dynamics and control 2 2010 exam_v6

8/2/2019 Dynamics and Control 2 2010 Exam_v6

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Course ID: MECH ENG 3028 Page 1 of 15

Examination for the Bachelor of Engineering

Semester 2, 2010

101888 Dynamics and Control 2MECH ENG 3028

Official Reading Time:Writing Time:Total Duration:

10 mins180 mins190 mins

Part / Section Questions Time Marks

Part 1

Part 2

Answer all 23 questions

Answer all 3 questions

90 mins

90 mins

50 marks50 Total

50 marks50 Total

Instructions

This is an Open Book examination. Answer Parts 1 and 2 in separate books. For Part 1 please write your solutions in the book, then mark the correct

answer in the Multiple Choice Exam Answer Sheet provided.

For Part 2 begin each section (A, B, C) on a new page. Examination materials must not be removed from the examination room.

Materials

Two pink books. Notes and text books are permitted. Calculator without remote communications capability is permitted. English language dictionary is permitted. Multiple Choice Exam Answer Sheet

DO NOT COMMENCE WRITING UNTIL INSTRUCTED TO DO SO


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PART 1 AUTOMATIC CONTROL 2 (Total 50 Marks)

Consider the control moment gyroscope from the tutorials and shown in Figure 1.The fully coupled 4 degree-of-freedom dynamics are extremely complicated. We aregoing to consider the simplified system when axis 3 (in Figure 1) is locked and axes1, 2 and 4 are free to rotate. Axis 1 will be driven at a constant speed by a DC motorwhich drives rotor D. It will be assumed that we can control the angular speed of axis2, which in practice may be achieved with a high authority (gain) control loop. Axis 4is completely free.

Figure 1: Photograph of the Control Moment Gyroscope

When the plane in which the rotor spins is rotated (by rotating axis 2) it generates agyroscopic torque through axis 4. Thus in this configuration the angular velocity of

axis 2, (rad/s)2 , may be seen as the control input, and the angle of axis 4, 4q , may

be seen as the control output.

Axis 4 - Free

Axis 3 - Locked

Axis 2 - Driven

Axis 1 ConstantSpeed


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Figure 2: Schematic of the reaction wheel system in the gyroscopic torquer mode

Before deriving the dynamics of the system a number of assumptions must be made.

These are: The motor shafts are assumed to be rigidly coupled to the bodies and infinitelystiff.

The mass centres of all the bodies comprising the system are at the centre ofrotor D, which is also the centre of all the gimbals.

There is no backlash in the motor/gearbox assembly. There is no friction in any of the axes. This is far from the truth as the axes

are quite lossy and the motors experience viscous losses.

With reference to Figure 2 we will make a few definitions to aid us in the derivation ofthe dynamics.

Fixed Mechanical Parameters

2, kg.m067.0=AzzJ is the moment of inertia for body A through its z-axis,

2, kg.m030.0=BzzJ is the moment of inertia for body B through its z-axis,

2, kg.m022.0=CzzJ is the moment of inertia for body C through its z-axis,

2, kg.m015.0=DxxJ is the moment of inertia for body D through its x-axis,

2, kg.m027.0=DyyJ is the moment of inertia for body D through its y-axis.

q4 4,

u=2

1y

x

z

J +J +J zz,A zz,B zz,C+ +Jxx,D

A

B

CD

Jyy,D


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Independent Variables

(rad/s)11 q= is the (constant) angular velocity of the rotor Drelative to body

C,

(rad/s)22 q

=is the angular velocity of body C(and D) relative to body B, andalso representsthe speed command sent to the servo-motor driving axis 2.

(rad)4q is the angle of body A relative to the earth centred frame,

(rad/s)44 q= is the angular velocity of body A relative to the earth centred

frame.

System differential equationsThe equations of motion of the system may be derived by considering the schematic

of the system in Figure 2. It can be shown that the torque balance for body A relativethe earth is given by

( ) ( ) )()( 21,4,,,, tJtqJJJJ DyyCzzBzzAzzDxx =+++ (1)where )(4 tq represents the angular acceleration of body A arising from the

gyroscopic torque input ( ) )(21, tJ Dyy , and is dynamically equivalent to a rigid bodyof inertia ( )CzzBzzAzzDxx JJJJ ,,,, +++ .

