Fourier Transforms - Lecture 9

1 Introduction

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′ ′ + k2y = 0

subject to the boundary conditions, y(x) = 0 for x = 0, b, results in a Strum-Liouvilleproblem with harmonic eigenfunctions and eigenvalues x = nπ/k. The eigenfunctions forma complete set that “span” the space 0 ≤ x ≤ a. Thus any function with a finite numberof discontinuities can be represented by these eigenfunctions for 0 ≤ x ≤ a. In general theeigenfunctions are harmonic and can be represented by eikx with k = nπ/b for example. Thespace could also be re-defined to be c ≤ x ≤ d. The limits c and d being the values of thefunction which produce the eigenfunctions. Thus in general;

f(x) =∞∑

n=−∞

An einx

However, to consider completeness in another way, suppose we consider a Laurant expansionof f(x)

f(x) =∞∑

n=−∞

αn zn

On the unit circle, z = eiθ (This defines a map of f(x) into the unit circle)

f(z) =∞∑

n=−∞

αn[eiθ]n

This is complete because a power series is formed. Now because of orthogonality we find;

2π∫

0

dx sin(mx) sin(nx) =

[

π δnm m 6= 00 m = 0

]

2π∫

0

dx cos(mx) cos(nx) =

[

π δnm m 6= 02π m = 0

]

2π∫

0

dx cos(mx) sin(nx) = 0

These integrals are used to obtain the expansion coefficients. Therefore,;

f(x) =∞∑

n=−∞

An einx = A0 + 2 [n∑

n=1

An cos(nx) +n∑

n=1

(i)An sin(nx)]

1

A

x

−A

π 2π−2π −π

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 2πwe need only consider the coordinate range 0 ≤ x ≤ 2π. Note the discontinuities within thisrange. The function, f(x) is odd, so we expect that an odd harmonic function will representthis f(x). Thus sin(nx) is uses.

f(x) =n∑

n=1

An sin(nx)

The expansion coefficients are;

An = (1/π)2π∫

0

dxA(x) sin(nx) = 4Anπ n odd

Here A(x) = ±A which is a constant.

f(x) = 4Aπ [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 func-tion is periodic, but this time it is an even function. We only need to consider the range,0 ≤ x ≤ 2L. Here;

f(x) =

[

Ax/L 0 ≤ x ≤ LA(2 − x/L) L ≤ x ≤ 2L

]

f(x) = A0 +∞∑

n=1

An cos(nkx)

The representation is in terms of the even functions cos(nkx). To be periodic in 2L thenk = π/L, with nkx = 2πn when x = 2L. Evaluate the coefficients;

An = (1/π)L∫

0

dx (Ax/L) cos(nkx) + (1/π)2L∫

L

dxA(2 − x/L) cos(nkx)

2

A

xL−L 2L−2L

Figure 2: The sawtooth function

For n = 0

A0(1/π)[L + [2x − x2/2L]2LL ] = AL/π

Otherwise;

An = −4ALπ3n2

f(x) = ALπ [1 − 4

π2

∑

n odd

cos(nπx/L)n2 ]

3 Fourier Series

The Fourier series representation of a function is an extremely useful tool. One reason isthat the harmonic oscillator model can be applied to almost all problems, at least as a firstapproximation. Suppose we have a force F (x) which has an equilibrium point at x = x0.At this point the force vanishes, F (x0) = 0. Thus for small displacements use a Taylorexpansion expand about the equilibrium point where F (x0) = 0.

