% Controllability % 👤 Sébastien Boisgérault

Control Engineering with Python

📖 Documents (GitHub)
©️ License CC BY 4.0
🏦 Mines ParisTech, PSL University

Symbols


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

🐍 Imports

::: slides :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

from numpy import *
from numpy.linalg import *
from numpy.testing import *
from scipy.integrate import *
from scipy.linalg import *
from matplotlib.pyplot import *

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::: notebook :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

from numpy import *
import matplotlib; matplotlib.use("nbAgg")
%matplotlib notebook
from matplotlib.pyplot import *

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::: hidden :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

# Python 3.x Standard Library
import gc
import os

# Third-Party Packages
import numpy as np; np.seterr(all="ignore")
import numpy.linalg as la
import scipy.misc
import matplotlib as mpl; mpl.use("Agg")
import matplotlib.pyplot as pp
import matplotlib.axes as ax
import matplotlib.patches as pa


#
# Matplotlib Configuration & Helper Functions
# --------------------------------------------------------------------------

# TODO: also reconsider line width and markersize stuff "for the web
#       settings".
fontsize = 10

width = 345 / 72.27
height = width / (16/9)

rc = {
    "text.usetex": True,
    "pgf.preamble": r"\usepackage{amsmath,amsfonts,amssymb}",
    #"font.family": "serif",
    "font.serif": [],
    #"font.sans-serif": [],
    "legend.fontsize": fontsize,
    "axes.titlesize":  fontsize,
    "axes.labelsize":  fontsize,
    "xtick.labelsize": fontsize,
    "ytick.labelsize": fontsize,
    "figure.max_open_warning": 100,
    #"savefig.dpi": 300,
    #"figure.dpi": 300,
    "figure.figsize": [width, height],
    "lines.linewidth": 1.0,
}
mpl.rcParams.update(rc)

# Web target: 160 / 9 inches (that's ~45 cm, this is huge) at 90 dpi
# (the "standard" dpi for Web computations) gives 1600 px.
width_in = 160 / 9

def save(name, **options):
    cwd = os.getcwd()
    root = os.path.dirname(os.path.realpath(__file__))
    os.chdir(root)
    pp.savefig(name + ".svg", **options)
    os.chdir(cwd)

def set_ratio(ratio=1.0, bottom=0.1, top=0.1, left=0.1, right=0.1):
    height_in = (1.0 - left - right)/(1.0 - bottom - top) * width_in / ratio
    pp.gcf().set_size_inches((width_in, height_in))
    pp.gcf().subplots_adjust(bottom=bottom, top=1.0-top, left=left, right=1.0-right)

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🏷️ Controllability

The system $\dot{x} = f(x,u)$ is controllable if

for any $t_0 \in \mathbb{R}$, $x_0 \in \mathbb{R}^n$ and $x_f \in \mathbb{R}^n$,
there are $t_f > t_0$ and $u: [t_0, t_f] \to \mathbb{R}^m$ such that
the solution $x(t)$ such that $x(t_0)=x_0$ satisfies

$$ x(t_f) = x_f. $$

🏷️ Admissible Trajectory

Let $x_r$ be a reference trajectory of the system state: $$ x_r(t), , t\in[t_0, t_f]. $$

It is admissible if there is a function $u_r(t)$, $t \in [t_0, t_f]$ such that the solution $x$ of the IVP $$ \dot{x} = f(x, u_r), ; x(t_0) = x_r(t_0) $$ is the function $x_r$.

🔍 Car

The position $d$ (in meters) of a car of mass $m$ (in kg) on a straight road is governed by

$$ m \ddot{d} = u $$

where $u$ the force (in Newtons) generated by its motor.

The car is initially at the origin of a road and motionless.

We would like to cross the end of the road (location $d_f > 0$) at time $t_f > 0$ and speed $v_f$.

Numerical values:

$m=1500 , \mbox{kg}$,
$t_f=10 , \mbox{s}$, $d_f=100 , \mbox{m}$ and $v_f=100 , \mbox{km/h}$.

Step 1 -- Trajectory Planning

We search for a reference trajectory for the state

$$x_r(t) = (d_r(t), \dot{d}_r(t))$$

such that:

$d_r(0)=0$, $\dot{d}_r(0) = 0$,
$d_r(t_f) = x_f$, $\dot{d}_r(t_f) = v_f$.

Step 2 -- Admissibility

We check that this reference trajectory is admissible, i.e. that we can find a control $u_r(t)$ such that the solution of the IVP is $x(t) = x_r(t)$ when $x(0) = x_r(t)$.

