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docs/Chapters/1-MODERN-CONTROL.md

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+# Modern Control
+
+Normally speaking, we know much about classical control, in the form
+of:
+
+$$
+\dot{x}(t) = ax(t) + bu(t) \longleftrightarrow sX(s) - x(0) = aX(S) + bU(s)
+$$
+
+With the left part being a derivative equation in continuous time, while the 
+right being its tranformation in the complex domain field.
+
+> [!NOTE]
+>
+> $$
+> \dot{x}(t) = ax(t) + bu(t) \longleftrightarrow x(k+1) = ax(k) + bu(k)
+> $$
+>
+> These are equivalent, but the latter one is in discrete time.
+>
+
+## A brief recap over Classical Control
+
+Be $Y(s)$ our `output variable` in `classical control` and $U(s)$ our 
+`input variable`. The associated `transfer function` $G(s)$ is:
+
+$$
+G(s) = \frac{Y(s)}{U(s)}
+$$
+
+### Root Locus
+<!-- TODO: write about Root Locus -->
+
+### Bode Diagram
+
+### Nyquist Diagram
+
+## State Space Representation
+
+### State Matrices
+A state space representation has 4 Matrices: $A, B, C, D$ with coefficients in 
+$\R$:
+- $A$: State Matrix `[x_rows, x_columns]`
+- $B$: Input Matrix `[x_rows, u_columns]`
+- $C$: Output Matrix `[y_rows, x_columns]`
+- $D$: Direct Coupling Matrix `[y_rows, u_columns]`
+
+$$
+\begin{cases}
+\dot{x}(t) = Ax(t) + Bu(t) \;\;\;\; \text{Dynamic of the system}\\
+      y(t) = C{x}(t) + Du(t) \;\;\;\; \text{Static of the outputs}
+\end{cases}
+$$
+
+This can be represented with the following diagrams:
+
+#### Continuous Time:
+![continuous state space diagram](..\Images\Modern-Control\state-space-time.png)
+
+---
+#### Discrete time:
+
+![discrete state space diagram](..\Images\Modern-Control\state-space-discrete.png)
+
+### State Vector
+This is a state vector `[x_rows, 1]`:
+$$
+x(t) =  \begin{bmatrix}
+            x_1(t)\\
+            \dots\\
+            x_x(t)
+        \end{bmatrix}
+
+\text{or} \:
+
+x(k) =  \begin{bmatrix}
+            x_1(k)\\
+            \dots\\
+            x_x(k)
+        \end{bmatrix}
+$$
+
+Basically, from this we can know each next step of the state vector, represented 
+as:
+
+$$
+x(k + 1) = f\left( 
+    x(k), u(k)
+\right) = Ax(k) + Bu(k)
+$$
+
+### Examples
+
+#### Cart attached to a spring and a damper, pulled by a force
+![cart being pulled image](../Images/Modern-Control/cart-pulled.png)
+
+##### Formulas
+- <span style="color:#55a1e6">Spring: $\vec{F} = -k\vec{x}$</span>
+- <span style="color:#ff8382">Fluid Damper: $\vec{F_D} = -b \vec{\dot{x}}$</span>
+- Initial Force: $\vec{F_p}(t) = \vec{F_p}(t)$
+- Total Force: $m \vec{\ddot{x}}(t) =  \vec{F_p}(t) -b \vec{\dot{x}} -k\vec{x}$
+
+> [!TIP]
+>
+> A rule of thumb is to have as many variables in our state as the max number
+> of derivatives we encounter. In this case `2`
+>
+> Solve the equation for the highest derivative order
+>
+> Then, put all variables equal to the previous one derivated:
+>
+> $$
+x(t) =  \begin{bmatrix}
+            x_1(t)\\
+            x_2(t) = \dot{x_1}(t)\\
+            \dots\\
+            x_n(t) = \dot{x}_{n-1}(t)
+        \end{bmatrix}
+\;
+\dot{x}(t) =  \begin{bmatrix}
+            \dot{x_1}(t) = x_2(t)\\
+            \dot{x_2}(t) = x_3(t)\\
+            \dots\\
+            \dot{x}_{n-1}(t) = \dot{x}_{n}(t)\\
+            \dot{x}_{n}(t) = \text{our formula}
+        \end{bmatrix}
+
+> $$
+> 
+
+Now in our state we may express `position` and `speed`, while in our 
+`next_state` we'll have `speed` and `acceleration`:
+$$
+x(t) =  \begin{bmatrix}
+            x_1(t)\\
+            x_2(t) = \dot{x_1}(t)
+        \end{bmatrix}
+\;
+\dot{x}(t) =  \begin{bmatrix}
+            \dot{x_1}(t) = x_2(t)\\
+            \dot{x_2}(t) = \ddot{x_1}(t)
+        \end{bmatrix}
+
+$$
+Our new state is then:
+$$
+\begin{cases}
+\dot{x}_1(t) = x_2(t)\\
+\dot{x}_2(t) = \frac{1}{m} \left( \vec{F}(t) - b x_2(t) - kx_1(t) \right)
+\end{cases}
+$$
+
+let's say we only want to check for the `position` and `speed` of the system, our
+State Space will be:
+$$
+A = \begin{bmatrix}
+0 & 1 \\
+- \frac{k}{m} & - \frac{b}{m} \\
+\end{bmatrix}
+
+B = \begin{bmatrix}
+0 \\
+\frac{1}{m} \\
+\end{bmatrix}
+
+C = \begin{bmatrix}
+1 & 0 \\
+0 & 1
+\end{bmatrix}
+
+D = \begin{bmatrix}
+0
+\end{bmatrix}
+$$
+
+let's say we only want to check for the `position` of the system, our
+State Space will be:
+$$
+A = \begin{bmatrix}
+0 & 1 \\
+- \frac{k}{m} & - \frac{b}{m} \\
+\end{bmatrix}
+
+B = \begin{bmatrix}
+0 \\
+\frac{1}{m} \\
+\end{bmatrix}
+
+C = \begin{bmatrix}
+1 & 0 
+\end{bmatrix}
+
+D = \begin{bmatrix}
+0
+\end{bmatrix}
+$$
+
+> [!TIP]
+> In order to being able to plot the $\vec{x}$ against the time, you need to 
+> multiply $\vec{\dot{x}}$ for the `time_step` and then add it to the state[^so-how-to-plot-ssr]
+>
+
+### Horner Factorization
+let's say you have a complete polynomial of order `n`, you can factorize 
+in this way:
+
+$$
+\begin{align*}
+p(s) &= s^5 + 4s^4 + 5s^3 + 2s^2 + 10s + 1 =\\
+
+&= s ( s^4 + 4s^3 + 5s^2 + 2s + 10) + 1 = \\
+
+&= s ( s (s^3 + 4s^2 + 5s + 2) + 10) + 1 = \\
+
+&= s ( s (s (s^2 + 4s + 5) + 2) + 10) + 1 = \\
+
+&= s ( s (s ( s (s + 4) + 5) + 2) + 10) + 1
+
+\end{align*}
+$$
+
+If you were to take each s with the corresponding number in the parenthesis, 
+you'll make this block:
+![horner factorization to diagram block](../Images/Modern-Control/horner-factorization.png)
+
+
+
+
+
+
+
+
+
+
+### Case Studies
+<!-- TODO: Complete case studies -->
+- PAGERANK 
+- Congestion Control
+- Video Player Control
+- Deep Learning
+  
+[^so-how-to-plot-ssr]: [Stack Exchange | How to plot state space variables against time on unit step input? | 05 January 2025 ](https://electronics.stackexchange.com/questions/307227/how-to-plot-state-space-variables-against-time-on-unit-step-input)

docs/Chapters/2-RELATION-TO-CLASSICAL-CONTROL.md

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+# Relation to Classical Control
+
+## A Brief Recap of Discrete Control
+Let's say we want to control something ***Physical***, hence intrinsically
+***time continuous***, we can model our control in the `z` domain and make our
+$G_c(z)$. But how do we connect these systems:
+
+![scheme of how digital control interconnects with classical control](../Images/Relation-to-classical-control/digital-control.png)
+
+#### Contraints
+- $T_s$: Sampling time
+- $f_s \geq 2f_m$: Sampling Frequency must be at least 2 times the max frequency 
+  of the system
+
+
+#### Parts of the system
+1. Take `reference` and `output` and compute the `error`
+2. Pass this signal into an `antialiasing filter` to avoid ***aliases***
+3. Trasform the `Laplace Transform` in a `Z-Transform` by using the following 
+   relation:\
+   $z = e^{sT}$
+4. Control everything through a `control block` engineered through 
+   `digital control`
+5. Transform the `digital signal` to an `analogic signal` through the use of a 
+    `holder` (in this case a `zero order holder`)
+6. Pass the signal to our `analogic plant` (which is our physical system)
+7. Take the `output` and pass it in `retroaction`
+
+### Zero Order Holder
+It has the following formula:
+$$
+ZoH = \frac{1}{s} \left( 1 - e^{sT}\right)
+$$
+
+#### Commands:
+- `c2d(sysc, Ts [, method | opts] )`[^matlab-c2d]: Converts `LTI` systems into 
+  `Discrete` ones 
+
+
+## Relation between $S(A, B, C, D)$ to $G(s)$
+
+### From $S(A, B, C, D)$ to $G(s)$
+Be this our  $S(A, B, C, D)$ system:
+$$
+\begin{cases}
+\dot{x}(t) = Ax(t) + Bu(t) \;\;\;\; \text{Dynamic of the system}\\
+      y(t) = C{x}(t) + Du(t) \;\;\;\; \text{Static of the outputs}
+\end{cases}
+$$
+
+now let's make from this a `Laplace Transform`:
+$$
+\begin{align*}
+& \begin{cases}
+    sX(s) - x(0)= AX(s) + BU(s) \\
+    Y(s) = CX(s) + DU(s)
+  \end{cases} \longrightarrow && \text{Normal Laplace Transformation}\\
+
+
+& \longrightarrow
+\begin{cases}
+    sX(s) = AX(s) + BU(s) \\
+    Y(s) = CX(s) + DU(s)
+  \end{cases} \longrightarrow && \text{Usually $x(0)$ is 0}\\
+
+
+& \longrightarrow
+\begin{cases}
+    X(s) \left(sI -A \right) =BU(s) \\
+    Y(s) = CX(s) + DU(s)
+  \end{cases} \longrightarrow && \text{$sI$ is technically equal to $s$}\\
+
+
+& \longrightarrow
+\begin{cases}
+    X(s) = \left(sI - A\right)^{-1}BU(s) \\
+    Y(s) = CX(s) + DU(s)
+  \end{cases} \longrightarrow && \\
+
+
+& \longrightarrow
+\begin{cases}
+    X(s) = \left(sI - A\right)^{-1}BU(s) \\
+    Y(s) = C\left(sI - A\right)^{-1}BU(s) + DU(s)
+  \end{cases} \longrightarrow && \text{Substitute for $X(s)$}\\
+
+
+& \longrightarrow
+\begin{cases}
+    X(s) = \left(sI - A\right)^{-1}BU(s) \\
+    Y(s) = \left(C\left(sI - A\right)^{-1}B + D\right)U(s)
+  \end{cases} \longrightarrow && \text{Group for $U(s)$}\\
+
+
+& \longrightarrow
+\begin{cases}
+    X(s) = \left(sI - A\right)^{-1}BU(s) \\
+    \frac{Y(s)}{U(s)} = \left(C\left(sI - A\right)^{-1}B + D\right)
+  \end{cases} \longrightarrow && \text{Get $G(s)$ from definition}\\
+
+\longrightarrow \;& G(s) = \left(C\left(sI - A\right)^{-1}B + D\right) && 
+\text{Formal definition of $G(s)$}\\
+
+
+\end{align*}
+$$
+
+#### Properties
+- Since $G(s)$ can be ***technically*** a matrix, this may represent a 
+`MIMO System`
+- The system is ***always*** `proper` (so it's denominator is of an order 
+  higher of the numerator)
+- If $D$ is $0$, then the system is `strictly proper` and ***realizable***
+
+- While each $S(A_i, B_i, C_i, D_i)$ can be transformed into a ***single***
+$G(s)$, this isn't true viceversa.
+
+- Any particular $S(A_a, B_a, C_a, D_a)$ is called `realization`
+- $det(sI - A)$ := Characteristic Polinome
+- $det(sI - A) = 0$ := Characteristic Equation
+- $eig(A)$ := Solutions of the Characteristic Equation and `poles` of the system
+- If the system is `SISO` and this means that $C \in \R^{1,x}$, 
+  $B \in \R^{x,1}$ and $D \in \R$, meaning 
+  that:  
+$$
+\begin{align*}
+  G(s) &= \left(C\left(sI - A\right)^{-1}B + D\right) =\\ 
+  &= \left(C \frac{Adj\left(sI - A\right)}{det\left(sI - A\right)}B + D\right) 
+  = && \text{Decompose the inverse in its formula}\\
+  &= \frac{n(s)}{det\left(sI - A\right)} \in \R
+\end{align*}
+$$
+
+> [!NOTE]
+> As you can see here, by decomposing the inverse matrix in its formula it's 
+> easy to see that the divisor is a `scalar`, a `number`. 
+> 
+> Moreover, because of how $B$ and $C$ are composed, the result of this Matrix 
+> multiplication is a `scalar` too, hence we can write this as a single formula.
+>
+> Another thing to notice, regardless if this is a `MIMO` or `SISO` system is 
+> that at the divisor we have all `eigenvalues` of A as `poles` by 
+> [definition](../Formularies/GEOMETRY-FORMULARY.md/#eigenvalues)
+> 
+
+### Transforming a State-Space into Another one
+We basically need to use some non singular `Permutation Matrices`:
+$$
+\begin{align*}
+&A_1, B_1, C_1, D_1 \\
+&A_2 = PAP^{-1} \\
+&B_2 = PB \\
+&C_2 = CP^{-1} \\
+&D_2 = D_1
+\end{align*}
+$$
+
+
+
+[^matlab-c2d]: [Matlab Official Docs | c2d | 05 January 2025](https://it.mathworks.com/help/control/ref/dynamicsystem.c2d.html)

