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HW2-solution(2)

# HW2-solution(2) - Flight Control Systems Homework#2 Dr...

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Unformatted text preview: Flight Control Systems Homework #2 Dr. Bishop - Fall 2007 Due Date: Thursday, September 13, 2007 Reading Assignment: Modem Control Systems: Chapter 2, Sect. 2.1 - 2.6, 2.7 (optional), 2.8 - 2.11 NOTE: The wading on algal—ﬂow gnplu in for your information only. You will not be mud on this “will. 1. Exercises: E228 2-3. Problems: P234, P251 4. Advanced Problems: AP2.5 5-6. Computer Problems: CP2.5, CP2.9 E238 (a) If 1 G(8)—m and H(8)—28+15, then the closed-loop transfer function of Figure E2.28(a.) and (b) (in Dorf 8: Bishop) are equivalent. (b) The closed-loop transfer function is 1 T“) = 92+17e+65' P2.34 The closed-loop transfer ﬁmction is Y(9) = 03(8)01(8)(02(9) + K5K6) 3(8) 1 - 03(8)(H1(8) + K6) + Ga(s)GI(8)(Gz(8) + K5K6)(H2(8) + K4) I P2.51 (a) The closed-loop transfer function is 14000 33 + 45.92 + 31003 + 14500 ' T(s) = (b) The poles of T(s) are 31 = —5 and 32.3 = —20 :1: 350. (c) The partial fraction expansion (with a step input) is 0.9655 _ 1.0275 0.0310 + 0.03907 + 0.0310 — 003907 N”): s 3+5+ s+20+j50 s+20—j50 0.9- f“ os- 01 as 0.5' Amplitude 0.4- 03- 02- am 0 0.2 0.4 0.6 as i {2 1.4 1.5 is 2 Time (secs) FIGURE P2.51 Step response. ((1) The step response is shown in Figure P2.51. The real root dominates the response. (e) The ﬁnal value of y(t) is yes = [1551.6st = 0.9655 . AP2.5 Considering the motion of each mass, we have M3533 + (13533 + kars = "3 + 03532 + 193232 M252 + (bz + b3)i‘2 + (k2 + k3)1‘2 = “2 + bsﬁi‘a + 13313 4- 5211 + 162531 M151 -|- an + b2)i1 + (k1 +k2)11 = 111 + b21532 + 16222 In matrix form the three equations can be written as M1 0 0 5‘1 bi + 112 -b2 0 1‘1 0 M2 0 5‘2 + -bz bz + ba -b3 1‘2 0 0 M3 5133 0 —b3 b3 i3 ’61 + k2 —k2 0 £1 pP2.5 The spacecraft simulations are shown in Figure CP2.5. We see that as J is decreased, the time to settle down decreases. Also, the overhoot from 10° decreases as J decreases. Thus, the performance seems to get better (in some sense) as J decreases. Noni-i (solid: miml mull-d]: MIMI swam-d) Sputnik ncua- my 1i" Ind 96Part (a) 3:1; b=8; k=10.8e+08; J:10.8e+08; num=k'[1 a]; den=J'[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[l]); 96 96 Part (D) and (c) t=[0:0.l:100]; 96 96 Nominal case f=10'pI/180; sysf=sys_cl'f; y=step(sysf.t); 96 96 Off-nominal use 8096 J:lO.8e+08'0.8; den=J'[‘l b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[‘l 1); sysf=sys_cl'f; y1=step(sysf,t); 96 96 Off-nominal case 5096 J:10.8e+08'0.5;den=1'[1 b 0 0]; sys=tf(num,den); sys_cl=feedback(sys,[‘l ]); sysf=sys_c|'f; y2=step(sysf,t); 96 plot(t.y'180/pl,t,y1‘lﬂO/pl,‘--‘,t,y2'180/pl,':'),gtld xlabel("l1me (sed') ylabel('Spacecraft attitude (deg)') title('Nomlnal (solid); Off-nominal 8096 (dashed); Off-nominal 5096 (dotted)') FIGURE CP2.5 Step responses for the nominal and off-nominal spaoemft parameters. CP2.9 (a,b) Computing the closed-loop transfer function yields 0(9) _82-I-28+1 T“)=Wi‘m- The poles are 8 = —3, —1 and the zeros are 8 = —1,—1. (c) Yes, there is one pole-zero cancellation. The transfer function (after pole-zero cancellation) is 5+1 9+3 ' ((1) Only after all pole-zero cancellations have occurred is the transfer function of minimal complexity obtained. T(s) = vol-m Map 0‘ M 0.! 1-02 H 7-1.5 M 1-Is H H5 0 hulk! 5 >> Transfer function: 09=U 1]:¢9=[12]:SY59 ='ﬂng.d9): sA2+2 s+1 nh=[1]; dh=[‘l 1]; sysh =tf(nh.dh): ----- — 9‘sysﬂeemacklsysgﬁyﬂﬂ 5A2 + 4 s + 3 WMSYS) 96 p : News) wows) -3 -l 1805 FIGURE CPZJ Pole-zero map. ...
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