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| \documentclass[12pt,notitlepage,fleqn]{article} |
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| \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} |
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| \begin{document} |
| |
| \section*{Řešení 7. zadané úlohy - Jakub Kákona} |
| |
| \begin{enumerate} |
| |
| \item |
| |
| Na základě fyzikálních procesů probíhajících v obvodu sestavíme stavové rovnice popisující systém: |
| |
| \begin{equation} |
| v(t) = V_c(t) + v_R(t) = V_c(t) + Ri_c(t) = v_c(t) + RC \frac{dv_c(t)}{dt} |
| \end{equation} |
| |
| Napětí na obou větvích obvodu ale musí být stejné, proto zároven platí: |
| |
| \begin{equation} |
| v(t) = V_L(t) + v_R(t) = Lv_L(t) + Ri_L(t) = L \frac{di_L(t)}{dt} + Ri_L(t) |
| \end{equation} |
| |
| Celkový proud obvodem pak je: |
| |
| \begin{equation} |
| i(t) = i_C(t) + i_L(t) = \frac{v(t)-v_c(t)}{R} + i_L(t) |
| \end{equation} |
| |
| \begin{eqnarray} |
| y(t) =& i(t), \\ |
| u(t) =& v(t), \\ |
| x_1(t) =& v_C(t), \\ |
| x_2(t) =& i_L(t), \\ |
| u(t) =& x_1(t) + RC \dot{x_1}(t), \\ |
| u(t) =& L \dot{x_2}(t) + R x_2(t), \\ |
| y(t) =& \frac{1}{R} u(t) - \frac{1}{R} x_1(t) + x_2(t) \\ |
| \end{eqnarray} |
| |
| Stavový popis přepíšeme do maticového tvaru: |
| |
| \begin{equation} |
| x(t) = |
| \left[ \begin{array}{cc} |
| - \frac{1}{RC} & 0 \\ |
| 0 & - \frac{R}{L} |
| \end{array} |
| \right] x(t) + |
| \left[ \begin{array}{c} |
| \frac{1}{RC} \\ |
| \frac{1}{L} |
| \end{array} |
| \right] u(t) |
| \end{equation} |
| |
| \begin{equation} |
| y(t) = |
| \left[ \begin{array}{cc} |
| - \frac{1}{R} & 1 \\ |
| \end{array} |
| \right] x(t) + |
| \frac{1}{R} u(t) |
| \end{equation} |
| |
| K určení impulzní odezvy potřebujeme přenos systému |
| |
| \begin{eqnarray} |
| H(s) =& C [sI - A]^{-1} B + D \\ |
| H(s) =& |
| \left[ \begin{array}{cc} |
| - \frac{1}{R} & 1 \\ |
| \end{array} |
| \right] |
| \left[ \begin{array}{cc} |
| s + \frac{1}{RC} & 0 \\ |
| 0 & s + \frac{R}{L} \\ |
| \end{array} |
| \right]^{-1} |
| \left[ \begin{array}{c} |
| \frac{1}{RC} \\ |
| \frac{1}{L} \\ |
| \end{array} |
| \right] + \frac{1}{R} \\ |
| H(s) =& - \frac{1}{R^2 C (s + RC)} + \frac{1}{L(s +\frac{R}{L})} + \frac{1}{R} |
| \end{eqnarray} |
| |
| Z přenosu zjistíme impulzní odezvu: |
| |
| \begin{equation} |
| h(t) = L^{-1} \left\{ H(s) \right\} |
| \end{equation} |
| |
| Do popisu systému dosadíme předpoklad $\frac{1}{RC} = \frac{R}{L}$ |
| |
| Tím získáme následující tvar stavového popisu: |
| |
| \begin{equation} |
| x(t) = |
| \left[ \begin{array}{cc} |
| - \frac{R}{L} & 0 \\ |
| 0 & - \frac{R}{L} |
| \end{array} |
| \right] x(t) + |
| \left[ \begin{array}{c} |
| \frac{R}{L} \\ |
| \frac{1}{L} |
| \end{array} |
| \right] u(t) |
| \end{equation} |
