hw 3 solutions - DCT-EE-EEEB 13:32_FIEE;H...

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Unformatted text preview: DCT-EE-EEEB 13:32 '_FIEE;H STFIEHU 'ENLiTNEEH'I NLi __5'H3_Htfi_ SEEM _ F'Len Jt‘e‘zi -W Homework Set #3: MAE 4720/7720 — Modern Control Fall 2008 Due October 7, 2003 2‘ D P.” Problem 1: Consider again the motion of an unpowered (gliding) reusable launch vehicle (RLV) first introduced in HW#1. The vehicle parameters are Vehicle mass m = 4000 slugs Wing reference area S = 1600 ft2 Zero—lift drag coefficient CD0 = 0.036 Induced-drag coefficient K = 0.22 Atmospheric density model parameters: pSL = 0.0023 77 slugs/fig )6 = 30,500 ft The (instantaneous) trajectory state we wish to linearize the system about is: Velocity W = 480 ft/s Flight-path angle y“ = -10 deg Down-range distance x’“ = 453,500 ft Altitude if“ = 10,000 ft Note that we can compute the required (reference) lift coefficient C; from the governing equation of motion for flightspath angle (3/) and the requirement that the reference flight- path angle y“ is constant. 19* a) Is this system controllable? Justify your answer. l O b) Suppose a ground—based tracking station measures line-of-sight range to the RLV, f and the elevation angle of the line-of-sight. Using simple trig relations, we can compute altitude (h) and down-range distance (x) from these two measurements. Assuming we know the reference states, WU), #0), h*(r), and x*(t), is this system observable? Problem 2: Is the folloWing system controllable and observable? Work the problem “by hand," and check your answer with Matlab. “[303 -4.6]X+[0]u y=[1 0]! 9.1 —6.7 4 DCT-EE-EEEB 13:32 '_FIEE;H STFIEHU 'ENLiTNEEH'I NLi __5'H3_Htfi_ SEEM _ PT 1213 Big Problem 3: Use the BER in Problem 2. 5 a) Compute the “open-loop" eigenvalues of the system, and determine the system’s --"""" damping ratio 4' and undamped natural frequency a)”. l 0 b) Suppose we want to increase the system's dainping ratio to §= 0.7071 and undamped ..--"' natural frequency to a)" = 11.3137 rad/s. Find the required feedback gain matrix K Using the “CCF transformation method” or T matrix. Show your work (work “by hand"). ' lb 0) Repeat part b, but use Ackermarm‘s formula, and work the problem “by hand.” “'6 d) Repeat part b, but use the “direct substitution method,” and work the problem “by L; hand." 5' e) Verify your solution using Matlab and the Mfiles “ackerm” or “placem.” ”if. f) Simulate the closed-loop response using Matlab or Simulink. Let the initial state be ill-e x(0) = [1.4 — 22.3]1r . Plot both states and the control vs. time. DCT-EE-EEEB 13:32 "MEuH &_HEHU'ENUTNEEHINU “5v3‘aaa‘5050 Homework Set #3: MAE 4720/7720 — Modern Control, Fall 2008 Solution Prob.1 % Mfila for Hw#3, Fall 2000, MAE 4?20/7720 $3. 