Question

Problem 2: Given the plant G,le)+2( +3) design a PI compensator Gc(s)-K Ш such the closed-loop unity feedback system has two dominant poles at s1.2 =-1 ±j. Using Matlab ritool (or simulink), simulate your closed loop system to show that the unit-step response of the system has PO ~ 4.3%, tr 2.35 sec, and 4 ะ 4.15 sec. Compute the closed-loop poles and zeros.

Answer 1

Given plant G_p(s) = 1/(s + 2)(s + 3) P1 compensator G_c(s) = k(s + z/s) Given dominant poles S_d = -1 plusminus j specific heat t_s = 4.15. P0 almostequalto 4.3%, tr almostequalto 2.35, e^-4 pi/squareroot 1 - zeta^2 = 4.3/100. -3 pi/squareroot 1 - zeta^2 = (-3.14) 3^2 pi^2/1 - zeta^2 = 9.9 zeta = squareroot 9.9/pi^2 + 9.9 zeta = 0.707 t_s = 4.15 4/zeta omega_n = 4.15 omega_n = 4/6.707 times 4.15 omega_n = 1.363. therefore Dominant pole S_d = -zeta omega_n plusminus j omega_n squareroot 1 - zeta^2 = -0.96 plusminus 0.96 almostequalto -1 + j given (dominant) therefore step - 1 Dominant pole from given and obtained are approx equal therefore We take S_d = -1 + j For design Let D = Vertical-bar S_d Vertical-bar & beta = angle S_d therefore S_d = -1 + j = 1.414 angle 135 therefore D = 1.414, beta = 135 degrees \r\n step 2 To