Lecture 5: Imaging Theory (3/6): Plane Waves and the Two-Dimensional Fourier Transform.

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# Lecture 5: Imaging Theory (3/6): Plane Waves and the Two-Dimensional Fourier Transform. - PowerPoint PPT Presentation

Lecture 5: Imaging Theory (3/6): Plane Waves and the Two-Dimensional Fourier Transform. . Review of 1-D Fourier Theory: Fourier Transform: x ↔ u F( u ) describes the magnitude and phase of the exponentials used to build f( x ). Consider u o , a specific value of u .

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Lecture 5: Imaging Theory (3/6): Plane Waves and the Two-Dimensional Fourier Transform.

Review of 1-D Fourier Theory:

Fourier Transform: x↔u

F(u) describes the magnitude and phase of the exponentials used to build f(x).

Consider uo, a specific value of u.

The integral sifts out the portion of f(x) that consists of exp(+i·2·uo·x)

Review: 1-D Fourier Theorems / Properties

If f(x) ↔ F(u) and h(x) ↔ H(u) ,

Performing the Fourier transform twice on a function f(x) yields f(-x).

Linearity: af(x) + bh(x) ↔ aF(u) + bH(u)

Scaling: f(ax) ↔

Shift: f(x-xo) ↔

Duality: multiplying by a complex exponential in the space domain results in a shift in the spatial frequency domain.

Convolution: f(x)*h(x) ↔ F(u)H(u)

Example problem

Find the Fourier transform of

f(x) = Π(x /4) – Λ(x /2) + .5Λ(x)

Find the Fourier transform of

Using the Fourier transforms of Π and Λ

and the linearity and scaling properties,

F(u) = 4sinc(4u) - 2sinc2(2u) + .5sinc2(u)

–2 1 0 1 2

Example problem: Alternative Answer.

Find the Fourier transform of

f(x) = Π(x /4) – 0.5((Π(x /3) * Π(x))

*

–1 -.5 0 .5 1

Using the Fourier transforms of Π and Λ

and the linearity and scaling and convolution properties ,

F(u) = 4sinc(4u) – 1.5sinc(3u)sinc(u)

Plane waves

 The period; the distance between successive maxima of the waves

Let’s get an intuitive feel for the plane wave

• defines the direction

of the undulation.

Lines of constant phase undulation

in the complex plane

A

B

q

c = 90 - q

D

q

C

y

Plane waves, continued.

q

1/u

Thus, similar triangles exist.

Taking a ratio,

x

L

1/v

As u and v increase, L decreases.

Plane waves, continued (2).

y

Frequency of the plane wave

Each set of u and v defines a complex plane wave with a different L and .

1/u

q

x

(cycles/mm)

L

1/v

q

• gives the direction of the undulation,

and can be found by

Plane waves: sine waves in the complex plane.

sin(10*p*x)

sin(10*p*x +4*pi*y)

Two-Dimensional Fourier Transform

Two-Dimensional Fourier Transform:

Where in f(x,y), x and y are real, not complex variables.

Two-Dimensional Inverse Fourier Transform:

 

amplitude basis functions

and phase of

required basis functions

Separable Functions

Two-Dimensional Fourier Transform:

What if f(x,y) were separable? That is,

f(x,y) = f1(x) f2(y)

Breaking up the exponential,

Separable Functions

Separating the integrals,

f(x,y) = cos(10px)*1

Fourier Transform

F(u,v) = 1/2 [d(u+5,0) + d(u-5,0)]

Imaginary [F(u,v)]

Real [F(u,v)]

v

v

v

u

u

f(x,y) = sin(10px)

Fourier Transform

F(u,v) = i/2 [d(u+5,0) - d(u-5,0)]

Imaginary [F(u,v)]

Real [F(u,v)]

v

v

v

u

u

f(x,y) = sin(40px)

Fourier Transform

F(u,v) = i/2 [d(u+20,0) - d(u-20,0)]

Imaginary [F(u,v)]

Real [F(u,v)]

v

v

v

u

u

f(x,y) = sin(20px + 10py)

Fourier Transform

F(u,v) = i/2 [d(u+10,v+5) - d(u-10,v-5)]

Imaginary [F(u,v)]

Real [F(u,v)]

v

v

v

u

u

Properties of the 2-D Fourier Transform

Linearity: a·f(x,y) + b·g(x,y) ↔ a·F(u,v) + b·G(u,v)

Scaling: g(ax,by) ↔

Let f(x,y) ↔ F(u,v) and g(x,y) ↔ G(u,v)

Properties of the 2-D Fourier Transform

Shift: g(x – a ,y – b) ↔

Let G(x,y) ↔ G(u,v)

L(x/16)L(y/16)

Real and even

Real{F(u,v)}= 256 sinc2(16u)sinc2(16v)

Imag{F(u,v)}= 0

Phase is 0 since

Imaginary channel is 0 and

F(u,v) > = 0 always

Log10(|F(u,v)|)

Shift: g(x – a ,y – b) ↔

L((x-1)/16) L(y/16)

Shifted one pixel right

Log10(|F(u,v)|)

Angle(F(u,v))

L((x-7)/16)L(y/16)

Shifted seven pixels right

Log10(|F(u,v)|)

Angle(F(u,v))

L((x-7)/16)L((y-2)/16)

Shifted seven pixels right, 2 pixels up

Log10(|F(u,v)|)

Angle(F(u,v))

Properties of the 2-D Fourier Transform

Convolution:

Let g(x,y) ↔ G(u,v) and h(x,y) ↔ H(u,v)

v

u

Image Fourier Space

v

u

Image Fourier Space

(log magnitude)

Detail

Contrast

5 %

10 %

20 %

50 %

2D Fourier Transform problem: comb function.

y

In one dimension,

-2 -1 0 1 2

In two dimensions,

y

x

2D Fourier Transform problem: comb function, continued.

Since the function does not describes how comb(y) varies in x, we can assume that by definition comb(y) does not vary in x.

We can consider comb(y) as a separable function,

where g(x,y)=gX(x)gY(y)

Here, gX(x) =1

Recall, if g(x,y) = gX(x)gY(y), then its transform is

gX(x)gY(y)  GX(u)GY(v)

y

x

2D Fourier Transform problem: comb function, continued (2).

y

gX(x)gY(y)  GX(u)GY(v)

So, in two dimensions,

x

g(x,y)

G(u,v)

v

u