Transcription

PARTCFourier Analysis.PartialDifferentialEquations (PDEs)C H A P T E R 1 1 Fourier AnalysisC H A P T E R 1 2 Partial Differential Equations (PDEs)Chapter 11 and Chapter 12 are directly related to each other in that Fourier analysis hasits most important applications in modeling and solving partial differential equations(PDEs) related to boundary and initial value problems of mechanics, heat flow,electrostatics, and other fields. However, the study of PDEs is a study in its own right.Indeed, PDEs are the subject of much ongoing research.Fourier analysis allows us to model periodic phenomena which appear frequently inengineering and elsewhere—think of rotating parts of machines, alternating electric currentsor the motion of planets. Related period functions may be complicated. Now, the ingeneousidea of Fourier analysis is to represent complicated functions in terms of simple periodicfunctions, namely cosines and sines. The representations will be infinite series calledFourier series.1 This idea can be generalized to more general series (see Sec. 11.5) andto integral representations (see Sec. 11.7).The discovery of Fourier series had a huge impetus on applied mathematics as well as onmathematics as a whole. Indeed, its influence on the concept of a function, on integrationtheory, on convergence theory, and other theories of mathematics has been substantial(see [GenRef7] in App. 1).Chapter 12 deals with the most important partial differential equations (PDEs) of physicsand engineering, such as the wave equation, the heat equation, and the Laplace equation.These equations can model a vibrating string/membrane, temperatures on a bar, andelectrostatic potentials, respectively. PDEs are very important in many areas of physicsand engineering and have many more applications than ODEs.1JEAN-BAPTISTE JOSEPH FOURIER (1768–1830), French physicist and mathematician, lived and taughtin Paris, accompanied Napoléon in the Egyptian War, and was later made prefect of Grenoble. The beginningson Fourier series can be found in works by Euler and by Daniel Bernoulli, but it was Fourier who employedthem in a systematic and general manner in his main work, Théorie analytique de la chaleur (Analytic Theoryof Heat, Paris, 1822), in which he developed the theory of heat conduction (heat equation; see Sec. 12.5), makingthese series a most important tool in applied mathematics.473

CHAPTER11Fourier AnalysisThis chapter on Fourier analysis covers three broad areas: Fourier series in Secs. 11.1–11.4,more general orthonormal series called Sturm–Liouville expansions in Secs. 11.5 and 11.6and Fourier integrals and transforms in Secs. 11.7–11.9.The central starting point of Fourier analysis is Fourier series. They are infinite seriesdesigned to represent general periodic functions in terms of simple ones, namely, cosinesand sines. This trigonometric system is orthogonal, allowing the computation of thecoefficients of the Fourier series by use of the well-known Euler formulas, as shown inSec. 11.1. Fourier series are very important to the engineer and physicist because theyallow the solution of ODEs in connection with forced oscillations (Sec. 11.3) and theapproximation of periodic functions (Sec. 11.4). Moreover, applications of Fourier analysisto PDEs are given in Chap. 12. Fourier series are, in a certain sense, more universal thanthe familiar Taylor series in calculus because many discontinuous periodic functions thatcome up in applications can be developed in Fourier series but do not have Taylor seriesexpansions.The underlying idea of the Fourier series can be extended in two important ways. Wecan replace the trigonometric system by other families of orthogonal functions, e.g., Besselfunctions and obtain the Sturm–Liouville expansions. Note that related Secs. 11.5 and11.6 used to be part of Chap. 5 but, for greater readability and logical coherence, are nowpart of Chap. 11. The second expansion is applying Fourier series to nonperiodicphenomena and obtaining Fourier integrals and Fourier transforms. Both extensions haveimportant applications to solving PDEs as will be shown in Chap. 12.In a digital age, the discrete Fourier transform plays an important role. Signals, suchas voice or music, are sampled and analyzed for frequencies. An important algorithm, inthis context, is the fast Fourier transform. This is discussed in Sec. 11.9.Note that the two extensions of Fourier series are independent of each other and maybe studied in the order suggested in this chapter or by studying Fourier integrals andtransforms first and then Sturm–Liouville expansions.Prerequisite: Elementary integral calculus (needed for Fourier coefficients).Sections that may be omitted in a shorter course: 11.4–11.9.References and Answers to Problems: App. 1 Part C, App. 2.11.1Fourier SeriesFourier series are infinite series that represent periodic functions in terms of cosines andsines. As such, Fourier series are of greatest importance to the engineer and appliedmathematician. To define Fourier series, we first need some background material.A function f (x) is called a periodic function if f ( x) is defined for all real x, except474

