ENGR202 2021 Q2 (ENGR5001)

ENGR202 2021 Q2Past paperOld spec ENGR2022:216 marks19 min

Answer ALL parts (a) - (c).

Formulas you may need

Hurwitz: for $a_0 s^3 + a_1 s^2 + a_2 s + a_3 = 0$ , stable if $\Delta_1 = a_1 > 0$ , $\Delta_2 = a_1 a_2 - a_0 a_3 > 0$ , $\Delta_3 = a_3\Delta_2 > 0$ (learn this)
Frequency response: $M = |G(s)|_{s = j\omega}$ , $\phi = \mathrm{Arg}(G(s))|_{s = j\omega}$ (learn this)
$|j\omega + a| = \sqrt{\omega^2 + a^2}$ , $\mathrm{Arg}(j\omega + a) = \tan^{-1}(\omega/a)$ (learn this)
First order: $\tau\dfrac{dx}{dt} + x = Ku$ , $X(s) = \dfrac{K}{\tau s + 1}U(s)$ (learn this)
Proportional control: $U = k_P(V - X)$ ; closed loop $\dfrac{k_P G}{1 + k_P G}$ (learn this)

(a)
Use the Hurwitz method to determine the stability condition of the following characteristic equation: $s^3 + 4s^2 + s + 2 = 0$ .
Delta_2 =
Delta_3 =
[3]
(b)
Determine the frequency response for the following model i.e. showing your working, derive equations for the magnitude and phase. $G(s) = \frac{1}{(s + 1)(s + 2)(s + 5)}$
[4]
(c)
A first order system has a time constant of 8 s and a steady state gain of 7. Design a proportional control system such that the closed loop time constant is just 4 s. For a step in the set point of magnitude 10, what is the steady state error of this control system? Show all your working.
k_P =
steady state error =
[9]