lecture 01 - fourier series to post

8/13/2019 Lecture 01 - Fourier Series to Post

1/19

MATH 5311 Advanced Engineering Math

Fourier Series

Deborah Koslover

[email protected]

RBN 4010

Section 11.1

Heat distribution in a metal plate,using Fourier's method

Joseph Fourier

1768-1830

Joseph Fourier French mathematician, physicist and Egyptologist

December 21, 1807 Presented paper to theParis Institute, On the Propagation of Heat in

Solid Bodies

Wrote periodic functions as infinite series of trigonometric functions, what

we now call a Fourier series.

Joseph Fourier


2/19

Applications of Fourier Analysis Vibration and Wave Analysis

Electrostatics Problems

Signal and Image Processing

Data Compression

Deducing chemical composition of stars

Optimizing the design of

telecommunications systems

Heat Flow

Optics

Acoustics

Antenna Design and Analysis

Electronic Filter Design and Analysis

X-ray Crystallography Analysis


3/19

Copyright 1997 American Physiological Society

Heart rate power spectrum from a

control subject (A) and a patient

suffering from diabetic neuropathy (B).

Frequency peaks are clearly blunted inthe latter (note different power scaling).A noninvasive, sensitive method for the

early diagnosis of autonomic

neuropathy in diabetes mellitus,


4/19

Periodic FunctionsDefinition: A periodic functionis a function that repeats its values onregularly spaced intervals.

Definition: The interval on which a function repeats is called the period.

Period

Mathematically, we say a function has period p if ( ) ( )f x p f x+ =

x

f (x)

x + p

Notice also that ( ) ( )2 ,f x p f x+ =

x + 2p

( )( ) ( )1 ,f x p f x+ =

x - p

and in fact

( ) ( ) ,f x np f x+ = where n is any nonzero integer.

So a function with period p also has a period of any multiple of p.

Periodic Functions

-6 -4 -2 2 4 6

-1

-0.5

0.5

1

Example: ( ) ( )cos 2f x x=

p =

and p = 2

and p = 3

Example: ( ) ( ) ( ) ( ) ( ) ( )cos , sin , cos 2 , sin 2 , cos 3 , sin 3 ,x x x x x x

all have period 2 .p =


5/19

Periodic Functions

Constant functions are periodic over arbitrary interval lengths.

-6 -4 -2 2 4 6

0.5

1

1.5

2Example:

( ) 1f x = has period 1, 12, 3.27, 2, or any other value you would liketo choose.


6/19

Functions with Period 2

Generally, functions with period 2 can be written in the form of aninfinite sum, as

( ) [ ]01

cos sinn nn

f x a a nx b nx=

= + +

This is called a Fourier series. The as and bs are called Fouriercoefficients.

To be able to use Fourier series to solve problems, we must find the

Fourier coefficients that go with any particular function.


7/19

Finding Fourier Coefficients

We will need the following trigonometric identities.

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 12 2

1 12 2

1 12 2

cos cos cos cos

sin sin cos cos for , positive integers

sin cos sin sin

nx mx n m x n m x

nx mx n m x n m x n m

nx mx n m x n m x

= + +

= +

= + +

Integrate each identity from to .

( ) ( )

( ) ( )

1 1sin sin

2 2n m x n m x

n mm

nn

m

= + +

+

( ) ( ) ( ) ( )1 12 2cos cos cos cosnx mx dx n m x dx n m x dx

= + +

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1sin sin

2

1sin sin

2

n m n mn m

n m n mn m

= + + +

+

0=


So if n m then ( ) ( )cos cos 0nx mx dx

=

What if n= m?

( ) ( ) ( ) ( )1 12 2cos cos cos cosnx nx dx n n x dx n n x dx

= + +

( ) ( ) [ ]2 1 12 2cos cos 2 cos 0nx dx nx dx dx

= +

( )1 12 2cos 2nx dx dx

= +

( ) 121 sin 24

nx xn

= +

( ) ( )( ) ( )121

sin 2 sin 24

n nn

= +

=


8/19


Summarizing,

( ) ( )cos cos 0nx mx dx

=

If n= m, then ( ) ( )cos cosnx nx dx

=Similarly we can show,

if n m, then ( ) ( )sin sin 0nx mx dx

=

if n= m, then ( ) ( )sin sinnx nx dx

=

if n m, then

( ) ( )sin cos 0nx mx dx

=

if n= m, then ( ) ( )sin cos 0nx nx dx

=

if n m, then

Using these results, we willbe able to find the Fourier

coefficients.

