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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( ωt + 3 2 π ) → 5 e j 3 2 π = 5(0- j ) = 0- 5 j 4 cos( ωt + 2 3 π ) → 4 e j 2 3 π = 4(- 1 2 + j √ 3 2 ) =- 2 + 3 . 46 j 4 cos( ωt + 1 3 π ) → 4 e j 1 3 π = 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 π/ 2. x ( t ) = 1 . 93 cos( ωt + π/ 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 π ( t + k 1 T 1 ) + sin( √ 2 π ( 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 can’t be represented by a fraction of two integers, there are...
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Solution2 - EE 200 – Fall 2009(Weber Homework 2 Solutions...

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