User:Egm6321.f09.team4.sls.equations

= Equation Scratch Pad =

Scratch #1

 * Both equations together:


 * $$ \begin{align}

x_{1}\left(t\right) &= C_{1}\cdot e_{11}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{1}\left(\tau\right) d\tau \right] + C_{2}\cdot e_{21}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{2}\left(\tau\right) d\tau \right] \\

x_{2}\left(t\right) &= C_{1}\cdot e_{12}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{1}\left(\tau\right) d\tau \right] + C_{2}\cdot e_{22}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{2}\left(\tau\right) d\tau \right]

\end{align} $$


 * Each equation in separate picture:


 * $$ \begin{align}

x_{1}\left(t\right) &= C_{1}\cdot e_{11}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{1}\left(\tau\right) d\tau \right] + C_{2}\cdot e_{21}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{2}\left(\tau\right) d\tau \right] \end{align} $$


 * $$ \begin{align}

x_{2}\left(t\right) &= C_{1}\cdot e_{12}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{1}\left(\tau\right) d\tau \right] + C_{2}\cdot e_{22}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} p_{2}\left(\tau\right) d\tau \right] \end{align} $$


 * Each equation after transformation to normalized eigenvectors:


 * $$ \begin{align}

x_{1}\left(t\right) &= C_{1}\cdot \text{exp}\left[ \int_{0}^{t} \bar{p}_{1}\left(\tau\right) d\tau \right] + C_{2}\cdot \text{exp}\left[ \int_{0}^{t} \bar{p}_{2}\left(\tau\right) d\tau \right] \end{align} $$


 * $$ \begin{align}

x_{2}\left(t\right) &= C_{1}\cdot \bar{e}_{12}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} \bar{p}_{1}\left(\tau\right) d\tau \right] + C_{2}\cdot \bar{e}_{22}\left(t\right)\cdot \text{exp}\left[ \int_{0}^{t} \bar{p}_{2}\left(\tau\right) d\tau \right] \end{align} $$

Egm6321.f09.team4.sls: Tue Oct 27, 2:25 PM (EST)

Scratch #2
$$ \begin{align} C_{1} \cdot e_{11}\left(t\right) \cdot &\text{ exp}\left[ \int_{0}^{t} p_{1}\left(\tau\right) d\tau \right] \\ &= C_{1} \cdot \text{exp}\left[ f_{1}\left(t\right) + f_{2}\left(t\right) \right] \end{align} $$

$$ \begin{align} \text{where:  } &\ \\ &f_{1}\left(t\right) = \int_{0}^{t} p_{1}\left(\tau\right) d\tau \\ &f_{2}\left(t\right) = \text{ln}\left[ e_{11}\left(t\right) \right] \end{align} $$

Scratch #3

 * LTV eigen-relation:


 * $$ \begin{align}

A \ \vec{\varepsilon}_{i} - \lambda_{i}\ \vec{\varepsilon} = \dot{\vec{\varepsilon}} \end{align} $$


 * $$ A $$ matrix in canonical form:


 * $$ \begin{align}

A^{2x2} = \left[ \begin{array}{cc} 0 & 1 \\                  -a_{0} & -a_{1} \\ \end{array} \right] \end{align} $$


 * $$ \begin{align}

A\left(t\right) = \left[ \begin{array}{cc} 0 & 1 \\                  -0.2 & t-0.5 \\ \end{array} \right] \end{align} $$


 * $$ \begin{align}

A\left(t\right) = \left[ \begin{array}{cc} -1-t^{2} & -2.5 \\ -0.7-2.5t & -2-t \\ \end{array} \right] \end{align} $$


 * $$ \begin{align}

x = \left[ \begin{array}{c} x_{1} \\ x_{2} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\dot{x} = A\ x \end{align} $$


 * $$ \begin{align}

x_{2} = \dot{x}_{1} \end{align} $$


 * $$ \begin{align}

\bar{\varepsilon}_{1} = \left[ \begin{array}{c} 1 \\ \bar{e}_{12} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\bar{\varepsilon}_{2} = \left[ \begin{array}{c} 1 \\ \bar{e}_{22} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\bar{p}_{1} = \ldots \end{align} $$


 * $$ \begin{align}

\bar{p}_{2} = \ldots \end{align} $$


 * $$ \begin{align}

\varepsilon_{1} = \left[ \begin{array}{c} 1 \\ e_{12} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{2} = \left[ \begin{array}{c} 1 \\ e_{22} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{1} = \left[ \begin{array}{c} 1 \\ p_{1} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{2} = \left[ \begin{array}{c} 1 \\ p_{2} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{1}\left(t\right) = \left[ \begin{array}{c} e_{11}\left(t\right) \\ e_{12}\left(t\right) \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{2}\left(t\right) = \left[ \begin{array}{c} e_{21}\left(t\right) \\ e_{22}\left(t\right) \end{array} \right] \end{align} $$


