Beobachtbarkeit

Steuerbarkeit

Modellierung

Diskretisierung

Zustandsraumdarstellung

Stabilität

Unterschied In/Homogen

Basis, Basiswechsel

Abbildungsmatrizen

Funktionen bestimmen

Geraden und Ebenen

Gruppen

lineare abbildung 

quadratische ergänzung

Baustatik 1

Anfangswertprobleme :)

Differentialgleichungen

basis 

QR-Algorithmus

QR-Zerlegung

Cholesky-Zerlegung

Bild & Urbild

Zweigstromanalyse

Transistor als Thermometer

Investment

Teilfolgen

Epsilon Delta Kriterium

prinzip der virtuellen verrückung 

Linear Kombination

Folgen, Reihen und Potenzreihen

Flächenpressung

Arbeit 

Cholesky Zerlegung

Polynomfunktionen beschreiben

Zu Maß-und Integrationstheorie: Majorisierte Konvergenz

Teilchenphysik und Festkörperphysik

Orthogonaltrajektorien

test

Taylorreihen

Regel von de L'Hopital

Mathe

KRAFT

Grenzwert Funktion 

Folgenkriterium

ϵ-δ Kriterium 

Beweis Stetigkeit 

Wahrheitstabelle

2x2 Matrix

Matrix invertieren

Basiswechselsatz

Lineare Abbildunge 

Vorderrad fahradgabel lagerkräfte berechnen 

Spannungsregelung

Jordan Normal Form

Test

Induktion 

Jordan-Normalform

Ansatz der Störfunktion zur Bestimmung der Partikulären Lösung

Kurvenintegral

dlg mithilfe inhomogenen linearen system lösen

Inverse einer Matrix

Gleichmäßige Stetigkeit

addition

Regel von L´Hospital

Lineare Abbildungen

dimension eines Untervektorraumes bestimmen

Stationäre Stoffübertragung

Polplan, Prinzip der virtuellen Verrückungen

komplexe Ungleichungen 

Bidualräume

Dualräume

Taylorpolynome

Kern und Basis

Diagonalisieren einer Matrix

Rekursive Folgen

Exponentialreihe

Darstellungsmatrizen

restglied

Basiswechsel

Gamma Funktion

Newton Verfahren

Randwertaufgabe

Numerische Integration: Trapez- und Simpsonregel

Funktionenfolgen

dualräume

Dualraum

Komplexe Nullstellen

Chemie

Natascha Kirchmann

Differentialquotient

Folgedefinition des Funktionslimes

Grenzwerte und Stetigkeit 

Komplexe Zahlen

Kern und Bild 

Direkte Summe 

Koordinaten 

Exponentielle Verzinsung

Lineare Verzinsung

Festverzinsliche Wertpapiere

Zustandsregelung

Darstellung des zustandsgeregelten Systems

Die Zustandsdifferentialgleichung mit den gegeben Werte ist:

\[ \left(\begin{array}{c}\dot{z}_{1} \\ \dot{z}_{2}\end{array}\right)=\left(\begin{array}{cc}0 &amp; 1 \\ -100 &amp; -1\end{array}\right)\left(\begin{array}{c}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{c}0 \\ \frac{1}{10}\end{array}\right) F_{1} \]

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\left(\begin{array}{c}\dot{z}_{1} \\ \dot{z}_{2}\end{array}\right)=\left(\begin{array}{cc}0 & 1 \\ -100 & -1\end{array}\right)\left(\begin{array}{c}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{c}0 \\ \frac{1}{10}\end{array}\right) F_{1} 
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\[\begin{align} \left(\begin{array}{l}\dot{z}_{1} \\ \dot{z}_{2}\end{array}\right)&amp;=\left(\begin{array}{cc}0 &amp; 1 \\ -100 &amp; -1\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right)\left(F_{\text {ext }}-F_{K}\right)\\ &amp;=\left(\begin{array}{cc}0 &amp; 1 \\ -100 &amp; -1\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right) F_{\text {ext }}-\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right) F_{K} \end{align}\]
\[\begin{align} \\\left(\begin{array}{l}\dot{z}_{1} \\ \dot{z}_{2}\end{array}\right)&amp;=\left(\begin{array}{cc}0 &amp; 1 \\ -100 &amp; -1\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right) F_{\text {ext }}-\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right)\left(\left(\begin{array}{ll}K_{1} &amp; K_{2}\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)\right)\\ &amp;=\left(\begin{array}{cc}0 &amp; 1 \\ -100 &amp; -1\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right) F_{\text {ext }}-\left(\begin{array}{l}\frac{1}{10}\left(K_{1} \cdot z_{1}+K_{2} \cdot z_{2}\right)\end{array}\right) \end{align}\]
\[ \left(\begin{array}{l}\dot{z}_{1} \\ \dot{z}_{2}\end{array}\right)=\left(\begin{array}{cc}0 &amp; 1 \\ -100 &amp; -1\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right) F_{\text {ext }}-\left(\begin{array}{c}\frac{K_{1}}{10} \cdot z_{1}+\frac{K_{2}}{10} \cdot z_{2}\end{array}\right) \]
\[ \left(\begin{array}{l}\dot{z}_{1} \\ \dot{z}_{2}\end{array}\right)=\left(\begin{array}{cc}0 &amp; 1 \\ -100-\frac{K_{1}}{10} &amp; -1-\frac{K_{2}}{10}\end{array}\right)\left(\begin{array}{l}z_{1} \\ z_{2}\end{array}\right)+\left(\begin{array}{l}0 \\ \frac{1}{10}\end{array}\right) F_{\text {ext }} \]

