exam 1 solutions

exam 1 solutions - mus—ee—sees 1e:ee P'llzL'H a HtHLI...

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Unformatted text preview: mus—ee—sees 1e:ee P'llzL'H a HtHLI IzNLithlelNLi eve 554 seen |-'.l:11 F 0 2467 A V3 '. 35- a MAE 4720/7720 — Modern Control, Fall 2008 Exam I — Closed Book (30 pts, 30% of the total points) Definitions/Short Answers: 1. (5 pts) True or False: After using the linear transfonnation z = Fit, the eigenvalues associated with “new” SSR (for state vector 1) are equal to the transformation matrix P multiplied by the eigenvalues of the “old” SSR (for state vector x). 2. (5 pts) Define the term observability for a dynamic system. 3. (5 pts). Describe one method for computing the state transition matrix, (DU). [an equation is not necessary, but may help]. Mi 4. (15 pts) Suppose we have a general nth-order nonlinear system with a single (scalar) control input, u: s = am) We wish to linearize the system about a nominal reference state x*(r) that results from the reference control u*(r). Show the form for the state equation for the linear model. Define fl terms, and give their respective dimensions. NUU—Ub—EUUH llfllblfl MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUHU I-‘. .-._.,_E¥&H_mw. h—LLWnNt- T_.Lu-éuU—_._J_m___._d=&. . d__ CtoSed Beak rx 1 LL Fdfie. :5th: rem‘n Canshnf- ) :23; Obi-Ervfibbl‘lrl'l‘j '- 4. 575% (‘5' oéIHM/Ew WEED Quay fiftie- Cad-n b6. Jflwmhflr £11 53 9‘9"“ fl. 'Q‘M‘f-L Maura! sac 0%fEfi-Jf‘ dike Wt); we“; *4: 5%: A1 a; COMPL-Jh‘fifl STM, E (1) = a *9 i) Ingflrififl SEn'as sated-:3“ '- . a. IL: 3.63 E'Zzt)..I +At+AET+Afi+_H r‘a Laplace. {Pia-+£1.an '. = fh'[.(S’-I fill)- C) NMMQH'Cway in figmie STM ODE .- ém = A £62) ,, £52): 3: palladium-«(4.39.935 Mai-had '- l:- Penal-f-) _E (‘3 wduhgmaé'ahj“ 5:. ML iv" = rm, M ) 3% a( hem-3: 3&2: A2; + Mu. WM 33?: irfiw , Su=u~u* A = MEI-er 9x A: fl E = n X, H" SX‘ mu 3 3c: Efiilar C NUU—Ub—EUUH 11:1: '31 MtL'H E: HtHLI IzNLillelelNLi {Zr-Yd BEE-4 DUHU |-‘ . [£15 MAE 4722017720 — Modern Control, Fall 2008 Exam I — Open Book (70 pts, 70% of the total points) Problem 1. A state equation is given below: , 0 l 0 x = x+ u [0 —0.6] [0.3] a) (5 pts) Compute the open-loop system eigenvalues. b) (10 pts) If the initial state vector is x(0) = [—1 2y, then compute the “free response” at time t = 0.1 see. That is, compute the state vector at 1.