docs/Chapters/3-CANONICAL-FORMS.md

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+# Canonical Forms
+
+In order to see if we are in one of these canonical forms, just write the 
+equations from the block diagram, and find the associated $S(A, B, C, D)$.
+
+> [!TIP]
+> In order to find a rough diagram, use 
+> [Horner Factorization](MODERN-CONTROL.md/#horner-factorization) to find 
+> $a_i$ values. Then put all the $b_i$ to the right integrator by shifting them
+> as many left places, starting from the rightmost, for the number of 
+> associated $s$
+
+## Control Canonical Form
+It is in such forms when: 
+$$
+A = \begin{bmatrix}
+- a_1 & -a_2 & -a_3 & \dots & -a_{n-1} &-a_n\\
+1 & 0 & 0 & \dots  & 0 &  0\\
+0 & 1 & 0 & \dots & 0 & 0\\
+\dots & \dots & \dots & \dots  & \dots \\
+0 & 0 & 0 & \dots & 1 & 0
+\end{bmatrix}
+
+B = \begin{bmatrix}
+1 \\ 0 \\ \dots \\ \dots \\ 0
+\end{bmatrix}
+
+C = \begin{bmatrix}
+b_1 & b_2 & \dots & b_n
+\end{bmatrix}
+
+D = \begin{bmatrix}
+0
+\end{bmatrix}
+$$
+
+## Modal Canonical Forms
+> [!CAUTION]
+> This form is the most difficult to find, as this varies drastically in cases 
+> of double roots
+>
+$$
+A = \begin{bmatrix}
+- a_1 & 0 & 0 & \dots & 0\\
+0 & -a_2 & 0 & \dots & 0\\
+0 & 0 & -a_3 & \dots & 0\\
+\dots & \dots & \dots & \dots & \dots \\
+0 & 0 & 0 & 0 & -a_n
+\end{bmatrix}
+
+B = \begin{bmatrix}
+1 \\ 1 \\ \dots \\ \dots \\ 1
+\end{bmatrix}
+
+C = \begin{bmatrix}
+b_1 & b_2 & \dots & b_n
+\end{bmatrix}
+
+D = \begin{bmatrix}
+0
+\end{bmatrix}
+$$
+
+## Observable Canonical Form
+<!--TODO: Correct here -->
+
+[^reference-input-pole-allocation]: [MIT | 06 January 2025 | pg. 2](https://ocw.mit.edu/courses/16-30-feedback-control-systems-fall-2010/c553561f63feaa6173e31994f45f0c60_MIT16_30F10_lec11.pdf)

