Page 116 - Mechatronics with Experiments

102 MECHATRONICS
where
1
2
a = lim s x(s) =− (2.146)
1
s→0 K D
d 2 1
a = lim [s x(s)] = (2.147)
2
s→0 ds K 2
D
1
a = lim (s + K ) x(s) =− (2.148)
D
3
s→−K D K 2
D
1 1 1 −K D t
x(t) =− ⋅ t + ⋅ 1(t) − ⋅ e (2.149)
K D K 2 K 2
D D
The position of mass under the derivative control due to a constant disturbance force is
1 1 1 −K D t
x(t) = t − 1(t) + e
K D K 2 K 2
D D
In steady state,
1
̇ x(t) =− (2.150)
K D
Since ̇ x (t) = 0.0 (desired velocity is zero), the error in steady-state
d
1
̇ e(t) = ̇ x (t) − ̇ x(t) = (2.151)
d
K D
which means that in steady state, the velocity error due to a step input constant force
disturbance will result in constant velocity of the mass, even though the desired velocity is
zero. The velocity error is inversely proportional to the velocity feedback gain.
This example shows that the derivative feedback control alone would not be able
to reject a constant disturbance acting on a second-order mass–force system. Derivative
feedback introduces damping into the closed loop system poles (Figure 2.45) and increases
the stability margin.
2.12.3 Integral Control
Now let’s consider the case where the control action is based on the integral of position
error,
t
u(t) = K I ∫ [x ( ) − x( )]d
d
0
K I
u(s) = [x (s) − x(s)]
d
s
Substituting this into mass-force model
K I
2
s x(s) = u(s) = [x (s) − x(s)]
d
s
3
s x(s) + K x(s) = K x (s)
I d
I
3
(s + K )x(s) = K x (s)
I
I d
The closed loop system poles are given by (Figure 2.46);
1
Δ (s) = 1 + K
cl I 3
s
Figure 2.46b shows the locus of CLS poles for various values of K as it takes on
I
values from 0 to ∞. The integral control alone would result in an unstable mass-force
system. It tends to destabilize the system. However, the main purpose of integral control