Models

Control Engineering with Python

Symbols

🐍	Code	🔍	Worked Example
📈	Graph	🧩	Exercise
🏷️	Definition	💻	Numerical Method
💎	Theorem	🧮	Analytical Method
📝	Remark	🧠	Theory
ℹ️	Information	🗝️	Hint
⚠️	Warning	🔓	Solution

🐍 Imports

from numpy import *
from numpy.linalg import *
from matplotlib.pyplot import *

🏷️ Ordinary Differential Equation (ODE)

The “simple” version:

\[ \dot{x} = f(x) \]

where:

State: \(x \in \mathbb{R}^n\)
State space: \(\mathbb{R}^n\)
Vector field: \(f:\mathbb{R}^n \to \mathbb{R}^n\).

🏷️ Vector Field

Visualize \(f(x)\) as an arrow with origin the point \(x\).
Visualize \(f\) as a field of such arrows.
In the plane (\(n=2\)), use quiver from Matplotlib.

🐍 Helper

We define a Q function helper whose arguments are

f: the vector field (a function)
xs, ys: the coordinates (two 1d arrays)

and which returns:

the tuple of arguments expected by quiver.

def Q(f, xs, ys):
    X, Y = meshgrid(xs, ys)
    fx = vectorize(lambda x, y: f([x, y])[0])
    fy = vectorize(lambda x, y: f([x, y])[1])
    return X, Y, fx(X, Y), fy(X, Y)

🔍 Rotation Vector Field

Consider \(f(x,y) = (-y, x).\)

def f(xy):
    x, y = xy
    return array([-y, x])

📈 Vector Field

figure()
x = y = linspace(-1.0, 1.0, 20)
ticks = [-1.0, 0.0, 1.0]
xticks(ticks); yticks(ticks)
gca().set_aspect(1.0)
quiver(*Q(f, x, y))

🏷️ ODE Solution

A solution of \(\dot{x} = f(x)\) is

a (continuously) differentiable function \(x:I \to \mathbb{R}^n,\!\)
defined on a (possibly unbounded) interval \(I\) of \(\mathbb{R}\),
such that for every \(t \in I,\)

\[\dot{x}(t) = dx(t)/dt = f(x(t)).\]

📈 Stream Plot

When \(n=2\), represent a diverse set of solutions in the state space with streamplot

figure()
x = y = linspace(-1.0, 1.0, 20)
gca().set_aspect(1.0)
streamplot(*Q(f, x, y), color="k")

🏷️ Initial Value Problem (IVP)

Solutions \(x(t)\), for \(t\geq t_0\), of

\[ \dot{x} = f(x) \]

such that

\[ x(t_0) = x_0 \in \mathbb{R}^n. \]

🏷️

The initial condition \((t_0, x_0)\) is made of

the initial time \(t_0 \in \mathbb{R}\) and
the initial value or initial state \(x_0 \in \mathbb{R}^n\).

The point \(x(t)\) is the state at time \(t\).

🏷️ Higher-Order ODEs

(Scalar) differential equations whose structure is

\[ y^{(n)}(t) = g(y, \dot{y}, \ddot{y}, \dots, y^{(n-1)}) \]

where \(n > 1\).

💎 Higher-Order ODEs

The previous \(n\)-th order ODE is equivalent to the first-order ODE

\[ \dot{x} = f(x), \, x \in \mathbb{R}^n \]

with

\[ f(y_0, \dots, y_{n-2}, y_{n-1}) := (y_1, \dots, y_{n-1}, g(y_0, \dots, y_{n-1})). \]

🗝️

The result is more obvious if we expand the first-order equation:

\[ \begin{array}{ccl} \dot{y}_0 &=& y_1 \\ \dot{y}_1 &=& y_2 \\ \vdots &\vdots& \vdots \\ \dot{y}_n &=& g(y_0, y_1, \dots, y_{n-1}) \end{array} \]

🧩 Pendulum

1. 🧠 🧮

Establish the equations governing the pendulum dynamics.

2. 🧠 🧮

Generalize the dynamics when there is a friction torque \(c = -b \dot{\theta}\) for some \(b \geq 0\).

