Handin 1
Bo Bernhardsson, K. J. Åström
Department of Automatic Control LTH,
Lund University
Bo Bernhardsson, K. J. Åström
Handin 1
Handin 1 - goals
Get some practice using the Matlab control system toolbox (or similar)
Get started with some control design
Bo Bernhardsson, K. J. Åström
Handin 1
Example - Double Integrator
Consider the double integrator
y=
1
u
s2
controlled with state-feedback + Kalman filter
u = −K x̂ = −K(sI − A + BK + LC)−1 Ly
Let’s place eigenvalues of
A − BK
and
A − LC
in Butterworth patterns
>Pm
>K
>Po
>L
=
=
=
=
butter(2,wm,’s’)
place(A,B,Pm)
butter(2,wo,’s’)
place(A’,C’,Po)’
Bo Bernhardsson, K. J. Åström
Handin 1
Results
Bode Diagram, C
40
Magnitude (dB)
60
50
0
-50
20
0
-100
-20
-150
-40
-135
45
-180
0
Phase (deg)
Phase (deg)
Magnitude (dB)
Bode Diagram P*C
100
-225
-45
-90
-270
0
1
10
10
Frequency (rad/sec)
2
0
3
10
10
10
1
10
Frequency (rad/sec)
2
3
10
10
2
Nichols Chart
1.5
40
0.5 dB
1 dB
Open-Loop Gain (dB)
20
0.5
0
-0.5
-1
3 dB
6 dB
0
-20
-40
-60
-1.5
-2
-2
0 dB
0.25 dB
1
-80
-270
-1.5
-1
-0.5
0
0.5
1
1.5
Bo Bernhardsson, K. J. Åström
-225
-180
-135
Open-Loop Phase (deg)
2
Handin 1
-90
Results
3
10
2
Max |C(iw)|
10
1
10
0
10 0
10
1
10
wgc rad/s
2
10
Trade off between closed loop bandwidth and controller gain
(Hmmmm, why is the slope 2?)
(Advanced hmmmm, why do LQG design with varying control
penalty ρ and Butterworth pole placement give the same results?
Answer to this later in the course.)
Bo Bernhardsson, K. J. Åström
Handin 1
Handin 1A- PI control of 1st order system
First order model: the archetype system
Normalizing input, output and time variables we have
G(s) =
1
s+1
Let’s use PI control
u = (kp + ki /s)(r − y)
This gives a 2nd order closed system.
Bo Bernhardsson, K. J. Åström
Handin 1
Handin 1A- PI control of 1st order system
Let’s design the closed loop polynomial to become
s2 + 2ζ0 ω0 s + ω02
we get
kp = 2ζ0 ω0 − 1
ki = ω02
Let’s assume a slow design is ok, say ω0 = 0.1, ζ0 = 0.5.
Bo Bernhardsson, K. J. Åström
Handin 1
Result - Rise time vs ω0
6
100
90
5
input load disturbance peak
80
risetime
70
60
50
40
30
20
4
3
2
1
10
0
0
2
4
omega0
6
8
10
0
0
Looks as expected.
Let’s check the control signal size also.
Bo Bernhardsson, K. J. Åström
Handin 1
2
4
ω0
6
8
10
Step Response - input signal size
10
9
control signal peak
8
7
6
5
4
3
2
1
0
2
4
omega0
6
8
10
Hmm, the behavior when ω0 is small is rather unexpected.
Let’s check the Bode and Nyquist diagrams.
Bo Bernhardsson, K. J. Åström
Handin 1
Nyquist Diagram ω0 =0.1
step response with G=1.1/(s+1), omega0=[0.1 0.3 1 3 10]
2
10
1.5
8
6
1
4
0.5
2
0
0
-0.5
-2
-1
-4
-1.5
-2
-2
-6
-1.5
-1
-0.5
0
0.5
1
1.5
2
-8
0
20
40
60
80
The design with ω0 = 0.1 has terrible robustness. The system
becomes unstable with 10 % model error
Practically useless!
Why? What to do?
Bo Bernhardsson, K. J. Åström
Handin 1
100
Handin 1
Exercise 1A Verify the previous figures. Use pole placement design to
do PI control of the system 1/(s + 1) for varying ω0 .
Use any method you like to find a PI-controller that achieves good
robustness and a gain-crossover frequency ωgc = 0.1, or describe
why this is not possible.
Bo Bernhardsson, K. J. Åström
Handin 1
Handin 1
Exercise 1B Consider the system P (s) =
s+1
. Design a controller
s2
with pole-placement where the observer poles and the controller poles
have ω0 = 10 and damping ratio ζ0 = 0.707. Plot the Nyquist curve of
the loop transfer function and the Gang of Four for the closed loop
systems obtained. Comment on the design.
Handin 1 is due Thursday Feb 25, 10.00
Bo Bernhardsson, K. J. Åström
Handin 1