3a sin2 φ

φ = 1 ka

F (k) = (12.45)

2

4π φ2

k =0 (12.46)

3

σk = k 2 =

2

(12.47)

a2

3

σx σk = = 0.5477 (12.48)

10

12.5.3 Gaussian function

The Gaussian function (Fig. 12.3) is the Fourier transform of itself. the

equations are:

√ x2

1

’2

exp ’ 2

f (x) = (σ 2π) (12.49)

4σ

x =0 (12.50)

σx = x2 = σ 2

2

(12.51)

√ ’1 k2 1

2 exp ’

F (k) = (σk 2π) ; σk = (12.52)

2 2σ

4σk

k =0 (12.53)

1

σk = k 2 = 2

2

(12.54)

4σ

1

σx σk = 2 (12.55)

So we see that “ of the functions given here “ the Gaussian wave function

attains the smallest product of variances in real and reciprocal space. In

12.6 Discrete Fourier transforms 323

fact, the Gaussian function has the smallest possible product of variances of

all (well-behaved) functions.

12.6 Discrete Fourier transforms

Now consider periodic functions f (x) with periodicity a:

f (x + na) = f (x), n ∈ Z. (12.56)

The function is assumed to be piecewise continuous and absolutely integrable

over the domain (0, a):

a

|f (x)| dx exists. (12.57)

0

The function f (x) can now be expressed in an in¬nite series of exponential

functions that are each periodic functions of x on (0, a):

2πinx

f (x) = Fn exp , (12.58)

a

n∈Z

with

’2πinx

a

1

Fn = f (x) exp dx. (12.59)

a a

0

This can also be written as

2πn

, n ∈ Z,

Fk eikx , k =

f (x) = (12.60)

a

k

a

1

f (x)e’ikx dx.

Fk = (12.61)

a 0

a

The validity can be checked by computing 0 [ k exp(ik x)] exp(’ikx) dx.

All terms with k = k vanish, and the surviving term yields aFk .

—

It is clear that f (x) is real when F’k = Fk . In that case the transform

can also be expressed in sine and cosine transforms:

∞

1 2πn

f (x) = 2 a0 + k (ak cos kx + bk sin kx), k= a, n = 1, 2, . .(12.62)

.,

a

1

ak = f (x) cos kx dx, (12.63)

2a 0

a

1

bk = f (x) sin kx dx. (12.64)

2a 0

324 Fourier transforms

12.7 Fast Fourier transforms

A special form of discrete Fourier transforms is the application to periodic

discrete functions known at regular intervals ”x = a/N , where N is the

number of intervals within the period a. The function values are fn = f (xn ),

with xn = na/N, n = 0, . . . , N ’ 1. The Fourier relations are now

N ’1

Fm e2πinm/N ,

fn = (12.65)

m=0

N ’1

1

fn e’2πinm/N .

Fm = (12.66)

N

n=0

These relations are easily veri¬ed by inserting (12.65) into (12.66) and re-

alizing that n exp[2πi(m ’ m)n/N ] equals zero unless m = m. In the

general case the arrays f and F are complex. One should view both arrays

as periodic: they can be shifted to another origin if required.

Consider the values fn as a variable in time t, de¬ned on a periodic interval

[0, T ) and discretized in N small time intervals ”t. Thus T = N ”t. The

data are transformed to a discretized frequency axis νm , with resolution

”ν = 1/T and maximum frequency determined by νmax = 1/2”t. The

latter follows from Nyquist™s theorem stating that two data points per period

of the highest frequency are su¬cient and necessary to reproduce a function

of time that is band-limited to νmax (i.e., which has no Fourier components

beyond νmax ).3 The transform contains N frequency components between

’νmax and νmax in steps of ”ν. Because of the periodicity of Fm in [0, n),

the negative frequencies are to be found for N/2 ¤ m < N : F’m =

FN ’m . In the special but common case that fn is real, the transform is

—

described by a complex array of length N/2 because F’m = FN ’m = Fm .

If a continuous function that contains frequencies ν > νmax is sampled at

intervals ”t = 1/2νmax , these higher frequencies are aliased or folded back

into the frequency domain [0, νmax ) and appear as frequencies 2νmax ’ν, i.e.,

the signal is mirrored with respect to νmax . Such a “false” signal is called

an alias. When noisy signals are sampled at intervals ”t, noise components