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)

B) Two complex conjugate zeros:

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

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