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Christian Risi
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docs/Chapters/1-MODERN-CONTROL.md
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# Modern Control
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Normally speaking, we know much about classical control, in the form
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of:
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$$
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\dot{x}(t) = ax(t) + bu(t) \longleftrightarrow sX(s) - x(0) = aX(S) + bU(s)
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$$
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With the left part being a derivative equation in continuous time, while the
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right being its tranformation in the complex domain field.
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> [!NOTE]
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>
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> $$
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> \dot{x}(t) = ax(t) + bu(t) \longleftrightarrow x(k+1) = ax(k) + bu(k)
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> $$
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>
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> These are equivalent, but the latter one is in discrete time.
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>
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## A brief recap over Classical Control
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Be $Y(s)$ our `output variable` in `classical control` and $U(s)$ our
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`input variable`. The associated `transfer function` $G(s)$ is:
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$$
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G(s) = \frac{Y(s)}{U(s)}
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$$
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### Root Locus
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<!-- TODO: write about Root Locus -->
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### Bode Diagram
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### Nyquist Diagram
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## State Space Representation
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### State Matrices
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A state space representation has 4 Matrices: $A, B, C, D$ with coefficients in
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$\R$:
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- $A$: State Matrix `[x_rows, x_columns]`
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- $B$: Input Matrix `[x_rows, u_columns]`
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- $C$: Output Matrix `[y_rows, x_columns]`
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- $D$: Direct Coupling Matrix `[y_rows, u_columns]`
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$$
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\begin{cases}
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\dot{x}(t) = Ax(t) + Bu(t) \;\;\;\; \text{Dynamic of the system}\\
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y(t) = C{x}(t) + Du(t) \;\;\;\; \text{Static of the outputs}
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\end{cases}
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$$
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This can be represented with the following diagrams:
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#### Continuous Time:
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---
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#### Discrete time:
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### State Vector
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This is a state vector `[x_rows, 1]`:
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$$
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x(t) = \begin{bmatrix}
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x_1(t)\\
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\dots\\
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x_x(t)
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\end{bmatrix}
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\text{or} \:
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x(k) = \begin{bmatrix}
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x_1(k)\\
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\dots\\
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x_x(k)
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\end{bmatrix}
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$$
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Basically, from this we can know each next step of the state vector, represented
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as:
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$$
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x(k + 1) = f\left(
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x(k), u(k)
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\right) = Ax(k) + Bu(k)
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$$
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### Examples
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#### Cart attached to a spring and a damper, pulled by a force
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##### Formulas
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- <span style="color:#55a1e6">Spring: $\vec{F} = -k\vec{x}$</span>
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- <span style="color:#ff8382">Fluid Damper: $\vec{F_D} = -b \vec{\dot{x}}$</span>
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- Initial Force: $\vec{F_p}(t) = \vec{F_p}(t)$
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- Total Force: $m \vec{\ddot{x}}(t) = \vec{F_p}(t) -b \vec{\dot{x}} -k\vec{x}$
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> [!TIP]
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>
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> A rule of thumb is to have as many variables in our state as the max number
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> of derivatives we encounter. In this case `2`
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>
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> Solve the equation for the highest derivative order
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>
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> Then, put all variables equal to the previous one derivated:
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>
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> $$
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x(t) = \begin{bmatrix}
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x_1(t)\\
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x_2(t) = \dot{x_1}(t)\\
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\dots\\
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x_n(t) = \dot{x}_{n-1}(t)
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\end{bmatrix}
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\;
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\dot{x}(t) = \begin{bmatrix}
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\dot{x_1}(t) = x_2(t)\\
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\dot{x_2}(t) = x_3(t)\\
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\dots\\
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\dot{x}_{n-1}(t) = \dot{x}_{n}(t)\\
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\dot{x}_{n}(t) = \text{our formula}
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\end{bmatrix}
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> $$
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>
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Now in our state we may express `position` and `speed`, while in our
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`next_state` we'll have `speed` and `acceleration`:
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$$
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x(t) = \begin{bmatrix}
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x_1(t)\\
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x_2(t) = \dot{x_1}(t)
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\end{bmatrix}
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\;
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\dot{x}(t) = \begin{bmatrix}
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\dot{x_1}(t) = x_2(t)\\