SEC. 11.1 Fourier Series475f (x)xpFig. 258. Periodic function of period ppossibly at some points, and if there is some positive number p, called a period of f (x),such thatf (x p) f (x)(1)for all x.(The function f (x) tan x is a periodic function that is not defined for all real x butundefined for some points (more precisely, countably many points), that is x p 2, 3p 2, Á .)The graph of a periodic function has the characteristic that it can be obtained by periodicrepetition of its graph in any interval of length p (Fig. 258).The smallest positive period is often called the fundamental period. (See Probs. 2–4.)Familiar periodic functions are the cosine, sine, tangent, and cotangent. Examples offunctions that are not periodic are x, x 2, x 3, ex, cosh x, and ln x, to mention just a few.If f (x) has period p, it also has the period 2p because (1) implies f (x 2p) f ([x p] p) f (x p) f (x), etc.; thus for any integer n 1, 2, 3, Á ,f (x np) f (x)(2)for all x.Furthermore if f (x) and g (x) have period p, then af (x) bg (x) with any constants a andb also has the period p.Our problem in the first few sections of this chapter will be the representation of variousfunctions f (x) of period 2p in terms of the simple functions(3)1,cos x,sin x,cos 2x,sin 2x, Á ,cos nx,sin nx, Á .All these functions have the period 2p. They form the so-called trigonometric system.Figure 259 shows the first few of them (except for the constant 1, which is periodic withany period).0π2π0cos x0πsin xπ2π0cos 2x2π0πsin 2xπ2πcos 3x2π0π2πsin 3xFig. 259. Cosine and sine functions having the period 2p (the first few members of thetrigonometric system (3), except for the constant 1)

476CHAP. 11 Fourier AnalysisThe series to be obtained will be a trigonometric series, that is, a series of the forma0 a1 cos x b1 sin x a2 cos 2x b2 sin 2x Áⴥ(4) a0 a (an cos nx bn sin nx).n 1a0, a1, b1, a2, b2, Á are constants, called the coefficients of the series. We see that eachterm has the period 2p. Hence if the coefficients are such that the series converges, itssum will be a function of period 2p.Expressions such as (4) will occur frequently in Fourier analysis. To compare theexpression on the right with that on the left, simply write the terms in the summation.Convergence of one side implies convergence of the other and the sums will be thesame.Now suppose that f (x) is a given function of period 2p and is such that it can berepresented by a series (4), that is, (4) converges and, moreover, has the sum f (x). Then,using the equality sign, we writeⴥ(5)f (x) a0 a (an cos nx bn sin nx)n 1and call (5) the Fourier series of f (x). We shall prove that in this case the coefficientsof (5) are the so-called Fourier coefficients of f (x), given by the Euler formulas(0)(6)(a)(b)1a0 2p1an p1bn p冮pf (x) dxⴚp冮p冮pf (x) cos nx dxn 1, 2, Áf (x) sin nx dxn 1, 2, Á .ⴚpⴚpThe name “Fourier series” is sometimes also used in the exceptional case that (5) withcoefficients (6) does not converge or does not have the sum f (x)—this may happen butis merely of theoretical interest. (For Euler see footnote 4 in Sec. 2.5.)A Basic ExampleBefore we derive the Euler formulas (6), let us consider how (5) and (6) are applied inthis important basic example. Be fully alert, as the way we approach and solve thisexample will be the technique you will use for other functions. Note that the integrationis a little bit different from what you are familiar with in calculus because of the n. Donot just routinely use your software but try to get a good understanding and makeobservations: How are continuous functions (cosines and sines) able to represent a givendiscontinuous function? How does the quality of the approximation increase if you takemore and more terms of the series? Why are the approximating functions, called the

