CHAPTER 4 FOURIER SERIES AND INTEGRALS
CHAPTER 4FOURIER SERIES AND INTEGRALS4.1FOURIER SERIES FOR PERIODIC FUNCTIONSThis section explains three Fourier series: sines, cosines, and exponentials eikx.Square waves (1 or 0 or 1) are great examples, with delta functions in the derivative.We look at a spike, a step function, and a ramp—and smoother functions too.Start with sin x. It has period 2π since sin(x 2π) sin x. It is an odd functionsince sin( x) sin x, and it vanishes at x 0 and x π. Every function sin nxhas those three properties, and Fourier looked at infinite combinations of the sines:Fourier sine seriesS(x) b1 sin x b2 sin 2x b3 sin 3x · · · bn sin nx (1)n 1If the numbers b1 , b2 , . . . drop off quickly enough (we are foreshadowing the importance of the decay rate) then the sum S(x) will inherit all three properties:Periodic S(x 2π) S(x)Odd S( x) S(x)S(0) S(π) 0200 years ago, Fourier startled the mathematicians in France by suggesting that anyfunction S(x) with those properties could be expressed as an infinite series of sines.This idea started an enormous development of Fourier series. Our first step is tocompute from S(x) the number bk that multiplies sin kx. Suppose S(x) bn sin nx. Multiply both sides by sin kx. Integrate from 0 to π: π π πS(x) sin kx dx b1 sin x sin kx dx · · · bk sin kx sin kx dx · · · (2)000On the right side, all integrals are zero except the highlighted one with n k.This property of “orthogonality” will dominate the whole chapter. The sines make90 angles in function space, when their inner products are integrals from 0 to π: πOrthogonalitysin nx sin kx dx 0 if n k .(3)0317
318Chapter 4 Fourier Series and IntegralsZero comes quickly if we integrate cos mx dx sin mx πm0 0 0. So we use this:11cos(n k)x cos(n k)x .(4)22Integrating cos mx with m n k and m n k proves orthogonality of the sines.Product of sinessin nx sin kx The exception is when n k. Then we are integrating (sin kx)2 12 12 cos 2kx: π π ππ11dx cos 2kx dx .(5)sin kx sin kx dx 222000The highlighted term in equation (2) is bkπ/2. Multiply both sides of (2) by 2/π: 2 π1 πSine coefficientsbk S(x) sin kx dx S(x) sin kx dx.(6)S( x) S(x)π 0π πNotice that S(x) sin kx is even (equal integrals from π to 0 and from 0 to π).I will go immediately to the most important example of a Fourier sine series. S(x)is an odd square wave with SW (x) 1 for 0 x π. It is drawn in Figure 4.1 asan odd function (with period 2π) that vanishes at x 0 and x π.SW (x) 1 π0π2π-xFigure 4.1: The odd square wave with SW (x 2π) SW (x) {1 or 0 or 1}.Example 1 Find the Fourier sine coefficients bk of the square wave SW (x).SolutionFor k 1, 2, . . . use the first formula (6) with S(x) 1 between 0 and π: " π2 π2 cos kx2 2 0 2 0 2 0(7), , , , , ,.sin kx dx bk π 0πkπ 1 2 3 4 5 60The even-numbered coefficients b2k are all zero because cos 2kπ cos 0 1. Theodd-numbered coefficients bk 4/πk decrease at the rate 1/k. We will see that same1/k decay rate for all functions formed from smooth pieces and jumps.Put those coefficients 4/πk and zero into the Fourier sine series for SW (x): 4 sin x sin 3x sin 5x sin 7xSquare waveSW (x) ···π1357(8)Figure 4.2 graphs this sum after one term, then two terms, and then five terms. Youcan see the all-important Gibbs phenomenon appearing as these “partial sums”
