EPE2200
Automatic Control
___________________________________________________________________________
Midterm Test 2 - <Trimester 1, Academic Year 20/21>
<Thursday>, <14/11/2022>
4:00pm – 5:30pm
(1.5 hours)
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Instructions to students:
1. This examination comprises TWO (2) Questions printed on TWO (2) pages
including the cover page.
2. You must answer ALL the questions.
3. All answers must be written in the answer book
At the end of the examination,
Please write your Student ID on each answer book used.
Failure to do so will mean that your work will not be identified.
If you have used more than one answer book, please tie the answer books together using the
strings provided.
The University reserves the right not to mark your
script if you fail to follow these instructions.
Examination - <Trimester, Academic Year>
Page X of Y
Question 1. Consider the system with the Open Loop Transfer Function given as
𝐺OL (𝑠) = 𝑘
𝑠 2 + 6𝑠 + 9
𝑠 2 + 2𝑠 + 1
Now answer the following:
(a) Draw the Root Locus of the system and show all the steps clearly.
[20 Marks]
Step 1: OL Poles & Zeros
OL Numerator Polynomial Eq.: 𝑠 2 + 6𝑠 + 9 = 0 ⇒ 𝑠 = −3, −3 & 𝑛𝑧 = 2
OL Denominator Polynomial Eq.: 𝑠 2 + 2𝑠 + 1 = 0 ⇒ 𝑠 = −1, −1 & 𝑛𝑝 = 2
We have 𝑛𝑝 = 2 branches and 𝑛𝑝 − 𝑛𝑧 = 0 asymptotes. That means each
branch will converge with an OL zero.
𝐼𝑚
xx
−3
−1
0
𝑅𝑒
Step 2: Break in/away points
From the C.E.
𝑠2 +2𝑠+1
𝐺𝑂𝐿 (𝑠) + 1 = 0 ⇒ 𝑘 = − 𝑠2 +6𝑠+9
𝑑𝑘
𝑑 𝑠 2 + 2𝑠 + 1
−
=
(
)=
𝑑𝑠 𝑑𝑠 𝑠 2 + 6𝑠 + 9
𝑑
𝑑
(𝑠 2 + 6𝑠 + 9) (𝑠 2 + 2𝑠 + 1) − (𝑠 2 + 2𝑠 + 1) (𝑠 2 + 6𝑠 + 9)
𝑑𝑠
𝑑𝑠
=
=0=
(𝑠 2 + 6𝑠 + 9)2
(𝑠 2 + 6𝑠 + 9)(2𝑠 + 2) − (𝑠 2 + 2𝑠 + 1)(2𝑠 + 6)
=
=
(𝑠 2 + 6𝑠 + 9)2
= 2(𝑠 + 3)2 (𝑠 + 1) − 2(𝑠 + 1)2 (𝑠 + 3) = 0 =
= 2(𝑠 + 3)(𝑠 + 1)(𝑠 + 3 − 𝑠 − 1) = 4(𝑠 + 3)(𝑠 + 1) = 0
Therefore, the roots are 𝑠 = −1 & − 3 which are located at the OL poles &
zeros, so, they are break away (−1) and break in (−3).
Examination - <Trimester, Academic Year>
Page X of Y
𝐼𝑚
xx
−3
−1
0
𝑅𝑒
The branches don’t cross the Im-axis, therefore, we don’t need to do the
other steps.
(b) Can this system become unstable? Explain your answer.
[5 Marks]
No, as the branches of the R-Locus don’t cross the Im-axis.
Question 2. Consider the system with the Open Loop Transfer Function given as
10
𝑠(𝑠 + 10)
It is required that the CL poles be located at 𝒔 = −𝟓, −𝟓 and that the static
velocity gain be 𝑲𝒗 = 𝟐𝟎. Now answer the following:
𝐺(𝑠) = 𝐺𝑐 (𝑠)
(a) What is the value of the controller gain, 𝐾𝑐 , for the uncompensated system?
[5 Marks]
To determine the value of 𝐾𝑐 , we will use the magnitude criteria. This is
given as
|𝐺𝑂𝐿 (𝑠)|𝑠=−5 = |
10𝐾𝑐
10𝐾𝑐
|
=|
|=1
(−5)(5)
𝑠(𝑠 + 10) 𝑠=−5
𝐾𝑐 = 2.5
(b) What is the value of the state velocity gain, 𝐾𝑣 , for the uncompensated system?
[5 Marks]
To determine the value of 𝐾𝑣 , we will use the formula
𝐾𝑣 = lim 𝑠𝐺𝑂𝐿 (𝑠) = lim 𝑠 ×
𝑠→0
Examination - <Trimester, Academic Year>
𝑠→0
10𝐾𝑐
10𝐾𝑐
=
= 𝐾𝑐 = 2.5
𝑠(𝑠 + 10) (0 + 10)
Page X of Y
(c) Design a Lag Compensator such that the design requirements are achieved.
[15 Marks]
From (b) the value of 𝐾𝑣 for the uncompensated system needs to be
increased by a factor of 8 to satisfy the design requirement. Therefore, for
𝐺𝑐 (𝑠) = 𝐾𝑐
𝑠+𝑎
𝑠+𝑏
We choose
𝑎
𝑏
= 8 ⇒ choose 𝑏 = 0.001 ⇒ 𝑎 = 0.008
then,
𝐺𝑐 (𝑠) = 𝐾𝑐
𝑠 + 0.008
𝑠 + 0.001
To determine the value of 𝐾𝑐 , we will use the magnitude criteria. This is
given as
|𝐺𝑂𝐿 (𝑠)|𝑠=−5 = |𝐾𝑐
𝑠 + 0.008
10𝐾𝑐
−4.992
10𝐾𝑐
×
|
=|
×
|=1
𝑠 + 0.001 𝑠(𝑠 + 10) 𝑠=−5
−4.999 (−5)(5)
𝐾𝑐 ≈ 2.5035
To determine the value of 𝐾𝑣 , we will use
𝐾𝑣 = lim 𝑠𝐺𝑂𝐿 (𝑠) = lim 𝑠 ×
𝑠→0
Examination - <Trimester, Academic Year>
𝑠→0
𝑠 + 0.008
10𝐾𝑐
×
≈ 20.03
𝑠 + 0.001 𝑠(𝑠 + 10)
Page X of Y