midterm 2 2013 solution

Department of Mechanical, Automotive, & Materials Engineering401 Sunset Avenue, Windsor

Ontario, Canada N9B 3P4

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92-321 Control TheorySummer 2013

Midterm Exam #2

Engineering is a professional faculty, and as a student of engineering, it is expected that you will behave in aprofessional manner during your exam, and abide by the following standards of conduct.

1. Unless explicitly allowed, a candidate must not give assistance to, or receive assistance from, or commu-nicate in any manner with any person other than the examiner or proctors. A candidate who is involvedin such activity may be subject to disciplinary procedures of the University.

2. The possession of any telecommunications device is strictly prohibited during an exam, and will be con-sidered as cheating, whether it is used or not. Students found with said devices may be subject to todisciplinary procedures of the University. If you are in possession of any telecommunications devices,identify yourself to the proctor and surrender them immediately for the duration of the exam. Pleasenote that the University is not responsible for lost or stolen items.

3. The possession of any unauthorized aids during an exam will be considered as cheating, whether it is usedor not. Students found with said materials may be subject to disciplinary procedures of the University.

4. Be prepared to hand in your exam to the proctors immediately after time expires. Students who delaycollection of the exams by the proctor (i.e., they must wait for you to finish writing, sign your name,etc.,) may be subject to downward adjustment of their grade.

5. Remain seated quietly until all the exams have been collected, not just your own. Recognize that otherstudents are still holding their exams and could be influenced by any discussions. There will be plenty ofopportunity to discuss the exam outside the hall.

6. Each student will be allowed a maximum of three inquiries regarding the exam material; further requestsfor clarification will be denied. If you have uncertainties about the exam, make sure you have readthe entire question, state your assumptions, and proceed. Do not waste the examiner’s time by askingquestions regarding the correctness of your solution.

7. Answer all questions, to the best of your ability.

1



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Question 1

Shown in Figure 1 is a model of an engine and engine mount bushing commonly used in the automotiveindustry. The rubber compounds in the bushing do not behave like a normal spring-damper, so the model ismodified by adding an additional spring and damper in series. This model is sometimes referred to as ‘frequencydependent’, because at low frequencies, the damper allows the secondary spring to move without compressing,lowering the total stiffness, but at higher frequencies, the decreased motion of the damper gives an effectivelystiffer mount. Notice that in order to model a mount like this, an additional coordinate (x) is needed to markthe location of the midpoint of the series spring-damper.

Use the following values in the problem:

Parameter Value Parameter Valuem 50 kg c2 100 N · s/mk 15000 N/m k2 5000 N/mc 100 N · s/m

a) Write the equation of motion of the engine mass. Hint: knowing the location of the midpoint (x) allowsthe compression in the secondary spring to be found, and therefore the force. Another hint: rememberthat all the spring and damper forces rely on the relative motion, i.e., the difference in the motion of eachside, and notice that the base will have motion u(t).

b) Next, consider the series spring-damper. Recognize that the spring and damper force must be equal.Setting the spring force (a function of x) equal to the damper force (a function of x) results in a firstorder differential equation describing the location of the midpoint. This equation can be combined withthe equation of motion of the engine mass.

c) Find the transmissibility transfer function G(s) = Y (s)U(s) relating the engine mass motion to the base motion.

Hint: take Laplace and gather your terms before substituting to eliminate x .d) What is the order of the resulting transfer function? Describe the possible types of unforced motion that

you could encounter with this type of system, assuming feasible physical parameters. Is it hard to givespecifics? Why or why not?

e) If the base motion is a sine wave with amplitude of U0 = 0.1 m and frequency of 1.5916 Hz, write theequation that describes the engine mass motion as a function of time, after all the transient terms havevanished.

a)my + c( y − u) + k(y − u) + k2(y − x) = 0

my + c y + (k+ k2)y = ku+ cu+ k2 x

b)k2(y − x) = c2( x − u)

