9 Transform Techniques in Physics

Transcription

1 9 Transform Techniques in Physics There is no branch of mahemaics, however absrac, which may no some day be applied o phenomena of he real world., Nikolai Lobachevsky ( ) 9. Inroducion Some of he mos powerful ools for solving problems in physics are ransform mehods. The idea is ha one can ransform he problem a hand o a new problem in a differen space, hoping ha he problem in he new space is easier o solve. Such ransforms appear in many forms. As we had seen in Chaper 3 and will see laer in he book, he soluions of linear parial differenial equaions can be found by using he mehod of separaion of variables o reduce solving parial differenial equaions (PDEs) o solving ordinary differenial equaions (ODEs). We can also use ransform mehods o ransform he given PDE ino ODEs or algebraic equaions. Solving hese equaions, we hen consruc soluions of he PDE (or, he ODE) using an inverse ransform. A schemaic of hese processes is shown below and we will describe in his chaper how one can use Fourier and Laplace ransforms o his effec. In his chaper we will explore he use of inegral ransforms. Given a funcion f (x), we define an inegral ransform o a new funcion F(k) as b F(k) f (x)k(x, k) dx. a Here K(x, k) is called he kernel of he ransform. We will concenrae specifically on Fourier ransforms, ˆf (k) f (x)e ikx dx, and Laplace ransforms F(s) f ()e s d. Transforms ODE Figure 9.: Schemaic indicaing ha PDEs and ODEs can be ransformed o simpler problems, solved in he new space and ransformed back o he original space. PDE AlgEq Inverse Transforms

2 358 parial differenial equaions 9.. Example - The Linearized KdV Equaion As a relaively simple example, we consider he linearized Korewegde Vries (KdV) equaion: u + cu x + βu xxx, < x <. (9.) The nonlinear counerpar o his equaion is he Koreweg-de Vries (KdV) equaion: u + 6uu x + u xxx. This equaion was derived by Diederik Johannes Koreweg (848-94) and his suden Gusav de Vries ( ). This equaion governs he propagaion of raveling waves called solions. These were firs observed by John Sco Russell (88-882) and were he source of a long debae on he exisence of such waves. The hisory of his debae is ineresing and he KdV urned up as a generic equaion in many oher fields in he laer par of he las cenury leading o many papers on nonlinear evoluion equaions. This equaion governs he propagaion of some small ampliude waer waves. Is nonlinear counerpar has been a he cener of aenion in he las 4 years as a generic nonlinear wave equaion. We seek soluions ha oscillae in space. So, we assume a soluion of he form u(x, ) A()e ikx. (9.2) Such behavior was seen in Chapers 3 and 6 for he wave equaion for vibraing srings. In ha case, we found plane wave soluions of he form e ik(x±c), which we could wrie as e i(kx±ω) by defining ω kc. We furher noe ha one ofen seeks complex soluions as a linear combinaion of such forms and hen akes he real par in order o obain physical soluions. In his case, we will find plane wave soluions for which he angular frequency ω ω(k) is a funcion of he wavenumber. Insering he guess (9.2) ino he linearized KdV equaion, we find ha da d + i(ck βk3 )A. (9.3) Thus, we have convered he problem of seeking a soluion of he parial differenial equaion ino seeking a soluion o an ordinary differenial equaion. This new problem is easier o solve. In fac, given an iniial value, A(), we have A() A()e i(ck βk3 ). (9.4) Therefore, he soluion of he parial differenial equaion is u(x, ) A()e ik(x (c βk2 )). (9.5) We noe ha his soluion akes he form e i(kx ω), where ω ck βk 3. A dispersion relaion is an expression giving he angular frequency as a funcion of he wave number, ω ω(k). In general, he equaion ω ω(k) gives he angular frequency as a funcion of he wave number, k, and is called a dispersion relaion. For β, we see ha c is nohing bu he wave speed. For β, he wave speed is given as v ω k c βk2. This suggess ha waves wih differen wave numbers will ravel a differen speeds. Recalling ha wave numbers are relaed o wavelenghs, k 2π λ, his means ha waves wih differen wavelenghs will ravel a differen speeds. For example, an iniial localized wave packe will no mainain is

3 ransform echniques in physics 359 shape. I is said o disperse, as he componen waves of differing wavelenghs will end o par company. For a general iniial condiion, we wrie he soluions o he linearized KdV as a superposiion of plane waves. We can do his since he parial differenial equaion is linear. This should remind you of wha we had done when using separaion of variables. We firs sough produc soluions and hen ook a linear combinaion of he produc soluions o obain he general soluion. For his problem, we will sum over all wave numbers. The wave numbers are no resriced o discree values. We insead have a coninuous range of values. Thus, summing over k means ha we have o inegrae over he wave numbers. Thus, we have he general soluion The exra 2π has been inroduced o be consisen wih he definiion of he Fourier ransform which is given laer in he chaper. u(x, ) A(k, )e ik(x (c βk2 )) dk. (9.6) 2π Noe ha we have indicaed ha A is a funcion of k. This is similar o inroducing he A n s and B n s in he series soluion for waves on a sring. How do we deermine he A(k, ) s? We inroduce as an iniial condiion he iniial wave profile u(x, ) f (x). Then, we have f (x) u(x, ) A(k, )e ikx dk. (9.7) 2π Thus, given f (x), we seek A(k, ). In his chaper we will see ha A(k, ) f (x)e ikx dx. This is wha is called he Fourier ransform of f (x). I is jus one of he so-called inegral ransforms ha we will consider in his chaper. In Figure 9.2 we summarize he ransform scheme. One can use mehods like separaion of variables o solve he parial differenial equaion direcly, evolving he iniial condiion u(x, ) ino he soluion u(x, ) a a laer ime. u(x, ) Fourier Transform A(k, ) Figure 9.2: Schemaic of using Fourier ransforms o solve a linear evoluion equaion. PDE ODE u(x, ) A(k, ) Inverse Fourier Transform The ransform mehod works as follows. Saring wih he iniial condiion, one compues is Fourier Transform (FT) as 2 A(k, ) f (x)e ikx dx. 2 Noe: The Fourier ransform as used in his secion and he nex secion are defined slighly differenly han how we will define hem laer. The sign of he exponenials has been reversed.

