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e) <math> s(s+1)+2s+4 = 0 \Rightarrow s^2+3s+4=0 </math> | e) <math> s(s+1)+2s+4 = 0 \Rightarrow s^2+3s+4=0 </math> | ||
<math> \therefore \omega_n^2 =4, \; 2\zeta \omega_n = 3 \Rightarrow \tau = \frac{1}{\zeta \omega_n} = \frac{3}{2}</math> | <math> \therefore \omega_n^2 =4, \; 2\zeta \omega_n = 3 \Rightarrow \tau = \frac{1}{\zeta \omega_n} = \frac{3}{2}</math> | ||
| + | |||
| + | f) <math> \frac{3}{4}</math> | ||
| + | |||
| + | g) since <math>\zeta > 0 \therefore \omega_n = 2 | ||
Revision as of 22:46, 1 August 2019
Automatic Control (AC)
Question 1: Feedback Control Systems
August 2017 (Published in Jul 2019)
Problem 1
a) $ \frac{C(s)}{R(s)} = \frac{4}{s(s+1)} $
b) $ \frac{B(s)}{E(s)} = \frac{2}{s+1}+\frac{4}{s(s+1)} = \frac{2s+4}{s(s+1)} $
c) $ \frac{C(s)}{R(s)} = \frac{\frac{4}{s(s+1)}}{1+\frac{2s+4}{s(s+1)}} $
d) $ 1+\frac{2s+4}{s(s+1)} = 0 $
e) $ s(s+1)+2s+4 = 0 \Rightarrow s^2+3s+4=0 $
$ \therefore \omega_n^2 =4, \; 2\zeta \omega_n = 3 \Rightarrow \tau = \frac{1}{\zeta \omega_n} = \frac{3}{2} $
f) $ \frac{3}{4} $
g) since $ \zeta > 0 \therefore \omega_n = 2 $
