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8/13/2019 Lecture 01 - Fourier Series to Post
1/19
MATH 5311 Advanced Engineering Math
Fourier Series
Deborah Koslover
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
8/13/2019 Lecture 01 - Fourier Series to Post
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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
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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,
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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 =
8/13/2019 Lecture 01 - Fourier Series to Post
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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.
8/13/2019 Lecture 01 - Fourier Series to Post
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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.
8/13/2019 Lecture 01 - Fourier Series to Post
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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=
Finding Fourier Coefficients
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
= +
=
8/13/2019 Lecture 01 - Fourier Series to Post
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Finding Fourier Coefficients
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=
= + +
Finding Fourier Coefficients
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
=
8/13/2019 Lecture 01 - Fourier Series to Post
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Finding Fourier Coefficients
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=
= + +
Integrate both sides of the equation from to .
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
=
Finding Fourier Coefficients
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=
= + +
Integrate both sides of the equation from to .
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
8/13/2019 Lecture 01 - Fourier Series to Post
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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
=
+ <
8/13/2019 Lecture 01 - Fourier Series to Post
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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).
-
( ) 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
= + = + = =
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).
-
( ) 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
= =
8/13/2019 Lecture 01 - Fourier Series to Post
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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).
-
( ) 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
=
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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)
8/13/2019 Lecture 01 - Fourier Series to Post
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Energy Spectrum for the Flute and Violin
8/13/2019 Lecture 01 - Fourier Series to Post
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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
=
8/13/2019 Lecture 01 - Fourier Series to Post
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Fourier Series for functions of period p= 2L
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
=
Fourier Series for functions of period p= 2L
( ) 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
=
8/13/2019 Lecture 01 - Fourier Series to Post
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1-1
Example
Find the Fourier series for on the interval( )f x x= 1 1x