Line 37: Line 37:
  
 
I) <math>\ddot{y}(t)+\dot{y}(t) = 4u(t)</math>
 
I) <math>\ddot{y}(t)+\dot{y}(t) = 4u(t)</math>
 +
 +
==Problem 2==
 +
<math> k_p = \lim_{s\rightarrow 0} G(s) = \infty</math>
 +
<math> k_v = \lim_{s\rightarrow 0} sG(s) = \frac{K}{6}</math>
 +
<math> e_ss = \lim_{s\rightarrow 0}sE(s) =

Revision as of 22:57, 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) $

Problem 2

$ k_p = \lim_{s\rightarrow 0} G(s) = \infty $ $ k_v = \lim_{s\rightarrow 0} sG(s) = \frac{K}{6} $ $ e_ss = \lim_{s\rightarrow 0}sE(s) = $

Alumni Liaison

Prof. Math. Ohio State and Associate Dean
Outstanding Alumnus Purdue Math 2008

Jeff McNeal