 18*5 in.,
angle A~ 28. Arc there two possible solutions'?
(ii) Solve a triangle ABC, given AB = 62 yards, AC = 83
yards, angle A = 76 28'. Are there two possible solutions?
36. Two rocks A and B lie due East and West of one another
and are seven miles apart. From a ship A bears S. 24 W. and
B bears S. 35 E.
How far is each rock from the ship ?
37. The flagstaff outside a certain coastguard station is
24 feet in height and stands at the summit of a perpendicular
cliff. From a boat the angles of elevation of the top and bottom
of the flagstaff are observed to be 30 63' and 26 44' respectively.
Find the distance of the boat from the foot of the cliff.
38. Two circles of radii 3 feet 6 inches and 2 feet 3 inches
have their centres 4 feet apart. What is the angle between their
common tangents?
p. T.
66 PBOBLBMS [CH.
39. P and Q are two points on a field such that Q is 66 yards
N. 38 15' E. from P. From Q a tree bears N. 74 21' W., and
from P the same tree bears N. 24 8' W.
(i) Find the distance of Q from the line joining P to the
tree,
(ii) Find the distance of the tree from Q.
40. In a triangle ABC, B = 110, C = 20, a = 158 ft.
Find the side 6, and the area of the triangle.
41. O is the centre of a circle, and AB a diameter 10 inches
long. C is a point on the circumference. The straight line AC
is 7*66 inches long. Find (i) the angle AOC, (ii) the length of
the minor arc AC.
42. The position of an inaccessible point C is required.
From A and B, the ends of a base-line 220 yards long, the
following bearings are taken :
From A the bearing of B is N. 70 30' E. and the bearing
of C is N. 30 20' E. From B the bearing of C is N. 59 40' W
Find the distances of C from A and B.
43. From a ship steaming N. 50 E. a lighthouse bears
N. 30* E. Six minutes later the lighthouse bears N. 10 E.
When will it be abeam and how near will the ship approach it ?
Speed of ship 10 knots.
44. In a simple steam engine the connecting rod CP is
15 feet 9 inches long and the crank 3 feet 4 inches. Find the
angle at P when the distance CO is 17 feet.
Fig. 24.
XIIl] PROBLEMS 67
45. A rugby football ground is 76 yards wide, and the goal
posts are 18 ft. 6 in. apart. Find the angle subtended by the
line joining the goal posts at (i) the middle point of the 25-yard
line, (ii) the point where the 25-yard line meets the touch line.
46. Two points A, B are taken on a level plain such that the
distance between them is 1 mile. A point C on the same level is
observed from A and B, and angles CAB and CBA are measured.
If CAB = 40 and CBA 100% calculate the angle ACB and the
lengths of CA and CB in miles.
47. If with the same base-line (AB) as in the last question a
point D is observed and DAB = 80 and DBA = 70, calculate AD
and DB.
48. What is the angle CAD in the figure of questions 46 and
47 ? By means of the triangle CAD, find the length of CD.
49. By means of the triangle CBD, find the length of CD.
(The answer should be the same as that of the last question.)
50. Using the results of the questions 46 to 49 above, show
how CD can be found in the shortest way when AB = 1 mile and
the angles at A and B are the same as before. Arrange the work
very neatly and verify by drawing to scale.
51. A base line AB is 2 miles long ; C and D are two points
on the same level as AB and the angles at A and B are observed.
Find CD if ABC - 76, ABD = 54, BAC=*43, BAD =113*.
Arrange the work as in the last exercise.
52. In the triangle CAD of the last question calculate the
angle ADC.
Now of A is known to be due West of B, find the bearing
of C from D.
The process above is called Triangulation and is used in
making maps of the country.
62
68
PBOBLEMS
[CH.
53. P is 800 yards due North of Q. If X is N. 73 25' E. of
P and N. 39 41' E. of Q; and Y is S. 79 20' E. of P and
N. 77 12' E. of Q, find the distance and bearing of X from Y.
Verify by drawing to scale.
54. The angle of elevation of a tower 100 feet high due
North of an observer was 50. What- will be its elevation when
the observer has walked due Bast 300 feet 1
55. The elevation of a balloon was observed at a certain
station to be 20 and its bearing was N.E. At a second station
4000 yards due South of the former one its bearing was N. 1 1 E,
Find its height.
56. The elevation of the top of a spire at one station A was
23 50' and the horizontal angle at the station between the spire
and another station B was 93 4'. The horizontal angle at B was
54 28' and the distance between the stations 416 feet. What
was the height of the spire ?
57. A ship was 2640 yards due South of* a lighthouse.
After the ship had sailed 800 yards N. 33 W. the angle of
elevation of the top of the lighthouse was 5 25'. Find its height.
58. If the jib of an ordinary crane is 15 feet long, the post
is 10 feet high and the tie is 7 feet long,
what angles do the jib and tie make with
the post ?
59. What must be the lengths of the
tie in the crane above when the jib makes
angles of 20, 40, 55 with the post]
60. What are the lengths of the di- Fig. 25.
agonals of a parallelogram having adjacent '
sides 3 in. and 5 in. long and the included angle 40 1
XIII]
PROBLEMS
69
61. If a parallelogram has two adjacent sides 10 cm. and
12 cm. long and the included angle 73, what is the length of the
diagonal passing through the intersection of the given sides ?
62. If the included angle were 107 (instead of 73), what
would the length of the diagonal be ?
63. From one corner O of a cube distances OA, OB, OC are
measured along the edges so that OA=2 in., OB = 3 in. and
OC = 4 in. Find the angles of the triangle ABC.
64. From one corner A of a regular tetrahedron (see Fig. 9)
distances AX = 5 cm., AY = 4 cm., A2 = 3 cm. are measured along
the edges AB, AC, AD. Find the angles of the triangle XYZ.
65. From a point A due West of a hill TN the angle of
elevation of the top T is found to be 43 26'. From B which is
300 feet W. of A its elevation is 27 39'. Find its height.
Fig. 26.
66. If C is 500 .feet 8.E. of B in question 65, what is the
angle of elevation of T from C ?
70
PROBLEMS
[OH.
67. A path AB straight down the steepest slope of a smooth
hillside makes an angle of 40 with
the horizontal. If another straight
path AC starting from the same point
goes down the slope and makes an
angle of 30 with the first path AB,
what is its angular slope L AGO?
AB
Find the ratios
AD
AC , , AC
AB and hen e AD
6
Fig. 27.
and ZACD.
68. If AB in the last question runs due South, in what
direction (give compass bearing) does AC run?
Note that angles CAB, CDB are not equal. CDB gives the bearing.
69. A hillside slopes towards the South at a gradient of 1 in
2. In what direction (compass bearing) must a path be made up
the hillside so that its gradient may be 1 in 5 1
70. If a hillside slopes at an angle of 42 towards the North,
what is the slope of a road running up it in a direction S. 73 E. 1
71. A weight R is suspended by three strings OR, OP, OQ
knotted together at O. If a parallelogram OBCA be drawn (as
in the figure) with CB, CA parallel to PO, QO, and the diagonal
OC in the same straight line as OR, then the tensions on the
strings OP, OQ, OR are proportional to the lengths of OA, OB, OC.
If the weight R is 10 Ibs. and the strings OP, OGl make angles
of 30 and 70 with OC, show that the tensions on these strings
are approximately 6 '08 and 9 -54 Ibs. weight
0///^////////^
Fig. 28.
XIllJ PROBLEMS 71
72. If R = 10 Ibs. and each of the strings OP, OQ makes an
angle of 35 with OC, what must the tensions on OP and OQ be?
73. If R=12 IbS. and tho tension on OP is 8 Ibs. weight
when L POQ = 120, what must be the tension on OQ, and what
is tho angle QOR?
74. What must the weight R be if the tension on OQ is 3 Ibs.
weight, the tension on OP is 5 Ibs. weight, and the angle POQ is
130 1
75. A station A is 1000 yards due West of another station
B and on the same level. A point C is N. 20 E. of B and
N. 50 B. of A and its elevation from B is 30. Find the height
of C above the level of AB and the angle subtended by AB at C.
APPENDIX A
DEFINITIONS AND RELATIONS BETWEEN RATIOS
I. If from any point P in one arm of an angle A a straight
line PN be drawn perpendicular to the other arm, then
PN
the ratio is called the sine of the angle A,
AN
~ ,, cosine ,,
PN
tangent
AP
^j secant
AP
cosecant
AN
cotangent
A N
II. From the definitions above:
(1) cosec A = . secA = , cot A =7 .
^ f sm A cos A tan A
PN
/rt . sin A AP PN . . cos A
(2) = . = -~ = tan A. Also-: - = cotA.
v ' cos A AN AN sin A
AP
. 9 /PN\ 3 /AN\ a PN 2 4-AN 2 ,AP 2
/O\ cin" A J. rrk A i i 4- I 1 -
\ / olll r\ T OLIft r\ I I ~ I I ~ "" -". o ~" "I _Q
by Pythagoras' Tlieorem. c
/. sin a A + cos 9 A SB 1.
APPENDIX A
73
TIT. If ABC is a triangle with a right angle at C, the angles
at A and B are complementary and B = 90 - A.
sin B = = cos A,
tan B = -= cot A,
BC
sec B = = cosec A.
Each ratio of an angle = the co-ratio of its complement.
(90
IV. If an angle POX be continually increased by keeping
OX in a fixed position and revolving OP, PN increases and
ON decreases until POX becomes 90, and then PN = OP and
ON vanishes.
M O N
If OP continue to revolve so that POX becomes an obtuse
angle, PN will begin to decrease and it is measured upwards from
OX as before, but ON will begin to increase and is now measured
in the opposite direction away from the angle.
Thus the sine of an obtuse angle (QOX) is the same as the
sine of its supplement (POX), while the cosine of an obtuse angle
is always negative but its measure is the same as the cosine of its
supplement.
APPENDIX B
SINE FORMULA
Let ABC be any triangle and O the centre of the circum-
scribing circle.
Draw a diameter BOAj and join AjC.
The angles at A and Aj are equal (in the same segment).
If R is the radius of the circle, then BA X = 2R.
Also BCAj = 90 (in a semicircle).
BC a
.*. = sin BAjC or ^-~ = sm A,
BA &r\
that is 2R = .
sin A
By joining AAi it may be shown in the same manner that
2R=-4-.
smC
By drawing a diameter through A (or C) it may also be shown
that
8inB
a
* i
in any triangle ~. - =
sin A sin B
c
sin C
APPENDIX C
COSINE FORMULA
Let ABC be any triangle with an acute angle A.
Draw BN perpendicular to AC. Let its length be p units.
Suppose AN is x units long. Then AC is b x units.
Then a 2 = p* + (b - a:) 2 (by Pythag.)
= 6 2 + c a - 2bx (for c* = j
= 6 2 -f c 2 26c cos A (for 05 = c cos A).
If the angle A he obtuse, with the same construction as before
ON will be 6 + x units long.
Then a* = p* + (b + xf (by Pythag. )
a; (for e = p 1 + a?)
26CCOSBAN
. ^ + c a _ 25 C cos A (for cos BAN = - cos A).
APPENDIX D
EXAMPLES OF ORDERLY ARRANGEMENT OF WORK
I. Solve the triangle ABC in which
C = 90
C = 90 }
A = 37 16' V.
a= 2-1 cm. J
B = 90 -37 16' ~ 62* 44'.
6 = a tan B = 2-1 x 1-3143
= 2 -7 60 cm.
acosec A = 2-1 x 1-6515
= 3-468 era.
2'lcm,
13143
21.
26286
13143
276003
16515
21
33030
16515
346815
APPENDIX D
77
II. Solution by dividing into right-angled triangles.
Giveu A = 44 23' j
6 = 7-6 cm. L
c = 5 cm. )
Draw BN perpendicular to AC.
AN = 5 cos 44 23'
= 5 x 0-7 147 =3-5735.
/. CN = 7-6-3-5735 = 4-0265 cm.
Also BN = 5 sin 44 23'
= 5 x 0-6994 = 3-497.
n .
But
But
BN
= -j
= 0-8688.
= 40 59'.
A = 44 23'.
3-497
3-497
4-026
0-5437
0-6048
T-9389
(Check A+B-f = 180 0'.)
CB =
12-23 + 16-21 = V2*R4
: 5-33 cm.
!a- 5-33 cm.
B - 94 38 .
0=40 59'.
78
APPENDIX D
III. Solution by formulae.
Given A = 64 23'
6 = 6-31 cm.
= 5*27 cm.
= 6-3P + 5-27 3 -2 x 6-31 x 5-27cos6423'
= 39-82 + 27-77 -28-75
-67-59-28-75
= 38-84.
/. a = 6-232 cm.
sin A 5-27 x sin 64 23'
2
6-31
5-27
cos 64*23'
Also
sin C = c -
6-232
.*. C = 49 42' or its supplement.
But C must be acute, for c is not
the longest side.
5-27
sin 64 23'
6-232
But
(Check A
0-49 42'.
A = 64 23',
B = 65 55'.
= 180 0'.)
0-3010
0-8000
0-7218
1-6358
1-4586
0-7218
1-9551
0-6769
0-7946
1-8823
Ja = 6-23cm.
<B = 65 55'.
*C=4942'.
ANSWERS
The lengths have in general been calculated to four significant
figures and the angles to the nearest minute. Slight discrepancies
will often appear if the methods of calculation differ from one
another.
CHAP. I
1. 180, 45, 30, 6, 18. 2. 22 30', 7 30', 33 45', 64 17'.
3. 53 25', 77 0', 58 34', 139 46'.
4. 67 50', 58 45', 60 35', 42 25', 51 C 9', 70 17', 10 3', 30 Q 11'.
5. 126 35', 103 0', 121 26', 40 14'. 6. 60, 23% 46 39'.
7. 36, 72, 72; 20, 80, 80; 45, 45, 90.
8. 72, 108, 72.
9. (1) 45, 135, 45. (2) 40, 140, 140. (3) 51f .
10. S. 31 W. 11. 90. 12. N. 13* EL
13. S. 45 B. 14. 24. 15. 37.
16. 5. 17. 60, 30. 18. |, *, .
19. ABP = 48. 20. x-y\
21. 60, 60, 90, 30. CK = 5cm. AK = 8*66 cm AKN is
greater than ACN. AKN decreases.
22. It varies, and is greatest when equidistant from the two
points.
23. The circumference of the circle passing through the ship and
the two points.
OHAP. II
13. 1-4826, 1-5051, 1-5108, 1-5080;
0-5317, 0-5362, 0-5384, 0-5373;
0-2946, 0-3476, 0-0673.
15. 25% 72, 17 30', 55" 12', 21 55', 74 50', 11 10', 66 39',
50" 9', 33 33'.
10. tan 35 = 0-7002. 17. 38" 40', 63 26', 74 3'.
80
ANSWERS
18.
20.
22.
25.
28.
31.
33.
34.
35.
36.
37.
38.
39.
40.
43.
46.
48.
50.
55-43 feet.
67 23', 22 37', 13 feet.
134-9 feet.
5-139.
2-690.
A = 38 40',
A = 34
19. 47-67 feet.
21. S. 35 32' B., 8-602 miles,
30 58'. 24. 5-095.
