modeling & analysis of the international space...

Modeling & Analysis ExerciseSpace Station Case Study

K. Craig 1

Modeling & Analysisof the

International Space Station


K. Craig 2

Physical System

Solar Alpha Rotary Joints


K. Craig 3

Physical System

SolarArray

SolarArray

OutboardBody

InboardBody

+x

StatorRotor

GearTrain


K. Craig 4

Physical System• SARJ allows controlled rotation about the x-axis and

constrains the other 5 DOF.• Mass moment of inertia of the inboard body is significantly

greater than the outboard body, so assume that the inboard body is mechanically grounded.

• Intermediate gear train inertia is small compared to the rotor inertia.

• Torsional flexibility of the gear train is important as is the torsional stiffness of the solar arrays.

• All system parameters are approximately constant over time.• Bearing friction and gear-train backlash can be neglected.• Environment noise can be neglected except for plume loads.


K. Craig 5

Motivation

Boss says design a system to keep the solar arrays pointed towards the sun as the station orbits the earth!

How do you do that? What do you do first?

Given:• physical dimensions of the station• motor• drive train


K. Craig 6

What Are the Steps You Take?Understand the

Physical System

Develop aMathematical Model

Predict theDynamic Behavior

Develop aPhysical Model


K. Craig 7

Physical Model

Vin

+

-

R L

im+

-

eb

Stator

Rotor

Statormechanically

grounded

Jm

JsaJobN:1Gear Ratio

K1 K2

B2B1

+x

′θθm θθob θθsa

θθm

Td


K. Craig 8

Parameters inPhysical Model

Symbol Meaning UnitsV

ininput voltage to armature volts

im armature current ampsR motor resistance ohmsL motor inductance henryseb back emf voltage voltsK

tmotor torque constant ft-lb/amp

Ke back emf constant volts-sec/radN gear ratio unitlessJm motor inertia slug-ft

2

θm motor position rad′θm motor position reflected through gear train rad

Job

outboard truss inertia slug-ft2

θob outboard body position radJ

sasolar array inertia slug-ft

2

θsa solar array position radK1 gear train and truss torsional stiffness ft-lb/radB1 gear train and truss torsional damping ft-lb-s/radK2 solar array torsional stiffness ft-lb/radB2 solar array torsional damping ft-lb-s/radT

dplume disturbance torque ft-lb

Tm

motor torque ft-lb′Tm motor torque reflected through gear train ft-lb


K. Craig 9

Simplifying Assumptions(a) one degree-of-freedom motion for each body(b) inboard inertia >> outboard inertia(c) gear train inertia << motor rotor inertia(d) lumped elements(e) neglect gear backlash(f) neglect bearing friction(g) constant parameters(h) neglect all noise except plume load


K. Craig 10

Mathematical Simplifications

(a), (b), (c): reduces the number and complexity of the differential equations

(d): leads to ordinary differential equations(e), (f): makes equations linear(g): leads to constant coefficients in the

differential equations(h): avoids statistical treatment


K. Craig 11

Gearing up for the Math Model

• What are the constitutive physical relations needed?• What are the equilibrium relations?• What are the compatibility relations?• What types of equations do we expect from this system?

linear vs. nonlinearODE vs. PDE

• What are the inputs to the system?


K. Craig 12

Constitutive Physical Relations:

T J B T K

V iR V Ldidt

= = =

= =

&& &θ θ θ T

Equilibrium Relations: Kirchhoff’s Current LawNewton’s Second law

Compatibility Relations: Kirchhoff’s Voltage Law


K. Craig 13

Free Body Diagrams

Jsa

+x

Td

B sa ob2& &θ θ−b g

K sa ob2 θ θ−a f

J sa sa&&θ

+θsa

J B K Tsa sa sa ob sa ob d&& & &θ θ θ θ θ+ − + − − =2 2 0b g a f


K. Craig 14

Job

+x

B ob m1& &θ θ− ′b g

K ob m1 θ θ− ′a f

Job ob&&θ

+θob

J B K B Kob ob ob m ob m ob sa ob sa&& & & & &θ θ θ θ θ θ θ θ θ+ − ′ + − ′ + − + − =1 1 2 2 0b g a f b g a f