( ) [ ]01

cos sinn nn

f x a a nx b nx=

= + +


Given a functionf, which has a period of 2, we want to find the Fourier

coefficients so that

Integrate both sides of the equation from to .

( ) 01

cos sinn nn

f x dx a dx a nx dx b nx dx

=

= + +

Assuming term-wise

integration is allowed.

0

1

sin cosn n

n

a ba x nx nx

n n

=

= +

( )( ) ( )( ) ( )( )01

sin sin cos cosn n

n

a ba n n n n

n n

=

= +

( ) ( )01

2 0 0 cos cosn n

n

a ba n n

n n

=

= +

02a =

Dividing both sides by 2( ) 0

1

2f x dx a

=


9/19


Now find the other coefficients. Multiply both sides by cos (mx).

( ) [ ]01

cos cos cos cos sin cosn nn

f x mx a mx a nx mx b nx mx=

= + +( ) 01

cos cos cos cos sin cosn nn

f x mx dx a mx dx a nx mx dx b nx mx dx

=

= + +

( ) [ ]0

1

cos sinn nn

f x a a nx b nx=

= + +


Recall ( ) ( )sin cos 0nx mx dx

=

( ) cos mf x mx dx a

=

Recall ( ) ( )cos cos 0nx mx dx

=

If n= m, then ( ) ( )cos cosnx nx dx

=

if n m, then

Dividing both sides by , we get

( )1

cos mf x mx dx a

=


Multiply both sides by sin (mx).

( ) [ ]01

sin sin cos sin sin sinn nn

f x mx a mx a nx mx b nx mx=

= + +( ) 01

sin sin cos sin sin sinn nn

f x mx dx a mx dx a nx mx dx b nx mx dx

=

= + +

( ) [ ]0

1

cos sinn nn

f x a a nx b nx=

= + +


Recall ( ) ( )sin cos 0nx mx dx

=

( )sin mf x mx dx b

=

Recall

Dividing both sides by , we get

( )1

sin mf x mx dx b

=

Now do the same thing with sine.

( ) ( )sin sin 0nx mx dx

=

if n= m, then ( ) ( )sin sinnx nx dx

=

if n m, then


10/19

Euler Equations

( )0 12

a f x dx

=

( )1

cosma f x mx dx

=

( )1

sinmb f x mx dx

=

Example: Find the Fourier series for the givenf (x), which is assumed tohave the period 2. Graph the partial sums up to that including cos (5x)

and sin (5x).

-

3 42-3-4 -2 -

First we start by finding a formula for the function.

Between and0, the function isf (x) = 0.

Between 0 and, the function is a line. It has slope m = -1

andy-intercept b = . Thus, f (x) = -x +

Or ( ) 0 if 0

if 0

xf x

x x

=

+ <


11/19


and sin (5x).

-

( ) 0 if 0

if 0

xf x

x x

=

+ <

Now apply the Euler equations.

( )01

2a f x dx

= ( ) ( )0

0

1 1

2 2f x dx f x dx

= + 0

0

1 10

2 2dx x dx

= + + 2 2

2

0

0

1 1

2 2 2 2 4

xx a

= + = + = =


and sin (5x).

-

( ) 0 if 0

if 0

xf x

x x

=

+ <

04

a =

( )1

cosma f x mx dx

= ( ) ( )0

0

1 1cos cosf x mx dx f x mx dx

= +

( )0

1cosx mx dx

= + 1cos

sinm

u x dv mx dx

du dx v mx

= + =

= =

Integrating by parts

0

0

1 1sin sin

xmx mx dx

m m

+= +

0

1sin mxdx

m

=

( )

1 if is evencos

-1 if is odd

1 if is even

-1 if is odd1

m

mm

m

m

m

=

=

Notice( )

2 20

1 1cos cos cos0mx m

m m

= =

( )( )21

1 1m

mam

= =


12/19


and sin (5x).

