A tutorial on Fourier Transforms

Transcription

1 Problem. A tutorial on Fourier Transforms This is intended as a warm up for a number of problems on waves. Answer all parts as briefly as possible. Define the fourier transform the physicist way: F (k) = e ikx f(x) f(x) = e ikx F (k) () Here integrals over wavenumbers k mean the following = dx x x This own little notation that you might try it is covenient since k x eikx =. Recall that the integral of a pure phase is a delta function: e ikx = 2πδ(k) e ikx = δ(x) (3) We use x k k k 2π (2) f(x) F (k) (4) g(x) G(k) (5) to indicate Fourier transform pairs. a is a postive constant of unit length. b is a positive constant of unit wavenumber k. ɛ is a small real constant (a) (Examples:) Consider the following most useful Fourier transforms that every serious physicist (experimentalist and theorist) knows by memory: 2a exp( x /a) + (ka) 2 (6) sin(ka/2) step(x/a) sinc(ka) a (ka/2) (7) 2πa 2 exp( x2 /(2a 2 )) exp( 2 a2 k 2 ) (8) θ(x)e bx b + ik Here step(x) = θ(x + /2) θ(x /2) (see the first panel of Fig. ) is the square wave function with integral one. Try to think of a way to remember these. For instance the second one is what comes out of a single slit diffraction of experiment. If you don t know how to do these integrals try to fix that. The table can be read either way, with the replacements k x and an additional factor of 2π, e.g. 2π + (xb) 2 (9) 2b e k /b (0)

2 In this case b has units /length and is the width in k-space. Prove that the first row of this table holds. Graph exp( x /a) and its Fourier transform for several values of ɛ, with ɛ /a, and a or ɛ 0. Argue that this Fourier transform is 2π times a Dirac sequence : 2π [ π ɛ ɛ 2 + k 2 ] = 2πδ ɛ (k) () Whenever you see a δ-function it must be remembered that it is shorthand for a Dirac sequence δ ɛ (x). The identity e ikx = 2πδ(k) (2) x is shorthand for a limiting process where the Fourier integral is cutoff in some way. Here we have explored the cutting it off like this lim ɛ 0 e ikx e ɛ x = 2πδ ɛ (k) (3) but it can be cutoff in many ways. (b) (Real/Imaginary/Even/Odd) Consider the following table: x If... then... f(x) is real F ( k) = (F (k)) f(x) is imaginary F ( k) F (k) f(x) is even (f( x) = f(x)) F ( k) = F (k) (i.e. F is even) f(x) is odd (f( x) = f(x)) F ( k) = F (k) (i.e. F is odd f(x) is real and even F (k) is real and even f(x) is real and odd F (k) is imaginary and odd f(x) is imaginary and even F (k) is imaginary and even f(x) is imaginary and odd F (k) is real and odd Table : Prove the first line and state what this means for the even and odd properties of the real and imaginary parts of F (k). Prove the sixth line as well. (c) (Shifting) Consider the Fourier transform pair f(k) F (k). We have the following properties f(x)e ik 0x F (k k 0 ) wavenumber shifting (4) f(x x 0 ) F (k)e ikx 0 spatial shifting (5) Prove the first of these. A Dirac sequence is any family of functions δ ɛ (x), labelled by a parameter ɛ 0, which satisfies δ ɛ(x) =, and approaches zero for x 0 and fixed. 2

3 (d) (Scaling) Consider the Fourier transform pair f(k) F (k). f(xa) F (k/a) a spatial scaling (6) f(x/b) b F (bk) wavenumber scaling (7) Prove the first of these and qualitatively describe the relation to the uncertainty principle. (e) (Derivatives) Show that x f(x) = F (k), (8) k=0 and show more generally that the moments of f(x) are related to to the taylor series of F (k) at the origin by ( ) n d f(x)( ix) n = F (k) (9) k=0 x The converse also holds the moments of F (k) are related to the derivatives of f(x) at the origin ( ) n d F (k)(ik) n = f(x) dx (20) x=0 k (f) (Analyticity/Asymptotic form) It follows from Eq. (20), that if a function s Fourier transform F (k) falls slower than /k n, then its n-th derivative will not generically exist. Generally analytic functions (which have all derivatives) will have a Fourier transform which decreases exponentially at large k. Consider the Fourier transform of How does the result corroborate this theorem? (g) (Convolutions) Consider the convolution of two functions (f g)(x) + (ka) 2 (2) dy f(y) g(x y) (22) Usually the convolution is used to provide a smeared version of the function f. For instance if g(x) is a normalized narrow gaussian: g(x) = 2πa 2 exp( x2 2a 2 ) (23) then convolution process, just replaces any function f(x) with a kind of average of all of its neighboring values. The figure below shows a step-like function f(x) convolved with a gaussian, g(x). g(x) has been called the kernel of the convolution. 3