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\dot{x_2}(t) = \ddot{x_1}(t)
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\end{bmatrix}
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$$
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Our new state is then:
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$$
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\begin{cases}
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\dot{x}_1(t) = x_2(t)\\
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\dot{x}_2(t) = \frac{1}{m} \left( \vec{F}(t) - b x_2(t) - kx_1(t) \right)
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\end{cases}
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$$
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let's say we only want to check for the `position` and `speed` of the system, our
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State Space will be:
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$$
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A = \begin{bmatrix}
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0 & 1 \\
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- \frac{k}{m} & - \frac{b}{m} \\
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\end{bmatrix}
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B = \begin{bmatrix}
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0 \\
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\frac{1}{m} \\
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\end{bmatrix}
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C = \begin{bmatrix}
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1 & 0 \\
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0 & 1
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\end{bmatrix}
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D = \begin{bmatrix}
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0
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\end{bmatrix}
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$$
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let's say we only want to check for the `position` of the system, our
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State Space will be:
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$$
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A = \begin{bmatrix}
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0 & 1 \\
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- \frac{k}{m} & - \frac{b}{m} \\
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\end{bmatrix}
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B = \begin{bmatrix}
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0 \\
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\frac{1}{m} \\
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\end{bmatrix}
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C = \begin{bmatrix}
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1 & 0
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\end{bmatrix}
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D = \begin{bmatrix}
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0
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\end{bmatrix}
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$$
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> [!TIP]
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> In order to being able to plot the $\vec{x}$ against the time, you need to
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> multiply $\vec{\dot{x}}$ for the `time_step` and then add it to the state[^so-how-to-plot-ssr]
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>
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### Horner Factorization
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let's say you have a complete polynomial of order `n`, you can factorize
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in this way:
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$$
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\begin{align*}
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p(s) &= s^5 + 4s^4 + 5s^3 + 2s^2 + 10s + 1 =\\
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&= s ( s^4 + 4s^3 + 5s^2 + 2s + 10) + 1 = \\
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&= s ( s (s^3 + 4s^2 + 5s + 2) + 10) + 1 = \\
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&= s ( s (s (s^2 + 4s + 5) + 2) + 10) + 1 = \\
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&= s ( s (s ( s (s + 4) + 5) + 2) + 10) + 1
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\end{align*}
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$$
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If you were to take each s with the corresponding number in the parenthesis,
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you'll make this block:
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### Case Studies
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<!-- TODO: Complete case studies -->
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- PAGERANK
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- Congestion Control
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- Video Player Control
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- Deep Learning
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[^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)
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160
docs/Chapters/2-RELATION-TO-CLASSICAL-CONTROL.md
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# Relation to Classical Control
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## A Brief Recap of Discrete Control
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Let's say we want to control something ***Physical***, hence intrinsically
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***time continuous***, we can model our control in the `z` domain and make our
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$G_c(z)$. But how do we connect these systems:
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#### Contraints
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- $T_s$: Sampling time
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- $f_s \geq 2f_m$: Sampling Frequency must be at least 2 times the max frequency
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of the system
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#### Parts of the system
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1. Take `reference` and `output` and compute the `error`
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2. Pass this signal into an `antialiasing filter` to avoid ***aliases***
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3. Trasform the `Laplace Transform` in a `Z-Transform` by using the following
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relation:\
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$z = e^{sT}$
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4. Control everything through a `control block` engineered through
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`digital control`
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5. Transform the `digital signal` to an `analogic signal` through the use of a
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`holder` (in this case a `zero order holder`)
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6. Pass the signal to our `analogic plant` (which is our physical system)
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7. Take the `output` and pass it in `retroaction`
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### Zero Order Holder
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It has the following formula:
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$$
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ZoH = \frac{1}{s} \left( 1 - e^{sT}\right)
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$$
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#### Commands:
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- `c2d(sysc, Ts [, method | opts] )`[^matlab-c2d]: Converts `LTI` systems into
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`Discrete` ones
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## Relation between $S(A, B, C, D)$ to $G(s)$
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### From $S(A, B, C, D)$ to $G(s)$
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Be this our $S(A, B, C, D)$ system:
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$$
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\begin{cases}
