WebSVN – svnkaklik – Path Comparison – /

Ignore whitespace Rev 1061 → Rev 1062

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+\documentclass[12pt,notitlepage,fleqn]{article}
+\usepackage[czech]{babel}
+\usepackage[pdftex]{graphicx}
+\usepackage{fancyhdr,multicol} %nastavení češtiny, fancy, grafiky, sloupce
+\usepackage[utf8]{inputenc} %vstupni soubory v kodovani UTF-8
+\usepackage[a4paper,text={17cm,25cm},centering]{geometry} %nastavení okrajů
+\usepackage{rotating}
+\begin{document}
+\section*{Řešení 12. zadané úlohy          -        Jakub Kákona}
+\begin{enumerate}
+\item
+Sestavíme minimální realizaci sytému v řiditelné formě
+\begin{equation}
+D_c = \lim _{ s \to \infty} H_1(s) = 0
+\end{equation}
+\begin{equation}
+C_c=[b_0, b_1, \cdots , b_{n-1} ] = \left[\begin{array}{cc}
+& 0 \\
+\end{array}\right]
+\end{equation}
+\begin{equation}
+A_c = \left[ \begin{array}{cccc}
+& 1 & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots  \\
+& 0 & \cdots & 1 \\
+-a_0 & -a_1 & \cdots & -a_{n-1} \\
+\end{array}
+\right] =
+ \left[ \begin{array}{cc}
+& 1 \\
+& 0 \\
+\end{array}
+\right]
+\end{equation}
+\begin{equation}
+B_c = \left[\begin{array}{c}
+\\
+\\
+\end{array}\right]
+\end{equation}
+Potřebujeme vybrat takové matice
+$F =  \left[\begin{array}{cc}
+a & b \\
+\end{array}\right]$,
+$K =  \left[\begin{array}{c}
+c \\
+d \\
+\end{array}\right]$, že vlastní čísla matic $A + BF$ a $A -KC$ budou mít zápornou reálnou část.
+\begin{equation}
+\det(sI - (A + BF)) = s^2 - sb - a \\
+F = \left[\begin{array}{cc}
+-1 & -2 \\
+\end{array}\right]
+\end{equation}
+\begin{equation}
+\det(sI - (A + KC)) = s^2 + sc + d \\
+K = \left[\begin{array}{c}
+\\
+\\
+\end{array}\right]
+\end{equation}
+Stabilizující regulátory $H_2$ mají tuto stavovou reprezentaci:
+\begin{eqnarray}
+\dot{x} =& Ax + Bu +K(y-(Cx + Du))\\
+u =& Fx +K'(q)(y-(Cx + Du))
+\end{eqnarray}
+Kde $K'(q)$ je stabilní ryzí matice.
+Po dosazení známých hodnot dostaneme všechny ryzí regulátory $H_2$ stabilizující systém $H_1$
+\begin{eqnarray}
+\dot{x} =&
+\left[ \begin{array}{cc}
+& 1 \\
+& 0 \\
+\end{array}
+\right]x
++
+\left[ \begin{array}{c}
+\\
+\\
+\end{array}
+\right]u
++
+\left[ \begin{array}{c}
+\\
+\\
+\end{array}
+\right](y-
+\left[ \begin{array}{cc}
+& 0 \\
+\end{array}
+\right]x)\\
+u =&
+\left[ \begin{array}{cc}
+& 2 \\
+\end{array}
+\right] + K'(q)(y-
+\left[ \begin{array}{cc} 1 & 0 \\ \end{array} \right]
+x)
+\end{eqnarray}
+\item
+\item
+\item
+\begin{enumerate}
+\item
+Stavová reprezentace systému pro případ  $\theta = x_1, \dot{\theta}=x_2$ stejná jako v případě úkolu 10.
