20082ee102_1_HW_6_solution

20082ee102_1_HW_6_solution - HW6 solution, SPRING 2008, EE...

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HW6 solution, SPRING 2008, EE 102 1. (i) φ 0 ( s ) = 2 s + 1 φ 0 ( t ) = 2 e - t φ 1 ( s ) = 2( s - 1) ( s + 1) 2 φ 1 ( t ) = 2(1 - 2 t ) e - t (ii) < φ 0 ( t ) 0 ( t ) > = 1 < φ 1 ( t ) 1 ( t ) > = 1 < φ 1 ( t ) 0 ( t ) > = 0 It’s orthonormal. (iii) < f ( t ) 0 ( t ) > = 2 / 4 < f ( t ) 1 ( t ) > = - 2 / 4 f 2 ( t ) = ( 2 / 4)( 2 e - t ) + ( - 2 / 4) 2(1 - 2 t ) e - t ) See last page for the figure. (iv) ¯ ² 2 = 0 . 2. (i) e - 1 ( t ) = e - iπt ,e - 1 ( t + T ) = e - ( t + T ) T = 2 For e 1 ( t ), T = 2 . e 0 ( t ) = e 0 = 1 . A constant function is periodic with any period. (ii) < e - 1 ( t ) ,e 0 ( t ) > = 0 = < e 0 ( t ) ,e 1 ( t ) > 1
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< e 1 ( t ) ,e 0 ( t ) > = 0 = < e 0 ( t ) ,e - 1 ( t ) > < e 1 ( t ) ,e 1 ( t ) > = < e 0 ( t ) ,e 0 ( t ) > = < e - 1 ( t ) ,e - 1 ( t ) > = 2 Normalize: e 1 ( t ) = e - iπt 2 ,e 0 = e iπt 2 , ( t ) e - 1 ( t ) = 1 2 (iii) F 1 = - 2 - π 2 F - 1 = - 2+ π 2 F 0 = 1 2 The figure is at last page. (iv)
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This note was uploaded on 04/13/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

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20082ee102_1_HW_6_solution - HW6 solution, SPRING 2008, EE...

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