Admissible Trajectory

Here, if $d_r$ is smooth and if we apply the control $u(t) = m\ddot{d}_r(t)$,

$$ m \frac{d^2}{dt^2} (d-d_r) = 0, $$ $$ (d-d_r)(0) = 0, ; \frac{d}{dt} (d-d_r)(0)= 0. $$

Thus, $d(t) = d_r(t)$ -- and thus $\dot{d}(t) = \dot{d}_r(t)$ -- for every $t\geq 0$.

Reference Trajectory

We can find $d_r$ as a third-order polynomial in $t$

$$ d_r(t) = \alpha t^3 + \beta t^2 + \gamma t + \delta $$

with

$$ \alpha = \frac{v_f}{t_f^2}- 2\frac{d_f}{t_f^3}, ; \beta = 3\frac{d_f}{t_f^2} - \frac{v_f}{tf},; \gamma=0, ; \delta=0. $$

(equivalently, with $u(t)$ as an affine function of $t$).

🐍 Constants

m = 1500.0
xf = 100.0
vf = 100.0 * 1000 / 3600 # m/s
tf = 10.0
alpha = vf/tf**2 - 2*xf/tf**3
beta = 3*xf/tf**2 - vf/tf

🐍 State & Input Evolution

def x(t):
    return alpha * t**3 + beta * t**2
def d2_x(t):
    return 6 * alpha * t + 2 * beta
def u(t):
    return m * d2_x(t)

🐍 💻 Simulation

y0 = [0.0, 0.0]
def fun(t, y):
    x, d_x = y
    d2_x = u(t) / m
    return [d_x, d2_x]
result = solve_ivp(
  fun, [0.0, tf], y0, dense_output=True
)

📊 Graph of the Distance

figure()
t = linspace(0, tf, 1000)
xt = result["sol"](t)[0]
plot(t, xt)
grid(True); xlabel("$t$"); title("$d(t)$")

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tight_layout()
save("images/car")

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::: slides :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

{.section data-background="images/car.svg" data-background-size="contain"}

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📊 Graph of the velocity

figure()
vt = result["sol"](t)[1]
plot(t, 3.6 * vt)
grid(True); xlabel("$t$")
title(r"$\dot{d}(t)$ km/h")

::: hidden :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

tight_layout()
save("images/car-speed")

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::: slides :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

{.section data-background="images/car-speed.svg" data-background-size="contain"}

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🧩 Non-admissible trajectory

Let $\dot{x} = A x + B u$ with $x\in \mathbb{R}^2$, $u \in \mathbb{R}$, $$ A = \left[ \begin{array}{cc} 0 & 1 \ 0 & 0 \end{array} \right], ; B = \left[ \begin{array}{cc} 0 \ 1 \end{array} \right]. $$

1. 🧠

Find a smooth reference trajectory $x_r(t)$, $t\in[0, 1]$ which is not admissible.

🔓 Non-admissible trajectory

1. 🔓

The first line of the vector equation $\dot{x} = A x + Bu$ is

$$ \dot{x}_1 = x_2. $$

Any trajectory $x_r$ that does not satisfy this equation is not admissible; for example

$$ x_r(t) := \left[ \begin{array}{c} 0 \\ 1 \end{array} \right]. $$

🧩 Pendulum

Consider the pendulum with dynamics:

$$ m \ell^2 \ddot{\theta} + b \dot{\theta} + mg \ell \sin \theta = u $$

1. 🧠 🧮

Find a smooth reference trajectory that leads the pendulum from the bottom configuration

$$ \theta(0)=0, ; \dot{\theta}(0) = 0 $$

to the top configuration

$$ \theta(t_f) = \pi, ; \dot{\theta}(t_f) = 0. $$

2. 🧠 🧮

Show that the reference trajectory is admissible and compute the corresponding input $u_r(t)$.

3. 💻 🧠

Simulate the result and visualize the solution.

What should theoretically happen at $t=t_f$ if $u(t)=0$ is applied when $t \geq t_f$? What does happen in reality ? Why ? How can we mitigate this issue?

Numerical Values:

$$ m = 1.0, , l = 1.0, , b = 0.1,, g = 9.81, , t_f = 10. $$

🔓 Pendulum

1. 🔓

Since the initial and final conditions form a set of 4 equations, we search for an admissible trajectory defined by a 3rd-order polynomial

$$ \theta_r(t) := a t_f^3 + bt_f^2 + c t_f +d $$

with unknown coefficients $a, b, c, d$.