docs/Chapters/4-REACHABILITY-AND-OBSERVABILITY.md

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+# Reachability and Observability
+
+## Reachability
+While in the non linear world, we can solve a `system`
+***numerically***, through an `iterative-approach`:
+$$
+\begin{align*}
+\dot{x}(t) &= f(x(t), u(t)) && t \in \R \\
+x(k+1) &= f(x(k), u(k)) && t \in \N
+\end{align*}
+$$
+
+In the linear world, we can do this ***analitically***:
+> [!TIP]
+> We usually consider $x(0) = 0$
+$$
+\begin{align*}
+x(1) &= Ax(0) + Bu(0) \\
+x(2) &= Ax(1) + Bu(1) &&= A^{2}x(0) + ABu(0) + Bu(1) \\
+x(3) &= Ax(2) + Bu(2) &&= A^{3}x(0) + A^{2}Bu(0) + ABu(1) + Bu(2) \\
+\dots \\
+x(k) &= Ax(k-1) + Bu(k-1) &&= 
+\underbrace{A^{k}x(0)}_\text{Free Dynamic} + 
+\underbrace{A^{k-1}Bu(0) + \dots + ABu(k-2) + Bu(k-1) }
+_\text{Forced Dynamic} \\[40pts]
+
+x(k) &= \begin{bmatrix}
+B & AB &  \dots & A^{k-2}B & A^{k-1}B
+\end{bmatrix} 
+
+\begin{bmatrix}
+u(k-1) \\ u(k-2) \\  \dots \\ u(1) \\ u(0)
+\end{bmatrix} 
+\end{align*}
+$$
+
+Now, there's a relation between the determinant and the matrix containing
+$A$ and $B$:
+
+$$
+\begin{align*}
+
+&p_c(s) = det(sI -A) = 
+s^n + a_{n-1} s^{n-1} +  a_{n-2}s^{n - 2} + \dots + a_1s + a_0 = 0 
+\\[10pt]
+
+&\text{Apply Caley-Hamilton theorem:} \\
+&p_c(A) = A^{n} + a_{n-1}A^{n_1} + \dots + a_1A + a_0 = 0\\[10pt]
+
+&\text{Remember $G(s)$ formula and multiply $p_c(A)$ for $B$:} \\
+&p_c(A)B = A^{n}B + a_{n-1}A^{n_1}B + \dots + a_1AB + a_0B = 0 \rightarrow \\
+
+&p_c(A)B = a_{n-1}A^{n_1}B + \dots + a_1AB + a_0B = -A^{n}B
+\end{align*} 
+$$
+
+All of these makes us conclude something about the Kallman 
+Controllability Matrix:
+$$
+K_c = \begin{bmatrix}
+B & AB &  \dots & A^{k-2}B & A^{k-1}B
+\end{bmatrix} 
+$$
+
+Moreover, $x(n) \in range(K_c)$ and $range(K_c)$ is said 
+`reachable space`.\
+In particular if $rank(K_c) = n \rightarrow range(K_c) = \R^{n}$ this 
+is `fully reachable` or `controllable`
+> [!TIP]
+> Some others use `non-singularity` instead of the $range()$ definition
+
+
+> [!NOTE]
+> On the Franklin Powell there's another definition to $K_c$ that 
+> comes from the fact that we needed to find a way to transform
+> ***any*** `realization` into the 
+> [`Control Canonical Form`](./CANONICAL-FORMS.md/#control-canonical-form)
+>
+
+## Observability
+This is the capability of being able to deduce the `initial state` by just
+observing the `output`.
+
+Let's focus on the $y(t)$ part:
+$$
+y(t) = 
+\underbrace{Cx(t)}_\text{Free Output} + 
+\underbrace{Du(t)}_\text{Forced Output}
+$$
+
+Assume that $u(t) = 0$:
+$$
+\begin{align*}
+& y(0) = Cx(0) && x(0) = x(0) \\
+& y(1) = Cx(1) && x(1) = Ax(0) && 
+\text{Since $u(t) = 0 \rightarrow Bu(t) = 0$} \\
+& y(2) = Cx(2) && x(2) = A^2x(0) \\
+& \vdots && \vdots \\
+&y(n) = Cx(n) && x(n) = A^nx(0) \rightarrow \\
+\rightarrow &y(n) = CA^nx(0)
+\end{align*}
+$$
+
+Now we have that: 
+$$
+\begin{align*}
+\vec{y} = &\begin{bmatrix}
+C \\
+CA \\
+\vdots \\
+CA^{n}
+\end{bmatrix} x(0) \rightarrow \\
+
+\rightarrow x(0) = &\begin{bmatrix}
+C \\
+CA \\
+\vdots \\
+CA^{n}
+\end{bmatrix}^{-1}\vec{y} \rightarrow \\
+
+\rightarrow x(0) = & \frac{Adj(K_o)}{det(K_o)} \vec{y}
+\end{align*} 
+$$
+
+For the same reasons as before, we can use Caley-Hamilton here too, also, we can see that if $K_o$ is `singular`, there can't be an inverse.
+
+As before, $K_o$ is such a matrix that allows us to see if there exists a 
+[`Canonical Observable Form`](CANONICAL-FORMS.md/#observable-canonical-form)
+
+The `non-observable-space` is equal to:
+$$
+X_{no} = Kern(K_o) : \left\{ K_ox = 0 | x \in X\right\}
+$$
+
+## Decomposition of these spaces
+The space of possible points is $X$ and is equal to 
+$X = X_r \bigoplus X_{r}^{\perp} = X_r \bigoplus X_{nr}$
+
+Analogously we can do the same with the observable spaces 
+$X = X_no \bigoplus X_{no}^{\perp} = X_no \bigoplus X_{o}$