| |
| \begin{equation} |
| y(t) = |
| \left[ \begin{array}{cc} |
| - \frac{1}{R} & 1 \\ |
| \end{array} |
| \right] x(t) + |
| \frac{1}{R} u(t) |
| \end{equation} |
| |
| K převodu do Kalmanova tvaru potřebujeme matici řiditelnosti a pozorovatelnosti |
| |
| \begin{equation} |
| C = \left[ B, AB \right] = |
| \left[ \begin{array}{cc} |
| \frac{R}{L} & - \frac{R^2}{L^2} \\ |
| \frac{1}{L} & - \frac{R}{L^2} |
| \end{array} |
| \right] |
| \end{equation} |
| |
| \begin{equation} |
| O = \left[ \begin{array}{c} |
| C \\ |
| AC \\ |
| \end{array} |
| \right] = |
| \left[ \begin{array}{cc} |
| - \frac{1}{R} & 1 \\ |
| \frac{1}{L} & - \frac{R}{L} |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Hodnost obou matic je 1. |
| |
| Určíme jádro matice O. |
| |
| \begin{eqnarray} |
| \left[ \begin{array}{cc} |
| - \frac{1}{R} & 1 \\ |
| \frac{1}{L} & - \frac{R}{L} |
| \end{array} |
| \right] |
| \left[ \begin{array}{c} |
| a \\ |
| b \\ |
| \end{array} |
| \right] =& 0 \\ |
| \frac{1}{R} + b =& 0 \\ |
| \frac{1}{L}a - \frac{R}{L} b =& 0 \\ |
| b =& 1\\ |
| a =& R |
| \end{eqnarray} |
| |
| Jádrem matice je proto jeden vektor $ker(O)=\left[ \begin{array}{c} |
| R \\ |
| 1 \\ |
| \end{array} |
| \right]$ |
| |
| Nyní můžeme sestavit transformační matici Q. |
| |
| \begin{equation} |
| Q = [v_1, Q_n] = |
| \left[ \begin{array}{cc} |
| R & 1 \\ |
| 1 & 0 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Nyní lze určit Kalmanovu formu matic |
| |
| \begin{equation} |
| \tilde{A}= Q^{-1} A Q = |
| \left[ \begin{array}{cc} |
| 0 & -1 \\ |
| -1 & R \\ |
| \end{array} |
| \right] |
| \left[ \begin{array}{cc} |
| -\frac{R}{L} & 0 \\ |
| 0 & -\frac{R}{L} \\ |
| \end{array} |
| \right] |
| \left[ \begin{array}{cc} |
| -\frac{R}{L} & 0 \\ |
| 0 & -\frac{R}{L} \\ |
| \end{array} |
| \right] |
| =\left[ \begin{array}{ccc} |
| \frac{R}{L} & 0 \\ |
| 0 & \frac{R}{L} \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| \begin{equation} |
| \tilde{B}= Q^{-1} B = |
| \left[ \begin{array}{c} |
| - \frac{1}{L} \\ |
| 0 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| \begin{equation} |
| \tilde{C}= CQ = |
| \left[ \begin{array}{cc} |
| 0 & - \frac{1}{R} \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| \begin{equation} |
| \tilde{D}= D = |
| \frac{1}{R} |
| \end{equation} |
| |
| Vlastní čísla matice $\tilde{A}$ jsou pak dvojnásobná s hodnotou $- \frac{R}{L}$ jedno z nich je pak řiditelné a nepozorovatelné a druhé neřiditelné a nepozorovatelné. |
| |
| Přenos systému zapsaný na základě Kalmanova tvaru je: |
| |
| \begin{eqnarray} |
| \tilde{H}(s) =& \tilde{C} [sI - \tilde{A}]^{-1} \tilde{B} + \tilde{D} \\ |
| \tilde{H}(s) =& |
| \left[ \begin{array}{cc} |
| 0 & - \frac{1}{R} \\ |
| \end{array} |
| \right] |
| \left[ \begin{array}{cc} |