B ob sa2& &θ θ−b g

K ob sa2 θ θ−a f


K. Craig 15

Gear Train Relations: θθ

m

m

m

m

NN

N

TT

NN N

′= ≡

′= ≡

2

1

1

2

1

Tm

N1

N2

θm

′Tm ′θm


K. Craig 16

Jm

+x

T K im t m=

J m m&&θ

+θm

J BN

KN

K im m m ob m ob t m&& & &θ θ θ θ θ+ ′ − + ′ − − =1 1

1 10b g a f

BNm ob11& &′ −θ θb g

KNm ob11′ −θ θa f


K. Craig 17

Circuit Diagram

+

-

+

-

R L

Vineb e Kb e m= &θ

V Ri Ldidt

Kinm

e m= + + &θ

im


K. Craig 18

Mathematical Model

J B K T

J B K B K

J BN

KN

K i

Ri Ldidt

K V

sa sa sa ob sa ob d

ob ob ob m ob m ob sa ob sa

m m m ob m ob t m

me m in

&& & &&& & & & &

&& & &

&

θ θ θ θ θ

θ θ θ θ θ θ θ θ θ

θ θ θ θ θ

θ

+ − + − − =

+ − ′ + − ′ + − + − =

+ ′ − + ′ − − =

+ + =

2 2

1 1 2 2

1 1

0

0

1 10

b g a fb g a f b g a fb g a f


K. Craig 19

Sinceθ θθ θ

θ θ

m m

m m

m m

N

N

N

= ′= ′= ′

& &&& &&

J B K T

J B K B K

J N B K NK i

Ri Ldidt

NK V

sa sa sa ob sa ob d

ob ob ob m ob m ob sa ob sa

m m m ob m ob t m

me m in

&& & &&& & & & &

&& & &

&

θ θ θ θ θ

θ θ θ θ θ θ θ θ θ

θ θ θ θ θ

θ

+ − + − − =

+ − ′ + − ′ + − + − =

′ + ′ − + ′ − − =

+ + ′ =

2 2

1 1 2 2

21 1

0

0

0

b g a fb g a f b g a fb g a f


K. Craig 20

MatLab / Simulink Block Diagram

theta_sa

To Workspace2

theta_ob

To Workspace1

theta_m_prime

To Workspace

Sum7

Sum6

Sum5

Sum4

Sum3

Sum2

Sum1

Sum

1/s

Integrator6

1/s

Integrator5

1/s

Integrator4

1/s

Integrator3

1/s

Integrator2

1/s

Integrator1

1/s

Integrator

Input:Td

Input: Vin

B2

Gain9

K2

Gain8

1/Job

Gain7

B1

Gain6

K1

Gain5

1/(N^2*Jm)

Gain4

N*Kt

Gain3

N*Ke/L

Gain2

1/Jsa

Gain10

R/L

Gain1

1/L

Gain


K. Craig 21

One More ApproximationIn practice, a servo amplifier operating in “torque mode” is used to run the DC motor.This compensates for the back-emf effect by using current feedback.It generates a current proportional to applied voltage.The motor model can be simplified to Tm = Ktim where im= KampVin and the motor current im now is an input rather than a state variable.Now, we need to solve the linear ODE’s so we can predict the behavior of the actual system....how do we do this?


K. Craig 22

State-Space Equations

′

′

L

N

MMMMMMMM

O

Q

PPPPPPPP

=− −

− − − −

− −

L

N

MMMMMMMMMMM

O

Q

PPPPPPPPPPP

′

′

&&&&&&&&&

&&

θθθθθθ

θθθθ

m

ob

sa

m

ob

sa

m m m m

ob ob ob ob ob ob

sa sa sa sa

m

ob

sa

m

KN J

KN J

BN J

BN J

KJ

K KJ

KJ

BJ

B BJ

BJ

KJ

KJ

BJ

BJ

0 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1

0 0

0 0

12

12

12

12

1 1 2 2 1 1 2 2

2 2 2 2

θθ

ob

sa

t

m

sa

m

d

NKN J

J

iT

&

L

N

MMMMMMM

O

Q

PPPPPPP

+

L

N

MMMMMMMMM

O

Q

PPPPPPPPP

LNM

OQP

0 00 00 0

0

0 0

01

2

&r r rr r rx Ax Bu

y Cx Du

= += +

y

y

y

i

T

m

ob

sa

m

ob

sa

m

d

1

2

3

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0

0 0

0 0

L

NMMM

O

QPPP

=L

NMMM

O

QPPP

′

′

L

N

MMMMMMM

O

Q

PPPPPPP

+L

NMMM

O

QPPPLNM

OQP

θθθθθθ

&&&

Note:current im is an inputsince im = KampVin


K. Craig 23

Parameter Numerical ValuesSymbol Value Units

im input ampsKt 8.26 ft-lb/ampN 283 unitlessJm 9.4E-3 slug-ft

2

′θm output radiansJob 400 slug-ft

2

θob output radiansJsa 7.0E5 slug-ft

2

θsaoutput radians

K1 9.28E6 ft-lb/radianB1 492 ft-lb-s/radianK2 2.76E5 ft-lb/radianB2 4.40E3 ft-lb-s/radianTd input ft-lb