 * $$ \begin{align}

\bar{\varepsilon}_{1,III}\left(t\right) = \left[ \begin{array}{c} 1 \\ \bar{e}_{12}\left(t\right) \end{array} \right] \end{align} $$


 * $$ \begin{align}

\bar{\varepsilon}_{2,III}\left(t\right) = \left[ \begin{array}{c} 1 \\ \bar{e}_{22}\left(t\right) \end{array} \right] \end{align} $$


 * $$ \begin{align}

\bar{\varepsilon}_{1,IV}\left(t\right) = \left[ \begin{array}{c} \bar{\bar{e}}_{11}\left(t\right) \\ 1 \end{array} \right] \end{align} $$


 * $$ \begin{align}

\bar{\varepsilon}_{2,IV}\left(t\right) = \left[ \begin{array}{c} \bar{\bar{e}}_{21}\left(t\right) \\ 1 \end{array} \right] \end{align} $$


 * $$ \begin{align}

p_{1}\left(t\right) \end{align} $$


 * $$ \begin{align}

p_{2}\left(t\right) \end{align} $$

Scratch #4
Scalar Polynomial Differential Operators (SPDO) :


 * For scalar system:


 * $$ \begin{align}

y^{\left(n\right)}+ &\alpha_{n}\left(t\right)y^{\left(n-1\right)} + \dots + \alpha_{2}\left(t\right)\dot{y}+\alpha_{1}\left(t\right)y = \\ &\beta_{m+1}\left(t\right)u^{\left(m\right)} + \dots + \beta_{2}\left(t\right)\dot{u} + \beta_{1}\left(t\right)u \end{align} $$


 * Define two SPDO's (one for each side):


 * $$ \begin{align}

D_{\alpha} &= \delta^{n} + a_{n}\left(t\right)\delta^{n-1} + \dots + a_{2}\left(t\right)\delta + a_{1}\left(t\right) \\ &= \left(\delta - \lambda_{n}\left(t\right)\right)\dots\left(\delta - \lambda_{2}\left(t\right)\right)\left(\delta - \lambda_{1}\left(t\right)\right) \end{align} $$


 * $$ \begin{align}

D_{\beta} &= \beta_{m+1}\left(t\right)\delta^{m}+\dots+\beta_{2}\left(t\right)\delta+\beta_{1}\left(t\right) \\ &= \beta_{m+1}\left(\delta - \gamma_{m}\left(t\right)\right)\dots\left(\delta - \gamma_{2}\left(t\right)\right)\left(\delta - \gamma_{1}\left(t\right)\right) \end{align} $$


 * Then, the scalar LTV system can be expressed as:


 * $$ \begin{align}

D_{\alpha} \cdot y = D_{\beta} \cdot u \end{align} $$

Vector Polynomial Differential Operators :


 * For the following MIMO system:


 * $$ \begin{align}

\dot{\vec{x}} &= A\left(t\right)\vec{x} + B\left(t\right)\vec{u} \\ \vec{y}      &= C\left(t\right)\vec{x} + D\left(t\right)\vec{u} \end{align} $$


 * Define a VPDO for $$A\left(t\right)$$:


 * $$ \begin{align}

\mathcal{P}_{A} = I \cdot \delta - A\left(t\right) \end{align} $$


 * This allows the LTV MIMO system to be rewritten as:


 * $$ \begin{align}

\mathcal{P}_{A}\cdot \vec{x} &= B\left(t\right)\vec{u} \\ \vec{y} &= C\left(t\right)\vec{x} + D\left(t\right)\vec{u} \end{align} $$

Scratch #5

 * Mode-Vector (see Wu - 1982):


 * $$ \begin{align}

m_{i}\left(t\right) = \exp \left[ \int_{t_{0}}^{t} p_{i}\left(\tau\right) \text{d}\tau\right] \cdot \vec{\varepsilon}_{i}\left(t\right) \end{align} $$


 * This is the mode vector associated with the i^{th} x-eigenpair of $$A\left(t\right)$$:


 * $$ \begin{align}

\left\{ p_{i}\left(t\right), \vec{\varepsilon}_{i}\left(t\right) \right\} \end{align} $$


 * The system described by $$A\left(t\right)$$ is stable if and only if:


 * $$ \begin{align}

\left|\left|m_{i}\left(t\right)\right|\right| < \infty \qquad \forall\ t>t_{0},\quad \forall\ i=1,\dots,n \end{align} $$


 * The system described by $$A\left(t\right)$$ is asymptotically stable if and only if the system is stable, and:


 * $$ \begin{align}

\left|\left|m_{i}\left(t\right)\right|\right| \to 0 \text{ as } t \to \infty,\quad \forall\ i=1,\dots,n \end{align} $$

