EE_387_200809SP_homework_05_solution

# EE_387_200809SP_homework_05_solution - 1. sa sb ra rb ra sa...

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1 Homework 05 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 sa sasa sa sasb sb sara ra sarb rb sb sbsa sa sbsb sb sbra ra sbrb rb ra rasa sa rasb sb rara ra rarb rb rb rbsa sa rbsb sb rbra ra rbrb rb Li Li Li Li λ =+++ 22 0 0 s s sasa sbsb ls m sbsa sasb rr rara rbrb lr m rarb rbra NN LL == + ℜℜ + () 21 1 3 rs m sarb rbsa m π θ πθ ππ −− < < −< < 2 2 10 m sara rasa sbrb rbsb m LLLL θπ << ==== + −< < 1 3 m sbra rasb m < < < r 2 1 0 2 3 , sarb rbsa , sbra rasb 00 m m 0 m 0 m m 0 m 0 , , sara rasa sbrb rbsb r -axis r -axis s sa sa sb sb ra ra rb rb 1.

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2 Homework 05 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 r r ms r r r sr s ms ls s s s i L dt i d L dt di L L i r v ω π θ + + + = 2 )) sgn(sin( ) ( ) ( dc 0 . . 0 c o . . 0 c o A 1 . 0 10 100 1 = = = r s r r i N N i rad/s 377 s 60 min 1 r 1 rad 2 min r 3600 = = r V )) sgn(sin( 368 . 2 377 1 . 0 10 7 . 98 2 )) sgn(sin( 3 r r = × = mH 7 . 98 10 1 2 100 10 100 10 50
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## This note was uploaded on 10/08/2009 for the course EE 387 taught by Professor Jefferymayer during the Spring '08 term at Penn State.

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EE_387_200809SP_homework_05_solution - 1. sa sb ra rb ra sa...

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