+\begin{eqnarray}
+\left[ \begin{array}{c}
+\dot{x}_1 \\
+\dot{x}_2 \\
+\end{array}
+\right] =&
+\left[ \begin{array}{cc}
+& 1 \\
+& -1 \\
+\end{array}
+\right]
+\left[ \begin{array}{cc}
+x_1 \\
+x_2 \\
+\end{array}
+\right]
+\left[ \begin{array}{cc}
+\\
+\\
+\end{array}
+\right]u
+\\
+y =&
+\left[ \begin{array}{cc}
+& 1 \\
+\end{array}
+\right]
+\left[ \begin{array}{cc}
+x_1 \\
+x_2 \\
+\end{array}
+\right]
+\end{eqnarray}
+\item
+Podmínka požaduje charakteristický polynom systému se stavovou zpětnou vazbou ve tvaru
+\begin{equation}
+(s + 1) (s + 1) = s^2 + 2s + 1
+\end{equation}
+\begin{equation}
+\det(sI - (A + BF)) = s^2 - s(1 - b) - a
+\end{equation}
+Kde $ F = \left[\begin{array}{cc}
+a & b \\
+\end{array}\right]$
+Porovnáním koeficientů obou polynomů dostaneme matici stavové zpětné vazby
+\begin{equation}
+F = \left[\begin{array}{cc}
+-1 & -1 \\
+\end{array}\right]
+\end{equation}
+\item
+Podmínka požaduje charakteristický polynom pozorovatele ve tvaru
+\begin{equation}
+(s + 5) (s + 5) = s^2 + 10s + 25
+\end{equation}
+Obecný charakteristický polynom pozorovatele je
+\begin{equation}
+\det(sI - (A + KC)) = s^2 - s(1 - a) + a +b
+\end{equation}
+Kde $ K = \left[\begin{array}{c}
+a \\
+b \\
+\end{array}\right]$
+Porovnáním koeficientů obou polynomů dostaneme matici stavové zpětné vazby
+\begin{equation}
+K = \left[\begin{array}{cc}
+\\
+\\
+\end{array}\right]
+\end{equation}
+\item
+Stavový popis systému systému s pozorovatelem a stavovou zpětnou vazbou je
+\begin{eqnarray}
+\left[ \begin{array}{c}
+\dot{x} \\
+\dot{\hat{x}} \\
+\end{array}
+\right] =&
+\left[ \begin{array}{cc}
+A & BF\\
+KC & A- KC + BF \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x_1 \\
+x_2 \\
+\end{array}
+\right]
++
+\left[ \begin{array}{c}
+B \\
+B \\
+\end{array}
+\right]r
+\\
+y =&
+\left[ \begin{array}{cc}
+C & DF \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+x \\
+\end{array}
+\right]
++
+Dr
+\end{eqnarray}
+Po dosazení dostaneme:
+\begin{eqnarray}
+\left[ \begin{array}{c}
+\dot{x} \\
+\dot{\hat{x}} \\
+\end{array}
+\right] =&
+\left[ \begin{array}{cccc}
+& 1 & 0 & 0 \\
+& -1 & -1 & -1 \\
+& 0 & -9 & 1 \\
+& 0 & -17 & -2 \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+\hat{x} \\
+\end{array}
+\right]
++
+\left[ \begin{array}{c}
+\\
+\\
+\\
+\\
+\end{array}
+\right]r\\
+y =&
+\left[ \begin{array}{cccc}
+& 0 & 0 & 0 \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+\hat{x} \\
+\end{array}
+\right]
+\end{eqnarray}
+Přenos tohoto systému pak je:
+\begin{equation}
+H(s)= C (sI - A)^{-1} B + D =
+\left[ \begin{array}{cccc}
+& 0 & 0 & 0 \\
+\end{array}
+\right]
+\left[ \begin{array}{cccc}
+s & -1 & 0 & 0 \\
+& s + 1 & 1 & 1 \\
+-9 & 0 & s+9 & -1 \\
+-16 & 0 & 17 & s + 2 \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+\\
+\\
+\\
+\\
+\end{array}
+\right] = \frac{1}{(s+1)^2}
+\end{equation}
+Charakteristický polynom systému je
+\begin{equation}
+\det(sI - A ) = (s + 5)^2 (s + 1)^2
+\end{equation}
+Vlastní čísla leží na záporné ose a systém je stabilní.
