dynamics and control of formation flying satellites in earth … · 2004-09-08 · dynamics and...

Alfriend/Barcelona 1

Dynamics and Control ofFormation Flying Satellites

In Earth OrbitLecture 2Terry Alfriend

Texas A&M University


Outline

• Example of the Error Analysis of a LEO satellite• Navigation Error Effect• Constellation Control• Impulsive and Continuous Control of Orbital

Elements

Lecture 2- Control Concepts and Navigation Error Effect


Error Analysis of a LEO satellite

Ref: Alfriend, K.T. and Lovell, T.A.,“Error Analysis of SatelliteFormations In Near Circular Low Earth Orbits,” Paper No. AAS03-651,2003 AAS/AIAA Astrodynamics Conference, Big Sky, MT, Aug. 2003.


Outline

• Background• Maneuver Strategy• Analytical Insights• Error Analysis• Conclusions


Background• Motivation for study was TechSat21

u 3 satellitesu Formations were designed to minimize differential J2 effects;

leader-follower, in-plane 2x1 ellipseu 550 km altitude; potential for significant differential dragu e<0.0001u One thruster, rotate satellite to fire, only out-of-plane and

tangential thrusting.u Need to maximize times between thruster firings, should be no less

than about 4 orbitsu Minimum desired separation was 30 m

• Need to determine the effects of the error sources onperformance

u Relative navigation, thrust magnitude, attitude.


Maneuver Strategy• Relative motion described by CW equations

u No differential inclinationu Circular orbit

• Impulsive thrusting• Control out-of-plane and in-plane separately

u One impulse for out-of-plane for maintenance, more if needed forreconfiguration

u Three impulses for in-plane, separated by 0.5 orbits.• Strategy developed using non-singular orbital elements and

Gauss’ variational equations

Use of CW equations is valid because this is an error analysis


Drag/ Cross Section Area Model

Satellite is thin parallelepiped that tracks the sun and slowlyrotates at orbit rate. The orientation is such that the velocityvector is nearly perpendicular to A2. Model deteriorates as sunapproaches orbit plane, but this is where differential drag effectis minimum.

A3

bSun

Orbit Normal (Z)

A2

A1

A = A1

re

u◊ re

y+ A

2

re

v◊ re

y+ A

3

re

w◊ re

y

A = 0.5A1

1 - cos b( ) sin2q + 0.5A2

1 + cos b( ) - 1 - cos b( )cos2qÈÎ ˘̊ + A3

sin b cosq

A = 0.5A1

1 - cos b( ) 4

2psin 2qdq

0

p / 2

Ú + 0.5A2

1

2p1 + cos b( ) + 1 - cos b( )cos2qÈÎ ˘̊ dq

0

2p

Ú

+ A3sin b 2

2pcosqdq

- p / 2

p / 2

Ú

A =A

1

p1 - cos b( ) + 0.5A

21 + cos b( ) +

A3

psin b


Drift Time StatisticsEstablish a deadband of ±D about desired positionLet t be the time in orbits to reach the deadband.The drift in the in-track direction due to differential drag is

y = -1.5

ad

n2

Ê

ËÁˆ

¯̃nt( )2

= -6pa

d

n2

Ê

ËÁˆ

¯̃t 2

t =2D

3kT 2 db

Ê

ËÁ

ˆ

¯˜

1/ 2

, k = 0.5rv2 CD

m

∂ A

∂b,db is differential attitude

ft t( ) =4K 2

2p s dbt 3exp -

K 4

2s db2 t 4

È

ÎÍÍ

˘

˚˙˙, K =

2D

3kT 2

È

ÎÍ

˘

˚˙

1/ 2

Assuming db is Gaussian the probability density function is


Drift Time Statistics

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 4 8 12 16 20 24Time - Orbits .

Tim

e to

Dea

dban

d D

istri

butio

n Fu

nctio

n

Deadband=5 mDeadband=10 mDeadband=20 m

Beta=30 deg


Relative Navigation Errors

d a = 4x + 2 &y / n

d = drift per orbit

d = -3pd a

s

d= 3ps d a

= 6p 2sx

- s &y / n( )2

+ 4sx

s &y 1 + rx&y( ) / nÈ

ÎÍ˘˚̇

1/ 2

Minimum drift occurs when r

x&y = -1. Typically - 1 < rx&y < 0

We had to use rx&y = 0 because the requirement was only

on relative position and velocity.

Requirements were

spos

= 10cm, sx

= 10 / 3cm

svel

= 0.5mm / s , sx

= 0.5 / 3mm / s


Relative Navigation Error Effect

0

0.8

1.6

2.4

3.2

4

4.8

5.6

6.4

7.2

0 0.1 0.2 0.3 0.4Relative Rate Standard Deviation - mm/s

Dri

ft R

ate S

tand

ard

Dev

iatio

n - m

/orb

it

'sig x=0 cm''sig x=2 cm''sig x=4 cm''sig x=6 cm''sig x=8 cm'Requirement

Relative velocity error is driver.


Relative Navigation Error Effect

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10Time to Reach Deadband - orbits

Tim

e to

Rea

ch D

eadb

and

Dist

ribu

tion

Func

tion

'D/sd=1'D/sd=2''D/sd=3''D/sd=4''D/sd=5'

Time to reach deadband is t = D / d

ft t( ) =2

pD

sd

Ê

ËÁˆ

¯̃1

t 2

ÊËÁ

ˆ¯̃

exp -D2

2sd2t 2

Ê

ËÁ

ˆ

¯˜


Maneuver Strategy

e = a,q ,i,q1,q

2,W( )

q = f + w , argument of latitude

q1

= ecoswq

2= esinw

d a =d a

a

dq = dq12 + dq

22( )1/ 2

Solve in-plane and out-of-plane separately. For in-plane use only tangential impulses. Three in-planeimpulses separated by 0.5 orbits are required. Needto find time of 1st maneuver and magnitude of thethree maneuvers (Dv’s).


Gauss’ Variational Equations

Dq1

= 2Dv

t

vc

cosq

Dq2

= 2Dv

t

vc

sinq

Da =2

nDv

t

Di =Dv

n

vc

cosq

DW =Dv

n

vcsin i

sinq

Dq = -cot isinq

vc

Dvn

Da = -2

nDv

1+ Dv

2+ Dv

3( )Dq

1= -

2

naDv

1- Dv

2+ Dv

3( )cosq1

Dq2

= -2

naDv

1- Dv

2+ Dv

3( )sinq1

tanq

1=

Dq2

Dq1

Dvn

= Di2 + DWsin i( )2ÈÎÍ

˘˚̇

1/ 2

tanqoop

=DWsin i

Di

D denotes the change in the element.


Strategy

• Separate into out-of-plane and in-plane.• Out-of-plane is one maneuver.• In-plane is three maneuvers, separated by 0.5 orbit.

u Time of first maneuver determined by required elementchanges.

u 4 unknowns, four equationsu Tangential thrust only


In-Plane Solution

Dv2

=na

4Dq -

Da

a

ÊËÁ

ˆ¯̃

Dv1

+ Dv3

= -na

4Dq +

Da

a

ÊËÁ

ˆ¯̃

Dq = Dq1cosq

1+ Dq

2sinq

1

&q =m p

r 2=

mp3

1 + q1cosq + q

2sinq( )2

�

d ˙ q = dn + 2n dq1 cos q + dq2 sinq( )dn = - 3n

2daa

d &q = d n + 2n dq1cosq + dq

2sinq( )

d n = -3n

2

d a

a

D denotes element change, d denotes element error


In Plane

�

dq = dq0 + Dn q1 - q0 + 2p( ) / n + Dn 1( ) 2p / n + Dn 2( )p / n + 2 dq1 cosq + dq2 sin q( )dqq 0

q1 + 2p

Ú

+2 Dq11( ) cosq + Dq2

1( ) sin qÊ Ë Á ˆ

¯ ˜ dqq1

q1 + 2p

Ú + 2 Dq12( ) cos q + Dq2

2( ) sin qÊ Ë Á ˆ

¯ ˜ dqq1 +p

q1 + 2p

Ú

�

Dq = Dq0 - 3Da2a

(q1 - q0) + 2p[ ] - 3pna

2Dv1 + Dv2( ) - 2D ˜ q 0

D ˜ q 0 = Dq1 sinq0 - Dq2 cos q0

�

Dv1 = na6p

Dq0 - 1.5Da q1 - q0( )[ ] - na8

3Da + Dq( ) - na3p

D ˜ q 0

Dv2 = na4

Dq - Da ( )

Dv3 = - na6p

Dq0 - 1.5Da q1 - q0( )[ ] + na8

Da - Dq( ) + na3p

D ˜ q 0

Da = Daa

Dq = Dq1 cos q1 + Dq2 sin q1

D ˜ q 0 = Dq1 sinq0 - Dq2 cos q0


Error Sources

Table 1 Baseline Error Sources

Error Source Pointing(deg)

Position(m)

Velocity(mm/s)

Thrust(%)

Value - 1s 1/3 0.1/31/2 0.5/31/2 5

Table 2 Baseline System Parameter Values

Parameter Deadband(m)

Altitude(km)

Density(kg/m3)

A1(m2)

A2(m2)

A3(m2)

b(deg)

mass(kg)

Value 5, 10,

15,20

550 3.183 x10-13 0.575 0.82 12.984 30 170


Approach

• All error sources selected from random number generator withspecified standard deviation

• Start in desired formation, generate differential drag from attitudeerror.

• When 75% of deadband reached initiate maneuver decision.• To compute desired Dv’s corrupt exact relative state with navigation

system errors. Actual Dv is corrupted in magnitude by thrustmagnitude error source and direction by attitude error. New attitudeerror generated after each maneuver.

• Once maneuver initiated not recalculated based on new estimated state.Option exists to recalculate last Dv to force estimated drift to zero.Results show not effective because of navigation error.


In-Track Time HistoryLeader-Follower - All error sources - 10 m Deadband

35

40

45

50

55

60

65

70

75

80

85

0 10 20 30 40 50 60 70 80 90 100 110Time - orbits

In-T

rack

Disp

lace

men

t - m Deadband

Deadband


In-Track Time HistoryLeader-Follower - All error sources - 20 m Deadband

30

40

50

60

70

80

90

0 10 20 30 40 50 60 70 80 90 100 110Time - orbits

In-T

rack

Disp

lace

men

t - m

Deadband

Deadband


In-Track Error Time History2x1 Ellipse - All error sources - 20 m Deadband

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 10 20 30 40 50 60 70 80 90 100 110

Time - orbits

In-T

rack

Err

or- m

deadband

deadband


Radial Error Time History2x1 Ellipse - All error sources - 20 m Deadband

-15

-10

-5

0

5

10

15

0 10 20 30 40 50 60 70 80 90 100 110

Time - orbits

Rad

ial E

rror

- m

deadband

deadband


Phase Plane Time History2x1 Ellipse - All error sources - 20 m Deadband

For time between maneuvers > 4 orbits D >20 m

-50

-40

-30

-20

-10

0

10

20

30

40

-100 -80 -60 -40 -20 0 20 40 60 80 100

In-Track Displacement - m

Rad

ial D

ispla

cem

ent -

m


Leader FollowerAverage Time Between Maneuvers

Leader Follower

0

2

4

6

8

10

12

14

16

18

4 8 12 16 20

Deadband - m

Ave

rage

Tim

e B

etw

een

Man

euve

rs -

or

bits

Att ErrorAll Error SourcesThrust Error


Leader FollowerAnnual Dv Orbit Maintenance

Leader Follower

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

4 6 8 10 12 14 16 18 20

Deadband - m

Ann

ual D

v -

m/s

Att Error

All Error Sources

Thrust Error


LF-CIPE Comparison

Average Time Between Maneuvers LF-CIPE Comparison

0

1

2

3

4

5

6

7

8

4 8 12 16 20

Deadband - m

Ave

. Tim

e B

etw

een

Man

euve

rs -

orbi

ts

LFCIPE

All Error sources


LF-CIPE ComparisonAnnual Dv Orbit Maintenance

LF-CIPE Comparison

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

4 8 12 16 20

Deadband - m

Ann

ual D

v -

m/s

LFCIPE

All Error sources


Distribution Function

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 5 10 15 20 25 30 35 40Time Between Maneuvers - orbits

Dist

ribu

tion

Func

tion

TheoryAll ErrorsNav Errors


Conclusions• The minimum acceptable deadband is about 20 meters. Even with a

20-meter deadband the deadband may be exceeded on occasion.• To guarantee that the motion stay within the 20-meter deadband will

require a new method for formation maintenance that responds faster.This will increase the fuel consumption.

• The relative navigation errors, particularly the relative velocity error,are the driver for the minimum acceptable deadband.

• Rel nav requirements should include a requirement on cross-correlation and on drift rate.

• The fuel, or Dv, required for orbit maintenance is minimal.• The 5% 1s thrust error has very little effect on the average time

between maneuvers or the annual fuel requirement.


Relative Navigation Analysis

• Differential carrier phase GPS (DCGPS) is the primarysource for relative navigation in LEO.

u May need to be augmented with intersatellite ranging.• MIT group led by Dr. How have demonstrated 5 cm

relative position accuracy and 0.5 mm/s relative velocity.

• Previous discussion indicated that a good Kalman filterwould minimize the semi-major axis error and give

• Dr. How has obtained

s

d= 3ps d a

= 6p 2sx

- s &y / n( )2

+ 4sx

s &y 1 + rx&y( ) / nÈ

ÎÍ˘˚̇

1/ 2

r

x&y ª -1

r

x&y ª -0.1

Why the discrepancy in ? r

x&y


What are the physics?

• The coupling between x and ydot is at orbit rate.• The relative navigation filter is getting relative

position measurements at a much faster rate,approximately every 10-30 seconds.

• Due to the fast measurement rate the filter isbasically seeing three double integrators withweak coupling due to the orbit rate.

Can we prove this analytically?


Solution

&&x - 2n&y - 3n2x = vx

&&y + 2n &x = vy

x = x, &x, y, &y( )T

&x = Ax + Bvv = N 0,s v( ),E vvT( ) = Qc

z = Hx + w ,Measurement equationw = N 0,s w( ),E wwT( ) = Rc

H =1 0 0 00 0 1 0

È

ÎÍ

˘

˚˙

Let Dt be time between measurements and make the timetransformation

Letting the equations of motion become

t =t

Dt,( )¢ =

ddt

, ddt

=1

Dtddt

e = nDt

¢¢x = 2e ¢y + 3e 2x + Dt( )2 n x

¢¢y = -2e ¢x + Dt( )2 n y

Two weakly coupled double integrators


ApproachExpand in a power series in e.

A = A0 + e A1 + e 2A2

A0 =A00 00 A00

ÊËÁ

ˆ¯̃,A00 =

0 10 0

ÊËÁ

ˆ¯̃

A1 =0 A11

-A11 0ÊËÁ

ˆ¯̃,A11 =

0 00 2

ÊËÁ

ˆ¯̃

A2 =A22 00 0

ÊËÁ

ˆ¯̃,A22 =

0 03 0

ÊËÁ

ˆ¯̃

We need the covariance so we need to solve the Riccati equation. Solvethe continuous problem, not the discrete.

AP + PAT + BQBT = PH T R-1HP 4x4 nonlinear matrix equation.

P = P0

+ e P1

+ K


Approach (cont)P = P0 + eP1 + e 2P2 + ...

Po =P0xx 00 P0yy

ÊËÁ

ˆ¯̃.P0xx = P0yy =

P11 P12

P12 P22

ÊËÁ

ˆ¯̃

Pn =Pnxx Pnxy

PnxyT Pnyy

ÊËÁ

ˆ¯̃, n = 1,2,...

P0A1T + P1A0

T + A0P1 + A1P0 = P0DP1 + P1DP0

P1 A0T - DP0( ) + A0 - P0D( ) P1 = - P0A1

T + A1P0( )D =

C 02

02 CÊËÁ

ˆ¯̃,C =

1s wd

2

1 00 0

ÊËÁ

ˆ¯̃

At 0th order the problem is decoupled and P0xx=P0yy


Solution

s x2 = s y

2 = 2s vd1/2s wd

3/2 Dt

s &x2 = s &y

2 = 2s vd3/2s wd

1/2 Dt 3

rx&y = -n s wd

s vd

ÊËÁ

ˆ¯̃

1/2

sd

= 3ps d a= 6p 2s

x- s &y / n( )2

+ 4sx

s &y 1 + rx&y( ) / nÈ

ÎÍ˘˚̇

1/ 2

sd

= 3ps d a=

6pn

s &y =6p 21/ 4

ns

vd3/ 4s

wd1/ 4Dt1/ 2

For the values used by Dr. How we obtained

which confirms the concept that the filter is dominated bythe measurement rate and the primary way to reduce thesemi-major axis error is to reduce the process noise. Theagreement between the above equations and the resultsobtained by How and his students is excellent.

r

x&y ª -0.07


Conclusions• Due to the fast measurement rate the relative navigation

filter basically sees three weakly coupled doubleintegrators.

• The semi-major axis error resulting from the DCGPSrelative navigation is almost strictly a function of thevelocity estimate error.

• To reduce the semi-major axis error we need to reduce theprocess noise, which is a function of the model error andmany errors in processing the GPS signals.

• The correlation coefficient will not approach (-1) asexpected unless the process noise is reduced several ordersof magnitude.


Results

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0 1 2 3 4 5 6Position Error - cm

Cor

rela

tion

Coe

ffici

ent

sigv=10^(-7)sigv=5*10^(-7)sigv=10^(-6)sigv=5*10^(-6)sigv=10^(-5)

Dt=30


Constellation Control

Objective: Design a control law that counters theout-of-plane drift due to J2 and minimizes systemfuel consumption and balances the fuelconsumption over all the satellites in theformation


Constellation ControlProjected Circular Orbit

• Fuel rqmt. to negate differentialnodal precession effect for Sat 1 istwice that for Sats 2 and 3.

• To maximize lifetime we wantequal fuel consumption for eachsatellite. r

1

2 3

Position at equator

a

y

z

xi = 0.5r sin nt + ai( )yi = r cos nt + a i( )zi = r sin nt + ai( )

Control Problemr - Dr( ) £ r £ r + Dr( )

Minimize

J = uij dtj =1

3

Â0

T

Úi =1

N

Â

umj dtj =1

3

Â0

TÚ ª unj dt

j =1

3

Â0

TÚ ,"m,n , m π n

T is large, a few months


Hill’s Initial Conditionsn Chief’s orbit is eccentric: e=0.005n Formation established using inclination difference only.

-0.6 -0.4 -0.2 0 0.2 0.4 0.6

-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

Along-Track (km)

Out-of-Plane (km)

Relative Orbits in the y-z plane, (2 orbits shown)

-0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

Along-Track (km)

Out-of-Plane (km)

150 Relative Orbits


Circular Projection Relative Orbit

Sat #1and 4 have max di

and zero dW

Sat #3 and 6 have maxdW but zero di

Sat # 1 and 4 will spendmax fuel and sat # 3 and6 will spend min fuel.

Snapshot when the chief is at the equator.Pattern repeats every orbit of the Chief

y

z

1

34

2

5

6

a


Constellation Control Concept

Balance the fuelconsumption over acertain period byrotating all the deputiesby an additional rate

Snapshot when the chief is atthe equator.

y

z

1

3

4

2

5

6

a

&a


Modified Hill’s Equations

&&x - 2nxy

&y - 3nxy2 x = u

x

&&y + 2nxy

x = uy

&&z + n 2z = uz

+ 2An cosa sin[nt]

A =3

2r J

2 n

0

Re

R

Ê

ËÁˆ

¯̃

2

sin2 i

Assume no in-track driftcondition satisfied.

n Æ n + &aThe near-resonance in the z-axis is detuned byintroducing an additional rotation.


Constellation Control Saves Fuel

xR

= 0.5r sin(q + a )

yR

= r cos(q + a )

zR

= r sin(q + a )

Ideal Trajectory

&a

Ideal Control for perfectcancellation of the disturbanceand for

ux

= 2n &a xR

uy

= -n &a yR

uz

= -2n &a zR

- 2An cosa sinqOptimize over time and aninfinite number of satellites

J =

1

4np 20

2p / n

Ú ux

2 + uy

2 + uz

2( )0

2p

Ú dt da0

&a

opt= -

A

3r, A =

3

2r J

2 n

0

Re

R

Ê

ËÁˆ

¯̃

2

sin2 i

J

opt=

2

3A2n

J

Brute force= A2n


Theoretical Results

0 20 40 60 80 100 120 140 160 180 2000

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8x 10-5

Time (days)

cost (m

2/sec

3)

a(0)= 0a(0)= p/4a(0)= p/2a(0)= 3p/4

a=7100 km, i=70 deg, r=500m


Simulation Results

0 20 40 60 80 100 120 140 160 1800

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6x 10-5

Time (days)

Cost (m

2/sec

3)

a=7100 km, i=70 deg, e=0.005,r=500m

0 20 40 60 80 100 120 140 160 1800

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6x 10-5

Time (days)

Cost (m

2/sec

3)

Equivalent to 28 m/sec/yr/sat

0 20 40 60 80 100 120 140 160 1800

2

4

6x 10-5

Time (days)

Cost (m

2/sec

3)

a0 = 0,p

a0 = p/2,3p/2

Equivalent to 52 m/sec/yr


Simulation Results

-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 02.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

4.2x 10-4

adot (deg/day)

Total Cost (m

2/sec

3/yr)


Formation Control Comments

• Use of orbital elements in formation design and formationcontrol offers distinct advantages over the Cartesian frame.However, relative navigation measurements will be in theCartesian frame.

• The control problem should address the real problem, notthe mathematically easy problem.

• With current technology formation control usingcontinuous control is probably not feasible due to the lowthrust levels.

dynamics and control of formation flying satellites in earth … · 2004-09-08 · dynamics and...

Documents