docs/Chapters/5-STATE-FEEDBACK.md

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+# State Feedback
+
+## State Feedback
+When you have your $G(s)$, just put poles where you want them to be and compute the right $k_i$.
+
+What we are doing is to put $u(t) = -Kx(t)$, feeding back the `state` to the
+`system`.
+
+> [!WARNING]
+> While $K$ matrix is controllable, the A matrix is equivalent to our plant, so 
+> it's impossible to change $a_i$ without changing the `system`
+
+> [!CAUTION]
+> We are doing some oversimplifications here, but this can't be used as formal 
+> proof for whatever we are saying. To see more, look at 
+> [Ackerman's formula](https://en.wikipedia.org/wiki/Ackermann%27s_formula).
+>
+> However is it possible to put a fictional reference input to make everything 
+> go back to a similar proof for the 
+> characteristic equation[^reference-input-pole-allocation]
+
+
+While it is possible to allocate poles in each form, the Canonical Control
+one makes it very easy to accomplish this task:
+
+$$
+A = \begin{bmatrix}
+- a_1 \textcolor{#ff8382}{-k_1}& -a_2 \textcolor{#ff8382}{-k_2}
+& -a_3 \textcolor{#ff8382}{-k_3}& \dots 
+& -a_{n-1}\textcolor{#ff8382}{-k_{n-1}}
+ &-a_n \textcolor{#ff8382}{-k_n}\\
+1 & 0 & 0 & \dots  & 0 &  0\\
+0 & 1 & 0 & \dots & 0 & 0\\
+\dots & \dots & \dots & \dots  & \dots \\
+0 & 0 & 0 & \dots & 1 & 0
+\end{bmatrix}
+$$
+
+This changes our initial considerations, however. The new Characteristi Equation
+becomes: 
+$$
+det\left(sI - (A - BK)\right)
+$$
+
+## State Observer
+What happens if we have not enough sensors to get the state in 
+`realtime`?
+
+> [!NOTE]
+> Measuring the state involves a sensor, which most of the times is 
+> either inconvenient to place because of space, or inconvenient 
+> economically speaking
+
+In this case we `observe` our output to `estimate` our state 
+($\hat{x}$). THis new state will be used in our $G_c$ block:
+
+$$
+\dot{\hat{x}} = A\hat{x} + Bu + L(\hat{y} - y) = A\hat{x} + Bu + 
+LC(\hat{x} - x) 
+$$
+
+Since we are estimating our state, we need the estimator to be fast,
+**at least 6 times fater** than our `plant`, and $u = -K\hat{x}$.
+
+Let's compute the error we introduce in our system:
+<!--TODO: See error formula-->
+$$
+\begin{align*}
+e &= x - \hat{x} \rightarrow \\
+\rightarrow \dot{e} &= \dot{x} - \dot{\hat{x}}  \rightarrow\\
+
+&\rightarrow Ax - BK\hat{x} - A\hat{x} + BK\hat{x} - LCe \rightarrow\\
+&\rightarrow -Ae - LCe \rightarrow\\
+&\rightarrow eig(-A-LC)
+\end{align*} 
+$$
+
+## Making a Controller
+So, this is how we will put all of these blocks saw until now.
+![block diagram](..\Images\State-Feedback\block-diagram.png)
+
+From this diagram we can deduce immediately that to us our input is $y$
+and our output is $u$. By treating the blue part as a `black box`, we can
+say that:
+$$
+\begin{cases}
+\dot{\hat{x}} = A\hat{x} - BK\hat{x} + LC\hat{x} - Ly \\
+u = -K\hat{x}
+\end{cases}
+$$
+
+So our $S(A_c, B_c, C_c, D_c)$ is: 
+
+$$
+\begin{align*}
+&A_c = A_p - B_pK + LC_p \\
+&B_c = -L \\
+&C_c = -K \\
+&D_c = 0 \\
+\end{align*}
+$$
+
+and so, you just need to substitute these matrices to the equation of the 
+$G_p(s)$ and you get you controller