4 36 parial differenial equaions Applying he ransform on he parial differenial equaion, one obains an ordinary differenial equaion saisfied by A(k, ) which is simpler o solve han he original parial differenial equaion. Once A(k, ) has been found, hen one applies he Inverse Fourier Transform (IFT) o A(k, ) in order o ge he desired soluion: u(x, ) 2π 2π A(k, )e ikx dk 9..2 Example 2 - The Free Paricle Wave Funcion A(k, )e ik(x (c βk2 )) dk. (9.8) The one dimensional ime dependen Schrödinger equaion. A more familiar example in physics comes from quanum mechanics. The Schrödinger equaion gives he wave funcion Ψ(x, ) for a paricle under he influence of forces, represened hrough he corresponding poenial funcion V(x). The one dimensional ime dependen Schrödinger equaion is given by i hψ h2 2m Ψ xx + VΨ. (9.9) We consider he case of a free paricle in which here are no forces, V. Then we have i hψ h2 2m Ψ xx. (9.) Taking a hin from he sudy of he linearized KdV equaion, we will assume ha soluions of Equaion (9.) ake he form Ψ(x, ) φ(k, )e ikx dk. 2π [Here we have oped o use he more radiional noaion, φ(k, ) insead of A(k, ) as above.] Insering he expression for Ψ(x, ) ino (9.), we have i h dφ(k, ) e ikx dk h2 φ(k, )(ik) 2 e ikx dk. d 2m Since his is rue for all, we can equae he inegrands, giving dφ(k, ) i h d h2 k 2 φ(k, ). 2m As wih he las example, we have obained a simple ordinary differenial equaion. The soluion of his equaion is given by i hk2 φ(k, ) φ(k, )e 2m. Applying he inverse Fourier ransform, he general soluion o he ime dependen problem for a free paricle is found as Ψ(x, ) ik(x hk φ(k, )e 2m ) dk. 2π

5 ransform echniques in physics 36 We noe ha his akes he familiar form Ψ(x, ) 2π where he dispersion relaion is found as φ(k, )e i(kx ω) dk, The wave speed is given as ω hk2 2m. v ω k hk 2m. As a special noe, we see ha his is no he paricle velociy! Recall ha he momenum is given as p hk. 3 So, his wave speed is v p 2m, which is only 3 half he classical paricle velociy! A simple manipulaion of his resul will clarify he problem. We assume ha paricles can be represened by a localized wave funcion. This is he case if he major conribuions o he inegral are cenered abou a cenral wave number, k. Thus, we can expand ω(k) abou k : Since p hk, we also see ha he dispersion relaion is given by ω hk2 2m p2 2m h Ē h. ω(k) ω + ω (k k ) (9.) Here ω ω(k ) and ω ω (k ). Insering his expression ino he inegral represenaion for Ψ(x, ), we have Ψ(x, ) φ(k, )e i(kx ω ω (k k)...) dk, 2π We now make he change of variables, s k k, and rearrange he resuling facors o find Ψ(x, ) 2π 2π ei( ω +k ω ) φ(k + s, )e i((k +s)x (ω +ω s)) ds φ(k + s, )e i(k +s)(x ω ) ds e i( ω +k ω ) Ψ(x ω, ). (9.2) Group and phase velociies, v g dω Summarizing, for an iniially localized wave packe, Ψ(x, ) wih wave numbers grouped around k he wave funcion,ψ(x, ), is a ranslaed version of he iniial wave funcion up o a phase facor. In quanum mechanics we are more ineresed in he probabiliy densiy for locaing a paricle, so from Ψ(x, ) 2 Ψ(x ω, ) 2 we see ha he velociy of he wave packe is found o be ω dω dk hk kk m. This corresponds o he classical velociy of he paricle (v par p/m). Thus, one usually defines ω o be he group velociy, v g dω dk v p ω k. dk,

6 362 parial differenial equaions and he former velociy as he phase velociy, v p ω k Transform Schemes These examples have illusraed one of he feaures of ransform heory. Given a parial differenial equaion, we can ransform he equaion from spaial variables o wave number space, or ime variables o frequency space. In he new space he ime evoluion is simpler. In hese cases, he evoluion was governed by an ordinary differenial equaion. One solves he problem in he new space and hen ransforms back o he original space. This is depiced in Figure 9.3 for he Schrödinger equaion and was shown in Figure 9.2 for he linearized KdV equaion. Figure 9.3: The scheme for solving he Schrödinger equaion using Fourier ransforms. The goal is o solve for Ψ(x, ) given Ψ(x, ). Insead of a direc soluion in coordinae space (on he lef side), one can firs ransform he iniial condiion obaining φ(k, ) in wave number space. The governing equaion in he new space is found by ransforming he PDE o ge an ODE. This simpler equaion is solved o obain φ(k, ). Then an inverse ransform yields he soluion of he original equaion. Schrödinger Equaion for Ψ(x, ) Ψ(x, ) Ψ(x, ) Fourier Transform Inverse Fourier Transform φ(k, ) φ(k, ) ODE for φ(k, ) This is similar o he soluion of he sysem of ordinary differenial equaions in Chaper 3, ẋ Ax. In ha case we diagonalized he sysem using he ransformaion x Sy. This lead o a simpler sysem ẏ Λy, where Λ S AS. Solving for y, we invered he soluion o obain x. Similarly, one can apply his diagonalizaion o he soluion of linear algebraic sysems of equaions. The general scheme is shown in Figure 9.4. Figure 9.4: This shows he scheme for solving he linear sysem of ODEs ẋ Ax. One finds a ransformaion beween x and y of he form x Sy which diagonalizes he sysem. The resuling sysem is easier o solve for y. Then, one uses he inverse ransformaion o obain he soluion o he original problem. ODE ẋ Ax A Transform: x Sy, Λ S AS Λ ODE ẏ Λy x() y() Inverse Transform: x S y Similar ransform consrucions occur for many oher ype of problems. We will end his chaper wih a sudy of Laplace ransforms, which are

7 ransform echniques in physics 363 useful in he sudy of iniial value problems, paricularly for linear ordinary differenial equaions wih consan coefficiens. A similar scheme for using Laplace ransforms is depiced in Figure 9.3. In his chaper we will begin wih he sudy of Fourier ransforms. These will provide an inegral represenaion of funcions defined on he real line. Such funcions can also represen analog signals. Analog signals are coninuous signals which can be represened as a sum over a coninuous se of frequencies, as opposed o he sum over discree frequencies, which Fourier series were used o represen in an earlier chaper. We will hen invesigae a relaed ransform, he Laplace ransform, which is useful in solving iniial value problems such as hose encounered in ordinary differenial equaions. 9.2 Complex Exponenial Fourier Series Before deriving he Fourier ransform, we will need o rewrie he rigonomeric Fourier series represenaion as a complex exponenial Fourier series. We firs recall from Chaper?? he rigonomeric Fourier series represenaion of a funcion defined on [ π, π] wih period 2π. The Fourier series is given by f (x) a 2 + (a n cos nx + b n sin nx), (9.3) n where he Fourier coefficiens were found as a n π b n π π π π π f (x) cos nx dx, n,,..., f (x) sin nx dx, n, 2,.... (9.4) In order o derive he exponenial Fourier series, we replace he rigonomeric funcions wih exponenial funcions and collec like exponenial erms. This gives f (x) a ( e [a inx + e inx ) ( e n + inx e inx )] b n n ( ) an ib n e inx + 2 a 2 + n n ( an + ib n 2 2i ) e inx. (9.5) The coefficiens of he complex exponenials can be rewrien by defining c n 2 (a n + ib n ), n, 2,.... (9.6) This implies ha c n 2 (a n ib n ), n, 2,.... (9.7) So far he represenaion is rewrien as f (x) a 2 + n c n e inx + n c n e inx.