4 f(x) (f*g)(x) - - (i) Working only in coordinate space show that if g(y) =. The integral of f is y unchanged by the convolution process. (ii) Show that the Fourier transform of the convolution is a product of Fourier transforms. (f g)(x) F (k)g(k) (24) (iii) By working in fourier space, show using the convolution theorem and Eq. (8) that if g(y) =, then the integral of f is unchanged by the convolution process. y (iv) Compute the fourier transform of ( ) 2 sin(ka/2) (25) (ka/2) by using the convolution theorem. You can check your result from the its integral in coordinate space. Describe qualitatively (using the convolution theorem) what are the functions B n (x) which are defined by Fourier transform of (sinc(ka)) n ( ) n sin(ka/2) B n (x) (26) ka/2 These functions B n (x) are known as B splines and are important for numerical work. Note: the higher the n, the faster it falls in k-space, the smoother the function. Fig. shows the first couple B splines. (v) Consider the convolution of a smooth function f(x) with a normalized gaussian of width a which is small compared to the scales of f(x). By working in k-space, show that (f g)(x) f(x) + f (x) a2 (27) 2 Qualitatively, what does f (x) the term do? 4

5 Figure : The first four B-splines, B... B 4, laid out like a two lines of a book. The second one is continuous but has discontinuous derivatives. The fourth one is the the cubic bspline, which has continuous second derivatives but discontinuous third derivatives. (h) (Correlations) Closely related to the convolution of two functions (but usually rather distinct in physical situations like in statistical mechanics) is the correlation of two functions: Corr(f, g)(x) dy f(x + y) g (y), (28) which is is relevant when we want to quantify over what range of lengths, x, a physical observable f is influenced by value g. Often g is a real function and the star is unnecessary. Show that the correlation function satisfies Corr(f, g) F (k)(g(k)) Correlation-Theorem (29) and thus the fourier transform of auto-correlation function is the power spectrum F (k) 2 Corr(f, f) F (k) 2 Wiener-Khinchin Theorem (30) 5

6 For this exercise you will need to recognize that the Fourier transform of g (x) (what I sometimes call G (k)) is not quite (G(k)). (i) (Parseval) Finally prove Parsevals theorem dx f(x) 2 = 2π F (k) 2 Parseval s Theorem (3) which says that the power can be computed either in coordinate or momentum space. 6

7 Solution: (g.iv). The n-th Bspline is the n-th convolution of the step function with itself. The first convolution of the step function with itself is the triangle function in Fig.. (g.v). In Fourier space the convolution is F (k)g(k) (32) Since g(x) is very sharp (a is small), in Fourier space G(k) is wide. We can expand G(k) G(k) 2 a2 k 2 (33) Fourier transforming back using yielding 2 x (ik) 2 (34) (f g)(x) = + f (x) a2 2 (35) 7

8 Problem 2. Transmitted and reflected wave forms A string is divided into two parts with mass density µ on the left, and mass density µ 2 on the right. Consider a wave form, y in (x v t), traveling from the left to the right, which scatters off the change in density (or impedance). (a) Show that when y in (t, x) is a traveling plane wave, A e ikx wt, the reflected and transmitted waves respectively take the form: with k = ω/v 2, and where the wave impedance is Z = T µ. B =A r, e i( kx ωt), (36) C =A t e i(k x ωt), (37) r = Z 2 Z Z + Z 2, (38) t = 2Z Z + Z 2. (39) (b) Continue working with the plane waves of part (a). Show that the energy transported per time by the incoming wave plane wave of part (a), equals the sum of the energies transported per time of the outgoing plane waves (i.e. the reflected and transmitted waves). (c) Now take a general real wave form y in (x vt), and assume that the wave form y in (x vt) has no appreciable support after a distance L, i.e. y in (x) 0 for x > L. Determine the wave form of the transmitted and reflected waves in coordinate space in terms of y in (x vt). Show by integration of these forms that at late times t L/v the total energy in the transmitted and reflected waves equals the energy in the incoming wave. 8

