Question

4. A lead compensator with a transfer function Ge(s) = K(s+0.5)/(s+3) has been designed for a Space vehicle with the transfer function 1/s? such that at the dominant closed loop poles are located at -1 +/-jl. (1) What is the angle deficiency of the uncompensated system at the designed point provided by the location of the dominant poles? Show that the compensator provides the necessary lead angle at the designed point to satisfy the root locus angle criterion. (iii) What value of K will satisfy the root locus magnitude criterion and move the dominant closed-loop poles to the desired location? (iv) Draw an approximate root locus of the feedforward transfer function of the compensated system. (4 + 5 + 5 + 6 = 20 pts) Lead Space vehicle

Answer 1

soll !-. 6) of The desired Gls) = 1 pole is . The poles s=-11j The Angle deliciency is AT Od = 180-20 where O= 180-raño (+) = 135° 3 Odz 180-270 = -900 LPP) Gus) = K (5+0.5) (5+3) The Angle Contribution is Cant (0.5) -Pañl (6:3) = 45° The gain is tond using magnitude Gitesia * S * = 1 Rz4 I (N) @S7-11) The towald teansta Junction is Gsx4 (S) ; k (stos) (8+3) s² Root locus las poles at s=0,0,03 and zeros at s=-0.3 | na3, main b) The number of lost locus blanches are kanmaa. c) The Angle of Asymptotes is 'PA is PA= 180 (2K+1) K=0,1 => XA= 90, 2700 d) The Centeiod to Ya' on Real are's is -TA= Spoles -Ezeos = [oro+3]– 0.5 = 1.25 » TA= 725 nam n- m

e The break point is found by dk zo where K = - $(5+3) B (Stors) = 2545²7 95+6) zo z) s20, -11 0:485 570-5)2 Hence the break point is szo- The locus is deaun below * -1.25 -05 m (1 o = 180-20, +9-03 LU T as Q= 135° g= 180- Paño CasJ= 116.5 ą= Tan (!)=26.50 Osta 180- 270° + 116.5°-26-5° = Hence the poin -II 15 is on go the locy