A warm-up question:
open-loop control
Reference input
r
controller
plant
Gain=1
Gain=1
output
y
Time delay: T/2 seconds
r(t)
1
Time (s)
0
y(t)=?
y(t)
1
Time (s)
0
T/2
Closed-loop control
output
Reference input
plant
controller
r
y
e
Gain=1
Gain=1
Time delay=0;
T/2 seconds
Gain=1
sensor
r(t)
1
Time (s)
0
1) Time delay =0, y=? 2) Time delay in plant =T/2, y=?
3) If time delay (=T/2) exists in sensor not in plant, y=?
Plot “y” in the above three conditions
Closed-loop control
4) The gain of controller is reduced to 1/2
Reference input
r
output
plant
controller
e
y
Gain=1/2
Gain=1
Time delay: T/2 seconds
Gain=1
r(t)
sensor
1
Time (s)
0
……(final value=?)
y(t)=?
in first 3 periods
0
T/2
T
3T/2
Plot “y”
2T 5T/2
3T
Time (s)
Solutions
output
Reference input
controller
plant
r
y
Gain=1
Gain=1
Time delay=0;
T/2 seconds
Gain=1
sensor
y(t)
1) Time delay=0
1/2
Time (s)
1
2) Time delay=T/2
in plant
0
T/2
T
3T/2
2T 5T/2
3T
Time (s)
0
T/2
T
3T/2
2T 5T/2
3T
Time (s)
1
3) Time delay=T/2
in sensor
Explaination for Question 2)
2) At the first T/2 seconds, due to the time delay, the signal
from the input cannot be sent to the output y. It means y=0
during the first T/2. The error e=r-y=1-0=1 during the first T/2
seconds.
Due to time delay, after T/2 seconds, the error signal (e=1)
can be sent to the output y, then y becomes 1 during T/2~T.
e =1-1=0 during T/2~T.
After another T/2, the error signal (e=0) is sent to the output
y, then y becomes 0 during T~3/2T.
The similar action repeats…..
In math, r-y1=y2, r-y2=y3, r-y3=y4………………
r=1 and y1=0, then you may find all the values for y2,y3,….
Explaination for Question 4)
Solutions
Reference input
e
plant
controller
output
r
y
Gain=1/2
Gain=1
Time delay=T/2 seconds
PID
P=1/2
Gain=1
4)
r(t)
sensor
1
Time (s)
0
1/2
1/4
y(t)
0
T/2
T
3/8 5/16
11/32
3T/2
2T 5T/2
……1/3 (final)
3T
Time (s)
General Conclusions
In a closed-loop control system:
1. Time delay may make control systems unstable.
In Question 1), there is no time delay, the output is ½.
In Question 2), there is time delay, so the ouput has sustainable
oscillation that is unstable.
2. Regarding controller, large gain may cause instability.
In Qeustion 2), the gain of controller is 1, the system is not stable.
The ouput has sustainable oscillation that is unstable.
If the gain of controller is reduced to ½ as shown in Question 4), the
steady-state output has a limited value 1/3.
3. Regarding controller, large gain may increase accuracy.
In Question 4), the controller gain is ½, the steady-state error is 11/3=2/3.
In Question 1), the controller gain is 1, the steady-state error is 11/2=1/2.