8 364 parial differenial equaions Re-indexing he firs sum, by inroducing k n, we can wrie f (x) a 2 + k c k e ikx + n c n e inx. Since k is a dummy index, we replace i wih a new n as f (x) a 2 + n c n e inx + n c n e inx. We can now combine all of he erms ino a simple sum. We firs define c n for negaive n s by c n c n, n, 2,.... Leing c a 2, we can wrie he complex exponenial Fourier series represenaion as f (x) c n e inx, (9.8) n where c n 2 (a n + ib n ), n, 2,... c n 2 (a n ib n ), n, 2,... c a 2. (9.9) Given such a represenaion, we would like o wrie ou he inegral forms of he coefficiens, c n. So, we replace he a n s and b n s wih heir inegral represenaions and replace he rigonomeric funcions wih complex exponenial funcions. Doing his, we have for n, 2,.... c n 2 (a n + ib n ) [ π f (x) cos nx dx + i π 2 π π π π π ( e f (x) inx + e inx ) dx + 2π π 2 π 2π π ] f (x) sin nx dx i π f (x) 2π π ( e inx e inx ) dx 2i f (x)e inx dx. (9.2) I is a simple maer o deermine he c n s for oher values of n. For n, we have ha c a 2 π f (x) dx. 2π π For n, 2,..., we find ha c n c n π f (x)e 2π inx dx π f (x)e inx dx. π 2π π Therefore, we have obained he complex exponenial Fourier series coefficiens for all n. Now we can define he complex exponenial Fourier series for he funcion f (x) defined on [ π, π] as shown below.

9 ransform echniques in physics 365 Complex Exponenial Series for f (x) defined on [ π, π]. f (x) n c n e inx, (9.2) c n π f (x)e inx dx. (9.22) 2π π We can easily exend he above analysis o oher inervals. For example, for x [ L, L] he Fourier rigonomeric series is wih Fourier coefficiens f (x) a ( 2 + a n cos nπx L n a n L b n L L L L L f (x) cos nπx L f (x) sin nπx L + b n sin nπx ) L dx, n,,..., dx, n, 2,.... This can be rewrien as an exponenial Fourier series of he form Complex Exponenial Series for f (x) defined on [ L, L]. f (x) n c n e inπx/l, (9.23) c n L f (x)e inπx/l dx. 2L L (9.24) We can now use his complex exponenial Fourier series for funcion defined on [ L, L] o derive he Fourier ransform by leing L ge large. This will lead o a sum over a coninuous se of frequencies, as opposed o he sum over discree frequencies, which Fourier series represen. 9.3 Exponenial Fourier Transform Boh he rigonomeric and complex exponenial Fourier series provide us wih represenaions of a class of funcions of finie period in erms of sums over a discree se of frequencies. In paricular, for funcions defined on x [ L, L], he period of he Fourier series represenaion is 2L. We can wrie he argumens in he exponenials, e inπx/l, in erms of he angular frequency, ω n nπ/l, as e iωnx. We noe ha he frequencies, ν n, are hen defined hrough ω n 2πν n nπ L. Therefore, he complex exponenial series is seen o be a sum over a discree, or counable, se of frequencies. We would now like o exend he finie inerval o an infinie inerval, x (, ), and o exend he discree se of (angular) frequencies o a

10 366 parial differenial equaions coninuous range of frequencies, ω (, ). One can do his rigorously. I amouns o leing L and n ge large and keeping n L fixed. We firs define ω π L, so ha ω n n ω. Insering he Fourier coefficiens (9.24) ino Equaion (9.23), we have f (x) n n n c n e inπx/l ( 2L ( ω 2π L L L L ) f (ξ)e inπξ/l dξ e inπx/l ) f (ξ)e iωnξ dξ e iωnx. (9.25) Now, we le L ge large, so ha ω becomes small and ω n approaches he angular frequency ω. Then, f (x) lim ω,l 2π ( L 2π n L ( f (ξ)e iωξ dξ ) f (ξ)e iωnξ dξ e iωnx ω ) e iωx dω. (9.26) Definiions of he Fourier ransform and he inverse Fourier ransform. Looking a his las resul, we formally arrive a he definiion of he Fourier ransform. I is embodied in he inner inegral and can be wrien as F[ f ] ˆf (ω) f (x)e iωx dx. (9.27) This is a generalizaion of he Fourier coefficiens (9.24). Once we know he Fourier ransform, ˆf (ω), hen we can reconsruc he original funcion, f (x), using he inverse Fourier ransform, which is given by he ouer inegraion, F [ ˆf ] f (x) 2π ˆf (ω)e iωx dω. (9.28) We noe ha i can be proven ha he Fourier ransform exiss when f (x) is absoluely inegrable, i.e., f (x) dx <. Such funcions are said o be L. We combine hese resuls below, defining he Fourier and inverse Fourier ransforms and indicaing ha hey are inverse operaions of each oher. We will hen prove he firs of he equaions, (9.3). [The second equaion, (9.32), follows in a similar way.]

11 ransform echniques in physics 367 The Fourier ransform and inverse Fourier ransform are inverse operaions. Defining he Fourier ransform as F[ f ] ˆf (ω) f (x)e iωx dx. (9.29) and he inverse Fourier ransform as F [ ˆf ] f (x) 2π ˆf (ω)e iωx dω. (9.3) hen and F [F[ f ]] f (x) (9.3) F[F [ ˆf ]] ˆf (ω). (9.32) Proof. The proof is carried ou by insering he definiion of he Fourier ransform, (9.29), ino he inverse ransform definiion, (9.3), and hen inerchanging he orders of inegraion. Thus, we have F [F[ f ]] 2π 2π 2π 2π F[ f ]e iωx dω [ f (ξ)e iωξ dξ ] e iωx dω f (ξ)e iω(ξ x) dξdω [ ] e iω(ξ x) dω f (ξ) dξ. (9.33) In order o complee he proof, we need o evaluae he inside inegral, which does no depend upon f (x). This is an improper inegral, so we firs define D Ω (x) and compue he inner inegral as Ω Ω e iωx dω e iω(ξ x) dω lim Ω D Ω(ξ x). 8 y We can compue D Ω (x). A simple evaluaion yields D Ω (x) Ω Ω eiωx ix e iωx dω Ω Ω eixω e ixω 2ix 2 sin xω. (9.34) x A plo of his funcion is in Figure 9.5 for Ω 4. For large Ω he peak grows and he values of D Ω (x) for x end o zero as shown in Figure Figure 9.5: A plo of he funcion D Ω (x) for Ω 4. x

12 368 parial differenial equaions 9.6. In fac, as x approaches, D Ω (x) approaches 2Ω. For x, he D Ω (x) funcion ends o zero. We furher noe ha lim D Ω(x), x, Ω 8 y and lim Ω D Ω (x) is infinie a x. However, he area is consan for each Ω. In fac, D Ω (x) dx 2π. We can show his by recalling he compuaion in Example 8.42, Then, sin x x dx π Figure 9.6: A plo of he funcion D Ω (x) for Ω 4. 4 x D Ω (x) dx 2 sin xω x dx 2 sin y dy y 2π. (9.35) Anoher way o look a D Ω (x) is o consider he sequence of funcions sin nx f n (x) πx, n, 2,.... Then we have shown ha his sequence of funcions saisfies he wo properies, 2 lim f n(x), x, n f n (x) dx. This is a key represenaion of such generalized funcions. The limiing value vanishes a all bu one poin, bu he area is finie. Such behavior can be seen for he limi of oher sequences of funcions. For example, consider he sequence of funcions f n (x) {, x > n, n 2, x < f racn Figure 9.7: A plo of he funcions f n (x) for n 2, 4, x This is a sequence of funcions as shown in Figure 9.7. As n, we find he limi is zero for x and is infinie for x. However, he area under each member of he sequences is one. Thus, he limiing funcion is zero a mos poins bu has area one. The limi is no really a funcion. I is a generalized funcion. I is called he Dirac dela funcion, which is defined by. δ(x) for x. 2. δ(x) dx.

