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==Problem 1==
 
==Problem 1==
a) <math>\frac{C(s)}{R(s)} = \frac{4}{s(s+1)}
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a) <math>\frac{C(s)}{R(s)} = \frac{4}{s(s+1)}</math>
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b) <math>\frac{B(s)}{E(s)} = \frac{2}{s+1}+\frac{4}{s(s+1)} = \frac{2s+4}{s(s+1)}</math>
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c) <math>\frac{C(s)}{R(s)} = \frac{\frac{4}{s(s+1)}}{1+\frac{2s+4}{s(s+1)}}</math>
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d) <math> 1+\frac{2s+4}{s(s+1)} = 0 </math>
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e) <math> s(s+1)+2s+4 = 0 \Rightarrow s^2+3s+4=0 </math>
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  <math> \therefore \omega_n^2 =4, \; 2\zeta \omega_n = 3 \Rightarrow \tau = \frac{1}{\zeta \omega_n} = \frac{3}{2}</math>

Revision as of 22:43, 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} $

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