SEC. 11.1 Fourier Series477partial sums of the series, in this example always zero at 0 and p? Why is the factor1 n (obtained in the integration) important?EXAMPLE 1Periodic Rectangular Wave (Fig. 260)Find the Fourier coefficients of the periodic function f (x) in Fig. 260. The formula is(7) kif p x 0kif0 x pf (x) bandf (x 2p) f (x).Functions of this kind occur as external forces acting on mechanical systems, electromotive forces in electriccircuits, etc. (The value of f (x) at a single point does not affect the integral; hence we can leave f (x) undefinedat x 0 and x p.)From (6.0) we obtain a0 0. This can also be seen without integration, since the area under thecurve of f (x) between p and p (taken with a minus sign where f (x) is negative) is zero. From (6a) we obtainthe coefficients a1, a2, Á of the cosine terms. Since f ( x) is given by two expressions, the integrals from pto p split into two integrals:Solution.an 1p冮pf (x) cos nx dx ⴚp 1p c冮0( k) cos nx dx ⴚp1p c k冮pk cos nx dx d0psin nx 0sin nx k d 0nnⴚp0because sin nx 0 at p, 0, and p for all n 1, 2, Á . We see that all these cosine coefficients are zero. Thatis, the Fourier series of (7) has no cosine terms, just sine terms, it is a Fourier sine series with coefficientsb1, b2, Á obtained from (6b);bn p冮1pf (x) sin nx dx ⴚp 1pc1冮0( k) sin nx dx ⴚpp ck冮pk sin nx dx d0cos nx pcos nx 0 k d.nnⴚp0Since cos ( a) cos a and cos 0 1, this yieldsbn k2k[cos 0 cos ( np) cos np cos 0] (1 cos np).npnpNow, cos p 1, cos 2p 1, cos 3p 1, etc.; in general, 1cos np b1for odd n,and thusfor even n,1 cos np b2for odd n,0for even n.Hence the Fourier coefficients bn of our function areb1 4kp,b2 0,b3 4k3p,b4 0,b5 4k Á, .5pFig. 260. Given function f (x) (Periodic reactangular wave)

478CHAP. 11 Fourier AnalysisSince the an are zero, the Fourier series of f (x) is4k(8)p(sin x 13 sin 3x 15 sin 5x Á ).The partial sums areS1 4kpS2 sin x,4kpasin x 1sin 3xb .3etc.Their graphs in Fig. 261 seem to indicate that the series is convergent and has the sum f (x), the given function.We notice that at x 0 and x p, the points of discontinuity of f (x), all partial sums have the value zero, thearithmetic mean of the limits k and k of our function, at these points. This is typical.Furthermore, assuming that f (x) is the sum of the series and setting x p 2, we havep4k11Áfa b k 2p a1 3 5 b .Thus1 13 15 17p Á .4This is a famous result obtained by Leibniz in 1673 from geometric considerations. It illustrates that the valuesof various series with constant terms can be obtained by evaluating Fourier series at specific points.䊏Fig. 261. First three partial sums of the corresponding Fourier series

