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Polcz Péter honlapja

Tartalomjegyzék

CCS 2017 fall. Homework 2. Solutions

Teljes Matlab script kiegészítő függvényekkel.

file:   ccs2017_hf2_mo.m
author: Peter Polcz <ppolcz@gmail.com>
Created on 2017. October 04.
Reviewed on 2017. October 30.
Output:
┌d2017_10_26_hf2_mo
│   - Persistence for `d2017_10_26_hf2_mo` reused (inherited) [run ID: 9200, 2017.10.30. Monday, 18:38:05]
│   - Script `d2017_10_26_hf2_mo` backuped

Acetone - human body

A = [
    -6 0 0
    4 0 0
    2 0 0
    ];
B = [ 1 ; 0 ; 0 ];
C = [ 0 0 1 ];

On = obsv(A,C)
Cn = ctrb(A,B)

rank(On)
rank(Cn)

orth(sym(Cn))
null(sym(On))

Hs = minreal(tf(ss(A,B,C,0)))

syms s t
clear H
H(s) = simplify(C/(s*eye(3) - A)*B)

h(t) = ilaplace(H)

disp 'NEM BIBO STABIL'
Output:
On =
     0     0     1
     2     0     0
   -12     0     0
Cn =
     1    -6    36
     0     4   -24
     0     2   -12
ans =
     2
ans =
     2
ans =
[ 1,             0]
[ 0, (2*5^(1/2))/5]
[ 0,     5^(1/2)/5]
ans =
 0
 1
 0

Hs =
 
      2
  ---------
  s^2 + 6 s
 
Continuous-time transfer function.

H(s) =
2/(s*(s + 6))
h(t) =
1/3 - exp(-6*t)/3
NEM BIBO STABIL

Kalman decomposition

Generate a decomposable system

a = 3;
b = 2;
c = 1;
d = 2;
n = a+b+c+d;
r = 1;
m = 1;

A_ = [
    randn(a,a) zeros(a,b) randn(a,c) zeros(a,d)
    randn(b,a) randn(b,b) randn(b,c) randn(b,d)
    zeros(c,a) zeros(c,b) randn(c,c) zeros(c,d)
    zeros(d,a) zeros(d,b) randn(d,c) randn(d,d)
    ];
B_ = [
    randn(a,r)
    randn(b,r)
    zeros(c,r)
    zeros(d,r)
    ];
C_ = [
    randn(m,a) zeros(m,b) randn(m,c) zeros(m,d)
    ];

D = zeros(m,r);

System $(\bar A, \bar B, \bar C, D)$ is in Kalman decomposed form. The transfer function is reducible to a 3rd order system.

sys = ss(A_,B_,C_,D);
H = minreal(tf(sys))
Output:
H =
 
      5.27 s^2 + 1.079 s + 9.61
  ---------------------------------
  s^3 + 1.639 s^2 + 1.662 s + 3.532
 
Continuous-time transfer function.

Transform the system with a random transformation matrix $T$.

T = orth(rand(n));
A = T * A_ * T';
B = T * B_;
C = C_ * T';

sys = ss(A,B,C,D);

Kalman decomposition. Solution

1. Controllability staircase form

Controllability matrix

Cn = ctrb(A,B);

Transformation matrix, which generates the controllability staircase form.

[S1,~,~] = svd(Cn)
Output:
S1 =
  Columns 1 through 7
    0.1851   -0.1260    0.6605    0.3331   -0.0259    0.0654   -0.4868
   -0.4454   -0.3309    0.5066    0.1294    0.0792   -0.0298    0.6392
    0.0858    0.5118    0.3553   -0.5421    0.5260   -0.0505    0.0978
   -0.3697    0.2123    0.3263   -0.3468   -0.5290    0.1723   -0.2998
   -0.7716    0.1020   -0.2560    0.0252    0.1330   -0.0619   -0.2651
    0.0504    0.6169    0.0561    0.3454   -0.4374   -0.4296    0.2901
   -0.1453    0.1048    0.0647    0.3751    0.4385   -0.4544   -0.2988
   -0.0656    0.4070   -0.0381    0.4444    0.1876    0.7535    0.1060
  Column 8
   -0.4009
    0.0543
   -0.1472
    0.4383
   -0.4860
   -0.1854
    0.5788
    0.1302

The transformation.