0 confitanta g = 32.174: % ft/s“2 S = 1600: % ft“2 m = 4000: 0 slugs rhoSL = 0.002377; 0 slmgfi/fit“3 beta = 30500; 0 ft 000 = 0.036; K = 0.22: 2 Ref 000005 {V*, gam”, etmm) V = 400; 0 ft/S gam = -lO*pi/100; 0 rad x I -53500; 0 ft h = 10000; 0 ft 'IhD = rhoSL*exp(-hfbeta); 0 filugg/ft"3 Qbar = 0.5*rho*V“2; 0 psi CL . m*g*cos(gam)/(Qbar*5); 0 CD - C00 + K*CL“2; lift cowff L0 maintain gamma” 0 partiala f0: gliding Eq5 of motion all = -rhoSL*exp(—h/beta)*v*s*(CDO+K*CL“2)/m; 0 Vth/V .312 . —g*CQE[gam); Fa mot/gamma a13 = 0; 0 Vdfltfx a14 I O.5*rhoSL/beta*exp(-h/beta)*V“2*S*(CDO+K*CL“2)/m; 0 VdoL/h a21 = D.5*rhoSL*exp(-h/beta)*S*CL/m+g*cos(gam)/V“2: 0 gamdct/V a22 = 9*5in(gam)/V; in gamdot/cgwuuna a23 = O; 0 gamdmtfx a24 = -0.5*thSL/beta*exp(-h/beta)*V*5*CL/m7 *5 Wmet/h a31 = 505(gam): % Kdot/V a32 = -V*sin(gam); 0 Hdotfgamma a33 n 0; 0 xdot/x a34 = 0; 0 det/h a41 = sin(gam); 0 hdwt/V a42 = V*cos(gam); 0 hdot/gamma E43 = 0; J tht/K add m 0; 0 0000/0 bll = -rhoSL*exp(-h/beta)*V‘2*S*K*CL/m; 0 VdntICL 1321 = 0.5*rhoSL*exp(—h/beta)*V*S/m; % gamthx’CL b31 = 0; % xdmu/CL b41 = 0; 0 tht/CL _ PTUJ DCT-EE-EEEB 13:32 "MELH ETHEHU'ENUTNEEHINU __573_HHZ_SBHU _ PJU4 % Define 83R 0 states xetate w [ V gamma 3 h j' a control 0 m CL A - [ all alE a13 ald ; a21 a22 a23 a24 ; a31 a32 a33 a34 ; adl a42 a43 add 1 B F [ bll : b21 } b31 ; b4l] C = [ 0 0 l O : 0 O 0 l 1; D n [ 0 ' 0 ]; % controllebilihy/ebsexvabllity cheek Q = Ctrb(ArB); rank_Q = rank(Q) N I obsvtA,C): rank_N = rank(NJ A = -0.0235 -31.6852 0 0.0002 0.0003 —0.01 16 0 -0.0000 0.9848 83.3511 0 0 01736 472.7077 0 0 B = -13.9415 0.1644 0 0 rank_Q = 4 9 controllable (firll rank = n = 4) rank_N a 4 9 observable (full rank = n = 4) Recall that the states of the linearized system are (3:: = [ 5V 5;! 5h cir1T ; so if we know the reference states (W, y", 11*, x*) and measured states I: and x, then we can get the perturbation states 6?: and fix, which are required for the observer. LIL; | —EJ—EUUH 15: dd WtLH E: HtHLI IzNLi 1 NI:le 1 NH b'r’d BEE-4 blfl‘jlfl I" . Ub- -- 0 Was 2 A: 4 6113 = _. G, 7 4' c: - l c: ConMMdMH-V OAeglg Pink '94? Q = E 5' A3] A13 - 3.3 “4.6:Ha] .. ‘""9"'/ ' 9.. -(,.1 4 "' 1M 1*? (3:03 +341 mnkCQ):2.:h‘/ 4- —2.e.'3 if. Cam‘fr‘odmufl 21::— Obmaltvr GAao/c [21M c170 N: [ I __ o . -4. CA - [L ][::: #6:} u; [3.3 r4419] IN 1 L I D ] _ FflnJ¢{N): n/ ' [No‘l't' dag”): -4.e M] W'- ~>> A: E .,_3 :3 [5: {W} 33> C‘- [l a? 2:) Q: C‘H‘ECAJB) >> PankCQ) >3 N‘- obsv(A,C) >> MnkCN) LIL' | —Ej—EUUH 15: dd P‘llzL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUEM |-‘ . [db Prob. 2 Matlab solutiom }}A=[3.3 -4.6;9.1-6.7]; }}B==[0; 4]; :5} Q = otrb(A,B) Q = 0 —lB.4000 4.0000 -26.8000 3'3"“ rank(Q) ans = 2 -} rank(Q) 2 n = 2 “full rank", so system is controllable 32+} N = obsv(A,C) N = 1.0000 0 3.3000 4.6000 2»: rank(N) HHS E 2 9' rankCN) = n = 2 “full rank”, so system is observable LIL' | —Ej—EUUH 15: dd MtL'H E: HtHLI IzNLillelelNLi {Zr-Yd HES-4 DUEM |-‘ . k1? P1205 3 61) Oflen-Loaf; BEBEAUM: AchhI—A) = aha *3" 4* ~94 )\[email protected] = )3 + 3,4} 4- :9.75 = o .1 )‘m. : "If? :54.