Chapter 3 Fourier Analysis

Chapter 3 Fourier Analysis

Introduction to orthogonal polynomials Representing a periodic function as an infinite sum of sine and cosine functions.

Finding the signal in the noise.

f (x) defined on [ , ] and f (x + 2) = f (x)

f (x)

=

a0 2

+

a1

cos x

+

a2

cos 2x

+

a3

cos 3x ? ? ?

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

(3.1)

3.1 A reminder of some wave basics

Solutions of the wave equation take the form:

u(x, t) = A sin 2 (x vt) + .

(3.2)

Note

that

@2u @t2

=

!2u, so particle at any point x executes simple harmonic motion

with frequency !. Here is the wavelength of the wave and this angular frequency

! is determined by ! = kv with k 2 the "wave number" of the wave.

If the endpoints of the medium in which the wave oscillates are fixed, such that

u(0, t) = u(L, t) then can take only a discrete set of values:

= 2L, L, 2L/3, L/2, . . . .

(3.3)

We then have only the discrete set of states, oops, I mean modes:

h n

i

un(x, t) = A sin L (x vt) + .

(3.4)

In this situation n = 1 is the fundamental mode of the string (let's say it's a string) and

n = 2, 3, . . . are harmonics. Modern players tune to A at 440 Hz. With v = 330m/s

for the sound wave this corresponds to a wavelength of the A sound wave in air of

about 0.78 m.

24

3.2 Fourier Series

Now we want to represent a given periodic function as a (possibly infinite) sum of

sine and cosine functions. This bears on the analysis of which modes are present in a

given signal. E.g., if the string is struck in a particular way do you excite the "pure

tone" or do you excite some ugly mix of many dierent notes. Such analysis can be

applied to nay function but we will begin by looking at functions defined in ( , ]

and periodic with period 2, i.e., f (x + 2) = f (x) for all real x.

The Fourier series of f is then defined to take the form (3.1). In 1807 Fourier

showed that this could be done for an arbitrary function f , although his paper was

rejected due to lack of rigor, since he did not prove that the series on the right-hand

side converged to the function on the left (see also below). For now we assume the

series converges to f (x). Then, by multiplying both sides by cos(mx) (resp. sin(mx))

and integrating from to we pick out the coe cient am (resp. bm) and find:

1Z

am

=

f (x) cos (mx) dx;

1

Z

m = 0, 1, 2, . . .

bm =

f (x) sin (mx) dx; m = 1, 2, . . .

(3.5) (3.6)

These are the Euler-Fourier formulae. The series that results when the coe cients thereby obtained from a given f are substituted back in to Eq. (3.1) represents the periodic extension of f from ( , ] to ( 1, 1). I.e., even if f is not initially periodic, if you calculate these integrals and stick them in the series definition then the resulting function will be 2-periodic.

Note that if f has a finite discontinuity on ( , ] then we have to split the integral into two bits: from to the discontinuity and from the discontinuity to .

3.2.1 Proof of Fourier coe cient relations and a first go at an orthogonal function expansion

To systematically determine an, we multiply equation (3.1) by cos mx and integrate over [ , ] w.r.t. x. (notation within parenthesis will be presented shortly)

Z

f (x) cos mxdx = (f (x), cos mx) =

a0 2

Z

X 1 Z

cos mxdx + an

n=1

X 1 Z

cos nx cos mxdx + bn

n=1

sin nx cos mxdx

(3.7)

We look first at the second term on the right side:

X 1 Z

an cos nx cos mxdx

n=1

25

if n 6= m

= X 1 an Z cos (n + m)x + cos (n

2

n=1

m)xdx

=

X 1 an

Z

d sin (n + m)x

+

sin (n

m)x

2

n=1

n+m

nm

=

X 1

an

sin

(n

+

m)x

+

sin

(n

m)x

2

n=1

n+m

nm

=0

if n = m

Z

= an=m cos2 nxdx

Z

1 + cos 2nx

= an=m

2

dx

x

sin 2nx

= an=m 2 + 4n

= an=m 2

2

= an=m

.

