Question

Consider the electro-mechanical feedback control system shown in Figure 3. The voltage Ea(s) - Liea(t)) is generated by an amplifier whose transfer function is Ga(s) -5 The position sensor has a transfer function H(s) 1 and the pre-compensator transfer function is pot X (s) Ea(s) The "Electro-Mechanical System" block, is X(s) Ea(s) 5.05s3 101s2 +505.2s 100 R(s) Amplifier, |Ea(S)Electro-MechanicalX(S) Controller, Gc(s) K, pot Ga(s) System, G(s) Encoder H(s) Figure 3: Electro-mechanical control system for Question 3 Consider a proportional controller Ge(s)- K. Find the range of K that will maintain closed-loop stability. Still with Ge(s) input R(s). You may assume second order dominance, but you must comment on the validity of that assumption after you find K Compute the settling time and steady-state error to a unit step for the closed-loop system using the controller you designed in (b) Design a lead / lag compensator for Ge(s) so that 1) the overshoot is less than 20%, 2) the settling time is less than 1s, and 3) the steady-state error to a unit step is less than 0.01. During your design, you may assume that the closed-loop system will be second order dominant. State the resulting Gc(s) and be sure to comment on the validity of the second order dominance assumption (a) (b) K, find a value for K that will produce 20% overshoot for a step (c) (d)

Answer 1

The open loop transfer function is, 5K (5) G. (s)G(s)G(s)=3 055 +101s2 +505.2s+100 25K 5.05s101s +505.2s +100 (a) Obtain the characteristic equation of the system 25K 1+ 5.05s3 101s2 +505.2s+100 0 5.05s3101s2 +505.2s+100 25K 0 Construct the Routh array. s3 5.05 505.2 101 10025K 0 10025K 101x 505.2 5.05 (100 +25K) 101 =500.2 1.25K For the stability of the system, all elements in the first column of the Routh array should be greater than zero. 100+25K>0 K<-4 500.2 -1.25K >0 K < 400.16 Thus, the condition on K for stability is K < 400.16

(b) The open loop transfer function is, 25K G(s)G(s)G(s)=3 055 +10 1s2 +505.2s +100 25K (s+0.206 (s8.48)(s+11.31) S S Neglect the pole nearer to zero to approximate the system to a second order system. 25K G. (s)G (s)G(s)(s+8.48)(s+11.31) (s +8.48)(s +11.31) Obtain the transfer function of the closed loop system. 25K (s +8.48)(s+11.31) T(s) 25K 1 + (s+8.48) (s +11.31 25K (s+8.48 s 11.3125K 25K s19.79s95.9 +25K An overshoot of 20% corresponds to a damping ratio of 0.4559 Compare the transfer function with the standard second-order transfer function,- s22c,saw n a 25K 25a19.79 19.79 19.79 21.7 24 2x.4559 21.7 2 K 25 25 K =18.8

(c) Consider the transfer function of the closed loop system 25K TS) 19.79s +95.9+25K 25 x18.8 s219.79s95.9 +25 x18.8 470 219.79s95.9+470 Determine the settling time 4 Tr Ça, 0.4559x21.7 4 T 0.404 s Determine the steady-state error to a unit step input 1 е. 1+limG(s) Here G(s) is the open-loop transfer function SS S0 25K G. (s)G.(s)G(s)(s+8.48)(5+11.31) 470 (s+8.48) (s+11.31) S 1 470 1 lim (s+8+11.31) 48) (s s0 1 1 4.9 e=0.169

(d) An overshoot of 20% corresponds to a damping ratio of 0.4559 The settling time is less than 1 s 4 <1 T S So>4 n 8.774 Consider a lead/lag compensator Sa G.(s) S Consider the open-loop transfer function 470 Sa G. (s)G(s)G(s) s bs8.48(s +11.31) Select a as 11.31 that results in pole-zero cancellation. S11.3 470 G.(s)G.(s)G(s) (s+8.48) (s+11.31) s+b 470 (s+b)(s+8.48) Determine the closed loop transfer function 470 (s +b)(s+8.48) 470 1+ b)(s+8.48) S 470 (s b)(s8.48)+470 470 (8.48+b)s 8.48b+ 470 The characteristic equation of the system is, s(8.48 bs +8.48b+470 = 0

Compare the transfer function with the standard second-order transfer function.- s2 a8.48b470 25a8.48b So,>48.48+b <8 b<0.48 = 8.48b+ 470 >8.7742 a, 8.774 b-46.3 The value satisfying both the conditions is, b 0.4 Thus, the designed lag compensator is S11.3 G.(s)= S+0.4