2 c⃝BP Minaker PhD 2013



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c) Taking Laplace �ms2+ cs+ k+ k2

�y = (k+ cs)u+ k2 x

k2(y − x) = c2s(x − u)

k2 y + c2su= (k2+ c2s)x

k2

k2+ c2sy +

c2s

k2+ c2su= x

�ms2+ cs+ k+ k2

�y = (k+ cs)u+ k2

�k2

k2+ c2sy +

c2s

k2+ c2su�

�ms2+ cs+ k+ k2− k2

2

k2+ c2s

�y =�

k+ cs+k2c2s

k2+ c2s

�u�

(k2+ c2s)(ms2+ cs+ k+ k2)− k22

�y =�(k2+ c2s)(k+ cs) + k2c2s

�u

G(s) =(k2+ c2s)(k+ cs) + k2c2s

(k2+ c2s)(ms2+ cs+ k+ k2)− k22

d) The resulting system is third order. We should expect there will be either one negative real root, and twocomplex roots with negative real parts (i.e., underdamped), or three negative real roots (overdamped).The first case indicates an oscillatory response, the second purely exponential decay. It is hard to givespecifics because the system is third order, and the roots are usually found using numerical methods.

e) Setting s = iω, where ω= 1.5916 ∗ 2π= 10

G(s) =(k2+ c2s)(k+ cs) + k2c2s

(k2+ c2s)(ms2+ cs+ k+ k2)− k22

G(10i) =(5000+ 1000i)(15000+ 1000i) + (5000)(1000i)

(5000+ 1000i)(−5000+ 1000i + 20000)− 25000000

G(10i) = 1.473− 0.0910i

|G(10i)|= 1.476

∠G(10i) =−0.061725

y = 0.147sin(10t − 0.0617)




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Question 2

Consider the schematic diagram of the water tank shown in Figure 2. The water outlet is closed, and the inletis fitted with a valve that adjusts the flow rate of the water entering the tank. The valve is made of a smallpiston that slides inside a sleeve. As the piston slides, it uncovers an opening in the sleeve that allows water toflow into the sleeve and on into the tank. The motion of the piston is restrained by a small spring and damper.Assume that the volume flow rate of the water into the tank is proportional to the distance that the valve opens:qi = bx . A force F is applied to the piston to open the valve. Using the following values in the problem:

Parameter Value Parameter Valuetank area At 2m2 valve damping c 20Ns/mvalve mass m 100g valve sensitivity b 1L/s/mm

valve stiffness k 2N/mm

a) Write the equation of motion for the valve. Assume that any forces on the valve due to the water floware negligible.

b) Write the volume flow balance equation for the tank (the water can be treated as incompressible).c) Relate the height of the water in the tank to the force acting on the valve using the state space form

shown below. Let v be the velocity of the valve. Fill in the ai, j , bi , ci .hvx

=a11 a12 a13

a21 a22 a23a31 a32 a33

hv

x

+b1

b2b3

F

h=�

c1 c2 c3

�hvx

d) Find the eigenvalues of the state space representation, and describe what they tell you about its behaviour.

Ex: is there any oscillatory behavior? If so, what are the natural frequencies and damping ratios? Howlong will the transient behaviour last?

e) Find the eigenvectors of the state space representation, and describe what they tell you about its be-haviour. Hint: try h= 1, or v = 1 when solving.

a)mv + cv + kx = F

v =− k

mx − c

mv +

1

mF

x = v




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b)qi − qo = V

bx − 0= At h

h=b

Atx

c) hvx

=0 0 b

At

0 − cm− k

m0 1 0

hv

x

+01m

0

F

h=�

1 0 0�hv

x

d)

det[Is− A] = 0

A=

0 0 bAt

0 − cm− k

m0 1 0

=0 0 0.5

0 −200 −200000 1 0

[Is− A] =

s 0 −0.50 s+ 200 200000 −1 s

det[Is− A] = (s)[(s+ 200)(s) + 20000] = (s)(s2+ 200s+ 20000) = 0

s = 0, s =−200±p40000− (4)(20000)

2=−100± 100i

One rigid body mode, and one damped oscillatory vibration.

