Fourier Series and Integrals

Transcription

1 Fourier Series and Integrals Fourier Series et f(x) beapiece-wiselinearfunctionon[, ] (Thismeansthatf(x) maypossessa finite number of finite discontinuities on the interval). Then f(x) canbeexpandedina Fourier series f(x) = a + a n cos nπx + b n sin nπx, (a) or, equivalently, with n= f(x) = c n e inπx/, (a n ib n )/ n<, c n = (a n + ib n )/ n>, a / n =. AusefulschematicformoftheFourierseriesis (b) f(x) = n (a n ĉ n + b n ŝ n ). () This emphasizes that the Fourier series can be viewed as an expansion of a vector f in Hilbert space, in a basis that is spanned by the ĉ n (cosine waves of different periodicities) and the ŝ n (sine waves). To invert the Fourier expansion, multiply Eq. (a) by cos nπx nπx or sin and integrate over the interval. For this calculation, we need the basic orthogonality relation of the basis functions: cos nπx mπx cos dx = δ mn, (3) and similarly for the sin functions. Intuitively, for n m, thereisdestructiveinterference of the two factors in the integrand, while for n = m, thereisconstructiveinterference. Thus by multiplying Eq. (a) by cos nπx and integrating, we obtain f(x)cos mπx [ ] dx = a + a n cos nπx + b nπx n cos mπx dx, (4) from which we find a n = f(x)cos nπx dx b n = f(x)sin nπx dx. (5)

2 The inversion of the Fourier series can be viewed as finding the projectionsof f along each basis direction. Schematically, therefore, the inversion can be represented as f ĉ m = n (a n ĉ n + b n ŝ n ) ĉ m = a n δ nm = a m, (6a) which implies a m = f ĉ n. These steps parallel the calculation that led to Eq. (5). Thisschematic representationemphasizes that the Fourier decomposition of a function is completely analogous to the expansion of a vector in Hilbert space in an orthogonal basis. The components of the vector correspond to the Fourier amplitudes defined in Eq. (5). Examples:. Square Wave f(x) (6b) h x Clearly f(x) isanoddfunction:f(x) = f( x). Hence a n = f(x)cos nπx dx = n. As a result, the spectral information of the square wave is entirely contained in the b n s. These coefficients are b n = f(x)sin nπx dx = h sin nπx h dx = ( cos nπ), nπ (7a) from which we find b n = { 4h/nπ n odd, n even. Thus the square wave can be written as a Fourier sine series f(x) = 4h π ( sin πx + 3πx sin 3 + 5πx sin 5 + (7b) ). (8) There are a number of important general facts about this expansion worth emphasizing:

3 Since f(x) isodd,onlyodd powers of the antisymmetric basis functions appear. At the discontinuities of f(x), the Fourier series converges to the mean of the two values of f(x) oneithersideofthediscontinuity. Apictureoffirstfewtermsoftheseriesdemonstratesthenature of the convergence to the square wave; each successive term in the series attempts to correct for the overshoot present in the sum of all the previous terms (a) one term (b) two terms (c) three terms (d) ten terms The amplitude spectrum decays as n ;thisindicatesthatasquarewavecanbewellrepresented by the fundamental frequency plus the first few harmonics. By examining the series for particular values of x, usefulsummationformulaemay sometimes be found. For example, setting x = in the Fourier sine series gives: f(x=/) = h = 4h π ( sin π + 3 sin 3π + 5 sin 5π + ). (9a) and this leads to a nice (but slowly converging) series representation for π 4 : π ( 4 = ). (9b) Alternatively we may represent the square wave as an even function. h f(x) x

4 Now f(x) =+f( x). By symmetry, then, b n = n, andonlythea n s are non-zero. For these coefficients we find a n = f(x)cos nπx dx = f(x)cos nπx dx [ = h / cos nπx ] dx cos nπx / dx [ = h nπ/ ] nπ cos udu cos udu nπ = h nπ = nπ/ [ sin u nπ/ sin u nπ nπ/ { 4h nπ ( )(n )/ n odd n even. Hence f(x) canbealsorepresentedasthefouriercosineseries f(x) = 4h π ] ( cos πx 3πx cos 3 + 5πx ) cos. () 5 This example illustrates the use of symmetry in determining a Fourier series, even function cosine series odd function sine series no symmetry both sine and cosine series Thus it is always simpler to choose an origin so that f(x) hasadefinitesymmetry,sothat it can be represented by either a sin or cosine series. Periodic Parabola The periodic parabola is the periodic extension of the function x,intherange[ π, π], to the entire real line. Given any function defined on the interval [a, b], the periodic extension may be constructed in a similar fashion. In general, we can Fourier expand any function on afiniterange;thefourierserieswillconvergetotheperiodic extension of the function. 3π π π 3π Figure : The periodic parabola 4