13 ransform echniques in physics 369 Before reurning o he proof ha he inverse Fourier ransform of he Fourier ransform is he ideniy, we sae one more propery of he Dirac dela funcion, which we will prove in he nex secion. Namely, we will show ha δ(x a) f (x) dx f (a). Reurning o he proof, we now have ha Insering his ino (9.33), we have e iω(ξ x) dω lim Ω D Ω(ξ x) 2πδ(ξ x). F [F[ f ]] 2π 2π [ ] e iω(ξ x) dω 2πδ(ξ x) f (ξ) dξ. f (ξ) dξ. f (x). (9.36) Thus, we have proven ha he inverse ransform of he Fourier ransform of f is f. 9.4 The Dirac Dela Funcion In he las secion we inroduced he Dirac dela funcion, δ(x). P. A. M. Dirac (92-984) inroduced As noed above, his is one example of wha is known as a generalized he δ funcion in his book, The Principles of Quanum Mechanics, 4h Ed., Oxford funcion, or a disribuion. Dirac had inroduced his funcion in he 93 s Universiy Press, 958, originally published in his sudy of quanum mechanics as a useful ool. I was laer sudied in a general heory of disribuions and found o be more han a simple ool used by physiciss. The Dirac dela funcion, as any disribuion, only makes sense under an inegral. in 93, as par of his orhogonal- iy saemen for a basis of funcions in a Hilber space, < ξ ξ > cδ(ξ ξ ) in he same way we inroduced discree orhogonaliy using he Kronecker dela. Two properies were used in he las secion. Firs one has ha he area under he dela funcion is one, δ(x) dx. Inegraion over more general inervals gives b a δ(x) dx {, [a, b],, [a, b]. (9.37) The oher propery ha was used was he sifing propery: δ(x a) f (x) dx f (a). This can be seen by noing ha he dela funcion is zero everywhere excep a x a. Therefore, he inegrand is zero everywhere and he only conribuion from f (x) will be from x a. So, we can replace f (x) wih f (a) under

14 37 parial differenial equaions Properies of he Dirac δ-funcion: δ(x a) f (x) dx f (a). δ(ax) dx δ(y) dy. a δ( f (x)) dx n j δ(x x j ) f (x j ) dx. (For n simple roos.) These and oher properies are ofen wrien ouside he inegral: δ((x a)(x b)) δ(ax) a δ(x). δ( x) δ(x). δ( f (x)) j for f (x j ), f (x j ). [δ(x a) + δ(x a)]. a b δ(x x j ) f (x j ), he inegral. Since f (a) is a consan, we have ha δ(x a) f (x) dx f (a) δ(x a) f (a) dx δ(x a) dx f (a). (9.38) Anoher propery resuls from using a scaled argumen, ax. In his case we show ha δ(ax) a δ(x). (9.39) As usual, his only has meaning under an inegral sign. So, we place δ(ax) inside an inegral and make a subsiuion y ax: L δ(ax) dx lim δ(ax) dx L L al lim δ(y) dy. (9.4) L a If a > hen However, if a < hen δ(ax) dx a δ(ax) dx a al δ(y) dy. δ(y) dy a δ(y) dy. The overall difference in a muliplicaive minus sign can be absorbed ino one expression by changing he facor /a o / a. Thus, δ(ax) dx δ(y) dy. (9.4) a Example 9.. Evaluae (5x + )δ(4(x 2)) dx. This is a sraigh forward inegraion: + )δ(4(x 2)) dx (5x (5x + )δ(x 2) dx 4 4. The firs srep is o wrie δ(4(x 2)) 4 δ(x 2). Then, he final evaluaion is given by (5x + )δ(x 2) dx (5(2) + ) Even more general han δ(ax) is he dela funcion δ( f (x)). The inegral of δ( f (x)) can be evaluaed depending upon he number of zeros of f (x). If here is only one zero, f (x ), hen one has ha δ( f (x)) dx f (x ) δ(x x ) dx. This can be proven using he subsiuion y f (x) and is lef as an exercise for he reader. This resul is ofen wrien as δ( f (x)) f (x ) δ(x x ), again keeping in mind ha his only has meaning when placed under an inegral.

15 ransform echniques in physics 37 Example 9.2. Evaluae δ(3x 2)x2 dx. This is no a simple δ(x a). So, we need o find he zeros of f (x) 3x 2. There is only one, x 2 3. Also, f (x) 3. Therefore, we have δ(3x 2)x 2 dx 3 δ(x 2 3 )x2 dx 3 ( ) Noe ha his inegral can be evaluaed he long way by using he subsiuion y 3x 2. Then, dy 3 dx and x (y + 2)/3. This gives δ(3x 2)x 2 dx 3 ( y + 2 δ(y) 3 ) 2 dy 3 ( ) More generally, one can show ha when f (x j ) and f (x j ) for j, 2,..., n, (i.e.; when one has n simple zeros), hen δ( f (x)) n j f (x j ) δ(x x j). Example 9.3. Evaluae 2π cos x δ(x 2 π 2 ) dx. In his case he argumen of he dela funcion has wo simple roos. Namely, f (x) x 2 π 2 when x ±π. Furhermore, f (x) 2x. Therefore, f (±π) 2π. This gives δ(x 2 π 2 ) [δ(x π) + δ(x + π)]. 2π Insering his expression ino he inegral and noing ha x π is no in he inegraion inerval, we have 2π cos x δ(x 2 π 2 ) dx 2π cos x [δ(x π) + δ(x + π)] dx 2π 2π cos π 2π. (9.42) Example 9.4. Show H (x) δ(x), where he Heaviside funcion (or, sep funcion) is defined as {, x < H(x), x > H(x) Figure 9.8: The Heaviside sep funcion, H(x). x and is shown in Figure 9.8. Looking a he plo, i is easy o see ha H (x) for x. In order o check ha his gives he dela funcion, we need o compue he area inegral. Therefore, we have H (x) dx H(x). Thus, H (x) saisfies he wo properies of he Dirac dela funcion.

