Two-Mass, Three-Spring Dynamic System Investigation Case Studyengineering.nyu.edu/mechatronics/Control_Lab/Criag/Craig_RPI/2002… · Dynamic System Investigation Case Study. Mechatronics

MechatronicsTwo-Mass, Three-Spring Dynamic System

K. Craig1

Two-Mass, Three-SpringDynamic System Investigation

Case Study


K. Craig2

Physical System Physical Model MathematicalModel

ModelParameter

Identification

ActualDynamicBehavior

ComparePredictedDynamicBehavior

MakeDesign

Decisions

DesignComplete

Measurements,Calculations,

Manufacturer's Specifications

Assumptionsand

Engineering Judgement

Physical Laws

ExperimentalAnalysis

Equation Solution:Analytical and Numerical

Solution

Model Adequate,Performance Adequate

Model Adequate,Performance Inadequate

Modify or

Augment

Model Inadequate:Modify

Dynamic System Investigation

Which Parameters to Identify?What Tests to Perform?


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Objective

• The objective was to design, build, and demonstrate a dynamic system that:– Would demonstrate dynamic system behavior

below, at, and above a system resonance– Would demonstrate the physical significance of

transfer function poles and zeros– Would show clearly the relationship between the

time domain and frequency domain– Would show the difference between collocated and

non-collocated control systems


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Physical System Picture


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Physical System Schematic

M1 M2

Infrared PositionSensor

Rack and Pinion

Motor withEncoder

Springs

Linear Bearings

Infrared PositionSensor

Connecting Bar

Guideway

Two-Mass Three-Spring Dynamic System


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Physical System Components

• Translating masses (2)• Tension springs (3)• Linear bearings (3) and guide-way• Rack and pinion gear system• Infrared position sensors (2)• Permanent magnet brushed DC motor with

optical encoder• PWM servoamp and power supply


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Physical Model Assumptions

• Springs– three tension springs are identical– linear springs– neglect mass of springs– neglect structural damping in springs– constant spring rate– springs are always in tension


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• Masses– two masses are rigid bodies– one degree of freedom (translation) for each mass– neglect air damping on masses– Coulomb friction and viscous damping in linear

bearings supporting masses and rack– Coulomb friction: independent of position and

velocity– Viscous damping: coefficient is constant


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• Structure– rigid and fixed– independent of system motions– straight track– level track (gravity acts perpendicular to track)


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• Rack and Pinion Gear System– neglect elasticity of gear teeth– neglect backlash of gear teeth– neglect friction of gear teeth– rigid connection to motor shaft and to translating

mass


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• Motor / Amplifier– PWM servoamp operates in current mode– neglect amplifier dynamics– Coulomb friction and viscous damping in motor– Coulomb friction: independent of angular position

and angular velocity– Viscous damping: coefficient is constant


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• Sensors– Optical encoder: 2000 counts per revolution with

quadrature decoding– Analog infrared position sensor

• time constant: 2.5 ms• range: 0.9 m• output: 0 to 10 V


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Diagram of Physical Model

M1 M2

K K K

Two-Mass Three-Spring Dynamic SystemPhysical Model

X1 X2

θ rp

Jmotor BmotorTfriction Jpinion

Tm

B1 Ff1 B2 Ff2


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Model Parameter Identification

J m rp p p≈12

2

• Translating Masses– m1 = 1.6725 kg (includes mass of rack)– m2 = 1.275 kg

• Pinion Gear– rp = 0.0127 m– mp = 0.0092 kg

• Infrared Position Sensors– Calibration Curves


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0 1 2 3 4 5 6 7 8 9 10-80

-60

-40

-20

0

20

40

60

80

100Sensor II X2(mm)= -0.286*V13+2.6671*V12-21.2174*V1+125.0637

Sensor Volts

Pos

ition

(mm

), "

+" w

hen

mov

es a

way

from

the

mot

or

Sensor I (near the motor) Sensor II

( )


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• Motor– Jm = 0.0078 oz-in-s2 = 5.5080E-5 kg-m2

– Bm = 0.20 oz-in/krpm = 1.3487E-5 N-m-s– Tf m = 0.19 in-lb = 0.0215 N-m– Kt = 11.8 oz-in/A = 8.3326E-2 N-m/A

• Springs– K = 494 N/m

• PWM Servo-Amplifier– Ka = 1 A/V


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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50.04

0.045

0.05

0.055

0.06

0.065

Time(sec)

Ampl

itude

(m)

Mx Bx Kx

nx

x

B m

n

n

ln

+ + =

=FHGIKJ

=+

=

+

2 0

1

22

1

1

2 2

δ

ζδ

π δ

ζω

a f

• Friction in System– Mass M1– Viscous Damping– B = 1.8 N-s/m

Approximate Exponential Decay


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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

Time(sec)

Ampl

itude

(m)

Mx Bx Kx

nx

x

B m

n

n

ln

+ + =

=FHGIKJ

=+

=

+

2 0

1

22

1

1

2 2

δ

ζδ

π δ

ζω

a f

Approximate Exponential Decay• Friction in System– Mass M2– Viscous Damping– B = 1.8 N-m/s


