Question

Problem 4) (20 Pts.) A Proportional controller is simply a gain block. In figure below, it is the block with gain 2nd order underdamped plant as shown. Kc which is behind the a) Simplify below block diagram to obtain the overall feedback system transfer funion)R(G) b) Choose Kc so that the overall feedback system transfer function G(s) has 50% overshoot due to a step input (called quarter decay ratio tuning) d) The feedback system transfer function Gs)- is faster than the open loop plant Gp(s) to show this, calculate and write down the rise times of both plant Gp(s) and feedback system G(s) for the gain Kc obtained in part (b). (Note: You need to find and nfor both cases.) To see how the controller achieves this speedup, find the transfer function form R(s) to controller d) output M(s): R0(Use Kc as found in part b.) e) What is the controller output m(t) at t-0 +(just after unit step is applied at input)? f Now, in your opinion, how is the controller achieving a higher speed as compared to the open loop operation of the plant? M(s) Controller R(s)+ Ko C(s) s2 +s1 Plant Gp(s)

Answer 1

(a) simplified block diagram C(S)/R(S) = K_c/s^2 + s + 1 + K_c (b) Peak overshoot = 50% M_p = 0.5 So M_p = e^-pi Es/squareroot 1 - E_s 0.5 = E^-pi E_s^2/squareroot 1 - E_s^2 On solving above equation we get E_s = 0.2155 and w_n = 2.32 after comparing with standard 2^nd order equation i.e T.F = C(S)/R(S) = w_n^2/s^2 + 2Ew_ns + w^2_n From this, we get w^2_n = 1 + k_c k_c = 4.38 \r\n (c) Considering open loop plant G(s) rise time(z_1) = pi - phi/w_d for open loop system G_p(s) = 1/s^2 + s + 1 Comparing this with general equation, we get 2^eQw_n = 1 and w_n = 1 E = 0.5 tau_1 = pi - cos^-1(0.5)/w_n squareroot 1 - E_s^2 {w_d = w_n squareroot 1 - E^2} = pi - pi/3/1/2 = 2 pi/3 times 2 tau_1 = 4 pi/3 \r\n Now, consider closed loop system having transfer function