Fourier Transforms - Lecture 9

Transcription

1 1 Introduction Fourier Transforms - Lecture 9 Previously we used the complete set of harmonic functions to represent another function, f(x), within limits in a Cartesian coordinate space. This is because the solutions to the ode; y + k 2 y = subject to the boundary conditions, y(x) = for x =, b, results in a Strum-Liouville problem with harmonic eigenfunctions and eigenvalues x = nπ/k. The eigenfunctions form a complete set that span the space x a. Thus any function with a finite number of discontinuities can be represented by these eigenfunctions for x a. In general the eigenfunctions are harmonic and can be represented by e ikx with k = nπ/b for example. The space could also be re-defined to be c x d. The limits c and d being the values of the function which produce the eigenfunctions. Thus in general; f(x) = n= A n e inx However, to consider completeness in another way, suppose we consider a Laurant expansion of f(x) f(x) = n= α n z n On the unit circle, z = e iθ (This defines a map of f(x) into the unit circle) f(z) = n= α n [e iθ ] n This is complete because a power series is formed. Now because of orthogonality we find; [ ] π δnm m dx sin(mx) sin(nx) = m = [ ] π δnm m dx cos(mx) cos(nx) = m = dxcos(mx) sin(nx) = These integrals are used to obtain the expansion coefficients. Therefore,; f(x) = n= A n e inx = A + 2 [ n n=1 A n cos(nx) + n (i)a n sin(nx)] n=1 1

2 A π π x A Figure 1: The square wave function. Note the discontiunities 2 Example1 Suppose the square wave function as shown in Figure 1. Since the function is periodic in we need only consider the coordinate range x. Note the discontinuities within this range. The function, f(x) is odd, so we expect that an odd harmonic function will represent this f(x). Thus sin(nx) is uses. f(x) = n n=1 A n sin(nx) The expansion coefficients are; A n = (1/π) dxa(x) sin(nx) = 4A nπ Here A(x) = ±A which is a constant. n odd f(x) = 4A π [sin(x) + (1/3)sin(3x) + (1/5)sin(5x) + ] = 4A π n=1 sin(nx) n As another example suppose the saw-tooth function shown in Figure 2. Again the function is periodic, but this time it is an even function. We only need to consider the range, x 2L. Here; [ ] Ax/L x L f(x) = A(2 x/l) L x 2L f(x) = A + A n cos(nkx) n=1 The representation is in terms of the even functions cos(nkx). To be periodic in 2L then k = π/l, with nkx = n when x = 2L. Evaluate the coefficients; A n = (1/π) L dx (Ax/L) cos(nkx) + (1/π) 2 2L L dxa(2 x/l) cos(nkx)

3 A 2L L L 2L x Figure 2: The sawtooth function For n = A (1/π)[L + [2x x 2 /2L] 2L L ] = AL/π Otherwise; A n = 4AL π 3 n 2 f(x) = AL π [1 4 π 2 n odd cos(nπx/l) n 2 ] 3 Fourier Series The Fourier series representation of a function is an extremely useful tool. One reason is that the harmonic oscillator model can be applied to almost all problems, at least as a first approximation. Suppose we have a force F(x) which has an equilibrium point at x = x. At this point the force vanishes, F(x ) =. Thus for small displacements use a Taylor expansion expand about the equilibrium point where F(x ) =. F(x) = F(x ) + (x x )F (x ) + F(x) (x x )F (x ) The force in the small displacement approximation is linear and leads to harmonic oscillations about the equilibrium point. Higher orders may be obtained by perturbation expansions. Let x = x x and write the acceleration of the motion. d 2 x dt 2 = α 2 x + β 2 x 2 + = 3

