midtem_2_2014_soln

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92-321 Control Theory

Fall 2014

Midterm Exam 2

Engineering is a professional faculty, and as a student of engineering, it is expected that you will behave in a

professional 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 involved

in 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 to

disciplinary 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. Please note

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 used

or 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 delay

collection 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 other

students are still holding their exams and could be influenced by any discussions. There will be plenty of

opportunity to discuss the exam outside the hall.

6. Each student will be allowed a maximum of three inquiries regarding the exam material; further requests

for clarification will be denied. If you have uncertainties about the exam, make sure you have read the

entire question, state your assumptions, and proceed. Do not waste the examiner’s time by asking questions

regarding the correctness of your solution.

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

1

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Question 1

Consider the three tank system shown in Figure 1. Liquid flows into tank A, with a volume flow rate of qi , and

between the first two tanks through an interconnecting tube. Both the flow between the tanks, and the flow

out of tank B can be treated as laminar, and therefore linearly proportional to pressure. The flow out of tank B

falls freely, and collects in tank C on the floor below.

Table 1: Two tank system parameters

Parameter Value

A A 9.81 m2

R A 2000 m−1s−1

A B 4.91 m2

R B 2000 m−1s−1

AC 2.45 m2

a) Write the state space equations that describe the level of liquid in each tank. Hint: use conservation of

volume.

b) Find the eigenvalues of the system matrix, and describe the information that you retrieve.

c) Find the eigenvectors of the system matrix, and describe the information that you retrieve.

2 cBP Minaker PhD PEng 2014

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qi

h A

A A

q A− B

R A

A B

hC

AC

R B

q B−C

Figure 1: Three tank system


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Solution

a) Write the conservation equations.

A Ah A = qi − qo

A Ah A = qi − g

R A

(h A− h B)

h A = qi

A A

− g

A A R A

h A + g

A A R A

h B

A B

h B

= qi −

qo

A Bh B = g

R A

(h A− h B)− g

R B

h B

h B = g

A B R A

h A−

g

A B R A

+ g

A B R B

h B

AC hC = qi − qo

hC = g

AC R B

h B

h A

h B

hC

=− g

A A R A

g A A R A

0 g

A B R A − g

A B R A + g

A B R B

00

g AC R B

0

h A

h B

hC

+

1 A A

00

qi

h A

h B

hC

=

−0.0005 0.0005 0

0.001 −0.002 0

0 0.002 0

h A

h B

hC

+

0.102

0

0

qi

b) Find the eigenvalues

det[ Is− A] = 0

[ Is− A] =

s + 0.0005 −0.0005 0

−0.001 s + 0.002 0

0 −0.002 s

det[ I s− A] = ( s + 0.0005)( s + 0.002)( s)− (−0.0005)(−0.001)( s) = 0

det[ I s− A] = s

( s2 + 0.0025 s + 0.000001− 0.0000005

= 0


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s = −0.0025± 0.00252 − 4(0.0000005)

2

s = 0,−2.2808e− 03,−2.1922e− 04

There are three roots, but one is zero, indicating that the system will never come to steady state. This

is obvious by inspection of the system. The other two roots are negative real, and so indicate stable

exponential decay, but with very different speeds. The fast one has as time constant of 438 seconds or 7.3

minutes, the slow one 4562 seconds or 76 minutes.c) To find the eigenvectors

[ Is− A] X = 0

[ I s− A] X =

s + 0.0005 −0.0005 0

−0.001 s + 0.002 0

0 −0.002 s

h1

h2

h3

= 0

Substitute first eigenvalue

s = 0

[ Is− A] X =

0.0005 −0.0005 0

−0.001 0.002 0

0 −0.002 0

h1

h2

h3

= 0

Last eqn says h2 = 0 and first says h1 = h2, and h3 can be anything, so set it at one.

