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20105ee211A_1_HW4Solutions

# 20105ee211A_1_HW4Solutions - EE 211A Fall Quarter 2010...

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1 EE 211A Digital Image Processing I Fall Quarter, 2010 Handout 13 Instructor: John Villasenor Homework 4 Solutions Textbook problems: 5.3 T V AUA = = - - 3 1 1 3 2 1 2 1 3 2 3 1 1 3 2 1 = - + 3 4 8 0 8 3 4 8 4 1 = - + 3 2 0 2 3 2 Basis images are obtained from outer products of columns of * T A : [ ] = = 1 3 3 3 4 1 1 3 1 3 4 1 0 , 0 A , [ ] - - = - = 3 1 3 3 4 1 3 1 1 3 4 1 1 , 0 A , - - = 3 3 1 3 4 1 0 , 1 A , . 3 3 3 1 4 1 1 , 1 - - = A Check: 1 , 1 1 , 0 0 , 0 ) 3 2 ( 2 ) 3 2 ( A A A - + + + - + + + + - - + + - + + - + - + = 3 3 6 3 2 3 2 3 3 2 2 3 3 2 3 3 2 6 3 3 2 3 2 3 2 3 3 6 4 1 = 2 1 3 2 .

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2 1. a. We know that ) , ( l k v must be Hermitian ) , ( * ) , ( l k v l k v - - = . Since we are dealing with a 2D DFT, ) , ( l k v must also be periodic, leading to ) , ( * ) , ( l N k N v l k v - - = . For 4 = N , * ) 1 , 1 ( * ) 3 , 3 ( F v v = = , * ) 2 , 1 ( * ) 2 , 3 ( G v v = = , etc.
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20105ee211A_1_HW4Solutions - EE 211A Fall Quarter 2010...

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