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Unformatted text preview: DCTEEEEEB 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
Induceddrag 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 Flightpath angle y“ = 10 deg
Downrange 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 lineofsight range to the RLV,
f and the elevation angle of the lineofsight. Using simple trig relations, we can compute altitude (h) and downrange 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 DCTEEEEEB 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 “openloop" 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 closedloop response using Matlab or Simulink. Let the initial state be ille x(0) = [1.4 — 22.3]1r . Plot both states and the control vs. time. DCTEEEEEB 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 DCTEEEEEB 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 BEE4 blﬂ‘jlﬂ I" . Ub
 0
Was 2 A: 4 6113 =
_. G, 7 4'
c:  l c: ConMMdMHV 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 {ZrYd HES4 DUEM ‘ . [db Prob. 2 Matlab solutiom }}A=[3.3 4.6;9.16.7]; }}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 {ZrYd HES4 DUEM ‘ . k1? P1205 3 61) OﬂenLoaf; 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
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[.7 LIL'  —Ej—EUUH 15: dd MtL'H E: HtHLI tNLillelelNLi {ZrYd HES4 DUEM ‘ . UH Pﬂofs 3 (b)
DE‘SiFCaF 0105224.: Loaf Pea/2r hawa '5' = £5.70 '7]
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T T
DH #1. " Flhao £634 bitGA. 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
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97‘: T? [4' —ze.3][1 0] : ['52 4'] Bile 4 ._ /<" [oh‘13» “I'ﬂuj T4
K: [me—19.75“ lb3J+J "0'05“ 05.]: [“B'NZ? 3”] LIL'  —Ej—EUUH 15: dd MtL'H E: HtHLI tNLillelelNLi {ZrYd HES4 DUEM ‘ . EH [9305 3c = AmerMum's. "FmMM“, K = [a \jM“¢/A) Who(E 9501) = Ah+ MIA+oczI (11:2,)
A ; 33 4.6 2' ’35.; “1%, [136 0 I
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' —o.u§43 D “@3641 0.2; New: #5196
a» [a w M “4.54:. 238 LIL'  —Ej—EUUH 15: dd MtL'H E: HtHLI tNLillelelNLi {ZrYd HES4 DUEM ‘ . 11:1 P110354! Dired— was'li‘lwlu'oh: {3+ A“: AEK Ac; = 3.3. *416:
sat41c, "6.741% Maw} CoMPw‘lL EIBEAuwgLEJ' 0,0 Ac; Ae+ [AI~A¢L] ... ae+["‘3'3 4"” l 9.: + 4k. k + 5:: +419 1 ' 1311:
= A +[3,4+4k_1)>~ +(l97F13AkT):
W W
=“' :11: clan": I33
=Z'S'ruh =Ié
SeamanD '. 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 [42.9
43.4 = ' .'. K= [1‘ £43.]: [mama 1:5] ‘/ DCTEEEEEB 13: 34 Prob. 3 MECH 8: FIEREI ENGINEERING e) Matlab solution: a: A= [ 3.3 4.5; 9.16.7]; }}B=[0;4]; s: 1amCL = [ —8+j*3 ; 8j*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 StateSpace ‘I'n Werlepeee1 Feedback gain matrix —It Cl
Cluck TO \Muﬂmaees or, use is} K = aeker(A,B,lamCL) 5T3 3384—3113111 desired closedloop poles (eigenvalues) Note: enter A, B, C, and D matrices in SSR block; use C = eye(2) = 2x2 identity matrix
since we must have “fullstate” 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 EICTEBEEIE’IB 13:34 NECH 8c FIEREI ENGINEERING 573 BHJ'EMHU _ I‘flid Closedwloop simulation results: states 1 and 2 vs time slate 1. x1 EICTEBEEIE’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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