9 Solution: (a) The general solution on the left is with harmonic time dependence e iωt is y L (t, x) = Ae ikx iωt + Be ikx ωt (40) with k = ω/v µ with v = T/µ The general solution on the right with harmonic time dependence is y R (t, x) = Ce ik 2x iωt + De ik 2x ωt (4) with k 2 = w/v 2 µ 2. The frequency dependence of the two waves must be the same in order to satisfy the boundary conditions (i.e. that the string is continuous) at all times. The tension is the same on both sides, otherwise the midlle point would move to the right or left. There is no wave coming from the right and one can set D = 0. At the middle point we have continuity A + B = C (42) We also have a continuity of derivative T y x = T y L x (43) R which says that (if dy/dx is positive) that the upward force by the right balances the downward force by the left, otherwise the middle point would have a net force leading (since it has negligible mass) to infinite accelearation. T A(ik) + T B( ik) = T C(ik 2 ) (44) Solving these equations equation using k 2 = µ 2 /µ k leads to the desired result with Z = T µ. r = B A =Z Z 2 Z + Z 2 (45) t = C A = 2Z Z + Z 2 (46) (b) Here we simply use the expression for the Poynting flux from lecture. S x = T y y x t (47) For a plane wave e ikx iωt this evaluates to S x = Z 2 ω2 A 2 (48) 9

10 For a reflected wave e ikx iωt this evaluates to S x = Z 2 ω2 A 2 (49) Drawing the pillbox around the central point (see figure) we should have Leading to a requirement that S x A + S x B = S x C (50) A 2 = B 2 + C 2 Z 2 Z (5) i.e. ( ) = r 2 + t 2 Z2 Z (52) So to verify energy conservation we need that which obviously holds. (c) By assumption = (Z Z 2 ) 2 (Z + Z 2 ) 2 + 4Z Z 2 (Z + Z 2 ) 2 (53) y A (t, x) = 2π A(k)eik(x vt) (54) Instead of integrating over k we can integrate over frequency ω = vk. Changing variables y A (t, x) = where Â(ω) = A(k)/v. Let us denote y 0 (x) = y A (0, x) = dω 2π Â(ω) eiω((x/v) t) (55) = 2π A(k)eikx (56) dω 2π Â(ω)ei(ω/v)x (57) which is the shape of the undisturbed incoming wave at a fixed time (t = 0). The incoming wave is y A (t, x) = y 0 (x vt) (58) The reflected wave is y B (t, x) = The transmitted wave takes the form y C (t, x) = = 2π B(k)eik( x vt) (59) 2π C(k)eik(x v 2t) dω 2π Ĉ(ω)eiω((x/v 2) t) 0 (60) (6)

11 Our fourier analysis of (a) shows that ˆB(ω) = râ(ω) (62) And the relation between the amplitudes at the same frequency reads Ĉ(ω) = tâ(ω) (63) This follows because the boundary conditions are specified at all times at x = 0. The incoming wave, the reflected wave, and the transmitted wave at x = 0 and arbitrary time take the form dω y A (t, 0) x=0 = 2π Â(ω)e iωt (64) dω y B (t, 0) x=0 = 2π ˆB(ω)e iωt (65) dω y C (t, 0) x=0 = 2π Ĉ(ω)e iωt (66) Requiring continuity for instance reads which will follow if which should be compared to Eq. (42). Thus the reflected wave is simply y A (t, 0) + y B (t, 0) = y C (t, 0) (67) Â(ω) + ˆB(ω) = Ĉ(ω) (68) y B (t, x) = r y 0 ( x vt) (69) while the transmitted wave form is Pulling out a factor of v y C (t, x) = t dω (2π)Â (ω) eiω((x/v 2) t) dω y C (t, x) = t (2π)Â (ω) ei(ω/v)(v(x/v 2) t) ( ) v = t y 0 (x v 2 t) v 2 (70) (7) (72) Then we can integrate the energy in these waves. The energy density is ɛ = 2 µ( ty) T ( xy) 2 (73) Then note that for any wave which moves entirely in one direction v y t = y x (74)

