Question

Design Problems: (1) A robotic system is described by the transfer function P(s)=- 100 s(s +9.7)(s + 51.2) Use the root locus method to design a lead controller that achieves a closed-loop step response with P.0.5 2.5 %, and a settling time T, < 0.25s (using the 2% criterion). Also, the steady-state error to a unit ramp should be ess < 0.15. (2) This system is open-loop unstable: P(S) = 500 (5 - 1)(s + 10) Using the root locus method, design a controller (lead, lag or lead/lag) that first of all yields a stable closed-loop system. Further, the closed-loop's step response should have P.O. < 58 % and T, < 2.3s (using the 2% criterion).

Answer 1

Sol? Pls) = loo s(s+ 9.7) (57512) 2 .5 from the given specifications - DIS-E2 e = 2.5 . By taking Log on both sides. These a los com 3 330961 and Is= 4 = 0.25 a {wn = 16 and wn = 21025 The dominant poles are soz -{wn I wndres 5 60 -0.25 L { un Sa= -16 I 13.6 J . - 51.2 The open loop poles one at S=0, -9.7-51.2. The Angle deficiency of 'se to open loop poles is Ody = 180-0-0-03 ---7136J 0,2 180– Tan ( 1305) Goz has % = 180- Ton' (link Rob t og = Ton (s. 6) - Oder = -95.6154 since a single hood Compensata Cam Compensate ony up to -50° . Here we need two lead controlles Contributing 95.6154 = 47.80 Sach. 6.3 2 pole - Zero, I PSZ =47.8 SA is Angle Biscata of BSO

-- -- - - - - .... .. . . . . . 2 1854 = \n30 = 12690° i PSA = 1852= 43,8 = 2390 k E in the BSP - Ton (698-23.9) = 13.6 1316 war 139.6 n 2 A Ž -16C 10 P » RS= 13.1793 BS ne pole is at S= 13.1793+ 16=2 9.1793 13.6 llly in Aesc Ton (69.8 –23.9-496) = 2= -0.8795 to The zero is at 3793-0.8795 2 998 The Dead Compensation is I Get = KTS22998)" (5+29.1793)2 The zuo of Compensator is 16-0.8795= 15.1205 The Lead Compensatol is Geis) = K (st 15. 1205) 2 15+29.1793)? The gain is found using magnitude Giteia i k (st 15.20 53² 100 5+29.1793}}.* 5C5+9,7) (5+512) @so-16+ 13:6J > K-229 G (s) = 229 (5+15.60s)? 16+29.1193)2