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4 2. The open-loop system with transfer function G(s)- is stable without a controller. s2 +6s +10 However, improvement of the

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Answer #1

MATLAB is used to plot the uncompensated step response.

s = tf('s');
G = 4/(s^2+6*s+10);
figure;step(G);grid on;

Step Response 0.4 System: G Settling time (seconds): 1.66 0.35 0.3 0.25 E 0.2 0.15 0.1 0.05 0 1.5 Time (seconds) 0 0.5 2 2.5

It is observed that the settling time is 1.66 seconds. and peak overshoot is zero.

Now we need to design a compensator that settled in 1.66/2 = 0.83 seconds and maintains zero overshoot.

won wn 1t.82 O-83- clased loo need to evaluatThe chasadaushe quation wih The chasadesishc 2 2 2-3

MATLAB code is given below for plotting the compensated step response.

code:

C = (0.91*(s+6.3891));
figure;step(feedback(C*G,1));grid on;

Step Response 0.8 System: untitled1 Settling time (seconds): 0.78 0.6 0.5 0.4 0.3 0.2 0.1 0 0 0 0.2 0.40.6 0.81.2 1.4 1.6 1.8

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