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\begin{document}
\title{Dynamic Modeling of an Inverted Pendulum Self-Transportation System}
\author[1]{Brianna Hillman}%
\affil[1]{University of Alabama}%
\vspace{-1em}
\date{\today}
\begingroup
\let\center\flushleft
\let\endcenter\endflushleft
\maketitle
\endgroup
\selectlanguage{english}
\begin{abstract}
There is a growing need for smaller, more economic transportation
systems. Personal transportation systems such as hover boards and
Segways have made their way onto the market. The two-wheeled
configuration of the Segway is a two-wheeled inverted pendulum system.
This paper derives the equations of motion of an inverted wheeled
pendulum system and a double inverted wheeled pendulum system. The
equations are derived using the Euler-Lagrange equations. The first
configuration derived is of the single inverted pendulum system. The
second configuration derived adds on another inverted pendulum system
similar to the first configuration, along with more variables and
constraints. The results of this paper provide valuable information
about the equations of motion of a Segway. The methods described in this
paper lay the groundwork for numerous possible future studies that may
investigate the motion of a two-wheeled pendulum system.~%
\end{abstract}%
\sloppy
\pagebreak
\section*{List of Symbols}
\begin{tabular}{cp{0.6\textwidth}}
$M_{w}$ & Mass of the wheel \\
$L_{p}$ & Length of the Pendulum \\
$r_{g}$ & Radius of the wheel \\
$K_{g}$ & Internal gear ratio \\
$K_{t}$ & Motor torque constant \\
$K_{m}$ & Back emf constant constant \\
$V_{in}$ & Motor input velocity range \\
$R$ & Motor armature resistance \\
$g$ & Gravitational constant of Earth \\
$c$ & Viscous friction of motor shaft and wheel\\
$G$ & Elastic modulus of beam \\
$J$ & Second moment of area of beam \\
$L$ & Motor armature inductance \\
$t_i , t_f$ & Initial time and final time \\
\end{tabular}\\
\section*{Introduction}
Small electric self-transportation systems were thought to become a more popular transportation system throughout all the world. The Segway, or the commercial version of the two-wheeled inverted pendulum system, has achieved great interest in the past few years. Developed from the self-balancing \textit{iBot} wheelchair, the Segway was birthed at the University of Plymouth by BAE Sytems and Sumitomo Precision Products \cite{wikipedia}.
These systems are ideal because of their reduced pollution and traffic volume. Self-transportation systems are also small enough for a single individual to use in much smaller spaces compared to the spaces of a small car. These systems can also be used for the elderly and disabled as a means of increased mobility. Although a fantastic idea, the complex dynamics of these type of systems make riding a hover board or Segway extremely difficult for someone with limited mobility. The control of a Segway is majorly affected by even small disturbances are presented. With that being said, it is advantageous to study the dynamics of a Segway's system in order to develop a system that is easier to control, especially for those with mobility issues.
The stabilization of a Segway is quite challenging. Variables to take into account when it comes to controlling the two-wheeled pendulum system include: the system is "run" by mechanical systes, a torque is applied to rotate each wheel, the position of the person on the system (although for this case, the dynamics of motion will be evaluated without a body present on the sytem), and the tilting angle of the system just to name a few. Because of the numerous variables to consider when discussing the motion of a two-wheeled inverted pendulum system, a geometric approach is also applied to help understand the way motion is generated.
The main objective of this report is to present the derivation of the equations of motion of a single wheeled inverted pendulum system, along with the equations of motion of a two-wheeled inverted pendulum system performed using the Euler-Lagrange equations of motion.
