211A_1_final2010solution

211A_1_final2010solution - EE211A Fall Quarter, 2010...

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EE211A Digital Image Processing I Fall Quarter, 2010 1 Final Exam Solutions 1. KL Transform (25 points) Consider a three-dimensional random vector u that has the following three equiprobable realizations: 0 1 2 2 1 1 1 , 2 , 1 1 1 2 u u u = = = . Let the KL transform of this random vector u be denoted by v. (a) (20 points) Find the transform coefficient energies. In other words find ( 29 ( 29 ( 29 2 2 2 0 , 1 , and 2 . E v E v E v (b) (5 points) Find the KL transform basis vector corresponding to the transform coefficient with the largest energy. Solution (a) All three of these vectors are completely symmetric with respect to the 1 1 1           vector, with a relatively small angle between 1 1 1           and any of the realizations (the angle can be approximated by taking the dot product and noting that the cosine of the angle is quite close to 1), Thus, 1 1 1           (after normalization) will be the principal component. The energy along this component can be found by taking the dot product of the realization with the normalized component and then squaring the result: ( 29 ( 29 1 1 1 4 2 1 1 1 2 1 1 3 3 3 1     = + + =       ( 29 2 16 Energy 5.333 0 3 E v = = = There is no directional preference to any of the other two vectors, so the remaining energy will be split evenly ( 29 ( 29 2 2 1 2 0.333 E v E v = =
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EE211A Digital Image Processing I Fall Quarter, 2010 2 (b) The principal component is the vector [ ] 1 1 1 1 3 as seen above with the most energy.
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EE211A Digital Image Processing I Fall Quarter, 2010 3 2. Image Transform (25 points) Consider the image ( 29 , u m n of size 256 by 256 consisting of an array of regular
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This note was uploaded on 12/27/2011 for the course EE211A 211A taught by Professor Villasenor during the Spring '11 term at UCLA.

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211A_1_final2010solution - EE211A Fall Quarter, 2010...

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