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</center> | </center> | ||
==Problem 1== | ==Problem 1== | ||
| − | + | A) <math>\frac{C(s)}{R(s)} = \frac{4}{s(s+1)}</math> | |
| − | + | B) <math>\frac{B(s)}{E(s)} = \frac{2}{s+1}+\frac{4}{s(s+1)} = \frac{2s+4}{s(s+1)}</math> | |
| − | + | C) <math>\frac{C(s)}{R(s)} = \frac{\frac{4}{s(s+1)}}{1+\frac{2s+4}{s(s+1)}}</math> | |
| − | + | D) <math> 1+\frac{2s+4}{s(s+1)} = 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 </math> | |
| + | |||
| + | H) two poles, type 2 | ||
| + | |||
| + | I) <math>\ddot{y}(t)+\dot{y}(t) = 4u(t)</math> | ||
Revision as of 22:51, 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 $
H) two poles, type 2
I) $ \ddot{y}(t)+\dot{y}(t) = 4u(t) $
