The Fourier Transform CS/BIOEN 4640: Image Processing Basics March - - PowerPoint PPT Presentation

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The Fourier Transform CS/BIOEN 4640: Image Processing Basics March - - PowerPoint PPT Presentation

The Fourier Transform CS/BIOEN 4640: Image Processing Basics March 20, 2012 Why Study Spectral Methods? Is often more efficient method for filtering than convolution Helps us understand image sampling, filtering, and aliasing Used


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SLIDE 1

The Fourier Transform

CS/BIOEN 4640: Image Processing Basics March 20, 2012

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SLIDE 2

Why Study Spectral Methods?

◮ Is often more efficient method for filtering than

convolution

◮ Helps us understand image sampling, filtering, and

aliasing

◮ Used in image compression ◮ Important for solving PDEs in image processing ◮ Fourier transform used in MRI

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SLIDE 3

Harmonic Analysis

◮ Decompose a function into

basic waves called “harmonics”

◮ Any signal can be written as a

summation of harmonics

◮ Think of sound waves and

music: harmonics are pure tones

1 1/2 1/3 1/4 1/5 1/6 1/7

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SLIDE 4

Sine and Cosine Functions

Harmonics are given by sine and cosine functions

1 2 3 4 5 6 7 −1.0 −0.5 0.0 0.5 1.0 sin cos

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SLIDE 5

Wave Properties: Frequency

1 2 3 4 5 6 −1.0 −0.5 0.0 0.5 1.0 sin(t) sin(2t) sin(3t)

Frequency is how many times a wave repeats.

cos(ωx)

and

sin(ωx) ω is the frequency

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SLIDE 6

Wave Properties: Amplitude

1 2 3 4 5 6 −3 −2 −1 1 2 3 sin(t) 2 sin(t) 3 sin(t)

Amplitude is the height of the wave.

a·cos(x)

and

a·sin(x) a is the amplitude

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SLIDE 7

Wave Properties: Phase

1 2 3 4 5 6 −1.0 −0.5 0.0 0.5 1.0 sin(t) sin(t − pi/4) sin(t − pi/2)

Phase is the horizon- tal shift of the wave.

cos(x−φ) and sin(x−φ) φ is the phase shift

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SLIDE 8

Who’s This Guy?

Jean Baptiste Joseph de Fourier (1768 - 1830)

While studying heat conduc- tion, Fourier discovered that functions could be decom- posed into summations

  • f

cosine waves with different amplitudes and frequencies. Trivia fact: Fourier also first described the greenhouse ef- fect!

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SLIDE 9

Fourier Series

Definition

Consider a function g(x) that is periodic on [0, 2π

ω0]

It’s Fourier series is given as

g(x) =

  • k=0

[ Ak cos(kω0x) + Bk sin(kω0x)] ,

where Ak, Bk are constants called the Fourier coefficients.

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SLIDE 10

Complex Numbers

Definition

A complex number is an ordered pair of real numbers,

z = (a, b), with a the real part and b the imaginary

part. Also written as

z = a + ib,

where i =

√ −1 is the imaginary unit.

The set of complex numbers is denoted C.

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SLIDE 11

Complex Numbers as 2D Coordinates

Im Re b a+bi a The real and imaginary parts of

z = a + ib are the coordinates: Re{z} = a Im{z} = b

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SLIDE 12

Complex Number Arithmetic

Take two complex numbers z1 = (a1, b1) and

z2 = (a2, b2).

◮ Addition

z1 + z2 = (a1 + a2, b1 + b2)

◮ Multiplication

z1 · z2 = (a1 + ib1) · (a2 + ib2) = (a1a2 − b1b2) + i(a1b2 + a2b1)

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SLIDE 13

Conjugation and Absolute Value

Consider a complex number z = (a, b)

◮ Conjugation: Simply negate the imaginary part:

z∗ = a − ib

◮ Absolute value: Same as 2D vector length:

|z| =

  • a2 + b2

Also given by |z| = √z · z∗

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SLIDE 14

Euler’s Representation of Complex Numbers

Im Re φ r

◮ A complex number can be

given as an angle φ and a radius r

◮ Think 2D polar coordinates ◮ Exponential form:

reiφ = r cos(φ) + i (r sin(φ))

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SLIDE 15

Operations in Euler’s Notation

Take z1 = r1eiθ1 and z2 = r2eiθ2.

◮ Multiplication:

z1 · z2 = r1r2ei(θ1+θ2)

◮ Conjugation:

z∗

1 = r1e−iθ1

◮ Absolute value:

|z1| = r1

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SLIDE 16

Fourier Integral

For periodic functions we have the Fourier series:

g(x) =

  • k=0

[ Ak cos(kω0x) + Bk sin(kω0x)] ,

But for nonperiodic functions we need a continuum of

  • frequencies. So, our Fourier series becomes an integral:

g(x) = ∞ Aω cos(ωx) + Bω sin(ωx) dω

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SLIDE 17

Computing Fourier Coefficients

Fourier coefficients describe how much a particular frequency ω contributes to the function g. They are computed by just multiplying and integrating with cos/sin waves:

Aω = A(ω) = 1 π ∞

−∞

g(x) · cos(ωx) dx Bω = B(ω) = 1 π ∞

−∞

g(x) · sin(ωx) dx

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SLIDE 18

Fourier Transform

Now, let’s put the B coefficient (the sine part) into the imaginary part of a complex number

G(ω) = π 2

  • A(ω) − i · B(ω)
  • =

1 √ 2π ∞

−∞

g(x) ·

  • cos(ωx) − i · sin(ωx)
  • dx

= 1 √ 2π ∞

−∞

g(x) · e−iωx dx