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</center>
 
</center>
 
==Problem 1==
 
==Problem 1==
a) <math>\frac{C(s)}{R(s)} = \frac{4}{s(s+1)}</math>
+
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>
+
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>
+
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>
+
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>
+
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>
+
F) <math> \frac{3}{4}</math>
  
g) since <math>\zeta > 0  \therefore \omega_n = 2
+
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 21:51, 1 August 2019


ECE Ph.D. Qualifying Exam

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) $

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Ryne Rayburn