4 The constants, α 2 = F (x ) and β 2 = F (x ). Then choose to represent the solution by a superposition of oscillations. To first order the harmonic form is e iαt ; x = n=1 A n e inαt Substitute this into the force equation ; (nα) 2 A n e int + α 2 A n e int = β 2 A n A m e i (n + m)t α 2 A n e int (1 n 2 ) = A n e int β 2 A m e imt In a perturbation series collect terms in powers of e iαt. n = A α 2 = β 2 A 2 Choose A = n = 1 A 1 α 2 (1 1) = 2β 2 (A 1 A ) = A = n = 2 A 2 α 2 (1 4) = β 2 [A A + A 2 A + A 2 1] A 2 = βa2 1 3α This series can be be continued to obtain higher powers of e iαt. 4 Half wave Rectifier The filter circuit of a half wave rectifier is illustrated in Figure 3. The equations governing the current flow are; I R + L di i dt + I 1 R L = V (t) I 2 /C = dv dt I = I 1 + I 2 R di dt 4

5 Ro I1 V(t) I C I2 L R Vo L Figure 3: The electronic circuit of a half wave rectifier V(t) π π t Figure 4: The output voltage from the diode inserted into the filter circuit above These equations must be solved simultaneously, and are usually done in frequency space not time space, by assuming harmonic time dependence. The filtered current is represented by I 1 with output voltage, V (t). The equation for I 1 is; d 2 I 1 dt 2 + [ 1 R C + R + R L ] di 1 dt R + R L CR I 1 = 1 LR C V The solution in frequency space is assumed to be of the form, I 1 = I sin(ωt + φ). We return to this in a moment. First the harmonic voltage (current) for the half-wave rectifier of Figure 4 is obtained by assuming; [ ] V sin(ωt) t π/ω V (t) = π/ω t /ω V (t) = A n cos(nωt) A n = V π A = 2/ω /ω A n = For n odd A n = 2V ω π(n 2 1) V (t) = 2V πω dt cos(nωt) V (t) n=even 1 n 2 1 cos(ωt) 5

6 [ ] n = 2/ω sin(ωt) V n (t) = n odd 4V ωπ(n 2 1) sin(ωt) Then substitute into the filter equation the input voltage, V (t) = V n (t) above, The output current will be I(t) = I sin(ωt + φ). I [ n 2 ω 2 sin(nωt + φ) + nαω cos(nωt + φ) + β sin(nωt + φ)] = γ V sin(nωt) α = R 1 C + R + R L L β = R + R L LR C γ = 1 LR C tan(φ) = αω n 2 ω 2 β γv I = n 2 ω 2 cos(φ) + αnω sin(φ) β cos(φ) 5 Temperature of a circular plate We are to find the steady-state temperature distribution of a circular plate with insulated faces and whose circumference is held at at temperature, T (r, θ). As previously derived, the steady-state temperature distribution is a solution to Laplace s equation. Solve this in Cylindrical coordinates. There will be a solution of the form; T = Ar k sin(kθ) + Br k cos(kθ) The solution is periodic in θ = so we develop a Fourier series; T = A k r k sin(kθ) + B k r k cos(kθ) Find the coefficients at r = r A k = 1 πr k B k = 1 πr k dθ sin(kθ) T(r, θ) dθ cos(kθ) T(r, θ) Substitute for the value to T(r o, θ) and integrate the above to get the solution. 6

7 6 Expansion of the delta function We seek a Fourier representation of the δ function with initial boundaries ±L and k = nπ/l so k = π/l; δ(x x ) = L π (Cn dk) e ikx C(k) = LC n π = 1 L dt f(t) e ikt L Formally take L and let dk. Thus we obtain; f(x) = C(k) = 1 dk C(k) e ikx dt f(t) e ikt This forms the basis of the Fourier integral theorem and will be discussed later. However, for the moment consider by substitution; f(t) = dt f(t ) 1 dke ik(t t ) Thus we find that the δ function can be represented by; δ(t t ) = 1 dk e ik(t t ) Below are some useful forms for the delta function which you should verify. δ(ax) = (1/a)δ(x) delta(x 2 a 2 ) = (1/2a)[δ(x a) + δ(x + a)] xδ (x) = δ(x) 7 Vibrations A mass, M, is attached to a spring with spring constant, k. The system is damped with resistance R. The geometry is shown in Figure 5. Newton s equations are; M d2 x dt 2 + R dx dt + k x = For a steady-state solution x = x e iωt. Upon substitution, one obtains; 7

