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16.346 Astrodynamics
Fall 2008
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Lecture 19
The State Transition Matrix
#9.5-9.6
Linearization of the Equations of Motion
• Deviations from reference
• Equations of motion
r(t) = rref (t) + δ (t)
v(t) = vref (t) + ν (t)
drref
dδ
= vref
=ν
dt
dt
=⇒
dν
dvref
= g(r) − g(rref ) = G(rref )δ = G(t)δ
= g(rref )
dt
dt
�
∂ g ��
δ + · · · = g(rref ) + G(rref )δ + O(δ 2 )
g(r) = g(rref ) +
∂ r �r=rref
dr
=v
dt
dv
= g(r)
dt
since
• State vector representation
� � �
�� �
d δ
O
I
δ
=
G(t) O
ν
dt ν
� �� � �
��
�
x(t)
F(t)
dx
= F(t)x
dt
or
The State Transition Matrix
• Define
�
�
Φ(t, t0 ) = x1 (t) x2 (t) . . . x6 (t)
�
�
Φ(t0 , t0 ) = x1 (t0 ) x2 (t0 ) . . . x6 (t0 ) = I
• Matrix differential equation
d
Φ(t, t0 ) = F(t)Φ(t, t0 )
dt
• Fundamental property
with
Φ(t0 , t0 ) = I
x(t) = Φ(t, t0 )x(t0 )
Symplectic Matrices
• Definition
An even-dimensional matrix A is symplectic if
T
A JA = J
where
�
O
J=
−I
I
O
�
Note: Since J2 = −I , the J matrix is analogous to the imaginary
algebra.
• Inverse of a Symplectic Matrix
Postmultiply by A−1 and premultiply by J to obtain
A−1 = −JA T J
16.346 Astrodynamics
Lecture 19
√
−1 in complex
The State Transition Matrix is Symplectic
• Symplectic Property of Φ(t, t0 )
dΦT
dΦ
d T
Φ (t, t0 )JΦ(t, t0 ) =
JΦ + ΦT J
dt
dt
dt
= ΦT FT JΦ + ΦT JFΦ
= ΦT [FT J + JF]Φ
�
�
T
O
T G(t) − G (t)
=Φ
Φ
O
I−I
=O
since the gravity-gradient matrix G = G T is symmetric. Therefore
Φ T (t, t0 )JΦ(t, t0 ) = constant = Φ T (t0 , t0 )JΦ(t0 , t0 ) = I T JI = J
• Inverse of Φ(t, t0 )
�
Φ1 (t, t0 )
Φ(t, t0 ) =
Φ3 (t, t0 )
From the partitions
the inverse is
−1
Φ
Φ2 (t, t0 )
Φ4 (t, t0 )
�
ΦT
4 (t, t0 )
(t, t0 ) = Φ(t0 , t) =
−ΦT
3 (t, t0 )
�
−ΦT
2 (t, t0 )
T
Φ1 (t, t0 )
�
Fundamental Perturbation Matrices
For the discussion of linear deviations from a reference orbit, define
�
�
�
�
δr(t)
δr0
x(t) =
and
x0 =
δv(t)
δv0
so that
�
�
⎡
�
�
�
δr0
δr(t)
= Φ(t, t0 )
δ v0
δv(t)
�
�
�
δ r0
δr(t)
= Φ(t0 , t)
δv(t)
δv0
where
⎤
⎢ ∂r
⎢ ∂r
⎢ 0
Φ(t, t0 ) = ⎢
⎢ ∂v
⎣
∂r0
∂r ⎥
⎥
∂v0 ⎥
⎥
∂v ⎥
⎦
∂v0
⎡
⎤
⎢ ∂ r0
⎢ ∂r
where Φ(t0 , t) = ⎢
⎢ ∂v
⎣
0
∂r
∂r0
∂v
∂v0
∂v
�
�
�
R(t)
R(t)
=
�
V(t) V(t)
ref
⎥
⎥
⎥
⎥
⎦
� T
V (t)
=
� T (t)
−V
ref
since
Φ(t, t0 )Φ(t0 , t) = I
16.346 Astrodynamics
and Φ(t0 , t) = Φ−1 (t, t0 )
Lecture 19
−RT (t)
� T (t)
R
�
Navigation Matrix
Let t0 ≤ t ≤ t1 and define
�
�
R(t)
Φ(t, t0 ) = �
V(t)
Φ(t, t0 )
R(t)
V(t)
Guidance Matrix
�
and
Φ(t, t1 )
�
� � (t) R (t)
R
Φ(t, t1 ) = � V (t) V (t)
Then
dR(t)
dR (t)
= V(t)
= V (t)
R(t0 ) = O
R (t1 ) = O
dt
dt
dV(t)
dV (t)
V (t1 ) = I
V(t0 ) = I
= G(t)R(t)
= G(t)R (t)
dt
dt
The differential equations for the “tilde” matrices are the same but with initial conditions
� (t ) = I
R
1
�
V (t ) = O
� (t ) = I
R
0
�
V(t ) = O
1
0
Furthermore,
� −1
� R
� =
V
C
Hence
�
�
Φ(t, t1 )�
t=t0
=Φ
−1
�
∂v
��
=
∂r �
v1 =constant
�
�
(t, t0 )
�
t=t1
=
⇒
−1
C =V R
�
∂v ��
=
∂r �r1 =constant
�
�
T
�
� (t )
R (t )
V (t1 )
R
0
0
=
� T (t )
� (t )
V (t )
−V
V
1
0
0
�
−RT (t1 )
� T (t )
R
1
Differential Equation for the C Matrix
C R = V
and
C−1 = R V−1
or
C−1 V = R
Differentiate the first expression:
d V
dR
dC =
R + C
dt
dt
dt
=⇒
dC R + C V = GR
dt
Finally, postmultiply by R−1 to obtain
dC
2
+ C = G
dt
Since G is symmetric, then C is symmetric. Because C (t1 ) is infinite, it is better to
use the equation for the inverse matrix.
dR
dC−1 dV
V + C−1
=
dt
dt
dt
=⇒
dC−1 V + C−1 GR = V
dt
Hence
dC−1
+ C−1 GC−1 = I
dt
with
C−1 (t1 ) = O
Note: For the constant gravity case we have G = O so that C−1 = (t − t1 )I .
16.346 Astrodynamics
Lecture 19
Midcourse Orbit Corrections
#11.6
• Fixed-Time-of-Arrival Correction
�
� � � (t)
δr(t)
R
δx(t) = Φ(t, t1 ) δx(t1 ) =⇒
= �
δv(t)
V (t)
Hence
R (t)
V (t)
��
0
δv(tA )
�
δr(t) = R (t) δv(tA )
δv(t) = V (t) δv(tA )
Eliminate δv(tA )
δv(t) = V (t)R −1 (t) δr(t) = C (t) δr(t)
Velocity correction
∆v(t) = δv(t+ ) − δv(t− ) = C (t) δr(t) − δv(t− )
• Variable-Time-of-Arrival Correction
�
� � ��
�
� (t) R (t)
δr(t)
R
δr(tA )
= �
δv(t)
δv(tA )
V (t) V (t)
�
�
Multiply through by −C (t) I
��
�
�
�
� � (t) R (t)
�
� δr(t)
�
� R
δr(tA )
−C (t) I
= −C (t) I
� (t) V (t)
δv(t)
δv(tA )
V
Hence, [using the starred form of Eqs. (9.57)], we have
� (t) + V
� (t)] δr(t )
−C (t) δr(t) + δv(t) = [−C (t)R
A
�
��
�
= −R −T (t)
δv(t) = C (t) δr(t) − R −T (t) δr(tA )
or
1. Choosing δr(tA )
rp (tA + δt) = rp (tA ) + vp (tA ) δt
r(tA + δt) = r(tA ) + v(tA ) δt
=⇒
δr(tA ) = r(tA ) − rp (tA ) = −vr (tA ) δt
Then
r(tA + δt) = rp (tA + δt)
where
vr (tA ) = v(tA ) − vp (tA ) . Hence
δv(t) = C (t) δr(t) + R −T (t)vr (tA ) δt
��
�
�
= w(t)
2. Choosing δt to minimize |∆v (t)|
δtA = −
∆v · w
w·w
16.346 Astrodynamics
and
or
∆v (t) = ∆v(t) + w(t) δt
�
wwT �
Min ∆v = I − T
∆v = M ∆v
w
w
�
��
�
Projection operator M
Lecture 19