\section*{Theory}
The equations of motion used comes from Hamilton's extended principle. According to Kim, "Hamilton formulated a variational method for dynamics, based upon the concept of stationary action, with action represented as the integral over time of the Lagrangian of the system" \cite{Kim_2013}. Hamilton's principle is used vastly in the fields of mathematics and engineering. The (extended) Hamilton's Principle is shown below:
\begin{equation}
\int_{t_i}^{t_f} (\delta L + \delta W^{ext})\,dt = 0
\end{equation}
where \textit{L} represents the Lagrangian while \textit{$W^{ext}$} represents the non-conservative forces, such as friction for example, acting on the system. We can define the Lagragian as:
\begin{equation}
\label{eqn:drag}
L(\dot x, x, t) = T(\dot x, x, t) - U(\dot x, x, t)
\end{equation}
where the kinetic energy \textit{T} and the potential energy \textit{U} are given by
\begin{equation}
T(\dot x, x, t)= \frac{1}{2} m [\dot x(t)]^2
\end{equation}
\begin{equation}
U(\dot x, x, t)= mgh
\end{equation}
where \textit{m} is the mass of the object, \textit{g} is the gravitational constant and \textit{h} is the height of the object from "ground level". Consequently, the most general form of Lagrange's Equation(s) of motion is given by
\begin{equation}
\frac {d}{dt} ( \frac{\partial L}{\partial \dot q_i} ) - \frac{\partial L}{\partial q_i} = Q_i
\end{equation}
where $q_i$ are the time-dependent functions and $Q_i$ are the generalized forces acting on the system.
\subsection*{Configuration 1}
The simplest form on the type of system under investigation is the inverted pendulum model. It is beneficial to understand the dynamics of a pendulum balanced on a moving wheel to help better understand the dynamics of a Segway.
Suppose there is a uniform mass-less pendulum attached to a base. This base is attached to a wheel with mass $M_w$ and radius $r_g$. The base is free to rotate at an angle $\alpha$ with the ground \textit{Z} frame. The pendulum can rotate relative to the base at an angle of $\theta (t)$ defined by the equation
\begin{equation}
\theta(t)= \frac{P {L_p}^2}{2GJ}
\end{equation}
where \textit{G} is the elastic modulus of the beam, \textit{J} is the second moment of area of the beam and \textit{P} is the load acting on the end of the pendulum in the direction perpendicular to the auxiliary \textit{z} axis. The auxiliary axis \textit{z} is defined normal to the base. In this case, the beam is the pendulum. The equation for the load is given as
\begin{equation}
P(t) = {L_p}^2 Mg sin(\alpha (t))
\end{equation}
and thus, the equation for $\theta (t)$ becomes:
\begin{equation}
\theta(t)= \frac{M g {L_p}^2 sin(\alpha (t))}{2GJ}
\end{equation}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Config1/Config1}
\caption{{Inverted Pendulum System - Configuration 1
{\label{886255}}%
}}
\end{center}
\end{figure}
There is also a point mass \textit{M} attached to the end of the pendulum, shown in the figure above.The wheel is equipped with a DC motor connected to the wheel through a shaft; the gear ratio between the motor shaft and the wheel radius is $K_g$. When the motor runs, the shaft drives the wheel. As the wheel moves, it rotates an encoder wheel from which the angular position of the wheel $\beta (t)$ can be obtained. Another encoder in place where the base is mounted records the angular position of the base $\alpha (t)$. The wheel is assumed to have pure rolling contact with the ground; thus $x_c (t) = r_g \beta (t)$, or the displacement of the wheel. When the motor runs, the torque created at the output shaft is translated to a linear force \textit{F}, which results in the wheel's motion. The equivalent input force acting on the wheel and generated by the torque from the DC motor can be expressed as
\begin{equation}
F(t)= \frac{K_g K_t }{r_g} i(t)
\end{equation}
where \textit{i(t)} is the current flowing through the motor, $r_g$ is the radius of the wheel $K_g$ is the gear ratio between the wheel and motor gear and $K_t$ is the motor torque constant. If the motor armature inductance is assumed small, then the current can be related to the input voltage of the DC motor $V_{in}$ as shown below. $K_m$ is the back electromotive force (emf) constant. Also, note that the viscous friction of the motor shaft and wheel is denoted by \textit{c}. Appendix A lists the numerical value for all constants.
\begin{equation}
R i(t)+ \frac{K_g K_m\dot x(t) }{r_g} = V_{in}(t)
\end{equation}
To make the derivation less complicated, a new variable $\gamma$ is introduced.
\begin{equation}
\gamma(t)= \theta(t) + \alpha(t)
\end{equation}
This is due to the fact that we are only concerned with and/or only need one reference frame for the derivations of Configuration 1.
