lecture 5: imaging theory (3/6): plane waves and the two-dimensional fourier transform. review of...

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)

dxxu xui

2e)f()F(

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) ↔ )F(e 2 uuxi o

a

u

aF

||

1

)(F)(fe 2o

xui uuxo

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)

Can you explain this movie via the convolution theorem?

Example problem

Find the Fourier transform of

Example problem: Answer.

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)

f(x) = Π(x /4) – Λ(x /2) + .5Λ(x)

Example problem: Alternative Answer.

Find the Fourier transform of

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)

f(x) = Π(x /4) – 0.5((Π(x /3) * Π(x))

–2 1 0 1 2 –1 -.5 0 .5 1*

Plane wavesLet’s get an intuitive feel for the plane wave )(2e vyuxi

Lines of constant phase undulationin the complex plane

The period; the distance between successive maxima of the waves

defines the direction of the undulation.

Plane waves, continued.

uvu

v111

1L

22

22

11

L

22

22vu

vuvu

uv

Thus, similar triangles exist. ABC ~ ADB. Taking a ratio,

y

1/v

x

L

1/uA B

C

D

Plane waves, continued (2).

22

1L

vu

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

As u and v increase, L decreases.

L

1 22 vu

(cycles/mm)

u

v

v

uarctan

/1

/1arctanθ

1/u

1/v

gives the direction of the undulation, and can be found by

Frequency of the plane wave

L

y

x

sin(2**x)

cos(2**x)

Plane waves: sine and cosine waves

sin(10**x)

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

Plane waves: sine waves in the complex plane.

Two-Dimensional Fourier Transform

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

Two-Dimensional Inverse Fourier Transform:

)(2e ),(f),F( dxdyyxvu vyuxi

)(2e ),F(),( dudvvuyxf vyuxi

amplitude basis functionsand phase of required basis functions

Two-Dimensional Fourier Transform:

Separable Functions

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

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

)(2e ),(f),F( dxdyyxvu vyuxi

Two-Dimensional Fourier Transform:

)(221 e )()(),F( dxdyyfxfvu vyuxi

)(22

)(21 e )(e)(),F( dxdyyfxfvu vyiuxi

Breaking up the exponential,

Separable Functions

)(22

)(21 e )(e)(),F( dxdyyfxfvu vyiuxi

dyyfdxxfvu vyiuxi )(22

)(21 e )(e)(),F(

Separating the integrals,

)()(),( 21 vFuFvuF

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

Fourier Transform

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

u

v

v

Real [F(u,v)]

x

y

-50 0 50

-50

0

50-0.5

0

0.5Real [F(u,v)]

u

vImaginary [F(u,v)]

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

Fourier Transform

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

u

v

v

Real [F(u,v)]

x

y

-50 0 50

-50

0

50-0.5

0

0.5Real [F(u,v)]

u

vImaginary [F(u,v)]

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

Fourier Transform

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

u

v

v

Real [F(u,v)]

x

y

-50 0 50

-50

0

50-0.5

0

0.5Real [F(u,v)]

u

vImaginary [F(u,v)]

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

Fourier Transform

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

u

v

v

Real [F(u,v)]

x

y

-50 0 50

-50

0

50-0.5

0

0.5Real [F(u,v)]

u

vImaginary [F(u,v)]

Properties of the 2-D Fourier Transform

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

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

Scaling: g(ax,by) ↔

b,

aG

|ab|

1 vu

Log display often more helpful

Properties of the 2-D Fourier Transform

Let G(x,y) ↔ G(u,v)

Shift: g(x – a ,y – b) ↔ )ba(2e),(G vuivu

(x/16y/16) Real and even

Real{F(u,v)}= 256 sinc2(16u)sinc2(16v) Imag{F(u,v)}= 0

Log10(|F(u,v)|)

Phase is 0 sinceImaginary channel is 0 andF(u,v) > = 0 always

((x-1)/16) (y/16) Shifted one pixel right

Log10(|F(u,v)|) Angle(F(u,v))

Shift: g(x – a ,y – b) ↔ )ba(2e),(G vuivu

Log10(|F(u,v)|) Angle(F(u,v))

((x-7)/16y/16) Shifted seven pixels right

Log10(|F(u,v)|) Angle(F(u,v))

((x-7)/16y-2)/16) Shifted seven pixels right, 2 pixels up

Properties of the 2-D Fourier Transform

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

Convolution:

),(H),(G)(h)(g

)(h ),g()(h)(g

vuvux,yx,y

dd,yxx,yx,y

u

v

Image Fourier Space

u

v

Image Fourier Space (log magnitude)

DetailDetail

ContrastContrast

10 %5 % 20 % 50 %

2D Fourier Transform problem: comb function.

In two dimensions,

n

nyy )δ()comb(

n

nyy )δ()comb(-2 -1 0 1 2

y

y

x

……In one dimension,

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).

gX(x)gY(y) GX(u)GY(v)

So, in two dimensions,

y

x

n

nvuy

vuy

),()comb(

)(comb)δ()comb(1

u

v

G(u,v)

g(x,y)

2D FT’s of Delta Functions: Good Things to Remember

(“bed of nails” function)

m n

numxyx ),())III(III(

y

x u

v

(x)

(u)(y)

(v)y v

ux

Note the 2D transforms of the 1D delta functions:

Example problem: Answer.

Find the Fourier transform of

Using the Fourier transforms of Π and Λ

and the linearity and scaling properties,

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

f(x) = Π(x /4) – Λ(x /2) + .5Λ(x)