A1 = S1' * A * S1
B1 = S1' * B
C1 = C * S1
Output:
A1 =
  Columns 1 through 7
   -1.8089    0.6452    0.1483    1.5986    0.8907    0.0490    0.1295
   -0.1648   -0.3673   -2.1454    0.5573    0.3674    0.7582   -0.6061
    0.1214    0.9705    0.5429   -1.3994   -0.7883   -0.6553    0.0855
   -0.0008   -0.0089    0.0035   -0.1199   -1.3464    0.8948    0.0355
    0.0020    0.0121    0.0126    0.2101    0.4802   -0.8108   -0.1284
    0.0000    0.0000    0.0000   -0.0000    0.0000   -2.4875    0.2476
   -0.0000   -0.0000   -0.0000    0.0000   -0.0000    0.1428    1.3063
   -0.0000   -0.0000   -0.0000   -0.0000   -0.0000    0.1130    0.0064
  Column 8
   -0.7829
    0.3596
   -0.4607
    0.0171
   -0.3724
    0.2728
   -0.7555
    0.5906
B1 =
   -0.2095
    1.3140
    0.1525
    2.5173
    0.2153
    0.0000
    0.0000
    0.0000
C1 =
  Columns 1 through 7
   -1.3935    0.0503    0.0561    1.7861    1.8917    0.0022    0.3711
  Column 8
   -0.3938

Order of the controllable subsystem

ab = rank(Cn);

2. Observability staircase form

2.1. Observability staircase form of the controllable subsystem

On1 = obsv(A1(1:ab,1:ab), C1(:,1:ab));
[~,~,S21] = svd(On1);

2.2. Observability staircase form of the uncontrollable subsystem

On2 = obsv(A1(ab+1:end,ab+1:end), C1(:,ab+1:end));
[~,~,S22] = svd(On2);

Second transformation matrix to generate the observability staircase form for both controllable and uncontrollable subsystems.

S2 = blkdiag(S21, S22)
Output:
S2 =
  Columns 1 through 7
    0.5406   -0.2665   -0.3286   -0.5400    0.4870         0         0
   -0.1069    0.3103    0.6597   -0.0538    0.6740         0         0
   -0.1265    0.8059   -0.2901   -0.4789   -0.1454         0         0
   -0.5703   -0.0446   -0.5775    0.2685    0.5168         0         0
   -0.5958   -0.4258    0.1979   -0.6357   -0.1429         0         0
         0         0         0         0         0   -0.0042    0.9914
         0         0         0         0         0   -0.6858   -0.0978
         0         0         0         0         0    0.7278   -0.0865
  Column 8
         0
         0
         0
         0
         0
   -0.1305
   -0.7212
   -0.6804

Transformation

A2 = round(S2' * A1 * S2,10)
B2 = round(S2' * B1,10)
C2 = round(C1 * S2,10)
Output:
A2 =
  Columns 1 through 7
   -1.7228    0.0510    0.0959         0         0   -0.2628         0
    1.6409    0.2415    0.8973         0         0    0.1467         0
   -0.3040   -2.1160   -0.1580         0         0    0.7482         0
    1.3084   -0.3290   -0.6989    0.0519   -0.1450    0.6167    0.9176
   -1.0943   -0.9964   -0.0245    1.5653    0.3143    0.2081    1.2303
         0         0         0         0         0    1.3011         0
         0         0         0         0         0    0.1434   -2.5055
         0         0         0         0         0    0.7488    0.2298
  Column 8
         0
         0
         0
   -0.7288
    0.0807
         0
    0.0408
    0.6139
B2 =
   -1.8370
    0.3824
   -0.5197
    0.5085
    2.0316
         0
         0
         0
C2 =
  Columns 1 through 7
   -2.9116   -0.4531   -0.1822         0         0   -0.5412         0
  Column 8
         0

Kalman decomposition using minreal

Using minreal the transfer function which generates the Kalman decomposition can be retained easily, however, this does not work all the times. The uncontrollable subsystem is not always decomposed (since the uncontrollable subsystem is already redundant).

[H,U] = minreal(sys);

A1 = round(U*A/U,10)
B1 = round(U*B,10)
C1 = round(C/U,10)
Output:
5 states removed.
A1 =
  Columns 1 through 7
   -1.1425   -1.6553    0.2030         0         0    0.5666   -0.5119
   -0.0107    0.0245    2.5022         0         0   -0.1765    0.1594
    0.5187   -0.4917   -0.5213         0         0    0.0771   -0.0697
   -0.2197    0.4070    1.4025    0.3210   -1.5640   -0.0442    0.2659
   -1.3625   -0.2481    0.6331    0.1463    0.0452    0.8996    0.0772
         0         0         0         0         0    0.6169         0
         0         0         0         0         0   -0.7573    1.3011
         0         0         0         0         0   -0.1920   -0.0812
  Column 8
         0
         0
         0
    1.2256
   -0.9628
         0
         0
   -2.5085
B1 =
    0.7434
    1.1530
   -1.3816
   -2.0339
    0.4989
         0
         0
         0
C1 =
  Columns 1 through 7
    1.2907    0.6553   -2.5731         0         0   -0.4015    0.3628
  Column 8
         0

End of the script.

Output:
└ 0.47728 [sec]