HW6_ece220_2007_solution

HW6_ece220_2007_solution - Homework 6 Solution(ECE220 Fall...

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Homework 6 Solution (ECE220 - Fall 2007) 1. (40 points) Fourier Series for analog signal An analog image is a 2D signal which has limited support in the independent variables I ( x, y ). Let I ( x, y ) be defined over a square region | x | < L 2 and | y | < L 2 . For any given point y we can find the Fourier series of I ( x, y ) pretending that the image periodically repeats itself outside the period. We can do the same fixing x and pretending the image repeats itself periodically. (a) (5 points) Using this argument extend the concept of Fourier series in 2D an indicate under what conditions we show that an image can be represented as: I ( x, y ) = X k 1 = -∞ X k 2 = -∞ a k 1 ,k 2 e j 2 π L k 1 x e j 2 π L k 2 y Holds in the sense that the error: e N ( x, y ) = I ( x, y ) - N X k 1 = - N N X k 2 = - N a k 1 ,k 2 e j 2 π L k 1 x e j 2 π L k 2 y is such that: lim N →∞ Z L 2 - L 2 Z L 2 - L 2 | e N ( x, y ) | 2 dxdy = 0 Solution: (b) (10 points) Suppose that the white level is 1 and the black level is 0. Calculate the Fourier series of an image that is a white square which covers one quarter of the area of the image and that is perfectly centered in the image (the center is the (0,0) point and the image is a perfectly square image as before with | x | < L 2 and | y | < L 2 ). Solution: a k 1 ,k 2 = 1 T x T y Z T y Z T x I ( x, y ) e - j 2 π Tx k 1 x e - j 2 π Ty k 2 y dxdy = 1 L 2 Z L 2 - L 2 Z L 2 - L 2 I ( x, y ) e - j 2 π L k 1 x e - j 2 π L k 2 y dxdy = 1 L 2 Z L 4 - L 4 1 e - j 2 π L k 1 x dx Z L 4 - L 4 1 e - j 2 π L k 2 y dy = 1 L 2 - L 2 jπk 1 e - j 2 π L k 1 x L 4 x = - L 4 - L 2 jπk 2 e - j 2 π L k 2 y L 4 y = - L 4 = 1 L 2 L πk 1 " e j π 2 k 1 - e - j π 2 k 1 2 j # L πk 2 " e j π 2 k 2 - e - j π 2 k 2 2 j #
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= sin( π 2 k 1 ) πk 1 sin( π 2 k 2 ) πk 2 = 1 4 sinc π 2 k 1 · sinc π 2 k 2 · (c) (5 points) Since the image is real, what is the relationship between the coefficients?
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