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: $(1+j) - (1-j) = 90^{\circ}$

angle to poles: $(1+j) - 0 = 45^{\circ}$

               $(1+j) - (-2) = 3+j = 18.43^{\circ}$


Arrival angle of 1+j: $180^{\circ} + 45^{\circ} + 18.43^{\circ} - 90^{\circ} = 153.43^{\circ}$

similar for 1-j: $180^{\circ} - 45^{\circ} - 18.43^{\circ} + 90^{\circ} = -153.43^{\circ}$

C) $s^2 \; K+1 \;\;\;\;\; 2K$

  $s^1 \; 2-2K$

  $s^0 \; -2K$


set $2-2K = 0 \Rightarrow K =1$, so system stable at $0<K<1$

when $K=1$, for $s^2$ row, $2s^2+2 = 0 \Rightarrow \omega_n = 1$

## Problem 4

A) $\angle G(s)_{|s=-2+j2\sqrt{3}} = \frac{2}{(-2+j2\sqrt{3})(-1+j2\sqrt{3})} = -(180^{\circ}-tan^{-1}(\frac{2\sqrt{3}}{2}))-(180^{\circ}-tan^{-1}(2\sqrt{3})) = 60^{\circ}+73.9^{\circ}-360^{\circ} = -226.1^{\circ}$

$\phi = 46.1^{\circ}$

B) $K_v^{G} = \lim_{s\rightarraow0}sG(s) = 2$

$\alpha_2 = \frac{K_v^{G_d}}{K_v^{G}} = \frac{20}{2} = 10$

10% rule: zeros location: $\frac{-1}{\tau_2} = -0.1$ and poles location: $\frac{-1}{\alpha_2\tau_2} = -0.01$

PI portion $G_c = \frac{s+0.01}{s+0.1}$

C) $\phi = 46.1^{\circ} + 3^{\circ} = 49.1^{\circ}$

$\theta = cos^{-1}(0.5) = 60^{\circ}$

zero location: $\frac{1}{\tau} = \frac{\omega_n cos(\frac{\theta+\phi}{2})}{\frac{cos(\theta-\phi)}{2}} = \frac{4cos(\frac{109.1^{\circ}}{2})}{cos(\frac{10.9}{2})} = 2.3032$

pole location: $\frac{1}{\alpha\tau} = \frac{\omega_n \frac{cos(\theta-\phi)}{2}}{cos(\frac{\theta+\phi}{2})} = \frac{4cos(\frac{10.9}{2})}{cos(\frac{109.1^{\circ}}{2})} = 6.9470$

$G_c(s) = \frac{(s+2.3032)(s+0.01)}{(s+6.9470)(s+0.1)}$

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010