4.1 Fourier Series for Periodic Functions319include more terms. Away from the jumps, we safely approach SW (x) 1 or 1.At x π/2, the series gives a beautiful alternating formula for the number π: 11114 1 1 1 11 ···so that π 4 · · · . (9)π 1 3 5 71357The Gibbs phenomenon is the overshoot that moves closer and closer to the jumps.Its height approaches 1.18 . . . and it does not decrease with more terms of the series!Overshoot is the one greatest obstacle to calculation of all discontinuous functions(like shock waves in fluid flow). We try hard to avoid Gibbs but sometimes we can’t.4 sin x sin 3x π134 sin xDashedπ 1 ππ4 sin xsin 9x ··· π19overshoot SW 15 terms:Solid curvexxπ2Figure 4.2: Gibbs phenomenon: Partial sums N1bn sin nx overshoot near jumps.Fourier Coefficients are BestLet me look again at the first term b1 sin x (4/π) sin x. This is the closest possibleapproximation to the square wave SW , by any multiple of sin x (closest in the leastsquares sense). To see this optimal property of the Fourier coefficients, minimize theerror over all b1 : π π2The error is (SW b1 sin x) dx The b1 derivative is 2 (SW b1 sin x) sin x dx.00The integral of sin2 x is π/2. So the derivative is zero when b1 (2/π)This is exactly equation (6) for the Fourier coefficient. π0S(x) sin x dx.Each bk sin kx is as close as possible to SW (x). We can find the coefficients bkone at a time, because the sines are orthogonal. The square wave has b2 0 becauseall other multiples of sin 2x increase the error. Term by term, we are “projecting thefunction onto each axis sin kx.”Fourier Cosine SeriesThe cosine series applies to even functions with C( x) C(x):Cosine series C(x) a0 a1 cos x a2 cos 2x · · · a0 n 1an cos nx.(10)
320Chapter 4 Fourier Series and IntegralsEvery cosine has period 2π. Figure 4.3 shows two even functions, the repeatingramp RR(x) and the up-down train UD(x) of delta functions. That sawtoothramp RR is the integral of the square wave. The delta functions in UD give thederivative of the square wave. (For sines, the integral and derivative are cosines.)RR and UD will be valuable examples, one smoother than SW , one less smooth.First we find formulas for the cosine coefficients a0 and ak . The constant term a0is the average value of the function C(x): π1 π1a0 Averagea0 C(x) dx C(x) dx.(11)π 02π πI just integrated every term in the cosine series (10) from 0 to π. On the right side,the integral of a0 is a0 π (divide both sides by π). All other integrals are zero: ππsin nxcos nx dx 0 0 0.(12)n00In words, the constant function 1 is orthogonal to cos nx over the interval [0, π].The other cosine coefficients ak come from the orthogonality of cosines. As withsines, we multiply both sides of (10) by cos kx and integrate from 0 to π: π π π πC(x) cos kx dx a0 cos kx dx a1 cos x cos kx dx ·· ak(cos kx)2 dx ··0000You know what is coming. On the right side, only the highlighted term can benonzero. Problem 4.1.1 proves this by an identity for cos nx cos kx—now (4) has aplus sign. The bold nonzero term is akπ/2 and we multiply both sides by 2/π: 2 π1 πCosine coefficientsak C(x) cos kx dx C(x) cos kx dx . (13)C( x) C(x)π 0π πAgain the integral over a full period from π to π (also 0 to 2π) is just doubled.2δ(x) 6RR(x) x π0π-2πRepeating Ramp RR(x)Integral of Square Wavex2δ(x 2π) 6Up-down U D(x) π0 2δ(x π)?π2π-x 2δ(x π)?Figure 4.3: The repeating ramp RR and the up-down UD (periodic spikes) are even.The derivative of RR is the odd square wave SW . The derivative of SW is U D.
