Tartalomjegyzék

Script d2018_01_16_LPV_zero_dynamics

Teljes Matlab script kiegészítő függvényekkel.

File: d2018_01_16_LPV_zero_dynamics.m
Directory: projects/3_outsel/2018_01_10_LPV_inversion
Author: Peter Polcz (ppolcz@gmail.com)

Created on 2018. January 16.

Inherited from

LPV decomposition (version 3)

File: d2018_01_13_Kalman_decomp_LPV_v3.m
Directory: projects/3_outsel/2018_01_10_LPV_inversion
Author: Peter Polcz (ppolcz@gmail.com)

Created on 2018. January 13.

Inherited from

Kalman decomposition

File: Kalman_decomposition.m
Directory: demonstrations/oktatas/ccs/2017fall
Author: Peter Polcz (ppolcz@gmail.com)

Created on 2018. January 13.

Inhereted from:

CCS 2017 fall. Homework 2. Solutions

File: d2017_10_26_hf2_mo.m
Directory: demonstrations/oktatas/ccs/2017fall
Author: Peter Polcz (ppolcz@gmail.com)

Created on 2017. October 04.
Reviewed on 2017. October 30.

Nice matrices

Given the following LPV system:

$$ \begin{aligned} &\Sigma: \left\{\begin{aligned} &\dot x = A(\rho) x + B(\rho) u,~~~ \rho \in \mathcal P \\ &y = C x \\ \end{aligned}\right. \\ &\begin{aligned} \text{where: } & A(\rho) = A_0 + A_1 \rho \in \mathbb{R}^{n\times n} \\ & B(\rho) = B_0 + B_1 \rho \in \mathbb{R}^{n\times r} \\ & C \in \mathbb{R}^{m\times n} \\ & D = 0_{m\times r} \end{aligned} \end{aligned} $$

load LPV_18_01_13_Time211859

A_fh = @(rho) A0 + A1*rho;
B_fh = @(rho) B0 + B1*rho;
C_fh = @(rho) C0 + C1*rho;
D_fh = @(rho) D0 + D1*rho;

A = {A0, A1};
B = {B0, B1};
C = {C0, C1};
D = {D0, D1};

AA = [A{:}];
BB = [B{:}];

\begin{equation} {\LARGE(1) \quad} A(\rho) = \left(\begin{array}{ccccccc} 1.0 & -1.1 & 2.7 & -0.068 & 1.0 & -0.39 & -0.68 \\ 0 & -1.210^{-3} & -0.67 & 0 & 1.6 & 0 & -0.82 \\ 0 & 0.17 & -0.7 & 0 & 1.1 & 0 & -0.55 \\ -0.73 & 2.1 & -3.4 & 0.3 & -3.4 & 2.2 & 0.93 \\ -0.73 & 0.82\rho +0.93 & -1.6 & -1.1 & -2.0 & 0.33 & 0.93 \\ 0 & 1.0 & -1.6 & 1.0 & -1.0 & -1.7 & 0 \\ -1.5 & 2.2 & -4.6 & -2.2 & -6.5 & 0.67 & 3.1 \\ \end{array}\right) \end{equation} \begin{equation} {\LARGE(2) \quad} B(\rho) = \left(\begin{array}{cc} 0.98 & 0.88 \\ 0 & 0 \\ 0 & 0 \\ 0.85\rho -0.5 & -0.44 \\ 0.15 & 0.13 \\ -0.23 & -0.21 \\ 0.29 & 0.26 \\ \end{array}\right) \end{equation} \begin{equation} {\LARGE(3) \quad} C(\rho) = C = \left(\begin{array}{ccccccc} 0 & -1.7 & 2.6 & -0.23 & 0.23 & -2.2 & 0 \\ 0 & -0.4 & 0.61 & -1.3 & 1.3 & 1.0 & 0 \\ \end{array}\right) \end{equation}

Nálam az $E_c$ most az $Im(B)$.

Im_B = IMA([B{1} B{2}]);

\begin{equation} {\LARGE(4) \quad} \mathrm{Im}\big(B(\rho)\big) = \left(\begin{array}{cc} -0.838 & -0.392 \\ 0 & 0 \\ 0 & 0 \\ 0.424 & -0.906 \\ -0.125 & -0.0586 \\ 0.201 & 0.0941 \\ -0.25 & -0.117 \\ \end{array}\right) \end{equation}

Ker_C = INTS(KER(C{1}), KER(C{2}));

\begin{equation} {\LARGE(5) \quad} \mathrm{Ker}(C) = \left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & -0.88 & 0 & 0 & 0 \\ 0 & -0.386 & -0.585 & 0 & 0 \\ 0 & 0.129 & -0.377 & -0.707 & 0 \\ 0 & -0.129 & 0.377 & -0.707 & 0 \\ 0 & 0.21 & -0.611 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right) \end{equation}

We compute $\mathcal V^*$ and $\mathcal R^*$.