When solving the equations of motion, several factors of kinetic energy must be taken into effect. The kinetic energy will consist of a rotational term of the point mass about the wheel's center, a translational term for the point mass along the \textit{X} axis, a rotational term for the wheel about it's center and a translational term for the wheel along the \textit{X} axis.
The kinetic energy of the wheel is described as
\begin{equation}
T_w = \frac{1}{2} M_w {V_w}^2 + \frac{1}{2} I_w {\omega_w}^2
\end{equation}
where \textit{$V_w$} is the velocity of the wheel, \textit{$I_w$} is the inertia of the wheel and \textit{$\omega_w$} is the angular velocity of the wheel. The values for each of these is shown below. Note that the wheel has a no-slip condition and the angular velocity is directly dependent on linear velocity.
\begin{equation}
V_w = \dot x (t)
\end{equation}
\begin{equation}
I_w = \frac{1}{2} M_w {r_g}^2
\end{equation}
\begin{equation}
\omega_w = \dot \beta (t)
\end{equation}
Therefore, the kinetic energy of the wheel becomes
\begin{equation}
T_w = \frac{1}{2} M_w {\dot x(t)}^2 + \frac{1}{2}( \frac{1}{2} M_w {r_g}^2 ) {\frac{{\dot x(t)}^2}{{r_g}^2}}
\end{equation}
and reduces further to
\begin{equation}
T_w = \frac{3}{4} M_w {\dot x(t)}^2
\end{equation}
Both the translational and rotational terms of the wheel are included in the equations shown. Next, the kinetic energy of the point mass is denoted as
\begin{equation}
T_m = \frac{1}{2} M {V_m}^2
\end{equation}
The velocity of the point mass is denoted as $V_m$. No rotational term is included here because the mass is a point and and the pendulum itself has no mass. The velocity of the point mass can be derived from its position. The position of the point mass is given below, followed by it's velocity.
\begin{equation}
x_m = x_c (t) + L_p sin(\gamma (t))
\end{equation}
\begin{equation}
y_m = r_g + L_p cos(\gamma (t))
\end{equation}
\begin{equation}
{V_m}^2 = (\frac{d}{dt} (x_m))^2 + (\frac{d}{dt}(y_m))^2
\end{equation}
Thus, the kinetic energy of the point mass becomes
\begin{equation}
T_m = \frac{M (\dot x_c (t))^2}{2} + \frac{M L_p^2 (\dot \gamma (t))^2}{2} + M L_p \dot x_c (t) \dot \gamma (t) cos (\gamma (t))
\end{equation}
This results in a total kinetic energy of the system defined as:
\begin{equation}
T = \frac{3 M_w {\dot x(t)}^2}{4} + \frac{M (\dot x_c (t))^2}{2} + \frac{M L_p^2 (\dot \gamma (t))^2}{2} + M L_p \dot x_c (t) \dot \gamma (t) cos (\gamma (t))
\end{equation}
Only the point mass has a potential energy; therefore
\begin{equation}
U = M g (r_g + L_p cos(\gamma (t))
\end{equation}
\subsection*{Configuration 2}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Config2/Config2}
\caption{{X-Y motion of two-wheeled inverted pendulum system (left) and side view
of separate pendulums (right) - Configuration 2
{\label{984725}}%
}}
\end{center}
\end{figure}
The equations of motion of the Segway, or two-wheeled inverted pendulum system, are derived from the detailing shown in the figure above. Configuration 2 has a similar setup to Configuration 1; however, the wheel(s) can now move in two-dimensions. There is also a second pendulum and point mass with the same physical parameters as the first pendulum system added to this system. The two pendulums are attached to two wheels, which are at each end of rigid, mass-less bases. Each base is free to rotate independently. The two pendulums are placed along the rigid bases at a distance \textit{d} from both the center of gravity (CG) of each wheel, respectively. As the system moves in two-dimensions, the bases make an angle of $\phi$ with the \textit{X} axis of the ground frame. An auxiliary coordinate system \textit{x-y} is placed along the direction of the rigid bases. The angles of pendulums relative to their respective bases are defined in the same manner as Configuration 1.