16 372 parial differenial equaions 9.5 Properies of he Fourier Transform We now reurn o he Fourier ransform. Before acually compuing he Fourier ransform of some funcions, we prove a few of he properies of he Fourier ransform. Firs we noe ha here are several forms ha one may encouner for he Fourier ransform. In applicaions funcions can eiher be funcions of ime, f (), or space, f (x). The corresponding Fourier ransforms are hen wrien as ˆf (ω) f ()e iω d, (9.43) or ˆf (k) f (x)e ikx dx. (9.44) ω is called he angular frequency and is relaed o he frequency ν by ω 2πν. The unis of frequency are ypically given in Herz (Hz). Someimes he frequency is denoed by f when here is no confusion. k is called he wavenumber. I has unis of inverse lengh and is relaed o he wavelengh, λ, by k 2π λ. We explore a few basic properies of he Fourier ransform and use hem in examples in he nex secion.. Lineariy: For any funcions f (x) and g(x) for which he Fourier ransform exiss and consan a, we have F[ f + g] F[ f ] + F[g] and F[a f ] af[ f ]. These simply follow from he properies of inegraion and esablish he lineariy of he Fourier ransform. [ ] d f 2. Transform of a Derivaive: F ik dx ˆf (k) Here we compue he Fourier ransform (9.29) of he derivaive by insering he derivaive in he Fourier inegral and using inegraion by pars. [ ] d f F dx d f dx eikx dx [ lim f (x)e ikx] L ik f (x)e ikx dx. L L (9.45) The limi will vanish if we assume ha lim x ± f (x). The las inegral is recognized as he Fourier ransform of f, proving he given propery.

17 ransform echniques in physics 373 [ d n ] f 3. Higher Order Derivaives: F dx n ( ik) n ˆf (k) The proof of his propery follows from he las resul, or doing several inegraion by pars. We will consider he case when n 2. Noing ha he second derivaive is he derivaive of f (x) and applying he las resul, we have [ d 2 ] f F dx 2 This resul will be rue if [ ] d F dx f [ ] d f ikf dx lim f (x) and lim x ± ( ik) 2 ˆf (k). (9.46) f (x). x ± The generalizaion o he ransform of he nh derivaive easily follows. 4. Muliplicaion by x: F [x f (x)] i d dk ˆf (k) This propery can be shown by using he fac ha dk d eikx ixe ikx and he abiliy o differeniae an inegral wih respec o a parameer. F[x f (x)] i d dk x f (x)e ikx dx f (x) d dk ( ) i eikx f (x)e ikx dx dx i d dk ˆf (k). (9.47) This resul can be generalized o F [x n f (x)] as an exercise. 5. Shifing Properies: For consan a, we have he following shifing properies: f (x a) e ika ˆf (k), (9.48) f (x)e iax ˆf (k a). (9.49) Here we have denoed he Fourier ransform pairs using a double arrow as f (x) ˆf (k). These are easily proven by insering he desired forms ino he definiion of he Fourier ransform (9.29), or inverse Fourier ransform (9.3). The firs shif propery (9.48) is shown by he following argumen. We evaluae he Fourier ransform. F[ f (x a)] f (x a)e ikx dx. Now perform he subsiuion y x a. Then, F[ f (x a)] e ika f (y)e ik(y+a) dy f (y)e iky dy e ika ˆf (k). (9.5)

18 374 parial differenial equaions The second shif propery (9.49) follows in a similar way. 6. Convoluion of Funcions: We define he convoluion of wo funcions f (x) and g(x) as ( f g)(x) f ()g(x ) dx. (9.5) Then, he Fourier ransform of he convoluion is he produc of he Fourier ransforms of he individual funcions: F[ f g] ˆf (k)ĝ(k). (9.52) e ax2 /2 We will reurn o he proof of his propery in Secion Fourier Transform Examples Figure 9.9: Plos of he Gaussian funcion f (x) e ax2 /2 for a, 2, 3. x In his secion we will compue he Fourier ransforms of several funcions. Example 9.5. Find he Fourier ransform of a Gaussian, f (x) e ax2 /2. This funcion, shown in Figure 9.9 is called he Gaussian funcion. I has many applicaions in areas such as quanum mechanics, molecular heory, probabiliy and hea diffusion. We will compue he Fourier ransform of his funcion and show ha he Fourier ransform of a Gaussian is a Gaussian. In he derivaion we will inroduce classic echniques for compuing such inegrals. We begin by applying he definiion of he Fourier ransform, ˆf (k) f (x)e ikx dx e ax2 /2+ikx dx. (9.53) The firs sep in compuing his inegral is o complee he square in he argumen of he exponenial. Our goal is o rewrie his inegral so ha a simple subsiuion will lead o a classic inegral of he form eβy2 dy, which we can inegrae. The compleion of he square follows as usual: a 2 x2 + ikx a [ x 2 2ik ] 2 a x [ a 2 x 2 2ik a x + ( a 2 x ik a ( ik ) 2 ( ik ) ] 2 a a ) 2 k2 2a. (9.54) We now pu his expression ino he inegral and make he subsiuions y x ik a and β a 2. ˆf (k) e k2 2a e ax2 /2+ikx dx e a 2 (x ik a ) 2 dx e k2 ik a 2a e βy2 dy. (9.55) ik a

19 ransform echniques in physics 375 One would be emped o absorb he ik a erms in he limis of inegraion. However, we know from our previous sudy ha he inegraion akes place over a conour in he complex plane as shown in Figure 9.. In his case we can deform his horizonal conour o a conour along he real axis since we will no cross any singulariies of he inegrand. So, we now safely wrie ˆf (k) e k2 2a e βy2 dy. The resuling inegral is a classic inegral and can be performed using a sandard rick. Define I by 4 Then, I 2 I e βy2 dy. e βy2 dy e βx2 dx. Noe ha we needed o change he inegraion variable so ha we can wrie his produc as a double inegral: I 2 e β(x2 +y 2) dxdy. y x z x ik a Figure 9.: Simple horizonal conour. 4 Here we show π e βy2 dy β. Noe ha we solved he β case in Example 5., so a simple variable ransformaion z βy is all ha is needed o ge he answer. However, i canno hur o see his classic derivaion again. This is an inegral over he enire xy-plane. We now ransform o polar coordinaes o obain I 2 2π 2π π β [e βr2] e βr2 rdrdθ e βr2 rdr The final resul is goen by aking he square roo, yielding π I β. π β. (9.56) We can now inser his resul o give he Fourier ransform of he Gaussian funcion: 2π ˆf (k) a e k2 /2a. (9.57) Therefore, we have shown ha he Fourier ransform of a Gaussian is a Gaussian. Example 9.6. Find he Fourier ransform of he Box, or Gae, Funcion, { b, x a f (x), x > a. This funcion is called he box funcion, or gae funcion. I is shown in Figure 9.. The Fourier ransform of he box funcion is relaively easy o compue. I is given by ˆf (k) f (x)e ikx dx The Fourier ransform of a Gaussian is a Gaussian. a y Figure 9.: A plo of he box funcion in Example 9.6. a b x

