midterm 1 2013 solution

Mechanical, Automotive, & Materials Engineering

92-321 Control TheorySummer 2013

Midterm Exam #1

Engineering is a professional faculty, and as a student who wishes to become a professional engineer, itis expected that you will behave in a professional manner during your exam, and abide by the followingstandards of conduct.

Standards of conduct:

1. The possession of any telecommunications device during an exam will be viewed by the proctorsas cheating, whether it is used or not. Students found with said devices will be subject to a signif-icant downward adjustment of their grade. If you are in possession of any telecommunicationsdevices, identify yourself to the proctor and surrender them immediately for the duration of theexam.

2. The possession of any unauthorized aids, (i.e., material other than the approved course text, anda calculator) during an exam will be viewed by the proctors as cheating, whether it is used or not.Students found with said materials will be subject to a significant downward adjustment of theirgrade.

3. Each student will be allowed a maximum of three inquiries regarding the exam material; furtherrequests for clarification will be denied. If you have uncertainties about the exam, make sure youhave read the entire question, state your assumptions, and proceed. Do not waste the examinerstime by asking questions regarding the correctness of your solution.

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

5. Remain seated quietly until all the exams have been collected, not just your own. Recognize thatother students are still holding their exams and could be influenced by any discussions. Therewill be plenty of opportunity to discuss the exam outside the hall.

6. Answer all questions to the best of your ability.

1 c©B.P. Minaker, 2013


Question 1

Consider the schematic diagram of the automotive suspension system in shown Figure 1. This modelcan be used to investigate the high frequency motion of the suspension, called ‘wheel hop’. Becausethe vehicle chassis is relatively much more massive than the other components, at high frequency itsmotion will be very small; it can safely be treated as stationary. The suspension can be modelled as afour bar mechanism, consisting of four bodies; the (fixed) vehicle chassis, the upper control arm, thelower control arm, and the wheel assembly. Note also you can assume that the upper control arm isrelatively much less massive than the other components; its inertia can be neglected and assumed tobe approximately zero.

The lower control arm is attached to the chassis by a hinge at one end (A) and the wheel assemblyby a hinge at the other end (B). A spring and co-axial damper also attach the lower control arm to thechassis. The tire is modelled as a vertical spring, between the wheel assembly and the ground. Allowthat the tire may be disturbed by bumps on the ground of size u(t).

Notice that in the state shown, the parallel geometry of the control arms results in a purely verticaltranslation of the wheel assembly, and the spring-damper is oriented vertically. Assume that any mo-tions are small deviations from the position shown, and note that in this model the effects of gravitycan be safely ignored.

Use the following values in the problem:

Parameter Value Parameter Valueks 30 kN/m cs 2 kN · s/mkt 150 kN/m m 20 kgIA 1.0 kg ·m2 l1 0.5 ml2 0.2 m

a) Using the free body diagrams given, find the transfer function G(s) = Y (s)U(s) relating the bump size

to the vertical motion of the wheel assembly. Hint: use the angle θ to define the motion of thelower control arm, and note that y ≈ l1θ . Use a similar approach to find the deflection of thesuspension spring.

b) What is the order of the resulting transfer function? Describe the motion you would expect fromthis system. Find the roots to show how you know.

c) If the system was disturbed from rest, how many oscillations would you expect before the ampli-tude of the motion was reduced by 95%?



Solution

Part a)

Consider the summation of forces in the y-direction on the wheel assembly

ΣFy = Ft − By = my

Ft = kt(u− y)

Simplifymy + kt y + By = ktu

Consider the summation of moments about A on the lower control arm.

ΣM = l1By − l2Fs = IAθ

Fs = ks l2θ + cs l2θ

Simplify

By =1

l1

�

IAθ + l2(cs l2θ + ks l2θ)�

my + kt y +1

l1

�

IAθ + l2(cs l2θ + ks l2θ)�

= ktu

Now using kinematics of small motionsy = l1θ

my + kt y +1

l21

�

IA y + l2(cs l2 y + ks l2 y)�

= ktu

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) = 0 Therefore, the transfer function is

Y (s)U(s)

=kt

�

IAl21+m

�

s2+�

l2l1

�2css+

�

�

l2l1

�2ks

�

+ kt

Part b)

The resulting system is second order. I expect that it will be underdamped and oscillatory.

Y (s)U(s)

=150000

�

10.25+ 20

�

s2+�

0.20.5

�22000s+

�

�

0.20.5

�230000

�

+ 150000

Y (s)U(s)

=150000

24s2+ 320s+ 154800

s =−320±

p

3202− 4(24)(154800)(2)(24)

s =−6.67± 80.0i

ωn = 80.3 rad/s = 12.8 Hz

ζ= 6.67/80.3= 0.083



Part c)

The time constant isτ= 1/6.67= 0.15 sec

Three time constant is3τ= 0.45 sec

One wavelengthλ= 1/12.8= 0.0781 sec

So we will see0.45/0.0781= 5.76 c ycles



Question 2

Consider the block diagram in Figure 2.

a) Reduce the block diagram to its simplest form, and find the transfer function that is associatedwith the block 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 impulse function. If the initial conditions are

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

Solution

Part a)

See Figure 3.

Part b)

The transfer function of the simplified block diagram is

G(s) =8s+ 80

25s2+ 266s

The roots are s = 0 and s = −266/25. The system is neutrally stable; it has a rigid body mode. It willnot return to zero on its own when forced. It has first order numerator dynamics.

Part c)

The differential equations that corresponds to the transfer function is

25 y + 266 y = 80u+ 8u

or some linear multiple.

Part d)

G(s) =�

8

25s

��

s+ 10

s+ 266/25

�

U(s) = 1

Y (s) = G(s)

Y (s) =�

8

25

��

A

s+

B

s+ 266/25

�

A(s+ 266/25) + Bs = s+ 10

A+ B = 1, A(266/25) = 10

A= 250/266, B = 16/266



Y (s) =�

8

(25)(266)

��

250

s+

16

s+ 266/25

�

y(t) =�

8

(25)(266)

�

�

250+ 16e−266t/25�

y(t) = 0.30+ 0.019e−10.6t



ks, cs

l1

l2

B

kt

u(t)

C

A

IAy(t)

By

Ft

Cy ≈ 0

Ay By

θ

Fs

Ax Bx

m

Figure 1: Vehicle suspension schematic, and FBDs of the lower control arm and the wheel assembly(note that only the vertical forces acting on the wheel assembly are shown in the FBD)

y1s

8

3

2

u

5s

Figure 2: A block diagram



y1s

8

3

2

u

5s

yu

y1s

8

3

2

u

5s

y1s

u825

2ss+10

8s+8025s2+266s

Figure 3: A block diagram



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 asa 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 besimplified

Σ ~MG = IG~α

orΣ ~MO = IO~α


midterm 1 2013 solution

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