F (x) = F (x0) + (x − x0)F′(x0) + · · ·

F (x) ≈ (x − x0)F′(x0)

The force in the small displacement approximation is linear and leads to harmonic oscillationsabout the equilibrium point. Higher orders may be obtained by perturbation expansions.Let ∆x = x − x0 and write the acceleration of the motion.

d2∆xdt2

= α2 ∆x + β2 ∆x2 + · · · = 0

3

The constants, α2 = F ′(x0) and β2 = F ′ ′(x0). Then choose to represent the solution by asuperposition of oscillations. To first order the harmonic form is eiαt;

∆x =∞∑

n=1

An einαt

Substitute this into the force equation ;

−∑

(nα)2An eint + α2∑

An eint = −β2∑

AnAmei(n + m)t

∑

α2 An eint (1 − n2) = −∑

An eintβ2∑

Ameimt

In a perturbation series collect terms in powers of eiαt.

n = 0

A0α2 = −β2A2

0 Choose A0 = 0

n = 1

A1α2(1 − 1) = −2β2(A1A0) = 0 A0 = 0

n = 2

A2α2(1 − 4) = −β2[A0A0 + A2A0 + A2

1]

A2 =βA2

13α

This series can be be continued to obtain higher powers of eiαt.

4 Half wave Rectifier

The filter circuit of a half wave rectifier is illustrated in Figure 3. The equations governingthe current flow are;

I R0 + LdIidt

+ I1 RL = V (t)

I2/C = dVdt

− R0dIdt

I = I1 + I2

4

I

I I

1

2RL

V(t)

L

Ro

C Vo

Figure 3: The electronic circuit of a half wave rectifier

π 2π−π

V(t)

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 byI1 with output voltage, V0(t). The equation for I1 is;

d2I1

dt2+ [ 1

R0C+ R0 + RL]dI1

dt− R0 + RL

CR0I1 = 1

LR0CV0

The solution in frequency space is assumed to be of the form, I1 = I0 sin(ωt + φ). Wereturn to this in a moment. First the harmonic voltage (current) for the half-wave rectifierof Figure 4 is obtained by assuming;

V (t) =

[

V0 sin(ωt) 0 ≤ t ≤ π/ω0 π/ω ≤ t ≤ 2π/ω

]

V (t) =∑

An cos(nωt)

An = V0π

2π/ω∫

0

dt cos(nωt) V (t)

A0 = 2/ω

An = 0 For n odd

An = − 2V0ωπ(n2 − 1)

V (t) = −2V0πω

∞∑

n=even

1n2 − 1

cos(ωt)

5

Vn(t) =

[

n = 0 2/ω sin(ωt)

n odd − 4V0

ωπ(n2 − 1)sin(ωt)

]

Then substitute into the filter equation the input voltage, V (t) = Vn(t) above, The outputcurrent will be I(t) = I0 sin(ωt + φ).

I0[−n2ω2 sin(nωt + φ) + nαω cos(nωt + φ) + β sin(nωt + φ)] = γ V0 sin(nωt)

α = 1R0C

+ R0 + RLL

β = R0 + RLLR0C

γ = 1LR0C

tan(φ) = αωn2ω2 − β

I0 =−γV0

n2ω2 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 insulatedfaces and whose circumference is held at at temperature, T0(r0, θ). As previously derived,the steady-state temperature distribution is a solution to Laplace’s equation. Solve this inCylindrical coordinates. There will be a solution of the form;

T = Ark sin(kθ) + Brk cos(kθ)

The solution is periodic in θ = 2π so we develop a Fourier series;

T =∑

Ak rk sin(kθ) +∑

Bk rk cos(kθ)

Find the coefficients at r = r0

Ak = 1πrk

0

2π∫

0

dθ sin(kθ) T (r0, θ)

Bk = 1πrk

0

2π∫

0

dθ cos(kθ) T (r0, θ)

Substitute for the value to T (ro, θ) and integrate the above to get the solution.

6

6 Expansion of the delta function

We seek a Fourier representation of the δ function with initial boundaries ±L and k = nπ/Lso ∆k = π/L;

δ(x − x′) = Lπ

∑

(Cn dk) eikx

C(k) = LCnπ = 1

2π

L∫

−L

dt f(t) e−ikt

Formally take L → −∞ and let dk → 0. Thus we obtain;

f(x) =∞∫

−∞

dk C(k) eikx

C(k) = 12π

∞∫

−∞

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′) 12π

∞∫

−∞

dkeik(t−t′)

Thus we find that the δ function can be represented by;

δ(t − t′) = 12π

∞∫

−∞

dk eik(t−t′)

