SOL14

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Signals and Systems Fall 2012 Solution to Homework Assignment #14 1. ( ) { } 2 ( ) 3 2 j F x t j X j ω ω ω ω = = (a) ( ) ( ) 1 y t x t = ( ) ( ) j Y j e X j ω ω ω = ( ) ( ) ( ) 2 2 1 z t y t x t = = ( ) ( ) ( ) 2 2 2 1 1 2 2 j j j Z j Y e X ω ω ω ω = = ( ) { } ( ) ( ) 2 2 2 2 2 1 2 1 2 3 2 j j j j F x t e ω ω ω ω + = + + (b) ( ) ( ) dx t y t dt = { } ( ) { } ( ) ( ) ( ) ( ) 2 2 3 2 j F dx dt F y t j X j j j ω ω ω ω ω = = = + + (c) ( ) ( ) ( ) ( ) 1 1 2 2 jt jt x t e y t x t cos t e = = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 1 1 2 2 1 1 1 2 1 3 1 2 1 3 1 2 Y j X j X j j j j j j j ω ω ω ω ω ω ω ω ω = + + + = + + + + + + + ( ) ( ) { } ( ) ( ) ( ) ( ) ( ) 2 4 3 2 2 3 cos 6 15 18 9 j j F x t t j j j j ω ω ω ω ω ω + = + + + + 2. (a) ( ) ( ) 2 2 2 4 1 , 2 a a F j e e a a ω ω π ω π = = = ( ) 2 2 2 4 1 2 a t t a f t e e π π = =
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(b) ( ) ( ) 1 1 j j e e F j F j j j ω ω ω ω ω ω = = + + ( ) ( ) ( ) 1 1 t s f t e u t = ( ) ( ) ( ) ( ) 1 1 t s f t e u t + = (c) ( ) 1 j F j j ω ω ω = + ( ) ( ) ( ) ( ) d d t t t s s f t e u t e u t e t t δ = = + ( ) ( ) ( ) t s f t e u t t δ = + 3. (a) < < < < = otherwise 2 1 1 2 0 1 1 ) ( t t t x 0 2 t x ( t ) 1 -2 -1 1 Let ( ) 1, 1 2 1 2 0, otherwise t y t < < = ( ) ( ) ( ) sin 2 2 Y j ω ω ω = Then ( ) ( ) ( ) 1.5 1.5 x t y t y t = + + ( ) ( ) ( ) 1.5 1.5 j j
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