+Pozorovatelnost celkového systému:
+\begin{equation}
+O = \left[ \begin{array}{c}
+C \\
+C A\\
+\end{array}
+\right]=
+\left[ \begin{array}{cccc}
+& 0 & 0 & 0 \\
+& 1 & 0 & 0 \\
+& -1 & -1 & -1 \\
+-25 & 1 & 27 & 2 \\
+\end{array}
+\right]
+\end{equation}
+Hodnost matice pozorovatelnosti je 4 a systém je  pozorovatelný.
+Ověříme řiditelnost systému
+\begin{equation}
+C = \left[ \begin{array}{cc}
+B & AB \\
+\end{array}
+\right]=
+\left[ \begin{array}{cccc}
+& 1 & -2 & 3 \\
+& -2 & 4 & -4 \\
+& 1 & -2 & 3 \\
+& -2 & 3 & -4 \\
+\end{array}
+\right]
+\end{equation}
+Hodnost matice řiditelnosti je 2 a není úplně řiditelný.
+\end{enumerate}
+\item
+\begin{enumerate}
+\item
+Podmínka požaduje charakteristický polynom systému se stavovou zpětnou vazbou ve tvaru
+\begin{equation}
+(s + 0,5 +0,5j) (s + 0,5 + 0,5j) = s^2 + s + 0,5
+\end{equation}
+Obecný charakteristický polynom je
+\begin{equation}
+\det(sI - (A + BF)) = s^2 + sb + a - 1
+\end{equation}
+Kde $ F = \left[\begin{array}{cc}
+a & b \\
+\end{array}\right]$
+Porovnáním koeficientů obou polynomů dostaneme matici stavové zpětné vazby
+\begin{equation}
+F = \left[\begin{array}{cc}
+\frac{3}{2} & 1 \\
+\end{array}\right]
+\end{equation}
+\item
+Podmínka požaduje charakteristický polynom systému se stavovou zpětnou vazbou ve tvaru
+\begin{equation}
+(s + \alpha +j) (s + \alpha - j) = s^2 + \alpha s + \alpha ^2 + 1
+\end{equation}
+Obecný charakteristický polynom  pozorovatele je
+\begin{equation}
+\det(sI - (A - KC)) = s^2 + sa + b - 1
+\end{equation}
+Kde $ K = \left[\begin{array}{c}
+a \\
+b \\
+\end{array}\right]$
+Porovnáním koeficientů obou polynomů dostaneme matici stavové injekce
+\begin{equation}
+K = \left[\begin{array}{c}
+\alpha \\
+\alpha ^2 + 2 \\
+\end{array}\right]
+\end{equation}
+\item
+Stavový systém s pozorovatelem a stavovou zpětnou vazbou využívající pozorovatelem odhadované stavy je
+\begin{eqnarray}
+\left[ \begin{array}{c}
+\dot{x} \\
+\dot{\hat{x}} \\
+\end{array}
+\right] =&
+\left[ \begin{array}{cc}
+A & BF\\
+KC & A- KC + BF \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+\hat{x} \\
+\end{array}
+\right]
++
+\left[ \begin{array}{c}
+B \\
+B \\
+\end{array}
+\right]r
+\\
+y =&
+\left[ \begin{array}{cc}
+C & DF \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+\hat{x} \\
+\end{array}
+\right]
++
+Dr
+\end{eqnarray}
+Po dosazení dostaneme:
+\begin{eqnarray}
+\left[ \begin{array}{c}
+\dot{x} \\
+\dot{\hat{x}} \\
+\end{array}
+\right] =&
+\left[ \begin{array}{cccc}
+& 1 & 0 & 0 \\
+& 0 & -\frac{3}{2} & -1 \\
+\alpha & 0 & 2 \alpha & 1 \\
+\alpha ^2 + 2 & 0 & - \alpha ^2 - \frac{5}{2}  & -1 \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+\hat{x} \\
+\end{array}
+\right]
++
+\left[ \begin{array}{c}
+\\
+-1 \\
+\\
+-1 \\
+\end{array}
+\right]r\\
+y =&
+\left[ \begin{array}{cccc}
+& 0 & 0 & 0 \\
+\end{array}
+\right]
+\left[ \begin{array}{c}
+x \\
+\hat{x} \\
+\end{array}
+\right]
+\end{eqnarray}
+\end{enumerate}
+\end{enumerate}
+\end{document}

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