The rotor speed is assumed constant and equal to RPM4001 = . Hint, when using

this parameter, do not forget to convert to rad/s.


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Now answer the following:

1. Using Equation (1), derive the transfer function between the angular velocity

(in rad/s) command, 2 , and the angular position (in rad) of body A, 4q . The

transfer function is given by [4 marks]

a. ( ) 2,,,,2

2,

1

4 6.80)()(

sJJJJsJ

ssq

CzzBzzAzzDxx

Dyy =+++

=

b.( ) 13.1

44.8

)(

)(2

1,,,,,2

1,

2

4

=

+++

=

sJJJJJs

J

s

sq

DyyCzzBzzAzzDxx

Dyy

c.( ) 2,,,,

2

1,

2

4 44.8

)(

)(

sJJJJs

J

s

sq

CzzBzzAzzDxx

Dyy =

+++

=

d.( )

6.80)(

)(

,,,,

1,

2

4 =+++

=

CzzBzzAzzDxx

Dyy

JJJJ

J

s

sq

e. None of the above.

2. From this, determine the locations of the open loop poles. The open looppoles are given by [2 marks]a. rad/s6.80,6.80 =s

b. rad/s0,0=s

c. rad/s98.8,98.8 iis +=

d. rad/s06.1,06.1 +=s


3. Discuss what the poles represent and why they are at the locations found inQuestion 2. [2 marks]a. The plant is inherently stable, hence both poles are real and on the LHS of

the s-plane.b. The two poles are both integrators, and arise since there is no velocity or

displacement term in the differential equation.c. The system has no damping and hence is marginally stable with two

imaginary (conjugate poles).d. The system is clearly unstable, with one real stable pole and one real

unstable pole.e. None of the above.

4. With reference to the number of integrators in the transfer function, is the plant [1 mark]a. Type 0?b. Type 1?c. Type 2?d. Type 3?e. None of the above.


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5. Is the open loop plant [1 mark]a. Stable?b. Marginally stable?c. Unstable?d. Globally, asymptotically stable?e. None of the above.

6. Calculate the open-loop zero(s) between the input 2 , and the output 4q .

The zeros are: [2 marks]a. There are no zeros.b. rad/s0=s

c. rad/s0,0=s

d. rad/s81.932.1 is +=


7. Using Equation (1), it can be shown that the state equation, for the two states

angle 4q and the angular velocity 4 , and input 2 , is given by [3 marks]

a.

( ))2(

6.80

0

00

100

00

102

4

4

2

,,,,

1,

4

4

4

4

+

=

+++

+

=

q

JJJJ

Jqq

CzzBzzAzzDxx

Dyy

b.

( )

)2(44.8

0

01

000

01

002

4

4

2

,,,,

1,

4

4

4

4

+

=

+++

+

=

q

JJJJ

Jqq

CzzBzzAzzDxx

Dyy

c.

( ))2(

44.8

0

00

100

00

104

2

4

4

,,,,

1,

2

4

2

4

+

=

+++

+

=

q

JJJJ

Jqq

CzzBzzAzzDxx

Dyy

d.

( ))2(

44.8

0

00

100

00

102

4

4

2

,,,,

1,

4

4

4

4

+

=

+++

+

=

q

JJJJ

Jqq

CzzBzzAzzDxx

Dyy


8. What is the minimum number of states required to model the system given inEquation (2) (given in Question 7) assuming our plant output is the angular

velocity 4 ? [1 mark]

a. Noneb. Onec. Twod. Three



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9. Using Equation (1), the state output equation, given that the system output is

the angular position, 4q , of body A is given by, [2 marks]

a. [ ] )3(001)( uxDuCx +=+=ty b. [ ] )3(010)( uxDuCx +=+=ty

c. [ ] )3(0044.8)( uxDuCx +=+=ty d. [ ] )3(44.801)( uxDuCx =+=ty e. None of the above.

Hint: At this stage the transfer function selected in Question 1 should be the same

as [ ] DBAsIC += 1)(sG .

10. Is the system represented by Equations (2) and (3) a minimal realisation of the

system given that the system output is the angular position 4q ? [1 mark]a. Nob. Yesc. Only sometimesd. Dont knowe. None of the above.