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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

t

Ampl

itude

(m)

Approximate Linear Decay

• Friction in System– Mass M1 + Rack/Pinion + Motor– Coulomb Damping– Ff1-eff = 2.4 N

Mx Kx F

F Kf

f

+ =

=

22

4∆


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Mathematical Model: FBD’s

T K i K K Vm t t a in= =

+θ

Fc

J Tθ

rp

Bmθ

Tf sgn θb g

o

Motor Rotor + Pinion

J J JT motor pinion= +

J B T K K V F rT m f t a in c psgnθ θ θ+ + = −b gMo∑ = 0


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x1

M1

M x1 1Fstatic Fstatic

Fc

Kx1K x x2 1−a f

B x1 1

F xf1 1sgna f

Mass M1

M x B x Kx Kx F F xc f1 1 1 1 1 2 1 12 sgn+ + = + − a fFx∑ = 0


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x2

M2

M x2 2Fstatic Fstatic

K x x2 1−a f −Kx2

B x2 2

F xf 2 2sgna f

Mass M2

M x B x Kx Kx F xf2 2 2 2 2 1 2 22 sgn+ + = − a fFx∑ = 0


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Mathematical Model:Equations of Motion

J B T K K V F rT m f t a in c psgnθ θ θ+ + = −b g [1]

M x B x Kx Kx F F xc f1 1 1 1 1 2 1 12 sgn+ + = + − a f [2]

M x B x Kx Kx F xf2 2 2 2 2 1 2 22 sgn+ + = − a f [3]

x r

x r

x r

p

p

p

1

1

1

=

=

=

θ

θ

θ

Kinematic Relations:


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Mathematical Model:Equations of Motion

Solve for Fc in equation [1] using the kinematic relations:

F J xr

B xr

Tr

K Kr

Vc Tp

mp

f

p

t a

pin= − − − +

sgn12

12

θb g

Substitute this expression into equation [2]:

M Jr

x B Br

x Kx

Kx F xT x

rK K

rV

T

p

m

p

ff

p

t a

pin

1 2 1 1 2 1 1

2 1 11

2+LNM

OQP

+ +LNM

OQP

+

= − − +sgnsgn

a f a f


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Mathematical Model:Equations of Motion Summary

M x B x Kx Kx F xf2 2 2 2 2 1 2 22 sgn+ + = − a fM x B x Kx Kx F x K K

rVeff eff f eff

t a

pin1 1 1 1 1 2 1 12− − −+ + = − +sgna f

M M Jr

B B Br

F F Tr

effT

p

effm

p

f eff ff

p

1 1 2

1 1 2

1 1

−

−

−

= +

= +

= +


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Mathematical Model:Linear Model

M x B x Kx Kx K Kr

Veff efft a

pin1 1 1 1 1 22− −+ + = +

M x B x Kx Kx2 2 2 2 2 12+ + =

Laplace Transform:

M s B s K KK M s B s K

X sX s

K Kr Veff efft a

p in1

21

22

2

1

2

22 0

− −+ + −− + +

LNM

OQPLNMOQP =L

NMMO

QPP

a fa f


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Mathematical Model:Linear Model - Transfer Functions

X s

K Kr

V K

M s B s KM s B s K K

K M s B s K

X s

M s B s K K Kr

V

KM s B s K K

K M s B s K

t a

pin

eff eff

eff efft a

pin

eff eff

12

22

12

1

22

2

2

12

1

12

1

22

2

0 22

2

2

02

2

a f

a f

=

−

+ ++ + −− + +

=

+ +

−+ + −− + +

− −

− −

− −


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Mathematical Model:Linear Model - Transfer Functions

X sV s

K Kr

M s B s K

D s

X sV s

K Kr

K

D s

in

t a

p

in

t a

p

12

22

2

2a fa f

c h

a fa f

=

FHGIKJ + +

=

FHGIKJ

( )

( )

D s M M s M B M B s

M M K B B s B B K s Keff eff eff

eff eff eff

( ) = + + +

+ + + + +− − −

− − −

1 24

1 2 2 13

1 2 1 22

1 222 2 3

a f a fa f a f


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Mathematical Model:Linear Model - State Space Equations

q Aq Buy Cq Du= += +

State Variables: State Space Equations:q xq xq xq x

1 1

2 1

3 2

4 2

====

qqqq

KM

BM

KM

KM

KM

BM

qqqq

K Kr M V

yy

qqqq

eff

eff

eff eff

t a

p eff in

1

2

3

4

1

1

1 1

2

2

2

1

2

3

4

1

1

2

1

2

3

4

0 1 0 02 0

0 0 0 10 2

2

0

00

1 0 0 00 0 1 0

0

L

N

MMMM

O

Q

PPPP=

− −

− −

L

N

MMMMM

O

Q

PPPPP

L

N

MMMM

O

Q

PPPP+

L

N

MMMMM

O

Q

PPPPP

LNMOQP =LNM

OQPL

N

MMMM

O

Q

PPPP+

−

−

− − −

0LNMOQPVin


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Mathematical Model:Transfer Functions and State Space Equations

X sV s

s ss s s s

X sV s s s s s

A

C

in

in

12

4 3 2

24 3 2

3 2503 4 5887 2518 72 3449 1265 7 1414 1 284460

1259 32 3449 1265 7 1414 1 284460

0 1 0 0489 45 0 93313 244 72 0

0 0 0 1387 45 0 774 90 14118 0

1 0 0 00 0 1 0

( )( )

. . .. . .

