Question

6. Given the following closed-loop system, the objective is to design a controller D(s) such that the closed-loop poles are placed at -V3+j. (a) Show that this objective cannot be achieved by choosing a proportional control alone. (b) Design a controller of the form K(s-a) to achieve the objective. [Hint: You could use the root locus method to introduce a zero at a such that -V3 + j are on the locus.] r(t) y(t) D(s) + s+2 s(s+1)

Answer 1

Page Q- + Yl) ED(S) 3+2 Stati) blocks Cascades multiplied are Baric i brain of Grilon bizt for uc Unity feedback system ROSA 6103 Y(S) CLTF = y (5) airs) It bols) R(3) CLTF meam Closed loop transfer function and its denominator will give poles. (a when we Us Controller proportional D(s).=k meary 1 8 by formula! CLTF = k (5+2) s(5+1) K (5+2) s(+1) 1 + k (5+2) 5 5+1) 515+1) + K 13+2) SLUIT)

Date: Page: CITF = controller is k (3+2) 52 +5+ ks +2K so pole equation for proportional 52 +5+ks +2k a) 52 + ("+ k) + 2K whill he requirand poles are at €qin s= BIJ (5 + 13-0) (521st Formula (a+b) (a-6)= a ?- 62 and J?=-1 + 1 = 0 15+53) - 52 + 3 +25ſ This is 52 + 25 G + 4 =0 Comparing this with equation (al Mequired equal 2 K = 4 tk = 26 K = 2 K = 3.461 k = 2.46 time hoo distinct so K=2 of value then At same K not possible coefficient of 5 = 2.46 not mattch Hence S would not maken. And of go would then coefficient Controller - 3 t By proportional possible. poles not

When D (5) - K(3-2) OLTF = K 15-2) (5+2) Sesti) By formula CLTF- (Griven in Basic) of this Question k (sd) (3+2)/s(s+1) K 15-2) (5+2) 5(5+1) 1 + k (s-d) (5+2) Sist) alus s(5+1) + K (5-2) (5+2) Sestr) - K (5-2) (5+2) 32+ s + k[s²+32-32-22 SCALE k (5-0) (5+2) 52(1+k) + 5 + 2 Sk - Sko - 2ko

New Globe K(sta) (5+2) 52(1+k) +5 (1+2K-kd) -zka Date: Page k (5-2) (5+2) 1 +K 32 + s (1+2K-K) Divide by in Numerator - 2k2 Itk & Denominators 1+k Now Compare this with 32 +2513 + 4 6 263 = 1 + 2K-KO ciz 1+k 2 4 - 2ka trig 1 +K 231+ - 1 + 2 K-KO NATARAJ, 621 SCALE 283 + 283K-2k-1 = put this Nalue of -k2 -kd in canci, 4 283 + 2 rz K -26-1 1 tk 2 2+2K = 2 rok-2k 4K - 263k = 283-1-2 +2√3-1 K = 0.46 4 0.86567 (put this in eyni 4-283 2 4 = - 2 x 0.86567 xd 1+0.86567 - 4.3103 2 K = 0.865 - Arywer