8 k x = M R Figure 5: Mechanical vibrations of a damped oscillator k x = x 1 2 = k M M k R R Figure 6: A set of coupled, mechanical oscillators ω 2 M x + iωr x + k x = ω = ir 2m ± [k/m ( R 2M )2 ] 1/2 x = x e R/(2M) e iωt Now suppose a set of coupled oscillators as illustrated in Figure 6. Keep the masses, resistances, and spring constants the same to simplify the exposition. The equations for the motion are; M d2 x 1 dt 2 + R dx 1 dt + k x 1 + k(x 1 x 2 ) = M d2 x 2 dt 2 + R dx 2 dt + k x 2 + k(x 2 x 1 ) = Add and subtract the equations to decouple them. Let y + = x 1 + x 2 and y = x 2 x 1. M d2 y dt 2 + R dy dt + 3k y = M d2 y + dt 2 + R dy + dt + k y + = For steady-state solutions assume y ± y ± e iω ±t 8

9 ω + = ir/(2m) ± [k/m ( R 2M )2 ] 1/2 ω = ir/(2m) ± [3k/M ( R 2M )2 ] 1/2 The two frequencies are the natural modes of the system. There will be a natural node for each degree of freedom, but they can be degenerate. To simplify, return to the single oscillator and drive the system with a frequency ω. Assume the natural mode of the system is given by, ω = k/m. The equation of motion is; M d2 x dt + R dx dt + k x = F sin(ωt) The steady-state solution is x = x sin(ωt + φ). There is a phase factor, φ, in this solution indicating that while the system oscillates with the driving frequency, ω, it is out of phase with respect to the driving frequency. Substitute the solution into the equation of motion and equate terms in sin(ω t) and cos(ωt). This results in; tan(φ) = (R/M)ω ω ω F x = cos(φ) M[(ω ω ) 2 + (Rω/M) 2 ] Resonance occurs when ω = ω. At this frequency the phase factor is π/2 and the oscillation is 9 out of phase with the driving frequency. The amplitude depends only of the resistance, R, in the equation of motion and the energy input into the system given by F x is completely expended in the resistance when averaging over a cycle. 8 Fourier Integral When we developed the delta function, a representation of a function by an integral over e ±iω was demonstrated by taking appropriate limiting values in a harmonic expansion of the function. Suppose the general case; f(α) = dβ K(α, β) g(β) In the above K(α, β) is the kernel of the integral transformation. This is really an integral equation as we wish to obtain, g(β) knowing f(α) and K(α, β). Thus g(β) is to be found from an inverse transformation operating on f(α), and this can happen for a number of different kernels. We investigate kernels having the harmonic forms, sin(α β), cos(α β), and e iα β. This yields the transforms and their inverses. 9

10 Form 1 f(α) = 2/π g(β) = 2/π Form 2 f(α) = 2/π g(β) = 2/π Form 3 f(α) = 2/π g(β) = 2/π dβ g(β) sin(αβ) dα f(α) sin(α β) dβ g(β) cos(αβ) dα f(α) cos(αβ) dβ g(β) e i(αβ) i(α β) dα f(α) e In order to use these forms we will need a function that satisfies Dirichlet conditions, and f(x) must have the properties; 1. f(x) has a finite number of maxima and minima 2. f(x) has a finite number of discontinuities 3. f(x) has no infinities 4. dx f(x) converges Write; f(x) = k= C k = (1/) Substitute to obtain; C k e ikx dxf(x) e ikx f(x) = k= (1/) dx f(x e ik(x x) 1