4.1 Fourier Series for Periodic Functions321Example 2 Find the cosine coefficients of the ramp RR(x) and the up-down UD(x).SolutionThe simplest way is to start with the sine series for the square wave: 4 sin x sin 3x sin 5x sin 7x ··· .SW (x) π1357Take the derivative of every term to produce cosines in the up-down delta function:Up-down seriesUD(x) 4[cos x cos 3x cos 5x cos 7x · · · ] .π(14)Those coefficients don’t decay at all. The terms in the series don’t approach zero, soofficially the series cannot converge. Nevertheless it is somehow correct and important.Unofficially this sum of cosines has all 1’s at x 0 and all 1’s at x π. Then and are consistent with 2δ(x) and 2δ(x π). The true way to recognize δ(x) isby the test δ(x)f (x) dx f (0) and Example 3 will do this.For the repeating ramp, we integrate the square wave series for SW (x) and add theaverage ramp height a0 π/2, halfway from 0 to π: π π cos x cos 3x cos 5x cos 7x ··· .(15)Ramp series RR(x) 24 12325272ak drop off like 1/k 2 . They could beThe constant of integration is a0 . Those coefficients computed directly from formula (13) using x cos kx dx, but this requires an integrationby parts (or a table of integrals or an appeal to Mathematica or Maple). It was mucheasier to integrate every sine separately in SW (x), which makes clear the crucial point:Each “degree of smoothness” in the function is reflected in a faster decay rate of itsFourier coefficients ak and bk .No decay1/k decay1/k2 decay1/k4 decayr k decay with r 1Delta functions (with spikes)Step functions (with jumps)Ramp functions (with corners)Spline functions (jumps in f )Analytic functions like 1/(2 cos x)Each integration divides the kth coefficient by k. So the decay rate has an extra1/k. The “Riemann-Lebesgue lemma” says that ak and bk approach zero for anycontinuous function (in fact whenever f (x) dx is finite). Analytic functions achievea new level of smoothness—they can be differentiated forever. Their Fourier seriesand Taylor series in Chapter 5 converge exponentially fast.The poles of 1/(2 cos x) will be complex solutions of cos x 2. Its Fourier seriesconverges quickly because r k decays faster than any power 1/k p . Analytic functionsare ideal for computations—the Gibbs phenomenon will never appear.Now we go back to δ(x) for what could be the most important example of all.
322Chapter 4 Fourier Series and IntegralsExample 3 Find the (cosine) coefficients of the delta function δ(x), made 2π-periodic.Solution The spike occurs at the start of the interval [0, π] so safer to integrate from π to π. We find a0 1/2π and the other ak 1/π (cosines because δ(x) is even): π11δ(x) dx Average a0 2π π2π 11 πCosines ak δ(x) cos kx dx π ππThen the series for the delta function has all cosines in equal amounts:Delta functionδ(x) 11 [cos x cos 2x cos 3x · · · ] .2π π(16)Again this series cannot truly converge (its terms don’t approach zero). But we can graphthe sum after cos 5x and after cos 10x. Figure 4.4 shows how these “partial sums” aredoing their best to approach δ(x). They oscillate faster and faster away from x 0.Actually there is a neat formula for the partial sum δN (x) that stops at cos Nx. Startby writing each term 2 cos θ as eiθ e iθ :δN 11[1 2 cos x · · · 2 cos Nx] 1 eix e ix · · · eiN x e iN x .2π2πThis is a geometric progression that starts from e iN x and ends at eiN x . We have powersof the same factor eix . The sum of a geometric series is known:Partial sumup to cos N x1δN (x) 11 ei(N 2 )x e i(N 2 )x1 sin(N 12 )x. 2πeix/2 e ix/22πsin 12 x(17)This is the function graphed in Figure 4.4. We claim that for any N the area underneathδN (x) is 1. (Each cosine integrated from π to π gives zero. The integral of 1/2π is1.) The central “lobe” in the graph ends when sin(N 12 )x comes down to zero, andthat happens when (N 12 )x π. I think the area under that lobe (marked by bullets)approaches the same number 1.18 . . . that appears in the Gibbs phenomenon.In what way does δN (x) approach δ(x)? The terms cos nx in the series jump around1at each point x 0, not approaching zero. At x π we see 2π[1 2 2 2 · · · ] andthe sum is 1/2π or 1/2π. The bumps in the partial sums don’t get smaller than 1/2π.The right test for the delta function δ(x) is to multiply by a smoothf (x) ak cos kx and integrate, because we only know δ(x) from its integrals δ(x)f (x) dx f (0):Weak convergenceof δN (x) to δ(x) π πδN (x)f (x) dx a0 · · · aN f (0) .(18)In this integrated sense (weak sense) the sums δN (x) do approach the delta function !The convergence of a0 · · · aN is the statement that at x 0 the Fourier series of asmooth f (x) ak cos kx converges to the number f (0).