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\dot{x}(t) = Ax(t) + Bu(t) \;\;\;\; \text{Dynamic of the system}\\
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y(t) = C{x}(t) + Du(t) \;\;\;\; \text{Static of the outputs}
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\end{cases}
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$$
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now let's make from this a `Laplace Transform`:
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$$
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\begin{align*}
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& \begin{cases}
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sX(s) - x(0)= AX(s) + BU(s) \\
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Y(s) = CX(s) + DU(s)
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\end{cases} \longrightarrow && \text{Normal Laplace Transformation}\\
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& \longrightarrow
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\begin{cases}
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sX(s) = AX(s) + BU(s) \\
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Y(s) = CX(s) + DU(s)
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\end{cases} \longrightarrow && \text{Usually $x(0)$ is 0}\\
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& \longrightarrow
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\begin{cases}
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X(s) \left(sI -A \right) =BU(s) \\
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Y(s) = CX(s) + DU(s)
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\end{cases} \longrightarrow && \text{$sI$ is technically equal to $s$}\\
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& \longrightarrow
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\begin{cases}
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X(s) = \left(sI - A\right)^{-1}BU(s) \\
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Y(s) = CX(s) + DU(s)
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\end{cases} \longrightarrow && \\
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& \longrightarrow
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\begin{cases}
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X(s) = \left(sI - A\right)^{-1}BU(s) \\
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Y(s) = C\left(sI - A\right)^{-1}BU(s) + DU(s)
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\end{cases} \longrightarrow && \text{Substitute for $X(s)$}\\
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& \longrightarrow
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\begin{cases}
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X(s) = \left(sI - A\right)^{-1}BU(s) \\
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Y(s) = \left(C\left(sI - A\right)^{-1}B + D\right)U(s)
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\end{cases} \longrightarrow && \text{Group for $U(s)$}\\
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& \longrightarrow
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\begin{cases}
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X(s) = \left(sI - A\right)^{-1}BU(s) \\
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\frac{Y(s)}{U(s)} = \left(C\left(sI - A\right)^{-1}B + D\right)
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\end{cases} \longrightarrow && \text{Get $G(s)$ from definition}\\
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\longrightarrow \;& G(s) = \left(C\left(sI - A\right)^{-1}B + D\right) &&
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\text{Formal definition of $G(s)$}\\
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\end{align*}
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$$
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#### Properties
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- Since $G(s)$ can be ***technically*** a matrix, this may represent a
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`MIMO System`
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- The system is ***always*** `proper` (so it's denominator is of an order
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higher of the numerator)
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- If $D$ is $0$, then the system is `strictly proper` and ***realizable***
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- While each $S(A_i, B_i, C_i, D_i)$ can be transformed into a ***single***
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$G(s)$, this isn't true viceversa.
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- Any particular $S(A_a, B_a, C_a, D_a)$ is called `realization`
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- $det(sI - A)$ := Characteristic Polinome
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- $det(sI - A) = 0$ := Characteristic Equation
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- $eig(A)$ := Solutions of the Characteristic Equation and `poles` of the system
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- If the system is `SISO` and this means that $C \in \R^{1,x}$,
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$B \in \R^{x,1}$ and $D \in \R$, meaning
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that:
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$$
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\begin{align*}
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G(s) &= \left(C\left(sI - A\right)^{-1}B + D\right) =\\
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&= \left(C \frac{Adj\left(sI - A\right)}{det\left(sI - A\right)}B + D\right)
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= && \text{Decompose the inverse in its formula}\\
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&= \frac{n(s)}{det\left(sI - A\right)} \in \R
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\end{align*}
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$$
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> [!NOTE]
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> As you can see here, by decomposing the inverse matrix in its formula it's
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> easy to see that the divisor is a `scalar`, a `number`.
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>
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> Moreover, because of how $B$ and $C$ are composed, the result of this Matrix
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> multiplication is a `scalar` too, hence we can write this as a single formula.
|
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>
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> Another thing to notice, regardless if this is a `MIMO` or `SISO` system is
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> that at the divisor we have all `eigenvalues` of A as `poles` by
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> [definition](../Formularies/GEOMETRY-FORMULARY.md/#eigenvalues)
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>
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### Transforming a State-Space into Another one
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We basically need to use some non singular `Permutation Matrices`:
|
||||
$$
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\begin{align*}
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&A_1, B_1, C_1, D_1 \\
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&A_2 = PAP^{-1} \\
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&B_2 = PB \\
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&C_2 = CP^{-1} \\
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&D_2 = D_1
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\end{align*}
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$$
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||||
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||||
[^matlab-c2d]: [Matlab Official Docs | c2d | 05 January 2025](https://it.mathworks.com/help/control/ref/dynamicsystem.c2d.html)
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67
docs/Chapters/3-CANONICAL-FORMS.md
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docs/Chapters/3-CANONICAL-FORMS.md
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# Canonical Forms
|
||||
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In order to see if we are in one of these canonical forms, just write the
|
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equations from the block diagram, and find the associated $S(A, B, C, D)$.
