Given the transfer function G(s) = b/s^2 + as + b a. Evaluate percent overshoot,
ID: 2081252 • Letter: G
Question
Given the transfer function G(s) = b/s^2 + as + b a. Evaluate percent overshoot, settling time, and peak time for the following values: a = 4, b = 25. Also, plot the poles. b. For the system of exercise 2.2a, calculate the values of a and b so that the damping ratio remains the same but the natural frequency is increased two times over that of exercise 2.2a, and repeat exercise 2.2a. c. For the system of task 2a, calculate the values of a and b so that the damping ratio remains the same but the natural frequency is increased four times over that of exercise 2.2a, and repeat exercise 2.2a.Explanation / Answer
%step response K=45%
clc;
clear;
s=tf('s');
a=4;
b=25;
sys=(b)/(s^2+a*s+b^2)
stepinfo(sys)
%constant danping ratio and two times the natural frequency%
a=4;
B=50;
sys=(b)/(s^2+a*s+B^2)
stepinfo(sys)
%constant danping ratio and two times the natural frequency%
a=4;
C=2*B;
sys=(b)/(s^2+a*s+C^2)
stepinfo(sys)
RESULT:
Transfer function:
25
---------------
s^2 + 4 s + 625
ans =
RiseTime: 0.0444
SettlingTime: 1.9110
SettlingMin: 0.0158
SettlingMax: 0.0711
Overshoot: 77.7100
Undershoot: 0
Peak: 0.0711
PeakTime: 0.1257
Transfer function:
25
----------------
s^2 + 4 s + 2500
ans =
RiseTime: 0.0215
SettlingTime: 1.9504
SettlingMin: 0.0022
SettlingMax: 0.0188
Overshoot: 88.1820
Undershoot: 0
Peak: 0.0188
PeakTime: 0.0628
Transfer function:
25
-----------------
s^2 + 4 s + 10000
ans =
RiseTime: 0.0106
SettlingTime: 1.9491
SettlingMin: 2.9528e-004
SettlingMax: 0.0048
Overshoot: 93.9089
Undershoot: 0
Peak: 0.0048
PeakTime: 0.0314
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