Lecture 4 fourier series

8/13/2019 Lecture 4 fourier series

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1/30/2014

MATH 5311 – Advanced Engineering Math

Fourier Cosine and Sine Transforms

Section 11.8

Fourier Cosine Transform

0

cos f x A xd

0

2cos A f x x dx

Recall the Fourier cosine integral

Substitute A(ω ) into the Fourier cosine integral.

0 0

2cos cos f x f x x dx x d

0 0

2 2cos cos f x x dx x d

ˆ

C f

Fourier CosineTransform

0

2ˆ

cosC C f f x x dx f x

F

Inverse FourierCosine Transform

0

2 ˆ ˆ

cos -1C C C f x f x d f

F

Page534



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Fourier Sine Transform

Fourier SineTransform

0

2ˆ

sinS S f f x x dx f x

F

Inverse FourierSine Transform

0

2 ˆ ˆ

sin -1S S S f x f x d f

F

Page 535

Linearity of Fourier transforms

Theorem – Let a and b be constants and f and g be functions withFourier cosine transforms. Then

C C C af x bg x a f x b g x F F F

Proof – The Fourier cosine transform of is af x bg x

0

2cosC af x bg x af x bg x x dx

F

0

2cos cosaf x x bg x x dx

0 0

2 2cos cosaf x x dx bg x x dx

0 0

2 2cos cosa f x x dx b g x x dx

C C a f x b g x F F



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Fourier transforms of derivatives

Theorem – Let f be a function with a Fourier sine transform and assume

lim 0 x f x

2' 0C s f x f f x

F F Then

Proof

0

2' ' cosC f x f x x dx

F

cos '

sin

u x dv f x dx

du x dx v f x

0

0

2cos sin f x x f x x dx

0

2 20 sin f f x x dx

20 s f f x

F

Fourier transforms of derivatives

2

2

2' 0

'

2'' ' 0

2'' 0

C s

S C

C C

s S

f x f f x

f x f x

f x f f x

f x f f x

F F

F F

F F

F F



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Fourier Transform, Discrete and FastFourier Transforms

Section 11.9

Deborah Koslover

[email protected]

RBN 4010

Complex form of the Fourier Integral

0

cos sin f x A x B x d 1

sin B f d

1cos A f d

Substitute A and B into the Fourier integral.

Simplify. Note – The sine and cosine can be brought into the integralsbecause they have different variables.

0

1 1cos cos sin sin f x f dx f d x d

Combine inner integrals into one integral.

0

1cos cos sin sin f x f x x d d

( )

( ) x

mailto:[email protected]

mailto:[email protected]



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0

1cos cos sin sin f x f x x d d

Use the trigonometric identity cos cos sin sin cos A B A B B A

0

1cos f x f x d d

Notice that cos f x d g is a function of ω .

and further that it is an even function of ω .

cos cos

cos

g f x d f x d

f x d g

So

0

1 12

f x g d g d

1cos

2 f x d d


1cos

2 f x f x d d

sinh f x d The new function is an odd function of ω .

sin sin

sin

h f x d f x d

f x d h

So 0 sin

2 2i i

h d f x d d

Therefore,

1cos 0

2 f x f x d d

1cos sin

2 2i

f x f x d d f x d d



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1cos sin

2 2i

f x f x d d f x d d

Simplify.

1cos sin

2 f x f x d d f i x d d

Combine into one integral.

1cos sin

2 f x f x i x d d

Using Euler’s equation, cos sin ii e , the integral become

1

2i x

f x f e d d

This is the complex Fourier integral .

Fourier Transform

12

i x f x f e d d

Beginning with the Fourier integral

Breakup the constant and i xe

1 12 2

i i x f x f e d e d

Call ˆ

f

Fourier transform 1ˆ

2i f f e d f x

F

Inverse Fouriertransform

1 ˆ ˆ

2i x -1 f x f e d f

F



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Fourier Transform This function, f (t ), isalmost a sine wave.

Its Fourier transform,peaks at 1000 Hz.

ˆ

f

Function Its Fourier transform

Spectral density and total energy

The Fourier transform, , is also called the spectral density or spectrumbecause it measures the intensity of f (t ) in the frequency interval betweenω and ω + Δ ω .