20 376 parial differenial equaions Figure 9.2: A plo of he Fourier ransform of he box funcion in Example 9.6. This is he general shape of he sinc funcion. y x We can rewrie his as Here we inroduced he sinc funcion, a a be ikx dx b ik eikx a 2b k a sin ka ˆf (k) 2ab 2ab sinc ka. ka sinc x sin x x. sin ka. (9.58) A plo of his funcion is shown in Figure 9.2. This funcion appears ofen in signal analysis and i plays a role in he sudy of diffracion. We will now consider special limiing values for he box funcion and is ransform. This will lead us o he Uncerainy Principle for signals, connecing he relaionship beween he localizaion properies of a signal and is ransform.. a and b fixed. In his case, as a ges large he box funcion approaches he consan funcion f (x) b. A he same ime, we see ha he Fourier ransform approaches a Dirac dela funcion. We had seen his funcion earlier when we firs defined he Dirac dela funcion. Compare Figure 9.2 wih Figure 9.5. In fac, ˆf (k) bd a (k). [Recall he definiion of D Ω (x) in Equaion (9.34).] So, in he limi we obain ˆf (k) 2πbδ(k). This limi implies fac ha he Fourier ransform of f (x) is ˆf (k) 2πδ(k). As he widh of he box becomes wider, he Fourier ransform becomes more localized. In fac, we have arrived a he imporan resul ha 2. b, a, and 2ab. e ikx 2πδ(k). (9.59) In his case he box narrows and becomes seeper while mainaining a consan area of one. This is he way we had found a represenaion of he Dirac dela funcion previously. The Fourier ransform approaches a consan in his limi. As a approaches zero, he sinc funcion approaches one, leaving ˆf (k) 2ab. Thus, he Fourier ransform of he Dirac dela funcion is one. Namely, we have δ(x)e ikx. (9.6) In his case we have ha he more localized he funcion f (x) is, he more spread ou he Fourier ransform, ˆf (k), is. We will summarize hese noions in he nex iem by relaing he widhs of he funcion and is Fourier ransform.

21 ransform echniques in physics The Uncerainy Principle, x k 4π. The widhs of he box funcion and is Fourier ransform are relaed as we have seen in he las wo limiing cases. I is naural o define he widh, x of he box funcion as x 2a. The widh of he Fourier ransform is a lile rickier. This funcion acually exends along he enire k-axis. However, as ˆf (k) became more localized, he cenral peak in Figure 9.2 became narrower. So, we define he widh of his funcion, k as he disance beween he firs zeros on eiher side of he main lobe as shown in Figure 9.3. This gives 2ab y k 2π a. Combining hese wo relaions, we find ha x k 4π. Thus, he more localized a signal, he less localized is ransform and vice versa. This noion is referred o as he Uncerainy Principle. For general signals, one needs o define he effecive widhs more carefully, bu he main idea holds: x k c >. We now urn o oher examples of Fourier ransforms. { e Example 9.7. Find he Fourier ransform of f (x) ax, x, x <, a >. The Fourier ransform of his funcion is ˆf (k) f (x)e ikx dx e ikx ax dx a ik. (9.6) Nex, we will compue he inverse Fourier ransform of his resul and recover he original funcion. π a π a Figure 9.3: The widh of he funcion sin ka 2ab ka is defined as he disance beween he smalles magniude zeros. More formally, he uncerainy principle for signals is abou he relaion beween duraion and bandwidh, which are defined by f 2 and ω ω ˆf 2 f 2 ˆf, respecively, where f 2 2 f () 2 d and ˆf 2 2π ˆf (ω) 2 dω. Under appropriae condiions, one can prove ha ω 2. Equaliy holds for Gaussian signals. Werner Heisenberg (9-976) inroduced he uncerainy principle ino quanum physics in 926, relaing uncerainies in he posiion ( x) and momenum ( p x ) of paricles. In his case, x p x 2 h. Here, he uncerainies are defined as he posiive square roos of he quanum mechanical variances of he posiion and momenum. x Example 9.8. Find he inverse Fourier ransform of ˆf (k) a ik. The inverse Fourier ransform of his funcion is f (x) 2π ˆf (k)e ikx dk e ikx 2π a ik dk. This inegral can be evaluaed using conour inegral mehods. We evaluae he inegral e I ixz a iz dz,

22 378 parial differenial equaions R R ia ia y y C R C R Figure 9.4: Conours for invering ˆf (k) a ik. R R x x using Jordan s Lemma from Secion According o Jordan s Lemma, we need o enclose he conour wih a semicircle in he upper half plane for x < and in he lower half plane for x > as shown in Figure 9.4. The inegraions along he semicircles will vanish and we will have f (x) 2π ± 2π { { C e ikx a ik dk e ixz a iz dz, x < 2π 2πi Res [z ia], x >, x < e ax, x >. (9.62) Noe ha wihou paying careful aenion o Jordan s Lemma one migh no rerieve he funcion from he las example. Example 9.9. Find he inverse Fourier ransform of ˆf (ω) πδ(ω + ω ) + πδ(ω ω ). We would like o find he inverse Fourier ransform of his funcion. Insead of carrying ou any inegraion, we will make use of he properies of Fourier ransforms. Since he ransforms of sums are he sums of ransforms, we can look a each erm individually. Consider δ(ω ω ). This is a shifed funcion. From he shif heorems in Equaions (9.48)-(9.49) we have he Fourier ransform pair e iω f () ˆf (ω ω ). Recalling from Example 9.6 ha e iω d 2πδ(ω), we have from he shif propery ha F [δ(ω ω )] 2π e iω. The second erm can be ransformed similarly. Therefore, we have F [πδ(ω + ω ) + πδ(ω ω ] 2 eiω + 2 e iω cos ω. f () a Example 9.. Find he Fourier ransform of he finie wave rain. { cos ω f (), a, > a. For he las example, we consider he finie wave rain, which will reappear in he las chaper on signal analysis. In Figure 9.5 we show a plo of his funcion. A sraigh forward compuaion gives Figure 9.5: rain. A plo of he finie wave ˆf (ω) f ()e iω d

23 ransform echniques in physics a a a a a [cos ω + i sin ω ]e iω d a cos ω cos ω d + i a sin((ω + ω )a) ω + ω a sin ω sin ω d [cos((ω + ω )) + cos((ω ω ))] d + sin((ω ω )a) ω ω. (9.63) 9.6 The Convoluion Operaion In he lis of properies of he Fourier ransform, we defined he convoluion of wo funcions, f (x) and g(x) o be he inegral ( f g)(x) f ()g(x ) d. (9.64) In some sense one is looking a a sum of he overlaps of one of he funcions and all of he shifed versions of he oher funcion. The German word for convoluion is falung, which means folding and in old exs his is referred o as he Falung Theorem. In his secion we will look ino he convoluion operaion and is Fourier ransform. Before we ge oo involved wih he convoluion operaion, i should be noed ha here are really wo hings you need o ake away from his discussion. The res is deail. Firs, he convoluion of wo funcions is a new funcions as defined by 9.64 when dealing wi he Fourier ransform. The second and mos relevan is ha he Fourier ransform of he convoluion of wo funcions is he produc of he ransforms of each funcion. The res is all abou he use and consequences of hese wo saemens. In his secion we will show how he convoluion works and how i is useful. Firs, we noe ha he convoluion is commuaive: f g g f. This is easily shown by replacing x wih a new variable, y x and dy d. The convoluion is commuaive. (g f )(x) g() f (x ) d g(x y) f (y) dy f (y)g(x y) dy ( f g)(x). (9.65) The bes way o undersand he folding of he funcions in he convoluion is o ake wo funcions and convolve hem. The nex example gives a graphical rendiion followed by a direc compuaion of he convoluion. The reader is encouraged o carry ou hese analyses for oher funcions. Example 9.. Graphical Convoluion of he box funcion and a riangle funcion. In order o undersand he convoluion operaion, we need o apply i o specific

