ACT3701 Assignment 01 Solutions Semester 1 2024

Automatic Control Systems

Assessment 1 Answers

Semester 1 2024
18 APRIL 2024
DUE DATE




0027 65 934 4052

,
,
, 1.

𝑑3 𝑦(𝑡) 𝑑2 𝑦(𝑡) 𝑑𝑦(𝑡) 𝑑𝑟(𝑡)
3
+ 2 2
+5 + 6𝑦(𝑡) = 3 + 𝑟(𝑡) … … … … . (1)
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡

𝑦(0) = 1𝑚, 𝑑𝑦(0)/𝑑𝑡 = 10𝑚𝑠 −1 , 𝑑2 𝑦(0)/𝑑𝑡 2 = 0 𝑚𝑠 −2 , 𝑟(𝑡) = 𝑒 −2𝑡


Equation (1) can be wrtten as;

𝑦 ′′′ (𝑡) + 2𝑦 ′′ (𝑡) + 5𝑦 ′ (𝑡) + 6𝑦(𝑡) = 3𝑟 ′ (𝑡) + 𝑟(𝑡)


Taking Laplace Transform on both sides of the differential equation;

ℒ{𝑦 ′′′ (𝑡) + 2𝑦 ′′ (𝑡) + 5𝑦 ′ (𝑡) + 6𝑦(𝑡)} = ℒ{3𝑟 ′ (𝑡) + 𝑟(𝑡)}

[𝑠 3 𝑌(𝑠) − 𝑠 2 𝑦(0) − 𝑠𝑦 ′ (0) − 𝑦 ′′ (0)] + 2[𝑠 2 𝑌(𝑠) − 𝑠𝑦(0) − 𝑦 ′ (0)]

𝑑
+5[𝑠𝑌(𝑠) − 𝑦(0)] + 6𝑌(𝑠) = 3. ℒ{𝑒 −2𝑡 } + ℒ{𝑒 −2𝑡 }
𝑑𝑡


Substituting the initial conditions we have;

[𝑠 3 𝑌(𝑠) − 𝑠 2 (1) − 𝑠(10) − 0] + 2[𝑠 2 𝑌(𝑠) − 𝑠(1) − 10] + 5[𝑠𝑌(𝑠) − 1] + 6𝑌(𝑠)

−3 1
= +
(𝑠 + 2)2 𝑠 + 2


𝑠 3 𝑌(𝑠) − 𝑠 2 − 10𝑠 + 2𝑠 2 𝑌(𝑠) − 2𝑠 − 20 + 5𝑠𝑌(𝑠) − 5 + 6𝑌(𝑠)

−3 1
= 2
+
(𝑠 + 2) 𝑠+2


−3 1
(𝑠 3 + 2𝑠 2 + 5𝑠 + 6)𝑌(𝑠) − 𝑠 2 − 12𝑠 − 25 = +
(𝑠 + 2)2 𝑠 + 2