ˆ

f

Δω

The total signal energy is given by and 2ˆ

E f d 2ˆ

f d

represents the signal energy in the band between ω and ω + Δ ω .



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Linearity of the Fourier transform

The Fourier transform is linear.

2 23

5 / 202 2

1 1and

3 2 3 10 xe

e e x

F F

Example – Suppose one wishes to find the Fourier transform of25

2 2

210

3 xe

x

and one notes on page 536 of the textbook that

af x bg x a f x b g x F F F

Then

2 25 5

2 2 2 2

2 1

10 2 103 3 x x

e e x x

F F F

2 23 3

/ 20 / 2012 10

2 3 310e e

e e

Fourier transform of derivatives

Let f ( x) be continuous and f ( x) → 0 as | x| → . Further let ' f x dx

Then 2

'

''

f x i f x

f x f x

F F

F F

To apply, let’s take a detour and define a Dirac delta function .

It is not a real function, but a generalization of a function, called ageneralized function.Informally, the Dirac delta function, δ ( x) is a function which is zeroeverywhere except at x = 0 where it is .

The function, δ ( x-a ) is zero everywhere except at x = a where it is .

Additionally, it has the properties that

1 and x a dx f x x a dx f a



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Example Solve the following differential equation using Fourier transforms.

'' 9 cos5 f x f x x

'' 9 cos5 f x f x x F F Fourier transform-both sides

'' 9 cos5 f x f x x F F F Linearity

2 9 cos5 f x f x x - F F F Fourier of 2 nd derivative

2 9 5 52

f x f x x x

- F F Fourier of cosine

2 9 5 52

f x f x x x

- F F

2 9 5 52

f x x x

- F Factor

2 2

5 5

2 9 9

x x f x

F - -

Divide

-12 2

5 5

2 9 9

x x f x

F

- - Inverse Fourier

Example '' 9 cos5 f x f x x



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Example

-1

2 2

5 5

2 9 9

x x f x

F - -

2 2

5 51 12 29 9

i x i x x x f x e d e d

- -

2 2

5 512 9 92

i x x x f x e d

- -

Definition

Simplify

5 51 1 1 1

2 34 2 34

i x i x f x e e - -

5 51 1 1

34 2 2

i x i xe e -

1cos5

34 x

-

'' 9 cos5 f x f x x

1 1cos

2 2 2

i ii i e e

e e

Convolution

The convolution of functions f and g is defined by f g

f g f p g x p dp f x p g p dp

The Convolution Theorem – Suppose f ( x) and g ( x) are piecewisecontinuous, bounded and absolutely integrable on the x-axis. Then

2 f g f g F F F



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If one of the functions has one peakat x 0, the convolution will shift theother function by x 0 and then blurthe outline by an amount thatdepends on the width of the peak.The amplitude will also be affected

by the value of the peak.

Convolution

Signal Processing Input-Output



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Discrete Fourier Transform

2π

( x0, f ( x0))

( x1, f ( x1))N equally spaced points

Find a function of the form 1

0

N inx

nn

f x c e that fits all the points

1

00 0

0

N inx

nn

f f x c e

1

11 10

N

inxnn

f f x c e

1

( 1)1 1

0

N inx N

N N nn

f f x c e

N equations and

N unknowns 0 1 2 1, , , , N c c c c

Solve for cn 1

( )

0

1 N inx k

n k k

c f e N


1( )

0

1 N inx k

n k k

c f e N

Define1

( )

0

ˆ N

inx k n n k

k

f Nc f e

0 1n N

Since the points are equally spaced2

k x k N

21

0

ˆi N nk

N n k

k

f f e

Let2 i N w e

1

0

ˆ N

nk n k

k

f f w

1

0

N inx

nn

f x c eCompare to



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Discrete Fourier Transform 2 i N w e

1

0

ˆ N

nk n k

k

f f w 0 1n N

To reconstruct the original signal, we find the inverse of F N

0

1

1

ˆ

ˆ

ˆ

ˆ

N

f

f f

f

0

1

1 N

f

f f

f

0 0 0 0

0 1 2 1

0 2 4 2( 1)

0 1 1 2 1 1 1

N

N N

N N N N

w w w w

w w w w F w w w w

w w w w

andˆ

N f F f

We can write this as a matrix equation by letting

1ˆ

N f F f


Problems with this technique.

