ch7 control.pdf-Manipulator Control Ther...
ch7_control.pdf-Manipulator Control There are numerous approaches
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ch7 control.pdf-Manipulator Control There are nume...
ch7_control.pdf-Manipulator Control There are numerous approaches
ch7 control.pdf-Manipulator Control...
ch7_control.pdf-Manipulator Control There are numerous approaches
Page 1
7
Manipulator Control
There are numerous approaches to the control of manipulators,
all
of which are
model
dependent
. Throughout this chapter we assume that the robot dynamic model given
by
D
(
q
q
+
C
(
q,
˙
q
q
+
G
(
q
)=
u
+
J
T
f
n
τ
n
(1)
is accurate and that control torques can be generated by the robot actuators with-
out errors. Unless otherwise specified, external gripper forces
f
n
and torques
τ
n
are
assumed to be zero.
D
,
C
and
G
are as described in (178) through (183) of the
dynamics section.
We will study a number of simple schemes for
position
control.
Issues of force
or ”impedance” control will not be addressed and can be found in the bibliography
handed out at the beginning of the course.
7.1
Computed-Torque Control
The robot dynamics terms can be completely cancelled by computing the robot ac-
tuator forces and torques for a given acceleration
w
. Indeed, let
u
=
D
(
q
)
w
+
C
(
q,
˙
q
q
+
G
(
q
)
.
(2)
Then, after substituting
u
in (1), we obtain that
¨
q
=
w
(3)
which is in
decoupled, double integrator form
.
The evolution of each of the components of
q
is completely analogous to the dis-
placement of a linearly constrained unit mass particle acted upon by a force equal
to the component of
w
. Therefore, one could use a simple control law, such as PD
control, and let
w
i
=
k
pi
(
q
di
q
i
)
k
v
i
˙
q
i
(4)
to obtain that
ˆ
q
i
(
s
)=
k
pi
s
2
+
k
vi
s
+
k
pi
ˆ
q
di
(
s
)
.
(5)
92


Page 2
Choose
k
pi
=
ω
2
0
i
,
k
vi
=2
ρ
i
ω
0
i
, for desired characteristic frequency and damping
coefficient.
To track bounded acceleration trajectories exponentially, set
w
i
q
di
+
k
pi
(
q
di
q
i
)+
k
vi
q
di
˙
q
i
)
(6)
where
q
d
,t
[0
,T
] is given.
Note that computed-torque requires that
D
(
q
),
C
(
q,
˙
q
q
and
G
(
q
) be computed
on-line
, at every control loop iteration, so it is very demanding computationally.
Furthermore, modelling errors (inertial paramters, friction, actuator dynamics and
saturation), will prevent exact cancellation of the robot dynamics to the nice form (3).
An analysis of the stability of computed-torque control in the presence of dynamics
errors is presented in Vidyasagar and Spong. A block diagram of the computed-torque
position control scheme is shown in Figure 7.1.
Figure 1: Computed Torque Position Control
93


Page 3
7.2
Feedforward Position Control
The robot dynamical equation (1) can be put in standard state-space form.
With
x
=[
q
T
˙
q
T
]
T
IR
2
n
defined as the state, (1) becomes
d
dt
q
˙
q
=
˙
q
¨
q
=
˙
q
D
(
q
)
-
1
[
C
(
q,
˙
q
q
+
G
(
q
)]
+
0
D
(
q
)
-
1
u
(7)
or
˙
x
=
f
(
x, u
)
.
(8)
Suppose now that a desired nominal trajectory
{
x
d
(
t
):
t
[0
,T
]
}
is generated
by a path planning program (with techniques as described in (3.3.2), for example).
Using the Newton-Euler approach, the nominal torque
{
u
d
(
t
):
t
[0
,T
]
}
necessary
to execute the trajectory can be computed. It is clear that ˙
x
d
=
f
(
x
d
,u
d
).
The state-space equation (8) can be linearized about
{
x
d
(
t
):
t
[0
,T
]
}
,
{
u
d
(
t
):
t
[0
,T
]
}
by a formal Taylor series expansion. With
x
=
x
d
+
δx
,
u
=
u
d
+
δu
, we
have that, for any
t
[0
,T
],
˙
x
d
+
˙
δx
=
f
(
x
d
+
δx, u
d
+
δu
)
(9)
=
f
(
x
d
,u
d
)+
∂f
∂x
(
x
d
,u
d
)
δx
+
∂f
∂u
(
x
d
,u
d
)
δu
(10)
Therefore, to the first order, for any
t
[0
,T
],
˙
δx
=
∂f
∂x
(
x
d
,u
d
)
δx
+
∂f
∂u
(
x
d
,u
d
)
δu
(11)
or
˙
δx
=
A
(
t
)
δx
+
B
(
t
)
δu .
(12)
Equation (11) above is called the
variational
equation of (8) about the trajectory
{
x
d
(
t
):
t
[0
,T
]
}
, corresponding to the the input
{
u
d
(
t
):
t
[0
,T
]
}
.
If we assume that
A
(
t
)
=
A
0
,
B
(
t
)
=
B
0
do not change much over the trajectory of
the robot, we can design the controller by state-feedback,
i.e.
, by letting
δu
=
F
0
δx
,
leading to
˙
δx
=(
A
(
t
)
B
(
t
)
F
0
)
δx
=(
A
0
B
0
F
0
)
δx .
(13)
94


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