lecture 5: imaging theory (3/6): plane waves and the two-dimensional fourier transform
DESCRIPTION
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 . - PowerPoint PPT PresentationTRANSCRIPT
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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)
dxxu xui
2e)f()F(
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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)
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Can you explain this movie via the convolution theorem?
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Example problem
Find the Fourier transform of
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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)
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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*
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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.
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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
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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
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sin(2**x)
cos(2**x)
Plane waves: sine and cosine waves
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sin(10**x)
sin(10**x +4*pi*y)
Plane waves: sine waves in the complex plane.
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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:
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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,
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Separable Functions
)(22
)(21 e )(e)(),F( dxdyyfxfvu vyiuxi
dyyfdxxfvu vyiuxi )(22
)(21 e )(e)(),F(
Separating the integrals,
)()(),( 21 vFuFvuF
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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)]
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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)]
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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)]
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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)]
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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
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Log display often more helpful
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Properties of the 2-D Fourier Transform
Let G(x,y) ↔ G(u,v)
Shift: g(x – a ,y – b) ↔ )ba(2e),(G vuivu
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(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
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((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
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Log10(|F(u,v)|) Angle(F(u,v))
((x-7)/16y/16) Shifted seven pixels right
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Log10(|F(u,v)|) Angle(F(u,v))
((x-7)/16y-2)/16) Shifted seven pixels right, 2 pixels up
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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
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u
v
Image Fourier Space
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u
v
Image Fourier Space (log magnitude)
DetailDetail
ContrastContrast
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10 %5 % 20 % 50 %
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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,
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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
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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)
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2D FT’s of Delta Functions: Good Things to Remember
(“bed of nails” function)
m n
numxyx ),())III(III(
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y
x u
v
(x)
(u)(y)
(v)y v
ux
Note the 2D transforms of the 1D delta functions:
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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)