To be meaningful, one needs many sample points. Imagine a 1000 by1000 matrix.

Unwieldy, computationally intensive. N 2 operations

Need a less computational intensive technique.



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Fast Fourier Transform Presented in a paper by J.W. Cooley and J.Tukey, An algorithm for machine calculation of

complex Fourier Series, 1965.

J. Tukey1915-2000

J. W. Cooleyb. 1926

Uses the results of the Discrete FourierTransform, but in a divide and conquer way.One obtains the same results but with N log N operations

To use this method, one must start with 2 points. p

The vector is broken into an even piece and an odd piece.n f

,0 ,2 ,4 , 2 / 2

,1 ,3 ,5 , 1 / 2

ˆ ˆ ˆ ˆ ˆ

, , , ,ˆ ˆ ˆ ˆ ˆ

, , , ,

even ev ev ev ev N N even

odd od od od od N N odd

f f f f f F f

f f f f f F f

,0 ,2 ,4 , 2 ,1 ,3 ,5 , 1, , , , , , , ,even ev ev ev ev N odd od od od od N f f f f f f f f f f

The Fourier transform is then written.

Fast Fourier Transform

,0 ,2 ,4 , 2 ,1 ,3 ,5 , 1, , , , , , , ,even ev ev ev ev N odd od od od od N f f f f f f f f f f

,0 ,2 ,4 , 2 / 2

,1 ,3 ,5 , 1 / 2

ˆ ˆ ˆ ˆ ˆ

, , , ,ˆ ˆ ˆ ˆ ˆ

, , , ,

even ev ev ev ev N N even

odd od od od od N N odd

f f f f f F f

f f f f f F f

One then obtains using the formulasˆ

f

, ,

, ,

ˆ ˆ ˆ

1, 2, , / 2 1ˆ ˆ ˆ

1, 2, , / 2 1

nn ev n od n

nn M ev n od n

f f f n N

f f f n N



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Fast Fourier Transform One can continue cutting the vector in half until one is left with

just 2 by 2 matrices.

This process greatly reduces the number of computations

needed.


Partial Differential EquationsBasic Concepts

Section 12.1



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Partial Derivatives

Definition - Consider a function f of three variables, f ( x, y, z )If y and z are held constant and only x is allowed to vary, the partial

derivative of f with respect to x is denoted by or and is defined

by the limit

f x

x f

0

, , , , , ,limh

f x y z f x h y z f x y z

x h

Similar definitions can be give for the partial derivatives of f withrespect to y and to z .

Example : Given 2 2, , 2 cos( ) f x y z xy x yz xyz xyz

2 2, ,2 cos( )

f x y z xy x yz xyz xyz

x x x x

2 22 cos( ) y x yz x yz x xyz x x x

22 2 cos( ) cos( ) y yz x yz x xyz xyz x x x

22 2 sin( ) cos( ) y xyz yz x xyz xyz xyz x

Partial Derivatives

22 2 sin( ) cos( ) y xyz yz xyz xyz xyz

2 3 22 2 sin( ) 2 cos( ) y xyz xy z xyz yz xyz



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Partial Differential Equations

Definition – A partial differential equation (PDE) is an equation containing

partial derivatives of the dependent variable.

Example – In each of the following equations, u is the dependent variableand x, y and t are independent variables.

2 2

2 2

2 2 22

2 2

22

2

, ,0

, ,,

, , ,, 2 3 4

, , ,0

x

u x t u x t c

t xu x y u x y

f x y x y

u x y u x y u x y x y x e x x y y

u x y u x y u x y

x x y


We want to solve a PDE in a certain domain.

Definition – A domain is an open, connected set of points (usually in ( x,y),( x,y,z ) or ( x,y,z,t ))

Definition – An open sent of points is one that does not include itsboundary.

Open set Not 0pen set

(Closed set)Not 0pen set



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Partial Differential Equations Definition – A domain is an open, connected set of points (usually in ( x,y),( x,y,z ) or ( x,y,z,t ))

Definition – A connected set is one where any two points can be joined bya path without leaving the set.

Connected set Not connected set


Definition – A domain is an open, connected set of points (usually in ( x,y),( x,y,z ) or ( x,y,z,t ))

Domain

Not domain, not connected

Not domain, not open

Not domain, not connectedand not open

Lecture 4 fourier series

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