12 where the minus sign is for a right moving wave (x vt) and the plus sign is for a left moving wave ( x vt). Thus for a right mover or a leftmover (but not for superposition) and thus µ( t y) 2 = T ( x y) 2 (75) ɛ = T ( x y) 2 (76) The total energy density in a wave which is either right or left moving is for example E = = dx T ( x y(±x vt)) 2 (77) du T ( u y(u)) 2 (78) where here and below u is a dummy integration variable short for u = ±x vt We look well before and after the reflection so that the waves consist of well separated either right or left movers. The total energy in A is thus The energy in B is similarly The energy in C is Now it is convenient to define E A = E B = du T ( u y 0 (u)) 2 (79) du r 2 T ( u y 0 (u)) 2 (80) [ ( )] 2 v E C = dx t 2 T x y 0 (x v 2 t) (8) v 2 u = v v 2 (x v 2 t) (82) and then accounting for the change in measure dx du etc, we have E C = = du t 2 v v 2 T [ u y 0 (u)] 2 (83) du t 2 Z 2 Z T [ u y 0 (u)] 2 (84) where in passing to the second line we have used v/v 2 = Z 2 /Z. We see that since we have as could be hoped. = r 2 + t 2 Z 2 Z (85) E A = E B + E C (86) 2

13 Problem 3. Propagation of waves (a) Here I want to describe a wave-packet which moves to the right t = 0. equation is a second order differential equation: The wave v 2 2 t y 2 xy = 0. (87) Thus in order to specify the problem, I need to specify the initial amplitude and the initial velocity y(t, x) t=0 and t y(t, x) t=0. (88) If the amplitude is with g(x) = y(0, x) = Re[g(x)e ik 0x ], (89) 2πa 2 exp( x2 /(2a 2 )), (90) then what should be the initial conditions t y(t, x) for a right moving wave? (b) What is the Fourier transform Y 0 (k) of the wave initial wave packet y(0, x) of part (a)? (The function y(0, x) is real and symmetric. You should check that your Fourier transform corroborates Table.) (c) Repeat the argument (given in class) that if the solution to a wave equation is y(t, x) = 2π Y 0(k)e i(kx ω(k)t) (9) then the center of the wave packet (as a function of time) is at x c (t) = v g t v g (k 0 ) dω (92) k=k0 provided the spatial extent envelope g(x), is much larger than /k 0, i.e. k 0 a. This is intended as a warmup a problem below. 3

14 Solution: (a) The right mover should be a function of g(x vt) (93) So we need to have x g = v g t (94) (b) The real part of is y(0, x) = g(x) cos(k 0 x) = 2 g(x)eik 0x + 2 g(x)e ik 0x (95) Fourier transforming we find where (c) Then Y (0, k) = 2 G(k k 0) + 2 G(k + k 0) (96) y(t, x) = G(k) = exp( k 2 a 2 /2) (97) 2π Y (0, k)eik(x ω(k)t) (98) Y (0, k) is sharply peaked near k = k 0 and k = k 0. We will integrate over these two regions separately as they are widely separated (see picture) : y(t, x) = Y (0, k)eik(x ω(k)t) + Y (0, k)eik(x ω(k)t) (99) k k 0 2π k k }{{} 0 2π }{{} I + I Near k = k 0 the frequency is expanded ω(k) ω 0 + dω k (00) k0 where ω 0 ω(k 0 ) and k = (k k 0 ). The overall phase is where the group velocity The wave from the integration near k 0 is then (kx ω(k)t) (k 0 x ω 0 t) + k(x v g t) (0) v g = dω (02) k0 I + = e ik 0x ω 0 t d k 2π 2 G( k)e i k(x v gt) (03) 4

15 The first integral evaluates to I + = 2 ei(k 0x ω 0 t) g(x v g t) (04) The second integral I in Eq. (99) results from expanding the phase near k = k 0. We expect apriori that this will just produce the complex conjugate of I +. This is indeed the case. The frequency ω(k) is odd for any real wave equation. (You may recall that we found a dispersion curve of the form ω(k) = vk + αk at small k). The group velocity v g = dω/ is thus an even function of k. Defining k = k + k 0 now ω(k) =ω( k 0 ) + dω k (05) k= k0 The phase near k = k 0 is thus Then the second integral in Eq. (99) = ω 0 + v g k (06) kx ω(k)t = ( k 0 x + ω 0 t) + k(x v g t) (07) I =e i(k 0x ω 0 t) d k 2π 2 G( k)e i k(x v gt) = 2 g(x v gt) e i(k 0x ω 0 t) (08) (09) The combination of I + + I is y(t, x) = cos(k 0 x ω 0 t) g(x v g t) (0) showing that the overall wave packet (envelope) moves with the group velocity, while the phase moves with the phase velocity v 0 = ω 0 /k 0 : At t = 0 we recover the original wave form y(t, x) = cos(k 0 (x v 0 t)) g(x v g t) () y(0, x) = cos(k 0 x) g(x). (2) 5