4.1 Fourier Series for Periodic Functionsδ10 (x)height 21/2πδ5 (x) π323height 11/2πheight 1/2ππ height 1/2π0Figure 4.4: The sums δN (x) (1 2 cos x · · · 2 cos Nx)/2π try to approach δ(x).Complete Series: Sines and CosinesOver the half-period [0, π], the sines are not orthogonal to all the cosines. In fact theintegral of sin x times 1 is not zero. So for functions F (x) that are not odd or even,we move to the complete series (sines plus cosines) on the full interval. Since ourfunctions are periodic, that “full interval” can be [ π, π] or [0, 2π]:Complete Fourier seriesF (x) a0 n 1an cos nx bn sin nx .(19)n 1On every “2π interval” all sines and cosines are mutually orthogonal. We find theFourier coefficients ak and bk in the usual way: Multiply (19) by 1 and cos kx andsin kx, and integrate both sides from π to π: π 11 π1 πa0 F (x) dx ak F (x) cos kx dx bk F (x) sin kx dx. (20)2π ππ ππ πOrthogonality kills off infinitely many integrals and leaves only the one we want.Another approach is to split F (x) C(x) S(x) into an even part and an oddpart. Then we can use the earlier cosine and sine formulas. The two parts areC(x) Feven (x) F (x) F ( x)2S(x) Fodd (x) F (x) F ( x).2(21)The even part gives the a’s and the odd part gives the b’s. Test on a short squarepulse from x 0 to x h—this one-sided function is not odd or even.
324Chapter 4 Fourier Series and Integrals Example 4 Find the a’s and b’s if F (x) square pulse 1 for 0 x h0 for h x 2πSolution The integrals for a0 and ak and bk stop at x h where F (x) drops to zero.The coefficients decay like 1/k because of the jump at x 0 and the drop at x h: hh1Coefficients of square pulsea0 average1 dx 2π 02π sin kh1 cos kh1 h1 hbk .(22)cos kx dx sin kx dx ak π 0π 0πkπkIf we divide F (x) by h, its graph is a tall thin rectangle: height h1 , base h, and area 1.When h approaches zero, F (x)/h is squeezed into a very thin interval. The tallrectangle approaches (weakly) the delta function δ(x). The average height is area/2π 1/2π. Its other coefficients ak /h and bk /h approach 1/π and 0, already known for δ(x):F (x) δ(x)h1 sin kh1ak hπ khπandbk1 cos kh 0 as h 0. (23)hπkhWhen the function has a jump, its Fourier series picks the halfway point. Thisexample would converge to F (0) 12 and F (h) 12 , halfway up and halfway down.The Fourier series converges to F (x) at each point where the function is smooth.This is a highly developed theory, and Carleson won the 2006 Abel Prize by provingconvergencefor every x except a set of measure zero. If the function has finite energy F (x) 2 dx, he showed that the Fourier series converges “almost everywhere.”Energy in Function Energy in CoefficientsThere is an extremely important equation (the energy identity) that comes fromintegrating (F (x))2 . When we square the Fourier series of F (x), and integrate from π to π, all the “cross terms” drop out. The only nonzero integrals come from 12and cos2 kx and sin2 kx, multiplied by a20 and a2k and b2k : π Energy in F (x) π (a0 ak cos kx bk sin kx)2 dx π(24)(F (x))2dx 2πa20 π(a21 b21 a22 b22 · · · ). πThe energy in F (x) equals the energy in the coefficients. The left side is like thelength squared of a vector, except the vector is a function. The right side comes froman infinitely long vector of a’s and b’s. The lengths are equal, which says that theFourier transform from function to vector is like an orthogonal matrix. Normalizedby constants 2π and π, we have an orthonormal basis in function space.What is this function space ? It is like ordinary 3-dimensional space, except the“vectors” are functions. Their length f comes from integrating instead of adding:f 2 f (x) 2 dx. These functions fill Hilbert space. The rules of geometry hold:
4.1 Fourier Series for Periodic FunctionsLength f2 (f, f ) comes from the inner product (f, g) 325f (x)g(x) dxOrthogonal functions (f, g) 0 produce a right triangle: f g2 f2 g2I have tried to draw Hilbert space in Figure 4.5. It has infinitely many axes. Theenergy identity (24) is exactly the Pythagoras Law in infinite-dimensional space.cos kxv2k 1 π Isin kxv2k πsin xv2 π:f A0 v0 A1 v1 B1 v2 · · ·function in Hilbert spacef 90(vi, vj ) 01v0 2πR2 A20 A21 B12 · · ·cos xv1 πFigure 4.5: The Fourier series is a combination of orthonormal v’s (sines and cosines).Complex Exponentials ck eikxThis is a small step and we have to take it. In place of separate formulas for a0 and akand bk , we will have one formula for all the complex coefficients ck . And the functionF (x) might be complex (as in quantum mechanics). The Discrete Fourier Transformwill be much simpler when we use N complex exponentials for a vector. We practicein advance with the complex infinite series for a 2π-periodic function:F (x) c0 c1 eix c 1 e ix · · · Complex Fourier series cn einx (25)n If every cn c n , we can combine einx with e inx into 2 cos nx. Then (25) is thecosine series for an even function. If every cn c n , we use einx e inx 2i sin nx.Then (25) is the sine series for an odd function and the c’s are pure imaginary.To find ck, multiply (25) by e ikx (not eikx ) and integrate from π to π: π π π πF (x)e ikx dx c0 e ikx dx c1 eix e ikx dx · · · ckeikxe ikxdx · · · π π π πThe complex exponentials are orthogonal. Every integral on the right side is zero,except for the highlighted term (when n k and eikx e ikx 1). The integral of 1 is2π. That surviving term gives the formula for ck : πFourier coefficientsF (x)e ikx dx 2πckfor k 0, 1, . . .(26) π
326Chapter 4 Fourier Series and IntegralsNotice that c0 a0 is still the average of F (x), because e0 1. The orthogonalityof einx and eikx is checked by integrating, as always. But the complex inner product(F, G) takes the complex conjugate G of G. Before integrating, change eikx to e ikx :Complex inner product π(F, G) F (x)G(x) dx πOrthogonality of einx and eikx i(n k)x π πei(n k)xedx 0.i(n k) π π(27)Example 5 Add the complex series for 1/(2 eix ) and 1/(2 e ix ). These geometricseries have exponentially fast decay from 1/2k . The functions are analytic.1 eix e2ix ·· 2481 e ix e 2ix ··248 1 cos x cos 2x cos 3x ··248When we add those functions, we get a real analytic function:4 2 cos x1(2 e ix ) (2 eix )1 ix ixix ix2 e2 e(2 e )(2 e )5 4 cos x(28)This ratio is the infinitely smooth function whose cosine coefficients are 1/2 k . Example 6 Find ck for the 2π-periodic shifted pulse F (x) Solution1 for s x s h0 elsewhere in [ π, π]The integrals (26) from π to π become integrals from s to s h: s hs h11 e ikx ikxck 1·edx e iks2π s2π ik s1 e ikh2πik.(29)Notice above all the simple effect of the shift by s. It “modulates” each c k by e iks . Theenergy is unchanged, the integral of F 2 just shifts, and all e iks 1:Shift F (x) to F (x s) Multiply ck by e iks .(30)Example 7 Centered pulse with shift s h/2. The square pulse becomes centeredaround x 0. This even function equals 1 on the interval from h/2 to h/2:Centered by s h2ck eikh/21 sin(kh/2)1 e ikh .2πik2πk/2Divide by h for a tall pulse. The ratio of sin(kh/2) to kh/2 is the sinc function: 1 Fcenteredkh1/h for h/2 x h/2ikx e sincTall pulse0elsewhere in [ π, π]h2π2 1That division by h produces area 1. Every coefficient approaches 2πas h 0.The Fourier series for the tall thin pulse again approaches the Fourier series for δ(x).
