<a href="./?plugin=related&amp;page=%BC%F8%B6%C8%2F%C0%A9%B8%E6%B9%A9%B3%D8%C6%C3%CF%C02013">授業/制御工学特論2013

[lecture #1] 2013.9.5 outline of the lecture, review of classical and modern control theory (1/3)†

outline of this lecture
- syllabus
- map
  map_v1.0.pdf
- evaluation
  - mini report #1 ... 10%
  - mini exam #1 ... 10%
  - mini report #2 ... 10%
  - mini exam #2 ... 10%
  - final report ... 60%

review

... missed (derivation of G(s) = 1/(ms^2+cs+k) by Laplace transformation from given equation of motion)

whiteboard #3

whiteboard #4

[lecture #2] 2013.9.12 CACSD introduction with review of classical and modern control theory (2/3)†

introduction of Matlab and Simulink
- Basic usage of MATLAB and Simulink used for 情報処理演習及び考究II/Consideration and Practice of Information Processing II: Advanced Course of MATLAB

How to define open-loop system

s = tf('s');
G1 = 1 / (s+1);
G2 = 1 / (s^2 + 0.1*s + 1);

SSR

A = [-0.3, -1; 1, 0];
B = [1; 0];
C = [0, 1];
D = 0;
G3 = ss(A, B, C, D);

Bode plot

bode(G1, 'b-', G2, 'g', G3, 'r--');
grid on;

open-loop stability can be checked by
1. poles of TF
```
roots(G2.den{:})
```
2. eigenvalues of A-matrix in SSR
```
eig(G3.a)
```
3. also by simulation
  mod0912_1.mdl

closed-loop stability

L = 1/(s^3+1.5*s^2+1.5*s+1); % example of open-loop system
roots(L.den{:}) % confirm the open-loop system is stable

graphical test by Nyquist stability criterion and Bode plot with GM(gain margin) and PM(phase margin)
```
nyquist(L)
bode(L)
```

numerical test by closed-loop system

clp_den = L.den{:} + L.num{:};
roots(clp_den)

simulation
mod0912_2.mdl

a = 1
ver
t = [1 2 3]
pwd
ls
foo
pwd
bar
s = tf('s')
G1 = 1 / (s+1);
G2 = 1 / (s^2 + 0.1*s + 1);
G1
G2
A = [-0.3, -1; 1, 0];
B = [1; 0];
C = [0, 1];
D = 0;
G3 = ss(A, B, C, D);
G3
bode(G1, 'b-', G2, 'g', G3, 'r--');
grid on;
G2
G2.den
G2.den{:}
roots(G2.den{:})
G3.a
eig(G3.a)
mod0912_1
L = 1/(s^3+1.5*s^2+1.5*s+1); % example of open-loop system
roots(L.den{:}) % confirm the open-loop system is stable
nyquist(L)
nyquist(L*1.25)
nyquist(L)
clp_den = L.den{:} + L.num{:};
roots(clp_den)
ex0912_2
mod0912_2
L = 1.5*L

[lecture #3] 2013.9.19 CACSD introduction with review of classical and modern control theory (3/3)†

LQR problem
ARE and quadratic equation
J.pdf
(semi)-positive definiteness
example
mod0919.mdl

mod0919
A = [1, 2, 3; 4, 5, 6; 7, 8, 9]
eig(A)
B = [1; 1; 1]
Uc = ctrb(A, B)
det(Uc)
B = [1; 0; 0]
Uc = ctrb(A, B)
det(Uc)
help are
P = are(A, B/R*B', Q)
Q = eye(3)
R = 1
P = are(A, B/R*B', Q)
P = are(A, B*inv(R)*B', Q)
P - P'
eig(P)
x(0)
x0 = [1; 1; 1]
x0'*P*x0
F = R\B'*P
C = eye(3)
D = [0; 0; 0]
J

[lecture #4] 2013.9.26 Intro. to robust control theory (H infinity control theory) 1/3†

Typical design problems
1. robust stabilization
2. performance optimization
3. robust performance problem (robust stability and performance optimization are simultaneously considered)
H infinity norm
- definition
- example
H infinity control problem
- definition
- application example : reference tracking problem
  - relation to the sensitivity function S(s) (S(s) -> 0 is desired but impossible)
  - given control system
    ex0926_1.m

controller design with H infinity control theory
ex0926_2.m

s = tf('s')
G = 1/(s+1)
norm(G, 'inf')
G = s/(s+1)
norm(G, 'inf')
G = 1/(s^2+0.1*s+1)
bodemag(G)
norm(G, 'inf')
bodemag(G, 'b', ss(10.0125), 'r--')
G = 1/(s^2+0.5*s+1)
norm(G, 'inf')
bodemag(G, 'b', ss(2.0656), 'r--')
ex0926_1
ex0926_2
eig(K_hinf)

