Question

3. The open-loop transfer function of a unity feedback system is GS) ss+2) The desired system response to a step input is specified as peak time t,-1 sec and overshoot M,-5% Determine whether both specifications can be met simultaneously by selecting the right value of K

Answer 1

Given t_p = pi/w_l sec % pi_1 = 5/100 = 0.05 = e^pi 4/pi - E^2 a(s) = k/s (s. 12) closed loop transfer function implies C_1 (s)/1 + C (s) = S k/(s + 2)/1 + k/s (s + 2) C(s)/R (s) = k/s(s + 2) + K = (k/s^2 + 2s + k) so on comparing with standard transfer function = w_n^2/s^7 2q w_n + w_n^2 here w_n^2 = k 2 & w_n = 2 2 & w_n = 2 & w_n = 1 (w_n) = squareroot k = (1/E_2) from (1) e implies 1 ln (% M_1) Vertical-bar /squareroot (ln (% M_p))^2 + pi^2 M_p = 0.05 so w_n = squareroot K = (1/E) = 1/0.6901 K = (1/E)^2 = 2.099 so k = 2.1 here both are not possible at the same time. From root locus % overshoot is 5 for gain nearly 2.1 as i solved in paper. so k = 2.1 for %Mp = 5: but for Tp = 1sec k will vary.

ROOT locus for open loop transfer function

here both are not possible at the same time. From root locus % overshoot is 5 for gain nearly 2.1 as i solved in paper.

so k=2.1 for %Mp= 5;

but for Tp= 1sec k will vary.