| s - \frac{R}{L} & 0 \\ |
| 0 & s - \frac{R}{L} \\ |
| \end{array} |
| \right]^{-1} |
| \left[ \begin{array}{c} |
| -\frac{1}{L} \\ |
| 0 \\ |
| \end{array} |
| \right] + \frac{1}{R} \\ |
| \tilde{H}(s) =& \frac{1}{R} |
| \end{eqnarray} |
| |
| Impulzní odezva pak je: |
| |
| \begin{equation} |
| \tilde{h}(t) = L^{-1} \left\{ \tilde{H}(s) \right\} = \frac{1}{R} \sigma(t) |
| \end{equation} |
| |
| \item |
| |
| Zadanou matici přepíšeme do tvaru: |
| |
| \begin{equation} |
| H(s)= \frac{1}{d(s)} N(s) = \frac{1}{s(s+2)} |
| \left[ \begin{array}{ccc} |
| (s-1)(s+2) & 0 & s(s-2) \\ |
| 0 & (s+1)(s+2) & 0 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Matici $N(s)$ je třeba převést do Smithova tvaru |
| |
| \begin{equation} |
| S_N(s)= |
| \left[ \begin{array}{ccc} |
| \epsilon _1 (s) & 0 & 0 \\ |
| 0 & \epsilon _2 (s) & 0 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Kde $ \epsilon _i (s) = \frac{D_i (s)}{D_{i-1} (s)}$ |
| |
| \begin{equation} |
| D_0(s) = 1 , D_1(s) = 1, D_2(s) = (s+1)(s+2) |
| \end{equation} |
| |
| Smithova forma matice má pak tvar. |
| |
| \begin{equation} |
| S_N(s)= |
| \left[ \begin{array}{ccc} |
| 1 & 0 & 0 \\ |
| 0 & (s+1)(s+2) & 0 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| To ještě doupravíme na Smith-McMillanovu formu: |
| |
| \begin{equation} |
| SM_H(s)= \frac{1}{d(s)} S_N(s)= |
| \left[ \begin{array}{ccc} |
| \frac{\epsilon _1 (s)}{\psi _1 (s)} & 0 & 0 \\ |
| 0 & \frac{\epsilon _2 (s)}{s} & 0 \\ |
| \end{array} |
| \right]= |
| \left[ \begin{array}{ccc} |
| \frac{1}{s(s+2)} & 0 & 0 \\ |
| 0 & \frac{s+1}{s} & 0 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Póly $H(s)$ zjistíme z kořenů polynomu |
| \begin{equation} |
| P_H(s) = \psi _1(s) \psi _2 (s) = s^2 (s+2). |
| \end{equation} |
| |
| Tedy 0, 0, -2. |
| |
| A nuly jsou kořeny polynomu |
| |
| \begin{equation} |
| Z_H(s) = \epsilon _1(s) \epsilon _2 (s) = s+1. |
| \end{equation} |
| |
| Nula proto je v -1. |
| |
| \item |
| |
| Budeme potřebovat "systémovou matici" |
| |
| \begin{equation} |
| P(s)= |
| \left[ \begin{array}{cc} |
| sI-A & B \\ |
| -C & D \\ |
| \end{array} |
| \right]= |
| \left[ \begin{array}{cccc} |
| s-1 & -1 & 0 & 1 \\ |
| 0 & s-1 & -1 & 0 \\ |
| 0 & 0 & s-1 & 0 \\ |
| 0 & 0 & -1 & 1 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Tuto matici ale potřebujeme spíše ve Smithově tvaru. |
| |
| \begin{equation} |
| S_P(s)= |
| \left[ \begin{array}{cccc} |
| 1 & 0 & 0 & 0 \\ |
| 0 & 1 & 0 & 0 \\ |
| 0 & 0 & 1 & 0 \\ |
| 0 & 0 & 0 & (s-1)^3 \\ |
| \end{array} |
| \right] |
| \end{equation} |
| |
| Invariantní nuly najdeme řešením polynomu $z^I _P (s) = \epsilon _1 (s) \epsilon _2 (s) \epsilon _3 (s) \epsilon _4 (s) = (s-1)^3$. Invariantní nulou jeproto trojnásobná 1. |
| |
| \end{enumerate} |
| \end{document} |