A similar process is followed for Configuration 2; however, now the rotation about the center of gravity \textit{CG} has to be taken into account. This is done by adding an additional term onto the position of the masses and the wheels. The position equations for the masses and wheels are shown below, respectively.
\begin{equation}
x_{m,A} = x + L sin\gamma_A cos\phi + d sin\phi
\end{equation}
\begin{equation}
y_{m,A} = y + L sin\gamma_A sin\phi - d cos\phi
\end{equation}
\begin{equation}
z_{m,A} = r_g + L cos\gamma_A
\end{equation}
\begin{equation}
x_{m,B} = x + L sin\gamma_B cos\phi - d sin\phi
\end{equation}
\begin{equation}
y_{m,B} = y + L sin\gamma_B sin\phi + d cos\phi
\end{equation}
\begin{equation}
z_{m,B} = r_g + L cos\gamma_AB
\end{equation}
\begin{equation}
x_{w,A} = x - 2d cos\phi
\end{equation}
\begin{equation}
y_{w,A} = y + 2d sin\phi
\end{equation}
\begin{equation}
x_{w,B} = x + 2d cos\phi
\end{equation}
\begin{equation}
y_{w,B} = y - 2d sin\phi
\end{equation}
The velocities of the masses and wheels are taken in a manner similar to Configuration 1. Now, there will be two kinetic energies for mass and two kinetic energies for wheels. The potential energy for Configuration 2 is shown below.
\begin{equation}
U = Mg(r_g + L_p cos\gamma_A) + Mg(r_g + L_p cos\gamma_B)
\end{equation}
\section {Results}
\subsection*{Configuration 1}
From this point the Lagrangian is simply $L = T - U$. Once the Lagrangian is taken, the proper substitutions are made and small angle approximations are assumed. The equations for \textit{F(t)} and $\gamma (t)$ are substituted here. Below are the equations of motion relating the wheel displacement $x_c(t)$ and base angle $\alpha (t)$ to the input voltage $V_{in} (t)$.
\begin{equation}
(M + \frac{3}{2}M_w) \ddot x(t) + M L_p (1 + \frac{M g L_p^2}{2GJ}) \ddot \alpha (t)
= \frac{K_g K_t V_{in}(t)}{R r_g} + \frac{K_g^2 K_m K_t x(t)}{R r_g^2}
\end{equation}
\begin{equation}
M L_p^2 (1 + \frac{M g L_p^2}{2GJ}) \ddot \alpha(t) + M L_p \ddot x_c (t) - M g L_p (1 + \frac{M g L_p^2}{2GJ}) \alpha (t) = -c (1 + \frac{M g L_p^2}{2GJ}) \dot\alpha (t)
\end{equation}
Note that there is a non-conservative force in the second equation of motion. This is a viscous friction force caused by the wheel and shaft. For the state variables
\[\begin{bmatrix}
x \\
\dot x \\
\alpha \\
\dot\alpha
\end{bmatrix}\]
the state space matrices, $A_1$ and $B_1$, for Configuration 1 are shown below, respectively.
\[\begin{bmatrix}
0 & 1 & 0 & 0 \\
\frac{-2 K_g^2 K_m K_t}{(2M+3M_w) R r_g^2} & 0 & \frac{-2M g \sigma_1}{(2M+3M_w)} & \frac{-2c}{(2M+3M_w)ML_p^2} \\
0 & 0 & 0 & 1 \\
\frac{2 K_g^2 K_m K_t}{(2M+3M_w) L_p R r_g^2 \sigma_1} & 0 & \frac{g}{L_p} & \frac{-c}{ML_p^2}
\end{bmatrix}\]
\[\begin{bmatrix}
0 \\
\frac{2 K_g K_t}{(2M+3M_w) R r_g} \\
0 \\
\frac{-2 K_g K_t}{(2M+3M_w) L_p R r_g \sigma_1}
\end{bmatrix}\]
where
\begin{equation}
\sigma_1 = 1 + \frac{M g {L_p}^2}{2GJ}
\end{equation}
Initial conditions and parameters were given for Configuration 1. The parameters can be found in Appendix A. There is an initial pendulum angle and base angle of $ 0^{\circ}$ and the initial wheel displacement is of 0 m. The numerical solution of the state space matrices is given below.