ωn =p

1002+ 1002 = 141.4 rad/s

ζ= 0.707

τ= 1/(ζωn) = 0.01 sec

The frequency is quite high, 22.5 Hz, and so is the damping ratio. We expect the motion to decay veryquickly, less than 1/20th of a second.




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e)s = 0

[Is− A]X = 00 0 −0.50 200 200000 −1 0

hv

x

= 0

hvx

=10

0

The rigid body mode corresponds entirely to the h entry - the system can be stationary at any value of h,but not any non-zero value of x or v. The system will come to rest at any value of the water level, butonly with the valve closed.

s =−100+ 100i

[Is− A]X = 0−100+ 100i 0 −0.50 100+ 100i 200000 −1 −100+ 100i

hv

x

= 0

Let v = 1100+ 100i + 20000x = 0

x =−0.005− 0.005i

(−100+ 100i)h− 0.5(−0.005− 0.005i) = 0

h= 2.5000× 10−05i

hvx

=2.5000× 10−05i

1−0.005− 0.005i

The oscillatory mode consists of high frequency, high velocity low amplitude motion of the valve, thathas almost zero effect on the height of the water in the tank. Note that the equation for the flow impliesthat the flow can be in either direction, which would not be true in reality. However, the sinusoidalcomponent of the solution would be superimposed on top of flows resulting from non-zero applied forcesat the valve.




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..

y(t)

.

m

.

k2, c2

.

k, c

.

u(t)

.

x(t)

Figure 1: Frequency dependent mount model

..

h

.

qi

.

k, c

.

m

.

x(t)

.

F(t)

Figure 2: A tank with input flow valve




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Useful Information

Laplace transforms

F(s) f(t)1 δ(t)

1s

us(t)

1s+a

e−at

bs2+b2 sin bt

ss2+b2 cos bt

b(s+a)2+b2 e−at sin bt

s+a(s+a)2+b2 e−at cos bt

sF(s)− f (0) d fd t

s2F(s)− s f (0)− d fd t

��t=0

d2 fd t2

For a transfer function of the form

G(s) =c3s3+ c2s2+ c1s+ c0

s4+ d3s3+ d2s2+ d1s+ d0

A possible state space representation is

A=

0 1 0 00 0 1 00 0 0 1−d0 −d1 −d2 −d3

B =

0001

C =�

c0 c1 c2 c3

�8 c⃝BP Minaker PhD 2013



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The equations of motion for translationΣF = maG

The equations of motion for rotation for planar problems can be written using A, an arbitrary point as areference

ΣMA = IAα+ rG/A×maA

orΣMA = IGα+ rG/A×maG

If the point A is chosen as either the centre of mass, G, or a fixed point, O, the equations can be simplified

ΣMG = IGα

orΣMO = IOα

The magnitude of a complex number �� c + di

a+ bi

��=p

c2+ d2pa2+ b2

The angle of a complex number

∠�

c + di

a+ bi

�= tan−1�

d

c

�− tan−1�

b

a

�The determinant of a matrix A, is computed by choosing any row or column, proceeding along this row orcolumn, multiplying each entry of the row or column by the determinant of the ‘minor’ (i.e, the matrix, withthe row and column of the current entry ommitted.) Additionally, the sign of every other entry is reversed.This relies on the recognition that the determinant of a 2x2 matrix is the product of the main diagonal, less theproduct of the off-diagonal. For example, using the first column

A=

a11 a12 a13a21 a22 a23a31 a32 a33

det[A] = a11�a22a33− a23a32

�− a21�a12a33− a13a32

�+ a31�a12a23− a13a22

�


midterm 2 2013 solution

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