Question

Space vehicles using wings to maneuver while reentering the earth's atmosphere present an interesting control problem. The following figure indicates the block diagram of the pitch-rate control system of such a system. Select the values of K1=0 and K2 variable, obtain the Root-Locus plot of this system. Also determine the gain value for the damping ratio 5 = 0.707. Compute the step response of the closed-loop system at that value of gain K. Compare it with the step response of an equivalent second-order system based on dominant poles Use Routh-Hurwitz criterion to determine the range of K2 for stability. Desired pitch rate Kyls Vehicle Dynamics Actual pitch rate G(s) C(s) R(s) Rate Sensor H(S=1 (0.05s + 1} G(S) = (8 + 5s + 25) (10s + 1)(0.01s + 1)

Answer 1

ANSWER:

Solution given function Ges): 10.053 ty? 1 s?+53 +25) (105 +1)(0.013 +1) given block diagram vei tsu Da kллыг pie un rate o De sired pitch rate kils to 7 Actual pitch yo Gess to RES) Rate sensor HUP! damping ratio - ų = 0.707. Giselooss +1² (3455+25) (105+1)(0.01541) (5+20) 18054453125) 1st0.1) 154.100) Yoo 10% (5+2014 13'+53 +25) (540.1) (54100)

boss (32 +403 +400) 13+557232/3+10 + 100.15) 6(s) 13'+ 403 +400) (34101²+100.153 +593 +503 +500.55²+255²4 150 + 2502.55) (s? +40 3+400) ( 544 105.133+510.53²+2552.55 +250) K210.0253²+ 3410) 60): - (347105153 +510.554+2552.55+250) RH Criteria Chavarcler stics en I+G (HCS) (39+105,133 +510.53+2552,55 +256) 4k2 (0.02534+5+10) -0 s4 +105.13}+s? 1510.5+ kq 10.025)) +S(255265+k2 )+(150+1042) 510.5+k2 (0.025) 1250 +1062) 105.1 2552.5 +k2 ) s! $1101.05 - 1:6275K2 (250 +1042) 105.1 265.1(250+10K2) - (255-2.5 +K2) (5 1101-05-1.675k) 105-1 205.1(250+10ke ) - (2552.5+k2 )14862-0.0159 kq) 30 =) To become system stable

51101.05-1-6275k20 kze .slolos T. 6275 K2 2 313981 105.1 (250+10k 2) - (2552.5 tkz ) (486.2-0.0159 k2 )) ) 1051K2+26275 +0.0159 k2?-124102575 +62 (0.0159x2552.5 - 486.2) >6 =) 0.01594 + 60515 kq-1214750.5.0 K2=(-39992:12, 1910.36) ka > 1910:36 19110136<ka<31398 . For root locky: poly -0.11 -1001 -2.5 +43J 705-20 total nouot poly terminate at.oo = P-z =4-1= 3 i no of Angle B/w Asymplotes = 3 Asymploty = 180 229 +1) - ) 60°,120180° . P-Z -0.1-100-2.5x2720 centroid = -85.1

" o Boot locks will lic on the plane left side of tola 1 odd no. ot pole and zero Angle of deperture - fan (48) At 1-2.5.#4135) -tañ' ; Dt tañ' l ho vil +90° 0-215 -88.66 +90 +2.46 = 0.215 - 3.8 = 3.5850 At 1-2-5.4.3j) » -3585e Langle of deporture)