ch06 - SIX Transient Response Stability SOLUTIONS TO CASE...

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S I X Transient Response Stability SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Stability Design via Gain From the antenna control challenge of Chapter 5, () 32 76.39 151.32 198 76.39 K Ts ss s K = ++ + Make a Routh table: s 3 1 198 s 2 151.32 76.39 K s 1 29961.36 76.39 151.32 K 0 s 0 76.39 K 0 From the s 1 row, 392.2. K < From the s 0 row, 0 K < . Therefore, 0 392.2. K < < UFSS Vehicle: Stability Design via Gain ( ) ( ) ( ) 1 1 2 1 2 432 31 2 1 3 0.125 0.437 2 21 . 2 90 . 1 9 3 1 0.25 0.10925 3.483 3.465 0.60719 0.25 0.10925 3.483 3.465 0.60719 0.25 0.10925 1 3.483 3.465 0.25 2.42 s G s s G G Gs s G sss s GK G sK G s s K Gs s s s K −+ = + = +− −− = +++ =− + = + == + + + 1 88 0.10925 s K +
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6-2 Chapter 6: Transient Response Stability s 4 1 3.465 0.10925 K 1 s 3 3.483 ( ) 1 0.25 2.4288 K + 0 s 2 () 1 1 45.84 4 3.483 K −− 0.10925 K 1 0 s 1 ( ) 11 1 4.2141 26.42 0.25 45.84 KK K +− 0 0 s 0 0.10925 K 1 0 0 1 For stability :0 26.42 K << ANSWERS TO REVIEW QUESTIONS 1. Natural response 2. It grows without bound 3. It would destroy itself or hit limit stops 4. Sinusoidal inputs of the same frequency as the natural response yield unbounded responses even though the sinusoidal input is bounded. 5. Poles must be in the left-half-plane or on the j ω axis. 6. The number of poles of the closed-loop transfer function that are in the left-half-plane, the right-half-plane, and on the j axis. 7. If there is an even polynomial of second order and the original polynomial is of fourth order, the original polynomial can be easily factored. 8. Just the way the arithmetic works out 9. The presence of an even polynomial that is a factor of the original polynomial 10. For the ease of finding coefficients below that row 11. It would affect the number of sign changes 12. Seven 13. No; it could have quadrantal poles. 14. None; the even polynomial has 2 right-half-plane poles and two left-half-plane poles. 15. Yes 16. Det 0 s −= IA
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Solutions to Problems 6-3 SOLUTIONS TO PROBLEMS 1. s 5 1 5 1 s 4 3 4 3 s 3 3.667 0 0 s 2 4 3 0 s 1 2.75 0 0 s 0 3 0 0 2 rhp; 3 lhp 2. The Routh array for () 532 658 2 0 Ps s s s s =+ + ++ is: s 5 1 6 8 s 4 θε 5 20 s 3 5 ε 20 s 2 5 20 s 10 θ 1 20 The auxiliary polynomial for row 4 is ( ) 2 52 0 Qs s = + , with ( ) 10 = , so there are two roots half-plane. The balance, one root must be in the left half-plane. on the j ω -axis. The first column shows two sign changes so there are two roots on the right
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6-4 Chapter 6: Transient Response Stability 3. s 5 1 4 3 s 4 1 4 2 s 3 ε 1 0 s 2 14 2 0 s 1 2 21 4 + 0 0 s 0 2 0 0 3 rhp, 2 lhp 4. s 5 1 3 2 s 4 1 3 2 s 3 2 3 ROZ s 2 3 4 s 1 1/3 s 0 4 () ( ) ( ) Even 4 :4 ;Rest 1 :1rhp;Total 5 :1rhp;4 j j ω 5. s 4 1 5 6 s 3 4 8 0 s 2 3 6 0 s 1 6 0 0 ROZ s 0 6 0 0 () ( ) Ev en2:2 ;Re s t2:2 lhp ;To ta l :2 ;2 jj ωω
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Solutions to Problems 6-5 6. s 6 1 6 1 6 s 5 1 0 1 s 4 6 0 6 s 3 24 0 0 ROZ s 2 ε 6 s 1 144/ 0 s 0 6 () ( ) Even 4 :2rhp;2lhp;Rest 2 :1rhp;1lhp;Total:3rhp;3lhp 7. Program: %–det ([si() si(); sj() sj()])/sj() %Template for use in each cell.
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ch06 - SIX Transient Response Stability SOLUTIONS TO CASE...

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