PDE Refresher II : Fourier Transform Indraneel Kasmalkar (Neel), - - PowerPoint PPT Presentation

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PDE Refresher II : Fourier Transform Indraneel Kasmalkar (Neel), - - PowerPoint PPT Presentation

Jan 13, 2023 358 likes •1.84k views

PDE Refresher II : Fourier Transform Indraneel Kasmalkar (Neel), ICME (ineel@stanford.edu) IMPORTANT IMPORTANT Refresher classes are NOT comprehensive. The best way to refresh you memory is to solve problems . Problem sets available

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PDE Refresher II: Fourier Transform

Indraneel Kasmalkar (Neel), ICME

(ineel@stanford.edu)

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IMPORTANT

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  • Refresher classes are NOT comprehensive.
  • The best way to refresh you memory is to solve problems.
  • Problem sets available on the refresher course website.
  • Most of them are short, easy questions.
  • Need feedback to structure third PDE class:
  • http://freesuggestionbox.com/pub/oztvrxm
  • What you feel good about, what makes you uncomfortable, any

topics you hate, or do not understand. The first few phrases that pop up in your mind when you think PDEs.

IMPORTANT

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SLIDE 4
  • Refresher classes are NOT comprehensive.
  • The best way to refresh you memory is to solve problems.
  • Problem sets available on the refresher course website.
  • Most of them are short, easy questions.
  • Need feedback to structure third PDE class:
  • http://freesuggestionbox.com/pub/oztvrxm
  • What you feel good about, what makes you uncomfortable, any

topics you hate, or do not understand. The first few phrases that pop up in your mind when you think PDEs.

IMPORTANT

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SLIDE 5
  • Refresher classes are NOT comprehensive.
  • The best way to refresh you memory is to solve problems.
  • Problem sets available on the refresher course website.
  • Most of them are short, easy questions.
  • Need feedback to structure third PDE class:
  • http://freesuggestionbox.com/pub/oztvrxm
  • What you feel good about, what makes you uncomfortable, any

topics you hate, or do not understand. The first few phrases that pop up in your mind when you think PDEs.

IMPORTANT

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SLIDE 6
  • Refresher classes are NOT comprehensive.
  • The best way to refresh you memory is to solve problems.
  • Problem sets available on the refresher course website.
  • Most of them are short, easy questions.
  • Need feedback to structure third PDE class:
  • http://freesuggestionbox.com/pub/oztvrxm
  • What you feel good about, what makes you uncomfortable, any

topics you hate, or do not understand. The first few phrases that pop up in your mind when you think PDEs.

IMPORTANT

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SLIDE 7
  • Refresher classes are NOT comprehensive.
  • The best way to refresh you memory is to solve problems.
  • Problem sets available on the refresher course website.
  • Most of them are short, easy questions.
  • Need feedback to structure third PDE class:
  • http://freesuggestionbox.com/pub/oztvrxm
  • What you feel good about, what makes you uncomfortable, any

topics you hate, or do not understand. The first few phrases that pop up in your mind when you think PDEs.

IMPORTANT

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What is a Fourier Transform?

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It is a different perspective of looking at functions

What is a Fourier Transform?

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The Traditional Input-Output Perspective

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The Traditional Input-Output Perspective

x f(x) 0.1 0.01 2 4 r r2

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Fourier Perspective: As a sum of special functions

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Fourier Perspective: As a sum of special functions

g(x) = Aeix + Be-2ix + Ce2.7ix

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Fourier Perspective: As a sum of special functions

g(x) = Aeix + Be-2ix + Ce2.7ix

The coefficients are more important than the output value.

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Fourier Perspective: As a sum of special functions

g(x) = Aeix + Be-2ix + Ce2.7ix

The coefficients are more important than the output value.

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What are these special functions?

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What are these special functions?

ζ ranging over all real numbers.

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Why are they special?

ζ ranging over all real numbers.

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Why are they special?

ζ ranging over all real numbers.

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Why are they special?

ζ ranging over all real numbers.

DIFFERENTIATION IS SIMPLER!

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Why are they special?

ζ ranging over all real numbers.

g(x) = Aeix + Be-2ix + Ce2.7ix

DIFFERENTIATION IS SIMPLER!

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Why are they special?

ζ ranging over all real numbers.

g(x) = Aeix + Be-2ix + Ce2.7ix g'(x) = iAeix + -2iBe-2ix + 2.7iCe2.7ix

DIFFERENTIATION IS SIMPLER!

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Why are they special?

ζ ranging over all real numbers.

g(x) = Aeix + Be-2ix + Ce2.7ix g'(x) = iAeix + -2iBe-2ix + 2.7iCe2.7ix So easy to get new coefficients!!

DIFFERENTIATION IS SIMPLER!

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Infinite sum of special functions

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Infinite sum of special functions

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Infinite sum of special functions

More formally, as an integral:

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Infinite sum of special functions

More formally, as an integral:

The coefficients A(ζ) can be thought of as a function. This function is the Fourier transform of ƒ:

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Infinite sum of special functions

More formally, as an integral:

The coefficients A(ζ) can be thought of as a function. This function is the Fourier transform of ƒ:

  • r
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Formal definition of Fourier Transform

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Formal definition of Fourier Transform

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The coefficients Multiplying by inverse special function to (sort of) isolate the coefficient

Formal definition of Fourier Transform

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With the above definition, hopefully you get back ƒ as sum of Fourier coefficients times special functions.

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With the above definition, hopefully you get back ƒ as sum of Fourier coefficients times special functions.

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Fourier Transform

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Fourier Transform Inverse Transform

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Fourier Transform Inverse Transform

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Other Forms

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Other Forms

If ƒ is a discrete sequence:

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Other Forms

If ƒ is a discrete sequence: Discrete Fourier Transform

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Other Forms

If ƒ is a discrete sequence: Discrete Fourier Transform

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Other Forms

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Other Forms

If ƒ is 2π-periodic function:

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Other Forms

If ƒ is 2π-periodic function: Fourier Series

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Other Forms

If ƒ is 2π-periodic function: Fourier Series

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Other Forms

If ƒ is 2π-periodic function: Fourier Series !!

Only true for smooth functions.

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Differentiation

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Differentiation

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Differentiation

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Proof:

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Proof:

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Proof:

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Proof:

...Integration by parts

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Proof:

...Integration by parts

... ƒ decays

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Proof:

...Integration by parts

... ƒ decays

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But this should not be surprising

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But this should not be surprising

g(x) = Aeix + Be-2ix + Ce2.7ix g'(x) = iAeix + -2iBe-2ix + 2.7iCe2.7ix

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But this should not be surprising

g(x) = Aeix + Be-2ix + Ce2.7ix g'(x) = iAeix + -2iBe-2ix + 2.7iCe2.7ix

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But this should not be surprising

g(x) = Aeix + Be-2ix + Ce2.7ix g'(x) = iAeix + -2iBe-2ix + 2.7iCe2.7ix

Caution: Just an analogy. Fourier Transforms are integrals, not discrete sums.

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Shifting

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Shifting

g(x) = ƒ(x+a)

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Shifting

g(x) = ƒ(x+a)

g

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Proof Intuition: use the analogy

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Proof Intuition: use the analogy

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Proof Intuition: use the analogy

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Proof Intuition: use the analogy

New factor in coefficient

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Shifting

g(x) = ƒ(x+a)

g

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Example Fourier Transforms

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Example Fourier Transforms

0 function

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Example Fourier Transforms

0 function 0 function

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

How would I have even guessed this??

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

This is one of those special functions!

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Example Fourier Transforms

This is one of those special functions!

?

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Fourier Transform is coefficients of special functions.

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Fourier Transform is coefficients of special functions. How would you draw the Fourier Transform?

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CONFUSION ZONE

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FOURIER TRANSFORM CANDIDATE A

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FOURIER TRANSFORM CANDIDATE A

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FOURIER TRANSFORM CANDIDATE A

is equal to 1 for ζ = 3.2, and 0 for everything else.

