4_HW4Solutions

4_HW4Solutions - EE 211A Fall Quarter, 2011 Instructor:...

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Unformatted text preview: EE 211A Fall Quarter, 2011 Instructor: John Villasenor Digital Image Processing I Handout 13 Homework 4 Solutions Textbook problems: 5.3 1 ⎤ྏ ⎡ྎ2 3⎤ྏ 1 ⎡ྎ 3 ⎥ྏ ⎢ྎ ⎥ྏ ⎢ྎ 3 ⎦ྏ ⎣ྏ1 2⎦ྏ 2 ⎣ྏ 1 1 ⎡ྎ8 + 4 3 8 ⎤ྏ = ⎢ྎ ⎥ྏ 4 ⎣ྏ 0 8 − 4 3 ⎦ྏ ⎡ྎ2 + 3 2 ⎤ྏ = ⎢ྎ ⎥ྏ 2 − 3 ⎦ྏ ⎣ྏ 0 V = AUAT = 1 ⎡ྎ 3 ⎢ྎ 2 ⎣ྏ − 1 − 1⎤ྏ ⎥ྏ 3 ⎦ྏ Basis images are obtained from outer products of columns of A*T : 1 ⎡ྎ 3 ⎤ྏ 1 ⎡ྎ 3 3 ⎤ྏ ⎢ྎ ⎥ྏ 3 1 = ⎢ྎ ⎥ྏ , 4 ⎣ྏ 1 ⎦ྏ 4 ⎣ྏ 3 1 ⎦ྏ 1 ⎡ྎ 3 ⎤ྏ 1 ⎡ྎ− 3 3 ⎤ྏ A0,1 = ⎢ྎ ⎥ྏ − 1 3 = ⎢ྎ ⎥ྏ , 4 ⎣ྏ 1 ⎦ྏ 4 ⎣ྏ − 1 3 ⎦ྏ 1 ⎡ྎ− 3 − 1⎤ྏ A1,0 = ⎢ྎ ⎥ྏ , 4 ⎣ྏ 3 3 ⎦ྏ 1 ⎡ྎ 1 − 3 ⎤ྏ A1,1 = ⎢ྎ ⎥ྏ. 4 ⎣ྏ− 3 3 ⎦ྏ A 0, 0 = [ [ Check: (2 + 3) A0,0 + 2 A0,1 + (2 − 3) A1,1 1 ⎡ྎ6 + 3 3 − 2 3 + 2 − 3 2 3 + 3 + 6 − 2 3 + 3 ⎤ྏ ⎢ྎ ⎥ྏ 4 ⎣ྏ 2 3 + 3 − 2 − 2 3 + 3 2 + 3 + 2 3 + 6 − 3 3 ⎦ྏ ⎡ྎ2 3⎤ྏ = ⎢ྎ ⎥ྏ . ⎣ྏ1 2⎦ྏ = 1 1. a. We know that v(k , l ) must be Hermitian ⇒ v(k , l ) = v * (−k ,−l ) . Since we are dealing with a 2D DFT, v(k , l ) must also be periodic, leading to v(k , l ) = v * ( N − k , N − l ) . For N = 4 , v(3,3) = v * (1,1) = F * , v(3,2) = v * (1,2) = G * , etc. Since v(2,0) = v * (2,0) , this coefficient does not get represented twice in the array. Since it is not given, it is unknown. The values which can be filled in are shown n ⇒ ABCD E F GH m⇓ ? * I* * ? I * E H G F* b. For a given (k , l ) , v(k , l ) is the sum over m, n of u (m, n) multiplied by the DFT kernel e −j 2π ( mk + nl ) N v(k , l ) will be real when this kernel is real. In general, e jθ is real only if θ = integer multiple of π ⇒ mk + nl must be equal to a multiple of N / 2 . This occurs for k = l = 0, N , 2 N k = 0, l = , 2 N k = , l = 0. 