X 1 =

0

0

1

This makes sense. The first two tanks can reach steady values with a constant in flow, but tank three can

only continue to increase. The second eigenvalue:

s = −2.28e− 03

[ Is− A] X =

−0.00178 −0.0005 0

−0.001 −0.000281 0

0 −0.002 −0.00228

h1

h2

h3

= 0

Set h1 = 1, and solve first equation for h2, then third for h3.

X 2 =

1.00

−3.56

3.12


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The last eigenvalue

s = −2.19e− 04

[ Is− A] X =

0.000281 −0.0005 0

−0.001 0.00178 0

0 −0.002 −0.000219

h1

h2

h3

= 0

Set h1 = 1, and solve first equation for h2, then third for h3.

X 3

=

1.00

0.562

−5.12

In the zero mode, the vector indicates that tank C will continue to fill, even when the other tanks are at

a constant level. In the fast mode, tank B empties while tanks A and C fill, with tank C filling about three

times faster than tank A. In the slow mode, tanks A and B empty into tank C, with A training about twice

as fast as B, and C filling about 5 times faster than A drains.


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Question 2

A truck is towing a trailer carrying a ride-on lawnmower. A simple schematic drawing is shown in Figure 2.

The trailer bounces vertically due to imperfections in the road, causing the lawnmower to bounce on the trailer.

Assume the trailer and the lawnmower both have simple suspensions that can be modelled as linear springs,

and that damping and friction can be neglected. Ignore any pitching motion of the trailer or the lawnmower,

and the effects of the mass or stiffness of the tires. Allow that the combined stiffness of the trailer springs is k t ,

and the combined stiffness of the lawnmower springs is km.

Table 2: Trailer and lawnmower parametersParameter Value

mt 250 kg

mm 150 kg

kt 5000 N/m

km 7500 N/m

a) Write the four state space equations describing the vertical motion of the trailer and the lawnmower. Use

x t , vt , as the states of the trailer, and x m, vm as the states of the mower. Put the states in the following

order: [ x m, x t , vm, vt ]. Allow that the input causing the motion is the vertical location of the road x 0, andthat the output of interest is x m.

b) Convert the resulting state space to a transfer function that relates the position of the lawnmower to the

vertical location of road. See the attached useful information sheet for expressions.

c) Find the two natural frequencies of the system. Hint: you will need the relations: r = s2, r 2 = s4.

d) If the road can be approximated as a sinusoid with an amplitude of 0.05 m, and a wavelength of λ = 5 m,

and the truck is travelling with a forward speed of s = 10 m/s, find an expression for the resulting vertical

motion of the lawnmower as a function of time. Hint: ω = 2π sλ


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x t

x m

springs with total stiffness ‘kt ’

x 0

springs with total stiffness ‘km’

Figure 2: A trailer and lawnmower


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Solution

a) The equations of motion

˙ x m = vm

˙ x t = vt

mm ˙ vm + km( x m − x t ) = 0

˙ vm = km

mm

x t − km

mm

x m = 0

mt ˙ vt + km( x t − x m) + kt ( x t − x 0) = 0

˙ vt = km

mt

x m − km

mt

x t − kt

mt

x t = kt

mt

x 0

˙ x m˙ x t˙ vm

˙ vt

=

0 0 1 0

0 0 0 1

− km

mm

km

mm0 0

km

mt− km+kt

mt0 0

x m x t vm

vt

+

0

0

0kt

mt

x 0

˙ x m˙ x t˙ vm

˙ vt

=

0 0 1 0

0 0 0 1

−50 50 0 0

30 −50 0 0

x m x t vm

vt

+

0

0

0

20

x 0

x m =

1 0 0 0

x m x t vm

vt

b) Converting to transfer function

[ I s− A] =

s 0 −1 0

0 s 0 −1

50 −50 s 0

−30 50 0 s

det[ Is− A] = s

s 0 −1

−50 s 0

50 0 s

− 1

0 s −1

50 −50 0

−30 50 s

det[ I s− A] = s( s3 + 50 s)− 1(−50 s2 − 502 + 50(30)) = s4 + 100 s2 + 1000

BC =

0

0

0

20

1 0 0 0

=

0 0 0 0

0 0 0 0

0 0 0 0

20 0 0 0


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[ Is− A + B C ] =

s 0 −1 0

0 s 0 −1

50 −50 s 0

−10 50 0 s

det[ Is− A + B C ] = s( s3 + 50 s)− 1(−50 s2 − 502 + 50(10)) = s4 + 100 s2 + 2000