5 The Fourier expansion of the repeated parabola gives b n = a = π a n = π π π π n, x dx = 3, x cos nx dx =( ) n 4 n. Thus we can represent the repeated parabola as a Fourier cosine series Notice several interesting facts: f(x) =x = π 3 +4 n= ( ) n n cos nx. () The a term represents the average value of the function. For this example, this average is non-zero. Since f is even, the Fourier series has only cosine terms. There are no discontinuities in f, butfirstderivativeisdiscontinuous;thisimplies that the amplitude spectrum decays as /n.ingeneral,the smoother thefunction, the faster the decay of the amplitude spectrum as a function of n. Consequently, progressively fewer terms in the Fourier series are needed to representthewaveformto afixeddegreeofaccuracy. Setting x = π in the series gives π = π 3 +4 n= n= ( ) n n cos nπ, from which we find the cool formula n = π 6. () The approach given above provides an indirect, but simple, way to sum inverse powers of the integers. Simply Fourier expand the function x k on the interval [ π, π] andthenevaluate the series at x = π from which n= n k can be computed. The sum n= n z ζ(z), is called the Riemann zeta function, andbythisfourierseriestrickthezetafunctioncanbe evaluated for all positive integer values of z. In summary, a Fourier series represents a spectral decomposition of a periodic waveform into a series of harmonics of various frequencies. From the amplitudes of these harmonics we can gain understanding of the physical process underlying the waveform. For example, 5

6 one perceives in an obvious way the different tonal quality of an oboe and a violin. The origin of this difference are the distinct Fourier spectra of these two instruments. Even if the same note is being played, the different shapes of the two instruments(stringforthe violin; air column for the oboe) lead to very different contributions of higher harmonics. This translates into distinct respective Fourier spectra, and hence, very different musical tones. Fourier Integrals For non-periodic functions, we generalize the Fourier series to functions defined on [, ] with.toaccomplishthis,wetakeeq.(5), a n = f(x)cos nπx dk, and now let k = nπ/ ndk.thenas,weobtain A(k) a n = and, by following the same procedure, B(k) b n = Now the original Fourier series becomes f(x) = n = n π a n cos nπx f(x)coskx dx, f(x)sinkx dx. + b n sin nπx A(k) B(k) cos kx + sin kx [A(k)coskx + B(k)sinkx] dk. (3a) (3b) By using the exponential form of the Fourier series, we have the alternative, but more familiar and convenient Fourier integral representation of f(x), f(x) = f(k)e ikx dk. (4) Where the (arbitrary) prefactor is chosen to be for convenience, as then the same prefactor appears in the definition of the inverse Fourier transform. In symbolic form, the Fourier integral can be represented as f = f k ê k. continuous k 6

7 To find the expansion coefficients f(k), we proceed in precisely the same manner as in the case of Fourier series. That is, multiply f(x) byoneofthebasisfunctionsandintegrateover asuitablerange.thisgives f(x)e ik x dx = f(k)e i(k k )x dx dk. (5) However, the integral over x is just the Dirac delta function. To see this, we write e i(k k )x dx = lim e i(k k )x dx = sin(k k ). (6a) k k The function on the right-hand side is peaked at k = k,withtheheightofthepeakequal to, andawidthequaltoπ/. As,thepeakbecomesinfinitelyhighandnarrow in such a way that the integral under the curve remains a constant. This constant equals. Thuswewrite e i(k k )x dx =δ(k k ). (6b) Using this result in Eq. (5), we obtain f(k) = f(x)e ikx dx. (7) Schematically, f ê k = k f k ê k ê k f k = f ê k. These developments can be extended in a straightforward way to multidimensional functions. The Fourier transform has a number of important features that are very useful for a variety of physical applications: If f(x) isnon-zerooversomerangex about x =,thenf(k) isnon-zeroinarange /X about k =. Whilethisprecedingstatementisimprecise,theinverserelation between k and x is physically quite useful. Parseval relation: f(x) dx = f(k) dk. Differentiation: F[f (n) (x)] = (ik) n F[f(x)]. Here F[f(x)] denotes the Fourier transform of f(x) andthesuperscript(n) denotesthen th derivative. Thus the Fourier transform of the n th derivative of f(x) isjustthefouriertransformoff(x) itselftimes(ik) n. Translation: F[(f(x+a)] = f(x+a) e ikx dx = e ika F[f(x)]. This statement is extremely useful for translationally invariant systems. 7