24 38 parial differenial equaions f (x) g(x) x funcions. We will firs do his graphically for he box funcion {, x, f (x), x > and he riangular funcion g(x) { x, x,, oherwise Figure 9.6: A plo of he box funcion f (x) and he riangle funcion g(x). g( ) Figure 9.7: A plo of he refleced riangle funcion, g( ). g(2 ) Figure 9.8: A plo of he refleced riangle funcion shifed by 2 unis, g(2 ). x 2 as shown in Figure 9.6. Nex, we deermine he conribuions o he inegrand. We consider he shifed and refleced funcion g( x) in Equaion 9.64 for various values of. For, we have g(x ) g( x). This funcion is a reflecion of he riangle funcion, g(x), as shown in Figure 9.7. We hen ranslae he riangle funcion performing horizonal shifs by. In Figure 9.8 we show such a shifed and refleced g(x) for 2, or g(2 x). In Figure 9.8 we show several plos of oher shifs, g(x ), superimposed on f (x). The inegrand is he produc of f () and g(x ) and he inegral of he produc f ()g(x ) is given by he sum of he shaded areas for each value of x. In he firs plo of Figure 9.9 he area is zero, as here is no overlap of he funcions. Inermediae shif values are displayed in he oher plos in Figure 9.9. The value of he convoluion a x is shown by he area under he produc of he wo funcions for each value of x. Plos of he areas of he convoluion of he box and riangle funcions for several values of x are given in Figure 9.8. We see ha he value of he convoluion inegral builds up and hen quickly drops o zero as a funcion of x. In Figure 9.2 he values of hese areas is shown as a funcion of x. y y y y y y y y y Figure 9.9: A plo of he box and riangle funcions wih he overlap indicaed by he shaded area. The plo of he convoluion in Figure 9.2 is no easily deermined using he graphical mehod. However, we can direcly compue he convoluion as shown in he nex example.

25 ransform echniques in physics 38 Example 9.2. Analyically find he convoluion of he box funcion and he riangle funcion. The nonvanishing conribuions o he convoluion inegral are when boh f () and g(x ) do no vanish. f () is nonzero for, or. g(x ) is nonzero for x, or x x. These wo regions are shown in Figure 9.2. On his region, f ()g(x ) x. ( f g)(x).5 Figure 9.2: A plo of he convoluion of he box and riangle funcions. 2 x g(x) 2 f (x) 2 x Figure 9.2: Inersecion of he suppor of g(x) and f (x). Isolaing he inersecion in Figure 9.22, we see in Figure 9.22 ha here are hree regions as shown by differen shadings. These regions lead o a piecewise defined funcion wih hree differen branches of nonzero values for < x <, < x <, and < x < 2. g(x) 2 f (x) 2 x Figure 9.22: Inersecion of he suppor of g(x) and f (x) showing he inegraion regions. The values of he convoluion can be deermined hrough careful inegraion. The resuling inegrals are given as ( f g)(x) f ()g(x ) d x (x ) d, < x < (x ) d, < x < x x x 2 (x ) d, < x < 2 2 (x + )2, < x < 2, < x < [ (x ) 2 ] < x < 2 (9.66)

26 382 parial differenial equaions A plo of his funcion is shown in Figure Convoluion Theorem for Fourier Transforms In his secion we compue he Fourier ransform of he convoluion inegral and show ha he Fourier ransform of he convoluion is he produc of he ransforms of each funcion, F[ f g] ˆf (k)ĝ(k). (9.67) Firs, we use he definiions of he Fourier ransform and he convoluion o wrie he ransform as F[ f g] ( f g)(x)e ikx dx ( ) f ()g(x ) d e ikx dx ( ) g(x )e ikx dx f () d. (9.68) We now subsiue y x on he inside inegral and separae he inegrals: ( ) F[ f g] g(x )e ikx dx f () d ( ) g(y)e ik(y+) dy f () d ( ) g(y)e iky dy f ()e ik d. ( ) ( ) f ()e ik d g(y)e iky dy. (9.69) We see ha he wo inegrals are jus he Fourier ransforms of f and g. Therefore, he Fourier ransform of a convoluion is he produc of he Fourier ransforms of he funcions involved: F[ f g] ˆf (k)ĝ(k). Example 9.3. Compue he convoluion of he box funcion of heigh one and widh wo wih iself. Le ˆf (k) be he Fourier ransform of f (x). Then, he Convoluion Theorem says ha F[ f f ](k) ˆf 2 (k), or ( f f )(x) F [ ˆf 2 (k)]. For he box funcion, we have already found ha So, we need o compue ˆf (k) 2 sin k. k ( f f )(x) F [ 4 k 2 sin2 k] 2π ( 4 k 2 sin2 k ) e ikx dk. (9.7)

27 ransform echniques in physics 383 One way o compue his inegral is o exend he compuaion ino he complex k-plane. We firs need o rewrie he inegrand. Thus, ( f f )(x) 2π π π π 4 k 2 sin2 ke ikx dk k 2 [ cos 2k]e ikx dk [ k 2 ] 2 (eik + e ik ) e ikx dk k 2 [ e ikx 2 (e i( k) + e i(+k) ) ] dk. (9.7) We can compue he above inegrals if we know how o compue he inegral I(y) π e iky Then, he resul can be found in erms of I(y) as k 2 dk. ( f f )(x) I(x) [I( k) + I( + k)]. 2 We consider he inegral e iyz C πz 2 dz over he conour in Figure We can see ha here is a double pole a z. The pole is on he real axis. So, we will need o cu ou he pole as we seek he value of he principal value inegral. Recall from Chaper 8 ha e CR iyz dz πz 2 e iyz Γ R πz 2 dz + ɛ R e iyz πz 2 dz + e iyz C ɛ πz 2 dz + R ɛ e iyz πz 2 dz. y C ε Γ R R ε ε R x Figure 9.23: Conour for compuing P e iyz πz 2 dz. The inegral e iyz C dz vanishes since here are no poles enclosed in he conour! R πz 2 The sum of he second and fourh inegrals gives he inegral we seek as ɛ and R. The inegral over Γ R will vanish as R ges large according o Jordan s Lemma provided y <. Tha leaves he inegral over he small semicircle. As before, we can show ha lim f (z) dz πi Res[ f (z); z ]. ɛ C ɛ Therefore, we find e I(y) iyz [ e iyz ] P πz 2 dz πi Res πz 2 ; z. A simple compuaion of he reside gives I(y) y, for y <. When y >, we need o close he conour in he lower half plane in order o apply Jordan s Lemma. Carrying ou he compuaion, one finds I(y) y, for y >. Thus, I(y) { y, y >, y, y <, (9.72)

28 384 parial differenial equaions We are now ready o finish he compuaion of he convoluion. We have o combine he inegrals I(y), I(y + ), and I(y ), since ( f f )(x) I(x) 2 [I( k) + I( + k)]. This gives differen resuls in four inervals: ( f f )(x) x [(x 2) + (x + 2)], x < 2, 2 x [(x 2) (x + 2)] 2 + x 2 < x <, 2 x [(x 2) (x + 2)] 2 x, < x < 2, 2 x [ (x 2) (x + 2)], x > 2. (9.73) 2 A plo of his soluion is he riangle funcion, ( f f )(x), x < x, 2 < x < 2 x, < x < 2, x > 2, (9.74) which was shown in he las example. Example 9.4. Find he convoluion of he box funcion of heigh one and widh wo wih iself using a direc compuaion of he convoluion inegral. The nonvanishing conribuions o he convoluion inegral are when boh f () and f (x ) do no vanish. f () is nonzero for, or. f (x ) is nonzero for x, or x x +. These wo regions are shown in Figure On his region, f ()g(x ). Figure 9.24: Plo of he regions of suppor for f () and f (x ) f (x ) x + f () x x 2 Thus, he nonzero conribuions o he convoluion are { x+ { ( f f )(x) d, x 2, 2 + x, x 2, x d, 2 x, 2 x, 2 x. Once again, we arrive a he riangle funcion. In he las secion we showed he graphical convoluion. For compleeness, we do he same for his example. In figure 9.25 we show he resuls. We see ha he convoluion of wo box funcions is a riangle funcion.