\[\begin{bmatrix}
0 & 1 & 0 & 0 \\
-0.0298 & 0 & -4.72 & -0.0935 \\
0 & 0 & 0 & 1 \\
0.0456 & 0 & 15.3 & -0.0973
\end{bmatrix}\]
\[\begin{bmatrix}
0 \\
0.104 \\
0 \\
-0.162
\end{bmatrix}\]
\subsection*{Configuration 2}
Configuration 2 uses the same system parameters as Configuration 1, along with some initial conditions. Initial pendulum angles are $ 0^{\circ}$, initial base angles are $ 0^{\circ}$ and the initial \textit{CG} position is $(0,0)$. The complete derivation of the equations of motion for Configuration 2 can be found in the Appendix. The linearized equations of motion for Configuration 2 with all angles assumed small are shown below. For simplification purposes, the functions will be written as $\ddot x$ instead of $\ddot x(t)$. The functions that do have (t) symbolize input functions.
\begin{equation}
\begin{aligned}
\ddot x = \frac{1}{2M+3M_w}[ -ML_p \dot\phi \dot\gamma_A \phi\gamma_A + ML_p \dot\phi^2 + ML_p \ddot\phi\phi + ML_p \dot\gamma_A^2 \gamma_A + ML_p \dot\phi \dot\gamma_A \phi - ML_p \ddot\gamma_A \\
+ ML_p \dot\phi \dot\gamma_B \phi + ML_p \dot\phi^2 \gamma_B + ML_p \dot\gamma_B \gamma_B + ML_p \dot\phi \dot\gamma_B \phi - ML_p \ddot\gamma_B + \frac{K_g K_t V_{in}(t)}{R r_g} - \frac{K_g^2 K_m K_t \dot x}{R r_g^2} ]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\ddot y = \frac{1}{(2M+3M_w)} [-ML_p \dot\phi \dot\gamma_A + ML_p \dot\phi^2 \phi\gamma_A - ML_p \ddot\phi\gamma_A + ML_p \dot\gamma_A^2 \phi\gamma_A - ML_p \phi\gamma_A - ML_p \ddot\gamma_A \phi
- ML_p \dot\phi\dot\gamma_B + ML_p \dot\phi^2 \phi\gamma_B \\
- ML_p \ddot\phi\gamma_B + ML_p \gamma_B^2 \phi\gamma_B - ML_p \dot\phi\dot\gamma_B - ML_p \ddot\gamma_B \phi + \frac{K_g K_t V_{in}(t)}{R r_g} - \frac{K_g^2 K_m K_t}{Rr_g}]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\ddot\phi = \frac{1}{2Md+12M_w d^2}[-2ML_p^2 \dot\phi\dot\gamma_A\gamma_A - ML_p^2 \ddot\phi \gamma_A^2 + ML_p d\dot\gamma_A^2 \gamma_A - ML_P d\ddot\gamma_A - ML_p \dot x \dot\gamma_A \phi\gamma_A + ML_p \dot x \dot\phi + ML_p \ddot x \phi - ML_p \dot y \dot\gamma_A \\
+ ML_p \dot y \dot\phi\phi\gamma_A - ML_p \ddot y \gamma_A - 2ML_p^2 \dot\phi\dot\gamma_B \gamma_B - ML_p^2 \ddot\phi \gamma_B^2 - ML_p d \dot\gamma_B^2 + ML_p d \ddot\gamma_B - ML_p \dot x \dot\gamma_B \phi\gamma_B + ML_p \dot x \dot\phi + ML_p \ddot x \dot\phi - ML_p \dot y \dot\gamma_B \\
+ ML_p \dot y \dot\phi\phi\gamma_B - ML_p \ddot y \gamma_B - ML_p \dot x \dot\phi^2 - ML_p \dot x \dot\phi\dot\gamma_A \phi - ML_p \dot y \dot\phi^2 \phi\gamma_A + ML_p \dot y \dot\phi\dot\gamma_A - ML_p \dot x \dot\phi^2 \gamma_B \\
- ML_p \dot x \dot\phi\dot\gamma_B \phi - ML_p \dot y \dot\phi^2 \phi\gamma_B + ML_p \dot y \dot\phi\dot\gamma_B]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\ddot\gamma_A = \frac{1}{ML_p^2}[-ML_p d \dot\phi\dot\gamma_A \gamma_A - ML_p d \ddot\phi + ML_p \dot x \dot\gamma_A \gamma_A + ML_p \dot x \dot\phi\phi - ML_p \ddot x + ML_p \dot y \dot\gamma_A \phi\gamma_A - ML_p \dot y \dot\phi - ML_p \ddot y \phi \\