We will now define a notation to expedite this process

1Z

(f, g) =

f (x)g(x)dx

using this notation to summarize the findings of for the previous section

1

Z

cos nx cos mxdx

=

(cos nx, cos mx)

=

0 1

if n 6= m; if n = m.

Further conclusions can be written with this notation. First, we have: 1Z sin nx cos mxdx = (sin nx, cos mx) = 0

This is true because

LHS =

1

Z [sin (n + m)x + sin (n

2

m)x] dx

if n = m

=

1

Z sin 2nxdx

2

26

(3.8) (3.9) (3.10)

 11

=

cos 2nx

2 2n

=

1 4n

[cos

(2n)

cos ( 2n)] = 0

if n 6= m

1 Z cos (n + m)x cos (n m)x

=

d

dx

2

n+m

nm

1 cos (n + m) cos (n m) cos ( (n + m)) cos ( (n m))

=

+

=0

2 n + m

nm

n+m

nm

Next, we have:

1

Z

sin nx sin mxdx = (sin nx, sin mx) =

0 1

if n 6= m; if n = m.

if n = m if n 6= m

1Z

=

sin2 nxdx = 1

1 Z

=

[cos (n m)x

2

1 Z

sin (n

m)x

=

d

2

nm

cos (n + m)x] dx

sin (n + m)x

n+m

=0

(3.11)

Careful

to

record

factors

of

1

,

Equation

3.7

can

now

be

rewritten

using

this

notation

as:

1 Z f (x) cos mxdx =

a0 2

(1,

cos

mx)

+

X 1 (an

cos

nx,

cosmx)

+

X 1 (bn

sin

nx,

cos

mx)

n=1

n=1

(3.12)

When m 6= 0, the first term = 0. All terms in the last summation and all terms

with the exception of the one where the summation index n hits the value m in the

penultimate summation are zero as can be seen in equations (3.9) and (3.10). We are

left with an equation for am for m 6= 0

1Z

f (x) cos mxdx = am(cos mx, cos mx) = am

(3.13)

27

When m = 0, the first term is equal to a0 [(1,1) = 2]. Because n 1, both summations are zero as they become integrals [ , ] of trigonometric functions with periods which are rational fractions of 2 centered on x = 0. We thus must explicitly define the

coe cient a0 in terms of equation (3.12) with m = 0.

1Z

f (x)dx = a0

(3.14)

Note that it is the desire to maintain this equation specifically due to its attractive

conformity to the pattern of an equations (3.13) that requires us to define the Fourier

series with a coe

cient

of

1 2

for

a0.

We next wish to find the term bm within the sin series. To do this, we will similarly

multiply our function by sin mx and integrate over the symmetric interval [ , ].

1Z f (x) sin mxdx = (f (x), sin mx)

a0 2

(1,

sin

mx)

+

X 1 (an

cos

nx,

sin

mx)

+

X 1 (bn

sin

nx,

sin

mx)

n=1

n=1

(3.15)

From term

our previous discussions, is zero (whether m = 0 or

the not)

mfriodmdltehteerfamctisthkantoRwn

to sin

be always 0. The first mxdx = 0. This leaves

us with the last summation which is 0 for all terms except m = n.

1Z

f (x) sin mxdx = bm(sin mx, sin mx) = bm

(3.16)

To conclude, we will summarize our results from equations 3.13, 3.14, and 3.16. For determining the coe cients necessary to write an arbitrary function in terms of sine and cosine functions in the form of equation 3.1, we have found the equations

am = (f (x), cos mx)

(3.17)

bm = (f (x), sin mx)

(3.18)

a0 = (f (x), 1)

(3.19)

These three equations (3.17 - 3.19) represent a succinct and general method used to decompose a function into its Fourier series.

3.2.2 Example 1

Suppose

f (x) = x [ , ] f (x + 2) = f (x)

28

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download