5-154. 27. 1-288.
127-5. 30. 38 40'.
32. A = 58, B = 32.
21',
B = 47 43',
A = 00 49',
B = 767', 6-42-08, c=43-34.
A =17 19', a = 1-269, c = 4-262.
= 36 52', c = 15-5.
B = 5018', c= 194-9.
41. 248-6 feet.
44. 167-3 feet.
47. 1-82 cm.
23.
26.
29.
B = 5120'.
B = 55 39'.
a =23-10, c
6=0-9048,
= 34-34.
o = l-856.
A = 53 8', B
A = 39 42',
54 feet.
69-88 yards.
11 2 37'.
42.
45.
1003 feet.
28-8 in.
88-18, 59-62 feet.
49. 451-3 yards, 12 48'.
3341'
51. 42-88 sees. 52. 193-0 feet.
54. 93-26 feet. 55. 57-82 feet.
57. S-165 cm., 54 44', 70 32'.
53. 396-2 feet,
56. 45, 54 44'.
ANSWERS
81
11.
13.
14.
16.
CHAP. Ill
0-5764, 0-5779, 0-5771, 0-5766; 0-9559, 0-6839, 0-1888,
0-9898.
25, 71, 54 30', 16 48', 16 51', 78 23', 75 57', 46* 4'.
30. 15. 23 35', 48 36', 25 28'.
4-54, 4-571, 4-581, M45, 8-416, 2-210.
17.
6=67-10, a = 33
51, A =26 32'.
18.
(1) B
= 17 46',
a = 20, 6 = 6-407.
(2) A:
= 64 37',
a =9-395, 6 = 4-457.
(3) A
= 47 47',
a = 77-91, 6 = 70-68.
(4) B
= 78 41',
a = 3-791, 6 = 18-95.
19.
2 18'.
20. 34'.
21. 46', 1 26'.
22.
23 ft. 11 in.
23. 0-755 in.
24. 4-23 mi.
25.
6-47 in
26. 32 14'.
27. 3 ft 1 in.
28.
(1) 0-771, 1-845.
(2) 1-909, 0-595.
CHAP. IV
2.
0-515, 0-777.
3. 60% 41
4", 68-9*.
6.
0-5592,
0-5534, 0*
5519; 0'552G(7).
7.
0-8809,
0-9592, 0-2866, 0-1417, 0-5999,
0-9764.
8.
(1) 8-91. (2)
7-27. (3) 8-30.
(4) 1-003.
9.
44-65 feet,
10. 14-85 mi.
11. 1736yarda
12.
12-6 feet
13. 4-77 in.
14. 69-09 mi.
15.
1-054 feet.
16. 7-5 cm.
17. 9-34 feet
18.
5 ft. 6-1
. in.
19. 3064 mi., 19260
mi 20. 10-86.
21.
Half the length of the Equator. Half.
CHAP. V
5.
cosec A
= , secA
o . A 6
as T , COt A SB - ,
6 a
6.
Angle
56* 27'
28 14' 16* 25'
73' 2V
cosec
M999
2-1140 8-5383
1-0437
sec
1-8094'
M351 1-0425
3-4903
cot
0-6631
1-8624 3-3941
0-2991
P. T.
6
82 ANSWERS
7.
As angle increases, sec increases, cosec and cot
decrease.
8.
22-65, 33-48,
42-55 ft. 9. 32-86 ft.
10.
1191 ft.
11.
10 21 cm.
12. 11 -33 ft.
13.
47-91 in.
14.
156-7 ft.
15. 6-22 mi.
16.
49-05 in.
17.
18-70 ft.
18. 11-63 in.
19.
1 in 14.
20.
20-27 ft.
21. A = 36 52', B =
53 8'.
22.
A = 77, B = l
3,
23.
sin A = cos B,
cot A = tan B, sec A = cosec
B.
24.
22 37'.
25. 16 16'.
Oft
O*7 4- 01-1 A 2ir>5 A i nricfl A 1
zo,
&/. tan A, sin A 4- cos A A.
cos A
a
00
sin A c a
aO 9
cos A b b
c
a a + c ,
and sin* A -f cos* A = -r -f -r = -=- = = 1.
c 2 c 9 c 9 c 2
30. sin 90 = 1, cos 90 = 0, tan 90 =00 (the nearer 90, the
larger), sin 0* = 0, cos = 1, tan = 0.
32. sine and cosine must be less than 1. Tangent may have
any value.
335. sec 2 A -tan 2 A 1. 36. cosec 2 A - cot 3 A = 1.
CHAP. VI
etc.
a o o
n . x p a .
3. sin A = - = 7 = - etc -
a b c
4. B = 3652', A = 53 8', 6 = 15, aj=16, #-9, c = 25cm.
5. A = 6723', B = 23 37', a = 15 6, #=14-4, y = 2-5, c = 16-9cm.
6. ^>a=ajsina, y = rctana, AC = a3seca; CN = jt? tan a = a; sin a tan a.
7. a cos B = x, b sin A = />, c cos B = a, a cos A = j, p tan B = y,
p cot B = .r, p sec A = a, p cosec A = fc, b sin B = y.
8. x sin ap 9 x tan a = t/, p cot a = AN, y cos a = p, p tan a = CN.
ANSWERS 83
9. A = 3541', 6 = 4-700, c = 1141.
10. B = 60 7', 6 - 34-98, c = 40-34.
11. B - 45 29', 5 = 1-711, a = 1 -683.
12. A = 59 25', B = 3035', c = 28-7.
13. A = 33 22', B - 50 38', c - 1 16-9.
14. B = 1747', a = 48-33, c = 50-75.
15. A = 55 23', a = 987 -5, 6 = 681-6.
16. B = 30, A =60, a = 9-873.
17. A = 11 32', B = 7828', 6=16-42.
18. A =19 28', B = 7032', 6=ax2-828.
19. 821-5 ft. 20. S. 55 47' W. 21. 67-02 ft.
22. 177-5 ft. 23. 057'. 24. 8-274 cm.
25. 2-157 cm. 26. 30 14'. 27. 7-779 cm.
28. 9-398 cm. 29. 8-402 cm., 42 48'. 30. 181-4 yds.
31. 6-824 in. 32. 53 8'. 33. 83-25 in.
34. 3-441 cm. 35. 4-253 cm. 36. 4-330, 5cm.
37. 7-694, 8-090 cm. 38. 8'83 mi. per hr.
39. 54 ft. in., 60 ft. 4-5 in. 40. 69 41'.
41. 36 52', 66 25'. 42. 50 12'. 43. 27 54'.
44. 34 58'. 45. 8-201 mi. S. 52 26' E.
46. 52 0', 68 40', 75 24'. 47. 15-64 cm.
48. 16 42', 17 28'. 49. 20 ft. 1 in. 50. 747 fb.
51. 4650 fath. 52. 90, 66 25', 9-165 in.
53. AO = 11-92, BO = 8-39, AC = 15-56, BC = 13-05 inches.
54. 1-595. 55. 18-03 mi. S. 71 43' W.
56. 5-362, 7-332 mi. 57. 83-39 ft. 58. 1132 ft.
59. 027'. 60. 4-189 mi. 61. 51-5 ft., 354-6 ft.
62. 190ft. 63. 76-9 yds. 64. S. 70 E.
65. 140, 141 ft. 66. 11-18 mi. S. 23 37' W.
67. 5-342 in., 0-395 in.
68. 4-35 mi. N. 38 2# E. or S. 38 28' E. 69. 30 2V.
70. 297/9, 268-0, 29-2 ft. 71. 5 mi. N. 53 8' W.
72. 9-063, 4-226 mi. 73. 2-588 mi.
74. 16-73 knots, 18*99 mi.
75. 2-908 in., C = 68 12', A = 74 56'. 76. 65-8 ft.
84 ANSWEBS
77. 63-74 a 78. 0-515, 0-4415 mi. 79. 111*53'.
80. 30 58'. 81. 3-215 mi. 82. 54-63, 64-97 ft.
83. 1125 ft. 84. 1-884, 5-73, 6-85 mi.
85. 65 23', 53 54', 62 43'. 86. 3-535 in., 45, 54 44'.
87. 59 ; , 25 22', 16 36'. 88. 686-9, 1069 ft., 13 10'.
89. 11 48'. 90. 382yds.
CHAP. VII
1. 50 sq. cm., 13297-68 sq. yds.
2. 25 sq. cm., 6648*84 sq. yds. 3. 7-5, 6, 13*5 sq. cm.
4. Half the product of the number of cm. in BC and AN gives
number of sq. cm. in area.
5. 3-214, 4-50 cm., 11-25 sq. cm.
6. Product of number of cm. in base and number of cm. in
perpendicular height gives number of sq. cm. in area.
7. Area = xy sin A. 8. \xy sin A.
9. 5 sq. cm., 9 sq. cm., 16-9 sq. cm., 118 sq. cm.
10. 28-7 sq. cm. 11. 259-8 sq. cm.
12. 237-8, 273-7, 282-8, 289-3 sq. cm
13. 363-2, 346-4, 327-4 sq. cm, 14. 36 yds.
15. 53 8', 126 52', 4-268 cm. 16. 78 42', 22 36'.
17. Sq. pyramid, 273'2 sq. cm. Tetrahedron, 173*2 sq. oin.
18. Segment, 25-67 - 21-22 = 4-45 sq. cm. 19. 15-3 sq. cm.
20. 89-04 sq. cm.
OHAP. VIII
2. 10* = N/IO = 3-162. 3. 10* = -yiO=N/3 7 162 = 1-779.
4. 10* = 5-623.
7. 3-7 = 10 0668J , 3-75 = 10, 3-758 = 10 OW48 , 7-64 = 10 08881 ,
7-649 = lO " 8838 , 9-8 = 10 0>9912 , 9-805 = lO 4 .
8. 2, 7-6, 2-75, 2-76, 2-757, 6-16, 6-17, 6-167, 4-486, 7-0.
9. 4-318. 11. 8-572, 8-179, 2-999, 6-451, 8-164, 9-936.
12. 2-940. 13. 2-397, 1-263, 3-855 4 , 1-408, 2 752, 3-766.
16. Logs are 2-6660, 1-8986, 1-2119, 2-8581, 1-8581, 3-4934,
4-8825, 5-3109, 1-8664, 2-3294.
ANSWERS
85
17. 710-1, 33, 3050, 888, 278000, 7115, 337-2, 26-92.
18.
(1) 37-62.
(4) 149000.
(7) 804-2.
(10) 317800000.
(13) 463-1.
(16) 155-2.
(19) 1-006.
(3) 1143.
(6) 87000.
(9) 2637000.
(12) 1-375.
(15) 1-395.
(18) 2-867.
(2) 1390.
(5) 9243.
(8) 1023.
(11) 5-690.
(14) 266-1.
(17) 4083.
(20) 9-779.
20. Logs are 2-0835, T-0835, 1-5518, 4-5518.
22. Logs are 2-5378, T-8860, 3-6656, 4-0433, T-9987, 1-9573,
2-6385, 1-7350.
24. (1) 0-003762.
(4) 0-001390.
25. (1) 0-005690. *
(4) 137-5.
26. 3-684.
28. \/0-0()2 = (10 !
29. (1) 0-2.
(4) 0-2.
(7) 0-8747.
(10) 0-1968.
(2) 0-01490. (3) 0-01143.
(5) 92-43. (6) 1-023.
(2) 46-31. (3) 0-003758.
(5) 0-1031. (6) 10-28.
27. 17-88, 4-236, 10-45, 49-91, 6-223.
= 1 Q2-6505 =: Q-04472.
(2) 0-06325. (3) 0-02.
(5) 0-09283. (6) 0-4309.
(8) 0-3849. (9) 0-7830.
MISCELLANEOUS EXERCISES IN LOGARITHMS
1.
1888.
2.
35-62.
3.
6-811.
4.
0-7492.
5.
305-8.
6.
31620.
7.
5-473.
8.
2-290.
9.
0-3098.
10.
0-0003321.
11.
0-08680.
12,
0-3607.
13.
0-0412.
14.
0-03728.
15.
6-725.
16.
6092.
17.
0-1460.
18.
0-6015.
19.
1531.
20.
64-92.
21.
9-177.
22.
165-8.
23.
13-36.
24.
18-84.
25.
40-15.
26.
1-501.
27.
98-88.
28.
0-5399.
29.
96-22.
30.
0-4846.
31.
0-461.
32.
16-13.
33.
10-93.
34.
110-4
35.
0-4066.
36.
1-749.
63
86 ANSWERS
CHAP. IX
A. L 9-539 cm., 65 C 12', 54 48'.
4. 7-5 cm., 53 8', 73 44'.
5. 4-903 cm., B = 54 40', C = 101 45'.
6. Drop perpendicular on to either of the given sidea.
7. (1) c = 17-35 cm., A = 72 24', B = 66H'.
(2) 6=2-51, A = 65 11', = 53 25'.
(3) a = 19-39, B = 7130', C = 8413'.
8. a = 8-888, B = 103, C=17.
10. a = 15-69, B = 3039', C = 2229'.
12. (1) a = 15-92, B- 115 39', C = 26 33'.
(2) c= 129-7, A -102 4', B *= 19 33'.
(3) c = 53-78, A = 20 13', B - 24 31'.
(4) a = 5-977, C = 46 20', B = 3517'.
(5) 6 = -6333, C = 1250', A = 5 23'.
B. 1. C = 67 23', a = 8-666, b = 9-333.
2. AN = 8, NB = 6, NO = 3-333.
3. If AL be the perp., find angle C, AL, BL, CL, CA, CB.
4. Draw perp. from either end of the given side.
5. From B or C. If BN perp., find C, ON, NB, NA, AB, AC,
6. (1) C = 7524', c = 4-925, 6 = 4-458.
(2) C = 4259', a = 23-54, 6 = 22-52.
(3) C = 4150', a=1106, c = 150-9.
(4) B = 12510', 6 = 50-69, c = 40-83.
(5) B = 1954', a = 63-71, c = 79-86.
C. 1. a; = 0-5 cm., A = 55 47', B = 82 49', C = 4124'.
4. (1) A = 44 3', B = 5237', C = 83'20'.
(2) A = 104 29', B = 4634', C = 2857'.
(3) A = 51 19', B = 11029', = 18 12'.
(4) A = 38 13', B = 60, = 81 47'.
(5) Perp. from B, AN = 1 -827. A r = 58 32', B = 93 23',
= 28 8'.
ANSWERS 87
D. 1. B = 48 36' or 131 24', C = 94' 32' or 11 44', o = 13-29
or 2-708.
3. Only from B. 4. B = 30, C = 113 8', c = 18-39.
5. When the side opposite the given angle is less than the
other given side.
K 1. B = 5222', = 73 15', a =19-96.
2. C = 27 15', a = 9-058, b = 5-907.
3. B = 59 24', 120 36', C = 78 55', 17 45', c = 30-44, 9-439.
4. A = 33 19', B - 92 38', C = 54 3'.
5. A - 45 52', C = 70 42', c = 120-3.
6. A =75 59', 5 = 1-462, c= 3-521.
7. A = 70 36', B = 62 5', c = 148-8.
8. A = 51 45', B = 5432', C = 7343'.
9. A = 55 50', 124 10 ; , B = 86 47', 18 27', b = 149-6, 47-72.
10. B = 5612 / , a = 7-104, c = 8-529.
11. A =131 21', B-280', = 20 39'.
12. = 48 44', 6 = 15-59, c = ll-91.
