Question

Consider the sontrol system shown in the figure below: R(S) + E(s) C(s) K (s + 4)(s + 6) g) Sketch the uncompensated system root locus showing all details. (5 Points) h) Find the dominant closed loop poles of the uncompensated system to operate with a 16.3% overshoot and peak time tp = 0.7255 (make sure to show this point on the Root Locus) (5 Points) (s+z) Now we want to design a PI compensator of the form to increase the peak time by a factor of 2 and increase the percent overshoot by a factor of 2 i) Find from the given specifications given above the desired dominant closed loop specification. (5 Points) j) Find the location of the zero and gain K at the zero. (15 Points) k) Write down the compensated open-loop transfer function. (5 Points) S

Answer 1

Soln:- RS) E(S) cs (5+4X5+6) g) GS) = K (5+4) (StG) = -5 -6-4 2 or +90 Breakaway point: K= -($+4) 5+6) as to 7 - +6)+(5+4)] => dk 25 +10 zo → S =-5 Root locus Imis) Re is -6 -4

th Dominant closed loop poles: - Overshoot = 16.3% Peak time, tp = 0.7255 Nowi e-Trefl er 0.163 T²4² (1-92 Ilu 0.163) - 3.2906 52(043.2906) 3.2906 → ę z 0.5 2 3.2906 VT?+3.2906 Thus, § 20.5 Now, tpz I = 0.7255 Wr 1-G² or, wn= ㅠ 0.7255 71-0,52 5 Dominant poles - - Elin IJmond 1-2 =-0.5X5+j 5-0,52 – 2.5+j 4,33 z -2.5t

↑ Im(s) -2.5+j4.33 +433 -6 -4 --4 -2.5 Ress) 1-433 X -2.5-4-33 clearly the desired poles do not lie on the root for locus. Thus we have to use a controller. 1) Now we have to design PI compensator of the form (StZ) Desired specifications: - tp = 2 tp Overshoot'= 2 Hovershoot ori e-tis/JT-5E - (0.163X2) (en (2XO:163))= 1.2563 2 Tg2 (1-32 ş ş?(+1.2563) = 1.2563 1.2563 » ę z 2 Z z 0,336 TT? +1.2563

top 2x0,7255 Wn 5-4² or, Wn = T 2X0.7255 1-0.3362 2.298 closed loops: Dominant & wnt j written 0.336X 2.298 + j2 2 298 1-0,3362 0.7721 + j 2.164

K K j Gis) C+2) 6+4) 5+6) S Gs). (5+4) (S+6) We first calculate angle deficiency -0.772+j2.1641 ø 24 X -6 -4 Angle deficiency z 180-81-62 Ø, 2 4-0.7721 2 tant (2.164 = 33,838 tame 2.164 6 -0.7721 1)= 22,486

Angle deficiency - 180-6,-02 180-33.8381 - 22.486 - 123.676 Thus angle of 123.676 must be Contributed by PI Controller. 0.7121+ I 12,164 oktor Z B-6, - tant (2.164 tant / 2.164 0.74211 Z-0 7721 -123.676 tant/2.164 70.364 IZ-0.7721 or tant/2.164 Z-0.772) 53.312 ori 2:164 z 1.342 Z-0.7721 ori Z= 2:164 1.342 +0.7721 22,384

Thus the controller is K(S+2.384) S W Gain k at the zero, 1 K(S+ 2.384) S 5+4) (5+) S=2-384 -0.77+j2.164 K= 1-0.77+j2-104114.0.714) 2-164116-0.7745214 2.384-6.77+j2.1641 12.4 K = 2.2969% 3.887x 5.66 18.71 2.7 k) Compensated open loop transfer function 18.71 (5+2.384) S (5+45+ 6)