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20105ee211A_1_HW5Solutions

20105ee211A_1_HW5Solutions - EE 211A Fall Quarter 2010...

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1 EE 211A Digital Image Processing I Fall Quarter, 2010 Handout 18 Instructor: John Villasenor Homework 5 Solutions Solutions: 1. a. ( 29 ( 29 1 * 0 2 2 1 1 2 1 2 0 1 0 2 M T u k k k R u u M - = - = = - + - - . 2 1 1 1 4 - - = b. | | I R u λ - , 266 . 4 1 2 9 0 2 = + - = = λ λ λ 0.234. To find , 0 φ = + - - = - - C C C C 266 . 4 266 . 4 2 1 4 1 2 1 1 1 4 - = 266 . 0 C normalized - = 257 . 0 966 . 0 0 φ = 966 . 0 257 . 0 1 φ c. . 966 . 0 257 . 0 257 . 0 966 . 0 * 1 * 0 * - = = T T T φ φ φ d. . 514 . 0 932 . 1 452 . 0 189 . 2 1 * 1 0 * 0 - - = = - = = u v u v T T φ φ . 1 2 966 . 0 257 . 0 452 . 0 257 . 0 966 . 0 189 . 2 ) 1 ( ) 0 ( 0 1 0 0 0 u v v = - = - - = + φ φ e. v R = ] [ 2 1 * 1 1 * 0 0 T T v v v v +
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2 = = + - - 234 . 0 0 0 26 . 4 264 . 0 99 . 0 99 . 0 73 . 3 204 . 0 989 . 0 989 . 0 79 . 4 2 1 = diag ) ( k λ . f. The unitary DFT matrix for 2 = N is . 1 1 1 1 2 1 - 0 w = - - 1 2 1 1 1 1 2 1 = . 12 . 2 707 . 0 1 w = - - 0 2 1 1 1 1 2 1 = .
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