Question

(d) Consider the system shown in Figure 3(d), it is desired to determine the values of K1, K2 and K3 such that the dominant closed-loop poles are located at s-1+j2 design a PID controller by 9] Plant PID R(s) Y(s) (s+K, Xs+K,) 2 K X s2s1.25 S Figure Q3(d)

Answer 1

RCS) k (S1kg) (Stky) (sama YCS) sinas | S +8+1.25 7,25) G(s) ---- 6 = Closed loop transfer function YOS) = GG(3) BC) TG,(s) GCS HCS) = (15 (F15) (156Heuren) 1+ ( KLETK) (CM) ) Y ) ak, (Stky) (s+k) s(+8+125) +254stk) (Stky) ROS s. Charactergic eq is s345?+). 2554 9ky (s?++k;)stýk) -0 5+ [+2k)s*+(1.25 +884(+kk)]Stakkk =0 - (1) Güven pules are located at S=1921 .. (8+1-21) ( 5+1+21) is a rout 04 (1) = 6+1)2 + (aj)) = st1725 +4 = (stas+5) is a rout of") :1t is anut it must divide u)

Cobg syntetic division II e O - (1 +2K) -2 0 6 +2k12 (1.25 taxi(Kytky)) okyky -26kg-1). -5 -5 (+264) (5Kiky +2kyky469 -1,15) (2 Kik7x3 +5–104) => (2K4ką takpky – 4k7.75) 50 -42 and Cakyky kz +5-10k )=0 — 2) Solve U2 and (2), to get ky and K and Kg. We condt some because given with characters stic equation is 319 order a poles but the in, cue need to find 3 unknown.