Below are some useful forms for the delta function which you should verify.δ(ax) = (1/a)δ(x)

delta(x2 − a2) = (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 withresistance R. The geometry is shown in Figure 5. Newton’s equations are;

M d2xdt2

+ R dxdt

+ k x = 0

For a steady-state solution x = x0 eiωt. Upon substitution, one obtains;

7

x = 0

R

Mk

Figure 5: Mechanical vibrations of a damped oscillator

R

Mk

R

k kM

x = 01 2x = 0

Figure 6: A set of coupled, mechanical oscillators

−ω2M x0 + iωR x0 + k x0 = 0

ω = iR2m ± [k/M − ( R

2M )2]1/2

x = x0 e−R/(2M) eiωt

Now suppose a set of coupled oscillators as illustrated in Figure 6. Keep the masses, re-sistances, and spring constants the same to simplify the exposition. The equations for themotion are;

M d2x1

dt2+ R dx1

dt+ k x1 + k(x1 − x2) = 0

M d2x2

dt2+ R dx2

dt+ k x2 + k(x2 − x1) = 0

Add and subtract the equations to decouple them. Let y+ = x1 + x2 and y− = x2 − x1.

Md2y−dt2

+ Rdy−dt

+ 3k y− = 0

Md2y+

dt2+ R

dy+

dt+ k y+ = 0

For steady-state solutions assume y± − y0± eiω±t

8

ω+ = iR/(2m) ± [k/M − ( R2M )2]1/2

ω− = iR/(2m) ± [3k/M − ( R2M )2]1/2

The two frequencies are the natural modes of the system. There will be a natural nodefor each degree of freedom, but they can be degenerate. To simplify, return to the singleoscillator and drive the system with a frequency ω. Assume the natural mode of the systemis given by, ω0 = k/M . The equation of motion is;

M d2xdt

+ R dxdt

+ k x = F0 sin(ωt)

The steady-state solution is x = x0 sin(ωt + φ). There is a phase factor, φ, in this solutionindicating that while the system oscillates with the driving frequency, ω, it is out of phasewith respect to the driving frequency. Substitute the solution into the equation of motionand equate terms in sin(ω t) and cos(ωt). This results in;

tan(φ) =(R/M)ωω − ω0

x0 = − F0 cos(φ)M [(ω − ω0)

2 + (Rω/M)2]

Resonance occurs when ω = ω0. At this frequency the phase factor is π/2 and the oscillationis 90 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 completelyexpended 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 overe±iω was demonstrated by taking appropriate limiting values in a harmonic expansion of thefunction. 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 integralequation as we wish to obtain, g(β) knowing f(α) and K(α, β). Thus g(β) is to be foundfrom an inverse transformation operating on f(α), and this can happen for a number ofdifferent kernels. We investigate kernels having the harmonic forms, sin(α β), cos(α β), andeiα β. This yields the transforms and their inverses.

9

Form 1

f(α) =√

2/π∞∫

0

dβ g(β) sin(αβ)

g(β) =√

2/π∞∫

0

dα f(α) sin(α β)

Form 2

f(α) =√

2/π∞∫

0

dβ g(β) cos(αβ)

g(β) =√

2/π∞∫

0

dα f(α) cos(αβ)

Form 3

f(α) =√

2/π∞∫

0

dβ g(β) ei(αβ)

g(β) =√

2/π∞∫

0

dα f(α) e−i(α β)

In order to use these forms we will need a function that satisfies Dirichlet conditions, andf(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=−∞

Ck eikx

Ck = (1/2π)2π∫

0

dx f(x) e−ikx

Substitute to obtain;

f(x) =∞∑

k=−∞

(1/2π)2π∫

0

dx′ f(x′ eik(x′−x)

10

Interchange the sum and integral and identify;

δ(x − x′) =∑

k

Ck eikx

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 therea non-zero values in the limit. Thus;

e−ik′x′

= Ck′(2L)

Then let ∆n = (L/π)∆k so that;

∑

k

Ck eikx =∑

k

(L/π)∆k(e−ikx′

2L )(eikx)