11 Interchange the sum and integral and identify; δ(x x ) = k C k e ikx Multiply by e ik x and integrate over x from L to +L. e ik x = k lim L 2 sin([k k ]L) k k Choose (k k ) = nπ/l which produces a set of eigenfunctions. Only when k = k is there a non-zero values in the limit. Thus; e ik x = C k (2L) Then let n = (L/π) k so that; C k e ikx = (L/π) k( e ikx 2L )(eikx ) k k δ(x x ) = 1 dk e ik(x x ) 9 Integral evaluation by the inversion theorem Suppose an integral of the form; dα sin(αa) cos(αx) α Apply a Fourier Cosine transform. Choose Then; f(α) = 2/π g(β) = 2/π f(α) = dβ g(β) cos(αβ) dα f(α) cos(αβ) [ π/2 < α < a α a ] g(β) = a dα cos(αβ) = sin(β a)/β 11

12 a C 2 C 1 b Figure 7: The contour for evaluation of the integral dα sin(βa) cos(βx) β = f(α) From the example there is a relation between the Fourier integral theorem and contour integration. Look at the integral f(λ) = b a dρ e iλρ φ(ρ) Here, λ is real. The inversion theorem gives; [ ] φ(r) a < r < b (1/()) dλ e iλr f(λ) = Otherwise Thus write; J = J = dλ e iλρ f(λ) dλ e iλρ b a dρ e iλρ φ(ρ ) This is divided into an integral over λ and one in the complex plane over ρ. When the order of integration is interchanged the negative value of λ (integral from ) is closed in the upper half plane for ρ and the value for positive λ is closed in the lower half plane for ρ. Figure 7 shows the contour of integration. This yields J = c1 φ(s) i(s ρ) c2 φ(s) i(s ρ) = φ(s) i(s ρ) Which gives the result of the integral theorem. 1 Convolution The integral; 12

13 J(t) = (1/ ) dα h(α) g(t α) is called the convolution or Faltung of the functions f and g over the interval (, ). Consider the transforms; I(t) = (1/ ) h(ω) = (1/ ) f(t) = (1/ ) g(ω) = (1/ ) dω h(ω) e iωt dt I(t) e iωt dω g(ω) e iωt dt f(t) e iωt Then make the substitution into the integral above; J(t) = (1/ ) dα h(α)[(1/ ) dω f(ω) e iω[t α] ] Interchange order of integration and use the inverse transforms; J(t) = (1/ ) dω I(ω) f(ω) e iωt Substitution of t = yields the identity; (1/ ) dω I(ω) f(ω) = 1/ ) dα h(α) g( α) In more detail, suppose the transform of I(t) which is h(ω) above, can be considered as a product of two transforms, F(ω)G(ω) ; and ; I(t) = (1/ ) f(t) = (1/ ) g(t) = (1/ ) dω e iωt F(ω)G(ω) dω e iωt F(ω) dω e iωt G(ω) Substitute these forms and rearrange the terms to show; I(t) = (1/ ) du f(u) g(t u) 13

14 Parseval s theorem is obtained in a similar way. Suppose the integral of two functions. The result is an integral of the product of their transforms. du f(u) g(u) = du F(u) G( u) 11 Transforms of a derivative Consider a function of the form g(t) and we wish the transformation of d(r) g(t) f (r) (ω) = (1/ ) Integrating by parts; dt e iωt d (r) g(t) dt (r) dt (r). This is f (r) (ω) = [(1/ ) d(r 1) g(t) dt (r 1) e iωt (1/ ) dt (iω)e iωt d (r 1) g(t) dt (r 1) This allows the insertion of the boundary conditions. If the derivatives vanish as t then; f (r) (ω) = ( iω) r f(ω) In the case where the derivatives do not vanish, the values of the integrand at the surfaces must be included. 12 Vibration of a membrane Suppose the vibration of a membrane of uniform density. The membrane lies in the (x, y) plane and vibrates in the z direction. It is under a tension T per unit length. Consider a small element as shown in Figure 8. The displacement in the vertical direction, z(x, y), is due to the change in x as shown in the figure. T[ z(x + x/2, y) z(x x/2, y) x ]dy Then approximate sin(θ) Tan(θ) = z/ x and rewrite the above as; 14