1. Angle
CHAP. X
36 52' 53 8' 60 120' 126 52' 143 8'
8 (5-9999 say) 6 55 6 8cm.
25 23 22 12 11 9cm.
ON
AN
The cosine of each of these obtuse angles is negative.
2. Supplementary angles. O f 5, 0-6, 0*8. Cosine of an
obtuse angle = (cosine of its supplement).
3. PN = 6, 8, 8-66, 8-66, 8, 6 cm. Supplementary angles.
0-8660, 0-8000, 0-5999.
4. Sine = y|, tangent = \ 2 -, secant = T 6 7 , cosecant = |-|,
cotangent = y\.
5. sin. 0-6428, 0-8387, 0-9694, 0-2275, 0-4638.
cos -0-7660, -0-5446, -0-2456, -0-9738, -0-8859.
6. tan -0-8391, -1-5399, -3-9474, -0-2336, -0-5235.
88 ANSWERS
7. cot -1-1918, -06494, -0-2533, -4-2812, -1-9101
sec -1-3054, -1-8361, -4-0720, -1-0269, -1-1287.
cosec 1-5557, 1-1924, 1-0316, 4-3965, 2-1560.
8. 0-8988, 0-2756, 0-5948, -0-8949,
0-9802, -0-5233, 0-9324, -0-9951,
0-8811, -0-7988, 2-8478, -0-1646,
1-9040, 1-0234, -1-5037, 1-0807.
9. 50 and 130.
10. 64 10' or 115 50', 128 30', 16 37' or 163 23',
82 58', 97 2', 19 28' or 160 32',
69, 111, 103 30',
61 28' or 118* 32', 71 14', 108 46'.
11. The sine is positive. The cosine is positive for acute,
negative for obtuse angles.
CHAP. XI
1. C = 60, b = 15-32 cm., c = 13-47 cm.
4. The ratio of two sides = the ratio of sines of opposite angles.
6. 8-830. 7. 6-979, 8-412, 3-034, 2-966, 5-964.
sin A 3
8. _ = _. 9. CN=68mA = asinB.
sin B 10
11. (1) 8-333. (2) 8-50. (3) 4-069.
(4) 14-77. (5) 42-61. (6) 74 49', 7-139.
(7) 1-515. (8) 3-780.
(9) 36 52' or 143 8'. (10) 48 28' or 131 32'.
(11) 49 23' or 130 37'. (12) 30 24'.
12. 30, 8-134 cm., 6-580 cm. 13. 200 . ~ -^~- = 335-5.
' ' sin 20
14. 93-26 feet. 15. 193 feet. 16. 57-82 feet.
CHAP. XII
1. 1-565, 2-572, 3-606, 4-564.
5. a 2 = 6 a + c a - 26c cos A, 6 2 = c 2 + a? - 2ca cos B,
7. 12-81, 14-78, 16-33, 17 40, 17-94.
ANSWERS 89
9. (1) 2-661. (2) 8-655. (3) 9-495. (4) 11-59.
(5) 3-465. (6) 13-39. (7) 38-92.
10. (1) A = 41 25', B = 55 46', = 82 49'.
(2) 44 25', 57 7', 78 28'.
(3) 53 8', 90, 36 52'. (4) 119 5', 30 51', 30 4',
11. 29 55'. 12. 108 13'. 14. 7'76 miles. 15. 83 6'.
CHAP. XIII
1. 1588, 551, 491 yds. 2. 2019, 3445 yds.
3. 1985 yda 4. 4-82 mi. 5. 1119, 817, 717 ft.
6. 11 -Oft. 7. 98-1 ft.
8. (1) 5 51', 88-0 in. (2) 14 29', 697 in.
(3) 11 58', 60-4 in.
9. (1) 64 10', 78-0 in. ; 115 50', 62-3 in.
(2) 46 7', 83-3 in. ; 133 53', 58-3 in.
10. 49-75 ft. 11. 12-97, 11-33 mi. 12. 403yds.
13. 4-54 cm. 14. 9-164, 63 55'. 15. 28 21'.
16. 3-23 ft., 19 48'. 17. 4-877 mi., N. 57 35' W., 3-982 mi.
18. 50-10, 100-6 ft. 19. 270-9 yds.
20. A =127 29', C = 2031', b = 18-9 cm.
21. c = 1 23 yds. Area - 4863 sq. yds. 22. 95 44'.
23. 393-6 yds., S. 32 10' E., 739-4 yds.
24. 5-29 mi., N. 74 26' E. or S. 74 26' W.
25. 3-351 in., 37 18', 75 42'. 26. 10-13 cm.
27. 58-48 sq. in., 7-463, 18*98 in. 28. 3-834, 4-819 mi.
29. 27-67, 35-47 ft. 30. 15-98 mi. 31. 667ft.
32. A = 35 41', B = = 72 9', AB = AC = 5-7 11 cm.
33. 3-7 cm. 34. 763-2, 1203 yds.
35. (1) = 36 47' or 143 13', B - 115 13' or 8 47', 6 = 35-66
or 6-019 in.
(2) a- 91-24 yds., B = 62 11', C = 41 21'.
36. 6-Q89, 7-459 mi. 37. 254-5 ft. 38. 36 25'.
39. 58-48, 76-08 ft. 40. 194 ft., 5240 sq. ft.
41. lOO'^in. 42. 168,142yds.
90 ANSWERS
43. 4 min. 36 sees., 0-642 miles. 44. 106 10'.
45. 14 4', 4 17'. 46. 1*532 mi., 1 mi.
47. 1-879, 1-970 mi. 48. l-210mi. 49. 1-212 mi.
51. 6-76 mi. 52. S. 41 E. to within 5'.
53. PX = 919-8, PY = 1959, QX = 1380, QY = 1973. XY from
triangle PXY= 1217, and from triangle QXY = 1214 yds.
54. 19 9'. 55. 511-2 yds. 56. 278-6 ft.
57. 192yds. 58. 23 4', 57 7'.
59. 6-564, 9-757, 12-37 ft. 60. 3-32, 7-55 in.
61. 17-73 in. 62. 13-18 in.
63. 75 38', 60 3', 44 19'. 64. 69 32', 63, 47 28'.
65. 351-9 ft. 66. 36 28'. 67. 33 50'. 68. S.37W,
69. N. 69 18' E. or W. 70. 14 45' or about 1 in 4.
72. 6-104 Ibs. wt. 73 13-8 Ibs. wt., 144 44'.
74. 3-84 Ibs. wt. 75. 742-3 yds., 28 16'.
CAMBRIDGE: PRINTED B* w. LEWIS, M.A., AT THE UNIVERSITY PRESS | 677.169 | 1 |
This best-selling book provides an accessible introduction to discrete mathematics, using an algorithmic approach that focuses on problem-solving techniques. The new edition weaves techniques of proofs into the text as a running theme. Each chapter has a special section dedicated to showing students how to attack and solve problems.
"synopsis" may belong to another edition of this title.
About the Author:
Richard Johnsonbaugh has a Ph.D. from the University of Oregon. He is professor of Computer Science and Information Systems, at DePaul University. He has 25 years of experience in teaching and research, including programming in general and in the C language. Dr. Johnsonbaugh specializes in programming languages, compilers, data structures, and pattern recognition. He is the author of two very successful books on Discrete Mathematics.
Book Description U.S.A.: Prentice Hall, 2004. Soft cover. Book Condition: New. 5th or later9166811812870
Book Description Prentice Hall. Book Condition: New. 01311768621176867
Book Description Book Condition: Brand New. New. Soft Cover International edition. Different ISBN and Cover image but contents are same as US edition. Customer Satisfaction guaranteed!!. Bookseller Inventory # SHAK123326 | 677.169 | 1 |
I am for this resolution. I am new so Im not sure how much to discuss. My first point is that math forms the basis of everyday life. It has practical and realistic applications in society. Without math and everything in relation to it, there would be no order or organization in the real word. These are just some of my points. I will elaborate in depth as soon as I get a debater
I welcome you to the debating world, my dear competitor. In my portion of the debate, I will discuss how advanced maths is not necessary for the majority of jobs and what students really need from the world of arithmetic.
In round 2 of this debate, I will be discussing how advanced maths is not necessary for the majority of jobs.
.
This has been my second round of the debate. Thank you for listening and I beg you to oppose.
"."
Geometry is considered the most practical of math skills, as it includes area and volume. optimization problems are used is industry all of the time.
Many people do use the algebraic concept of solving for x when working backwards to find unknown quantities, even if they don't use the formal notation. | 677.169 | 1 |
TriAGe: Preparation for University Calculus
A refresher of material covered in the Grade 12 Advanced Functions course, TriAGe will help prepare students who plan to enrol in an entry-level university calculus course by reinforcing fundamental skills in Trigonometry, Algebra and Geometry. Review and deepen your understanding of core concepts to build a strong foundation and increase your chances of success in calculus!
Topics that will be discussed include:
Algebraic manipulations used to simplify expressions and solve equations and inequalities.
2016 Summer Session Information
Online Session
All students will start with the online session of TriAGe. Access to the course material will begin on Monday, June 27, although you can register for the course anytime up to July 25. You will then work through the course at your own pace, but you should have a goal to complete the course by Friday, Aug. 19, because this is the date of the final evaluation (we suggest you give yourself at least 4 weeks to complete the online version of TriAGe).
On-Campus Classes
Students who may be struggling to complete the course material independently, or who would like some extra instruction, are welcome to join on-campus classes for TriAGe in August.
Final Evaluation
The final evaluation takes place from 2 p.m. to 4 p.m. on Friday, Aug. 19.
Note: The Department of Mathematics at Laurier is permitting successful completion of the TriAGe course to be considered as an equivalency to the Grade 12 Advanced Functions prerequisite for MA100, MA129 and MA110*. To be granted the equivalency, a final grade of 70% or better must be attained through completion of course assignments and the final evaluation. The equivalency cannot be used directly towards program admission requirements.
Registration Process
To enroll in TriAGe, complete the online registration form (opens May 30). Non-Laurier students are welcome. Please take note of the withdrawal policy below. | 677.169 | 1 |
Descrizione:
Su questo libro:
Riassunto: This leading mathematics text for elementary and middle school educators helps you quickly develop a true understanding of mathematical concepts. It integrates rich problem-solving strategies with relevant topics and extensive opportunities for hands-on experience. By progressing from the concrete to the pictorial to the abstract, Musser captures the way math is generally taught in elementary schools.
This title will give you all the essentials mathematics teachers need for teaching at the elementary and middle school levels:
Highlights algebraic concepts throughout the text and includes additional supporting information.
Provides enhanced coverage of order of operations, Z-scores, union of two events, Least Common Multiple, and Greatest Common Factor.
Focuses on solid mathematical content in an accessible and appealing way.
Offers the largest collection of problems (over 3,000!), worked examples, and problem-solving strategies in any text of its kind.
Includes a comprehensive, five-chapter treatment of geometry based on the van Hiele model.
Descrizione libro Wiley. PAPERBACK. Condizione libro: Fine. 0471701173 Codice libro della libreria HML11257 | 677.169 | 1 |
Algebra, Structure and Method: Student text
4.11 - 1251 ratings - Source
-5 alt; -2 and -2 alt;0 -1alt; -3 and -3 alt; 5 An aquot;andaquot; state- true | ment is true false J only if
both parts are true. 5 agt;2 7 agt;4 ... The last step in the solution of an equation may
contain the variable in either side without loss of clarity. ... The list at the bottom of
page 460 may help students translate word problems into inequalities correctly.
Title
:
Algebra, Structure and Method: Student text
Author
:
Mary P. Dolciani, Richard G. Brown, William L. Cole
Publisher
:
- 1988
ISBN-13
:
Continue
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This text demonstrates how numerical methods are used to solve the problems that physicists face. Each chapter discusses different types of computational problems, with exercises developed around problems of physical interest. Students are led from discussions of relatively elementary problems and simple numerical approaches, through derivations of more complex and sophisticated methods, often culminating in the solution to problems of significant difficulty. A disk is included with the | 677.169 | 1 |
A Source Book in Mathematics
4.11 - 1251 ratings - Source
The writings of Newton, Leibniz, Pascal, Riemann, Bernoulli, and others in a comprehensive selection of 125 treatises dating from the Renaissance to the late 19th century a€" most unavailable elsewhere. Grouped in five sections: Number; Algebra; Geometry; Probability; and Calculus, Functions, and Quaternions. Includes a biographical-historical introduction for each article.For example: If 1 lira of saffron is worth 7 lire of pizoli, What will 25 lire of this
same saffron be worth? Here are not mentioned ... Required to know how much
falls to each man so that no one shall be cheated. There are two merchants of
whomanbsp;...
Title
:
A Source Book in Mathematics
Author
:
David Eugene Smith
Publisher
:
Courier Corporation - 2012-05-07
ISBN-13
:
Continue
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Essential MATLAB for Engineers and Scientists
4.11 - 1251 ratings - Source
The essential guide to MATLAB as a problem solving tool This Features: a€c Includes MATLAB Version 7.2, Release 2006a a€c Numerous simple exercises provide hands-on learning of MATLABa€™s functions a€c A new chapter on dynamical systems shows how a structured approach is used to solve more complex problems. a€c Common errors and pitfalls highlighted a€c Concise introduction to useful topics for solving problems in later engineering and science courses: vectors as arrays, arrays of characters, GUIs, advanced graphics, simulation and numerical methods a€c Text and graphics in four colour a€c Extensive instructor support Essential MATLAB for Engineers and Scientists is an ideal textbook for a first course on MATLAB or an engineering problem solving course using MATLAB, as well as a self-learning tutorial for students and professionals expected to learn and apply MATLAB for themselves. Additional material is available for lecturers only at This website provides lecturers with: A series of Powerpoint presentations to assist lecture preparation Extra quiz questions and problems Additional topic material M-files for the exercises and examples in the text (also available to students at the booka€™s companion site) Solutions to exercises An interview with the revising author, Daniel Valentine Am Numerous simple exercises give hands-on learning Am A chapter on algorithm development and program design Am Common errors and pitfalls highlighted Am Concise introduction to useful topics for solving problems in later engineering and science courses: vectors as arrays, arrays of characters, GUIs, advanced graphics, simulation and numerical methods Am A new chapter on dynamical systems shows how a structured approach is used to solve more complex problems. Am Text and graphics in four colour Am Extensive teacher support on solutions manual, extra problems, multiple choice questions, PowerPoint slides Am Companion website for students providing M-files used within the bookThe essential guide to MATLAB as a problem solving tool This text presents MATLAB both as a mathematical tool and a programming language, giving a concise and easy to master introduction to its potential and power.
