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raccourcis clavier

Tout ce qu'il faut savoir sur la conception des systèmes de contrôle

Note

sp.Heaviside(t) is $u(t)$

snippets

import sympy
import sympy as sp
from symbol import symbols, apart, inverse_laplace_transform, simplify
from sympy.abc import s, t

Frequency domain

Common Laplace transform

Laplace Theorem

$n^{th}$ order linear, time-invariant (LTI) differential equation:

a_n \frac{d^n c(t)}{dt^n} + a_{n-1} \frac{d^{n-1} c(t)}{dt^{n-1}} + \cdots + a_0 c(t) = b_m \frac{d^m r(t)}{dt^m} + b_{m-1} \frac{d^{m-1} r(t)}{dt^{m-1}} + \cdots + b_0 r(t)

takes Laplace transform from both side

\begin{aligned} & a_n s^n C(s) + a_{n-1} s^{n-1} C(s) + \cdots + a_0 C(s) \text{ and init terms for } c(t) \\ & = b_m s^m R(s) + b_{m-1} s^{m-1} R(s) + \cdots + b_0 R(s) \text{ and init terms for } r(t) \\ \end{aligned}

assume initial conditions are zero

\begin{aligned} (a_n s^n + a_{n-1} s^{n-1} + \cdots + a_0)C(s) &= (b_m s^m + b_{m-1} s^{m-1} + \cdots + b_0)R(s) \\[8pt] \frac{C(s)}{R(s)} &= G(s) = \frac{b_m s^m + b_{m-1} s^{m-1} + \cdots + b_0}{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_0} \end{aligned}

Transfer function

$G(s)=\frac{C(s)}{R(s)}$

Q: $G(s) = \frac{1}{S+2}$ . Input: $u(t)$ . What is $y(t)$ ?

\begin{aligned} Y(s) &= G(s)\cdot u(s) \rightarrow Y(s)=\frac{1}{s(s+2)} = \frac{A}{s} + \frac{B}{s+2} = \frac{1}{2\cdot{s}} - \frac{1}{2\cdot{(s+2)}} \\ y(t) &= -\frac{1}{2}(1-e^{-2t})u(t) \end{aligned}

Transfer function with feedback is under form

\frac{G(s)}{1+G(s)H(s)}

\begin{align*} \dot{x} &= Ax + Bu \\ y &= Cx + Du \end{align*}

Given

G(s) = \frac{\sum_{i=1}^{n-1}b_is^i + b_{0}}{s^n + \sum_{i=1}^{n-1}a_is^{i} + a_{0}} = \frac{Y(s)}{U(s)}

We get controller canonical state space form:

\begin{aligned} \dot{x}(t) &= \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\\ 0 & 0 & 1 & \cdots & 0 & 0 \\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\\ 0 & 0 & 0 & \cdots & 1 & 0 \\\ 0 & 0 & 0 & \cdots & 0 & 1 \\\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-2} & -a_{n-1} \end{bmatrix} x(t) + \begin{bmatrix} 0 \\\ 0 \\\ \vdots \\\ 0 \\\ 0 \\\ 1 \end{bmatrix} u(t) \\\ y(t) &= \begin{bmatrix} b_0 & b_1 & \cdots & b_{n-2} & b_{n-1} \end{bmatrix} x(t). \end{aligned}

We get observer canonical state space form:

\begin{aligned} \dot{x}(t) &= \begin{bmatrix} -a_{n-1} & 1 & 0 & \cdots & 0 & 0 \\ -a_{n-2} & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -a_2 & 0 & 0 & \cdots & 1 & 0 \\ -a_1 & 0 & 0 & \cdots & 0 & 1 \\ -a_0 & 0 & 0 & \cdots & 0 & 0 \end{bmatrix} x(t) + \begin{bmatrix} b_{n-1} \\ b_{n-2} \\ \vdots \\ b_2 \\ b_1 \\ b_0 \end{bmatrix} u(t) \\ y(t) &= \begin{bmatrix} 1 & 0 & \cdots & 0 & 0 \end{bmatrix} x(t). \end{aligned}

See this for applications

to have all roots in open left hand plane is to have all coefficients of polynomial to be present and have same sign.

Tip

To find transfer function for a system given a step response graph, *look for time over around 63% of the final value$

Closed-loop transfer function

$T(s) = \frac{G(s)}{1+G(s)}$

\%OS = e^{\zeta \pi / \sqrt{1-\zeta^2}} \times 100 \%

If a unity feedback system has a feedforward transfer function $G(s)$ then transfer function $\frac{E(s)}{R(s)}$ can be derived as:

\begin{aligned} C(s) &= E(s)\cdot G(s) \\\ E(s) &= R(s) - C(s) \end{aligned}

For $G(s) = K$ we get $\frac{E(s)}{R(s)} = \frac{1}{1+G(s)}$

Closed-loop system characteristic equation

det(SI - (A-BK))

Closed loop pole exists when

1+KG(s)H(s) = 0

Nyqust frequency:

f_N = \frac{1}{2}f_s

Set the third pole to s=-2 to cancel a zero as third pole.

Note

sp.Heaviside(t) is $u(t)$

snippets

import sympy
import sympy as sp
from symbol import symbols, apart, inverse_laplace_transform, simplify
from sympy.abc import s, t