# Review of 1-D Fourier Theory:

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Review of 1-D Fourier Theory:
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)

Can you explain this movie via the convolution theorem?

Example problem Find the Fourier transform of

Find the Fourier transform of f(x) = Π(x /4) – Λ(x /2) + .5Λ(x) Using the Fourier transforms of Π and Λ and the linearity and scaling properties, F(u) = 4sinc(4u) - 2sinc2(2u) + .5sinc2(u)

Find the Fourier transform of f(x) = Π(x /4) – 0.5((Π(x /3) * Π(x)) * 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)

 The period; the distance between successive maxima of the waves
Plane waves Let’s get an intuitive feel for the plane wave  The period; the distance between successive maxima of the waves defines the direction of the undulation. Lines of constant phase undulation in the complex plane

Plane waves, continued. y x L Thus, similar triangles exist.
ABC ~ ADB. Taking a ratio, A B C D q 1/u x q c = 90 - q q L 1/v

Plane waves, continued (2).
As u and v increase, L decreases. y Frequency of the plane wave 1/u q x (cycles/mm) L Each set of u and v defines a complex plane wave with a different L and . 1/v q gives the direction of the undulation, and can be found by

Plane waves: sine and cosine waves
sin(2*p*x) cos(2*p*x)

sin(10*p*x) sin(10*p*x +4*pi*y)
Plane waves: sine waves in the complex plane. sin(10*p*x) sin(10*p*x +4*pi*y)

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

What if f(x,y) were separable? That is, f(x,y) = f1(x) f2(y)
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,

Separating the integrals,
Separable Functions Separating the integrals,

F(u,v) = 1/2 [d(u+5,0) + d(u-5,0)]
f(x,y) = cos(10px)*1 Fourier Transform F(u,v) = 1/2 [d(u+5,0) + d(u-5,0)] Real [F(u,v)] Imaginary [F(u,v)] v v v u u

F(u,v) = i/2 [d(u+5,0) - d(u-5,0)]
f(x,y) = sin(10px) Fourier Transform F(u,v) = i/2 [d(u+5,0) - d(u-5,0)] Real [F(u,v)] Imaginary [F(u,v)] v v v u u

F(u,v) = i/2 [d(u+20,0) - d(u-20,0)]
f(x,y) = sin(40px) Fourier Transform F(u,v) = i/2 [d(u+20,0) - d(u-20,0)] Real [F(u,v)] Imaginary [F(u,v)] v v v u u

F(u,v) = i/2 [d(u+10,v+5) - d(u-10,v-5)]
f(x,y) = sin(20px + 10py) Fourier Transform F(u,v) = i/2 [d(u+10,v+5) - d(u-10,v-5)] Real [F(u,v)] Imaginary [F(u,v)] v v v u u

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) ↔ ~~~~~~~~~~~~~~~~~~

Properties of the 2-D Fourier Transform
Let G(x,y) ↔ G(u,v) Shift: g(x – a ,y – b) ↔ ~~~~~~~~~~~~~~~~~~

L(x/16)L(y/16) Real and even Log10(|F(u,v)|)
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)|)

Shifted one pixel right
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))

Shifted seven pixels right
L((x-7)/16)L(y/16) Shifted seven pixels right Log10(|F(u,v)|) Angle(F(u,v))

Shifted seven pixels right, 2 pixels up
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
Let g(x,y) ↔ G(u,v) and h(x,y) ↔ H(u,v) Convolution: ~~~~~~~~~~~~~~~~~~

Image Fourier Space u v

Image Fourier Space (log magnitude) u v Detail Contrast

5 % 10 % 20 % 50 %

2D Fourier Transform problem: comb function.
In one dimension, y 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).
gX(x)gY(y)  GX(u)GY(v) So, in two dimensions, y x g(x,y) G(u,v) v u

2D FT’s of Delta Functions: Good Things to Remember
(“bed of nails” function) Terminology ok?

Note the 2D transforms of the 1D delta functions:
y v (v) (x) x u y v (y) (u) u x