15 T(x, y+ y/2) dx T(x x/2,y) dy T(x + x/2,y) dy T(x, y y/2) dx T dy x z θ T dy Figure 8: An element of a vibrating membrane T 2 z x 2 dxdy There is a contribution to the tension from the displacement for both the x and y directions. This tension (force) equals the mass of the element times its acceleration. T [ 2 z x z y 2 ] = x M y 2 z t 2 = σ 2 z t 2 The above is the 2-D wave equation with v = T/σ. Apply a Fourier transform to find the solution in 2-D. Z(α, β, t) = (1/) dx dy z(x, y, t) e i(αx+βy) The boundary conditions are fixed so that z =, x z = and z = at the boundaries. Apply the Fourier transform by multiplying by e i(αx + βy) and integrating over x and y. This is y done by parts applying the boundary conditions which sets these surface terms to zero. Thus; (1/) dx Set c 2 = T/σ to obtain; dy [ 2 z x z y 2 ] e i(αx+βy) = (α 2 + β 2 )Z 15

16 d 2 Z dt 2 + c 2 [α 2 + β 2 ]Z = The solution is harmonic. Z = A cos(ωt) + B sin(ωt) ω 2 = c 2 (α 2 + β 2 ) Apply the initial conditions. Assume that at t = ; z = f(x, y) and z dt = g(x, y) Fourier transform these conditions to obtain F and G. Apply these to the transformed displacement, Z. Z = F cos(ωt) + G/ω sin(ωt) Then apply the inverse transformation. Choose ω > z = (1/) dα dβ Z e i(αx+βy) The functional form for Z contains transforms of the initial conditions. This requires the forms for F and G, and in general lead to a complicated integral for the inverse transform. If we attempt to use the convolution theorem we would need the Fourier transform of cos(ct α 2 + β 2 ) and sin(ct α 2 + β 2 ) c. However, these forms have divergent integrals. α 2 + β 2 Thus multiply the transforms by e ǫu with u = c α 2 + β 2 and take ǫ after the integration. The end result leaves the solution in terms of the convolution theorem; z(x, y, t) = 2cπ 1 1 2cπ dα dα dβ t [ f(α, β) [(ct) 2 (x α) 2 (y β) 2 ] 1/2] g(α, β) dβ [ [(ct) 2 (x α) 2 (y β) 2 ] 1/2] 13 Vibrations of a string A string of length x L is stretched with tension, T. The string is fixed so that, z(x) = for x =, L The equation of motion for an element, x, of the string is obtained in the same way as the section above for the 2-D membrane. 16

17 T 2 z x 2 σ 2 z t 2 As previously let the velocity, c 2 = T/σ. Apply a sin transformation so that the boundary conditions are satisfied at x =, l. In this case the wave vector, k = nπ/l. The result is; L dxt 2 z x 2 sin(kx) σ 2 dt 2 L dxz sin(kx) = Integrate the first term by parts. The first integration gives; [k z dx sin(kx)]l L dxkt 2 z x 2 cos(kx) The surface term vanishes in this case. Integrate again by parts, and again the surface term vanishes, leaving the integral; k 2 T L dxz sin(kx) = k 2 T Z Transform the second term in the above equation to obtain the equation for the transform; 2 Z t 2 + ω 2 Z = ω 2 = k 2 c. The solution for the transform is; Z = A sin(ωt) + B cos(ωt) Choose initial conditions such that z = and z t to the initial conditions. = f(x) at t =. Now apply the transform Z(k, ) = Z(k, ) t L = dx () sin(kx) = L dxf(x) sin(kx) = Z (k, ) This means we must choose B = and the solution becomes; Z = n Z /ω sin(ωt) Then invert this form to obtain; z(x, t) = 2 πc n sin([nπ/l]x) n sin([nπc/l]t) 17

