midtem_1_2014_soln

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

Ontario, Canada N9B 3P4

519-253-3000

www.uwindsor.ca

92-321 Control TheoryFall 2014

Midterm Exam 1

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. Please notethat 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 read theentire question, state your assumptions, and proceed. Do not waste the examiner’s time by asking questionsregarding the correctness of your solution.

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

1



519-253-3000

www.uwindsor.ca

Question 1

Consider the system shown in Figure 1. A sensitive instrument is mounted using a spring and damper in thehopes of reducing its vertical motion, denoted y(t). It is mounted on a base that vibrates with a displacementof u(t). (Hint: gravity can be safely ignored in this problem.) Use the values in Table 1 in the problem.

Table 1: Instrument dataParameter Value

c 300 Ns/mk 151700 N/mm 50.0 kg

a) Find the transfer function G(s) = Y (s)U(s) relating the vertical motion of the instrument to the base motion.

Hint: the force in the damper is fd = c(u− y). Use a similar approach to find the force in the spring.b) What is the order of the resulting transfer function? Describe the motion you would expect from this

system. Find the time constants, the natural frequency, and damping ratio as appropriate to show howyou know.

c) If the system was disturbed from rest, how many oscillations would you expect before the amplitude ofthe motion was reduced by 95%?

d) Use a partial fractions expansion to find the impulse response of the system (u(t) = δ(t)), when the initialconditions are zero.

c k

m

y(t)

u(t)

Figure 1: Instrument suspended on vibrating base

2 c©BP Minaker PhD PEng 2014



519-253-3000

www.uwindsor.ca

Solution

a) Consider the summation of forcesΣF = fd + fs = my

fd = c(u− y)

fs = k(u− y)

Simplifyk(u− y) + c(u− y) = my

my + c y + k y = ku+ cu

Now that we have the ODE, we can use Laplace Transforms to obtain the transfer function. Note: initialconditions are assumed to be zero: y(0) = y(0) = u(0) = u(0) = 0. Therefore, the transfer function is

(ms2 + cs+ k)Y (s) = (k+ cs)U(s)

G(s) =Y (s)U(s)

=cs+ k

ms2 + cs+ k

b) The resulting system is second order. Given the high stiffness, I expect that it will be underdamped andoscillatory.

Y (s)U(s)

=300s+ 151700

50s2 + 300s+ 151700

s =−300±

p

3002 − 4(50)(151700)(2)(50)

s = −3± 55i

ωd = 55 rad/s = 8.75 Hz

ωn =

√

√15170050

= 55.1 rad/s = 8.77 Hz

ζ=300

2p

(50)(151700)= 0.0545

The motion is very lightly damped at oscillates at around 9 Hz.c) The time constant is

τ= 1/3 sec

Three time constants is3τ= 1 sec

One wavelengthλ= 1/8.75= 0.114 sec

So we will see1/0.114= 8.75 c ycles




519-253-3000

www.uwindsor.ca

d) From the Laplace transform of u(t) = δ(t), U(s) = 1.

Y (s) =300s+ 151700

50s2 + 300s+ 151700=

6s+ 3034s2 + 6s+ 3034

Y (s) =6s+ 3034

(s+ 3)2 + 3025=

A(s+ 3)(s+ 3)2 + 3025

+Bp

3025(s+ 3)2 + 3025

A= 6

3A+ 55B = 3034

B =3034− 18

55= 54.84

y(t) = e−3t(6cos(55t) + 54.84 sin(55t))

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−40

−20

0

20

40

60

0

t [s]

y[m]

Figure 2: Vertical displacement vs. time




519-253-3000

www.uwindsor.ca

Question 2

Consider the block diagram in Figure 3.

a) Reduce the block diagram to its simplest form, and find the transfer function that is associated with theblock diagram.

b) Find the roots of the system, and describe how you might expect it to behave.c) Write the differential equation that corresponds to the transfer function.d) Suppose that the system is subjected to a unit ramp function u(t) = t. If the initial conditions are assumed

to be zero, find an expression for the resulting output as a function of time.

25

5s

2s+5

8

1s

yu 12

Figure 3: A block diagram

Solution

a) See Figure 4.b) The roots are:

s =−26±

p

262 − 4(1)(25)(2)(1)

=−26± 24

2

s = −25, s = −1

The system is stable; it has one very fast mode, and one very slow mode. It has first order numeratordynamics.

c) The differential equations that corresponds to the transfer function is

y + 26 y + 25y = 25u+ 5u

or some linear multiple.




519-253-3000

www.uwindsor.ca

2s+215s+ 25 1

s

yu

u y5s+25s2+26s+25

u y5s+25s2+21s

2s+5

8

1s

yu 1225

5s

12

Figure 4: A block diagram




519-253-3000

www.uwindsor.ca

d)

G(s) =5s+ 25

s2 + 26s+ 25

U(s) =1s2

Y (s) =5s+ 25

s2(s2 + 26s+ 25)=

As+

Bs2+

Cs+ 25

+D

s+ 1

As(s+ 25)(s+ 1) + B(s+ 25)(s+ 1) + Cs2(s+ 1) + Ds2(s+ 25) = 5s+ 25

A+ C + D = 0

26A+ B + C + 25D = 0

25A+ 26B = 5

25B = 25

B = 1

25A+ 26= 5

A= −21/25

C + D = 21/25

−26(21/25) + 1+ C + 25D = 0

−26(21/25) + 1+ 21/25− D+ 25D = 0

D = (46− (26)(21))/((−24)(25)) = 0.8333

C = 0.006667

y(t) = −21/25+ t + 0.006667e−25t + 0.8333e−t




519-253-3000

www.uwindsor.ca

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

0

2

4

0

t [s]

y[m]

Figure 5: Output vs. time




519-253-3000

www.uwindsor.ca

Useful Information

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

The equations of motion for translationΣ~F = m~aG

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

Σ ~MA = IA~α+ ~rG/A×m~aA

orΣ ~MA = IG ~α+ ~rG/A×m~aG

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

Σ ~MG = IG ~α

orΣ ~MO = IO ~α


midtem_1_2014_soln

Documents