“ = 0.1 sec for the case with zero input [i.e., u(r) = O fer t3 0]. Show your steps. c) (10 pts) Prove that the system is controllable. d) (10 pts) Pole—placement has been used to determine the closed-loop control law a = 4.53:1 —l.75x1. Compute the resulting closed-loop system eigenvalues with this control law. Problem 2. (10 pts) Consider again the system in Problem 1. Suppose we solve the optimal steadystate LQR problem with the following state and control weighting matrices: 10 0 Q ‘- l: 0 10] State—weighting matrix R = 20 State-weighting “matrix” (scalar) The resulting steady-state Riccati matn'x (i.e., solution to the ARE) is __ 23.79 17.68 17.68 23.31 Determine the optimal steady—state gain matrix I? for this LQR problem. Problem 3 is on the back of this page NUU—Ub—EUUH lUlbl MtEH & HtHU tNUthtHlNU 5T5 HH4 DUHU Problem 3. Consider the optimal control problem for rotating a satellite from an “at— rest" angular position to the vicinity of the “origin” (i.e., zero angular position and zero angular velocity). The two states are x1 = angular position (rad) and x2 = angular velocity (rad/s). The control um is the applied torque from firing small thrusters on the periphery of the satellite. We wish to minimize a combination of the control effort (thruster fuel) and the deviation of the states from the origin at t = T. Therefore, the optimal control problem is T J-uzdt 0 Minimize J = 103:? (T) + iooxfi (T) +% subject to: icl = r2 , i2 =u with initial conditions xl(0) = 1 rad, 15(0) = 0 with fixed end-time T = 30 see. a) (20 pts) Set up the complete two-point boundary value problem (2PBVP). b) (5 pts) Qualitatively describe the nature of the optimal thruster control, u*(t). “Extra Credit“ Solve the BPBVP and sketch the optimal solutions x1*(t), x2*(t), and #0). F. U4 5'35 554 bUHU |-'.l:1'::- J NUU-Ub-EUUH 11:11:51 MtL'H & HtHLI IzNLillelelNLi ‘ z . .. __. ‘I ,. _ -_- V; "I" M..- .. __.....‘. .- _LF.,_.._..J—__,_8 _ ___.._. . E ': Sear-fag Sflaon-I'. g/f'): I + A? 4- A3? 4. u- a D —'0:é 2- c: 0.36:. i 14 '1'- , A3 = { a o § . 0:94/3 => mg: I W" -’ ___ ‘09”? a 0.9413 2 1.9335” NUV—Ub— I EUUH lUlbl 1— . _,.___._H "Hi i P0205. |L)Ca_r_1;t: RIF-I‘M M313 an. upper— FDA?“ Erin. (5144) ’ 1 .5 Mr - “fife c: _.L,__ 5+ flute ..f L {.647 L667 — R0 3 s ‘ S+¢.é.. a _r____ . Sal-0,6 O Enaét '6 :0! MtEH & HtHU tNUthtHlNU J 5T5 HH4 DUHU F. NUU—Ub—EUUH lUlbl _ _..__.___ _ . \ PEM- m -...__....H 15) MtEH &IHtHU tNUthtHlNU 5T5 HH4 DUHU NUU—UbeUUH 11:1: {Zr-11’ MtL'H E: ‘leHLl IzNLillelelNLi aw .m..._._,_m‘ Pram I '- 1Lez5-r- : § 2 I = O | D _ 0-8 5 I! A3 [a ho.61[°-9 ' (“0.643 I 2,.» M = a D's ‘hu 5‘ J { 0-3 ‘fi-‘l—S OMJ: i g WCM3=2 =11 (Sauce; da+(fl)¢a)’ [5 $ ¢5m%i(d(£ i i g A) came and a. = «5:: {2:5 1.75] :s g 1 L—"_"-r:-«-—I ‘ (9:9,? Sysku Media}? 1' K ‘A:A~Bk’= o']_"—”‘. i a!" [a -c..é. 52.3 [25 H's-Ti if: a g = L2 «a J g E deffAT-fl: - 43+[2’D‘ ‘ Ciflfé’af-c’mf: Act. : 5?; 554 swam ! I ) .._——,—4..im_-—. Cmmdéabdé‘z; #le mew NUU—Ub—EUUH lUle m___w ._ ._- _ L; 843%? u 5m DPWM (7m; Mm‘fl‘iy IJ 42y mm ,0me.