δ(x − x′) = 12π

∞∫

−∞

dk eik(x−x′)

9 Integral evaluation by the inversion theorem

Suppose an integral of the form;

∞∫

0

dαsin(αa) cos(αx)

α

Apply a Fourier Cosine transform.

f(α) =√

2/π∞∫

0

dβ g(β) cos(αβ)

g(β) =√

2/π∞∫

0

dα f(α) cos(αβ)

Choose

f(α) =

[ √

π/2 0 < α < a0 α ≥ a

]

Then;

g(β) =a∫

0

dα cos(αβ) = sin(β a)/β

11

C

C

1

2

a

b

Figure 7: The contour for evaluation of the integral

∞∫

0

dαsin(βa) cos(βx)

β= f(α)

From the example there is a relation between the Fourier integral theorem and contour in-tegration. Look at the integral

f(λ) =b∫

a

dρ eiλρ φ(ρ)

Here, λ is real. The inversion theorem gives;

(1/(2π))∞∫

−∞

dλ e−iλr f(λ) =

[

φ(r) a < r < b0 Otherwise

]

Thus write;

J =∞∫

−∞

dλ e−iλρ f(λ)

J =∞∫

−∞

dλ e−iλρb∫

a

dρ′ eiλρ′ φ(ρ′)

This is divided into an integral over λ and one in the complex plane over ρ′. When the orderof integration is interchanged the negative value of λ (integral from −∞ → 0) is closed inthe 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.

10 Convolution

The integral;

12

J(t) = (1/√

2π)∞∫

−∞

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/√

2π)∞∫

−∞

dω h(ω) eiωt

h(ω) = (1/√

2π)∞∫

−∞

dt I(t) e−iωt

f(t) = (1/√

2π)∞∫

−∞

dω g(ω) eiωt

g(ω) = (1/√

2π)∞∫

−∞

dt f(t) e−iωt

Then make the substitution into the integral above;

J(t) = (1/√

2π)∞∫

−∞

dα h(α)[(1/√

2π)∞∫

−∞

dω f(ω) e−iω[t−α]]

Interchange order of integration and use the inverse transforms;

J(t) = (1/√

2π)∞∫

−∞

dω I(ω) f(ω) e−iωt

Substitution of t = 0 yields the identity;

(1/√

2π)∞∫

−∞

dω I(ω) f(ω) = 1/√

2π)∞∫

−∞

dα h(α) g(−α)

In more detail, suppose the transform of I(t) which is h(ω) above, can be considered as aproduct of two transforms, F (ω)G(ω) ;

I(t) = (1/√

2π)∞∫

−∞

dω eiωt F (ω)G(ω)

and ;

f(t) = (1/√

2π)∞∫

−∞

dω e−iωt F (ω)

g(t) = (1/√

2π)∞∫

−∞

dω e−iωt G(ω)

Substitute these forms and rearrange the terms to show;

I(t) = (1/√

2π)∞∫

−∞

du f(u) g(t− u)

13

Parseval’s theorem is obtained in a similar way. Suppose the integral of two functions. Theresult 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 ofd(r)g(t)

dt(r). This is

f (r)(ω) = (1/√

2π)∞∫

−∞

dt eiωt d(r)g(t)

dt(r)

Integrating by parts;

f (r)(ω) = [(1/√

2π)d(r−1)g(t)

dt(r−1) eiωt − (1/√

2π)∞∫

−∞

dt (iω)eiωt 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 surfacesmust 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 asmall 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 thefigure.

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

∆x/2+

y+∆y/2T(x, )

T(x ,y)−∆x/2

y−∆y/2

T(x ,y)

dx

dx

dy dy

T dy

T dy

∆∆

xz

θ

T(x, )

Figure 8: An element of a vibrating membrane

T ∂2z∂x2 dx dy

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 [∂2z

∂x2 + ∂2z∂y2 ] = ∆M

∆x ∆y∂2z∂t2

= σ ∂2z∂t2

The above is the 2-D wave equation with v =√

T/σ. Apply a Fourier transform to findthe solution in 2-D.

