ME 37500 Midterm Exam
February 14th, 2017
6:30 – 7:30 pm
CL50 224

Name: ________________________________ HWID:
Section: Deng / Shelton / Roll (circle one)
Instructions
(1) TURN OFF your cell phones and tablets. If you are using a smart watch, please also turn it off
and put it in your bags.
(2) Write your name and HWID at the top of each page.
(3) This is a closed book examination, but you are allowed one single-sided 8.5”×11” hand written
crib sheet.
(4) Only ME approve exam calculator: TI-30XIIs can be used
(5) You have one hour to work all four problems on the exam.
(6) If you use extra pages, be sure to write your NAME and HW ID on the top of each extra
page. Ensure that your pages are in the correct order and re-staple the packet together.
(7) Any work on the back of the page do not count and will not be graded.
(9) Use the solution procedure we have discussed: what are you given, what are you asked to find,
(10) You must write neatly and should use a logical format to solve the problems. You are encouraged
to really “think” about the problems before you start to solve them.
(11) A table of Laplace transform pairs and properties of Laplace transform is attached at the end of
this exam set.

Problem 1 (15 points) ________________________
Problem 2 (20 points) ________________________
Problem 3 (40 points) ________________________
Problem 4 (25 points) ________________________
TOTAL (***/100 points) ________________________

Name: HWID:
Page 2 of 8
PROBLEM 1 (15 points)

(A) For the closed-loop system shown below, find the characteristic equation. (5 pts.)

The transfer function is () = ()
() = ()1+() = 1+1+� 1
+
��

+

= (+)(+)(+)+. Thus, the characteristic
equation is ( + )( + ) + = 0. Alternatively, 2 + ( + ) + ( + ) = 0.
(B) Find the zeros of transfer function (). (5 pts.)
() = (3 + )(2 + )( + )2(2 + + )

The zeros are located at = 0, = −
3
, and = ±√.

(C) You are working with a feedback system that automatically maintains its two closed-loop poles
at = −2 ± 2, regardless of the number of system zeros. Assume there are currently no zeros,
and you must introduce a single zero into this system. If you want to minimize overshoot when a
unit step is applied at the reference input, should you place the zero at = −1, = −2, or
= −3? Why? (5 pts.)

Since the complex conjugate closed-loop poles are fixed, we desire the fastest possible zero to minimize
step response overshoot. Therefore, we choose = −3.

Name: HWID:
Page 3 of 8
PROBLEM 2 (20 points)
(A) Consider the following closed-loop transfer function:
() = ( + 2)
4 + 53 + 452 + 196
Based on the Hurwitz Criterion, is this system unstable? Why or why not? (5 pts.)

According to the Hurwitz Criterion, it is a necessary condition of stability that all coefficients in the
characteristic polynomial (transfer function denominator) be of the same sign, and non-zero. Since the
term is missing, the system is indeed unstable.
(B) Consider the following closed-loop transfer function:
() = ( − 1)
4 + 23 + 32 + 4 +
Use a Routh array to determine the values of that make the closed-loop system stable. (15
pts.)

Characteristic equation: () = 4 + 23 + 32 + 4 + = 0.
Constructing a Routh array:
4: 1 3
3: 2 4
2: 1
1: 4 − 2
0:
For closed-loop stability, all entries in the first column must be of the same sign. Thus, 0 < < 2.

Name: HWID:
Page 4 of 8
PROBLEM 3 (40 points)

Consider the following closed-loop system, in which is a non-negative integer ( = 0,1, 2, 3, … )

Let () = 0 for this non-unity feedback system, we have
() = ()1 + ()(),
and
() = () − () = 0 − () = − ()1 + ()().

(A) What is the minimum integer that causes the output () to completely reject a unit-step
disturbance ()? (15 pts.)
() = 25 + 251 + � 1� (2) � 25 + 25� = 25
−1
−1( + 25) + 252
() = = − = −25−1−1( + 25) + 252
To reject a step input, we need a Type 1 system, which requires at least 1 differentiator in the numerator
of (). Thus, ≥ 2.
Alternatively, using the final value theorem, with () = 1

,
() = lim→0 ⋅ 1 ⋅ 25−1−1( + 25) + 252 = lim→0 25−1−1( + 25) + 252
Since 0−1 = ∞, and 00 = 1, we need 0≥1 = 0 in the numerator as → 0 to drive the steady-state
output to zero. Hence, ≥ 2.
()
() ()