SEC. 11.1 Fourier Series479Derivation of the Euler Formulas (6)The key to the Euler formulas (6) is the orthogonality of (3), a concept of basic importance,as follows. Here we generalize the concept of inner product (Sec. 9.3) to functions.THEOREM 1Orthogonality of the Trigonometric System (3)The trigonometric system (3) is orthogonal on the interval p x p (hencealso on 0 x 2p or any other interval of length 2p because of periodicity); thatis, the integral of the product of any two functions in (3) over that interval is 0, sothat for any integers n and m,(a)冮p冮pcos nx cos mx dx 0(n m)ⴚp(9)(b)sin nx sin mx dx 0(n m)sin nx cos mx dx 0(n m or n m).ⴚp(c)冮pⴚpPROOFThis follows simply by transforming the integrands trigonometrically from products intosums. In (9a) and (9b), by (11) in App. A3.1,冮p冮pcos nx cos mx dx ⴚp121sin nx sin mx dx 2ⴚp冮p12cos (n m)x dx ⴚp冮p1cos (n m)x dx 2ⴚp冮pcos (n m)x dxⴚp冮pcos (n m)x dx.ⴚpSince m n (integer!), the integrals on the right are all 0. Similarly, in (9c), for all integerm and n (without exception; do you see why?)冮psin nx cos mx dx ⴚp12冮psin (n m)x dx ⴚp12冮psin (n m)x dx 0 0.䊏ⴚpApplication of Theorem 1 to the Fourier Series (5)We prove (6.0). Integrating on both sides of (5) from p to p, we get冮pf (x) dx ⴚp冮pⴥc a0 a (an cos nx bn sin nx) d dx.ⴚpn 1We now assume that termwise integration is allowed. (We shall say in the proof ofTheorem 2 when this is true.) Then we obtain冮pⴚpf (x) dx a0冮pⴥdx a aanⴚpn 1冮pcos nx dx bnⴚp冮psin nx dxb .ⴚp

480CHAP. 11 Fourier AnalysisThe first term on the right equals 2pa0. Integration shows that all the other integrals are 0.Hence division by 2p gives (6.0).We prove (6a). Multiplying (5) on both sides by cos mx with any fixed positive integerm and integrating from p to p, we have(10)冮pf (x) cos mx dx ⴚp冮pⴥc a0 a (an cos nx bn sin nx) d cos mx dx.ⴚpn 1We now integrate term by term. Then on the right we obtain an integral of a0 cos mx,which is 0; an integral of an cos nx cos mx , which is amp for n m and 0 for n m by(9a); and an integral of bn sin nx cos mx, which is 0 for all n and m by (9c). Hence theright side of (10) equals amp. Division by p gives (6a) (with m instead of n).We finally prove (6b). Multiplying (5) on both sides by sin mx with any fixed positiveinteger m and integrating from p to p, we get(11)冮pⴚpf (x) sin mx dx 冮pⴥc a0 a (an cos nx bn sin nx) d sin mx dx.ⴚpn 1Integrating term by term, we obtain on the right an integral of a0 sin mx, which is 0; anintegral of an cos nx sin mx, which is 0 by (9c); and an integral of bn sin nx sin mx, whichis bmp if n m and 0 if n m, by (9b). This implies (6b) (with n denoted by m). Thiscompletes the proof of the Euler formulas (6) for the Fourier coefficients.䊏Convergence and Sum of a Fourier SeriesThe class of functions that can be represented by Fourier series is surprisingly large andgeneral. Sufficient conditions valid in most applications are as follows.THEOREM 2Representation by a Fourier SeriesLet f (x) be periodic with period 2p and piecewise continuous (see Sec. 6.1) in theinterval p x p. Furthermore, let f (x) have a left-hand derivative and a righthand derivative at each point of that interval. Then the Fourier series (5) of f (x)[with coefficients (6)] converges. Its sum is f (x), except at points x0 where f (x) isdiscontinuous. There the sum of the series is the average of the left- and right-handlimits2 of f (x) at x 0.f (x)f (1 – 0)2The left-hand limit of f (x) at x 0 is defined as the limit of f (x) as x approaches x0 from the leftand is commonly denoted by f (x 0 0). Thus1f (1 0)0x1Fig. 262. Left- andright-hand limitsƒ(1 0) 1,ƒ(1 0) 12of the functionf (x) bx2if x 1x 2if x 1ƒ(x0 0) lim ƒ( x0 h) as h * 0 through positive values.h*0The right-hand limit is denoted by ƒ(x0 0) andƒ(x0 0) lim ƒ( x0 h) as h * 0 through positive values.h*0The left- and right-hand derivatives of ƒ(x) at x0 are defined as the limits off (x 0 h) f (x 0 0) handf (x 0 h) f (x 0 0) h,respectively, as h * 0 through positive values. Of course if ƒ(x) is continuous at x0, the last term inboth numerators is simply ƒ(x0).