Z(α, β, t) = (1/2π)∞∫

−∞

dx∞∫

−∞

dy z(x, y, t) ei(αx+βy)

The boundary conditions are fixed so that z = 0, ∂z∂x

= 0 and ∂z∂y

= 0 at the boundaries. Ap-

ply the Fourier transform by multiplying by ei(αx +βy) and integrating over x and y. This isdone by parts applying the boundary conditions which sets these surface terms to zero. Thus;

(1/2π)∞∫

−∞

dx∞∫

−∞

dy [∂2z

∂x2 + ∂2z∂y2 ] ei(αx+βy) = −(α2 + β2)Z

Set c2 = T/σ to obtain;

15

d2Zdt2

+ c2[α2 + β2]Z = 0

The solution is harmonic.

Z = A cos(ωt) + B sin(ωt)

ω2 = c2(α2 + β2)

Apply the initial conditions. Assume that at t = 0;

z = f(x, y) and ∂zdt

= g(x, y)

Fourier transform these conditions to obtain F and G. Apply these to the transformed dis-placement, Z.

Z = F cos(ωt) + G/ω sin(ωt)

Then apply the inverse transformation. Choose ω > 0

z = (1/2π)∞∫

−∞

dα∞∫

−∞

dβ Z e−i(αx+βy)

The functional form for Z contains transforms of the initial conditions. This requires theforms for F and G, and in general lead to a complicated integral for the inverse trans-form. If we attempt to use the convolution theorem we would need the Fourier transform

of cos(ct√

α2 + β2) andsin(ct

√

α2 + β2)

c√

α2 + β2. However, these forms have divergent integrals.

Thus multiply the transforms by e−ǫu with u = c√

α2 + β2 and take ǫ → 0 after the inte-gration. The end result leaves the solution in terms of the convolution theorem;

z(x, y, t) = − 12cπ

∞∫

−∞

dα∞∫

−∞

dβ ∂∂t

[f(α, β)

[(ct)2 − (x − α)2 − (y − β)2]1/2 ]−

12cπ

∞∫

−∞

dα∞∫

−∞

dβ [g(α, β)

[(ct)2 − (x − α)2 − (y − β)2]1/2 ]

13 Vibrations of a string

A string of length 0 ≤ x ≤ L is stretched with tension, T . The string is fixed so that,z(x) = 0 for x = 0, L The equation of motion for an element, ∆x, of the string is obtainedin the same way as the section above for the 2-D membrane.

16

T ∂2z∂x2 − σ∂2z

∂t2

As previously let the velocity, c2 = T/σ. Apply a sin transformation so that the boundaryconditions are satisfied at x = 0, l. In this case the wave vector, k = nπ/L. The result is;

L∫

0

dx T ∂2z∂x2 sin(kx) − σ ∂2

dt2L∫

0

dx z sin(kx) = 0

Integrate the first term by parts. The first integration gives;

[k ∂z∂dx

sin(kx)]L0 −L∫

0

dx kT ∂2z∂x2 cos(kx)

The surface term vanishes in this case. Integrate again by parts, and again the surface termvanishes, leaving the integral;

−k2TL∫

0

dxz sin(kx) = −k2T Z

Transform the second term in the above equation to obtain the equation for the transform;

∂2Z∂t2

+ ω2Z = 0

ω2 = k2c. The solution for the transform is;

Z = A sin(ωt) + B cos(ωt)

Choose initial conditions such that z = 0 and ∂z∂t

= f(x) at t = 0. Now apply the transformto the initial conditions.