[lecture #5] 2013.10.3 Intro. to Robust Control Theory (H infinity control theory) 2/3 cancelled†

[lecture #5] 2013.10.10 Intro. to Robust Control Theory (H infinity control theory) 2/3†

Typical design problems
1. robust stabilization
2. performance optimization
3. robust performance problem (robust stability and performance optimization are simultaneously considered)
connection between [H infinity control problem] and [robust stabilization problem]
- small gain theorem
- normalized uncertainty \Delta
- sketch proof ... Nyquist stability criterion
How to design robust stabilizing controller with H infinity control problem ?
- practical example : unstable plant with perturbation
  ex1010_1.m
- how to use uncertainty model (multiplicative uncertainty model)
  ex1010_2.m
- how to set generalized plant G ?
  ex1010_3.m
- simulation
  mod1010.mdl

ex1010_1
ex1010_2
WT]
WT
P0
P0_jw
ex1010_3
mod1010

[lecture #6] 2013.10.17 Intro. to robust control theory (H infinity control theory) (3/3)†

review
- robust stabilization ... (1) ||WT T||_inf < 1 (for multiplicative uncertainty)
- performance optimization ... (2) ||WS S||_inf < gamma -> min
- mixed sensitivity problem ... simultaneous consideration of stability and performance
a sufficient condition for (1) and (2) ... (*) property of maximum singular value
definition of singular value
mini report #1
meaning of singular value ... singular value decomposition (SVD)
proof of (*)
example
ex1017.m

j
A = [1, j; 0, 2]
A'
eig(A'*A)
sqrt(ans)
3+sqrt(5)
sqrt(3+sqrt(5))
sqrt(3-sqrt(5))
A
[U,S,V] = svd(A)
U'*U
U*U'
help svd
ex1017

whiteboard #3

... sorry for missing to take photo ... mini report #1: write by hand; submit at the beginning of the next lecture; You will have a mini exam #1 related to this report on 31st Oct.

[lecture #7] 2013.10.24 review of SVD, motivation of robust performance (robust performance problem 1/3), state space representation of generalized plant†

review of SVD
ex1024_1.m
motivation of robust performance
ex1024_2.m

ex1024_3.m
... SSR of generalized plant
ex1024_4.m

ex1024_1
ex1024_2
ex1024_3
ex1024_4

[lecture #8] 2013.10.31 Robust performance problem (2/3)†

review of mini report #1
review of the limitation of mixed sensitivity problem
a solution of conservative design
- example based on the one given in the last lecture
  ex1031_1.m
- a check of the conservativeness
  ex1031_2.m
mini exam #1
exam1.pdf

%-- 10/31/2013 1:02 PM --%
A = [j, 0; -j, 0]
A = [j, 0; -j, 1]
svd(A)
sqrt((3+sqrt(5))/2)
sqrt((3-sqrt(5))/2)
ex1024_2
ex1024_3
ex1024_4
ex1031_1

[lecture #9] 2013.11.14 Robust performance problem (3/3)†

map
review of #8
Q1 and Q2
ex1031_2.m
scaled H infinity control problem
ex1114_1.m
mini report #2

ex1024_2
ex1024_3
ex1024_4
ex1024_5
ex1031_1
gam
ex1024_3
gam
ex1031_2
help lft
ex1031_2
ex1031_1
ex1031_2

[lecture #10] 2013.11.21 Robust performance problem (1/3) (cont.)†

effect of scaling
mini report #2
practical design procedure
derivation of generalized plant in SSR for mixed sensitivity problem

ex1024_2
ex1024_3
gam
ex1024_4
ex1031_1
gam
ex1031_2
ex1114_1
gam

[lecture #11] 2013.11.28 Robust stabilization of inverted pendulum†

plant model for perturbed unstable poles
LFT(Linear Fractional Transformation)
a simple example
inverted pendulum (penddemo.m)
design example
references:
- How to control objects to design, to simulate and to experiment control system by using MATLAB/Simulink with an application of Inverted Pendulum
  - stabilization of 1-link inverted pendulum
- 情報処理演習および考究 II MATLAB コースホームページ (2005) (in Japanese)
  - 倒立振子の安定化 (in Japanese)

[lecture #12] 2013.12.5 Robust control design for a practical system : Active vibration control of a pendulum using linear motor (1/3)†

review of report #2
introduction of experimental setup

linear motor : Oriental motor EZC4D005M-A / stepping motor, 0.01mm = 1pulse, stroke 50mm (= 5000 pulse), thrust 70N, speed(max)600mm/s (but used as 400mm/s max with pulse_module.c which generates a pulse every 25us)
Potentio meter : Midori Precisions Model QP-2H / input:5V, output:0-4.17V for 90 deg (experimentally measured)

