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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 u o, a specific value of u. The integral sifts out the portion of f(x) that consists of exp(+i·2 ·u o ·x)

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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-x o ) ↔ 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) - 2sinc 2 (2u) +.5sinc 2 (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)) – – *

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Plane waves Let’s get an intuitive feel for the plane wave Lines of constant phase undulation in 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. Thus, similar triangles exist. ABC ~ ADB. Taking a ratio, y 1/v x L 1/u AB C D

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Plane waves, continued (2). Each set of u and v defines a complex plane wave with a different L and . As u and v increase, L decreases. (cycles/mm) 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: amplitude basis functions and 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) = f 1 (x) f 2 (y) Two-Dimensional Fourier Transform: Breaking up the exponential,

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Separable Functions Separating the integrals,

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

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

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F(u,v) = i/2 [ (u+20,0) - (u-20,0)] Fourier Transform f(x,y) = sin(40 x) u v v Real [F(u,v)] u v Imaginary [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(20 x + 10 y) u v v Real [F(u,v)] u v Imaginary [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) ↔

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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) ↔

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(x/16 y/16) Real and even Real{F(u,v)}= 256 sinc 2 (16u)sinc 2 (16v)Imag{F(u,v)}= 0 Log10(|F(u,v)|) Phase is 0 since Imaginary channel is 0 and F(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) ↔

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

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Log10(|F(u,v)|)Angle(F(u,v)) ((x-7)/16 y-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:

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u v Image Fourier Space

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u v (log magnitude) Detail Contrast

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10 %5 %20 %50 %

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2D Fourier Transform problem: comb function. In two dimensions, 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)=g X (x)g Y (y) Here, g X (x) =1 Recall, if g(x,y) = g X (x)g Y (y), then its transform is g X (x)g Y (y) G X (u)G Y (v) y x

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2D Fourier Transform problem: comb function, continued (2). g X (x)g Y (y) G X (u)G Y (v) So, in two dimensions, y x 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)

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y x u v (x)(x) (u)(u) (y)(y) (v)(v) yv 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)

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