EE_365_200708FA_homework_07_solution

EE_365_200708FA_homework_07_solution - 1. sc sa sb s EE 365...

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1 Homework 07 Solution © Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 s ϕ sa a s sb b s sc c s 2c o s ( ) 2 cos( 120 ) 2 cos( 120 ) sa s e is sb s e is sc s e is iI t t t ωϕ =+ = ++ ° = +− ° (, ) ) ) ) cos( ) 2 cos( ) cos( 120 ) 2 cos( 120 ) cos( 120 ) 2 cos( 120 ) 2 cos( ) cos( 120 )cos( 120 ) cos( 120 gs s gsa s gsb s gsc s ssse i s ss se i s i s s s ei s s s s tttt nI t n I t n I t In t t ϕϕϕ ϕω ω =++ = + + + + −° + + + ° + + ° + ° FF F F [ ] () 3 2 )cos( 120 ) o s ( ) s s s t t ϕωϕ ° + Maximum mmf occurs when the argument of the cos(·) is zero: (0 ) 0 s is s is t ϕϕ +⋅+ = ⇒ = Track a constant value of the wave: i s tK ++= s s Kt ⇒= + [] s s dd dt dt d dt =− + 1.
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2 Homework 07 Solution © Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 3 2 3 2 (0) 2 ( ) (0) 2 (0) (0) 2 ( ) sa s sb s sc s iI = = =− 1 2 1 2 (30 ) 2 (1) (30 ) 2 ( ) (30 ) 2 ( ) sa s sb s sc s °= 3 2 3 2 (60 ) 2 ( ) (60 ) 2 ( ) (60 ) 2 (0) sa s sb s sc s 1 2 1 2 (90 ) 2 ( ) (90 ) 2 ( 1) (90 ) 2 ( ) sa s sb s sc s 2c o s ( 3 0 ) o s ( 9 0 ) 2 cos( 150 ) sa s e sb s e sc s e t t t ω ° =+ ° ° a. b. 0 t = 30 t = ° 60 t = ° 90 t = ° 2. s ϕ sa a s sb b s sc c s
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3 Homework 07 Solution © Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 2 π π s ϕ 0 B gs 2 π π s 0 B gs c. d. 2. (cont.) 2c o s ( 3 0 ) o s ( 9 0 ) 2 cos( 150 ) sa s e sb s e sc s e iI t t t ω =− ° =+ ° ° s sa a s sb b s sc c s
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4 Homework 07 Solution © Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 r θ sa a s sb b s sc c s ra ra rb rb rc rc ra axis sa axis rc rcrc rb rcrb ra rcra sc r sb rcsb sa rcsa rc rc rbrc rb rbrb ra rbra sc rbsc sb rbsb sa rbsa rb rc rarc rb rarb ra rara sc rasc sb rasb sa rasa ra rc scrc rb scrb ra scra sc s sb scsb sa scsa sc rc sbrc rb sbrb ra sbra sc sbsc sb sbsb sa sbsa sb rc sarc rb sarb ra sara sc sasc sb sasb sa sasa sa i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L i L + + + + + = + + + + + = + + + + + = + + + + + = + + + + + = + + + + + = csc csc λ 22 rr rara rbrb rcrc lr mr lr m nn LLLL L === + = + s s sasa sbsb scsc ls ms ls m L + = + 1 2 cos( ) cos(0 120 ) sbsa ms sa sb ms ms LL L L α =− ° 1 2 cos(120 ) rcra rarc rbrc rcrb rarb rbra mr mr LLLLLLL L ====== ° = 1 2 scsa sasc sbsc scsb sasb sbsa ms LLLLLL L 3.
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5 Homework 07 Solution © Jeffrey Mayer 2005-2007. All rights reserved.
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This note was uploaded on 10/08/2009 for the course EE 365 at Penn State.

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EE_365_200708FA_homework_07_solution - 1. sc sa sb s EE 365...

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