10 Fourier Series - University College London

10 Fourier Series

10.1 Introduction

When the French mathematician Joseph Fourier (1768-1830) was trying to study the flow of heat in a metal plate, he had the idea of expressing the heat source as an infinite series of sine and cosine functions. Although the original motivation was to solve the heat equation, it later became obvious that the same techniques could be applied to a wide array of mathematical and physical problems.

In this course, we will learn how to find Fourier series to represent periodic functions as an infinite series of sine and cosine terms.

A function f (x) is said to be periodic with period T , if

f (x + T ) = f (x), for all x.

The period of the function f (t) is the interval between two successive repetitions.

10.2 Definition of a Fourier Series

Let f be a bounded function defined on the [-, ] with at most a finite number of maxima and minima and at most a finite number of discontinuities in the interval. Then the Fourier series of f is the series

1 f (x) = 2 a0 + a1 cos x + a2 cos 2x + a3 cos 3x + ...

+ b1 sin x + b2 sin 2x + b3 sin 3x + ...

where the coefficients an and bn are given by the formulae

1

a0

=

f (x)dx

-

1

an

=

f (x)cos(nx)dx

-

1

bn

=

f (x)sin(nx)dx

-

10.3 Why the coefficients are found as they are

We want to derive formulas for a0, an and bn. To do this we take advantage of some properies of sinusoidal signals. We use the following orthogonality conditions:

Orthogonality conditions

(i) The average value of cos(nx) and sin(nx) over a period is zero.

cos(nx)dx = 0

-

sin(nx)dx = 0

-

(ii) The average value of sin(mx)cos(nx) over a period is zero.

sin(mx)cos(nx)dx = 0

-

(iii) The average value of sin(mx)sin(nx) over a period,

sin(mx)sin(nx)dx =

-

if m = n = 0 0 otherwise

(iv) The average value of cos(mx)cos(nx) over a period,

2 if m = n = 0

cos(mx)cos(nx)dx = if m = n = 0

-

0 if m = n

Remark. The following trigonometric identities are useful to prove the above orthogonality conditions:

1 cos A cos B = [cos(A - B) + cos(A + B)]

2 1 sin A sin B = [cos(A - B) - cos(A + B)] 2 1 sin A cos B = [sin(A - B) + sin(A + B)] 2

Finding an We assume that we can represent the function by

1 f (x) = 2 a0 + a1 cos x + a2 cos 2x + a3 cos 3x + ...

+ b1 sin x + b2 sin 2x + b3 sin 3x + ... (1)

Multiply (1) by cos(nx), n 1 and integrate from - to and assume it is permissible to integrate the series term by term.

1

- f (x) cos(nx)dx = 2 a0

cos(nx) + a1

-

cos x cos(nx)dx + a2

-

cos 2x cos(nx)dx + ...

-

+ b1 sin x cos(nx)dx + b2 sin 2x cos(nx)dx + ...

-

-

= an, because of the above orthogonality conditions

1

an =

f (x) cos(nx)dx

-

Similarly if we multiply (1) by sin(nx), n 1 and integrate from - to , we can find the formula for the coefficient bn. To find a0, simply integrate (1) from - to .

10.4 Fourier Series

Definition. Let f (x) be a 2-periodic function which is integrable on [-, ]. Set

1

a0

=

f (x)dx

-

1

an

=

f (x)cos(nx)dx

-

1

bn

=

f (x)sin(nx)dx

-

then the trigonometric series

1

2 a0 + (ancos(nx) + bnsin(nx))

n=1

is called the Fourier series associated to the function f (x).

Remark. Notice that we are not saying f (x) is equal to its Fourier Series. Later we will discuss conditions under which that is actually true.

Example. Find the Fourier coefficients and the Fourier series of the squarewave function f defined by

f (x) =

0 if - x < 0 1 if 0 x <

and f (x + 2) = f (x)

Solution: So f is periodic with period 2 and its graph is:

Using the formulas for the Fourier coefficients we have

1

a0 =

f (x)dx

-

10

1

= ( 0dx +

-

1 = ()

1dx)

0

=1

1

an =

f (x) cos(nx)dx

-

10

1

= ( 0dx +

cos(nx)dx)

-

0

1 = (0 +

sin(nx) n

0

)

1 = (sin n - sin 0)

n

=0

1

bn =

f (x) sin(nx)dx

-

10

1

= ( 0dx +

sin(nx)dx)

-

0

=

1 (-

cos(nx)

)

n0

1 = - (cos n - cos 0)

n

= - 1 ((-1)n - 1), since cos n = (-1)n. n

0, if n is even =2

, if n is odd n

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