Question

System dynamics and control need a pro to solve this

Let a system have plant transfer function 10(s +20) Design a PD controller such that the closed-loop rise time is about 0.2 seconds and the overshoot is just under 25%.

Answer 1

Design specifications Rise time 0.2 sec Over shoot < 25% These control design specifications will be converted to desired closed loop dominant poles of the system. Rise time = pi - cos^-1 g/w_n squareroot 1 - g^2 = 0.2 - (1) & Over shoot = e^-gpi/squareroot 1-g^2 < 25/100 - (2) Solving for (2) we get g = 0.41 Substituting g = 0.41 in(1) we get w_n = 10.92 Therefore fix w_nat 11 rad/sec. Therefore Dominant pole location is given by S_domi = -g w_n plusminus j w_n squareroot 1 - g^2 S_domi = -4.51 plusminus j 10.03 Now we need to design a PD controller that yields Dominant poles given above. Lets compute angle deficiency. i.e. angle contributed by all poles & zero at S_domi From above geometry, theta_P_1 = 188 - tan^-1 (10.03/ Vertical-bar - 4.51 + 1 Vertical-bar ) = 109.28 degree theta_P_2 = 180 degree - tan^-1 (10.03/4.51 - 4) = 92.91 degree theta_P_3 = tan^-1 (10.03/10 - 4.51) = 61.3 degree theta_z = tan^-1 (10.03/20 - 4.51) = 32.92 degree Total angle contribution = -sigma theta_p + sigma theta_z = 32.92 degree - 109.28 - 92.91 - 61.3