midterm solution fall 04

midterm solution fall 04 - Solution to Midterm I Problem 2:...

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Solution to Midterm I Problem 2: For departure angle, we have φ 2 = φ dep , φ 1 = 135 o 3 =90 o 1 = 225 o , and ψ 2 =90 o +tan - 1 (3 / 7) = 113 . 2 o . Hence φ dep = ψ 1 + ψ 2 - φ 1 - φ 3 - 180 o = 113 . 2 o - 180 o = - 66 . 8 o . Forarr iva lang le ,wehave φ 1 =tan - 1 (5), φ 2 =45 o , φ 3 =tan - 1 (7 / 3) 1 = ψ arr 2 =90 o . Hence ψ arr = φ 1 + φ 2 + φ 3 - ψ 2 + 180 o = - 79 . 5 o .F o r -axis crossing, we examine the roots of a ( s )+ Kb ( s )= s 3 +4 s 2 +8 s + K [ s 2 - 2 s + 26] = s 3 +(4+ K ) s 2 +(8 - 2 K ) s +26 K =0 . By Routh table, we have zero row for s 1 ,i f ( K + 4)(8 - 2 K ) - 26 K =0 has a positive real root. It can be verified that K 0 =1 . 1322 is the only positive root. Hence the auxiliary polynomial is obtained as (4 + K 0 ) s 2 +26 K 0 =0 ⇐⇒ s = ± j 2 . 395 , which is the two points at which root locus across. Problem 3: For (i) we compute first departure angle φ dep = φ 1 .S in c e φ 2 =90 o ,and
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midterm solution fall 04 - Solution to Midterm I Problem 2:...

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