16 Problem 4. Time shift in a phase A string of tension T and mass per length µ is separated in two parts by a small ring of mass m. Consider the wave packet y in (x vt) as described in the previous problem (problem 3), incident upon the ring. At t = 0 the center of the wave incoming wave packet will arrive at the ring, i.e. at negative times the wave the is centered around v t. (a) First consider a plane wave A(k) e ikx wt. Shown that the reflected and transmitted amplitudes are B(k) =A(k) r(k) e i( kx ωt) (3) C(k) =A(k) t(k) e i(kx ωt) (4) where r(k) = k k ib t(k) = ib k ib (5) (6) where b = 2µ/m has the units of wavenumber. (b) Show that in general the reflected wave form is y trans (t, x) =b 0 y refl (t, x) = y in ( (x + vt)) + b dξ e bξ y in (x vt + ξ) (7) 0 dξ e bξ y in ( (x + vt) + ξ) (8) (c) When the mass m is very heavy, show that the reflection coefficient is r(k) e iφ(k) (9) where φ(k) b/k. The k-dependent phase has consequences. Show (using the Fourier techniques of the previous problem) that the center of the reflected wave packet is at y c = v(t t 0 ) (20) where the time shift t 0 = dφ v = b vk0 2 Argue that this result also follows from the stationary phase condition. (2) 6

17 Solution: (a) See also lecture. The left solution with frequency ω is While the right solution is y L = Ae ikx iωt + Be ikx iωt (22) y R = Ce ikx iωt (23) The boundary conditions are, first, continuity at x = 0 at all times A + B = C. (24) Second, we have F y net = ma y (t) for the ring. Drawing a free body diagram for the ring T dy L dx + T dy R dx = mÿ (25) Since the height of the ring is y R (t, 0) = Ce iωt (or you could use y L (t, 0) = (A + B)e iωt ) we have T A(ik) + T B(ik) + T (ikc) = mω 2 C (26) We solve these equations to find the expected result r(k) = k k ib t(k) = ib k ib (27) (28) (b) The incoming wave is of the form The reflected wave is Let us denote u = (x + vt) y A (t, x) = y B (t, x) = This then becomes a convolution 2π A(k)eik(x vt), (29) =y in (x vt). (30) y B (t, x) = y B (t, x) = = 2π r(k)a(k)eik( x vt). (3) 2π r(k)a(k)eiku. (32) du y in (u u )r(u ), (33) du y in (u )r(u u ), (34) 7

18 where r(u ) = e iku k k k ib. (35) This integral is easy to do using the table of Fourier transforms r(u ) = e k iku k k ib, (36) ( = e iku + ib ), (37) k ib k = δ(u ) + θ( u )be bu. (38) So Leading to 0 y B (t, x) = y in (u) + b du y in (u u ) e bu (39) y B (t, x) = y in ( (x + vt)) + b Similarly the transmitted wave is y C (t, x) = 0 dξ e bξ y in ( (x + vt) + ξ) (40) 2π t(k)a(k)e iku (4) but now u = x vt. So Here Leading to or y C (t, x) = u y in (u u ) t(u ) (42) t(u ) = e ib iku k k ib (43) =θ( u )be bu (44) 0 y C (t, x) = b y in (u u )e bu (45) y C (t, x) = b 0 e bξ y in ((x vt) + ξ) (46) (c) The amplitude reflection coefficient is r(k) = k k ib k k ( + i b k ) eiφ(k) (47) 8

19 with φ(k) = b/k. We note the phase in an odd function of k y R (t, x) = 2π eikx A(k)e iφ(k) e ik( x vt) (48) Then take the wave packet of the previous problem and expand near k 0 we have φ(k) + k( x vt) = φ(k 0 ) k 0 (x + vt) + k( (x + vt) + φ (k 0 )) (49) Following the steps of the previous problem one y R (t, x) = cos( k 0 (x + vt) + φ 0 ) g( (x + vt) + φ (k 0 )) (50) The center of the wave packet of the is given by the condition that the argument of g is zero Solving x c (t) vt + φ (k 0 ) = 0 (5) x c (t) = vt + φ (k 0 ) (52) So we see that the reflected wave is slightly ahead (more negative) than the naive expectation, vt, by b/vk0. 2 x c (t) = vt b (53) Perhaps one way can understand this counter intuitive result is that the leading edge of the incoming wave starts to reflect before the center of the wave packet reaches and reflects off the ring. vk 2 0 9