18 L dx f(x ) sin([nπ/l]x ) 14 Fourier transform of the diffusion equation Suppose at t = a plug is introduced in an infinite cylindrical pipe. We treat the pipe only in the dimension of length. The plug is elastic and can diffuse into the pipe. This is a common problem in gas pipelines when various gaseous or liquid materials use the same pipeline and a plug is inserted to separate the components. The plug obeys the diffusion equation; ρ t = D 2 ρ x 2 In the above equation ρ is the mass density, and D is the diffusion coefficient. the boundary conditions are that ρ = for all x when t <, and when x. At t = the plug is concentrated at one point so it is considered as a δ function. ρ(x, ) = M/A δ(x l) Take the Fourier transform ; R(k, t) = 1 The transformed equation is; R t = k 2 DR R = R e k2 D t dxρ(x, t) e ikx The initial condition determines R R (k, ) = 1 R = (M/A) e ikl dx (M/A)δ(x l) e ikx R(k, t) = (M/A) e ikl e k2 D t The inverse transform is; 18

19 ρ(x, t) = (M/A) 1 dk e ik(x l) k2 D t The integral can be obtained by completing the square. The result is; ρ(x, t) = (M/A) 1 2 πd t 2 l) e (x4d t The plug spreads as a Gaussian distribution with time. 15 Fourier transform of a power spectrum The EM power radiated through an area is given by the Poynting vector. P = E B µ In a radiation field B is perpendicular to E, and both are perpendicular to the wave vector, k. Assuming we write the magnitude of the area as da = r 2 dω where dω is the solid angle, the power radiated per unit solid angle is then; dp = r 2 S = r2 E 2 dω µ c In the above c is the velocity of the wave. Obviously, E and B are functions of time. Integrate over time to get the total energy radiated. dw dω = dt r2 µ c E(t) E(t) Use Parseval s theorem to write; dw dω = dω r2 µ c E(ω) E (ω) In the above E(t) and E(ω) are Fourier transform pairs. Finally one obtains; d 2 W dωdω = r2 µ c E(ω) E (ω) 16 Fast Fourier Transforms Define the Fourier transforms and the inverse; 19

20 Figure 9: A test of the FFT program for a signal of a fixed frequency H(f) = h(t) = dt h(t) e ift dt H(f) e ift We assume that H(f) = H( f) where f represents a frequency, f = ω. We define a two-sided spectral density by, P f (f) = H(f) 2 + H( f) 2 for f. The power is then; P = df P f. Then suppose h(t) is sampled at evenly spaced time intervals. h n = h(nδ) n =, 3, 2, 1,, 1, 2, 3,. The reciprocal of δ is the sampling rate. The Nyquist frequency is f c = 1/(2δ). If sin wave is sampled at the Nyquist frequency at its positive peak value, the next sample is at its negative peak value. This means 2 samples per cycle. Thus a function sampled with an interval δ, is bandwidth limited do frequencies smaller than f c = 1/(2δ). The entire information content of a signal can be recorded by sampling at a rate of δ 1, ie at twice the maximum frequency passed by an amplifier. Also any frequency outside the Nyquist frequency is over-laid (mapped) into the range f c < f < f c. A Fast Fourier transform (FFT) is a computation algorithm which uses evenly spaced sampling intervals to compute the Fourier transform of a set of discrete data. The number of points must be divided so that N is an integral power of 2. An example of a FFT where 2

21 FFT of PMT Background Power 9 Frequency (f = 1/2, = 1.33 x 1 ns/ch ) Figure 1: A FFT analysis of the noise from a digitized PMT signal the data is generated for a sin wave of fixed frequency is shown in Figure 9. The peaks in the power spectrum show the concentration of the power in the wave is concentrated at the frequency of the wave. This technique used to analyze electronic circuits to determine their response to various signals. An example of this is shown in Figure 1 which is an analysis of the noise on the output from a photomultiplier sensor after amplification. 21