Aufgabe 3.2 – Steuersequenz für Zustandsregler

 

Gegeben ist das zeitdiskrete System

{{\vec x}_{k+1}} = \left( {\begin{array}{*{20}{c}} 0&1 \\ a&b \end{array}} \right)\vec x+\left( {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right){u_k}

{y_k} = \left( {\begin{array}{*{20}{c}} 1&1 \end{array}} \right){{\vec x}_k}

Steuern Sie das System in möglichst wenigen Abtastschritten vom Anfangszustand

{\vec x_0} = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)

in den Zustand

{\vec x_{end}} = \left( {\begin{array}{*{20}{c}}{-3} \\ 3 \end{array}} \right).

Wie viele Schritte werden benötigt?

Geben Sie die Steuersequenz (also {u_0},{u_1}, \ldots) an.

Lösung 3.2

Steuerbarkeitsmatrix:

{A_d}{\vec b_d} = \left( {\begin{array}{*{20}{c}} 0&1 \\ a&b \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 1 \\ b \end{array}} \right)

{S_S} = \left( {\begin{array}{*{20}{c}} 0&1 \\ 1&b \end{array}} \right)

\det \left( {{S_S}} \right) = 0 \cdot b-1 = -1 \ne 0

Das System ist vollständig steuerbar. Die Modellordnung ist n = 2, es werden also zwei Abtastschritte benötigt.

Berechnung der Steuerschritte:

\left( {\begin{array}{*{20}{c}}{{u_{n-1}}} \\ \vdots \\ {{u_1}} \\ {{u_0}} \end{array}} \right) = S_S^{-1}\left( {{{\vec x}_{end}}-{A^n}{{\vec x}_0}} \right)

Hier:

\left( {\begin{array}{*{20}{c}}{{u_1}} \\ {{u_0}} \end{array}} \right) = S_S^{-1}\left( {{{\vec x}_{end}}-{A^2}{{\vec x}_0}} \right)

{A^2} = \left( {\begin{array}{*{20}{c}} 0&1 \\ a&b \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0&1 \\ a&b \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a&b \\ {ab}&{a+{b^2}} \end{array}} \right)

{A^2}{{\vec x}_0} = \left( {\begin{array}{*{20}{c}} a&b \\ {ab}&{a+{b^2}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}}{a+b} \\ {a+ab+{b^2}} \end{array}} \right)

{S_S} = \left( {\begin{array}{*{20}{c}} 0&1 \\ 1&b \end{array}} \right)\quad \Rightarrow \quad S_S^{-1} = \left( {\begin{array}{*{20}{c}}{-b}&1 \\ 1&0 \end{array}} \right)

\Rightarrow \quad \left( {\begin{array}{*{20}{c}}{{u_1}} \\ {{u_0}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}{-b}&1 \\ 1&0 \end{array}} \right)\left[ {\left( {\begin{array}{*{20}{c}}{-3} \\ 3 \end{array}} \right)-\left( {\begin{array}{*{20}{c}}{a+b} \\ {a+ab+{b^2}} \end{array}} \right)} \right]

\Rightarrow \quad \left( {\begin{array}{*{20}{c}}{{u_1}} \\ {{u_0}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}{-b}&1 \\ 1&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}}{-\left( {3+a+b} \right)} \\ {3-a-ab-{b^2}} \end{array}} \right)

\Rightarrow \quad \left( {\begin{array}{*{20}{c}}{{u_1}} \\ {{u_0}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}{b\left( {3+a+b} \right)+3-a-ab-{b^2}} \\ {-\left( {3+a+b} \right)} \end{array}} \right)

\Rightarrow \quad \left( {\begin{array}{*{20}{c}}{{u_1}} \\ {{u_0}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}}{3b+3-a} \\ {-\left( {3+a+b} \right)} \end{array}} \right)