Exam Preparation Notes

$H_0$ : Null hypothesis (default, uninteresting outcome).
$H1 \sim \neg H0$ : Alternative hypothesis.
Process:
1. Assume $H_0$ .
2. Under $H_0$ , compute statistic of interest (e.g., $Q$ ).
3. Compare statistic with its distribution under $H_0$ (e.g., $X^2(\text{df})$ ).
- $p$ -value: Probability of observing $Q$ or something more extreme if $H_0$ was true.

Probability Density Function (PDF): $p = fQ(Q{\text{data}})$
Cumulative Distribution Function (CDF): $FQ(Q{\text{data}}) = \int{-\infty}^{Q{\text{data}}} f_Q(Q) dQ$
$p$ -value $= 1 - FQ(Q{\text{data}})$
Reject $H_0$ if p < \alpha (e.g., 0.05).
- $\alpha$ : Probability of False Positives that we accept.

Motivation: Need "states" for control / fault detection / etc.
System representation:
- Input: $u(t)$
- Output: $y(t)$
- State: $x(t)$
- Equation: $y(t) = G(q) u(t)$
Controller design (e.g., LQR - Linear Quadratic Regulator) uses state feedback.

Change of coordinates: $\tilde{x}(t) = Px(t)$
Transformed state equation: $\tilde{x}(t+1) = PAP^{-1}\tilde{x}(t) + Bu(t)$
Transformed output equation: $y(t) = CP^{-1}\tilde{x}(t) + Du(t)$
Transfer function invariance: $G(z) = C(zI - A)^{-1}B + D = \tilde{C}(zI - \tilde{A})^{-1}\tilde{B} + \tilde{D}$

Under similarity transformation:
- $\tilde{A}=PAP^{-1}$
- $\tilde{B}=PB$
- $\tilde{C}=CP^{-1}$
- $\tilde{D}=D$
Extended Observability Matrix changes under coordinates
- $O = \begin{bmatrix} C \ CA \ CA^2 \ … \ CA^{r-1} \end{bmatrix}$

Goal: Given a learned I/O model, find a state-space representation.
Learned I/O model: $y(t) = G(q; \theta)u(t)$
$G(q; \theta) = \frac{b0 + b1q^{-1} + b2q^{-2} + … + bmq^{-m}}{1 + a1q^{-1} + a2q^{-2} + … + a_nq^{-n}}$
Objective: Solve $G(q) = C(qI - A)^{-1}B + D$

$G(q; \theta) = \frac{b0 + b1q^{-1} + b2q^{-2} + b3q^{-3}}{1 + a1q^{-1} + a2q^{-2} + a_3q^{-3}}$
State equation:
$x(t+1) = \begin{bmatrix} -a1 & -a2 & -a_3 \ 1 & 0 & 0 \ 0 & 1 & 0 \end{bmatrix} x(t) + \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} u(t)$
Output equation:
$y(t) = \begin{bmatrix} b1 & b2 & b3 \end{bmatrix} x(t) + [b0]u(t)$

Optimal state estimator is the Kalman Filter.
State-space model:
- State equation: $x(t+1) = Ax(t) + Bu(t) + w(t)$
- Output equation: $y(t) = Cx(t) + Du(t) + v(t)$
- $w(t)$ : Process noise.
- $v(t)$ : Measurement noise.
Rearrange output equation: $Cx(t) = y(t) - Du(t) - v(t)$

$y(t) = Cx(t) + Du(t) + v(t)$
$y(t+1) = C[Ax(t) + Bu(t) + w(t)] + Du(t+1) + v(t+1) = CAx(t) + CBu(t) + Du(t+1) + Cw(t) + v(t+1)$
Vector of outputs:
$\begin{bmatrix} y(t) \ y(t+1) \ y(t+2) \ … \ y(t+r-1) \end{bmatrix} = \begin{bmatrix} C \ CA \ CA^2 \ … \ CA^{r-1} \end{bmatrix} x(t) + \begin{bmatrix} D & 0 & 0 & … \ CB & D & 0 & … \ CAB & CB & D & … \ … & … & … & … \ CA^{r-2}B & CA^{r-3}B & CA^{r-4}B & … \end{bmatrix} \begin{bmatrix} u(t) \ u(t+1) \ u(t+2) \ … \end{bmatrix} + \begin{bmatrix} v(t) \ w(t) \ w(t+1) \ … \end{bmatrix}$

Define:
$Or = \begin{bmatrix} C \ CA \ CA^2 \ … \ CA^{r-1} \end{bmatrix}$ $Sr = \begin{bmatrix} D & 0 & 0 & … \ CB & D & 0 & … \ CAB & CB & D & … \ … & … & … & … \ CA^{r-2}B & CA^{r-3}B & CA^{r-4}B & … \end{bmatrix}$
Rewrite: $Or x(t) = yr(t) - Sr ur(t) - v_r(t)$
In theory: $x(t) = Or^{-1} (yr(t) - Sr ur(t) )$
But use Kalman Filter (KF) instead.

$Ax = b$ : 3 possibilities
1. Unique solution: $A^{-1}b$ , A is full column rank.
2. 0 solutions: $x$ solves $min ||Ax - b||^2$
3. $\infty$ solutions: $x$ solves $min ||x||^2$ s.t. $Ax = b$

State equation: $x(t+1) = Ax(t) + Bu(t) + w(t)$
Output equation: $y(t) = Cx(t) + Du(t) + v(t)$
Initialization: $\hat{x}_{0|0}$
Prediction:
- $\hat{x}{1|0} = A \hat{x}{0|0} + Bu(0)$
Update:
- $\hat{x}{1|1} = \hat{x}{1|0} + K (y(1) - C \hat{x}_{1|0} - Du(0))$
- $K$ : Kalman Gain.
Output prediction error: $y(1) - C \hat{x}_{1|0} - Du(0)$
Next step prediction: $\hat{x}{2|1} = A \hat{x}{1|1} + Bu(1)$
Continue with update step