ch06 - S I X Stability SOLUTIONS TO CASE STUDIES CHALLENGES...

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S I X Stability SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Stability Design via Gain From the antenna control challenge of Chapter 5, T(s) = 76.39K s 3 +151.32s 2 +198s+76.39K Make a Routh table: s 3 1 198 s 2 151.32 76.39K s 1 29961.36-76.39K 151.32 0 s 0 76.39K 0 From the s 1 row, K<392.2. From the s 0 row, 0<K. Therefore, 0<K<392.2. UFSS Vehicle: Stability Design via Gain G 3 =− K 1 2 3 = (0.25 s G + 0.10925) K 1 s 4 + 3.483 s 3 + 3.465 s 2 + 0.60719 s G T ( s ) = G 3 ( s ) 1 + G 3 ( s ) = s + 0.10925) K 1 s 4 + 3.483 s 3 + 3.465 s 2 + 0.25( K 1 + 2.4288) s + 0.10925 K 1
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204 Chapter 6: Stability s 4 1 3.465 0.10925K 1 s 3 3.483 0.25(K 1 +2.4288) 0 s 2 - 1 4 (K 1 - 45.84) 3.483 0.10925K 1 0 s 1 0.25 (K 1 + 4.2141)(K 1 - 26.42) K 1 - 45.84 0 0 s 0 0.10925K 1 0 0 For stability : 0 < K 1 < 26.42 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 ω 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 (s I - A ) = 0
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Solutions to Design Problems 205 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. s 5 1 4 3 s 4 -1 -4 -2 s 3 ε 1 0 s 2 1 4 -2 0 s 1 2 2 + 1 4 1 4 0 0 s 0 -2 0 0 3 rhp, 2 lhp 3. 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
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206 Chapter 6: Stability Even (4): 4 j ω; Rest(1): 1 rhp; Total (5): 1 rhp; 4 j ω 4. s 4 1 8 15 s 3 4 20 0 s 2 3 15 0 s 1 6 0 0 ROZ s 0 15 0 0 Even (2): 2 j ω ; Rest (2): 2 lhp; Total: 2 j ω ; 2 lhp 5. 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/ ε 0 s 0 -6 Even (4): 2 rhp; 2 lhp; Rest (2): 1 rhp; 1 lhp; Total: 3 rhp; 3 lhp 6. Program: den=[1 1 -6 0 1 1 -6] A=roots(den) Computer response : den = 1 1 -6 0 1 1 -6 A = -3.0000 2.0000 -0.7071 + 0.7071i -0.7071 - 0.7071i 0.7071 + 0.7071i 0.7071 - 0.7071i 7. Program:
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Solutions to Design Problems 207 %-det([si() si();sj() sj()])/sj() %Template for use in each cell. syms e %Construct a symbolic object for %epsilon. %%%%%%%%%%%%%%%%%%%%%%%%%%$$$$$$$$$%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% s5=[1 4 3 0 0] %Create s^5 row of Routh table. %%%%%%%%%%%%%%%%%%%%%%%%%%%$$$$$$$$$%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% s4=[-1 -4 -2 0 0] %Create s^4 row of Routh table. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% if -det([s5(1) s5(2);s4(1) s4(2)])/s4(1)==0 s3=[e. .. -det([s5(1) s5(3);s4(1) s4(3)])/s4(1) 0 0]; %Create s^3 row of Routh table %if 1st element is 0. else s3=[-det([s5(1) s5(2);s4(1) s4(2)])/s4(1). .. -det([s5(1) s5(3);s4(1) s4(3)])/s4(1) 0 0]; %Create s^3 row of Routh table %if 1st element is not zero. end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% if -det([s4(1) s4(2);s3(1) s3(2)])/s3(1)==0 s2=[e . ..
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This note was uploaded on 10/01/2008 for the course ENGR 3350 taught by Professor Esmailzadeh during the Spring '08 term at UOIT.

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ch06 - S I X Stability SOLUTIONS TO CASE STUDIES CHALLENGES...

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