11. Suppose we wanted the output in degrees, rather than radians. The state

output matrix for an angular position, (degrees)4q , of body A in degrees. Is

given by, [2 marks]a. [ ]0360/2 =C b. [ ]/1800 =C c. [ ]0/180 =C d. [ ]10 =C e. None of the above.


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12. Which of the block diagrams below represent the open-loop plant given byEquations (2) and (3)? [3 marks]

a.

b.

c.

d.


13. What Matlab code below would you use to build a state space model called,gimbal(assuming that the A, B, C and D matrices had been defined)?

[1 mark]a.gimbal = tf(A,B)

b.gimbal = zpk(A,B,C)c.gimbal = ss(A,B,C,D)d.gimbal = statespace(A,B,C,D)e. None of the above.

14. Assuming that you had built the state space model in Matlab, what code belowwould you use to determine the poles of the system? [1 mark]

a.eig(A)b.damp(A)c.damp(gimbal)

d.pole(gimbal)e. All of the above.

1/s 1/s y4 q44 q4

1

u=2

-8.44

1/s 1/s y4 q44 q4

1u=2 -8.44

1/s 1/s y4 q44 q4

11u=2 -8.44

1/s 1/s y4 q44 q4

1

11u=2 -8.44

-8.44


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15. The states used to define the state equations are not unique. Therefore, it is

possible to rewrite Equation (2) by reordering the states, ie [ ]Tq44 . Byderiving the (inverse) transformation matrix to achieve this:

[ ] [ ]TT qq

xx

T 441

'

44 = , calculate the transformed state matrix, A , for the

alternative states [ ]Tq44 . [3 marks]

a.

=

00

10A

b.

=

00

01A

c.

=

01

00A

d.

=

10

00A


16. Using the original state equations given by Equation (2), calculate thecontrollability matrix and determine how many (if any) of the plant states arecontrollable. Hint: Relate this to the block diagram in Question 11 above.

[3 marks]a. No states are controllable.b. One state is controllable.c. Two states are controllable.

d. One state is controllable and one is stabilisable.e. None of the above.

17. You are going to design a full state feedback controller for the plant. You havean initial choice of 4 sets of control gains which are listed below along with theresulting pole locations. Choose the set of gains that gives a 2% settling timeof less than 500ms and has less than 1% overshoot. [2 marks]

a. [ ] [ ]79.33.3021 == kkK resulting in rad/s16,16 =s .b. [ ] [ ]79.37.6021 == kkK resulting in rad/s1616,1616 iis += .

c. [ ] [ ]03.3021 == kkK resulting in rad/s16,16 iis += .d. [ ] [ ]948.090.121 == kkK resulting in rad/s4,4 =s e. None of the above.


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18. The feedback controller developed above will be stable but will not guaranteetracking. This can only be done with an integral (tracking) controller by adding

another state, = dtqqq desI )( ,44 , which is the integral of the error betweenthe desired angle, desq ,4 , and actual angle.

Assume that the desired closed loop poles are rad/s12,10,8 =s , thenthe control gains needed for this are an integral control gain, 7.113=IK , and

a feedback control gain [ ] [ ]555.307.35210 == kkK . The resultingclosed loop state equations are: [4 marks]

a. des

II

q

q

q

q

q

,44

4

4

4

44.8

0

0

001

96030296

010

+

=

b. des

II

qq

q

q

q

,44

4

4

4

10

0

00196030296

010

+

=

c. des

II

q

q

q

q

q

,44

4

4

4

0

44.8

0

001

44.800

010

+

=

d. des

II

q

q

q

q

q

,44

4

4

4

1

0

0

001

44.800

010

+

=


19. Determine the characteristic equation for the closed loop state matrix in thequestion above. [3 marks]

a. 084423 =+++ sss

b. 096029630 23 =+++ sss

c. 084430 23 =+++ sss

d. 0960296 23 =+++ sss


20. Using the output equation from Question 9, calculate the observability matrixof the system. Using this, we can see that [3 marks]a. Any estimates of the states will be extremely accurate.b. Any estimates of the states will be moderately accurate.c. Any estimates of the states will be extremely inaccurate.d. The states cannot be estimated at all.e. None of the above.