2 k =l = 2. a. ⎡ྎ0 ⎢ྎ0 ⎢ྎ ⎢ྎ0 ⎢ྎ ⎣ྏ0 0 0 0⎤ྏ 0 0 ⎤ྏ ⎡ྎ0 0 ⎥ྏ ⎢ྎ1 − j − 1 j ⎥ྏ 1 0 0⎥ྏ 1 ⎢ྎ ⎥ྏ ⎯ྎrow→ ⎯ྎ 0 1 0⎥ྏ 2 ⎢ྎ1 − 1 1 − 1⎥ྏ ⎥ྏ ⎢ྎ ⎥ྏ 0 0 0⎦ྏ 0 0 ⎦ྏ ⎣ྏ0 0 2 −1− j 0 − 1 + j ⎤ྏ ⎡ྎ 2 ⎢ྎ− 1 − j 0 −1+ j 2 ⎥ྏ 1 ⎢ྎ ⎥ྏ = v(k , l ) . ⎯ྎcolumn→ ⎯ྎ ⎯ྎ −1+ j 2 − 1 − j ⎥ྏ 4 ⎢ྎ 0 ⎢ྎ ⎥ྏ 2 −1− j 0 ⎦ྏ ⎣ྏ− 1 + j b. ⎡ྎ0 ⎢ྎ0 4 ⎢ྎ ⎢ྎ0 ⎢ྎ ⎣ྏ1 0 0 0 ⎤ྏ ⎡ྎ 0 ⎥ྏ ⎢ྎ 0 0 0 0 ⎥ྏ row X forms ⎯ྎ⎯ྎ⎯ྎ⎯ྎ⎯ྎ 4 ⎢ྎ → ⎢ྎ 0 0 0 0 ⎥ྏ ⎥ྏ ⎢ྎ 1 1 1 ⎦ྏ ⎣ྏ 2 0 0 0 ⎤ྏ 0 0 0 ⎥ྏ column X forms ⎥ྏ ⎯ྎ⎯ྎ⎯ྎ⎯ྎ⎯ྎ⎯ྎ → 0 0 0 ⎥ྏ ⎥ྏ 0 0 0 ⎦ྏ ⎡ྎ 4 ⎢ྎ 4 j ⎢ྎ ⎢ྎ −4 ⎢ྎ ⎣ྏ −4 j 0 0 0 ⎤ྏ 0 0 0 ⎥ྏ ⎥ྏ = v(k , l ). 0 0 0 ⎥ྏ ⎥ྏ 0 0 0 ⎦ྏ c. This is a complex corrugation in the diagonal direction. We expect that its DFT will contain only 1 nonzero component. 1 − j − 1 ⎤ྏ ⎡ྎ j ⎡ྎ0 0 ⎢ྎ 1 − j − 1 ⎥ྏ ⎢ྎ j ⎥ྏ 1 ⎢ྎ 1 ⎢ྎ0 0 row ⎯ྎ⎯ྎ→ j 1 ⎥ྏ 4 ⎢ྎ− j − 1 2 ⎢ྎ0 0 ⎢ྎ ⎥ྏ ⎢ྎ −1 j 1 − j ⎦ྏ ⎣ྏ ⎣ྏ0 0 0 0 j ⎤ྏ 1 ⎥ྏ ⎥ྏ ⎯ྎcolumn → ⎯ྎ⎯ྎ 0 − 1 ⎥ྏ ⎥ྏ 0 − j ⎦ྏ ⎡ྎ0 ⎢ྎ0 ⎢ྎ ⎢ྎ0 ⎢ྎ ⎣ྏ0 0 0 0⎤ྏ 0 0 0⎥ྏ ⎥ྏ = v(k , l ). 0 0 0⎥ྏ ⎥ྏ 0 0 j ⎦ྏ 3. a. ⎡ྎ0 ⎢ྎ0 ⎢ྎ ⎢ྎ0 ⎢ྎ ⎣ྏ0 0 0 0⎤ྏ 1 0 0⎥ྏ ⎥ྏ ⎯ྎcolumn→ ⎯ྎ ⎯ྎ 0 1 0⎥ྏ ⎥ྏ 0 0 0⎦ྏ 0.5 0.5 0⎤ྏ ⎡ྎ0 ⎢ྎ0 0.271 − 0.271 0⎥ྏ ⎢ྎ ⎥ྏ ⎢ྎ0 − 0.5 − 0.5 0⎥ྏ ⎢ྎ ⎥ྏ ⎣ྏ0 − 0.653 0.653 0⎦ྏ 0 − 0.5 0 ⎤ྏ ⎡ྎ 0.5 ⎢ྎ 0 0.146 0 − 0.354⎥ྏ row ⎢ྎ ⎥ྏ = v(k , l ). ⎯ྎ⎯ྎ→ ⎢ྎ− 0.5 0 0.5 0 ⎥ྏ ⎢ྎ ⎥ྏ − 0.354 0 0.854 ⎦ྏ ⎣ྏ 0 b. ⎡ྎ0 ⎢ྎ0 4 ⎢ྎ ྎ0 ⎢ྎ ⎣ྏ1 0 0 0⎤ྏ 0 0 0⎥ྏ ⎥ྏ ⎯ྎrows → ⎯ྎ ⎯ྎ 0 0 0⎥ྏ ⎥ྏ 1 1 1⎦ྏ 0 ⎡ྎ 1 ⎢ྎ− 1.307 0 ⎯ྎcolumn→ 4⎢ྎ ⎯ྎ ⎯ྎ ⎢ྎ 1 0 ⎢ྎ ⎣ྏ− 0.541 0 3 ⎡ྎ0 0 0 0⎤ྏ ⎢ྎ0 0 0 0⎥ྏ ⎥ྏ 4⎢ྎ ⎢ྎ0 0 0 0⎥ྏ ⎢ྎ ⎥ྏ ⎣ྏ2 0 0 0⎦ྏ 0 0⎤ྏ 0 0⎥ྏ ⎥ྏ = v(k , l ). 0 0⎥ྏ ⎥ྏ 0 0⎦ྏ ...
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This note was uploaded on 12/27/2011 for the course EE211A 211A taught by Professor Villasenor during the Fall '11 term at UCLA.

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