& : i k ‘ I?”st i 4 ’ = 20 (BMW) 1! ET: [0 0.3 3 g p‘ _ 23.79 17-63 E 114,8 2%.?! i :7 F I: l? --. J. o 0.3 21.7?" I Ga J 2-0) [ J[:7.ea 23.3: E i $6517!" 5 /( = [67.7072— 0.9324] i i 3‘ MtEH & HtHU tNUthtHlNU 5T5 HH4 DUHU F. NUU—Ub—EUUH 11:1: {Zr-11’ MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUHU I-‘. 1111 ___L__; _l_fi ,. Prints 3 (9 t on 9 n a» M g; = f c9 c9 3“ r I ' « I X '1 L9 =- “2' Li. X _ ‘ .. :1; x = 2' 39Mm. X1 = :9 = a: “- Egg J- _ z 5.“- ' 2 I T a. fig m - 0 (1‘) + 1009(7) + i- u 4% 3% ° qusm'f‘ "'0? £1 = x; J RICE) = I Pic! m i; = H. J K: (a) = a (mas-r) “=1: HIth ‘9}!534 Encep‘h'ma "T" a 30 Jet. ~— :2.) Sefi- “(f9 ZFEVF" = , '—" A Hand/finian : u = L + Air-)0 f-x { 2. H: £65 + A'xl + )xacc - _ naH Cbg‘l'fifie- 0:35,: t >‘ak -— a! 1 O #b A. '3: Can sham. A2- "=- figfl -: "=§ (\2:,oé'fl.anu. *2 M (35-) [1.5+ ¢[?CT)_] = to 3520‘) +100 35:51") i 34$ #:1- NUU—Ub—EUUH 1 Id: {Zr-11’ > v .“_._—r—..—m_d._____—.—.w.- “ A I 'P flab 3 C Ca aft" ) rm EGSHEETS 22-142 mm SHEETS EE—‘E-fi-T ’ 22—?! r: _. .f‘ Oflfimaflb: ._..... a a +- A; «a C) I Sq, , : é um: = ——)\3 Own; mam, Consw‘un-r- e {s : c, -; Aaa') +5\,T a!) C“ = loaxch) + 2aTw,(T)= (If .- | i K MtL'H E: HtHLI IzNLi 1 NI:le 1 NLi {Zr-Yd HES-4 DUEM |-‘ . 1 l \ H v-I-w- a----—- - ----- - - -- ---——————~.--p m. —|— I kg (15) -—— flzox,(1')t + 200 XLCT)+20Tx'i(T) 53"?! [ I I 4 i A2KH= 20X.CT)[T-tj + .2on2 (-1—) {57% my e7. a-fl :2 (Ag) “$575) 1'! :2 1715:211- Dec-f. NLlU—lflb—Elfilfiti 1E“ '32 NtL'H 3: HtHLI IzNLillelelNLi b'l’j titi4 blfi‘jlfi |—'. 13 .. I PM 3 ‘*EK‘W cradr+>k 9 .....J_.. A ___ . m.— r.) gala LPE‘JP we. km”: Aime):- 20x,CT)[T~-b]-+ZOOXZCT) Ifl+%m 7:1. M xi. = —)‘L 22 141 2:1] SHEETS 22-142 1013 SHEETS 22 1441 Eur: SHEETS x2 (15): Kz(0) 4: 4079mm + IDXMTJtL-ZaanxJ'flt I Infim )2. 3m gar-- x L @945- x,(-&) = x, (a) + xzaot —- to mavn‘ + g? .035 ‘- I00 2:; (T) 152' fix I! Ndw/ em LGde 3:0de 42‘ 15 =_ T XICT) = XI“) “" “#507 u :DX,CT)T3 4- %XI(T)T3 ; —too Kac-r) T’- a /% X2 '2 x1(0) ‘20 X:(T)T1 + ‘0 ka'dT)T1_200xacT)-T M= 7‘1“): 1) X1(a)=o KMT) =- 1 ~ ~33? x.cTJT3 - loo xECT)T?‘ 3“; CT) = —f0 X; CT) T?“ - 2.0a high-)1“ fl QE, (+12%? :ao'r‘ Sad-r) J H lez‘ {+2ao-r XLCT) D NEIU-E’IE-EEEB 18:53 MECH 8: F‘IEREI ENGINEERING STE 884 BEBE P.13 “h:_______fl___mw ________L___wmh_____m__mHm_____J_____.__u________nu..h_%2;_4§§%§r I i 7:30 32;: ! ,ra 190,00! 90, Don x,(T) I 9000 @159] with) 0 mm «1‘ -. j 'x'(3a) 1-. 0.11-2—l (“D-'4) “we Egg 1:2 (3o) = ~—D.$ysl (10*) r“Ml/5 :1. MM) = 4.4426004) [Tat] - orwééé»Z I El. 5 % OIL. mark-bl f: = H?“ = —)‘2« “*(t) = -—o.ooééé‘t‘ + 4—.44‘2M/o‘o75‘ Thymung Chav1fl91flft J F Q 1 } i i l I TDTHL P.13 ...
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exam 1 solutions - mus—ee—sees 1e:ee P'llzL'H a HtHLI...

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