+
()
+ +
() () 1

2
25
+ 25
Name: HWID:
Page 5 of 8
PROBLEM 3 (continued)
(B) If = 1, () = 1

, and () = 0, what is the sensitivity function for the steady-state output
to changes in parameter ? (15 pts.)
Given that,
() = ()1 + ()() = � 25 + 25�1 + �1� �21 � � 25 + 25�() = 25( + 25) + 252 (),
⇒ () = 25 + 25(1 + 2)(),
therefore,
= lim→0 2525(1 + 2) 1 = 11 + 2.
The sensitivity of to variation in is then,

=

= ⋅ 1 + 21 ⋅ (1 + 2)(0)− 1(2)(1 + 2)2 = − 22(1 + 2)

(C) Let
represent the sensitivity of steady-state output to feedback gain . Assume = 0.2
and
= − 1.6 when = 2. If increases to 2.1, what is the change in ? (10 pts.)

It is given that at a nominal value of = 2, the sensitivity is =2 = −1.6 and the nominal steady-state
output is = 15 = 0.2. We can alternately express the sensitivity as = limΔ→0 �Δ ��Δ

. If
moves from 2 to 2.1, the denominator is Δ

= 0.1
2
= 0.05. Since Δ is small, Δ ≈ ⋅ �Δ �.
Thus,
Δ ≈ (0.2) �−85� (0.05) = −0.016.

Name: HWID:
Page 6 of 8
PROBLEM 4 (25 points)
Consider the following unity feedback system:

where
1( )
( 1)( 2)
G s
s s
=
+ +

a) Let ( ) as bC s
cs d
+
=
+
, determine the coefficients a , b , c , and d using the direct pole placement
method to locate the roots such that the desired closed-loop characteristic polynomial is
2 3 2( 6)( 4 20) 10 44 120s s s s s s+ + + = + + + . (10 pts.)

2
3 2 3 2
( 1)( 2)( ) 1( ) ( 6)( 4 20)
(3 ) (2 3 ) (2 ) 10 44 120
1
3 10 7
2 3 44 21
2 120 106
G C G C CLD D N N D
s s cs d as b s s s
cs c d s c d a s d b s s s
c
c d d
c d a a
d b b
+ =
+ + + + + = + + +
+ + + + + + + = + + +
=
+ = ⇒ =
+ + = ⇒ =
+ = ⇒ =

Therefore, the controller is: 21 106( )
7
sC s
s
+
=
+

Name: HWID:
Page 7 of 8
PROBLEM 4 (continued)

b) It is desired that the controller cancel one of the poles of the plant at −1. Use the direct pole
placement method to design a controller ( )C s that satisfies this condition while putting the
closed-loop system poles at 1 1s = − and 2,3 2 4s j= − ± . Hint: You need to determine the
controller form (different from part (a)) and its coefficients. (15 pts.)

Let
( 1)( ) C
C
s NC s
D
+
=

2
2
( 1)( 2) 1( 1) ( 1)( 4 20)
( 2) 4 20
G C G C CL
C C
C C
D D N N D
s s D s N s s s
s D N s s
+ =
+ + + + = + + +
+ + = + +

Note that the controller has to be proper, and the orders of the polynomials on the left and right side of the equation
should match. let
( 1) ( 1)( ) C
C
s N s aC s
D bs c
+ +
= =
+

2
2 2
( 2)( ) 4 20
(2 ) (2 ) 4 20
1
2 4 2
2 20 16
s bs c a s s
bs b c s c a s s
b
b c c
c a a
+ + + = + +
+ + + + = + +
=
+ = ⇒ =
+ = ⇒ =

Therefore, the controller is:
16( 1)( )
2
sC s
s
+
=
+

Name: HWID:
Page 8 of 8

Laplace Transform Pairs and Properties
( ( )F s is the Laplace transform of ( )f t . Note: ( ) 0, 0f t t= ∀ < )
( ), 0f t t ≥ ( )F s Comments
( )tδ 1 Unit Impulse
1 1
s
Unit Step
t 2
1
s
Unit Ramp
21
2
t 3
1
s
Unit Parabolic
te α−
1
s α+
Exponential
tte α−
( )sin tω 2 2s
ω
ω+
Sinusoidal
( )cos tω 2 2
s
s ω+

( )sinte tα ω−

( ) (0 )sF s f −− First Derivative
( ) ( )nf t nth Derivative
( )df t T− ( )dsTe F s− Time delay
( )te f tα− ( )F s α+ Modulation/translation
n tt e α− ( ) 1
!
n
n
s α ++

(0 ) lim ( )sf sF s
+
→∞= Initial Value Theorem
( )f t converges to
a constant value as
t →∞
0( ) lim ( )sf sF s→∞ = Final Value Theorem

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