-

( ) 0 if 0

if 0

xf x

x x

=

+ <

04

a = ( )( )2

11 1

m

mam

=

Similarly, integrating by parts, we can show that1

mb

m=

So, the Fourier series forf (x) is

( ) [ ]

( )( )

0

1

21

cos sin

1 11 1 cos sin

4

m m

m

m

m

f x a a mx b mx

mx mxm m

=

=

= + +

= + +

2 1 2 1cos sin sin 2 cos3 sin3

4 2 9 3

1 2 1sin 4 cos5 sin5 ...

4 25 5

x x x x x

x x x

= + + + + +

+ + +

-3 -2 -1 1 2 3

0.5

1

1.5

2

2.5

-3 -2 -1 1 2 3

0.5

1

1.5

2

2.5

-3 -2 -1 1 2 3

0.5

1

1.5

2

2.5

-3 -2 -1 1 2 3

0.5

1

1.5

2

2.5

-3 -2 -1 1 2 3

0.5

1

1.5

2

-3 -2 -1 1 2 3

0.25

0.5

0.75

1

1.25

1.5

Partial Sums

-

( ) 2 1 2 1cos sin sin 2 cos3 sin 34 2 9 3

1 2 1sin 4 cos5 sin5 ...

4 25 5

f x x x x x x

x x x

= + + + + +

+ + +

5

2 1 2cos sin sin 2 cos 3

4 2 9

1 1 2 1sin 3 sin 4 cos 5 sin 5

3 4 25 5

p x x x x

x x x x

= + + + +

+ + + +

4

2 1cos sin sin 2

4 2

2 1 1cos 3 sin 3 sin 4

9 3 4

p x x x

x x x

= + + +

+ + +

3

2 1cos sin sin 2

4 2

2 1cos 3 sin 3

9 3

p x x x

x x

= + + +

+ +

2

2 1cos sin sin 2

4 2p x x x

= + + +1

2cos sin

4p x x

= + +0

4p

=


13/19

http://math.furman.edu/~dcs/java/square.html

Convergence of Fourier Series

Theorem: Letf (x) be periodic with period 2 and piecewise continuous

on the interval - x .

A function is piecewise continuous if it is only discontinuous at a finite

number of points.

Further, letf (x) have a left hand derivative and

a right hand derivative at every point of that interval.

( ) ( ) ( ) ( )0 0

lim and lim existh h

f x h f x f x h f x

h h+

+ +

Then the Fourier

is discontinuous. There the sum is the average of the left and right

hand limit off (x) atx0.

-

2-2 -

f (x)

-

2-2 -

[ ]01

cos sinn nn

a a nx b nx=

+ +

series off (x) converges. Its sum isf (x) except at pointsx0 wheref (x)


14/19

Energy Spectrum for the Flute and Violin


15/19

MATH 5311 Advanced Engineering Math

Functions of Any Period p = 2L

Section 11.2

Fourier Series for functions of period p= 2L

3 4 2 -3 -4 -2 -

How would we find the formula for the Fourier series for this function?

Rescale!

L 3L 4L2L-3L-4L -2L -L

Find comparable positions of each graph.

y

x

Use ratios. 2

2

L y

x= or x y

L

=


16/19


L 2Ly 2 x

x yL

=

Old system New system

( ) [ ]01

cos sinn nn

f x a a nx b nx=

= + +

( )01

2a f x dx

=

( )1

cosna f x nx dx

=

( )1

sinnb f x nx dx

=

dx dyL

=

0

1

cos sinn nn

n na a y b y

L L

=

= + +

f yL

( )f y

0

1

2

L

L

a f y dyL L

=

1cos

L

n

L

a f y n y dy

L L L

=

1

sin

L

n

L

b f y n y dyL L L

=

( )1

2

L

L

f y dyL

=

( )1

cos

L

L

nf y y dy

L L

=

( )1

sin

L

L

nf y y dy

L L

=


( ) 01

cos sinn nn

n nf x a a x b x

L L

=

= + +

( )01

2

L

L

a f x dxL

=

( )1

cos

L

n

L

na f x x dx

L L

=

( )

1sin

L

nL

nb f x x dx

L L

=


17/19

1-1

Example

Find the Fourier series for on the interval( )f x x= 1 1x

lecture 01 - fourier series to post

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