Title
:
Essential MATLAB for Engineers and Scientists
Author
:
Brian Hahn, Daniel Valentine
Publisher
:
Newnes - 2007-01-29
ISBN-13
:
Continue
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• Capture the essence of chapters at a glance with chapter overviews • Easily access learning objectives and references to exam board specifications, KS4 Programme of Study, Functional Skills Standards and Personal Learning and Thinking Skills (PLTS) for each chapter • Link maths concepts and help students to access functional and problem-solving scenarios • Raise standards by providing the right level of progression for every student by using the well-differentiated lesson plans • Involve the whole class in engaging activities and discussions using the Starter • Lead students into the main concepts and exercises with the Main Lesson Activity • Consolidate and summarise learning using the Plenary • Quickly access the answers to all questions in the corresponding Student Book • Plan ahead and save time using the ready-made Scheme of Work • Customise your lessons using Lesson Plans in Word format on the CD-Rom
Suitable for second year of teaching the linear course. For the first year, you will need AQA Linear Teacher's Pack Higher 1 (9780007489367) | 677.169 | 1 |
About the Book: The book `Fundamental Approach to Discrete Mathematics` is a required part of pursuing a computer science degree at most universities. It provides in-depth knowledge to the subject for beginners and stimulates further interest in the topic. The salient features of this book include: Strong coverage of key topics involving recurrence... more...
An account of elementary real analysis positioned between a popular mathematics book and a first year college or university text. This book doesn't assume knowledge of calculus and, instead, the emphasis is on the application of analysis to number theory. more...
* Contains a selection of articles exploring geometric approaches to problems in algebra, algebraic geometry and number theory * The collection gives a representative sample of problems and most recent results in algebraic and arithmetic geometry * Text can serve as an intense introduction for graduate students and those wishing to pursue research... more...
The appearance of this volume celebrates the ?rst decade of Magma, a new computeralgebrasystemlaunchedattheFirstMagmaConferenceonCom- tational Algebra held at Queen Mary and West?eld College, London, August 1993. This book introduces the reader to the role Magma plays in advanced mathematical research. Each paper examines how the computer can be used... more... | 677.169 | 1 |
Forum for Science, Industry and Business
The Aftermath of Calculator Use in College Classrooms
13.11.2012
Students may rely on calculators to bypass a more holistic understanding of mathematics, says Pitt researcher
Math instructors promoting calculator usage in college classrooms may want to rethink their teaching strategies, says Samuel King, postdoctoral student in the University of Pittsburgh's Learning Research & Development Center.
King has proposed the need for further research regarding calculators' role in the classroom after conducting a limited study with undergraduate engineering students published in the British Journal of Educational Technology.
"We really can't assume that calculators are helping students," said King. "The goal is to understand the core concepts during the lecture. What we found is that use of calculators isn't necessarily helping in that regard."
Together with Carol Robinson, coauthor and director of the Mathematics Education Centre at Loughborough University in England, King examined whether the inherent characteristics of the mathematics questions presented to students facilitated a deep or surface approach to learning. Using a limited sample size, they interviewed 10 second-year undergraduate students enrolled in a competitive engineering program. The students were given a number of mathematical questions related to sine waves—a mathematical function that describes a smooth repetitive oscillation—and were allowed to use calculators to answer them. More than half of the students adopted the option of using the calculators to solve the problem.
"Instead of being able to accurately represent or visualize a sine wave, these students adopted a trial-and-error method by entering values into a calculator to determine which of the four answers provided was correct," said King. "It was apparent that the students who adopted this approach had limited understanding of the concept, as none of them attempted to sketch the sine wave after they worked out one or two values."
After completing the problems, the students were interviewed about their process. A student who had used a calculator noted that she struggled with the answer because she couldn't remember the "rules" regarding sine and it was "easier" to use a calculator. In contrast, a student who did not use a calculator was asked why someone might have a problem answering this question. The student said he didn't see a reason for a problem. However, he noted that one may have trouble visualizing a sine wave if he/she is told not to use said King. "Are these tools really helping to prepare students or are the students using the tools as a way to bypass information that is difficult to understand? Our evidence suggests the latter, and we encourage more research be done in this area."
King also suggests that relevant research should be done investigating the correlation between how and why students use calculators to evaluate the types of learning approaches that students adopt toward problem solving in | 677.169 | 1 |
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Mathematical Modelling: A Way of Life - ICTMA 11
Overview
Mathematical modelling is often spoken of as a way of life, referring to habits of mind and to dependence on the power of mathematics to describe, explain, predict and control real phenomena. This book aims to encourage teachers to provide opportunities for students to model a variety of real phenomena appropriately matched to students' mathematical backgrounds and interests from early stages of mathematical education. Habits, misconceptions, and mindsets about mathematics can present obstacles to university students' acceptance of a ''models-and-modelling perspective'' at this stage of mathematics education. Without prior experience in building, interpreting and applying mathematical models, many students may never come to view and regard modelling as a way of life. The book records presentations at the ICTMA 11 conference held in Milwaukee, Wisconsin in 2003.
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This issue of the Rosicrucian Digest presents a compendium of materials that provide a solid introduction to the most important aspects of mystical Ancient Egypt.
Examines mathematical modelling as a way of life, referring to habits of mind and dependence on the power of mathematics to describe, explain, predict and control real phenomena
Encourages teachers to provide students with opportunities to model a variety of real phenomena appropriately matched to students' mathematical backgrounds and interests from early stages of mathematical education
Records presentations at the ICTMA 11 conference held in Milwaukee, Wisconsin in 2003
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Regrettably, our database of publications not necessarily identified details about mcdougal LamonS J. But all of us is obviously working to get as well as create brand-new info. Once you discover the info who am I?, it's fine to use the item with the type to include an assessment. | 677.169 | 1 |
Rated "Top 10" by the U.S. Department of Education, UCSMP is the first full mathematics curriculum to implement the NCTM Standards by emphasizing applications, reading and writing, problem solving, and technology. All major content strands are integrated throughout each level of this innovative six-year curriculum. Carefully refined through years of field testing and user feedback, UCSMP enables students to learn by doing today's mathematics in a variety of meaningful situations. Author : Sharon L. Senk ISBN : 0673459268 Language : English No of Pages : 935 Edition : 2nd Publication Date : 1998 Format/Binding : Hardcover Book dimensions : 10.1x8.3x1.6 Book weight : 0.04 | 677.169 | 1 |
Description
YourTeacher's COMPASS math app is a complete COMPASS math test prep course with your own personal math teacher!
Browse or search over 140 COMPASS math lessons. Every lesson features video examples, interactive practice, multiple-choice self-tests, worksheets, and more!
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-Interactive practice problems with built-in support
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-A Challenge Problem
(similar to how a teacher assigns a higher level problem which students must work on their own to prove mastery)
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YourTeacher's COMPASS math prep course covers the exact math you need to ace the COMPASS.
CHAPTER 1: INTEGERS
Addition and Subtraction
Multiplication and Division
Word Problems
Order of Operations
Evaluation
Absolute Value
CHAPTER 2: FRACTIONS
Divisibility Rules
Factors and Primes
Prime Factorization
Multiples and Least Common Multiples
Greatest Common Factor
Introduction to Fractions
Equivalent Fractions (Part I)
Lowest Terms
Equivalent Fractions (Part II)
Improper Fractions and Mixed Numbers
Least Common Multiple
Addition and Subtraction
Multiplication and Division
Order of Operations
Adding Mixed Numbers
Subtracting Mixed Numbers
Multiplying Mixed Numbers
Dividing Mixed Numbers
CHAPTER 3: DECIMALS
CHAPTER 4: EXPRESSIONS & EQUATIONS
CHAPTER 5: INEQUALITIES & FUNCTIONS
CHAPTER 6: LINEAR EQUATIONS & SYSTEMS OF EQUATIONS
CHAPTER 7: EXPONENTS & POLYNOMIALS
CHAPTER 8: FACTORING & RATIONAL EXPRESSIONS
CHAPTER 9: RADICALS
CHAPTER 10: RATIO, PROPORTION, & PERCENT
CHAPTER 11: ANGLES & TRIANGLES
CHAPTER 12: SIMILARITY
CHAPTER 13: RIGHT TRIANGLES & CIRCLES
CHAPTER 14: MEASUREMENT
CHAPTER 15: STATISTICS
(Wifi or 3G connection required)
COMPASS is the registered trademark of ACT, Inc. YourTeacher has no affiliation with ACT, Inc., and the YourTeacher COMPASS app is not approved or endorsed by ACT | 677.169 | 1 |
Algebra Survival Guide
4.11 - 1251 ratings - Source
Presents questions and answers covering the fundamentals of algebra along with step-by-step instructions for solving algebraic operations and a variety of practice problems.Presents questions and answers covering the fundamentals of algebra along with step-by-step instructions for solving algebraic operations and a variety of practice problems.
Title
:
Algebra Survival Guide
Author
:
Josh Rappaport, Sally Blakemore
Publisher
:
- 2000-01-01
ISBN-13
:
Continue
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Pattern in the Teaching and Learning of Mathematics
4.11 - 1251 ratings - Source
This highly illustrated book draws together the wide variety of studies in the learning of mathematics undertaken by the Pattern in Mathematics Research Group at the University of Leeds. Their purpose has been '... to provide structure and support to ... studies of children's perception, conception and use of pattern in learning mathematics'. Set up in 1992, they have embraced work across the whole curriculum, and through all the years of compulsory schooling. As each chapter of this book relates to a different study that was undertaken, the reader can dip in and select relevant material. At the same time, the editor has ensured continuity and progression, allowing the book to be approached as a whole: the early chapters are concerned with very young children; subsequent chapters deal with the primary and middle age ranges, and later ones relate to secondary school work. With individual chapters relating to number, algebra, shape, graphic relations and probability, this new volume provides guidance for teachers of pupils of all age groups. Patterns in mathematics are of immense importance; this book relates pattern to the teaching of mathematics through all years of school. Practical and original, it is closely tied to the National Curriculum. It is a source of new ideas for mathematic teachers at all levels.If random trials are taking place with a fixed probability of success at each trial,
then we model the behaviour by the binomial distribution ... However, many of the
same people, when faced with the toss of a specific coin, will have a favourite call
and will nearly always make that call. ... classes of pupils who were towards the
end of Year 10 (fifteen-year-old pupils) to what A level mathematics might be like.
Title
:
Pattern in the Teaching and Learning of Mathematics
Author
:
Anthony Orton
Publisher
:
A&C Black - 2004-10-01
ISBN-13
:
Continue
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Summary and Info
Most volumes in analysis plunge students into a challenging new mathematical environment, replete with axioms, powerful abstractions, and an overriding emphasis on formal proofs. This can lead even students with a solid mathematical aptitude to often feel bewildered and discouraged by the theoretical treatment. Avoiding unnecessary abstractions to provide an accessible presentation of the material, A Concrete Introduction to Real Analysis supplies the crucial transition from a calculations-focused treatment of mathematics to a proof-centered approach. Drawing from the history of mathematics and. Read more... Abstract: Offers the crucial transition from a calculations-focused treatment of mathematics to a proof-centered approach. Drawing from the history of mathematics and practical applications, this book uses problems from calculus to introduce themes of estimation, approximation, and convergence. Read more... | 677.169 | 1 |
ISBN 13: 9780007212439
Do Brilliantly! Revision Guide - KS3 Maths
This highly successful book continues to provide the most comprehensive appoach to test success with its unique combination of study support and test practice. It also contains the latest Test Marker's comments.
Written by chief markers and highly experienced teachers, Revision Guide KS3 Maths provides an authoritative but practical approach to revising for the KS3 National Tests. It is divided into short revision sessions organised by level so students can use the book wherever they are in their KS3 course.
'Check yourself' sections throughout ensure revision is active and effective. Test questions are provided at every level with model answers and practical advice from test markers on how to boost grades.
"synopsis" may belong to another edition of this title.
About the Author:
Kevin Evans is Head of Mathematics at Abbey Grange C of E High School, Leeds. He is a Senior Moderator and Senior Examiner for GCSE Mathematics for a major examining board and a Senior Marker for Key stage 3 Mathematics.
Keith Gordon is Head of Mathematics at Wath Comprehensive School, Rotherham and is Principal Examiner for GCSE Mathematics for a major examining board.647
Book Description Collins 01/0112439
Book Description Collins 01/0112439
Book Description Collins, 2006773136
Book Description Collins247774 | 677.169 | 1 |
Writing Proofs in Analysis
Authors: Kane, Jonathan M. Teaches how to write proofs by describing what students should be thinking about when faced with writing a proof Provides proof templates for proofs that follow the same general structure Blends topics of logic into discussions of proofs in the context where they are needed Thoroughly covers the concepts and theorems of introductory in Real Analysis including limits, continuity, differentiation, integration, infinite series, sequences of functions, topology of the real line, and metric spaces
This is a textbook on proof writing in the area of analysis, balancing a survey of the core concepts of mathematical proof with a tight, rigorous examination of the specific tools needed for an understanding of analysis. Instead of the standard "transition" approach to teaching proofs, wherein students are taught fundamentals of logic, given some common proof strategies such as mathematical induction, and presented with a series of well-written proofs to mimic, this textbook teaches what a student needs to be thinking about when trying to construct a proof. Covering the fundamentals of analysis sufficient for a typical beginning Real Analysis course, it never loses sight of the fact that its primary focus is about proof writing skills. This book aims to give the student precise training in the writing of proofs by explaining exactly what elements make up a correct proof, how one goes about constructing an acceptable proof, and, by learning to recognize a correct proof, how to avoid writing incorrect proofs. To this end, all proofs presented in this text are preceded by detailed explanations describing the thought process one goes through when constructing the proof. Over 150 example proofs, templates, and axioms are presented alongside full-color diagrams to elucidate the topics at hand. | 677.169 | 1 |
Student Study Guide for use with Elementary Statistics: A Step By Step Approach: A Step By Step Approach - new book
Allan G. Bluman is Professor of Mathematics at Community College of Allegheny County, near Pittsburgh. For the McKeesport and New Kensington Campuses of Pennsylvania State University, he has taught teacher-certification and graduate education statistics courses. Prior to his college teaching, he taught mathematics at a junior high school.Professor Bluman received his B.S. from California State College in California, Penn.; his M.Ed. from the University of Pittsburgh; and, in 1971, his Ed.D., also from the University of Pittsburgh. His major field of study was mathematics education.In addition to Elementary Statistics: A Step by Step Approach, Third Edition, and Elementary Statistics: A Brief Version, the author has published several professional articles and the Modern Math Fun Book (Cuisenaire Publishing Company). He has spoken and presided at national and local mathematics conferences and has served as newsletter editor for the Pennsylvania State Mathematics Association of Two-Year Colleges. He is a member of the American Statistical Association, the National Council of Teachers of Mathematics, and the Mathematics Council of Western Pennsylvania.Al Bluman is married and has two children. His hobbies include writing, bicycling, and swimming. Allan Bluman, Books, Science and Nature, Student Study Guide for use with Elementary Statistics: A Step By Step Approach: A Step By Step Approach Books>Science and Nature
Bluman, Allan G.:
[EAN: 9780073048260], Used, good, [PU: McGraw-Hill Higher Education], Mathematics|Probability & Statistics, Mathematics|Probability & Statistics | 677.169 | 1 |
New Essential Mathematics for GCSE
4.11 - 1251 ratings - Source
A revised edition of this text with explanations, worked examples and exam questions to cover GCSE Maths in one year.You will be charged interest on any amount which you have not paid back within
a specified period. At the end of each month the credit card company will send
you a statement. The statement tells you how much you owe and the minimumanbsp;...