3274.1 Fourier Series for Periodic FunctionsHilbert space can contain vectors c (c0 , c1 , c 1 , c2 , c 2 , · · · ) instead of functionsF (x). The length of c is 2π ck 2 F 2dx. The function space is often denotedby L2 and the vector space is 2 . The energy identity is trivial (but deep). Integratingthe Fourier series for F (x) times F (x), orthogonality kills every cn ck for n k. Thisleaves the ck ck ck 2 : π π F (x) 2dx (cn einx )(ck e ikx )dx 2π( c0 2 c1 2 c 1 2 ··) . (31) π πThis is Plancherel’s identity: The energy in x-space equals the energy in k-space. ikxFinally I want to emphasize the three big rules for operating on F (x) ck e :1.2.3.dFhas Fourier coefficients ikck (ene
318 Chapter 4 Fourier Series and Integrals Zero comes quickly if we integrate cosmxdx sinmx m π 0 0 0. So we use this: Product of sines sinnx sinkx 1 2 cos(n k)x 1 2 cos(n k)x. (4) Integrating cosmx with m n k and m n k proves orthogonality of the sines.
Part One: Heir of Ash Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 .
Gambar 5. Koefisien Deret Fourier untuk isyarat kotak diskret dengan (2N1 1) 5, dan (a) N 10, (b) N 20, dan (c) N 40. 1.2 Transformasi Fourier 1.2.1 Transformasi Fourier untuk isyarat kontinyu Sebagaimana pada uraian tentang Deret Fourier, fungsi periodis yang memenuhi persamaan (1) dapat dinyatakan dengan superposisi fungsi sinus dan kosinus.File Size: 568KB
TO KILL A MOCKINGBIRD. Contents Dedication Epigraph Part One Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Part Two Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18. Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26
(d) Fourier transform in the complex domain (for those who took "Complex Variables") is discussed in Appendix 5.2.5. (e) Fourier Series interpreted as Discrete Fourier transform are discussed in Appendix 5.2.5. 5.1.3 cos- and sin-Fourier transform and integral Applying the same arguments as in Section 4.5 we can rewrite formulae (5.1.8 .
DEDICATION PART ONE Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 PART TWO Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 .
straightforward. The Fourier transform and inverse Fourier transform formulas for functions f: Rn!C are given by f ( ) Z Rn f(x)e ix dx; 2Rn; f(x) (2ˇ) n Z Rn f ( )eix d ; x2Rn: Like in the case of Fourier series, also the Fourier transform can be de ned on a large class of generalized functions (the space of tempered
Expression (1.2.2) is called the Fourier integral or Fourier transform of f. Expression (1.2.1) is called the inverse Fourier integral for f. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to anoth
to denote the Fourier transform of ! with respect to its first variable, the Fourier transform of ! with respect to its second variable, and the two-dimensional Fourier transform of !. Variables in the spatial domain are represented by small letters and in the Fourier domain by capital letters. expressions, k is an index assuming the two values O