( )( )

.. . .

. . .

. . .

=+ +

+ + + +

=+ + + +

=− −

− −

L

N

MMMM

O

Q

PPPP

L

N

MMMM

O

Q

PPPP

=LNM

OQP

LNMOQP

B =

03.2503

0

D =00


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Mathematical Model:Transfer Functions and State Space Equations

− ± ⇒ =− ± ⇒

0 536 17 1 17 1. . .i rad / s = 0.03130.637 31.2i = 31.2 rad / s = 0.0204

ω ζω ζ

Poles:

X sV sin

1a fa f⇒ − ± ⇒ =0 706 27 8 27 8. . .i rad / s = 0.0254ω ζ

Zeros:X sV sin

2 a fa f⇒ None


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Frequency Response Plots: Analytical

XVin

1

Frequency (rad/sec)

Phas

e (d

eg);

Mag

nitu

de (d

B)Bode Diagrams

-70

-60

-50

-40

-30

-20

100

101

102

-150

-100

-50

0


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Frequency Response Plots: Analytical

Frequency (rad/sec)

Phas

e (d

eg);

Mag

nitu

de (d

B)

Bode Diagrams

-80

-60

-40

-20

100

101

102

-300

-200

-100

0

XVin

2


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Frequency Response Plots: Experimental

XVin

1

100

101

102

-70

-60

-50

-40

-30

-20

Frequency (rad/s)

Mag

nitu

de (d

B)

100

101

102

-200

-150

-100

-50

0

Frequency (rad/s)

Phas

e (d

egre

es)

EncoderNetwork Analyzer


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10-1

100

101

-60

-50

-40

-30

-20

Frequency (Hz)

Mag

nitu

de (d

B)

Bode Plot for Mass1 (Optical Encoder)

10-1

100

101

-200

-150

-100

-50

0

Frequency (Hz)

Phas

e (d

egre

es)

XVin

1

EncoderSine Sweep


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10-1

100

101

-70

-60

-50

-40

-30

-20

Frequency (Hz)

Mag

nitu

de (d

B)

Bode Plot for Mass1 (IR Sensor)

10-1

100

101

-250

-200

-150

-100

-50

0

Frequency (Hz)

Phas

e (d

egre

es)

XVin

1

IR SensorSine Sweep


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XVin

2

10-1

100

101

102

-55

-50

-45

-40

-35

-30

-25

Frequency (rad/s)

Mag

nitu

de (d

B)

10-1

100

101

-400

-300

-200

-100

0

Frequency (rad/s)

Phas

e (d

egre

es)

IR SensorNetwork Analyzer


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10-1

100

101

-80

-70

-60

-50

-40

-30

-20

Frequency (Hz)

Mag

nitu

de (d

B)

Bode Plot for Mass2 (IR Sensor)

10-1

100

101

-500

-400

-300

-200

-100

0

Frequency (Hz)

Phas

e (d

egre

es)

XVin

2

IR SensorSine Sweep


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Collocated and Non-CollocatedControl Systems

• Collocated Control System– All energy storage elements that exist in the system exist

outside of the control loop.– For purely mechanical systems, separation between sensor

and actuator is at most a rigid link.

• Non-Collocated Control System– At least one storage element exists inside the control loop– For purely mechanical systems, separating link between

sensor and actuator is flexible.


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Poles and Zeros of Transfer Functions:Physical Interpretation

• Complex Poles of a collocated control system and those of a non-collocated control system are identical.

• Complex Poles represent the resonant frequencies associated with the energy storage characteristics of the entire system.

• Complex Poles, which are the natural frequencies of the system, are independent of the locations of sensors and actuators.


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• At a frequency of a complex pole, even if the system input is zero, there can be a nonzero output.

• Complex Poles represent the frequencies at which energy can freely transfer back and forth between the various internal energy storage elements of the system such that even in the absence of any external input, there can be nonzero output.

• Complex Poles correspond to the frequencies where the system behaves as an energy reservoir.


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• Complex Zeros of the two control systems are quite different and they represent the resonant frequencies associated with the energy storage characteristics of a sub-portion of the system defined by artificial constraints imposed by the sensors and actuators.

• Complex Zeros correspond to the frequencies where the system behaves as an energy sink.


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• Complex Zeros represent frequencies at which energy being applied by the input is completely trapped in the energy storage elements of a sub-portion of the original system such that no output can ever be detected at the point of measurement.

• Complex Zeros are the resonant frequencies of a subsystem constrained by the sensors and actuators.

Documents

Two-Mass, Three-Spring Dynamic System Investigation Case Studyengineering.nyu.edu/mechatronics/Control_Lab/Criag/Craig_RPI/2002… · Dynamic System Investigation Case Study. Mechatronics