Title
:
New Essential Mathematics for GCSE
Author
:
David Kent
Publisher
:
Heinemann - 1997
ISBN-13
:
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Algebra Teacher's Activities Kit
150 Activities that Support Algebra in the Common Core Math Standards, Grades 6-12
Overview
Help your students succeed with classroom-ready, standards-based activities
The Algebra Teacher's Activities Kit: 150 Activities That Support Algebra in the Common Core Math Standards helps you bring the standards into your algebra classroom with a range of engaging activities that reinforce fundamental algebra skills. This newly updated second edition is formatted for easy implementation, with teaching notes and answers followed by reproducibles for activities covering the algebra standards for grades 6 through 12. Coverage includes whole numbers, variables, equations, inequalities, graphing, polynomials, factoring, logarithmic functions, statistics, and more, and gives you the material you need to reach students of various abilities and learning styles. Many of these activities are self-correcting, adding interest for students and saving you time.
This book provides dozens of activities that
Directly address each Common Core algebra standard
Engage students and get them excited about math
Are tailored to a diverse range of levels and abilities
Reinforce fundamental skills and demonstrate everyday relevance
Algebra lays the groundwork for every math class that comes after it, so it's crucial that students master the material and gain confidence in their abilities. The Algebra Teacher's Activities Kit helps you face the challenge, well-armed with effective activities that help students become successful in algebra class and beyond.
About Gary Robert Muschla, Judith A. Muschla, Erin Muschla-Berry
Unfortunately, each of our databases associated with books not really identified details about this author Gary Robert Muschla. Nevertheless all of us is obviously diligence to seek out and also include new details. Knowing the information about me, you can the item over the sort to add a critique. | 677.169 | 1 |
Math&Brain
4.11 - 1251 ratings - Source
Learning should be fun and easy with the right methods and creativities. By harnessing all the experiences with some advice from the professionals, I think MathaBrain can help others to learn faster and create your own attitude of learning and thinking.MathaBrain are the combination of simplicity, conceptual, creativity and the free way of thinking in finding a solution. By boosting our ability in memorizing at the same time developing creativity in finding solutions in learning process will grows very quickly. By keep a strong understanding in concept we will be able to create more creative solutions to the same problems. While looking to the problems in different ways make us easily understand which solutions will be the best. The same concept can be applied to any problems that we faced in our daily life. Creative thinking will be one of the most powerful tools to help us succeed in whatever we do since it will allow creating the effective solutions.next book or somewhere else maybe in some blogs or websites. So we need to
make sure that we can do the simples exercise from this book following the
instruction given in the manual section. Manual a€" important part in order to do the
anbsp;...
Title
:
Math&Brain
Author
:
Ibrahim Mat Nor
Publisher
:
Ibrahim Mat Nor - 2010-02-24
ISBN-13
:
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Main objective of this lecture is to present on Math Vocabulary. Knowledge of math vocabulary is an essential component of learning mathematics. In order to communicate math thinking clearly and coherently students need to learn and use appropriate math vocabulary. If we want students to use the language of mathematics precisely it is important that that we model appropriate language in context, both verbally and visually | 677.169 | 1 |
Linear Systems: LU Decomposition
In this LU decomposition instructional activity, students multiply matrices, identify the elimination matrices, and the inverse of a matrix. This three-page instructional activity contains approximately six problems. | 677.169 | 1 |
Caroline Duksta
Kathryn Mason
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This 3 page Study Guide was uploaded by Rooshna Ali on Monday April 4, 2016.
The Study Guide belongs to 20123 at Texas Christian University taught by Susan Staples in Winter 2016.
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Discrete Mathematics I in
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Reviews for Discrete Mathematics 2 Study Guide (Sections 10-12, 19, 20) Sets o Set- A repetition: free unordered collection of objects o Subset- If we have two sets, A and B. We say A is a subset of B, provided every element of A is in B Notation: A ⊆ B o Proper subset- If we have two sets, A and B. We say A is a subset of B, provided not every element of A is in B (Can't be exact same) Notation: A ⊂ B o Element of a set- when an object is in a set Notation: ∈ o Cardinality of a set- the number of of element in a set S Notation: | 677.169 | 1 |
Popular in Mathematics (M)
Reviews for HonorsCLASS NOTES WEEK 1 1 COMPLEX NUMBER SYSTEM In this class we will use the following common number systems 0 N 12345 natural numbers 0 Z 0 i1i2i3 i4 integers 0 Q a E Zb E Zb 7E 0 rational numbers 0 R real numbers 0 C complex numbers Note N C Z C Q C R C C where C means is contained in77 or is a subset of7 These number systems can be characterized by operations under which the number system is said to be closed7 closed under closed under 7 closed under 7 gtlt R is closed under limits C is algebraically closed as we will explain below Before explaining where C comes from let7s consider an important property of Q namely that it is algebraically closed over linear equations of the form axb0 Algebraically closed means that if ab E Q then there exists a solution x E Q to the above equation Now try to solve the quadratic equation 2 7 2 0 This has no rational solution Q is not algebraically closed over quadratic polynomials However we can insert a number xi into our number system to get a new number system consisting of numbers of the form a lnE where a b E Q This number system has 7 gtlt ab i cdx aic bid a b c d ac 2bd ad bc ab i ab cid i aci2bd bciad Cd i 0272d2 7amp7de 0272d2 39 ls xi a number or just a placeholder In some sense there is no di erencel Now we know that even all the real numbers aren7t algebraically closed over all polynomials We still can7t solve for instance 2 1 0 So taking the above strategy lets make up a number called i V71 Then we get a new number system C de ned as CabiabER For a complex number 2 abi a is called the real part and b is call the imaginary part note these are both real numbersl We also write a Rez and b lmz 1 2 CLASS NOTES WEEK 1 We can perform operations on complex numbers like so ab2 i 0d2 aic bjd27 a b2e d2 ac 7 bd be ad2 ab2 ab2e7d2 ae7bd be7ad ed2ed2e7d202d2 02d2239 11 Examples Number 127 p 5 in the handout asks us to calculate 2 271723 7 22 72 7 22 7 2 13 7 22 71 7 322 7 22 72 2239 7 62 7 6 78 7 42 Number 157 p 5 asks if k is an integer7 show that file 17 4k1 4k2 717 4k3 Note that 2390 121 222 7123 7239 and 2394 42 722 771 1 So 24k 24 1k 1 and the rest follow immediately lntroduction to proofs Number 297 p 5 Prove that there is no rational number x that satis es 22 2 Well7 suppose z pq7 where p21 are integers We can write p and q in lowest terms7 so they don7t have a common factor But 102212 27 so p2 2212 Then p must be even7 so p 27quot for some integer r Thus7 422 22127 so 12 222 That means q is even7 too7 which means p and q must share a common factor This is a contradiction 2 7239 2 GRAPHING COMPLEX NUMBERS We can use a Cartesian plane to plot complex numbers lf 2 z 2217 then the point 272 on the Cartesian plane represents 2 We call it the compleco plane We can de ne the absolute value or modulus of a complex number by measuring its length in the complex plane That is7 if z a b2 then the modulus is given by lzl Va2 b2 Eg 39 1 1374219165 In general7 Z 0 and 0 if and only if z 0 The distance between two complex numbers 21 and 22 is 21 7 22 lol 0 21 Examples of Graphs Basic example a circle is given by z l2 7 20 7 More examples Example 2 on p 9 asks us to nd the set of points z satisfying a lz2l 7 12711 b l2 71 Rez1 To graph a7 note what the equation is saying geometrically 2 is equidistant from 72 and 1 This gives a vertical line through the point 712 One could also get this by doing some algebra lz2l2 271l2 means that ifzab2 then a22b2a7 12b2 or a24a4a272a17 or 6a737 or a 712 To graph b7 let 2 a 1 b2 this gives a712b2 a 17 or a 712 b2 a 12 or 4a bz7 or a b24 This gives us a sideways parabola CLASS NOTES WEEK 1 3 Number 7f asks as to graph lzi1l lz1l 7 Thus the distance from 1 plus the distance from 71 is a constant this gives an ellipse with foci at i1 Letting z x y we have lzillz 77 lz1l2 49714l21ll21l2 implying x712y2 49714l21lx12y2 so 49714l21l4x 0 or 49 4x 14lz 1 Then 492 392 16x2 196 12 yz 196x2 392 196 19612 Thus 180z2 19612 2205 which is now more clearly the equation of an ellipse 22 Complex Conjugate The eomplem conjugate of a complex number 2 a be is 2 a 7 in For example 715 7175 7239 38 Properties 21i22 72 2122 2122 The proof of the second property takes a little bit of computation Let 21 a1 he 22 a2 be Then 2122 0102 7 5152 1152 azbi 1102 7 b1b2 7 1le agbl On the other hand 2122 11 7 12 7 0102 7 b1b2 7 0152 a2b1z39 This completes the proof Additionally we have lt220gt 22 22 Rezi 2 272 1 m2 2 372 lzll5l 222l2 These facts can be used to solve problems such as rationalizing the denominator 21 21 2172 22 225 l22l239 In particular 1 2 Z 7 My so that the reciprocal of a complex number 2 is parallel to the complex conjugate of 2 Some more useful properties From Problem 9 p 13 if 2 0 is a real number then lrzl This is easy to see directly if z a M then lrzl 771 Tb2 7 2a2 b2 rxaz b2 From Problem 10 p 13 lRezl g and llmzl g We7ll prove the rst the second is similar lf 2 a in then lRele lalz a2 S a b2 M2 take square roots to get the result From Problem 17 p 13 if a1a2 an are real coef cients then the polynomial z alzn l agzn z A can has roots that come in conjugate pairs real roots are their own conjugates To see this just take the conjugate of the polynomial using the properties mentioned above 4 CLASS NOTES WEEK 1 3 VECTORS AND POLAR FORMS A vector in the plane is given by two coordinates We can think of complex numbers as vectors in the plane Thus complex number addition can be performed by the parallelogram law77 note this is not the normal meaning of parallelogram law Take two vectors form a parallelogram the sum of the two vectors is the resulting diagonal across the parallelogram starting from the tail vertex common to the two vectors Since complex numbers can be treated as vectors in the plane we have the triangle inequality for complex numbers lzi 22l S lzil There is another side of this inequality namely llzll lzzll S 121 i 221 Exercise deduce the second inequality from the rst Example 1 on p 15 asks prove that the three distinct points 2122 and 23 lie on the same straight line if and only if 23 7 22 c22 7 21 for some real number c For the if77 direction let7s assume 23 7 22 c22 7 21 for some real number c Let 20 22 7 21 so that 22 20 21 Then 23 C20 22 c 120 21 so 22 and 23 are both on the line r20 21 through 21 For the only if77 direction if 21 2223 are collinear then there exists 20 such that 23 r320 21 and 22 r220 21 Thus 23 7 22 r3 7 r220 7 2 22 7 21 Let c 732 and were done A vector can also be given by its magnitude and direction We already have a notion of magnitude for a complex number its modulus If we identify a useful notion of direction we can characterize complex numbers in this way Cartesian coordinates zy can be converted to polar coordinates r 9 by solving the following equations z rcos97 y rsin9 For a complex number 2 2 iy we have T and we de ne arg2 9 Note however that 9 is only unique up to integer multiples of 27139 For example the values of argi are 7T 7T 7T 77i27T7i47T 2 2 2 so we write 7T argi 5 21 k E Z So arg2 has in nitely many values We wish to pick a de nite value The customary choice is in 7n7T and we call this Arg2 this is called the principal branch of the argument of 2 Any branch of arg2 is discontinuous along its branch cut77 having a jump of 27139 there If you wish you can use the notation arg 2 to mean the branch taking values in 77 27f eg arg7r Arg Note arg0 never makes sense since literally any angle would work We can now write complex numbers in polar form 2 z iy rcos9 isin9 rcis9 We will quickly see that the notation cis77 is unnecessary Examples 7b on p 22 asks us to convert 2 73 3239 into polar form Well r2 32 32 18 so r 3x2 So 2 7 i which points in the direction 9 37r4 So 73 3239 3x2cis37T4 7d on p 22 asks the same about 2 72 7 2239 for which r2 12 4 16 so r 4 Thus 231 77 7 5239 which corresponds to 9 77r6 CLASS NOTES WEEK 1 5 Adding complex numbers in polar form is inconvenient It can be done7 however Problem 217 p 24 if 7quotcis rlcisdl 772cis1927 we wish to nd 7 and 9 in terms of 771772191192 We have r2 71 cos 91 r2 cos 922 71 sin 91 r2 sin 922 7 cos2 91 2771772 cos 91 cos 92 7 cos2 92 7 sin2 91 2771772 sin 91 sin 92 7 sin2 92 7 7 2r1r2cos 91 cos 92 sin 91 sin 92 7 7 7 2771772 cos 1 7 92 We also have 71 sin 91 r2 sin 92 tan 71 cos 91 7 r2 cos 92 That7s all I will attempt here Now for multiplication we have it much better if 21 r1cos 91 739sin 91 and 22 r2cos 92 739sin 92 then 2122 r1r2cos 91 cos 92 7 sin 91 sin 92 7sin 91 cos 92 cos 91 sin 92 r1r2cos 1 92 739sint91 92 Thus7 mutiplying two complex numbers is the same as multiplying the moduli and adding the angles 2122ll21ll22l arg 2122 argzL arg22 Case study the powers of 739 are 17717717 These are just rotations of 7T2 around the complex plane Division is the inverse of multiplication and thus works in an analogous way 21 7 T1 coswl 7 92 S1n91 7 92 2 22 2 arg 71 argzL 7 argzz 22 21 22 31 Examples Problem 7h asks us to write 7W1 239 V3 7 in polar coordinates The modulus of the top is v 14 The modulus of the bottom is 2 So the modulus of the quotient is 142 The principal argument of the top is 737T47 and the principal argument of the bottom is 7r6 So the argument of the quotient is 737r477r6 797r12 7 27T12 7117T12 14 So the answer is Tcos7117r12 7sin7117r12 Problem 11 on p 23 says7 using the complex product 1 75 7 73947 derive the formula 7r4 4tan 115 i tan 11239 6 CLASS NOTES WEEK 1 Using Cartesian coordinate multiplication we just get 1 25 724 1 225 7 102212 1 224 i102 1 2576 i 480239 i 100 1 239476 i 480239 476 i 480239 4762 480 956 i 42 Thus arg1 2 4 arg5 7 2 arg956 7 42 using the product rule for arguments In particular then n4 4tan 1715 tan l71239 Since tan is an odd function the conclusion follows 4 THE COMPLEX EXPONENTIAL In order to rigorously de ne the complex exponential function 61 we are going to have to borrow from more advanced analysis one way or another For instance one could simply accept from the the ory of complex differentiation that there exists a function fz such that f z Alternatively lets just de ne I ez em y emcos y 2 sin We can easily verify that this is its own derivative with respect to each of the variables x and 22 that is 16 Z39Eiy y d7cosy 2s1ny isiny 2cosy 2cosy 2s1ny 1 This function has all the familiar properties of exponentials 621622 61112 611 76 612 21722 We also have consistent Taylor series expansions 22 3 24 5 1 E 22 x4 cosx17 gi 23 x5 s1n2zig 7 Just take the power series for em and apply it at x 2y 2 3 4 2y nini i e 712y 2 23X4X 2 4 3 5 n 1 yin 39 1 7i 1 2W4 HG 3W5 cosy 2siny This relation is referred to as Euler s Equation Note these functions are in fact analytic on the entire complex plane so these power series equations are literally true Justifying this assertion however requires some more advanced complex function theory CLASS NOTES WEEK 1 7 Now we have a new shorter polar representation 2 rem where r and 9 arg 2 Notice that 62 is 2W2 periodic That7s because 62quot 1 for every k E Z Famous example 5 1 0 This equation relates