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||||
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||||
> [!TIP]
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||||
> 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
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> as many left places, starting from the rightmost, for the number of
|
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> associated $s$
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||||
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||||
## Control Canonical Form
|
||||
It is in such forms when:
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$$
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A = \begin{bmatrix}
|
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- a_1 & -a_2 & -a_3 & \dots & -a_{n-1} &-a_n\\
|
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1 & 0 & 0 & \dots & 0 & 0\\
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0 & 1 & 0 & \dots & 0 & 0\\
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\dots & \dots & \dots & \dots & \dots \\
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0 & 0 & 0 & \dots & 1 & 0
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||||
\end{bmatrix}
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||||
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B = \begin{bmatrix}
|
||||
1 \\ 0 \\ \dots \\ \dots \\ 0
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\end{bmatrix}
|
||||
|
||||
C = \begin{bmatrix}
|
||||
b_1 & b_2 & \dots & b_n
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||||
\end{bmatrix}
|
||||
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||||
D = \begin{bmatrix}
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||||
0
|
||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
## Modal Canonical Forms
|
||||
> [!CAUTION]
|
||||
> This form is the most difficult to find, as this varies drastically in cases
|
||||
> of double roots
|
||||
>
|
||||
$$
|
||||
A = \begin{bmatrix}
|
||||
- a_1 & 0 & 0 & \dots & 0\\
|
||||
0 & -a_2 & 0 & \dots & 0\\
|
||||
0 & 0 & -a_3 & \dots & 0\\
|
||||
\dots & \dots & \dots & \dots & \dots \\
|
||||
0 & 0 & 0 & 0 & -a_n
|
||||
\end{bmatrix}
|
||||
|
||||
B = \begin{bmatrix}
|
||||
1 \\ 1 \\ \dots \\ \dots \\ 1
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||||
\end{bmatrix}
|
||||
|
||||
C = \begin{bmatrix}
|
||||
b_1 & b_2 & \dots & b_n
|
||||
\end{bmatrix}
|
||||
|
||||
D = \begin{bmatrix}
|
||||
0
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||||
\end{bmatrix}
|
||||
$$
|
||||
|
||||
## 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)
|
||||
140
docs/Chapters/4-REACHABILITY-AND-OBSERVABILITY.md
Normal file
140
docs/Chapters/4-REACHABILITY-AND-OBSERVABILITY.md
Normal file
@@ -0,0 +1,140 @@
|
||||
# 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
|
||||
$X = X_no \bigoplus X_{no}^{\perp} = X_no \bigoplus X_{o}$
|
||||
104
docs/Chapters/5-STATE-FEEDBACK.md
Normal file
104
docs/Chapters/5-STATE-FEEDBACK.md
Normal file
@@ -0,0 +1,104 @@
|
||||
# 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.
|
||||
|
||||
|
||||
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
|
||||
0
docs/Chapters/CONGESTION-AVOIDANCE.md
Normal file
0
docs/Chapters/CONGESTION-AVOIDANCE.md
Normal file
0
docs/Chapters/CRYPTOGRAPHY-WITH-STATE-OBSERVER.md
Normal file
0
docs/Chapters/CRYPTOGRAPHY-WITH-STATE-OBSERVER.md
Normal file
132
docs/Chapters/Examples/EXAMPLE-3.md
Normal file
132
docs/Chapters/Examples/EXAMPLE-3.md
Normal file
@@ -0,0 +1,132 @@
|
||||
# Example 3
|
||||
|
||||
## Double Mass Cart
|
||||
|
||||
|
||||
|
||||
### 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
|
||||
|
||||
|
||||
|
||||
### 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 $S(A, B, C, D)$:
|
||||
$$
|
||||
A = \begin{bmatrix}
|
||||
0 & 1\\
|
||||
-\frac{k}{m} & - \frac{b}{m}
|
||||
\end{bmatrix}
|
||||
|
||||
B = \begin{bmatrix}
|
||||
0 \\
|
||||
\frac{k + sb}{m}
|
||||
\end{bmatrix}
|
||||
|
||||
C = \begin{bmatrix}
|
||||
1 & 0
|
||||
\end{bmatrix}
|
||||
|
||||
D = \begin{bmatrix}
|
||||
0 & 0
|
||||
\end{bmatrix}
|
||||
$$
|
||||
0
docs/Chapters/FEEDBACK-LINEARIZATION.md
Normal file
0
docs/Chapters/FEEDBACK-LINEARIZATION.md
Normal file
0
docs/Chapters/KALLMAN-FILTER.md
Normal file
0
docs/Chapters/KALLMAN-FILTER.md
Normal file
@@ -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:
|
||||
|
||||
|
||||
---
|
||||
Discrete time:
|
||||
|
||||
|
||||
|
||||
### 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
|
||||
0
docs/Chapters/SMITH-PREDICTOR.md
Normal file
0
docs/Chapters/SMITH-PREDICTOR.md
Normal file
0
docs/Chapters/VIDEO-STREAMING.md
Normal file
0
docs/Chapters/VIDEO-STREAMING.md
Normal file
Reference in New Issue
Block a user