Scratch #6

 * Case 1:


 * $$ \begin{align}

\vec{\varepsilon}_{1} = \left[ \begin{array}{c} 1 \\ \bar{e}_{12} \end{array} \right] \end{align} $$, $$ \begin{align} \vec{\varepsilon}_{2} = \left[ \begin{array}{c} \bar{e}_{21} \\ 1 \end{array} \right] \end{align} $$


 * Case 2:


 * $$ \begin{align}

\vec{\varepsilon}_{1} = \left[ \begin{array}{c} \bar{e}_{11} \\ 1 \end{array} \right] \end{align} $$, $$ \begin{align} \vec{\varepsilon}_{2} = \left[ \begin{array}{c} 1 \\ \bar{e}_{22} \end{array} \right] \end{align} $$


 * Case 3:


 * $$ \begin{align}

\vec{\varepsilon}_{1} = \left[ \begin{array}{c} 1 \\ \bar{e}_{12} \end{array} \right] \end{align} $$, $$ \begin{align} \vec{\varepsilon}_{2} = \left[ \begin{array}{c} 1 \\ \bar{e}_{22} \end{array} \right] \end{align} $$


 * Case 4:


 * $$ \begin{align}

\vec{\varepsilon}_{1} = \left[ \begin{array}{c} \bar{e}_{11} \\ 1 \end{array} \right] \end{align} $$, $$ \begin{align} \vec{\varepsilon}_{2} = \left[ \begin{array}{c} \bar{e}_{21} \\ 1 \end{array} \right] \end{align} $$

Scratch #7

 * $$ \begin{align}

\varepsilon_{1} = \left[ \begin{array}{c} 1 \\ p_{1} \\ \dot{p}_{1}+p^{2}_{1} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{2} = \left[ \begin{array}{c} 1 \\ p_{2} \\ \dot{p}_{2}+p^{2}_{2} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{3} = \left[ \begin{array}{c} 1 \\ p_{3} \\ \dot{p}_{3}+p^{2}_{3} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{1} = \left[ \begin{array}{c} 1 \\ e_{12} \\ e_{13} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{2} = \left[ \begin{array}{c} 1 \\ e_{22} \\ e_{23} \end{array} \right] \end{align} $$


 * $$ \begin{align}

\varepsilon_{3} = \left[ \begin{array}{c} 1 \\ e_{32} \\ e_{33} \end{array} \right] \end{align} $$


 * $$ \begin{align}

p_{1} = e_{12} \end{align} $$


 * $$ \begin{align}

p_{2} = e_{22} \end{align} $$


 * $$ \begin{align}

p_{3} = e_{32} \end{align} $$


 * $$ \begin{align}

\left(n^{2}+n\right) \end{align} $$


 * $$ \begin{align}

\left(n^{2}-n\right) \end{align} $$


 * $$ \begin{align}

\left(n\right) \end{align} $$

Scratch #8

 * $$\begin{align}

\hat{\theta} = \arg\max_{\theta}\left[ \mathcal{L}\left(\theta\right) \right] \end{align} $$


 * $$ \begin{align}

\mathcal{L}\left(\theta\right) = \log\left[ p\left(Y|\theta\right) \right] \end{align} $$


 * $$ \begin{align}

\mathcal{L}\left(\theta\right) = \log\left[ \int p\left(X,Y|\theta\right) dX \right] \end{align} $$


 * $$ \begin{align}

\text{, where} \end{align} $$


 * $$ \begin{align}

Y_{k} = \frac{1}{\sqrt{N}} \sum_{t=0}^{N-1} y_{t} z_{k}^{-t} \end{align} $$


 * $$ \begin{align}

U_{k} = \frac{1}{\sqrt{N}} \sum_{t=0}^{N-1} u_{t} z_{k}^{-t} \end{align} $$


 * $$ \begin{align}

N_{k} = \frac{1}{\sqrt{N}} \sum_{t=0}^{N-1} n_{t} z_{k}^{-t} \end{align} $$


 * $$ \begin{align}

P\left(Y|\theta\right) = \frac{P\left(X,Y|\theta\right)}{P\left(X|Y,\theta\right)} \end{align} $$


 * $$ \begin{align}

\log\left[P\left(Y|\theta\right)\right] = \log\left[P\left(X,Y|\theta\right)\right] - \log\left[P\left(X|Y,\theta\right)\right] \end{align} $$


 * $$ \begin{align}

\log\left[P\left(Y|\theta\right)\right] = \log\left[P\left(X,Y|\theta\right)\right] - \log\left[P\left(X|Y,\theta\right)\right] \end{align} $$


 * $$ \begin{align}

Y_{k} = C\ X_{k} + V_{k} \end{align} $$

$$ \begin{align} V_{k} \end{align} $$