docs/Chapters/Examples/EXAMPLE-3.md

@@ -0,0 +1,132 @@
+# Example 3
+
+## Double Mass Cart
+
+![double mass cart](./../../Images/Examples/Example-3/double-cart.png)
+
+### Formulas
+
+- Resulting forces for cart 1:\
+$
+m_1 \ddot{p}_1 = k_2(p_2 - p_1) + b_2( \dot{p}_2 - \dot{p}_1) - 
+    k_1 p_1 - b_1 \dot{p}_1
+$
+
+- Resulting forces for cart 2:\
+$
+m_2 \ddot{p}_2 = F - k_2(p_2 - p_1) - b_2( \dot{p}_2 - \dot{p}_1) 
+$ 
+
+### Reasoning
+We now have 2 different accelerations. The highest order of derivatives is 2 for 
+2 variables, hence we need 4 variables in the `state`:
+
+$$
+x = \begin{bmatrix}
+x_1 = p_1\\
+x_2 = p_2\\
+x_3 = \dot{p}_1\\
+x_4 = \dot{p}_2
+\end{bmatrix}
+
+\dot{x} = \begin{bmatrix}
+\dot{x}_1 = \dot{p}_1 = x_3 \\
+\dot{x}_2 = \dot{p}_2 = x_4\\
+\dot{x}_3 = \ddot{p}_1 = 
+    \frac{1}{m_1} \left[ k_2(x_2 - x_1) + b_2( x_4 - x_3) - 
+    k_1 x_1 - b_1 x_3 \right]\\
+\dot{x}_4 = \ddot{p}_2 = 
+    \frac{1}{m_2} \left[ F - k_2(x_2 - x_1) - b_2( x_4 - x_3) \right]\\
+\end{bmatrix}
+$$
+
+Let's write our $S(A, B, C, D)$:
+$$
+A = \begin{bmatrix}
+0 & 0 & 1 & 0 \\
+0 & 0 & 0 & 1 \\
+% 3rd row
+- \frac{k_2 - k_1}{m_1} & 
+\frac{k_2}{m_1}  & 
+-\frac{b_2 + b_1}{m_1} & 
+\frac{b_2}{m_1} \\
+% 4th row
+\frac{k_2}{m_12} & 
+- \frac{k_2}{m_2} & 
+\frac{b_2}{m_2} & 
+- \frac{b_2}{m_2} \\
+\end{bmatrix}
+
+B = \begin{bmatrix}
+0 \\ 
+0 \\ 0 \\ 1
+\end{bmatrix}
+
+C = \begin{bmatrix}
+1 & 0 & 0 & 0 \\
+0 & 1 & 0 & 0
+\end{bmatrix}
+
+D = \begin{bmatrix}
+0
+\end{bmatrix}
+$$
+
+
+## Suspended Mass
+
+
+> [!NOTE]
+> For those of you the followed CNS course, refer to professor 
+> PDF for this excercise, as it has some unclear initial conditions
+>
+> However, in the formulas section, I'll take straight up his own
+
+![suspended mass](./../../Images/Examples/Example-3/suspended-mass.png)
+
+### Formulas
+
+- Resulting forces for mass:\
+$
+m \ddot{p} = -k(p - r) -b(\dot{p} - \dot{r})
+$
+
+### Reasoning
+
+
+$$
+x = \begin{bmatrix}
+x_1 = p \\
+x_2 = \dot{x}_1
+\end{bmatrix}
+
+\dot{x} = \begin{bmatrix}
+\dot{x}_1 = x_2 \\
+\dot{x}_2 = \frac{1}{m} \left[-k(x_1 - r) -b(x_2 - \dot{r}) \right]
+\end{bmatrix}
+$$
+
+<!-- TODO: Correct here looking from book -->
+> [!WARNING]
+> Info here are wrong 
+
+Let's write our $S(A, B, C, D)$:
+$$
+A = \begin{bmatrix}
+0 & 1\\
+-\frac{k}{m} & - \frac{b}{m} 
+\end{bmatrix}
+
+B = \begin{bmatrix}
+0 \\ 
+\frac{k + sb}{m}
+\end{bmatrix}
+
+C = \begin{bmatrix}
+1 & 0 
+\end{bmatrix}
+
+D = \begin{bmatrix}
+0 & 0
+\end{bmatrix}
+$$

docs/Chapters/MODERN-CONTROL.md

@@ -1,98 +0,0 @@
-# Modern Control
-
-Normally speaking, we know much about classical control, in the form
-of:
-
-$$
-\dot{x}(t) = ax(t) + bu(t) \longleftrightarrow sX(s) - x(0) = aX(S) + bU(s)
-$$
-
-With the left part being a derivative equation in continuous time, while the 
-right being its tranformation in the complex domain field.
-
-> [!NOTE]
->
-> $$
-> \dot{x}(t) = ax(t) + bu(t) \longleftrightarrow x(k+1) = ax(k) + bu(k)
-> $$
->
-> These are equivalent, but the latter one is in discrete time.
->
-
-## A brief recap over Classical Control
-
-Be $Y(s)$ our `output variable` in `classical control` and $U(s)$ our 
-`input variable`. The associated `transfer function` $G(s)$ is:
-
-$$
-G(s) = \frac{Y(s)}{U(s)}
-$$
-
-### Root Locus
-<!-- TODO: write about Root Locus -->
-
-### Bode Diagram
-
-### Nyquist Diagram
-
-## State Space Representation
-
-### State Matrices
-A state space representation has 4 Matrices: $A, B, C, D$ with coefficients in 
-$\R$:
-- $A$: State Matrix `[x_rows, x_columns]`
-- $B$: Input Matrix `[x_rows, u_columns]`
-- $C$: Output Matrix `[y_rows, x_columns]`
-- $D$: Direct Coupling Matrix `[y_rows, u_columns]`
-
-$$
-\begin{cases}
-\dot{x}(t) = Ax(t) + Bu(t) \;\;\;\; \text{Dynamic of the system}\\
-      y(t) = C{x}(t) + Du(t) \;\;\;\; \text{Static of the outputs}
-\end{cases}
-$$
-
-This can be represented with the following diagrams:
-
-Continuous Time:
-![continuous state space diagram](..\Images\Modern-Control\state-space-time.png)
-
----
-Discrete time:
-
-![discrete state space diagram](..\Images\Modern-Control\state-space-discrete.png)
-
-### State Vector
-This is a state vector `[x_rows, 1]`:
-$$
-x(t) =  \begin{bmatrix}
-            x_1(t)\\
-            \dots\\
-            x_x(t)
-        \end{bmatrix}
-
-\text{or} \:
-
-x(k) =  \begin{bmatrix}
-            x_1(k)\\
-            \dots\\
-            x_x(k)
-        \end{bmatrix}
-$$
-
-Basically, from this we can know each next step of the state vector, represented 
-as:
-
-$$
-x(k + 1) = f\left( 
-    x(k), u(k)
-\right) = Ax(k) + Bu(k)
-$$
-
-
-### Case Studies
-<!-- TODO: Complete case studies -->
-- PAGERANK 
-- Congestion Control
-- Video Player Control
-- Deep Learning