Überprüfung der Polstellen \( \lambda_{1 / 2}=-0,5 \pm 0,5 \mathrm{j} \)

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Bestimmung des charakteristischen Polynoms:

\[ \left|\left(\begin{array}{cc}0-\lambda &amp; 1 \\ -100-\frac{K_{1}}{10} &amp; -1-\frac{K_{2}}{10}-\lambda\end{array}\right)\right|=(-\lambda)\left(-1-\frac{K_{2}}{10}-\lambda\right)-\left(-100-\frac{K_{1}}{10}\right) \]
\[ \quad =\lambda^{2}+\lambda\left(1+\frac{K_{2}}{10}\right)+\left(100+\frac{K_{1}}{10}\right) \]

Wurzeln des charakterisitschen Polynoms sind die Polstellen des Systems:

\[ \lambda_{1 / 2}=-\frac{1+\frac{K_{2}}{10}}{2} \pm \sqrt{\left(\frac{1+\frac{K_{2}}{10}}{2}\right)^{2}-\left(100+\frac{K_{1}}{10}\right)} \]

<mjx-container class="MathJax" jax="SVG" display="true" justify="left"><svg alt="
\lambda_{1 / 2}=-\frac{1+\frac{K_{2}}{10}}{2} \pm \sqrt{\left(\frac{1+\frac{K_{2}}{10}}{2}\right)^{2}-\left(100+\frac{K_{1}}{10}\right)} 
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Koeffizientenvergleich mit den gewünschten Polen \( \lambda_{1 / 2}=-0,5 \pm 0,5 \mathrm{j} \) liefert:

Koeffizientenvergleich mit den gewünschten Polen <mjx-container class="MathJax" jax="SVG"><svg alt=" \lambda_{1 / 2}=-0,5 \pm 0,5 \mathrm{j} " style="vertical-align: -0.8ex" xmlns="http://www.w3.org/2000/svg" width="18.679ex" height="2.37ex" role="img" focusable="false" viewBox="0 -694 8256 1047.5" xmlns:xlink="http://www.w3.org/1999/xlink"><defs><path id="MJX-849-TEX-I-1D706" d="M166 673Q166 685 183 694H202Q292 691 316 644Q322 629 373 486T474 207T524 67Q531 47 537 34T546 15T551 6T555 2T556 -2T550 -11H482Q457 3 450 18T399 152L354 277L340 262Q327 246 293 207T236 141Q211 112 174 69Q123 9 111 -1T83 -12Q47 -12 47 20Q47 37 61 52T199 187Q229 216 266 252T321 306L338 322Q338 323 288 462T234 612Q214 657 183 657Q166 657 166 673Z"></path><path id="MJX-849-TEX-N-31" d="M213 578L200 573Q186 568 160 563T102 556H83V602H102Q149 604 189 617T245 641T273 663Q275 666 285 666Q294 666 302 660V361L303 61Q310 54 315 52T339 48T401 46H427V0H416Q395 3 257 3Q121 3 100 0H88V46H114Q136 46 152 46T177 47T193 50T201 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data-c="2C" xlink:href="#MJX-849-TEX-N-2C"></use></g><g data-mml-node="mn" transform="translate(4782.9,0)"><use data-c="35" xlink:href="#MJX-849-TEX-N-35"></use></g><g data-mml-node="mo" transform="translate(5505.1,0)"><use data-c="B1" xlink:href="#MJX-849-TEX-N-B1"></use></g><g data-mml-node="mn" transform="translate(6505.3,0)"><use data-c="30" xlink:href="#MJX-849-TEX-N-30"></use></g><g data-mml-node="mo" transform="translate(7005.3,0)"><use data-c="2C" xlink:href="#MJX-849-TEX-N-2C"></use></g><g data-mml-node="mn" transform="translate(7450,0)"><use data-c="35" xlink:href="#MJX-849-TEX-N-35"></use></g><g data-mml-node="TeXAtom" data-mjx-texclass="ORD" transform="translate(7950,0)"><g data-mml-node="mi"><use data-c="6A" xlink:href="#MJX-849-TEX-N-6A"></use></g></g></g></g></svg></mjx-container> liefert:

\[ \begin{array}{lll}-\frac{1+\frac{K_{2}}{10}}{2}&amp;=-0,5 &amp; \wedge \ \left(\frac{1+\frac{K_{2}}{10}}{2}\right)^{2}-\left(100+\frac{K_{1}}{10}\right)&amp;=-0,25 \\ 1+\frac{K_{2}}{10}&amp;=1 &amp; \wedge \quad\quad\quad\quad 0,5^{2}-100-\frac{K_{1}}{10}&amp;=-0,25 \\ \frac{K_{2}}{10}&amp;=0 &amp; \wedge \quad\quad\quad\quad\quad\quad\ \ \ 0,25-\frac{K_{1}}{10}&amp;=99,75 \\ K_{2}&amp;=0 &amp; \wedge \quad\quad\quad\quad\quad\quad\quad\quad\ \ \ -\frac{K_{1}}{10}&amp;=99,5 \\  K_{1}&amp;=-995\end{array} \]

Überprüfung durch die Bestimmung der Wurzeln des charakteristischen Polynoms ergibt die gewünschten Polstellen:

\[ \left|\left(\begin{array}{cc}0-\lambda &amp; 1 \\ -100-\frac{-995}{10} &amp; -1-\lambda\end{array}\right)\right|=(-\lambda)(-1-\lambda)-(-100+99,5) \]
\[ \quad =\lambda^{2}+\lambda+0,5 \]

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\left|\left(\begin{array}{cc}0-\lambda & 1 \\ -100-\frac{-995}{10} & -1-\lambda\end{array}\right)\right|=(-\lambda)(-1-\lambda)-(-100+99,5) 
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\quad =\lambda^{2}+\lambda+0,5 
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\[\begin{align} \lambda_{1 / 2}&amp;=-0,5 \pm \sqrt{0,25-0,5}\\&amp;=-0,5 \pm \sqrt{-0,25}\\&amp;=-0,5 \pm 0,5 \mathrm{j} \end{align}\]

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Gegeben ist ein kraftangeregter 1-Massenschwinger in Zustandsraumdarstellung mit der Zustandsdifferentialgleichung
\[ \left(\begin{array}{c} \dot{z}_{1} \\ \dot{z}_{2} \end{array}\right)=\left(\begin{array}{cc} 0 &amp; 1 \\ -\frac{c}{m} &amp; -\frac{d}{m} \end{array}\right)\left(\begin{array}{l} z_{1} \\ z_{2} \end{array}\right)+\left(\begin{array}{l} 0 \\ \frac{1}{m} \end{array}\right) F_{1} \]
und der Ausgangsgleichung
\[ z_{1}=\left(\begin{array}{ll} 1 &amp; 0 \end{array}\right)\left(\begin{array}{l} z_{1} \\ z_{2} \end{array}\right)+0 \cdot F_{1} \]
Bestimmen Sie die Rückführmatrix für \( m=10 \mathrm{~kg}, c=1000 \mathrm{~N} / \mathrm{m} \) und \( d=10 \mathrm{Nsec} / \mathrm{m} \) so, dass das System die Pole \( \lambda_{1 / 2}=-0,5 \pm 0,5 \mathrm{j} \) besitzt.

Dies ist eine interaktive Aufgabe zu: Zustandsraumdarstellung mit praktischen Tipps zum Lösen und einer Zusammenfassung der nötigen Theorie

Zeitdiskrete und kontinuierliche Systeme - Zustandsraumdarstellung | A

Aufgabenstellung:

Lösungsweg:

Lösung: