Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.5
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C.12 Applications of Optimal Control
Problem
Minimizing the fuel (cost function)
12.6 Rocket Trajectories
The governing equation of rocket motion is
J=
dv
m
= m F + Fext
dt
(8.22)
T
∫ β dt
(12.41)
o
subject to the differential constraints
where
m
v
F
Fext
dv1 cβ
=
cos φ
dt
m
dv 2 cβ
=
sin φ − g
dt
m
: rocket’s mass
: rocket’s velocity
: thrust produced by the rocket motor
: external force
It can be seen that
(12.43)
and also
c
β
m
F =
(12.42)
(8.23)
where
c
: relative exhaust speed
β = −dm / dt : burning rate
dx
= v1
dt
dy
= v2
dt
(12.44)
(12.45)
The boundary conditions
Let φ be the thrust attitude angle, that is the angle between
the rocket axis and the horizontal, as in the Fig. 8.4, then the
equations of motion are
t = 0 : x = 0, y = 0, v1 = 0, v 2 = 0
t = T : x not specified, y = y , v1 = v , v 2 = 0
Thus we have four state variables, namely x, y , v1 , v 2 and two
dv1 cβ
=
cos φ
dt
m
dv 2 cβ
=
sin φ − g
dt
m
(8.24)
Here v = (v1 , v 2 ) and the external force is the gravity force
only Fext = (0,− mg ) .
controls β (the rate at which the exhaust gases are emitted)
and φ (the thrust attitude angle). In practice we must have
bounds on β , that is,
0≤β ≤β
(12.46)
so that β = 0 corresponds to the rocket motor being shut
y
down and β = β corresponds to the motor at full power.
path of rocket
The Hamilton for the system is
r
v
cβ
⎛ cβ
⎞
cos φ + p 4 ⎜
sin φ − g ⎟
m
⎝ m
⎠
(12.47)
where p1 , p 2 , p 3 , p 4 are the adjoint variables associated
H = β + p1v1 + p 2 v 2 + p 3
φ
x
Fig. 8.4 Rocket Flight Path
(12.47) ⇒
The minimum fuel problem is to choose the controls, β and
φ , so as to take the rocket from initial position to a
prescribed height, say, y , in such a way as to minimize the
fuel used. The fuel consumed is
T
∫ β dt
o
where T is the time at which y is reached.
Chapter 12 Applications of Optimal Control
with x, y , v1 , v 2 respectively.
(8.25)
c
c
⎛
⎞
H = p1v1 + p 2 v 2 − p 4 g + β ⎜1 + p 3 cos φ + p 4 sin φ ⎟
m
m
⎝
⎠
c
c
⎛
⎞
If we assuming that ⎜1 + p 3 cos φ + p 4 sin φ ⎟ ≠ 0 , we
m
m
⎝
⎠
see that H is linear in the control β , so that β is bang-bang.
That is β = 0 or β = β . We must clearly start with β = β so
that, with one switch in β , we have
59
Introduction to Control Theory Including Optimal Control
Nguyen Tan Tien - 2002.5
________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
⎧⎪ β
⎪⎩0
β =⎨
0 ≤ t < t1
for
(12.48)
for t1 < t ≤ T
Now H must also be maximized with respect to the second
control, φ , that is, ∂H / ∂φ = 0 giving
− p3
cβ
cβ
sin φ + p 4
cos φ = 0
m
m
p& 1 = −
∂H
= − p1 = − A
∂v1
∂H
p& 1 = −
= − p2
∂v 2
(12.50)
subject to
d 2θ
2
+α
dθ
+ ω 2θ = u
dt
(12.51)
Boundary conditions
t = 0 : θ = θ 0 , θ& = θ 0′
t = T : θ = 0, θ& = 0
The adjoint variables satisfy the equations
∂H
=0
∂x
∂H
p& 1 = −
=0
∂y
T
0
dt
which yields tan φ = p 4 / p 3 .
p& 1 = −
∫
T = 1dt
⇒
p1 = A
Constraint on control: | u | ≤ 1 .
⇒
p2 = B
Introduce the state variables: x1 = θ , x 2 = θ& . Then (12.51)
becomes
⇒
p 3 = C − At
⇒
p 4 = D − Bt
x&1 = x 2
x& 2 = a x 2 − ω 2 x1 + u
(12.52)
As usual, we form the Hamiltonian
where, A, B, C, D are constant. Thus
H = 1 + p1 y 2 + p 2 (−a x 2 − ω 2 x1 + u )
D − Bt
tan φ =
C − At
where p1 , p 2 are adjoint variables satisfying
Since x is not specified at t = T , the transversality condition
is p1 (T ) = 0 , that is, A = 0 , and so
tan φ = a − bt
(12.49)
where a = D / C , b = B / C .
r
v
y=y
β =0
β =β
∂H
= ω 2 p2
∂x
∂H
p& 1 = −
= − p1 + ap 2
∂y
p& 1 = −
(12.54)
(12.55)
Since H is linear in u, and | u | ≤ 1 , we again immediately see
The problem, in principle, I now solved. To complete it
requires just integration and algebraic manipulation. With φ
given by (12.49) and β given by (12.48), we can integrate
(12.42) to (12.45). This will bring four further constants of
integration; together with a, b and the switchover time t1 we
have seven unknown constants. These are determined from
the seven end-point conditions at t = 0 and t = T . A typical
trajectory is shown in Fig. 12.5.
y
(12.53)
⎧+1 if
u=⎨
⎩− 1 if
p2 < 0
p2 > 0
To ascertain the number of switches, we must solve for p 2 .
From (12.54) and (12.55), we see that
&p& 2 = − p& 1 + a p& 2 = −ω 2 p 2 + a p& 2
that is,
null thrust arc
t = t1
&p&2 − a p& 2 + ω 2 p 2 = 0
(12.56)
The solution of (12.56) gives us the switching time.
maximum thrust arc
0
that the control u is bang-bang, that is, u = ±1 , in fact
x
Fig. 12.5 Typical rocket trajectory
12.7 Servo Problem
The problem here is to minimize
Chapter 12 Applications of Optimal Control
59