The initial conditions $\theta_r(0)=0$ and $\dot{\theta}_r(0)=0$ are equivalent to

$$ b = 0, ; c= 0. $$

The final condition $\theta_r(t_f) = \pi$ is equivalent to

$$ at_f^3 + bt_f^2 = \pi $$

and $\dot{\theta}_r(t_f) = 0$ to

$$ 3a t_f^2 + 2b t_f = 0. $$

The latter equation can be transformed into

$$ b = -\frac{3}{2} a t_f $$

and the former becomes

$$ at_f^3 - \frac{3}{2} a t_f^3 = \pi ; \Leftrightarrow ; a = -\frac{2\pi}{t_f^3} $$

and thus $b = \frac{3\pi}{t_f^2}.$

Finally, the following trajectory $(\theta_r(t),\dot{\theta}_t(t))$ is smooth and meets the required initial and final conditions:

$$ \theta_r(t) = -2\pi \left(\frac{t}{t_f}\right)^3 + 3\pi \left(\frac{t}{t_f}\right)^2 $$

$$ \dot{\theta}_r(t) = -\frac{6\pi}{t_f} \left(\frac{t}{t_f}\right)^2 + \frac{6\pi}{t_f} \left(\frac{t}{t_f}\right) $$

2. 🔓

By construction our reference trajectory meets the initial condition. There is a unique input that makes the solution follows this reference; its is defined by

$$ u_r(t) := m \ell^2 \ddot{\theta}_r(t) + b \dot{\theta}_r(t) + mg \ell \sin \theta_r(t) $$

where

$$ \ddot{\theta}_r(t) = -\frac{12\pi}{t_f^2} \left(\frac{t}{t_f}\right) + \frac{6\pi}{t_f^2}. $$

3. 🔓

m = 1.0
l = 1.0
b = 0.1
g = 9.81 
t0, tf = 0.0, 10.0

def theta_r(t):
    s = t/tf
    return -2*pi*s**3 + 3*pi*s**2
def dtheta_r(t):
    s = t/tf
    return -6*pi/tf*s**2 + 6*pi/tf*s

def d2theta_r(t):
    s = t/tf
    return -12*pi/(tf*tf)*s + 6*pi/(tf*tf)
def u_r(t):
    return (m*l*l*d2theta_r(t) + 
            b*dtheta_r(t) + 
            m*g*l*sin(theta_r(t)))

def fun(t, theta_dtheta):
    theta, dtheta = theta_dtheta
    d2theta = ((-b*dtheta - m*g*l*sin(theta) + u_r(t)) 
               / (m * l * l))
    return dtheta, d2theta
t_span = [t0, tf]
y0 = [0.0, 0.0]

r = solve_ivp(fun, t_span, y0, dense_output=True)
t = linspace(t0, tf, 1000)
solt = r["sol"](t)
thetat, dthetat = solt

figure()
plot(t, theta_r(t), "k--", label=r"$\theta_r(t)$")
plot(t, thetat, label=r"$\theta(t)$")
yticks([0, 0.5*pi, pi], ["$0$", r"$\pi/2$", r"$\pi$"])
title(r"Simulation of $\theta(t)$")
grid(True); legend()

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tight_layout()
save("images/pendulum-miss")

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::: slides :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

{.section data-background="images/pendulum-miss.svg" data-background-size="contain"}

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Theoretically, we should have $\theta(t_f) = \pi$ and $\dot{\theta}(t_f)=0$, but the simulation is quite far from that.

Since the top equilibrium is unstable, small (numerical) errors in the state may cause large deviations of the trajectory.

We may reduce the simulation error thresholds to alleviate this problem.

rhp = solve_ivp(fun, t_span, y0, 
                dense_output=True, 
                rtol=1e-6, atol=1e-9)
t = linspace(t0, tf, 1000)
solt_hp = rhp["sol"](t)
thetat_hp, dthetat_hp = solt_hp

figure()
plot(t, theta_r(t), "k--", label=r"$\theta_r(t)$")
plot(t, thetat, label=r"$\theta(t)$  (standard)")
plot(t, thetat_hp, label=r"$\theta(t)$ (low error)")
yticks([0, 0.5*pi, pi], ["$0$", r"$\pi/2$", r"$\pi$"])
title(r"Simulation of $\theta(t)$")
grid(True); legend()

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tight_layout()
save("images/pendulum-miss-less")

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::: slides :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

{.section data-background="images/pendulum-miss-less.svg" data-background-size="contain"}

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💎 Controllability of LTI systems

A system $\dot{x} = Ax + Bu$ is controllable iff

from the origin $x_0 = 0$ at $t_0=0$,
we can reach any state $x_f \in \mathbb{R}^n$.

