Solution6

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

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Unformatted text preview: EE 200 Fall 2009 (Weber) Homework 6 Solutions 1. a. H ( j ) = integraldisplay ( t )- . 1 e . 1 t u ( t ) e jt dt = integraldisplay ( t ) e jt dt- . 1 integraldisplay e . 1 t e jt dt = e j integraldisplay ( t ) dt- . 1 1- . 1- j e ( . 1 j ) t vextendsingle vextendsingle vextendsingle vextendsingle = 1- . 1- . 1- j (0- 1) = 1- . 1 . 1 + j = j . 1 + j b. | H ( j ) | 2 = H ( j ) H ( j ) = j . 1 + j- j . 1- j = 2 . 01 + 2 c. | H ( j ) | 2 grows asymptotically to a maximum of one as goes to . It reaches -3db below its maxium when | H ( j ) | 2 = 1 / 2. 2 . 01 + 2 = 1 2 2 2 = . 01 + 2 = . 1 d. x ( t ) = 10 + 20 cos(0 . 1 t ) + ( t- . 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 outputat their respective frequencys of zero and 0....
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This note was uploaded on 02/26/2010 for the course EE 30446 at USC.

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

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