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21. If you were to repeat the exercise in Question 20 using Matlab, the code wouldbe: [1 mark]

a. zeros(obsv(A,C))

b. eig(obsv(A,C))

c. rank(cond(A,B))

d. cond(obsv(A,C))e. All of the above.

22. You are required to build an observer. Given that we have a sensor that

measures the angular position, 4q , we need only estimate the true velocity, 4

, using a reduced order observer. Using an observer gain, 10=L , the stateequations to provide, 4 , a velocity estimate is given by [3 marks]

a.[ ]

[ ]

+=

+=

u

qx

u

qxx

c

cc

4

4

4

01010

44.810010

b.

[ ]

[ ]

+=

+=

u

qx

u

qxx

c

cc

4

4

4

00

44.80

c.

[ ]

[ ]

+=

+=

u

qx

u

qxx

c

cc

4

4

4

0110

44.8010

d.

[ ]

[ ]

+=

+=

uqx

u

qxx

c

cc

44

4

010

44.810010


23. By determining the poles of the reduced order observer, the 2% settling timeof the observer is found to be [2 marks]a. 10 secondsb. 2.5 seconds

c. 0.4 secondsd. 0.1 secondse. None of the above.


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PART 2 VIBRATIONS

Please take note: Miscellaneous Data can be found after problem 3 of thissection. It contains data that may be useful in doing the problems.

Answer Part 2 of the exam in a separate booklet to Part 1.

Q1. As shown in Figure 1a, a 2.5-kg slender bar of length 40 cm is pinned at oneend. A 3-kg slider of negligible size can be fixed to the bar at any desiredlocation.

You may assume that the mass moment of inertia of a slender bar about its

centre of mass is given by =1

122, and that only small angles of oscillation

are considered.

The mechanism being analysed is employed as a metronomein musicalpractice, and is used to set tempo.

a) Use the energy method to determine how far from the pinned end theslider should be situated for the system to have a period of oscillation of1 s. [17 Marks]

b) Explain briefly why gravitational potential energy (which must beconsidered in the compound pendulum shown in Figure 1a) can beneglected in the suspended mass-spring system shown in Figure 1b.

[3 Marks]

Hint: Recall the Taylors Series approximation: cos 1 0.5 2


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Q2. A simple clamped-free cantilever beam (as shown in Figure 2a) has the

following dimensions and material properties:

= 0.5 m

= 0.02 m

= 0.002 m = 2700 kg.m-3

= 68.9 GPa

a) Determine the lowest 2 natural bending mode frequencies of the

cantilever. [6 Marks]

b) The beam is excited into vibration by actuators, at a variable driving

frequency. Accelerometers are attached to the beam at distances of 0.2

m (Position A) and 0.5 m (Position B) from the clamped end. At the

lowest resonant frequency of the system, the displacement amplitudes

at A and Bare found to be in phase, and with an amplitude ratio of

= 2.5

Referring to Figure 2b, and the amplitude ratio, what can you say about

the excitation of the beam? [3 Marks]


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c) If the beam is end-loaded with a mass of 0.5 kg and negligible size, use

a single degree of freedom model of this modified system to determine

its lowest natural frequency. [7 Marks]

Hint: Calculate the effective stiffness and mass of the beam and lumped-

mass system

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.1 0.2 0.3 0.4 0.5 0.6Amplitude

Beam Position (m)

Figure 2b

Un-Loaded Cantilever Mode Shapes

Mode 1

Mode 2


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Q3. A light steel rod of circular cross-section is clamped at both ends, as shown in

Figure 3.

The bar has dimensions and material properties:

= 0.6 m

= 0.01 m

= 7850 kg.m-3

= 210 GPa

If the rod is loaded with 2 equal 10-kg masses of negligible size, at distances

of 0.2 m and 0.4 m from the left hand end (As shown in Figure 3), find anapproximate value for the fundamental frequency of the system using

Dunkerlys Method. [14 Marks]

You may neglect the rods mass.

DO NOT FORGET TO INSERT YOUR MULTIPLE CHOICE EXAM ANSWER

SHEET INTO THE EXAM BOOKLET FOR THE AUTOMATIC CONTROL PART.

END OF EXAMINATION PAPER

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dynamics and control 2 2010 exam_v6

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