ve of the most famous constants in mathematics Useful identities 0039 We 2 29 72399 sin9 lmew 6 76 22 T162191 T262192 T1T26i9192 w 7151 2 51101762 736192 r2 72399 rem re Another interesting identity is known as De Moivre s formula cos 9 239sin 9 cos 729 2 sin 729 This is easily proved using the facts about 62 we have already seen yew 620620 62001 times 6262020 Ema Now apply Euler7s Equation 41 Examples Problem 3c7 p 31 1 06 ew4f 66267 seam239 Problem 4b p 31 2 2239 2925M fmm i 2 26257r6 Problem 127 p 31 Use DeMoivre7s formula and the binomial formula to derive the following a sin 39 3 cos2 9 sin9 7 sin3 9 b sin 49 4 cos3 9 sin9 7 4 cos 9sin3 9 For the rst7 note that cos39 2 sin39 cos 9 2 sin 93 cos3 9 32 cos2 9sin 9 7 3cos 9 sin2 9 7 239sin3 9 Taking imaginary parts7 we get a For the second7 note that cos49 239sin49 cos 9 239sin 94 cos4 9 42cos3 9 sin9 7 6 cos2 9 sin2 9 7 4239 cos 9 sin3 9 sin4 9 Once again7 taking imaginary parts7 we get Problem 207 p 32 For 2 7E 17 we have the classic geometric series formula z 1 7 1 1z22z 271 8 CLASS NOTES WEEK 1 Let 2 6w cost 2sint9 Then we get by DeMoivre7s formula 1 cos cos 219 cos n0 2sin sin 261 sinnt9 62n19 7 1 6w 7 1 62n192 7 672n192 T 5202 7 57202 i 22 sinn 1 2 22sin 2 i sinn 1 2 T sin 2 Taking imaginary parts we get the formula 622202 cosn 2 2 sinn 2 cosn 2 2 sinn 2 sinn 2 sinn 1 2 s1n s1n2 s1nn SinwQ Taking real parts we get the formula 9 2 1 t9 2 11003910032911003n9 W s1n 2 We can also write Ema2 me2 62n102 7 Maw2 cosn 2 sinn 1 2 2 22 Ema2w 7 67202 7 6202 7 672n1292 4239 6202 7 67202 62n120 7 672n1202 4239 4239 sin 2 sinn 12 Tf39 So we can write 1 12 sin n 1cos cos2 cosn i W 5 POWERS AND ROOTS By DeMoivre7s formula if z rem rcos 9 2sin 9 then 2 Name r cos n19 2 sin n19 Now lets say we want to nd complex numbers C such that C 2 lf 2 rem then by the above formula we have that C Walen satis es C 2 But recall that 62 is 27T2 periodic That is z 28099 for any k E Z Thus C Weiw mln will work We need only take k 01 n 7 1 because after that the roots will repeat A complex number 2 rem therefore has n nth roots Wyen7 n T62lt027rn7 39 39 39 7 w62027rn71n39 Eg the 4th roots of 1 are 127 71 and 72 or 606i39r26M and elm2 In general 1 is also called unity77 so the nth roots of 1 are called the nth roots of unity 21m 2k 11 512k cosi2sini 201m71 n 77 We denote by run emfn the principal nth mat of unity and all the nth roots of unity are given by 2 n 1 Laumwmmwn CLASS NOTES WEEK 1 9 51 Facts and examples Example 1 on p 36 Show that 1 7 run 7 w 7 A 7 wg l 0 This one is interesting to see visually take all the nth roots of unity on the complex unit circle and add them together their center of mass is clearly the origin To see this analytically7 just use the standard geometric series trick 1711 7wn7wg77w271 17w 0 But as long as tun 3A 1 which is true for any n gt 1 this implies the result Problem 117 p 37 Solve z 715 25 This could be pretty tricky if we just jumped into it algebraically Lets see if we can use roots of unity to solve Divide both sides by 25 to get 17125 1 That means 1712 emf 5 for k 0172374 That means the solution is 1 z k01234 Pretty quick Problem 5c7 p 37 Find all values of i14 Well7 i 6quotquot27 so its 4th roots are eiquot82kquot4 k 0 17 2 3 Problem 5d7 p 37 Find all values of 1 7 33913 Well7 1 7 3i 26 M37 so its 3rd roots are W6i77r972k7r339 Quadratic formula Example 3 on p 36 Let abc be complex constants7 1 31 0 Show that the solutions of the equation 122 7 b2 7 c 0 are given by the usual quadratic formula 2 7 7bixb2 746w 7 2a 7 where vb2 7 4ac represents one of the two values of b2 7 4ac12 The solution here is the same as in the real number case It involves completing the square A very clean way to do it is to multiply by 4a7 then add b2 7 4ac the discriminant to both sides This gives 41222 7 4abz 7 b2 b2 7 4ac7 or 2oz 7 b2 b2 7 4ac Taking square roots7 we7re done 52 Introduction to Induction Problem 17 p 37 asks us to prove DeMoivre7s formula by induc tion That is7 we show that cos 9 7 i sin9 cos n9 7 i sin n9 for n 1727374757 The principle of induction is this First7 show that the statement holds for the rst n here for n 1 this is a trivial case Then7 show that if the statement holds for n7 then it also holds for n 71 Then it must hold for n 17 237 The case n 1 is trivial To show how the proof is supposed to proceed in a concrete way7 let7s show that the formula holds for n 2 Indeed7 we have cos 9 7 i sin92 cos 9 7 isin 9 cos 9 7 i sin 9 cos2 9 7 sin2 9 7 icos 9 sin9 7 sin 9 cos 9 cos 29 7 isin | 677.169 | 1 |
Pell's Equation
4.11 - 1251 ratings - Source
Pell's equation is part of a central area of algebraic number theory that treats quadratic forms and the structure of the rings of integers in algebraic number fields. It is an ideal topic to lead college students, as well as some talented and motivated high school students, to a better appreciation of the power of mathematical technique. Even at the specific level of quadratic diophantine equations, there are unsolved problems, and the higher degree analogues of Pell's equation, particularly beyond the third, do not appear to have been well studied. In this focused exercise book, the topic is motivated and developed through sections of exercises which will allow the readers to recreate known theory and provide a focus for their algebraic practice. There are several explorations that encourage the reader to embark on their own research. A high school background in mathematics is all that is needed to get into this book, and teachers and others interested in mathematics who do not have (or have forgotten) a background in advanced mathematics may find that it is a suitable vehicle for keeping up an independent interest in the subject.... and Xavier Saint-Raymond Problem Book for First Year Calculus by George W.
Bluman Exercises in Probability by T. Cacoullos Probability Through Problems by
Marek CapiA"ski and Tomasz Zastawniak An Introduction to Hilbert Space andanbsp;...
Title
:
Pell's Equation
Author
:
Edward J. Barbeau
Publisher
:
Springer Science & Business Media - 2006-05-04
ISBN-13
:
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Fully updated and thoughtfully reorganized to make reading and locating material easier for instructors and students, the Sixth Edition of this bestselling, classroom-tested text:
Adds more than 160 new exercises
Presents many new concepts, theorems, and examples
Includes recent major contributions to long-standing conjectures such as the Hamiltonian Factorization Conjecture, 1-Factorization Conjecture, and Alspach's Conjecture on graph decompositions
Supplies a proof of the perfect graph theorem
Features a revised chapter on the probabilistic method in graph theory with many results integrated throughout the text
At the end of the book are indices and lists of mathematicians' names, terms, symbols, and useful references. There is also a section giving hints and solutions to all odd-numbered exercises. A complete solutions manual is available with qualifying course adoption.
Graphs & Digraphs, Sixth Edition remains the consummate text for an advanced undergraduate level or introductory graduate level course or two-semester sequence on graph theory, exploring the subject's fascinating history while covering a host of interesting problems and diverse applications.
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This group will develop 1-3 lessons that use the statistical software FATHOM as a tool for understanding the mathematics and the application of that mathematics in contexts that require data analysis, probability and statistical concepts. The goal is to write lessons and problems of outstanding caliber, having deep mathematical content but accessible at both the high school and middle school levels. The group will explore the use of the Math Forum's Problem of the Week environment, PCMI Problem, as a way to involve students with the concepts relevant. | 677.169 | 1 |
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Date Published:2012.
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Homework Helpers: Basic Math and Pre-Algebra will help build a solid mathematical foundation and enable students to gain the confidence they need to continue their education in mathematics. Particular attention is placed on topics that students traditionally struggle with the most. The topics are explained in everyday language before the examples are worked. The problems are solved clearly and systematically, with step-by-step instructions provided. Problem-solving skills and good habits, such as checking your answers after every problem, are emphasized along with practice problems throughout, and the answers to all of the practice problems are provided. Homework Helpers: Basic Math and Pre-Algebra is a straightforward and easy-to-read review of arithmetic skills. It includes topics that are intended to help prepare students to successfully learn algebra, including: Working with fractions Understanding the decimal system Calculating... | 677.169 | 1 |
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A series of performance assessments and rubrics designed to help students create a system of linear equations (in 2 and 3 variables) to solve for the number of bicycles, tricycles, tandems in a shop.
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Mathematics > General
Education > General
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They Cycle Shop is curriculum embedded, high school algebra tasks with instructional support. The instructional support includes a unit with learning activities and formative assessments. The unit is designed to help educators understand and implement tasks that are embedded in Common Core-aligned curricula. Student solution examples and grading explanations are provided with the goal to encourage analysis of student and teacher work to understand what alignment looks like. This unit is well organized and and can easily be used in the classroom. It also provides scaffolding for students with disabilities and English language learners.
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A series of performance assessments and rubrics designed to help students create a system of linear equations (in 2 and 3 variables) to solve for the number of bicycles, tricycles, tandems in a shop. | 677.169 | 1 |
Barron's Math Workbook for the GED
4.11 - 1251 ratings - Source
Adults preparing to take the GED High School Equivalency Test, and who feel the need for extra help in math, will find what they are looking for in this self-teaching workbook. The text has been updated in this edition to reflect latest changes in exam format and to familiarize students with use of the Casio fx-260 calculator. The book is filled with subject reviews, exercises, and worksheets covering arithmetic, measurement, geometry, algebra, number relations, and data analysis. All questions are answered, and a full-length diagnostic test plus four practice tests will help students discover their weak areas for concentrated study.At the end of the practice test there is a chart that lets you relate the questions you
missed to the topic in the book that deals ... Remember that in order for you to
pass the math section of the GED, you do not need to be able to answer every
question. ... Instructions on the correct use of grids were provided on pages 50-56
.
Title
:
Barron's Math Workbook for the GED
Author
:
Johanna Holm
Publisher
:
Barron's Educational Series - 2003
ISBN-13
:
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This guide is designed to help classroom teachers implement the Alabama Course of Study: Mathematics K-12. It is inclusive also of the objectives tested by the Stanford Achievement Tests and the Alabama Basic Competency Tests. One characteristic of the curriculum guide is that it clearly states what students should learn in each grade level. These objectives are called student outcomes and appear in the table of contents. The majority of the material, however, consists of classroom activities that lead students to the desired outcomes. The curriculum guide specifies what to teach as well as ways to teach it. Each section of the guide contains a specific table of contents, which lists what students should learn; activities; original sources; and student outcomes. Activities are preceded by a description of the area of mathematics covered, the specific student outcomes that are addressed in the activities, and notes for the teacher, which convey information important for effectively carrying out the activities. Topics include computation, estimation, approximation, powers, square roots, decimals, fractions, percent, integers, geometry, linear equations, inequalities, metric system, measurement, probability, statistics, and consumer mathematics. (KR) | 677.169 | 1 |
Section 3.4 Section ll
1. Given [0!) = Z'VX + 1. list the basic graph and describe the transformations in the
appropriate order. Use this to complete an accurate sketch of the function, labeling the
coordinates of at least 3 points.
Solution:
-_. H.
Section 3.1 (Version A) de
Section 5.4 Properties of Logarithms _.
Objectives
9 Apply the properties of logarithms to expand expressions involving the |ogarithm Of a
product/quotient/ power into a sum/difference of logarithms.
' APP'Y the properties of logarithms to condense a sun/
ijectives
0 Determine whether'a function is exponential. x . .
Identify the characteristics of exponential functions of the form f(x) - b , InClUdmg
the domain, range, intercept, asymptote, end behavior, and general graphs.
Determine the formula for an e
Section 2.4 line and sketch
Section 2.4 (Version A) lin
Section 1.5 (Version A)
Math 112 Section:
Instructors name:
Name:
1. Suppose the base and height of a triangle sum to 24 cm, and the area of the triangle is 70 square cm. If the base is larger than the
height, what are the dimensions (base and height) of
Section 3.1 describing wha
Enter the last four digits of your CatCard number:
On your Scantron, MARK TEST FORM -> 12 (POST)
Time allowed:
60 minutes
Exam content:
This exam covers all material covered in the lectures and
assignments from Class 21 to Class 28.
Exam questions:
All qu
Amanda is the best math teacher I've ever had and she made math so exciting! She was goofy but a ton of fun and I looked forward to going to class every day to learn from her! She was easily the best math teacher I've ever had.
Course highlights:
The highlight of the course was Amanda's goofy antics. I learned so much about math and not just what to do on each problem, but why that was what we had to do. She explained everything in such depth.
Hours per week:
3-5 hours
Advice for students:
Be ready to study if math isn't your strong suit, but the work is very helpful and rewarding. | 677.169 | 1 |
Essential Skills in Maths - Students' Book 4 (Essential Numeracy)
This series is ideal for comprehensive reinforcement of essential skills to improve results and to develop a more thorough understanding. The five books provide effective practice and consolidation of key mathematical skills for 11-16 year olds. | 677.169 | 1 |
Explore More Items Similar to A Textbook of Trigonometry for JEE Main & Advanced
Overview: A Textbook of Trigonometry for JEE Main & Advanced
Mathematics is one of the biggest hurdles for the Joint Entrance Exam (JEE) aspirants. To simplify the subject and to make it more interesting, Arihant Publications has brought forth A Textbook of Trigonometry for JEE Main & Advanced. The book aids you in preparing well for trigonometry, from the simplest of the formulas to the most difficult of questions. Often when the students close their text-books thinking that they have understood a chapter well, the exam problems prove otherwise. To get a firm and a clear hold of the concepts, this Textbook of Trigonometry for JEE is divided into four chapters - Trigonometric Functions and Identities, Trigonometric Equations and Inequalities, Properties and Solutions of Triangles, and Inverse Trigonometric Functions. The design is such that it will help the students to not just learn the basics with ease, but also solve the most complex questions with amazing speed.