det[ Is− A + BC ]− det[ I s− A] = 1000

G( s) = 1000

s4 + 100 s2 + 1000

Note the matching values in the numerator, and the constant term in the denominator. This means at very low frequencies ( s = iω, ω ≈ 0) that G ( s) will tend to one. This makes sense: if you drive over a bump

very slowly, the motion of the lawnmover will be the same as the road profile.

c) From the denominator of the transfer function

det[ I s− A] = s4 + 100 s2 + 1000 = r 2 + 100r + 1000 = 0

r = −100± 10000− 4000

2 =

−100± 77.46

2 = −88.73,−11.27

s =

r = 9.42i,3.36i

The purely imaginary roots indicate oscillatory response, with no decay. We have ignored damping, so

this makes sense. Converting units gives natural frequencies of 1.50 Hz, and 0.535 Hz.

d) Find the frequency response

ω = 2π10

5 = 4π rad/s

s = 4πi

G(iω) = 1000

ω4 − 100ω2 + 1000

G(4πi) = 1000

(4π)4 − 100(4π)2 + 1000 = 0.0986

|G(4πi)| = 0.0986

φ = tan−1

0.0

0.0986

= 0

x m = (0.0986)(0.05) sin(4πt) = 0.00478 sin(4πt) m


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Useful Information

Useful Information

Laplace transforms

F(s) f(t)

1 δ(t)

1 s u s(t)

1 s+a e−at

b s2+b2 sin bt

s s2+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 f

d t

s2 F ( s)− s f (0)− d f

d t

t=0

d2 f

d t2

State space

For a transfer function of the form

G( s) = c3 s

3 + c2 s2 + c1 s + c0

s4 + d3 s3 + d2 s

2 + d1 s + d0

A possible state space representation is

A =

0 1 0 0

0 0 1 0

0 0 0 1

−d0 −d1 −d2 −d3

B =

0

0

0

1

C =

c0 c1 c2 c3

G( s) = C [ I s− A]−1 B + D

= det[ Is− A + B C ]− det[ Is− A]

det[ Is− A] + D

= det[ Is− ( A− BC )]− det[ I s− A]

det[ Is− A] + D

Complex numbers

The magnitude of a complex number

c + d i

a + bi

=

c2 + d 2

a2 + b2

The angle of a complex number

∠c + d i

a + bi = t an−1

d

c−

t an−1

b

a

Determinants

The determinant of a matrix A is computed by choos-

ing any row or column, proceeding along this row or

column, multiplying each entry of the row or column

by the determinant of the ‘minor’ (i.e, the matrix, with

the 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

the product of the off-diagonal. For example, using the

first column

A =

a11 a12 a13

a21 a22 a23

a31 a32 a33

det[ A] =a11 (a22a33− a23a32)

− a21 (a12a33− a13a32)

+ a31 (a12a23− a13a22)


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Eigen analysis

det[ I s− A] = 0

[ Is− A] X = 0

For an eigenvalue s = a± bi , the natural frequency and

damping ratio are

ωn =

a2 + b2

ζ = −a

a2 + b2

a = −ζωn

b = ωd =

1− ζ2ωn

Frequency response

s = iω

ωr =

1− 2ζ2ωn

Dynamics

The equations of motion for translation

Σ F = maG

The equations of motion for rotation for planar prob-

lems can be written using A, an arbitrary point as a

reference

Σ M A = I Aα+ rG/ A×ma A

or

Σ M A = I G α+ 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

Σ M G = I G α

or

Σ M O = I O α

12 BP Mi k PhD PE 2014

midtem_2_2014_soln

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