10.FourierAnalysis

8/7/2019 10.FourierAnalysis

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T

T

2

2

22

21

t

y

vx

y

x

x!

x

xmaF!

Standing Waves Progressive Waves

tL

xnn[

Tcossin

vtxf

Stretched string

Newtons 2ndWave Equation


2/29

nnn

nn tx

Atxy H[

P

T

! cos

2sin,

g

!

!

1

cos2

sin,

n

nn

n

n tx

txy H[

P

T

Anywave motiononthe stringcanbe described bya sum of these modes!

Fourier Analysis

Normal Modes:

n = 1 n = 2 n = 3

JosephFourier, 1768-1830

Oneequation,infinityunknowns

thanksfornothin Joe!


3/29

!

!1

sin

nn L

xnBxy

T

Focus on x first!

How to find a specific Bn?

!

!

L

n

n

L

dxL

xn

L

xndx

L

xnxy

0 1

*

0

*

sinsinsinTTT

? AJUJUJU !coscossinsin

2

1

dx

L

x

L

xL

g

!

!

1

**

coscos2

TT

This reduces series to 1 term!

Mathematically

Multiplyby nth harmonic and integrate.


4/29

L

n

n

L

xnn

nn

L

L

xnn

nn

LB

01

*

*

*

*sinsin

2

g

!

!T

T

T

T

g

!

!1

*

*

*

*00sinsin

2n

n nnnn

Lnn

nn

LBT

TT

T

*nn {

*nn !

0! 0! all terms zero!

uh oh.0! 0!0

xgxf

xg

xf

cxcx d

d

pp

limlim

LHopitals Rule:

Iff(c)=0 andg(c)=0 then

x


5/29

1

cos

2

* T

T

LBn!

2

*Ln!

2

sin*

0

* LBdx

L

xnxy n

L

!

T

dxL

xnxy

LB

L

n

!0

sinT

!

!

1

sinn

nL

xnBxy

T

Any shape between 0 andL


6/29

n=1 n=2 n=3 n=4 n=5 n=6

Positivecontribution

zerocontribution

zerocontribution

zerocontribution

zerocontribution

zerocontribution

zerocontribution

Positivecontribution

zerocontribution

zerocontribution

zerocontribution

zerocontribution

Graphically.

0sinsin0

21 !

L

dxL

xn

L

xn TTorthogonal

n*=1

n*=2


7/29

0 L

dxL

xncx

LB

L

n

!0

sin2 T

Integrate by parts:

? A d!db

a

b

a

b

a

dxxgxfxgxfdxxgxf

xxx

c

00

ccT

T

T

T

c = slope


8/29

LL

nL

xn

n

L

L

xn

n

Lx

L

c

00

ccT

T

T

T

L

nL

xn

n

L

L

xnx

n

c

0

sicos2

T

T

T

T

? A000cos2 TT

nLn

Bn

TT

n

n

LBn cos!

g

!

! sin

nL

xnBxy

T


9/29

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

-3 -2 -1 0 1 2 3

x

y

1st


10/29

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

-3 -2 -1 0 1 2 3

x

y

1st

+ 2nd


11/29

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

-3 -2 -1 0 1 2 3

x

y

1st

+ 2nd

+ 3rd


12/29

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

-3 -2 -1 0 1 2 3

x

y

1st

+ 2nd

+ 3rd

+ 4th


13/29

-1.2

-0.8

-0.4

0

0.4

0.8

1.2

-3 -2 -1 0 1 2 3

x

y

1st

+ 2nd

+ 3rd

+ 4th

+ 5th


14/29


15/29

Barrys profile


16/29

0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 3 5 0 0

5 0

1 0 0

1 5 0

2 0 0

2 5 0

3 0 0

3 5 0

Barrys profile


17/29

0 50 100 150 20 0 250 300 350 400 0

20

40

60

80

10 0

12 0

Barrys 1st harmonic


18/29

0 5 0 1 0 0 1 5 0 2 0 0 2 5 0 3 0 0 3 5 0 4 0 0 0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

Barrys 1st and 2nd harmonic


19/29

0 50 1 00 150 20 0 250 300 350 400 -100

-5 0

0

50

10 0

15 0

20 0

25 0

Barrys 1st, 2nd, and 3rd harmonic


20/29


21/29

clearload f;f=f-172;plot(f)

B(200)=0;y(355)=0;for n = 1:200

for x = 1:355B(n)=B(n)+(2/355)*f(x)*sin(n*pi*x/355);

end

end

figureplot(B,'.')for x = 1:355

for n = 1:200

y(x)=y(x)+B(n)*sin(n*pi*x/355);endendfigureplot(y)


22/29

Frenchs Fourier Foible:

g

!

!

1

si

n

nLxnBxy T dxL

xnxyL

L

n !0

sin T

Good choice if boundaries are at 0: Not so good for other shapes..


23/29

More General:function periodic on intervalx = L to x = L.

g

!

g

!

!

11

0sincos

2 nn

n

n

L

xnB

L

xnA

Axy

TT

dxL

xnxy

LB

L

L

n

!

Tsi

1

dxL

xnxy

LA

L

L

n

!T

cos1

dxxyL

A

L

L

!1

0


24/29

Something hard:

0 L/8 L

H

x

!L

L

L

dxL

HdxL

A8

8

0

0011

80

HA !

!L

L

L

n dxLxn

Ldx

LxnH

LA

8

8

0

cos01cos1 TT

8

0

sin

L

n

L

xn

n

L

L

HA

!T

T

!8

sinT

T

n

n

HAn


25/29

0si

8

0

!

L

ndx

L

xnH

LB

T

8

0

cos

L

nL

xn

n

L

LB

!T

T

!8

c s1 TT

nn

Bn

g

!

g

!

!11

sin8

cos1cos8

sin8 nn L

xnn

n

H

L

xnn

n

HHxy

TT

T

TT

T


26/29

0 L

0

H

MATLAB summation of 1 to 100 terms.


27/29

Time dependence?

Make y(x) an initial condition: y(x,0) = y(x)

The normal modes will oscillate at [n:

g

!

g

!

!11

0 cossincoscos2

,n

nnnn

n

nn txkBtxkAA

txy [[

P

T[

vvk

2||

Given the wavelengths, the frequencies are known.

This is NOT a consequence of Fourier Analysis

2

2

22

2

t

y

vx

y

x

x!

x

xThis IS a consequence of the wave equation:


28/29

0 L

0

H

18,000,000 sinusoids in MATLAB:


29/29

Any well behaved repetitive function can bedescribed as an infinite sum of sinusoids withvariable amplitudes (a Fourier Series). On astretched string these correspond to the

normal modes. Fourier analysis can describearbitrary string shapes as well as progressivewaves and pulses.

10.FourierAnalysis

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