The author, Prof. Amit Agarwal, is renowned and has been teaching Mathematics to engineering students for long. With immense experience, the author knows the most common kind of errors made by the students. This seventh edition of A Textbook of Trigonometry for JEE Main & Advanced mentions such types of generic errors and tips on solving them. Most importantly, the book contains everything from the easiest to the most difficult kind of questions, making it ideal for both beginners and experts alike. Inclusion of fully solved question papers of JEE Main and Advanced of the previous years gives a clear insight to the students about what to expect from the oncoming exams. Get your copy of this latest edition of Arihant Publications, A Textbook of Trigonometry for JEE Main & Advanced in paperback format, and prepare methodically for best results.
Product Details
Language
English
Publication Date
May 7, 2014
Publisher
Arihant
Contributor(s)
Amit M. Agarwal
Binding
Paperback
Edition
Seventh
Page Count
338
ISBN 10
9351418464
ISBN 13
9789351418467
Dimensions and Weight
Product Weight
890 grams
Customer Reviews on A Textbook of Trigonometry for JEE Main & Advanced | 677.169 | 1 |
Designed as the bridge to cross the widening gap between mathematics and computer science, and planned as the mathematical base for computer science students, this maths text is written for upper-level college students who have had previous coursework involving proofs and proof techniques. The close tie between the theoretical and algorithmic aspects of graph theory, and graphs that lend themselves naturally as models in computer science, results in a need for efficient algorithims to solve any large scale problems. Each algorithm in the text includes explanatory statements that clarify individual steps, a worst-case complexity analysis, and algorithmic correctness proofs. As a result, the student will develop an understanding of the concept of an efficient algorithm. | 677.169 | 1 |
VSB Math
The Mathematics VSB Course will discuss topics in Algebra. The first lesson will lay the groundwork for all succeeding lessons, by familiarizing the user with definitions and notations that will be used throughout the | 677.169 | 1 |
Most of the equations governing the problems related to science and engineering are nonlinear in nature. As a result, they are inherently difficult to solve. Analytical solutions are available only for some special cases. For other cases, one has no easy means but to solve the problem must depend on numerical solutions.
Fluid Flow, Heat and Mass Transfer at Bodies of Different Shapes: Numerical Solutions presents the current theoretical developments of boundary layer theory, a branch of transport phenomena. Also, the book addresses the theoretical developments in the area and presents a number of physical problems that have been solved by analytical or numerical method. It is focused particularly on fluid flow problems governed by nonlinear differential equations. The book is intended for researchers in applied mathematics, physics, mechanics and engineering.
Addresses basic concepts to understand the theoretical framework for the method
Provides examples of nonlinear problems that have been solved through the use of numerical method
Focuses on fluid flow problems governed by nonlinear equations | 677.169 | 1 |
TAKE THE LIMIT WORDS BY: JOHN A. CARTER TUNE: "THE RUBBER TREE PLANT"
Say you really wanted to know, Where all of the y-values go, As you approach a value - a finite value of c. Then take the limit, Take the limit, And you will find out where all of those y-values go. So come on in from the left and Come in from the right And together you'll find What's the limit as x approaches, What's the limit as x approaches, What's the limit as x approaches c? Say you really wanted to know Where the slope of the secants will go, As you approach a fixed point on the graph of f. Then take the limit, Take the limit, And you will find out where the slope of the secants will go. So come on in from the left and Come in from the right And together you'll find What's the slope of the tangent at, What's the slope of the tangent at, What's the slope of the tangent at x? Say you really wanted to know, Where the graph off will go, As x increases without bound - toward infinity. Then take the limit, Take the limit, And you will find out where the graph of the function will go, So go on out to the left or Go out to the right And with either you'll find What's the horizontal asymptote, What's the horizontal asymptote, What's the horizontal asymptote of f?
1
DON'T THINK TWICE WORDS BY ALF ESTBERG AND DANE R. CAMP TUNE: "DON'T THINK TWICE, IT'S ALRIGHT"
It ain't no use to sit around and cry, guy You just didn't do your best, And any fool can tell you 'xactly why - - (sigh) You shoulda studied for the test! Now you wish there was something you could do or say To change that grade but there's just one way. You'll just have to do your homework everyday! Don't think twice, it's alright. Now I think it's time you turned on your light - - bright! Cuz the light will help you know What you've done wrong and how to get it right - - (right!) For your grade is way too low. So open your mind and go do the work, For the more that you do, the less you'll shirk, And no one cares that your teacher's a jerk. Don't think twice, it's alright. Now you're walking down the road to success. The future's in your hands For you've erased the memory of that mess - - (yes!) And you're squarely in command. So spread the word from here to Timbuktu, There ain't nothing anyone can do for you, That you can't do yourself....you know it's true. Don't think twice, it's alright.
2
BASIC GRAPHS WORDS BY: JOHN A. CARTER TUNE: "CAISSONS GO ROLLING ALONG"
Come join us as we graph. Use your arms and maybe laugh As we review the twelve curves you know. Quadratics, and the line, Y's a constant, and the sine, Y is e to the x - watch it grow, Here's the cube and the tangent too. Let's not forget all of the roots - square and cube and so on. Natural log and one over x, One on x squared is important too.
THE QUADRATIC FORMULA WORDS BY: JOHN A. CARTER TUNE: "AMAZING GRACE"
If you need to solve a quadratic Equation in any form, Set it equal to zero, And use this formula. You'll have the zeroes before you know. x equals the opposite of b Plus or minus the square root Of b squared minus 4a times c All over two times a.
The first thing you ask yourself, "is it even or odd degree?" In other words, do its ends do this or do they do one of these? (Hold arms up then down for even, alternate for odd.) (Refrain) The next thing you have to locate is the y-intercept. This ole point's not too hard to find, just sub zero in x. (Refrain) The next thing you have to find are the x-intercepts. Set each factor equal to zero, then solve each for x. (Refrain) The next thing you ask yourself, "Are these roots evens or odds?" Does it pass on through or 'sit bounce back from those x-intercept dots? (Refrain) The last thing you have to find where's the max/min in here? To answer this, you will need the Calculus. You'll learn this next year!
4
UNIT CIRCLE WORDS BY: J014N A. CARTER TUNE: "ON TOP OF OLD SMOKEY"
When you wrap a-round the unit circle, Given an arc's length find (cosine, sine) you will. When you wrap a zero, that's nothing I'll show, You're right where you start at one comma zero. When you wrap to one pi, you're half-way around. Negative one comma zero's the point where you're found. When you wrap a two pi, that's easy. Here's why: You're back where you started one, zero surprise. When you wrap to straight up, that's pi over two. One is your y here. Zero's your x-value. When you wrap on down to three pi over two, Negative one's now your y here and zero's x down here too. When you wrap on up to a pi over six, Root three over two's x and y is three-sixths. When you continue to pi over four, Root two over two's both x and y that's for sure. When you go on up to a pi over three, One-half is your x and y's one-half root three. When we go further to quadrants two, three And four all we need is to use symmetry.
5
SIERPINSKI'S GASKET WORDS BY: DANE R. CAMP TUNE: "MEIN HUT DER HAT DREI ECKEN" OR "MY HAT IT HAS THREE CORNERS"
My fractal has but three comers, Just count them: 1, 2, 3. And if you look at each comer, You cannot help but see, A fractal that has three comers... (Continue from line two.)
MORE MATH FOR ME! WORDS BY: DANE R. CAMP TUNE: "KING OF THE ROAD"
TI, you're heaven sent Straight from Texas Instruments. New cord and batteries, You make math class such a breeze. Two hours of plotting dots Gone with just one zoom-in box. Lose those anxieties! Get one of these. I know every button and every menu All of the functions and programming too! Every trick for getting around, And if you don't drop it, it won't make a sound. TI, now I know for sure, You're a primo calculator. You do the moron work so I have More math for me ... More math for me ... More math for me.
6
THE FINISHED PROOF WORDS BY: ALF ESTBERG'S GEOMETRY CLASS TUNE: "THE BRADY BUNCH THEME"
Here's the story of a lonely statement Who was proving congruent triangles on his own. All of them had angles and sides from the givens But they weren't done with the proof. Here's the story of a lovely reason Who knew SAS, ASA, and HL But was nothing without a statement And was in living hell. But 'til the one day when the statement met this reason And they knew they could live under the same roof'. That this problem would finally be complete And that's the way they became the finished proof. The finished proof, The finished proof, And that's the way they became the finished proof.
I LOVE TO LEARN MATHEMATICS WORDS BY: DANE R. CAMP TUNE: "TAKE ME OUT TO THE BALLGAME"
I love to learn math-e-ma-tics Like sol-ving e-qua-tions with x And all of the shapes in ge-o-me-try Or graph-ing a func-tion with tech-no-lo-gy For it's roots and pow-ers I live for Or a log-rithm now and then And with sine, co-sine and the tan-gent I'm in math Hea-ven!
7
SOHCAHTOA! WORDS BY: KATHY ANDERSON TUNE: "OKLAHOMA!"
SOHCAHTOA where we de-fine sine, cosine, and tan. And the ratios of each of those follow letters S, C, and T. SOHCAHTOA - opposite on hypotenuse is sine, And for cosine oh, it's the ratio, Of ad-ja-cent on hypotenuse. We know we can't forget tan, it's the opposite on a-jay-cent. And when we say LABEL YOUR SIDES, a trig-o-nom-e-try, CHOOSE YOUR FUNCTION. We're only saying thanks for your help SOHCAHTOA. SOHCAHTOA. SOHCAHTOA, SOHCAHTOA, okay!
CHEERS TO INTEGRATION WORDS BY: STUDENTS IN J. CARTER'S CALCULUS AB CLASS TUNE: "THE CHEERS THEME"
Integrating with trig identities takes everything you've got. Antidifferentiating sure can take a lot. Wouldn't you like to learn a way... ? Sometimes you want to know: Just how does one integrate? Wouldn't that be super great? You want to know how you can show You're better that any average Joe You want to know how you can integrate.
8
THE TWELVE DAYS OF ALGEBRA WORDS BY: JOHN A. CARTER TUNE: "THE TWELVE DAYS OF CHRISTMAS"
On the first day of algebra my teacher taught to me that math class can be fun. On the second day of algebra my teacher taught to me: variables and that math class can be fun. On the third day of algebra my teacher taught to me: combining like terms... On the fourth day of algebra my teacher taught to me: distributing... On the fifth day of algebra my teacher taught to me: solving equations... On the sixth day of algebra my teacher taught to me: how to find slope... On the seventh day of algebra my teacher taught to me: graphing of lines... On the eighth day of algebra my teacher taught to me: fitting lines to data... On the ninth day of algebra my teacher taught to me: probability... On the tenth day of algebra my teacher taught to me: exponential curves... On the eleventh day of algebra my teacher taught to me: quadratic functions... On the twelfth day of algebra my teacher taught to me: systems of equations...
9
WE WILL GRAPH YOU! WORDS BY: JOHN A. CARTER TUNE: "WE WILL ROCK YOU!"
Buddy, you're a man with a hard time graphing. All you need to do is find the m and the b. It's not too hard you see, You put your pencil on the b. Graphing's not as hard as you thought it might be, singing Chorus: We will, we will graph you! We will, we will graph you!
Now you've got a point on the y-intercept. All you need to do is find the rest of it. You need the slope to go on, That's rise over run. Delta y in delta x, boy it's fun, singing... (Chorus) Next, take the coefficient of the x baby. Find two more points and another one maybe. Go up or down first, Then go across. I dig graphing lines, I think it's boss, singing... (Chorus)
10
THE PROPERTY SONG WORDS BY: JOHN A. CARTER TUNE: "THIS OLD MAN"
This property, the Commutative Property, Tells us that we are free To change the order of a sum, Also in a multiplication. This property, the Associative Property, Tells us that you and me Can change the grouping when we multiply, Do it when you add, it'll make you look sly, This property, the Identity, Tells us that so obviously Anything times one will not change, Anything add zero will still remain. This property, the Inverse Property, Tells us that which we can see Multiply by the reciprocal to always obtain one, Add the opposite to anything to always leave none. This property, the Distributive Property, Talks to us about a quantity Which contains a sum and is being multiplied. Take the product with each term inside.
11
HOMEWORK CONNECTION WORDS BY: DANE R. CAMP TUNE: "RAINBOW CONNECTION"
How come there aren't any songs about homework And problems we tackle with pride? Homework completion can improve precision, 'Cause you must always show what you tried... Homework's important some may not believe it. I know they're wrong, wait and see. Someday they'll find it - the homework connection, The students, the teachers, and me. Who said to show your work Not just the answers On every assignment that's due? Somebody long ago Thought it'd be helpful And deep down you know that it's true! If math is frustrating Stop procrastinating And then in your heart you will see. Soon you will I] find it -the homework connection, The students, the teachers, and me. All of us want to do well And we know that it takes More than magic! Have you been half-asleep? And have you heard voices? The teacher is calling your name. The blank test you hand in Looks just like the homework The questions are one and the same. You've done this too many times to pretend that The secret is not plain to see. Now you have found it - the homework connection, The students, the teachers, and me.
12
COMPLEX NUMBERS WORDS BY: DANE R. CAMP TUNE: "OLD MACDONALD"
Complex numbers are lotsa fun. i-i-i-i-i 'Cause i's the square toot of negative 1. i-i-i-i-i With an (a + bi) here and an (a + bi) there. Here an (a + bi) , there an (a + bi) Everywhere an (a+ bi). Complex numbers are lotsa fun. i-i-i-i-i i squared is just negative one, i-i-i-i-i So powers of i are easily done. i-i-i-i-i With an i squared here and an i squared there. Here an i squared, there an i squared, Everywhere an i squared. i squared is just negative one. i-i-i-i-i Adding is just like a game. i-i-i-i-i Just com-bine terms that are the same. i-i-i-i-i With a like term here and a like term there. Here a like term, there a like term, Everywhere a like term. Adding is just like a game. i-i-i-i-i Mul-ti-pli-ca-tion is not toil. i-i-i-i-i Just ex-pand it by using FOIL. i-i-i-i-i With a product here and a product there. Here a product, there a product, Everywhere a product, product. Mul-ti-pli-ca-tion is not toil. i-i-i-i-i Di-vi-sion is really great. i-i-i-i-i Multiply both by the con-ju-gate. i-i-i-i-i With a conjugate here and a conjugate there, Here a conjugate, there a conjugate, everywhere a con-ju-gate. Di-vi-sion is really great. i-i-i-i-i
13
THE POLYNOMIAL DANCE WORDS BY: DANE R. CAMP TUNE: "HOKEY POKEY"
You put your right hand out, You put your left hand out, You can even flip direction Or jump straight up and down. Now that's a constant function And it makes you want to shout, "Polynomials can dance about!" You put your right hand up, You put your left hand down, You can alternate direction Or jump straight up and down. Now that's a linear function And it makes you want to shout, "Polynomials can dance about!" You put your right hand up, You put your left hand up, You can even flip direction Or jump straight up and down. Now that's a quadratic function And it makes you want to shout, "Polynomials can dance about!" You put your right hand up, You put your left hand down, You can alternate direction Or jump straight up and down. Now that's a cubic function And it makes you want to shout, "Polynomials can dance about!"
