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B) Two complex conjugate zeros:
 
B) Two complex conjugate zeros:
  
arrival angle of zero 1+j: angle to zeros <math>(1+j) - (1-j) = 90 \degree </math>
+
arrival angle of zero 1+j: angle to zeros <math>(1+j) - (1-j) = 90^{\circ} </math>
  
 
==Problem 4==
 
==Problem 4==

Revision as of 23:28, 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) = \lim_{s\rightarrow 0} \frac{sR(s)}{1+G(s)} = \frac{1}{1+\lim_{s\rightarrow 0}G(s)} + \frac{1}{\lim_{s\rightarrow 0}sG(s)} = \frac{1}{1+k_p} + \frac{1}{k_v} = 0.2 $ $ \therefore K = 30 $

Problem 3

A) The root locus plot

B) Two complex conjugate zeros:

arrival angle of zero 1+j: angle to zeros $ (1+j) - (1-j) = 90^{\circ} $

Problem 4

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman