
Now if we apply a drive signal to the plant that is proportional to E, then we can say that
our feedback mechanism invokes proportional error control in our system. The equation
of motion for such motion is given by:
(5)
=+
where
dt
d
θ =
&
and
2
2
dt
d θ
θ =
&&
is the motor torque,
, lJ is the inertia of the load reflected through the
gears and
is the inertia of the motors,
is friction and
is angle. We can also say
that:
(6)
=
where
is the motor current and
is a constant. Equating (4) and (5) gives:
(7)
m
=+
If we invoke proportional control, we can assume that we will apply a current that is
proportional to the error:
or we can say using equation (3) that
d
Ik
,
where
is a constant. Therefore:
(8) θθθθ
&&&
FJk
de
+=− )(
Here we have absorbed the constants
into
and rewritten
for
and
for
to simplify. Now if we assume that
d
θ
, then our equation of motion is:
(9) 0=++ θθθ
e
kFJ
&&&
If we assume a solution
θ
and let kk
e
= then (8) reduces to:
(10)
2
JqFqk
Solving for q gives:
(11)
2
q
−±−
=
We see that if:
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