We denote \(\omega\) the pendulum angular velocity:

\[\omega := \dot{\theta}.\]

3. 🧠 🧮

Transform the dynamics into a first-order ODE with state \(x = (\theta, \omega)\).

4. 📈

Draw the system stream plot when \(m=1\), \(\ell=1\), \(g=9.81\) and \(b=0\).

5. 🧠 🧮

Determine least possible angular velocity \(\omega_0 > 0\) such that when \(\theta(0) = 0\) and \(\dot{\theta}(0) = \omega_0\), the pendulum reaches (or overshoots) \(\theta(t) = \pi\) for some \(t>0\).

🔓 Pendulum

1. 🔓

The pendulum total mechanical energy \(E\) is the sum of its kinetic energy \(K\) and its potential energy \(V\):

\[ E = K + V. \]

The kinetic energy depends on the mass velocity \(v\):

\[ K = \frac{1}{2} m v^2 = \frac{1}{2} m \ell^2 \dot{\theta}^2 \]

The potential energy mass depends on the pendulum elevation \(y\). If we set the reference \(y=0\) when the pendulum is horizontal, we have

\[ V = mg y = - mg \ell \cos \theta \]

\[ \Rightarrow \; E = K+V = \frac{1}{2} m \ell^2 \dot{\theta}^2 - mg \ell \cos \theta. \]

If the system evolves without any energy dissipation,

\[ \begin{split} \dot{E} &= \frac{d}{dt} \left(\frac{1}{2} m \ell^2 \dot{\theta}^2 - mg \ell \cos \theta\right) \\ &= m \ell^2 \dot{\theta}\ddot{\theta} + m g \ell (\sin \theta) \dot{\theta} \\&= 0 \end{split} \]

\[ \Rightarrow \; m \ell^2 \ddot{\theta} + m g \ell \sin \theta = 0. \]

2. 🔓

When there is an additional dissipative torque \(c=-b\theta\), we have instead

\[ \dot{E} = c \dot{\theta} = - b\dot{\theta}^2 \]

and thus

\[ m \ell^2 \ddot{\theta} + b \dot{\theta} + m g \ell \sin \theta = 0. \]

3. 🔓

With \(\omega := \dot{\theta}\), the dynamics becomes

\[ \begin{array}{lll} \dot{\theta} &=& \omega \\ \dot{\omega} &=& - (b/m\ell^2) \omega -(g /\ell) \sin \theta \end{array} \]

4. 🔓

m=1.0; b=0.0; l=1.0; g=9.81
def f(theta_d_theta):
    theta, d_theta = theta_d_theta
    J = m * l * l
    d2_theta  = - b / J * d_theta
    d2_theta += - g / l * sin(theta)
    return array([d_theta, d2_theta])

📈

figure()
theta = linspace(-1.5 * pi, 1.5 * pi, 100)
d_theta = linspace(-5.0, 5.0, 100)
labels =  [r"$-\pi$", "$0$", r"$\pi$"]
xticks([-pi, 0, pi], labels)
yticks([-5, 0, 5])
streamplot(*Q(f, theta, d_theta), color="k")

5. 🔓

In the top vertical configuration, the total mechanical energy of the pendulum is

\[ E_{\top} = \frac{1}{2} m \ell^2 \dot{\theta}^2 - mg \ell \cos \pi = \frac{1}{2} m \ell^2 \dot{\theta}^2 + mg \ell. \]

Hence we have at least \(E_{\top} \geq mg \ell\).

On the other hand, in the bottom configuration,

\[ E_{\bot} = \frac{1}{2} m \ell^2 \dot{\theta}^2 - mg \ell \cos 0 = \frac{1}{2} m \ell^2 \dot{\theta}^2 - mg \ell. \]

Hence, without any loss of energy, the initial velocity must satisfy \(E_{\bot} \geq E_{\top}\) for the mass to reach the top position.

That is

\[ E_{\bot} = \frac{1}{2} m \ell^2 \dot{\theta}^2 - mg \ell \geq mg \ell = E_{\top} \]

which leads to:

\[ |\dot{\theta}| \geq 2 \sqrt{\frac{g}{\ell}}. \]