+ ML_p^2 \dot\phi^2 \dot\gamma_A \gamma_A - ML_p d \dot\phi \dot\gamma_A^2 \gamma_A + ML_p \dot x \dot\phi\dot\gamma_A \phi\gamma_A - ML_p \dot x \dot\gamma_A^2 \gamma_A + ML_p \dot y \dot\phi\dot\gamma_A - ML_p \dot y \dot\gamma_A^2 \phi\gamma_A -c \dot\gamma_A]
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\ddot\gamma_B = \frac{1}{ML_p^2} [ML_p d \dot\phi\dot\gamma_B \gamma_B + ML_p d \ddot\phi - ML_p \dot x \dot\gamma_B \gamma_B - ML_p \dot x \dot\phi\phi + ML_p \ddot x - ML_p \dot y \dot\gamma_B \phi\gamma_B + ML_p \dot y \dot\phi + ML_p^2 \dot\phi^2 \dot\gamma_B \gamma_B \\
- ML_p d \dot\phi \dot\gamma_B^2 \gamma_B + ML_p dot x \dot\phi\dot\gamma_B \phi + ML_p \dot x \dot\gamma_B^2 \gamma_B - ML_p \dot y \dot\phi\dot\gamma_B + ML_p \dot y \dot\gamma_B^2 \phi\gamma_B + ML_p \ddot x + ML_p \ddot y -c \dot\gamma_B]
\end{aligned}
\end{equation}
\pagebreak
\section*{Conclusion}
Unfortunately, I was not able to further solve the equations of motion of Configuration 2.Nevertheless, this report investigated the derivations of the equations of motion using the Euler-Lagrangian method for a single wheeled inverted pendulum and a double wheeled inverted pendulum. The immediate goal was to derive the equations and solve for them numerically given certain parameters an initial conditions. Knowing coding language such as Matlab or Mathematica would make solving these equations simpler if there is no knowledge of the "by hand" method. The equations of motion of a Segway show just how difficult it can be to stabilize a self-transportation system such as this.
\par\null\par\null
\section*{}
{\label{550205}}\par\null
\subsection*{}
{\label{992846}}\par\null
\pagebreak
\section*{Appendices}
\subsection*{Appendix A: System Parameters}\selectlanguage{english}
\begin{table*}
\caption{{\label{tab:table}System Parameters for Self-Transportation System}}
\begin{center}
\begin{tabular}{cccc}
\hline\
\textbf{Parameter}& \textbf{Symbol} & \textbf{Value} & \textbf{Units}\\
\hline
Mass of wheel & $M_{w}$ & $ 0.360$ & kg\\
\hline
Point Mass & $M$ & $ 0.5$ & kg\\
\hline
Length of pendulum & $ L_{p}$ & $0.6413$ & m\\
\hline
Radius of wheel & $r_{g}$ & $0.1$ & m\\
\hline
Internal gear ratio & $K_{g}$ & $ 3.7:1$ & -\\
\hline
Motor torque constant & $K_{t}$ & $0.00767$ & Nm/Amp \\
\hline
Back emf constant & $K_{m}$ & $0.00767$ & V.sec/rad \\\\
\hline
Motor input voltage range & $V_{in}$ & $\pm 5$ & V\\
\hline
Gravitational constant & $g$ & $9.81$ & $m/sec^{2}$ \\
\hline
Viscous friction of motor shaft and wheel & $c$ & $0.02$ & N.sec/m\\
\hline
Elastic modulus of Beam & $G$ & $79 \times 10^{12}$ & Pa\\
\hline
Second moment of Area of beam & $J$ & $1.6933 \times 10^{-11}$ & $m^4$\\
\hline
Moment armature inductance & $L$ & $0.18$ & mH\\
\hline
\end{tabular}
\end{center}
\end{table*}
\selectlanguage{english}
\FloatBarrier
\bibliographystyle{unsrt}
\bibliography{bibliography/converted_to_latex.bib%
}
\end{document}