114_1_sol_midterm_2011_Part2

# 114_1_sol_midterm_2011_Part2 - 1 1(8 v k v k = Thus 1 2 a b...

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EE114 Introduction to Speech and Image Processing Winter Quarter, 2011 Page 6 of 8 H π ( 29 H 0 ( 29 = 2 1 - a ( 29 1 + b ( 29 1 + a ( 29 1 - b ( 29 = 2 1 + b - a - ab = 2 1 - b + a - ab ( 29 - 1 + 3 b - 3 a + ab = 0 b = 1 + 3 a 3 + a OR a = 3 b - 1 3 - b

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EE114 Introduction to Speech and Image Processing Winter Quarter, 2011 Page 7 of 8 3. 1D DFT (25 points) For this problem, assume that the 1D DFT of length N is defined as follows ( 29 ( 29 1 2 0 1 N j kn N n v k x n e N π - - = = Given the following 8 N = transform pair: [ ] [ ] a b c d e f g h A B C D E F G H . What is the 16 N = transform of [ ] 0 0 0 0 0 0 0 0 a b c d e f g h in terms of , , , , , , and A B C D E F G H ? Answer: Denote [ ] 1 ( ) x n a b c d e f g h = and [ ] 2 ( ) 0 0 0 0 0 0 0 0 x n a b c d e f g h = From the problem statement, we know that ( 29 ( 29 7 2 8 1 1 0 1 8 j kn n v k x n e - = = = [ ] A B C D E F G H . The transform of x 2 ( n ) is ( 29 ( 29 ( 29 ( 29 ( 29 15 2 16 2 2 0 7 2 2 16 2 0 7 2 8 1 0 1 1 16 1 2 16 1 16 1 2 j kn n j k n n j kn n v k x n e x n e x n e v k - = - = - = = = = = where k is evaluated over the range from 0 to 15. Note that evaluating v 1 ( k ) outside the range [0, 7] involves using the periodic properties of DFTs – in other words,
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Unformatted text preview: 1 1 (8 ) ( ) v k v k + = . Thus, [ ] [ ] 1 2 a b c d e f g h A B C D E F G H A B C D E F G H ↔ EE114 Introduction to Speech and Image Processing Winter Quarter, 2011 Page 8 of 8 4. 2D Convolution (25 points) Consider the continuous 2D function ) , f( y x given below, which has value 1 where the dark regions are located and 0 elsewhere. Make a sketch showing the boundaries of non-zero regions of the self-convolution of ) , f( y x . Be sure to place your sketch on a set of marked axis such that relevant dimensions and positions of features are clearly indicated. Answer:...
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## This note was uploaded on 11/12/2011 for the course EE 114 taught by Professor Vanschaar during the Spring '11 term at UCLA.

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114_1_sol_midterm_2011_Part2 - 1 1(8 v k v k = Thus 1 2 a b...

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