Siemens Time Clock pcs 7 User Manual
Introduction
8
MPC Level
V 1.0, Beitrags-ID: 42200753
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Figure 2-1 Step response of a control loop with compensation (blue) and without
compensation (red), i.e. with integral behaviour.
0
10
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30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
mit
Ausgleich
Ausgleich
ohne
Ausgleich
Ausgleich
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
with
compensation
compensation
whithout
compensation,
i.e. with
integral
behaviour
compensation,
i.e. with
integral
behaviour
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
mit
Ausgleich
Ausgleich
ohne
Ausgleich
Ausgleich
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
0
10
20
30
40
50
60
70
80
90
100
0
0.5
1
1.5
2
2.5
3
Step Response
Time (sec)
Am
pli
tude
with
compensation
compensation
whithout
compensation,
i.e. with
integral
behaviour
compensation,
i.e. with
integral
behaviour
Unstable control loops cannot be stabilized without a controller. Therefore, switch-
ing a controller in such a loop to manual mode is not allowed for a longer time.
Hence, the recording of measurement data for the process identification (e.g. for
Hence, the recording of measurement data for the process identification (e.g. for
the PID tuner or the MPC configurator) via step experiments in open control loop is
not possible. The model type and the control algorithm of the MPC function block
are also inappropriate for unstable control loops. Therefore, the unstable part trans-
fer functions have to be stabilized by subordinated slave controllers before the ap-
plication of the MPC.