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Solution6 - EE 200 Fall 2009(Weber Homework 6 Solutions 1 a...

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EE 200 – Fall 2009 (Weber) Homework 6 Solutions 1. a. H ( ) = integraldisplay −∞ δ ( t ) - 0 . 1 e 0 . 1 t u ( t ) e jωt dt = integraldisplay −∞ δ ( t ) e jωt dt - 0 . 1 integraldisplay 0 e 0 . 1 t e jωt dt = e 0 integraldisplay −∞ δ ( t ) dt - 0 . 1 1 - 0 . 1 - e ( 0 . 1 ) t vextendsingle vextendsingle vextendsingle vextendsingle 0 = 1 - 0 . 1 - 0 . 1 - (0 - 1) = 1 - 0 . 1 0 . 1 + = 0 . 1 + b. | H ( ) | 2 = H ( ) H ( ) = 0 . 1 + - 0 . 1 - = ω 2 0 . 01 + ω 2 c. | H ( ) | 2 grows asymptotically to a maximum of one as ω goes to ±∞ . It reaches -3db below it’s maxium when | H ( ) | 2 = 1 / 2. ω 2 0 . 01 + ω 2 = 1 2 2 ω 2 = 0 . 01 + ω 2 ω = ± 0 . 1 d. x ( t ) = 10 + 20 cos(0 . 1 t ) + δ ( t - 0 . 2) The output resulting from the first two terms can be determined by evaluating the frequency response at their respective frequencys of zero and 0.1. Since the third term is a shifted impulse, the output from it is a shifted impulse response. y ( t ) = H ( j 0)10 + H ( j 0 . 1)20 cos(0 . 1 t ) + h ( t - 0 . 2) 1
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From part b, the magnitude of H ( j 0) is zero so the first term is filtered out. Using polar form, | H ( j 0 . 1) | =
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