Solution2

Solution2 - EE 200 Fall 2009 (Weber) Homework 2 Solutions...

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Unformatted text preview: EE 200 Fall 2009 (Weber) Homework 2 Solutions 1. Converting the magnitude and phase of the phasors to rectangular coordinates: 5 cos( t + 3 2 ) 5 e j 3 2 = 5(0- j ) = 0- 5 j 4 cos( t + 2 3 ) 4 e j 2 3 = 4(- 1 2 + j 3 2 ) =- 2 + 3 . 46 j 4 cos( t + 1 3 ) 4 e j 1 3 = 4( 1 2 + j 3 2 ) = 2 + 3 . 46 j Summing the phasors give 0 + j (- 5 + 4 3) = 0 + 1 . 93 j . The resulting phasor has a magnitude of 1.93 and an angle of / 2. x ( t ) = 1 . 93 cos( t + / 2)-1-2-3-4-5 1 2 3 4-2+3.46j 2+3.46j 0-5j 0+1.93j Im Re-1-2 2 1 2. a. For the function to be periodic x 1 ( t ) = x 1 ( t + T ) = sin(2 ( t + k 1 T 1 ) + sin( 2 ( t + k 2 T 2 )) where T 1 = 1 is the period of the first sine function and T 2 = 2 is the period of the second one, and there exist integer values k 1 and k 2 such that T = k 1 T 1 = k 2 T 2 . This says that a period T is equal to k 1 periods of the first sine function and also equal to k 2 periods of the second one. k 1 p 1 = k 2 p 2 k 1 k 2 = p 2 p 1 = 2 1 Since 2 is an irrational number and cant be represented by a fraction of two integers, there are...
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This note was uploaded on 02/26/2010 for the course EE 30446 at USC.

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Solution2 - EE 200 Fall 2009 (Weber) Homework 2 Solutions...

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