modeling & analysis of the international space...
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Modeling & Analysis ExerciseSpace Station Case Study
K. Craig 1
Modeling & Analysisof the
International Space Station
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K. Craig 2
Physical System
Solar Alpha Rotary Joints
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Physical System
SolarArray
SolarArray
OutboardBody
InboardBody
+x
StatorRotor
GearTrain
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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.
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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
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What Are the Steps You Take?Understand the
Physical System
Develop aMathematical Model
Predict theDynamic Behavior
Develop aPhysical Model
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Physical Model
Vin
+
-
R L
im+
-
eb
Stator
Rotor
Statormechanically
grounded
Jm
JsaJobN:1Gear Ratio
K1 K2
B2B1
+x
′θθm θθob θθsa
θθm
Td
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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
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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
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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
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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?
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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
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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
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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
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Gear Train Relations: θθ
m
m
m
m
NN
N
TT
NN N
′= ≡
′= ≡
2
1
1
2
1
Tm
N1
N2
θm
′Tm ′θm
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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
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Circuit Diagram
+
-
+
-
R L
Vineb e Kb e m= &θ
V Ri Ldidt
Kinm
e m= + + &θ
im
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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
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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
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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
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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?
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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
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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
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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
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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
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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
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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
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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
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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
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System Behavior at ωω = 15.4 rad/s
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
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System Behavior at ωω = 189.6 rad/s
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
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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
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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
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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!