Question

9. Plot the root locus for the following closed-loop control systems. K (s -1) (ST2)(S+4) and H(s) = 1 (a) Gle)s+2)(s (b) G(s) = K (s + 1) and H (s)-1 A formula you may need for calculating the break-in point N'(o)D(o)-N(σ)D'(o)-0

Answer 1

(a) G(s) = k(s - 1)/(s + 2)(s + 4) and H(s) = 1 sol, pole = -2, - 4 zero = 1 rightarrow No of branches N = P if P > Z therefore N = 2 rightarrow Angle of Asymptotes implies (2k + 1)180/P - Z, k = 0, 1, 2...theta = 180 degree rightarrow There is No - break away point rightarrow section of root locus lie on the real atis if the total no of open - loop poles and zero to the right of the section is odd G(s) = k(s + 1)/s^2 and H(s) = 1 pole = 0 (double, order) zero = -1 since here double order pole exist at origin then there break away point occur at origin Now 1 + G(s) H(s) = 0 1 + k(s + 1)/s^2 s^2 + k(s + 1) = 0 s^2/s + 1 = -k - dk/ds = (s + 1)2s - (1)s^2/(s + 1)^2 put dk/ds = 0