|0¢I age/HM? 2e F—f—ML 6-0“ ‘- 1371+ 4.:06: = 4,444] rial/.5 ‘/ ed = 4m“{1_'”i) : wan—(4%") = $7.51“ [.7 LIL' | —Ej—EUUH 15: dd MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUEM |-‘ . UH Pflofs 3 (b) DE‘SiFCaF 0105224.: Loaf Pea/2r haw-a '5' = £5.70 '7] (Fl-ad Cd” 1 ”-3137 hid/5 ' Claim—Loaf “575491 ; f + 2.33:4") + cunt“ = C) fit an... )‘“'+IG3~ + 12.3 :0 T T DH #1. "- Flhao £634 bit-GA. 34,6115 k, V l‘fl— (I: T ' Neflfi'x h Aflfldfifl $4.32 1 = = [E ABJT- (:4- 0 :feqfi‘] (flu hawk) SEE 2. -. Ofiéfg wiper CWIzfi'I‘C. £63: detIXIH'AB: “A" 4- 3+) + 19:75 1 o a 2% gt? 3: T: MW - WM W = [fl 0’] : [3:]. C: j _ o 494 34 I “’3‘“ 0 97‘: T? [4' —ze.3][1 0] : ['52- 4'] Bile 4 ._ /<" [oh-‘13» “I'fluj T4 K: [me—19.75“ lb-3J+J "0'05“ 05.]: [“B'NZ? 3”] LIL' | —Ej—EUUH 15: dd MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUEM |-‘ . EH [9305 3c = Amer-Mum's. "Fm-MM“, K = [a \jM“¢/A) Who-(E 9501) = Ah+ MIA+oczI (11:2,) A ; 3-3 -4-.6 2' ’35.; “1%, [13-6 0 I 61% J [9.1 FM] + lé[ 9.: earl] + a :28 A?! ”(I A “3- I 4:0” : 149.31. —5'7.,9¢ H446 2.3.8 m '1 _ -0r3&~‘+l 0-2.5 '- —o.u§43 D “@3641 0.2; New: #5196 a» [a w M “4.54:. 23-8 LIL' | —Ej—EUUH 15: dd MtL'H E: HtHLI tNLillelelNLi {Zr-Yd HES-4 DUEM |-‘ . 11:1 P110354! Dired— was'l-i‘lwl-u'oh: {3+ A“: A-EK Ac; = 3.3. *416: sat-41c, "6.741% Maw} CoMPw‘l-L EIBEAuwg-LEJ' 0,0 Ac; Ae+ [AI~A¢L] .-.. ae+["‘3'3 4"” l 9.: + 4k. k + 5:: +419 1 ' -1311: = A +[3,4+4k_1)>~ +(l9-7F-13AkT): W W =“' :11: clan": I33 =Z'S'r-uh =Ié Seaman-D '. k1 = H: —'3-’-I = 3‘ 15" If]. F zero—fl WW- H 19.75 ~J3.LkL _|i.4—kl “=- “a. 1 :23 a» k. = dig—LL: -3 [4-2.9 43.4 = ' .'. K= [1‘ £43.]: [mama 1:5] ‘/ DCT-EE-EEEB 13: 34 Prob. 3 MECH 8: FIEREI ENGINEERING e) Matlab solution: a:- A= [ 3.3 -4.5; 9.1-6.7]; }}B=[0;4]; s: 1amCL = [ —8+j*3 ; -8-j*8] lamCL = -3.0000 + SDOOOi -8.0000 - 8.0000i 3*} K = plaee(A,B,larnCL) K: -8.l429 3.1500 feedback gains f) Simulink model and simulation results: E Tu mmmz Tn Worlaaace State-Space ‘I'n Werlepeee1 Feedback gain matrix —|It Cl Cluck TO \Muflmaees or, use is} K = aeker(A,B,lamCL) 5T3 3384—3113111 desired closed-loop poles (eigenvalues) Note: enter A, B, C, and D matrices in SSR block; use C = eye(2) = 2x2 identity matrix since we must have “full-state” feedback. Enter the initial eenditions in the SSR block. A minus sign has been added to the gain matrix K since the control is u = -Kx _|-’.'11 EICT-EB-EEIE’IB 13:34 NECH 8c FIEREI ENGINEERING 573 BHJ'EMHU _ I-‘flid Closedwloop simulation results: states 1 and 2 vs time slate 1. x1 EICT-EB-EEIE’IB 13:34 NECH 8c FIEREI ENGINEERING 573 BHJ'EMHU _ I-‘JJ MAE 4720. Hw#3. Problem 3: Control w. time Cclntncll1 u TEITFIL F' . 13 ...
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