Advanced Automation†
[lecture #1] 2013.9.5 outline of the lecture, review of classical and modern control theory (1/3)†
- outline of this lecture
- syllabus
- map
- evaluation
- mini report #1 ... 10%
- mini exam #1 ... 10%
- mini report #2 ... 10%
- mini exam #2 ... 10%
- final report ... 60%
... missed (derivation of G(s) = 1/(ms^2+cs+k) by Laplace transformation from given equation of motion)
[lecture #2] 2013.9.12 CACSD introduction with review of classical and modern control theory (2/3)†
- introduction of Matlab and Simulink
- How to define open-loop system
- TF
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
- poles of TF
roots(G2.den{:})
- eigenvalues of A-matrix in SSR
eig(G3.a)
- also by simulation
- 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
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
- (semi)-positive definiteness
- example
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
- robust stabilization
- performance optimization
- robust performance problem (robust stability and performance optimization are simultaneously considered)
- H infinity norm
- 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
- controller design with H infinity control theory
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
- robust stabilization
- performance optimization
- 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
- how to use uncertainty model (multiplicative uncertainty model)
- how to set generalized plant G ?
- simulation
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
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
... 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
- motivation of robust performance
... SSR of generalized plant
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
- a check of the conservativeness
- mini exam #1
%-- 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
- 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:
[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
- to attenuate vibration due to pendulum oscillation
- robust stability against modelling error due to plant variation and non-linear dynamics etc.
- physical model
- frequency response experiment
[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
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
- control objective
- design example 1 : proportional control
- negative feedback always stabilizes the closed loop theoretically
- control experiment
- design example 2 : H infinity control (nominal performance, physical model is used as nominal plant)
- design example 3 : H infinity control (nominal performance, nominal plant is derived from freq. resp. exp.)
- design example 4 : H infinity control (robust performance)
- summary of design examples : comparison of designed controllers
- 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):
- bode diagram of controllers
- gain characteristic of closed-loop systems
- 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
- format of frdata.dat file
- 1st column: frequency (Hz)
- 2nd column: gain
- 3rd column: phase (deg)
- program sources for control experiment
- 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
- common header
[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
[lecture #15] 2013.12.26 Robust control design for a practical system : Active vibration control of a pendulum using linear motor (cont.)†
- preparation of your own controller(s)