[R,V] = CSA([A{1} A{2}], Im_B, Ker_C);

assert(rank(R) == 0 && rank([V Ker_C]) == rank(Ker_C));
assert(rank(INTS(V, Im_B)) == 0)

We ensure that $\mathcal R^* = \{0\}$ and that $\mathcal V^* \subseteq \mathrm{Ker}(C)$. Furthermore, we checked the strong invertibility condition $\mathcal V^* \cap \mathrm{Im}\big(B(\rho)\big) = \{0\}$.

\begin{equation} {\LARGE(6) \quad} \mathcal V^* = \left(\begin{array}{ccccc} -1 & 0 & 0 & 0 & 0 \\ 0 & -0.88 & 0 & 0 & 0 \\ 0 & -0.386 & -0.585 & 0 & 0 \\ 0 & 0.129 & -0.377 & -0.707 & 0 \\ 0 & -0.129 & 0.377 & -0.707 & 0 \\ 0 & 0.21 & -0.611 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right) \end{equation}

State transformation invariant feedback design

1. variáns

Let $T = \begin{pmatrix} {V^*}^\perp & L \end{pmatrix}^T$, where ${V^*}^\perp$ and $L$ are orthonormal bases for ${\mathcal V^*}^\perp \not\perp \mathcal L \subseteq \mathrm{Im}\big(B(\rho)\big)$, respectively.

L = ORTCO(Im_B);
T = [ ORTCO(V) L(:,1:end) ]';

\begin{equation} {\LARGE(7) \quad} T = \left(\begin{array}{cc|ccccc} 0 & 0.475 & -0.713 & 0.239 & -0.239 & 0.388 & 0 \\ 0 & 0 & 0 & 0.533 & -0.533 & -0.657 & 0 \\ 0.381 & 0 & 0 & 0 & -0.336 & 0.539 & -0.672 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0.932 & 0.227 & -0.283 \\ 0 & 0 & 0 & 0 & 0 & 0.78 & 0.626 \\ \end{array}\right) \end{equation}

tol = 1e-10;
prec = -log10(tol);

N = numel(A);
tA = cell(1,N);
tB = cell(1,N);
tC = cell(1,N);
tD = cell(1,N);

for i = 1:numel(A)
    tA{i} = round(T*A{i}/T, prec);
    tB{i} = round(T*B{i}, prec);
    tC{i} = round(C{i}/T, prec);
    tD{i} = round(D{i}, prec);
end

tA_fh = @(rho) tA{1} + tA{2}*rho;
tB_fh = @(rho) tB{1} + tB{2}*rho;
tC_fh = @(rho) tC{1} + tC{2}*rho;
tD_fh = @(rho) tD{1} + tD{2}*rho;

\begin{equation} {\LARGE(8) \quad} \bar A(\rho) = \left(\begin{array}{cc|ccccc} 0.99 & 0.93 & 0 & 0.098-0.2\rho & -0.15 & 0 & 0 \\ 3.2 & -1.3 & 0 & -0.44\rho -1.6 & 2.3 & 0 & 0 \\ \hline -9.8 & 8.8 & 4.3 & 3.0-0.28\rho & -3.2 & 9.3 & 4.6 \\ 0 & 0 & 0 & -1.210^{-3} & -0.67 & 1.8 & -0.51 \\ 0 & 0 & 0 & 0.17 & -0.7 & 1.2 & -0.34 \\ 1.4 & -0.91 & -0.7 & 0.77\rho -0.17 & 0.44 & -0.73 & -1.1 \\ 4.7 & -3.1 & -2.4 & -0.033 & -0.73 & -6.7 & -2.5 \\ \end{array}\right) \end{equation} \begin{equation} {\LARGE(9) \quad} \bar B(\rho) = \left(\begin{array}{cc} 0.2\rho -0.24 & -0.22 \\ 0.45\rho -0.19 & -0.17 \\ \hline 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \\ \end{array}\right) \end{equation} \begin{equation} {\LARGE(10) \quad} \bar C(\rho) = C = \left(\begin{array}{cc|ccccc} -3.6 & 1.2 & 0 & 0 & 0 & 0 & 0 \\ -0.85 & -2.1 & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right) \end{equation}

Published with MATLAB® R2017b, and sanitarized by BeautifulSoup (Python script using BSoup is written by Polcz).