SEC. 11.1 Fourier SeriesPROOF481We prove convergence, but only for a continuous function f (x) having continuous firstand second derivatives. And we do not prove that the sum of the series is f (x) becausethese proofs are much more advanced; see, for instance, Ref. 3C124 listed in App. 1.Integrating (6a) by parts, we obtain1an p冮pⴚpf (x) sin nx2f (x) cos nx dx nppⴚp冮1 nppf r (x) sin nx dx.ⴚpThe first term on the right is zero. Another integration by parts givesan f r (x) cos nxn 2pp2 ⴚp1n 2p冮pf s (x) cos nx dx.ⴚpThe first term on the right is zero because of the periodicity and continuity of f r (x). Sincef s is continuous in the interval of integration, we haveƒ f s (x) ƒ Mfor an appropriate constant M. Furthermore, ƒ cos nx ƒ 1. It follows thatƒ an ƒ 1n 2p2冮pf s (x) cos nx dx 2 ⴚp1n 2p冮pM dx ⴚp2Mn2.Similarly, ƒ bn ƒ 2 M n 2 for all n. Hence the absolute value of each term of the Fourierseries of f (x) is at most equal to the corresponding term of the seriesƒ a0 ƒ 2M a1 1 122 122 132 132 Ábwhich is convergent. Hence that Fourier series converges and the proof is complete.(Readers already familiar with uniform convergence will see that, by the Weierstrasstest in Sec. 15.5, under our present assumptions the Fourier series converges uniformly,and our derivation of (6) by integrating term by term is then justified by Theorem 3 ofSec. 15.5.)䊏EXAMPLE 2Convergence at a Jump as Indicated in Theorem 2The rectangular wave in Example 1 has a jump at x 0. Its left-hand limit there is k and its right-hand limitis k (Fig. 261). Hence the average of these limits is 0. The Fourier series (8) of the wave does indeed convergeto this value when x 0 because then all its terms are 0. Similarly for the other jumps. This is in agreementwith Theorem 2.䊏Summary. A Fourier series of a given function f (x) of period 2p is a series of the form(5) with coefficients given by the Euler formulas (6). Theorem 2 gives conditions that aresufficient for this series to converge and at each x to have the value f (x), except atdiscontinuities of f (x), where the series equals the arithmetic mean of the left-hand andright-hand limits of f (x) at that point.

482CHAP. 11 Fourier AnalysisPROBLEM SET 11.11–5PERIOD, FUNDAMENTAL PERIODThe fundamental period is the smallest positive period. Findit for1. cos x, sin x, cos 2x, sin 2x, cos px, sin px,cos 2px, sin 2px2px2px2pnx2. cos nx, sin nx, cos, sin, cos,kkk2pnxsink3. If f ( x) and g (x) have period p, show that h (x) af (x) bg(x) (a, b, constant) has the period p. Thusall functions of period p form a vector space.4. Change of scale. If f (x) has period p, show thatf (ax), a 0, and f (x b), b 0, are periodic functionsof x of periods p a and bp, respectively. Give examples.5. Show that f const is periodic with any period but hasno fundamental period.17.–π10. f (x) b cos2 xFOURIER SERIESFind the Fourier series of the given function f (x), which isassumed to have the period 2p. Show the details of yourwork. Sketch or graph the partial sums up to that includingcos 5x and sin 5x.12. f (x) in Prob. 613. f (x) in Prob. 914. f (x) x 2 ( p x p)15. f (x) x 2 (0 x – 1π221.1π20– 1π2ππ–ππif p x 0cos2 x if0 x p11. Calculus review. Review integration techniques forintegrals as they are likely to arise from the Eulerformulas, for instance, definite integrals of x cos nx,x 2 sin nx, eⴚ2x cos nx, etc.12–211–πGRAPHS OF 2p–PERIODIC FUNCTIONSSketch or graph f (x) which for p x p is given asπ018.6–10follows.6. f (x) ƒ x ƒ7. f (x) ƒ sin x ƒ , f (x) sin ƒ x ƒ8. f (x) eⴚƒ x ƒ, f (x) ƒ eⴚx ƒxif p x 09. f (x) bp x if0 x pπ–π22. CAS EXPERIMENT. Graphing. Write a program forgraphing partial sums of the following series. Guessfrom the graph what f (x) the series may represent.Confirm or disprove your guess by using the Eulerformulas.(a) 2(sin x 13 sin 3x 15 sin 5x Á) 2( 12 sin 2x 14 sin 4x 16 sin 6x Á)(b)1411 2 acos x cos 3x cos 5x Á bp2925(c)231p2 4(cos x 14 cos 2x 19 cos 3x 16cos 4x Á)23. Discontinuities. Verify the last statement in Theorem2 for the discontinuities of f (x) in Prob. 21.24. CAS EXPERIMENT. Orthogonality. Integrate andgraph the integral of the product cos mx cos nx (withvarious integer m and n of your choice) from a to aas a function of a and conclude orthogonality of cos mx

