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these periodic sine and cosine waves:

1 1

1 a + X a cos (n x=L) + X b sin (n x=L) :

f(x) = 2 0 (11.2)

n n

n=1 n=1

The factor 1=2 in the coe cient a0 has no special signi cance, it is used to simplify

subsequent expressions. To show that (11.2) is actually true is not trivial. Providing

this proof essentially amounts to showing that the functions cos (n x=L) and sin (n x=L)

actually contain enough \degrees of freedom" to describe f(x). However, since f(x) is a

function of a continuous variable x this function has in nitely many degrees of freedom

and since there are in nitely many coe cients an and bn counting the number of degrees

of freedom does not work. Mathematically one would say that one needs to show that the

set of functions cos (n x=L) and sin (n x=L) is a \complete set." We will not concern us

here with this proof, and simply start working with the Fourier series (11.2).

At this point it is not clear yet what the coe cients an and bn are. In order to derive

these coe cients one needs to use the following integrals:

ZL Z

2 (n x=L) dx = L sin2 (n x=L) dx = L

cos (n 1) (11.3)

;L ;L

ZL

cos (n x=L) cos (m x=L) dx = 0 if n=m (11.4)

;L

6

ZL

sin (n x=L) sin (m x=L) dx = 0 if n=m (11.5)

;L

6

ZL

cos (n x=L) sin (m x=L) dx = 0 all n m : (11.6)

;L

Problem b: Derive these identities. In doing so you need to use trigonometric identities

)) =2. If you have di culties deriving

such as cos cos = (cos( + ) + cos( ;

these identities you may want to consult a textbook such as Boas 11].

11.1. THE REAL FOURIER SERIES ON A FINITE INTERVAL 131

Problem c: In order to nd the coe cient bm , multiply the Fourier expansion (11.2) with

sin (m x=L), integrate the result from to L and use the relations (11.3)-(11.6)

;L

to evaluate the integrals. Show that this gives:

1 Z L f(x) sin (n x=L) dx :

bn = L (11.7)

;L

Problem d: Use a similar analysis to show that:

1 Z L f(x) cos (n x=L) dx :

an = L (11.8)

;L

In deriving this result treat the cases n = 0 and n = 0 separately. It is now clear why

6

the factor 1=2 is introduced in the a0 -term of (11.2) without this factor expression

(11.8) would have an additional factor 2 for n = 0.

There is a close relation between the Fourier series (11.2) and the coe cients given in

the expressions above and the projection of a vector on a number of basis vectors in linear

algebra as shown in section (10.1). To see this we will restrict ourselves for simplicity to

functions f(x) that are odd functions of x: f(;x) = (x), but this restriction is by no

;f

means essential. For these functions all coe cients an are equal to zero. As an analogue

of a basis vector in linear algebra let us de ne the following basis function un (x):

u (x) 1 sin (n x=L) : (11.9)

n

L

p

An essential ingredient in the projection operators of section (10.1) is the inner product

between vectors. It is also possible to de ne an inner product for functions, and for the

present example the inner product of two functions f(x) and g(x) is de ned as:

ZL

(f g) f(x)g(x)dx : (11.10)

;L

Problem e: The basis functions un(x) de ned in (11.9) are the analogue of a set of

orthonormal basis vectors. To see this, use (11.3) and (11.5) to show that

(un um ) = nm (11.11)

where nm is the Kronecker delta.

This expression implies that the basis functions un (x) are mutually orthogonal. If the

p

(un un ), expression (11.11) implies

norm of such a basis function is de ned as n ku k

that the basis functions are normalized (i.e. have norm 1). These functions are the

generalization of orthogonal unit vectors to a function space. The (odd) function f(x)

can be written as a sum of the basis functions un (x):

1

X

f(x) = cnun(x) : (11.12)

n=1

CHAPTER 11. FOURIER ANALYSIS

132

Problem f: Take the inner product of (11.12) with um (x) and show that cm = (um f).

Use this to show that the Fourier expansion of f(x) can be written as: f(x) =

P1 u (x) (u f). and that leaving out the explicit dependence on the variable x

n=1 n n

the result is given by 1

X

f= un (un f) : (11.13)

n=1

This equation bears a close resemblance to the expression derived in section (10.1) for the

projection of vectors. The projection of a vector v along a unit vector n was shown to be

^

Pv = n (^ v)

^n (10:2) again :

A comparison with equation (11.13) shows that un (x) (un f) can be interpreted as the

projection of the function f(x) on the function un (x). To reconstruct the function, one

must sum over the projections along all basis functions, hence the summation in (11.13).

It is shown in equation (10.12) of section (10.1) that in order to nd the projection of

the vector v onto the subspace spanned by a nite number of orthonormal basis vectors

one simply has to sum the projections of the vector v on all the basis vectors that span

P^ n

the subspace: Pv = i ni (^i v). In a similar way, one can sum the Fourier series (11.13)

over only a limited number of basis functions to obtain the projection of f(x) on a limited

number of basis functions:

X

n2

ffiltered = un (un f) (11.14)

n=n1

in this expression it was assumed that only values n1 n n2 have been used. The

projected function is called ffiltered because this projection really is a ltering operation.

Problem g: To see this, show that the functions un(x) are sinusoidal waves with wave-

length = 2L=n.

This means that restricting the n-values in the sum (11.14) amounts to using only wave-

lengths between 2L=n2 and 2L=n1 for the projected function. Since only certain wave-

lengths are used, this projection really acts as a lter that allows only certain wavelengths

in the ltered function.

It is the ltering property that makes the Fourier transform so useful for ltering data

sets for excluding wavelengths that are unwanted. In fact, the Fourier transform forms the

basis of digital ltering techniques that have many applications in science and engineering,

see for example the books of Claerbout 15] or Robinson and Treitel 51].

11.2 The complex Fourier series on a nite interval

In the theory of the preceding section there is no reason why the function f(x) should be

real. Although the basis functions cos(n x=L) and sin(n x=L) are real, the Fourier sum

(11.2) can be complex because the coe cients an and bn can be complex. The equation

of de Moivre gives the relation between these basis functions and complex exponential

functions:

ein x=L = cos(n x=L) + i sin(n x=L) (11.15)

11.2. THE COMPLEX FOURIER SERIES ON A FINITE INTERVAL 133

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