8 Convolution: F(f g) =f(k)g(k), where the notation f g means the convolution of f and g and is defined by f g f(x y)g(y) dy. (8) In the integrand, we may think of some physical attribute being propagated from to x y by the propagator function f(x y) andthenpropagatedfromx y to x by the propagator function g(y). The intermediate point y is integrated over, so that the convolution is a function only of x. Thuswemaywritef g = F (x). The convolution statement F(f g) =f(k)g(k) hasmanyimportantphysicalapplications. To show the convolution statement, we rewrite (8) in terms of the respective Fourier transforms F (x) = F (k)e ikx dk = = () 3/ = f(x y)g(y) dy f(k )e ik (x y) g(k ) e ik y dk dk dy e iy(k k ) f(k )g(k ) e ik x dk dk dy f(k )g(k )e ik x dk. (9) In going from the third to the last line, we integrate over y to give δ(k k ). This reduces the integral over dk and dk to a single integral over dk. Equating the integrands, we conclude that F (k) =f(k)g(k). Thus aconvolutioninrealspace reduces to multiplication in Fourier space. Examples. Rectangular Pulse f(t) = { t <T/ t >T/. Then f(ω) = T/ T/ e iωt dt = T sin ωt/ ωt/. () Notice that there is an inverse relationship between the width of f(t) and the width of the corresponding Fourier transform f(ω). The Fourier spectrum is concentrated at zero frequency; these conspire to give destructive interference exceptintherange t < T.. Damped Harmonic Wave f(t) =e iω t αt t>. 8

9 f(t) f(ω) T t T /T ω Figure : Rectangular pulse and its Fourier transform. The corresponding Fourier transform is or f(ω) = e αt+i(ω ω)t dt = [α i(ω ω)], (a) f(ω) [α +(ω ω) ] /. (b) This latter waveform is often called a orentzian. The relation between thedamped harmonic wave the its Fourier transform are shown below for the case ω =andα = f(t)=sint(ω t)e αt 4 3 /α f(ω). t α.5 3 ω ω 4 4 Figure 3: Damped harmonic wave and its Fourier transform. As the damping goes to zero, the width of the orentzian also vanishes. This embodies the fact that a less weakly-damped waveform is less contaminated withfrequencycomponents not equal to ω. Clearly, as the damping goes to zero, only the fundamental frequency remains, and the Fourier transform is a delta function. 3. Gaussian f(t) = t e t /t. () 9

10 To calculate this Fourier transform, complete the square in the exponential: f(ω) = = = t t t = π e t /t iωt dt e t /t iωt+ω t / ω t / dt e (t/ t iωt / ) ω t / dt e (u iωt / ) ω t / dt = e ω t /. (3) In the second-to-last line, we introduce u = t/ t,andforthelastlineweusethefact that e u du = π.thecrucialpointisthatthefouriertransformofagaussianisalso agaussian! Thewidthofthetransformfunctionequaltotheinverse width of the original waveform. 4. Finite Wave Train /T T f(ω) T. 3 ω ω. 4 4 Figure 4: Finite wave train and its Fourier transform. Then f(t) =cosω t = Re(e iω t ) f(ω) = T <t<t. e i(ω ω)t dt = e i(ω ω)t i(ω ω) = ei(ω ω)t/ sin(ω ω)t/ ω ω. (4) The behavior of the finite wavetrain is closely analogous to that of the damped harmonic wave. The shorter the duration of the wavetrain, the more its Fourier transform is contaminated with other frequencies beyond the fundamental.

11 Wavevector Quantization In certain applications, typically in solid state physics or otherproblemsdefinedonalattice, f(x) isdefinedonlyonadiscrete,periodicsetpoints.thediscreteness yields a condition of the range of possible wavevectors as well as the number of normal modes in the system. As atypicalexample,considerthetransverseoscillationsofaloadedstringoflength =4a which contains 5 point masses, with both endpoints held fixed. Thereareadiscretesetof normal mode oscillations for this system; these modes and their corresponding wavevectors k n = nπ are shown below. Also shown is the corresponding continuous waveform, sin k 4a nx (dashed). a a (a) k = 3a 4a (b) k = π/4a (c) k =/4a (d) k =3π/4a (e) k =4π/4a Notice that k = π and k =giveidenticaldisplacementpatternsforthediscretesystem. a Therefore in defining the Fourier transform, it is redundant to consider values of k such that k π. The range <k< π is called the first Brillouin zone in the context of solid-state a a physics. Thus when performing a Fourier analysis on a discrete system (either Fourier series or Fourier integral), one must account for the restriction imposed by the system discreteness in defining the appropriate range of k values.