29 ransform echniques in physics 385 f (x ) f () ( f g)(x) x Example 9.5. Show he graphical convoluion of he box funcion of heigh one and widh wo wih iself. Le s consider a slighly more complicaed example, he convoluion of wo Gaussian funcions. Figure 9.25: A plo of he convoluion of a box funcion wih iself. The areas of he overlaps of as f (x ) is ranslaed across f () are shown as well. The resul is he riangular funcion. Example 9.6. Convoluion of wo Gaussian funcions f (x) e ax2. In his example we will compue he convoluion of wo Gaussian funcions wih differen widhs. Le f (x) e ax2 and g(x) e bx2. A direc evaluaion of he inegral would be o compue ( f g)(x) This inegral can be rewrien as f ()g(x ) d ( f g)(x) e bx2 e (a+b)2 +2bx d. e a2 b(x ) 2 d. One could proceed o complee he square and finish carrying ou he inegraion. However, we will use he Convoluion Theorem o evaluae he convoluion and leave he evaluaion of his inegral o Problem 2. Recalling he Fourier ransform of a Gaussian from Example 9.5, we have π ˆf (k) F[e ax2 ] a e k2 /4a (9.75) and ĝ(k) F[e bx2 ] π b e k2 /4b. Denoing he convoluion funcion by h(x) ( f g)(x), he Convoluion Theorem gives ĥ(k) ˆf (k)ĝ(k) π e k2 /4a e k2 /4b. ab This is anoher Gaussian funcion, as seen by rewriing he Fourier ransform of h(x) as ĥ(k) π e 4 ( a + b )k 2 π e a+b 4ab k2. (9.76) ab ab

30 386 parial differenial equaions In order o complee he evaluaion of he convoluion of hese wo Gaussian funcions, we need o find he inverse ransform of he Gaussian in Equaion (9.76). We can do his by looking a Equaion (9.75). We have firs ha [ ] π F a e k2 /4a e ax2. Moving he consans, we hen obain F [e k2 /4a ] We now make he subsiuion α 4a, F [e αk2 ] a π e ax2. 4πα e x2 /4α. f () ˆf (ω) ω This is in he form needed o inver (9.76). Thus, for α a+b 4ab π ab ( f g)(x) h(x) e a+b x2. a + b we find Figure 9.26: Schemaic plo of a signal f () and is Fourier ransform ˆf (ω) Applicaion o Signal Analysis (a) ˆf (ω) Filering signals. (b) (c) p ω (ω) -ω ω ĝ(ω) Figure 9.27: (a) Plo of he Fourier ransform ˆf (ω) of a signal. (b) The gae funcion p ω (ω) used o filer ou high frequencies. (c) The produc of he funcions, ĝ(ω) ˆf (ω)p ω (ω), in (a) and (b) shows how he filers cus ou high frequencies, ω > ω. ω ω ω There are many applicaions of he convoluion operaion. One of hese areas is he sudy of analog signals. An analog signal is a coninuous signal and may conain eiher a finie, or coninuous, se of frequencies. Fourier ransforms can be used o represen such signals as a sum over he frequency conen of hese signals. In his secion we will describe how convoluions can be used in sudying signal analysis. The firs applicaion is filering. For a given signal here migh be some noise in he signal, or some undesirable high frequencies. For example, a device used for recording an analog signal migh naurally no be able o record high frequencies. Le f () denoe he ampliude of a given analog signal and ˆf (ω) be he Fourier ransform of his signal such he example provided in Figure Recall ha he Fourier ransform gives he frequency conen of he signal. There are many ways o filer ou unwaned frequencies. The simples would be o jus drop all of he high (angular) frequencies. For example, for some cuoff frequency ω frequencies ω > ω will be removed. The Fourier ransform of he filered signal would hen be zero for ω > ω. This could be accomplished by muliplying he Fourier ransform of he signal by a funcion ha vanishes for ω > ω. For example, we could use he gae funcion {, ω ω p ω (ω), (9.77), ω > ω as shown in Figure 9.27.

31 ransform echniques in physics 387 In general, we muliply he Fourier ransform of he signal by some filering funcion ĥ() o ge he Fourier ransform of he filered signal, ĝ(ω) ˆf (ω)ĥ(ω). The new signal, g() is hen he inverse Fourier ransform of his produc, giving he new signal as a convoluion: g() F [ ˆf (ω)ĥ(ω)] h( τ) f (τ) dτ. (9.78) Such processes occur ofen in sysems heory as well. One hinks of f () as he inpu signal ino some filering device which in urn produces he oupu, g(). The funcion h() is called he impulse response. This is because i is a response o he impulse funcion, δ(). In his case, one has h( τ)δ(τ) dτ h(). Anoher applicaion of he convoluion is in windowing. This represens wha happens when one measures a real signal. Real signals canno be recorded for all values of ime. Insead daa is colleced over a finie ime inerval. If he lengh of ime he daa is colleced is T, hen he resuling signal is zero ouside his ime inerval. This can be modeled in he same way as wih filering, excep he new signal will be he produc of he old signal wih he windowing funcion. The resuling Fourier ransform of he new signal will be a convoluion of he Fourier ransforms of he original signal and he windowing funcion. Windowing signals. Example 9.7. Finie Wave Train, Revisied. We reurn o he finie wave rain in Example 9. given by f () { cos ω h(), a, > a. a We can view his as a windowed version of f () cos ω obained by muliplying f () by he gae funcion {, x a g a (), x > a. (9.79) Figure 9.28: A plo of he finie wave rain. This is shown in Figure Then, he Fourier ransform is given as a convoluion, ĥ(ω) ( ˆf ĝ a )(ω) ˆf (ω ν)ĝ a (ν) dν. (9.8) 2π The convoluion in specral space is defined wih an exra facor of /2π so as o preserve he idea ha he inverse Fourier ransform of a convoluion is he produc of he corresponding signals. Noe ha he convoluion in frequency space requires he exra facor of /(2π). We need he Fourier ransforms of f and g a in order o finish he compuaion. The Fourier ransform of he box funcion was found in Example 9.6 as ĝ a (ω) 2 sin ωa. ω