🏷️ 💎 Kalman Criterion

The system $\dot{x} = Ax+Bu$ ($x\in \mathbb{R}^n$) is controllable iff:

$$ \mathrm{rank} , \left[B, AB, \dots, A^{n-1} B\right] = n $$

$[B, \dots, A^{n-1}B]$ is the controllability matrix.

🔍 Controllability Matrix

$$ A = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right], ; B = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] $$

🐍 Computation

def KCM(A, B):
    n = shape(A)[0]
    mp = matrix_power
    cs = column_stack
    return cs([mp(A, k) @ B for k in range(n)])

🐍 LTI System

n = 3 
A = zeros((n, n))
for i in range(0, n-1):
    A[i,i+1] = 1.0

B = zeros((n, 1))
B[n-1, 0] = 1.0

🐍 Rank Condition

C = KCM(A, B)

C_expected = [[0, 0, 1], [0, 1, 0], [1, 0, 0]]
assert_almost_equal(C, C_expected)

assert matrix_rank(C) == n

⚠️ Warnings

This implementation of KCM is not optimized: fewer computations are achievable using the identity:

$$ A^n B = A \times (A^{n-1}B). $$
Rank computations are subject to (catastrophic) numerical errors; sensitivity analysis or symbolic computations may alleviate the problem.

🧩 Fully Actuated System

Consider $\dot{x} = A x + Bu$ with

$x \in \mathbb{R}^n$,
$u \in\mathbb{R}^m$ and,
$\mathrm{rank} , B = n$.

1. 🧠

Show that $m \geq n$.

2. 🧮

Is the system controllable ?

3. 🧠 🧮

Given $x_0$, $x_f$ and $t_f > 0$, show that any smooth trajectory that leads from $x_0$ at time $t_0$ to $x_f$ at time $t_f$ is admissible.

🔓 Fully Actuated System

1. 🔓

The shape of $B$ is $n \times m$, thus

$$ \mbox{rank} , B \leq \min(n, m). $$

By assumption $\mbox{rank} , B = n$, thus $n \leq m$.

2. 🔓

The system is controllable since

$$ \mbox{rank} , [B, AB, \dots] \leq \mbox{rank} , B = n. $$

3. 🔓

Since $\mbox{rank} , B = n$, matrix $B$ contains a $n \times n$ invertible matrix $R$. Let's assume for the sake of simplicity that $R$ is made of the first $n$ columns of $B$ and let $$ S: = \left[ \begin{array}{c} R^{-1} \ 0 \end{array} \right] \in \mathbb{R}^{m \times n}. $$

Then by construction $B \times S = I \in \mathbb{R}^{n\times n}$.

If we define $u$ as a function of some auxiliary control $v \in \mathbb{R}^n$ by $$ u(t) = S v(t), $$ then $$ \dot{x} = A x + v. $$

To join $x_0$ at $t_0$ and $x_f$ at $t_f$, we can apply the control $$ v(t) := -A x_r(t) + \dot{x}_r(t) $$ where $$ x_r(t) := x_0 + \frac{t-t_0}{t_f-t_0} (x_f - x_0). $$

Indeed, that leads to $$ \dot{x}(t) = A x(t) - A x_r(t) + \dot{x}_r(t), ; x(t_0) = x_r(t_0) $$ or if we denote $e := x - x_r$, $$ \dot{e}(t) = Ae(t), ; e(t_0) =0 $$ and thus yields $x(t) = x_r(t)$ for any $t\geq t_0$.

🧩 Integrator Chain

$$\dot{x}n = u, , \dot{x}{n-1} = x_n, , \cdots ,, \dot{x}_1 = x_2.$$

1. 🧠 🧮

Show that the system is controllable

🔓 Integrator Chain

1. 🔓

We have

$$ A = \left[ \begin{array}{c} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0 \\ 0 & 0 & \cdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \vdots & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right], ; B =\left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{array} \right] $$

Thus,

$$ B =\left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{array} \right], ; AB =\left[ \begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \end{array} \right], , \dots, , A^{n-1} B = \left[ \begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \\ 0 \end{array} \right]. $$

The controllability matrix has full rank and the system is controllable.

🧩 Heat Equation

$d T_1/dt = u + (T_2 - T_1)$
$d T_2/dt = (T_1 - T_2) + (T_3 - T_2)$
$d T_3/dt = (T_2 - T_3) + (T_4 - T_3)$
$d T_4/dt = (T_3 - T_4)$

1. 🧠 🧮

Show that the system is controllable.