docs/Formularies/CONTROL-FORMULARY.md

@@ -0,0 +1,35 @@
+# Control Formulary
+## Settling time
+$
+T_s = \frac{\ln(a_{\%})}{\zeta \omega_{n}} 
+$
+
+- $\zeta$ := Damping ratio
+- $\omega_{n}$ := Natural frequency
+
+## Overshoot
+$
+\mu_{p}^{\%} = 100 e^{
+    \left(
+        \frac{- \zeta \pi}{\sqrt{1 - \zeta^{2}}}
+    \right)
+}
+$
+
+## Reachable Space
+$X_r = Span(K_c)$
+
+$X = X_r \bigoplus X_{r}^{\perp} = X_r \bigoplus X_{nr}$
+
+> [!TIP]
+> Since $X_{nr} = X_r^{\perp}$ we can find a set of perpendicular
+> vectors by finding $Ker(X_r^{T})$
+
+## Non Observable Space
+$X_no = Kern(K_o)$
+
+$X = X_no \bigoplus X_{no}^{\perp} = X_no \bigoplus X_{o}$
+
+> [!TIP]
+> Since $X_{o} = X_no^{\perp}$ we can find a set of perpendicular
+> vectors by finding $Ker(X_{no}^{T})$

docs/Formularies/GEOMETRY-FORMULARY.md

@@ -0,0 +1,49 @@
+# Geometry Formulary
+
+## Inverse of a Matrix
+$A^{-1} = \frac{1}{det(A)} Adj(A)$
+
+## Adjugate Matrix
+The adjugate of a matrix $A$ is the `transpose` of the `cofactor matrix`:\
+$Adj(A) = C^{T}$
+
+### $(i-j)$-minor (AKA $M_{ij}$)
+$M_{ij}$ := Determinant of the matrix $B$ got by removing the 
+***$i$-row*** and the ***$j$-column*** from matrix $A$
+
+### Cofactors
+$C$ is the matrix of `cofactors` of a matrix $A$ where all the elements $c_{ij}$
+are defined like this:\
+$
+c_{ij} = \left( -1\right)^{i + j}M_{ij}
+$
+
+## Eigenvalues
+By starting from the definition of `eigenvectors`:\
+$A\vec{v} = \lambda\vec{v}$
+
+As we can see, the vector $\vec{v}$ was unaffected by this matrix 
+multiplication appart from a scaling factor $\lambda$, called `eigenvalue`
+
+By rewriting this formula we get:\
+$
+\left(A - \lambda I\right)\vec{v} = 0 
+$ 
+
+This is solved for:\
+$\det(A- \lambda I) = 0$
+
+> [!NOTE]
+> If the determinant is 0, then $(A - \lambda I )$ is not invertible, so
+> we can't solve the previous equation by using the trivial solution (which
+> can't be taken into account since $\vec{v}$ is not $0$  by definition)
+
+## Caley-Hamilton
+Each square matrix over a `commutative ring` satisfies its own 
+characteristic equation $det(\lambda I - A)$
+
+> [!TIP]
+> In other words, once found the characteristic equation, we can 
+> substitute the ***unknown*** variable $\lambda$ with the matrix itself
+> (***known***),
+> powered to the correspondent power

docs/Formularies/PHYSICS-FORMULARY.md

@@ -0,0 +1,37 @@
+# Physics Formulary
+
+> [!TIP]
+>
+> You'll often see $\vec{v}$ and $\vec{\dot x}$, and $\vec{a}$ and 
+> $\vec{\ddot{x}}$.
+>
+> These are equal, but the latter forms, express better the relation
+> between state space variables
+
+## Hooke's Law (AKA Spring Formula)
+$\vec{F} = -k\vec{x}$
+- $k$: Spring Constant
+- $\vec{x}$: vector of spring stretch from rest position
+  
+## Fluid drag
+$\vec{F_D} = \frac{1}{2}b\vec{v}^2C_{D}A$
+- $b$: density of fluid
+- $v$: speed of object ***relative*** to the fluid
+- $C_D$: drag coefficient
+- $A$: cross section area
+
+### Stokes Drag
+$\vec{F_D} = -6 \pi R\mu \vec{v}$
+- $\mu$: dynamic viscosity
+- $R$: radius (in meters) of the sphere
+- $\vec{v}$: flow velocity ***relative*** to the fluid
+
+### Simplified Fluid Drag (Simplified Stokes Equation)
+$\vec{F_D} = -b \vec{v}$
+- $b$: simplified coefficient that has everything else
+- $\vec{v}$: flow velocity ***relative*** to the fluid
+##
+
+## Newton force
+
+