K. Craig 24

Mathematical AnalysisPoles: 0, 0

-1.86 ± 15.28i (ωn = 15.39 rad/s, ζ = 0.121)-4.58 ± 189.56i (ωn = 189.61 rad/s, ζ = 0.024)

Zeros: ′θ

θ

θ

m

m

ob

m

sa

m

i

i

i

-0.00296 ± 0.6188i, -6.12 ± 154.4i(ωn = 0.6188 rad/s, 154.56 rad/s)

-0.00314 ± 0.6279i(ωn = 0.6279 rad/s)

None


K. Craig 25

Frequency Response Plots: Input im

10-1

100

101

102

103

-400

-200

0

Frequency (rad/sec)

Gai

n dB

10-1

100

101

102

103

-360

0

360

θob

′θm

θsa

θob

′θm

θsa

φo


K. Craig 26

10 -1 100 101 102 103-200

-100

0

Frequency (rad/sec)

Gai

n dB

10 -1 100 101 102 103

-90

-180

0

φo

′θm

mi

0.62

15.39 189.6

154.6


K. Craig 27

10-1 100 101 102 103-200

-100

0

Frequency (rad/sec)

Gai

n dB

10-1 100 101 102 103

-180

-360

0

θob

mi

0.62

15.39189.6

φo


K. Craig 28

10-1

100

101

102

103

-400

-200

0

Frequency (rad/sec)

Gai

n dB

10-1

100

101

102

103

-180

0

180

θsa

mi

φo

15.39189.6


K. Craig 29

System Behavior at ωω = 0.62 rad/s

θsa

K2

B2

Jsa

J B K

KJ

BJ

sa sa sa sa

sa

sa sa sa

sa

sa

&& &&&& &

θ θ θ

θθ

θθ

+ + =

LNM

OQP = − −

LNMM

OQPPLNM

OQP

2 2

2 2

0

0 1 Natural Frequency: 0.62 rad/s


K. Craig 30


θsa

K2

B2

JsaJmN2+Job

θ θob m= ′

( )&& & &&& & &

&&&&&&

&

&

J N J B K B K

J B K B K

KJ

BJ

KJ

BJ

KJ N J

BJ N J

KJ N J

BJ N J

m ob ob ob ob sa sa

sa sa sa sa ob ob

sa

sa

ob

ob

sa sa sa sa

m ob m ob m ob m ob

sa

sa

ob

22 2 2 2

2 2 2 2

2 2 2 2

22

22

22

22

0 1 0 0

0 0 0 1

+ + + = +

+ + = +

L

N

MMMM

O

Q

PPPP=

− −

+ +−

+−

+

L

N

MMMMMM

O

Q

PPPPPP

θ θ θ θ θ

θ θ θ θ θ

θθθθ

θθθθob

L

N

MMMM

O

Q

PPPP

Natural Frequencies:0, 15.49 rad/s


K. Craig 31


θob

K1

B1

JobJmN2

′θm

K2

B2

N J B K B K

J B B K K B K

KN J

BN J

KN J

BN J

KJ

BJ

K KJ

B BJ

m m m m ob ob

ob ob ob ob m m

m

m

ob

ob

m m m m

ob ob ob ob

21 1 1 1

2 1 2 1 1 1

12

12

12

12

1 1 2 1 2 1

0 1 0 0

0 0 0 1

&& & &&& & &

&&&&&&

′ + ′ + ′ = +

+ + + + = ′ + ′

′′

L

N

MMMM

O

Q

PPPP=

− −

− + − +

L

N

MMMMMM

O

Q

PPPPPP

θ θ θ θ θ

θ θ θ θ θ

θθθθ

a f a f

a f a f

′′

L

N

MMMM

O

Q

PPPP

θθθθ

m

m

ob

ob

&

&

Natural Frequencies:15.38, 189.6 rad/s


K. Craig 32

Time Response: im = cos(0.6t)

0 5 10 15 20 250

0.002

0.004

0.006

0.008

0.01

0.012

0.014

0.016

0.018

0.02

Time (secs)

Am

plitu

de

θsa

′θm

θob


K. Craig 33

0 1 2 3 4 5-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

Time (secs)

Am

plitu

de

Time Response: im = cos(15.39t)

θsa

θob

′θm


K. Craig 34

Conclusion• What did we do?

Understood physical systemDeveloped physical modelDeveloped mathematical modelPredicted dynamic behavior

• Mathematical model based on justifiable approximations!

Simplify - remember to verify your model!Do the results make sense?Garbage in = Garbage out

• Next step: Use the model to design a control system which will satisfy the boss!

modeling & analysis of the international space...

Documents