SEC. 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansionsif f is continuous but f r df dx is discontinuous, 1 n 3if f and f r are continuous but f s is discontinuous, etc.Try to verify this for examples. Try to prove it byintegrating the Euler formulas by parts. What is thepractical significance of this?and cos nx (m n) for a p from the graph. For whatm and n will you get orthogonality for a p 2, p 3,p 4? Other a? Extend the experiment to cos mx sin nxand sin mx sin nx.25. CAS EXPERIMENT. Order of Fourier Coefficients.The order seems to be 1 n if f is discontinous, and 1 n 211.2483Arbitrary Period. Even and Odd Functions.Half-Range ExpansionsWe now expand our initial basic discussion of Fourier series.Orientation. This section concerns three topics:1. Transition from period 2p to any period 2L, for the function f, simply by atransformation of scale on the x-axis.2. Simplifications. Only cosine terms if f is even (“Fourier cosine series”). Only sineterms if f is odd (“Fourier sine series”).3. Expansion of f given for 0 x L in two Fourier series, one having only cosineterms and the other only sine terms (“half-range expansions”).1. From Period 2p to Any Period p 2LClearly, periodic functions in applications may have any period, not just 2p as in the lastsection (chosen to have simple formulas). The notation p 2L for the period is practicalbecause L will be a length of a violin string in Sec. 12.2, of a rod in heat conduction inSec. 12.5, and so on.The transition from period 2p to be period p 2L is effected by a suitable change ofscale, as follows. Let f (x) have period p 2L. Then we can introduce a new variable vsuch that f (x), as a function of v, has period 2p. If we set(1)(a) x p2pv,2pp(b) v p x xLso thatthen v p corresponds to x L. This means that f, as a function of v, has period2p and, therefore, a Fourier series of the form(2)ⴥLf (x) f a p vb a0 a (an cos nv bn sin nv)n 1with coefficients obtained from (6) in the last section1a0 2p(3)冮pⴚpfaLpvb dv,1bn p冮pⴚp1an p冮pfaⴚpLpLf a p vb sin nv dv.vb cos nv dv,

484CHAP. 11 Fourier AnalysisWe could use these formulas directly, but the change to x simplifies calculations. Sincev (4)pLx,dv we havepdxLand we integrate over x from L to L. Consequently, we obtain for a function f (x) ofperiod 2L the Fourier series(5)ⴥnpnpx bn sinxbf (x) a0 a aan cosLLn 1with the Fourier coefficients of f (x) given by the Euler formulas (p L in dx cancels1 p in (3))(0)(6)(a)(b)冮1a0 2L1Lan 1Lbn Lf (x) dxⴚL冮L冮Lf (x) cosⴚLf (x) sinⴚLn pxdxLn 1, 2, ÁnpxdxLn 1, 2, Á .Just as in Sec. 11.1, we continue to call (5) with any coefficients a trigonometric series.And we can integrate from 0 to 2L or over any other interval of length p 2L.EXAMPLE 1Periodic Rectangular WaveFind the Fourier series of the function (Fig. 263)Solution.0if 2 x 1f (x) d kif 1 x 10if1 x 2p 2L 4, L 2.From (6.0) we obtain a0 k 2 (verify!). From (6a) we obtainan 2冮12f (x) cosⴚ2npx2dx 2冮11k cosnpxⴚ12dx 2knpsinnp2.Thus an 0 if n is even andan 2k np ifn 1, 5, 9, Á ,an 2k np if n 3, 7, 11, Á .From (6b) we find that bn 0 for n 1, 2, Á . Hence the Fourier series is a Fourier cosine series (that is, ithas no sine terms)f (x) k2 2kpacosp2x 13cos3p2x 15cos5p2x Áb .䊏