PC : Dell PowerEdge840 (RTAI3.6.1/Linux kernel 2.6.20.21)
Parallel input and output board : CONTEC PIO-32/32T(PCI) / 32bit 200ns (connected to linear motor)
A/D : CONTEC AD12-16 (PCI) 12bit, 10us

Objective of control system
1. to attenuate vibration due to pendulum oscillation
2. robust stability against modelling error due to plant variation and non-linear dynamics etc.

physical model
frequency response experiment
freqresp.m

frdata_0_2.dat

frdata_0_3.dat

frdata_0_4.dat

[lecture #13] 2013.12.12 Robust control design for a practical system : Active vibration control of a pendulum using linear motor (2/3)†

mini exam #2
pendulum No.2 ... l = 8.5cm
pendulum2.jpg
due to the difficulty of the inverted and short pendulum, the control target has been changed to an non-inverted and longer pendulum (I'm sorry)

modelling based on frequency response experiment
frdata_0.5mm.dat

freqresp_fixed.m

control objective
design example 1 : proportional control
- negative feedback always stabilizes the closed loop theoretically
  check_pcont.m
- control experiment
  `cont.dat' file for P control
  
  `cont_order.dat' file for P control
  
  result_P.dat
  
  result_openloop.dat
  
  openloop.mp4
  
  ex1.mp4

design example 2 : H infinity control (nominal performance, physical model is used as nominal plant)
- m-files
  weight_ex2.m
  
  cont_ex2.m
```
>> weight_ex2
>> cont_ex2
```
- control experiment
  `cont.dat' file for ex2
  
  `cont_order.dat' file for ex2
  
  `cont.mat' file for ex2
  
  result_ex2.dat

design example 3 : H infinity control (nominal performance, nominal plant is derived from freq. resp. exp.)
- m-files
  nominal_ex3.m
  ... n4sid is not available IPC! (19 Dec.)
  weight_ex3.m
  
  cont_ex3.m
```
>> nominal_ex3
>> weight_ex3
>> cont_ex3
```
- control experiment
  `cont.dat' file for ex3
  
  `cont_order.dat' file for ex3
  
  `cont.mat' file for ex3
  
  result_ex3.dat
  
  ex3.mp4

design example 4 : H infinity control (robust performance)
- m-files
  weight_ex4.m
  
  cont_ex4.m
```
>> weight_ex4
>> cont_ex4
```
- control experiment
  `cont.dat' file for ex4
  
  `cont_order.dat' file for ex4
  
  `cont.mat' file for ex4
  
  result_ex4.dat
  
  ex4.mp4

summary of design examples : comparison of designed controllers
compare.m

report

design your controller(s) so that the system performance is improved compared with the design ~~example 3 (ex3)~~ example 2 (ex2)
Draw the following figures and explain the difference between two control systems (your controller and ~~ex3~~ ex2):
1. bode diagram of controllers
2. gain characteristic of closed-loop systems
3. time response of control experiment
Why is the performance of your system improved(or unfortunately decreased)?
- due date: 31th(Tue) Dec 17:00
- submit your report(pdf or doc) by e-mail to kobayasi@nagaokaut.ac.jp
- You can use Japanese
- maximum controller order is 20
- submit your cont.dat, cont_order.dat, and cont.mat to kobayasi@nagaokaut.ac.jp not later than 26th(Thu) Dec

program sources for frequency response experiment
freqresp.h

freqresp_module.c

freqresp_app.c
- format of frdata.dat file
  - 1st column: frequency (Hz)
  - 2nd column: gain
  - 3rd column: phase (deg)
program sources for control experiment
hinf.h

hinf_module.c

hinf_app.c
- format of result.dat file
  - 1st column: time (s)
  - 2nd column: potentio meter's output (V)
  - 3rd column: theta (rad)
  - 4th column: x (m)
  - 5th column: reference position for linear motor (number of pulse)
program sources for generating linear motor's pulse
pulse.h

pulse_module.c
common header
common.h

[lecture #14] 2013.12.19 Robust control design for a practical system : Active vibration control of a pendulum using linear motor (3/3)†

review of the design examples

[IMPORTANT] Due to unavailability of n4sid in IPC which is used in cont_ex3.m, please compare your controller and example 2 (not 3) in your report. The explanation of the report has been modified due to this change. See above.

preparation of your own controller(s)

participant list2013

freqresp_fixed
frdata
check_pcont
weight_ex2
freqresp_fixed
weight_ex2
cont_ex2
nominal_ex3
weight_ex4
cont_ex4
compare