3B SE

SE380: Introduction to Feedback Control

Concepts

• State Space Representation
• Linearization (Control Systems)
• Laplacian Matrix

Steps

Step 1: Study the system (inputs / outputs, actuators / sensors) Step 2: Define control specifications Step 3: Obtain a model of the system (first principles, data) Step 4: Simplify the model (ex: linearization) Step 5: Design the controller Step 6: Simulate the controlled system using the model

Notes are really hard to type out.

Lecture Notes

Summary here

The

09-16-2024

With the linearized equation, we have

$\{\begin{array}{l}\dot{x}=Ax+Bu\\ y=Cx+Du\end{array}$

Taking the Laplace transform, we get

$\{\begin{array}{l}sX(s)-x(0)=AX(s)+BU(s)\\ Y(s)=CX(s)+DU(s)\end{array}$

Solving for $X(s)$:

sIX(s) - AX(s) = BU(s) \\ X(s) = (sI - A)^{-1} B U(s) \end{align}$$ An inverse exists when the determinant is non-zero, i.e. $det(sI -A)) \neq 0$ Solving for $Y(s)$ after plugging $X(s)$: $$\begin{align} Y(s) = C (sI - A)^{-1} B U(s) + D U(s)\\ Y(s) = (C (sI - A)^{-1} B + D) U(s) \end{align}

$Y(s)=G(s)U(s)$ $G_{ij}(s)$ represents the influence of $U_j(s)$ on $Y_i(s)$.

Compute $y(t)$ response to $u(t)$.

1. $U(S)=L(u(t))$
2. $Y(s)=G(s)U(s)$
3. $y(t)=L^{-1}(Y(s))$

For single input single output systems (SISO):

Definition: A transfer function $G(s)$ is real rational if $G(s)=\frac{b_m{s}^m+\cdots +b_{1}s+b_0}{s^n+a_{n-1}{s}^{n-1}+\cdots +a_0}{a}_i,b_i\in \mathbb{R}$

$G(s)$ is proper if ${\lim }_{s\to \infty }G(s)$ exists. $\to$ $n\ge m$ $G(s)$ is strictly proper if ${\lim }_{s\to \infty }G(s)=0$ $\to$ $n>m$

$p\in \mathbb{C}$ is a pole of $G(s)$ if ${\lim }_{s\to p}|G(s)|=\infty$ $\to$ root of denominator $z\in \mathbb{C}$ is a zero of $G(s)$ if ${\lim }_{s\to z}|G(s)|=0$ $\to$ root of denominator

2024-09-18

\begin{align} G(s) &= \frac{Y(s)}{U(s)} \\ &= C (sI - A)^{-1} B + D \\ &=[1 \quad 0] \left(s \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix}0 & 1 \\ \frac{-g}{l} & 0 \end{bmatrix} \right)^{-1} \begin{bmatrix}0 \\ 1 \end{bmatrix} + 0 \\ &=[1 \quad 0] \left(\begin{bmatrix}s & -1 \\ \frac{g}{l} & 1 \end{bmatrix} \right)^{-1} \begin{bmatrix}0 \\ 1 \end{bmatrix}\\ &= \frac{1}{s^2 + \frac{g}{l}} \end{align}$$ $$G(s)= \frac{1}{s^2 + \frac{g}{l}}$$ m = 0, n = 2 It is both proper and strictly proper. We have no zeros, poles = $\pm J \sqrt{\frac{g}{l}}$ Example 2 $$\begin{align} &\dot{x}_{ts} = u_{ts} \\ &\dot{x}_{em} = u_{em} \\ &\dot{x}_i = x_{ts} - x_i, i.e. \{1 \dots 5\} \\ &\dot{x}_j = x_{em} - x_j, j \in \{6 \dots 8\} \\ \end{align}

$\dot{x}=\left[\begin{array}{c}{\dot{x}}_{ts}\\ {\dot{x}}_{em}\\ {\dot{x}}_1\\ \dots \\ {\dot{x}}_6\end{array}\right]=-Lx+\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\\ \dots & \dots \\ 0& 0\end{array}\right]u$

$y=\frac{1}{N}{1}^{T}x$

2024-09-20

Dropping Deltas

\dot{x} = u \\ y = x \\ \end{cases}

$u=\delta u+s$

$\frac{Y(s)}{U(s)}=G(s)=C(sI-A{)}^{-1}B+D=1(s{)}^{-1}+0=\frac{1}{s}$

$Y(s)=\frac{1}{s}U(s)$

y(t) &= \int_0^t u(\tau) d\tau \\ &= \int_0^t (u (\tau) - s(\tau)) d\tau \\ \end{align}$$ $$y(t) = L^{-1} \left\{ \sum_{i=1}^n \frac{A_i}{s - p_i} + \frac{Q}{s-jw} + \frac{\overline{Q}}{s+jw} \right\}$$ Assuming $Re(p_i) < 0$, then we have $A_i e ^{p_i t} = 0$ as $t \rightarrow \infty$. $$y(t) = L^{-1} \left\{\frac{Q}{s-jw} + \frac{\overline{Q}}{s+jw} \right\}$$ #### 2024-11-22 $$\begin{cases} \dot{x}_1 = x_2 \\ \dot{x_2} = u + x_2 \\ y = x_1\end{cases}$$