SEC. 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions485f(x)kf(x)–2k2x–k–2–101x2Fig. 263. Example 1EXAMPLE 2Fig. 264. Example 2Periodic Rectangular Wave. Change of ScaleFind the Fourier series of the function (Fig. 264) kif 2 x 0kif0 x 2f (x) cSolution.p 2L 4,L 2.Since L 2, we have in (3) v px 2 and obtain from (8) in Sec. 11.1 with v instead of x, that is,g(v) 4kpasin v 13sin 3v 15sin 5v Á bthe present Fourier seriesf (x) 4kpasinp2x 13sin3p21x 5sinx Áb .5p2䊏Confirm this by using (6) and integrating.EXAMPLE 3Half-Wave RectifierA sinusoidal voltage E sin vt, where t is time, is passed through a half-wave rectifier that clips the negativeportion of the wave (Fig. 265). Find the Fourier series of the resulting periodic function0u(t) cE sin vtSolution.if L t 0,if0 t Lp 2L 2p,vL pv.Since u 0 when L t 0, we obtain from (6.0), with t instead of x,a0 v2p 冮p vE sin vt dt 0Epand from (6a), by using formula (11) in App. A3.1 with x vt and y nvt,an vp冮p vE sin vt cos nvt dt 0vE2p 冮p v[sin (1 n) vt sin (1 n) vt] dt.0If n 1, the integral on the right is zero, and if n 2, 3, Á , we readily obtainan vE2pE2pc acos (1 n) vt(1 n) v cos (1 n)p 11 ncos (1 n) vt(1 n) v dp v0 cos (1 n)p 11 nb.If n is odd, this is equal to zero, and for even n we havean Ea22p 1 n 21 nb 2E(n 1)(n 1)p(n 2, 4, Á ).

486CHAP. 11 Fourier AnalysisIn a similar fashion we find from (6b) that b1 E 2 and bn 0 for n 2, 3, Á . Consequently,u(t) EpE 2sin vt 2Epa11cos 2vt cos 4vt Á b .1#33 #5䊏u(t)– π /ωπ /ω0tFig. 265. Half-wave rectifier2. Simplifications: Even and Odd FunctionsIf f (x) is an even function, that is, f ( x) f (x) (see Fig. 266), its Fourier series (5)reduces to a Fourier cosine seriesxⴥnpf (x) a0 a an cosxLn 1(5*)Fig. 266.Even function( f even)with coefficients (note: integration from 0 to L only!)xFig. 267.Odd function(6*)a0 1L冮Lf (x) dx,an 0冮2LLf (x) cos0npxdx,Ln 1, 2, Á .If f (x) is an odd function, that is, f ( x) f (x) (see Fig. 267), its Fourier series (5)reduces to a Fourier sine series(5**)ⴥnpf (x) a bn sinxLn 1( f odd)with coefficientsbn (6**)2L冮Lf (x) sin0npxdx.LThese formulas follow from (5) and (6) by remembering from calculus that the definiteintegral gives the net area ( area above the axis minus area below the axis) under thecurve of a function between the limits of integration. This implies(a)冮L冮Lg (x) dx 2ⴚL(7)(b)冮Lg (x) dxfor even g0h (x) dx 0for odd hⴚLFormula (7b) implies the reduction to the cosine series (even f makes f (x) sin (npx L) oddsince sin is odd) and to the sine series (odd f makes f (x) cos (npx L) odd since cos is even).Similarly, (7a) reduces the integrals in (6*) and (6**) to integrals from 0 to L. These reductionsare obvious from the graphs of an even and an odd function. (Give a formal proof.)