Z(k, 0) =L∫

0

dx (0) sin(kx) = 0

∂Z(k, 0)∂t

=L∫

0

dx f(x) sin(kx) = Z ′(k, 0)

This means we must choose B = 0 and the solution becomes;

Z =∑

n

Z ′/ω sin(ωt)

Then invert this form to obtain;

z(x, t) = 2πc

∑

n

sin([nπ/L]x)n sin([nπc/L]t)

17

L∫

0

dx′ f(x′) sin([nπ/L]x′)

14 Fourier transform of the diffusion equation

Suppose at t = 0 a plug is introduced in an infinite cylindrical pipe. We treat the pipeonly in the dimension of length. The plug is elastic and can diffuse into the pipe. This isa common problem in gas pipelines when various gaseous or liquid materials use the samepipeline and a plug is inserted to separate the components. The plug obeys the diffusionequation;

∂ρ∂t

= D∂2ρ∂x2

In the above equation ρ is the mass density, and D is the diffusion coefficient. the boundaryconditions are that ρ = 0 for all x when t < 0, and when x → ∞. At t = 0 the plug isconcentrated at one point so it is considered as a δ function.

ρ(x, 0) = M/A δ(x − l)

Take the Fourier transform ;

R(k, t) = 1√2π

∞∫

−∞

dx ρ(x, t) e−ikx

The transformed equation is;

∂R∂t

= −k2DR

R = R0 e−k2D t

The initial condition determines R0

R0(k, 0) = 1√2π

∞∫

−∞

dx (M/A)δ(x − l) e−ikx

R0 = (M/A)e−ikl√2π

R(k, t) = (M/A)e−ikl√2π

e−k2D t

The inverse transform is;

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ρ(x, t) = (M/A) 12π

∞∫

−∞

dk eik(x−l)−k2D t

The integral can be obtained by completing the square. The result is;

ρ(x, t) = (M/A) 12√

πD te−

(x − l)2

4d 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

µ0

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 = r2 dΩ where dΩ is the solid angle,the power radiated per unit solid angle is then;

dPdΩ

= r2|~S| = r2E2

µ0c

In the above c is the velocity of the wave. Obviously, E and B are functions of time. Inte-grate over time to get the total energy radiated.

dWdΩ

=∞∫

−∞

dt r2

µ0c~E(t) · ~E(t)

Use Parseval’s theorem to write;

dWdΩ

=∞∫

−∞

dω r2

µ0c~E(ω) · ~E∗(ω)

In the above ~E(t) and ~E(ω) are Fourier transform pairs. Finally one obtains;

d2WdΩ dω

= r2

µ0c~E(ω) · ~E∗(ω)

16 Fast Fourier Transforms

Define the Fourier transforms and the inverse;

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Figure 9: A test of the FFT program for a signal of a fixed frequency

H(f) =∞∫

−∞

dt h(t) ei 2πft

h(t) =∞∫

−∞

dt H(f) e−i2πft

We assume that H(f) = H(−f) where f represents a frequency, f = 2πω. We define atwo-sided spectral density by, Pf(f) = |H(f)|2 + |H(−f)|2 for 0 ≤ f ≤ ∞. The power is

then; P =∞∫

0

df Pf . Then suppose h(t) is sampled at evenly spaced time intervals.

hn = h(nδ) n = · · · ,−3,−2,−1, 0, 1, 2, 3, · · · .

The reciprocal of δ is the sampling rate. The Nyquist frequency is fc = 1/(2δ). If sinwave is sampled at the Nyquist frequency at its positive peak value, the next sample is atits negative peak value. This means 2 samples per cycle. Thus a function sampled withan interval δ, is bandwidth limited do frequencies smaller than fc = 1/(2δ). The entireinformation content of a signal can be recorded by sampling at a rate of δ−1, ie at twicethe maximum frequency passed by an amplifier. Also any frequency outside the Nyquistfrequency is over-laid (mapped) into the range fc < f < fc.

A Fast Fourier transform (FFT) is a computation algorithm which uses evenly spacedsampling intervals to compute the Fourier transform of a set of discrete data. The numberof points must be divided so that N is an integral power of 2. An example of a FFT where

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Frequency (f = 1/2 , = 1.33 x 10 )∆ ∆ ns/ch−9P

ower

FFT of PMT Background

Figure 10: 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 inthe power spectrum show the concentration of the power in the wave is concentrated at thefrequency of the wave. This technique used to analyze electronic circuits to determine theirresponse to various signals. An example of this is shown in Figure 10 which is an analysisof the noise on the output from a photomultiplier sensor after amplification.

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