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FOURIER TRANSFORM CANDIDATE A

is equal to 1 for ζ = 3.2, and 0 for everything else. 1 ζ = 3.2

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Does this work with our summation intuition?

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Does this work with our summation intuition?

is equal to 1 for ζ = 3.2, and 0 for everything else.

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Does this work with our summation intuition?

is equal to 1 for ζ = 3.2, and 0 for everything else.

f is the 'infinite sum' of coefficients times special functions

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Does this work with our summation intuition?

is equal to 1 for ζ = 3.2, and 0 for everything else.

f is the 'infinite sum' of coefficients times special functions

Integrand equals zero almost everywhere!

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Does this work with our summation intuition?

is equal to 1 for ζ = 3.2, and 0 for everything else.

f is the 'infinite sum' of coefficients times special functions

Integrand equals zero almost everywhere! After integral, ƒ(x) = 0?

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FOURIER TRANSFORM CANDIDATE A

is equal to 1 for ζ = 3.2, and 0 for everything else. 1 ζ = 3.2

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FOURIER TRANSFORM CANDIDATE A

is equal to 1 for ζ = 3.2, and 0 for everything else. 1 ζ = 3.2

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FOURIER TRANSFORM CANDIDATE B

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FOURIER TRANSFORM CANDIDATE B

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FOURIER TRANSFORM CANDIDATE B

Delta "function".

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FOURIER TRANSFORM CANDIDATE B

Delta "function". Not really a function.

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FOURIER TRANSFORM CANDIDATE B

Purely defined by its effect inside an integral.

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Does this work with our summation intuition?

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Does this work with our summation intuition?

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Does this work with our summation intuition?

f is the 'infinite sum' of coefficients times special functions

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Does this work with our summation intuition?

f is the 'infinite sum' of coefficients times special functions

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Does this work with our summation intuition?

f is the 'infinite sum' of coefficients times special functions

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Does this work with our summation intuition?

f is the 'infinite sum' of coefficients times special functions

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Does this work with our summation intuition?

f is the 'infinite sum' of coefficients times special functions

It Works!

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Differences

CANDIDATE A CANDIDATE B

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Differences

CANDIDATE A CANDIDATE B

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Differences

CANDIDATE A CANDIDATE B This is a function.

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Differences

CANDIDATE A CANDIDATE B This is a function. This is a distribution.

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Differences

CANDIDATE A CANDIDATE B This is a function. This is a distribution. Each point has equal weight, so integral is zero. Cannot capture the behavior we want.

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Differences

CANDIDATE A CANDIDATE B This is a function. This is a distribution. Each point has equal weight, so integral is zero. Cannot capture the behavior we want. All the weight concentrated at point 3.2. Integral behaves exactly how we want.

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Talk to each other

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Example Fourier Transforms

This is one of those special functions!

?

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Example Fourier Transforms

This is one of those special functions!

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Example Fourier Transforms

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Why did we do all this?

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Solve over the whole real line:

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Solve over the whole real line: FOURIER TRANSFORM!

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Solve over the whole real line: FOURIER TRANSFORM!

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Solve over the whole real line: FOURIER TRANSFORM!

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Solve over the whole real line: FOURIER TRANSFORM!

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Solve over the whole real line: FOURIER TRANSFORM!

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Solve over the whole real line: FOURIER TRANSFORM! Differentiation was converted to polynomial algebra!

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Solve over the whole real line:

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Solve over the whole real line: INVERSE FOURIER TRANSFORM!

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Solve over the whole real line: INVERSE FOURIER TRANSFORM!

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Solve over the whole real line: INVERSE FOURIER TRANSFORM!

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Solve over the whole real line: INVERSE FOURIER TRANSFORM!

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Solve over the whole real line: INVERSE FOURIER TRANSFORM!

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Solve over the whole real line: INVERSE FOURIER TRANSFORM!

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Questions?

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Questions?

Why did we not need boundary conditions?

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Solve over the whole real line:

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Solve over the whole real line: Fourier Transform has given you ONE solution. This solution is the one which does not grow exponentially.

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Thank you