Question

4. (20 points) Adapted from [DB AP5.7]: Consider the unity feedback system with variable gain K shown (a) Assume that the complex poles dominate and estimate the 2% settling time and percent overshoot (b) Using MATLAB, determine the actual settling times and percent overshoots of the full third- in Figure 2. We are constrained to keep 1000 s K 5000. in response to a unit step for K 1000, 2000, 3000, 4000, and 5000 order system for the values of K in part (a). Comment on any discrepancies between parts (a) and 1 of 2 G(s) 100 s+100 s(s+60) Figure 2: Block Diagram for Question 4 (c) For each value of K, use MATLAB to co-plot the step responses of the full system and the reduced system. Submit the five plots and any script you used to generate them

Answer 1

(a) The open loop transfer function is G(s) = k 100/s (s + 60) (s + 100) (i) for k = 1000 the characteristic equation 1 + K G(s) = 0 is s^3 + 160 s^2 + 6000 s + 10^5 = 0 the roots are + 15.57 and - 22.21 + 19.28 J the dominant poles are s_d = -sigma w_n plusminus w_n squareroot 1 - s^2 = -22.21 plusminus 19.28 J t_s = 4/sigma w_n = 4/22.21 = 0.181 sec and sigma = 0.7552 mu_p = e^-pi Vertical-bar squareroot 1 - s^2 = 0.0268 or 2.68% (ii) for k = 2000 the characteristic equation is s^3 + 160 s^2 + 6000 s + 10^5 times 2 the dominant poles are s = 17.62 plusminus 35.95 J t_s = 4/17.622 = 0.2270 seconds and sigma = 0.4401 mu_p = e^-pi /squareroot 1 - s^2 = 0.2144 or 21.44 % (iii) for k = 3000 the

ii) matlab:

clc;

clear all;

s=tf('s');

k=[1000 2000 3000 4000 5000];% gains

g=100/(s*(s+60)*(s+100));% open loop transfer function

step(feedback(k(1)*g,1));grid

legend('step response for k=1000')

figure;

step(feedback(k(2)*g,1));grid

legend('step response for k=2000')

figure;

step(feedback(k(3)*g,1));grid

legend('step response for k=3000')

figure;

step(feedback(k(4)*g,1));grid

legend('step response for k=4000')

figure;

step(feedback(k(5)*g,1));grid

legend('step response for k=5000')

the settling times and the over shoots are not exactly the same but the second order approximation is valid.

iii) matlab plot for original system and reduced system:

clc;

clear all;

s=tf('s');

k=[1000 2000 3000 4000 5000];% gains

g=100/(s*(s+60)*(s+100));% open loop transfer function

g1=865/((s+22.21+19.28*i)*(s+22.21-19.28*i));

g2=1603/((s+17.62+35.95*i)*(s+17.62-35.95*i));

g3=2277/((s+14.13+45.58*i)*(s+14.13-45.58*i));

g4=2909/((s+11.24+52.75*i)*(s+11.24-52.75*i));

g5=3508/((s+8.74+58.58*i)*(s+8.74-58.58*i));

step(feedback(k(1)*g,1),g1);grid

legend('step response actual','approximated')

figure;

step(feedback(k(2)*g,1),g2);grid

legend('step response actual','approximated')

figure;

step(feedback(k(3)*g,1),g3);grid

legend('step response actual','approximated')

figure;

step(feedback(k(4)*g,1),g4);grid

legend('step response actual','approximated')

figure;

step(feedback(k(5)*g,1),g5);grid

legend('step response actual','approximated')