Question

Problem 1: (20 points) Assume that a standard unity feedback system has the open- loop plant transfer function: G(S) s(s+3)(s +6) Use Root Locus Methods to design an analog compensator to meet the following specifications: • The step response settling time is less than 5 seconds. • The step response overshoot is less than 17%. • The steady-state error to a unit-ramp input is less than 10%.

Answer 1

2 Carmen Gls) system s(s+3)(5+6) time ts55 sec Setting OSY <177. Root locus method new spocifuation Step respons on Steady state erres to a unit - ramp input is less than -10 y. guen OSY. £ 17%. -39-32 zity. e 2-177 In 10-17) 2 1-32 32 3 3 = 3.139 + 2+ 3.139 = 3.139 1-32 13 =0.5 16. 4 ts=4 + 0.5X5 n 3 wr sd Dominant pale, - zwnzjundi- ?? =-0.58146j16 J-0.52 Sd - -D•8+1+381 Let oz Dzisal and ß= Lsd me By considering and demont pole at (-008 + 1938j), get,

S = -0.8 +1.58j - 1-64120 D=1-6 and B = 12.0 -2 t Step 2 Je find magutude and phase of als) at Sesi det Ad = / alsd) I and od=Lalsd) Comen that als) (also) s(s+3)s+6) (-0.8 +138j) (-0.8 +381 +37(-0.8+1-38f +6) 000444-169 Ad 1=0.044 and Od 5-167 Slep = 12.5 -3 Deleanlue the traspes function of PID contratte Comen that exs s 10y. 50! Let ess 0.08 for Accerat. Raz less erros content nelocity Thorfer fueltor of Gels) = (kp tki+kas, CALLY) Ekt4?kps+k; PID Controlle

Opeen loop trasfes fueter) of compensated system 6o(s) = Gels26(3) Gols) of Kd 9P+kas tki (3+3)(3+6) s 5->0 Elt 30 kd3 kp stki 5+(3+3)(3+6) Alteration valority error constant = Lt schols) of compensated system Kd s ²+kpstki st (3+3)(3+6) 0+o+ki But the accuration error constant of compensates Sigsten should be greater than or ki = 12-5 → Ki = 225 ki 18 3x6 to 12-5 equal 18 D Beatnature constant + 225 kd = =dindd to DAd sing =sin(-167) 16x60044- sin(120) 716) ? kd 3840 2 Propes lonel Constant, kp =-Sin[B78d) – 2 kilose Ad SunB. kp -Sin (120 - 167) 2/225) Los(120) 00044 Sn(120) kp= 160 1.6

turne How? bels) = kp s S Trasfer functHowy of PID controle kpt ki akds 160+225 +84.23 74-25+1605 +225 24.2 (3²41.9342167) Leep-t really the design . open loop thansfer Go(s) = Ge(s) Gls) function of compensated Sustem Go(s) = 84-2184195+267)x I s(3+3)+6 Sue hone sda = -0.8+138 0-922 (179.45 n 12-1862-1 17 h green Co(sd) Sree Golsd) =-1, the host lowus will pass through Sd and so the design Accepted is 6