Question

Consider the unity feedback system is given below R(S) C(s) G(s) with transfer function: G() = K(+2) s(s+ 1/s + 3)(+5) a) Sketch the root locus. Clearly indicate any asymptotes. b) Find the value of the gain K, that will make the system marginally stable. c) Find the value of the gain K, for which the closed-loop transfer function will have a pole on the real axis at (-0.5).

Answer 1

characteristics equations sls Musts)(5+5)+ Ks+2k si stos'tus't 205 +35+15) +ks+2K sl sta51 95'+235+15)+K5+2K "tast +95 +2351155+ks+2K S4+95 +235'+s(K+15)+2K. ñ 4 s - 23 2K 5 9 (+15) 0 5 192 k s = 207-8-15 o 192-K (K+15) 162K 9 0 (192-x) (192-K)K+15)-186 2K (192-6) 9 K=0 s'a SE K+15) - 18Kxq (192-K) terite to starpty standar TK = 192 102K2142 system will be margainally stable K 20; K: 192 CS Scanned with CamScanner

c) S-005 S: -0.5 X * -5-3 -0.5 u es 0.5 0. la 1:5 7 product of de length Pole's of lengle product of Zero's 0.5 x 0.5x2.5x405 K 1.875 109 CS Scanned with CamScanner

Given unity feed back system ای، و Cou Gus) - KLS12) s(5+1)(5+3)(5+5) a) sketch the root locus : From Given open loop transter finitions K(512) Gls) SCS) (5+3)(545) Zero's S=-2 poleso S=0) 52-1) S2-3, 52-5 pole - Kero diagram ping * -5 * - 3 T -2 0 treal RLD RLD lle RLD Root locus diagrame (ov) Roor loci exist centroid (6), Ex-2) Number of poles-alumber of hero 0-1-3-5-(-2) -9+2 4-1 (6) Q-335 3 CS Scanned with CamScanner

Angle of asymptores (0) =+(2+1) 160 P-Z. 0-(2911) 180 - +18012961) 3 At (920) 0 lfo 3 - 609 (90-1) 0 180(2+1) = 180° 3 19=2) 180(5) 300. 3 centroid 12-33) Ki- 6de 180° -5 -3 -1 Y-60° kone b) Find value of x at strable. may marginally .H4)21) Tif (18) RID 405) I + Gis) Hs) K(5+2) s(5+1)(5+3)(3+5) I + K (5+2) s(5+1)(5+3) (545) ki 5+2) S(+1XS+5)(3+5) +(372) CS Scanned with CamScanner