2. 🧠 🧮

Assume now that the four cells are organized as a square.

Is the system still controllable?
Why?
How could you solve this problem?

🔓 Heat Equation

1. 🔓

We have

$$ A = \left[ \begin{array}{rrrr} -1 & 1 & 0 & 0 \\ 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 1 & -1 \end{array} \right], ; B = \left[ \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \right]. $$

$$ A B = \left[ \begin{array}{r} -1 \\ 1 \\ 0 \\ 0 \end{array} \right], ; A^2 B = \left[ \begin{array}{r} 2 \\ -3 \\ 1 \\ 0 \end{array} \right], ; A^3 B = \left[ \begin{array}{r} -5 \\ 9 \\ -5 \\ 1 \end{array} \right]. $$

The controllability matrix

$$ \left[ \begin{array}{rrrr} 1 & -1 & 2 & -5 \\ 0 & 1 & -3 & 9 \\ 0 & 0 & 1 & -5 \\ 0 & 0 & 0 & 1 \end{array} \right] $$

is full rank, thus the system is controllable.

2. 🔓

If the system is organized as a square

$$ A = \left[ \begin{array}{rrrr} -2 & 1 & 1 & 0 \\ 1 & -2 & 0 & 1 \\ 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & -2 \end{array} \right], ; B = \left[ \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \end{array} \right]. $$

The controllability matrix is

$$ \left[ \begin{array}{rrrr} 1 & -2 & 6 & -20 \\ 0 & 1 & -4 & 16 \\ 0 & 1 & -4 & 16 \\ 0 & 0 & 2 & -12 \end{array} \right] $$

It is not full rank since the second and the third rows are equal. Thus the system is not controllable.

The lack of controllability is due to symmetry.

If the initial temperature in cell 2 and 3 are equal, since they play symmetric role in the system, their temperature will be equal for any subsequent time.

Hence, it will be impossible to reach a state with different temperature in cell 2 and 3, no matter what the input is.

To break this symmetry and restore controllability, we may for example try to add a second independent heat source sink in cell 2.

This leads to

$$ A = \left[ \begin{array}{rrrr} -2 & 1 & 1 & 0 \\ 1 & -2 & 0 & 1 \\ 1 & 0 & -2 & 1 \\ 0 & 1 & 1 & -2 \end{array} \right], ; B = \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 0 & 0\\ 0 & 0 \end{array} \right]. $$

and the controllability matrix

$$ \left[ \begin{array}{rrrr} 1 & 0 & -2 & 1 & 6 & -4 & -20 & 16 \\ 0 & 1 & 1 & -2 & -4 & 6 & 16 & -20\\ 0 & 0 & 1 & 0 & -4 & 2 & 16 & -12 \\ 0 & 0 & 0 & 1 & 2 & -4 & -12 & 16 \end{array} \right] $$

which has a full-rank.

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Control Engineering with Python

Symbols

🐍 Imports

🏷️ Controllability

🏷️ Admissible Trajectory

🔍 Car

Step 1 -- Trajectory Planning

Step 2 -- Admissibility

Admissible Trajectory

Reference Trajectory

🐍 Constants

🐍 State & Input Evolution

🐍 💻 Simulation

📊 Graph of the Distance

{.section data-background="images/car.svg" data-background-size="contain"}

📊 Graph of the velocity

{.section data-background="images/car-speed.svg" data-background-size="contain"}

🧩 Non-admissible trajectory

1. 🧠

🔓 Non-admissible trajectory

1. 🔓

🧩 Pendulum

1. 🧠 🧮

2. 🧠 🧮

3. 💻 🧠

🔓 Pendulum

1. 🔓

2. 🔓

3. 🔓

{.section data-background="images/pendulum-miss.svg" data-background-size="contain"}

{.section data-background="images/pendulum-miss-less.svg" data-background-size="contain"}

💎 Controllability of LTI systems

🏷️ 💎 Kalman Criterion

🔍 Controllability Matrix

🐍 Computation

🐍 LTI System

🐍 Rank Condition

⚠️ Warnings

🧩 Fully Actuated System

1. 🧠

2. 🧮

3. 🧠 🧮

🔓 Fully Actuated System

1. 🔓

2. 🔓

3. 🔓

🧩 Integrator Chain

1. 🧠 🧮

🔓 Integrator Chain

1. 🔓

🧩 Heat Equation

1. 🧠 🧮

2. 🧠 🧮

🔓 Heat Equation

1. 🔓

2. 🔓