SEC. 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions487SummaryEven Function of Period 2 . If f is even and L p, thenⴥf (x) a0 a an cos nxn 1with coefficients1a0 p冮p2an pf (x) dx,0冮pf (x) cos nx dx, n 1, 2, Á0Odd Function of Period 2p. If f is odd and L p, thenⴥf (x) a bn sin nxn 1with coefficients2bn pEXAMPLE 4冮pn 1, 2, Á .f (x) sin nx dx,0Fourier Cosine and Sine SeriesThe rectangular wave in Example 1 is even. Hence it follows without calculation that its Fourier series is aFourier cosine series, the bn are all zero. Similarly, it follows that the Fourier series of the odd function inExample 2 is a Fourier sine series.In Example 3 you can see that the Fourier cosine series represents u(t) E p 12 E sin vt. Can you provethat this is an even function?䊏Further simplifications result from the following property, whose very simple proof is leftto the student.THEOREM 1Sum and Scalar MultipleThe Fourier coefficients of a sum f1 f2 are the sums of the corresponding Fouriercoefficients of f1 and f2.The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f.EXAMPLE 5Sawtooth WaveFind the Fourier series of the function (Fig. 268)f (x) x p if p x pandf (x 2p) f (x).f (x)–ππxFig. 268. The function f(x). Sawtooth wave

488CHAP. 11 Fourier AnalysisyS20S35S2S1y–ππ0xFig. 269. Partial sums S1, S2, S3, S20 in Example 5We have f f1 f2, where f1 x and f2 p. The Fourier coefficients of f2 are zero, except forthe first one (the constant term), which is p. Hence, by Theorem 1, the Fourier coefficients an, bn are those off1, except for a0, which is p. Since f1 is odd, an 0 for n 1, 2, Á , andSolution.bn p冮2pf1 (x) sin nx dx 0p冮2px sin nx dx.0Integrating by parts, we obtainbn 2 x cos nx2pcnp 01n冮p02cos nx dx d cos np.nHence b1 2, b2 22 , b3 23 , b4 24 , Á , and the Fourier series of f (x) isf (x) p 2 asin x 12sin 2x 13sin 3x Á b .(Fig. 269)䊏3. Half-Range ExpansionsHalf-range expansions are Fourier series. The idea is simple and useful. Figure 270explains it. We want to represent f (x) in Fig. 270.0 by a Fourier series, where f (x)may be the shape of a distorted violin string or the temperature in a metal bar of lengthL, for example. (Corresponding problems will be discussed in Chap. 12.) Now comesthe idea.We could extend f (x) as a function of period L and develop the extended function intoa Fourier series. But this series would, in general, contain both cosine and sine terms. Wecan do better and get simpler series. Indeed, for our given f we can calculate Fouriercoefficients from (6*) or from (6**). And we have a choice and can take what seemsmore practical. If we use (6*), we get (5*). This is the even periodic extension f1 of fin Fig. 270a. If we choose (6**) instead, we get (5**), the odd periodic extension f2 off in Fig. 270b.Both extensions have period 2L. This motivates the name half-range expansions: f isgiven (and of physical interest) only on half the range, that is, on half the interval ofperiodicity of length 2L.Let us illustrate these ideas with an example that we shall also need in Chap. 12.

SEC. 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions489f (x)xL(0) The given function f (x)f1(x)–LxL(a) f (x) continued as an even periodic function of period 2Lf

Fourier Analysis This chapter on Fourier analysis covers three broad areas: Fourier series in Secs. 11.1-11.4, more general orthonormal series called Sturm-Liouville expansions in Secs. 11.5 and 11.6 and Fourier integrals and transforms in Secs. 11.7-11.9. The central starting point of Fourier analysis is Fourier series. They are infinite .