14
FRACTAL GEOMETRY WORDS BY: DANE R. CAMP TUNE: "YELLOW SUBMARINE"
Chorus: We love to do fractal geometry, Fractal Geometry, fractal Geometry, We love to do fractal geometry, Fractal geometry, fractal geometry, In the world that we all live, There's so much that we don't understand, But there's a tool that we now have, That will certainly lend us a hand. (Chorus) Technology helps us to find Lots of connections that were missed before. A country's coast and von Koch curve Can be shown to have the same structure. (Chorus) All the trees out in the woods Aren't Euclidean that's plain to see And the fungus on your feet, Have self-similar geometry. (Chorus)
15
THE CHAIN RULE WORDS BY: JOHN A. CARTER TUNE: "CLEMENTINE"
Here's a function in a function And your job here is to find The derivative of the whole thing With respect to x inside. Call the outside f of u And the inside u of x. Differentiate to find df/du And multiply by du/dx. Use the chain rule. Use the chain rule. Use the chain rule whene'er you find The derivative of a function compositionally defined. | 677.169 | 1 |
Graph of Exponential Functions - Graphing and sketching exponential functions: step by step tutorial. The properties such as domain, range, horizontal asymptotes and intercepts of the graphs of these functions are also examined in details. Free graph paper is available.
Exploring Histograms - The interactive data analysis tool in this investigation allows students to create their own sets of data and examine how various statistical functions such as mean, median, and standard deviation depend on the choice of data.
Graphing Calculator Help - select a calculator and after selecting a calculator, click a topic within the frame on the right side of the page for specific instructions
Graphing Calculator Toolbar 1.91 [for use with the Firefox browser] - Enter up to 5 equations into this toolbar, then view the graphs using Firefox native SVG support. The toolbar allows rescaling the x and y coordinates and redefining the domain of x
Online Graphing Calculator - GCalc: a basic, easy-to-use, well-balanced set of graphing functionality for algebra, pre-calculus, calculus and beyond. Click on the GCalc2 applet button to get the online calculator
Spreadsheet and Graphing Calculator - This tool can be used to investigate rational functions, or exponential functions. It can also be used to investigate any other functions that can be explored with a graphing calculator or spreadsheet software | 677.169 | 1 |
The EL-W535XBSL performs over 330 advanced scientific functions and utilizes WriteView Technology 4-line display and Multi-Line Playback to make scientific equations easier for students to solve. It is ideal for students studying general math algebra geometry trigonometry statistics biology chemistry and general science. | 677.169 | 1 |
Perspectives on School Algebra
This book confronts the issue of how young people can find a way into the world of algebra. It represents multiple perspectives which include an analysis of situations in which algebra is an efficient problem-solving tool, the use of computer-based technologies, and a consideration of the historical evolution of algebra. | 677.169 | 1 |
metics is not done by is the easiest subject and the most interesting one too if u understand the basic concepts. Once you're strong in the basics, the tougher sums will become easily solvable. When a chapter is taught, make sure that u understand the concept first and then proceed to the sum. Don't jump into sums directly. Try solving sums on your own. If you find that to be difficult go through the solution of the sum, nicely understand it , close the solution, do the sum now and repeat it until you get it right without getting stuck anywhere. Practice the sums , doesn't matter if u run out of sums, do them daily........ (This is what I did to get a centum)
Just see Mathematics as a friend... dont think to memories
the part or formula.. just try to understand the logic and how they framed formula..
Try to understand maths as pictorial format.. if u can understand the core of
maths.. surly u ll proceed further… | 677.169 | 1 |
ISBN 13: 9780007170913
KS3 Maths (Total Revision)
This text has been updated in line with the latest tests. It continues to provide a comprehensive approach to exam success with its combination of study support and exam practice. Written by chief markers and experienced teachers, the book provides an authoritative but practical approach to revising for the KS3 National Tests. It is divided into short revision sessions organized by level so students can use the book wherever they are in their KS3 course. "Check yourself" sections throughout ensure revision is active and effective. Test questions are provided at every level with model answers and practical advice from test markers on how to boost grades.
"synopsis" may belong to another edition of this title.
About the Author:
Kevin Evans is Head of Mathematics at Abbey Grange C of E High School, Leeds. He is a Senior Moderator and Senior Examiner for GCSE Mathematics for a major examining board and a Senior Marker for Key stage 3 Mathematics. Keith Gordon is Head of Mathematics at Wath Comprehensive School, Rotherham and is Principal Examiner for GCSE Mathematics for a major examining board.4250930915035
Book Description Collins 19/010913
Book Description Collins 19/010913
Book Description Collins 190913
Book Description Collins, 20048615
Book Description Collins, 2004. Paperback. Book Condition: Very Good. In VERY GOOD general condition, with some signs of previous use. Dispatched from the UK daily Another Croaking bargain from the Frog !!!. Bookseller Inventory # mon0001267813 | 677.169 | 1 |
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Tutorial
Published on Nov 7, 2014
Shop.TutorMeMath.net
This video by Fort Bend Tutoring shows the process of graphing piecewise functions. There are five (5) examples in all. The lesson is instructed by Larry "Mr. Whitt" Whittington. Intro/Outro by Perry "Lelo" Graham.
Evaluating Piecewise Functions
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About
Overview
Designed to enhance math skills of the reader in the field of drafting, this completely updated fourth edition of Practical Problems in Mathematics For Drafting and CAD presents a comprehensive overview of contemporary drafting problems, CAD drawings, and industry applications and practices. This text provides a variety of integrated math problems and CAD operations in order to facilitate critical thinking, problem solving, and basic mathematics literacy. Filled with real-world applications and designed to cover a range of skills and levels of difficulty, the fourth edition includes updated figures, illustrations, problem sets, examples, and solutions in order to give you the skills you need to succeed in the field of drafting.
Features and Benefits
The text features problems and values that reflect current and emerging drafting and CAD technologies.
What's New
The content has been updated and reorganized into 9 sections and 42 units, including a new section on Geometric Dimensioning and Tolerancing and new units on 3D- and coordinate geometry.
Problem sets have been reorganized into "Skill Problems," "Practical Problems," and "CAD Problems" in order to make it easier for instructors to assign specific types of problems and for students to study specific content areas.
This edition now includes an increased number of stepped out examples, with corresponding explanations and solutions. In addition, problems and answers have been revised in order to give students practice with realistic problem sets.
Figures and drawings have been improved in order to appeal to visual learners.
Learning Resource Bundles
Choose the textbook packaged with the resources that best meet your course and student needs.
Contact your Learning Consultant for more information.
Bundle: Text + CourseMate, 3 terms (19 months) Printed Access Card
ISBN-10: 1133219152 | ISBN-13: 9781133219156
List Price = $125.95
| CengageBrain Price = $125John LarkinConcetta Duval | 677.169 | 1 |
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Overview Presents an introduction to the concepts of geometry and offers a collection of geometry problems along with detailed answer explanations.
Publishers Description Geometry is one of the oldest mathematical subjects in history. Unfortunately, few geometry study guides offer clear explanations, causing many people to get tripped up or lost when trying to solve a proof—even when they know the terms and concepts like the back of their hand. However, this problem can be fixed with practice and some strategies for slicing through all the mumbo-jumbo and getting right to the heart of the proof.
Geometry Workbook For Dummies ensures that practice makes perfect, especially when problems are presented without the stiff, formal style that you'd find in your math textbook. Written with a commonsense, street-smart approach, this guide gives you the step-by-step process to solve each proof, along with tips, shortcuts, and mnemonic devices to make sure the solutions stick. It also gives you plenty of room to work out your solutions, providing you with space to breathe and a clear head. This book provides you with the tools you need to solve all types of geometry problems, including:
Congruent triangles
Finding the area, angle, and size of quadrilaterals
Angle-arc theorems and formulas
Touching radii and tangents
Connecting radii and chords
Parallel, perpendicular, and intersecting lines and planes
Slope, distance, and midpoint formulas
Line and circle equations
Handling rotations, reflections, and other transformations
Packed with tons of strategies for solving proofs and a review of key concepts, Geometry Workbook For Dummies is the ultimate study aid for students, parents, and anyone with an interest Geometry Workbook For Dummies by Mark Ryan today - and if you are for any reason not happy, you have 30 days to return it. Please contact us at 1-877-205-6402 if you have any questions.
More About Mark Ryan
Mark Ryan has taught pre-math through calculus for more than a decade. He is a member of the National Council of Teachers of Mathematics.
Mark Ryan has an academic affiliation as follows - University of Coventry, UK University of Birmingham University of Birm | 677.169 | 1 |
Students start with an assessment and receive immediate instructional feedback throughout. Step-by-step tutorials, which introduce each level, can be referred to during practice. Problems are broken down into small, easily understood steps.
The program is self-paced and self-monitored. Students advance as they demonstrate readiness. They may track their own improvement through progress-to-date and last session scores. Scores are kept in a record management system that allows teachers to view and print detailed reports.
Designed for students in U.S. grades 6 through 9 (age 10 and up), the program can also be used by ESL and adult students interested in improving their algebra skills. | 677.169 | 1 |
Algebra
There are certain mistakes that students frequently make while learning algebra. This workbook clearly explains these mistakes so students can avoid them. Examples then illustrate the correct way of working an algebra problem, and practice problems are provided.
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Every product created at Lorenz Educational Press is based on our commitment to support teachers providing a positive educational experience for students. We strive to produce products that embrace our mission to define learning areas vital to developing well-rounded students and empowering our future leaders.
Lorenz Educational Press is committed to supporting teachers providing a positive educational experience for students. Our mission is "Bridging the Gaps in Education" and it's the statement behind the development of all our products. We believe that there are important concepts that should be part of a student's complete education, but because of the rigors of teaching to the standards they are often not specifically addressed.
Our publications identify those areas, provide easy-to-use activities that can be integrated into core subject studies, and include tools for student assessment. We're creating supplementary materials that focus on such areas as critical thinking and listening skills; cross-curricular learning; music and movement; and team building, which still meet the National Standards objectives necessary to easily incorporate them into the classroom. | 677.169 | 1 |
A First Course in Probability
4.11 - 1251 ratings - Source
A First Course in Probability, Ninth Edition, features clear and intuitive explanations of the mathematics of probability theory, outstanding problem sets, and a variety of diverse examples and applications. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and business students. It assumes a background in elementary calculus.This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and business students. It assumes a background in elementary calculus.
Title
:
A First Course in Probability
Author
:
Sheldon M. Ross
Publisher
:
Pearson College Division - 2014
ISBN-13
:
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About
Overview
ELEMENTS OF MODERN ALGEBRA, Eighth Edition, with its user-friendly format, provides you with the tools you need to succeed in abstract algebra and develop mathematical maturity as a bridge to higher-level mathematics courses. Strategy boxes give you guidance and explanations about techniques and enable you to become more proficient at constructing proofs. A summary of key words and phrases at the end of each chapter help you master the material. A reference section, symbolic marginal notes, an appendix, and numerous examples help you develop your problem-solving skills.
Features and Benefits
Nearly 300 True/False statements that encourage the students to thoroughly understand the statements of definitions and results of theorems appear in this edition.
Descriptive labels and titles are used with definitions and theorems to indicate their content and relevance.
Strategy boxes appear to give guidance and explanation about techniques of proof. This feature forms a component of the bridge that enables students to become more proficient in their proof construction skills.
Symbolic marginal notes are used to help students analyze the logic in the proofs of theorems without interrupting the natural flow of the proof.
A reference system provides guideposts to continuations and interconnections of exercises throughout the text.
An appendix on the basics of logic and methods of proof is included to assist students with a weak background in logic.
Biographical sketches of great mathematicians whose contributions are relevant to the respective material conclude each chapter.
A summary of key words and phrases is included at the end of each chapter.
A list of special notations used in the book appears on the front endpapers.
What's New
Alerts that draw attention to counterexamples, special cases, proper symbol or terminology usage, and common misconceptions. Frequently these alerts lead to True/False statements in the exercises that further reinforce the precision required in mathematical communication.
More emphasis placed on special groups, such as the general linear and special linear groups, the dihedral group, and the group of units.
Moving some definitions from the exercises to the sections for greater emphasis.
Using marginal notes to outline the steps of the induction arguments required in the examples.
More than 200 new theoretical and computational exercises have been added.
Many new examples have also been added to this edition.
Meet the Author
Author Bio
Linda Gilbert | 677.169 | 1 |
Four Place Tables of Logarithms and Trigonometric Functions,
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Algebra and Trigonometry
Algebra and Trigonometry: A Straightforward Approach
In the middle ones put the sum of the digits in the two squares above each. Of these lines, the second gives the combinations with one syllable, the third the combinations with two syllables, ... [34] The text also indicates that Pingala was aware of the combinatorial identity: [35] Katyayana (c. 3rd century BCE) is notable for being the last of the Vedic mathematicians epub. The dominance of Baghdad and the influence of the Arab World was to last for the next 500 years. The scholars in the House of Wisdom came from many cultures and translated the works of Egyptian, Babylonian, Greek, Indian and Chinese astronomers and mathematicians. The Mathematical Treatise of Ptolemy was one of the first to be translated from the Greek into Arabic by Ishaq ben Hunayn (830-910) e-Study Guide for: College Algebra and Trigonometry by J. S. Ratti, ISBN 9780321296429. Now the angle of depression to the car is 35 degrees read Four Place Tables of Logarithms and Trigonometric Functions, With Auxiliary Tables (chiefly to Three Figures) of Squares, Square Roots, Cubes, Cube ... Natural Logarithms, Radians, and Con online. A slower-paced introduction to calculus for students who require additional preparation for calculus. This sequence presents the same calculus topics as Mathematics 9, together with all the necessary pre-calculus topics. Students successfully completing this sequence will be prepared for Mathematics 0100. Placement in this course requires permission of the instructor. A one-semester introduction to calculus recommended for students who wish to learn the basics of calculus for application to social sciences or for cultural appreciation as part of a broader education epub. Algebra with pizzazz worksheet, solve with radicals calculator, easy ways to solve discriminant, dividing algebraic expressions. MATLAB CHEMICAL ENGINEERING EXAMPLE SIMPSON, adding, subtracting, dividing and multiplying integers, algebraic expressions solver, review worksheet adding subtracting multiplying dividing rational expressions, question and answer for algebra tutorial, prentice hall Practice workbook pre- algebra, free download accounting principles 8th edition A Treatise On Surveying: Containing the Theory and Practice : To Which Is Prefixed a Perspicuous System of Plane Trigonometry : The Whole Clearly ... Particularlry Adapted to the Use of S. Next draw a line AD from A perpendicular to BC. From this we can determine the following trig ratios for the special angles 30º and 60º: For any other angle θ, you can calculate approximately the values of sin θ, cos θ, tan θ by using a scientific calculator. Make sure you set the mode on your calculator to DEG if the angle is measured in degrees or RAD if the angle is measured in radians online.
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