52AND A CERTAIN NUMBER OF ITS TIME DERIVATIVES. FOR THIS PURPOSE THE...
1.5
2
and a certain number of its time derivatives. For this purpose the corresponding transfer1
function of the system under consideration is employed0.5
0
b0
+
b1
·s+
b2
·s2
YK
(
s)
x
Kpd
in m/s
y
Kpd
in m/s
−0.5
−1
Ud
(
s) =
cT
y
sI−Ay
−1
by
=
N(
s)
. (44)−2
−1.5
Obviously, the numerator of the control transfer function contains a second degree polynomial0
5
10
15
t in s
in s, leading to two transfer zeros. This shows that the considered outputyK
(
t)
represents anon-flat output variable that makes computing of the feedforward term more difficult. A pos-Fig. 4. Desired trajectories for the cage motion: desired and actual position in horizontalsible way for calculating the desired input variable is given by a modification of the numeratordirection (upper left corner), desired and actual position in vertical direction (upper rightof the control transfer function by introducing a polynomial ansatz for the feedforward actioncorner), actual velocity in horizontal direction (lower left corner) and actual velocity in verticalaccording todirection (lower right corner).YKd
(
s)
. (45)Ud
(
s) =
kV0
+
kV1
·s+
. . .+
kV4
·s4
For its realisation the desired trajectoryyKd
(
t)
as well as the first four time derivatives areand its first three time derivatives are considered. Including the equations of motion (12)available from a trajectory planning module. The feedforward gains can be computed fromyields the following set of equationsa comparison of the corresponding coefficients in the numerator as well as the denominatorpolynomials ofyKd
(
t) =
yS
(
t) +
1
2
κ2
(
3−κ)
·v1
(
t)
, (49)b0
+
. . .+
b2
·s2
kV0
+
. . .+
kV4
·s4
y˙Kd
(
t) =
y˙S
(
t) +
1
2
κ2
(
3−κ)
·v˙1
(
t)
, (50)YKd
(
s) =
N(
s)
y¨Kd
(
t) =
y¨S
(
t) +
1
2
κ2
(
3−κ)
·v¨1
(
t) =
y¨K
(
v1
(
t)
, ˙yS
(
t)
, ˙v1
(
t)
,ud
(
t)
,κ)
, (51)kVj
·s+
. . .+
bV6
·s6
+
bV1
...yKd
(
t) =
...yK
(
v1
(
t)
, ˙yS
(
t)
, ˙v1
(
t)
,ud
(
t)
, ˙ud
(
t)
,κ)
. (52)=
bV0
a0
+
a1
·s+
. . .+
s4
(46)Solving equation (49) to (52) for the system states results in the desired state vector,i=
0, . . . ,n=
4 . (47)yKd
(
t)
,ud
(
t)
, ˙ud
(
t)
,κ)
ySd
(
yKd
(
t)
, ˙yKd
(
t)
, ¨yKd
(
t)
,...ai
=
bVi
v1d
(
y˙Kd
(
t)
, ¨yKd
(
t)
,...yKd
(
t)
,ud
(
t)
, ˙ud
(
t)
,κ)
xd
(
t) =
yKd
(
t)
,ud
(
t)
, ˙ud
(
t)
,κ)
y˙Sd
(
y˙Kd
(
t)
, ¨yKd
(
t)
,... . (53)This leads to parameter-dependent feedforward gainskVj
=
kVj
(
κ)
. It is obvious that duev˙1d
(
y˙Kd
(
t)
, ¨yKd
(
t)
,...yKd
(
t)
,ud
(
t)
, ˙ud
(
t)
,κ)
the higher numerator degree in the modified control transfer function a remaining dynamicsmust be accepted. Lastly, the desired input variable in the time domain is represented byThis equation still contains the inverse dynamicsud
(
t)
and its time derivative ˙ud
. Substitutingud
for equation (48) and ˙ud
(
t)
for the time derivative of (48), which can be calculated analyti-yKd
(
t)
,y(4)
Kd
(
t)
,κy˙Kd
(
t)
, ¨yKd
(
t)
,...ud
(
t) =
ud
. (48)To obtain the desired system states as function of the output trajectory the output equation