# ELEN90055讲解、辅导Java,c++语言编程

Control Systems
Midsemester Test
Semester 2, 2021
Instructions
This test consists of 2 questions, with marks as indicated, summing to 27. You have one
12.50pm, so you have time for scanning and upload. For full marks, complete all questions
This test is ＆open book＊ and you may refer to any subject materials. The lecturer will
be available on the usual zoom channel until 1.05pm if you need clarification - use the
chat function in zoom to avoid distracting others. You are not allowed to communicate
or collaborate with, or seek/provide assistance from/to, any other persons, from the start
time of the test until the late submission time is over. The questions are randomised.
Late submissions are permitted until 1.30pm Melbourne time but a penalty may apply.
Submissions after that time will not be accepted. In case of potential technical issues,
download a copy of this question sheet onto your computer or device as soon as you start.
If you have technical issues, take a screenshot of the error message.
1. (2 + 5 + 4 + 3 = 14 marks) Consider a system given by the time-domain differential
equation
y“ ? y步
y
+ 6y步 ? 6 ln |y| = u步? u,
where y 6= 0 is the output and u is the input. Let (y‘, u‘) denote an equilibrium point.
(a) Find an expression for y‘ in terms of u‘.
(b) Find the linearised (i.e. incremental) model of this system if u‘ = 0 and y‘ =
variables 汛y, 汛u and their time-derivatives.
(c) Find the simplified transfer function G(s) for your linearised model as a ratio
of two coprime polynomials. Plot any poles and zeros (using x＊s and o＊s respec-
tively) in the complex plane, and check that G(s) is BIBO stable (do not define
BIBO stability).
(d) Suppose u‘ and 汛u are both = 0 for all t ≡ 0?. It is observed that when the
incremental initial condition 汛y(0
?) is not exactly zero, the output y(t) does
not approach the equilibrium value y‘ = 1 as t ↙ ﹢. Explain briefly why this
happens even though G(s) is BIBO stable.
1
2. (3 + 2 + 5 + 3 = 13 marks) Consider a plant with transfer function
G(s) =
1
s2 + 4
You wish to use a controller C(s) = q(s + 1) + k/s to stabilise this plant in a unity
feedback loop, where k and q are real parameters to be chosen.
(a) Find a simplified expression for the closed-loop transfer function H(s) from the
reference signal to the plant output. Express your answer as a ratio of two
polynomials.
(b) Assuming closed-loop stability, show that this controller achieves a steady-state
error of zero when the reference is a step function.
(c) The characteristic equation is
s3 + qs2 + (4 + q)s+ k = 0.
Find conditions on q and k for the closed-loop system to be stable.
(d) Referring to suitable transfer functions, explain briefly why making q large would
improve how well the output follows the reference, but also increases the sensi-
tivity of the output to high-frequency measurement noise.