#+TITLE: Vibration control of a rotating Stewart platform
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#+LATEX_HEADER_EXTRA: \author[1,3] {T. Dehaeze}
#+LATEX_HEADER_EXTRA: \author[1,2] {C. Collette}

#+LATEX_HEADER_EXTRA: \affil[1] {Precision Mechatronics Laboratory\NewLineAffil University of Liege, Belgium \NewAffil}
#+LATEX_HEADER_EXTRA: \affil[2] {BEAMS Department\NewLineAffil Free University of Brussels, Belgium \NewAffil}
#+LATEX_HEADER_EXTRA: \affil[3] {European Synchrotron Radiation Facility \NewLineAffil Grenoble, France e-mail: \textbf{thomas.dehaeze@esrf.fr}}

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* Abstract                                                           :ignore:
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* Introduction
<<sec:introduction>>


* Theory
<<sec:theory>>

** Rotating Positioning Stage

# Description of the system

- $k$: Actuator's Stiffness [N/m]
- $m$: Payload's mass [kg]
- $\omega_0 = \sqrt{\frac{k}{m}}$: Resonance of the (non-rotating) mass-spring system [rad/s]
- $\Omega = \dot{\theta}$: rotation speed [rad/s]


#+name: fig:rotating_xy_platform
#+caption: Figure caption
#+attr_latex: :scale 1
[[file:figs/rotating_xy_platform.pdf]]


** Equation of Motion

Let's express the kinetic energy $T$ and the potential energy $V$ of the mass $m$ (neglecting the rotational energy):
#+name: eq:energy_inertial_frame
\begin{subequations}
  \begin{align}
    T & = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \\
    R & = \frac{1}{2} c \left( \dot{x}^2 + \dot{y}^2 \right) \\
    V & = \frac{1}{2} k \left( x^2 + y^2 \right)
  \end{align}
\end{subequations}

The Lagrangian is the kinetic energy minus the potential energy:
#+name: eq:lagrangian_inertial_frame
\begin{equation}
L = T-V = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) - \frac{1}{2} k \left( x^2 + y^2 \right)
\end{equation}

The external forces applied to the mass are:
\begin{subequations}
  \begin{align}
    F_{\text{ext}, x} &= F_u \cos{\theta} - F_v \sin{\theta}\\
    F_{\text{ext}, y} &= F_u \sin{\theta} + F_v \cos{\theta}
  \end{align}
\end{subequations}


From the Lagrange's equations of the second kind eqref:eq:lagrange_second_kind, the equation of motion eqref:eq:eom_mixed is obtained.

#+name: eq:lagrange_second_kind
\begin{equation}
  \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_j} \right) = \frac{\partial L}{\partial q_j}
\end{equation}

#+name: eq:eom_mixed
\begin{subequations}
  \begin{align}
    m\ddot{x} + kx = F_u \cos{\theta} - F_v \sin{\theta}\\
    m\ddot{y} + ky = F_u \sin{\theta} + F_v \cos{\theta}
  \end{align}
\end{subequations}

Performing the change coordinates from $(x, y)$ to $(d_x, d_y, \theta)$:
\begin{subequations}
  \begin{align}
    x & = d_u \cos{\theta} - d_v \sin{\theta}\\
    y & = d_u \sin{\theta} + d_v \cos{\theta}
  \end{align}
\end{subequations}

Gives
#+name: eq:oem_coupled
\begin{subequations}
  \begin{align}
    m \ddot{d_u} + (k - m\dot{\theta}^2) d_u &= F_u + 2 m\dot{d_v}\dot{\theta} + m d_v\ddot{\theta} \label{eq:du_coupled} \\
    m \ddot{d_v} + (k \underbrace{-\ m\dot{\theta}^2}_{\text{Centrif.}}) d_v &= F_v \underbrace{-\ 2 m\dot{d_u}\dot{\theta}}_{\text{Coriolis}} \underbrace{-\ m d_u\ddot{\theta}}_{\text{Euler}} \label{eq:dv_coupled}
  \end{align}
\end{subequations}

We obtain two differential equations that are coupled through:
- *Euler forces*: $m d_v \ddot{\theta}$
- *Coriolis forces*: $2 m \dot{d_v} \dot{\theta}$

Without the coupling terms, each equation is the equation of a one degree of freedom mass-spring system with mass $m$ and stiffness $k- m\dot{\theta}^2$.
Thus, the term $- m\dot{\theta}^2$ acts like a negative stiffness (due to *centrifugal forces*).

** Constant Rotating Speed
To simplify, let's consider a constant rotating speed $\dot{\theta} = \Omega$ and thus $\ddot{\theta} = 0$.

#+NAME: eq:coupledplant
\begin{equation}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
\frac{1}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2}
\begin{bmatrix}
  ms^2 + (k-m{\omega_0}^2) & 2 m \omega_0 s \\
  -2 m \omega_0 s          & ms^2 + (k-m{\omega_0}^2) \\
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}

# Explain each term

#+NAME: eq:coupled_plant
\begin{equation}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
\frac{\frac{1}{k}}{\left( \frac{s^2}{{\omega_0}^2} + (1 - \frac{{\Omega}^2}{{\omega_0}^2}) \right)^2 + \left( 2 \frac{{\Omega} s}{{\omega_0}^2} \right)^2}
\begin{bmatrix}
  \frac{s^2}{{\omega_0}^2} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} & 2 \frac{\Omega s}{{\omega_0}^2} \\
  -2 \frac{\Omega s}{{\omega_0}^2}          & \frac{s^2}{{\omega_0}^2} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \\
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}

When the rotation speed is null, the coupling terms are equal to zero and the diagonal terms corresponds to one degree of freedom mass spring system.
#+NAME: eq:coupled_plant_no_rot
\begin{equation}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
\frac{\frac{1}{k}}{\frac{s^2}{{\omega_0}^2} + 1}
\begin{bmatrix}
  1 & 0 \\
  0 & 1
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}

# Campbell Diagram

When the rotation speed in not null, the resonance frequency is duplicated into two pairs of complex conjugate poles.
As the rotation speed increases, one of the two resonant frequency goes to lower frequencies as the other one goes to higher frequencies (Figure [[fig:campbell_diagram]]).

#+name: fig:campbell_diagram
#+caption: Campbell Diagram
[[file:figs/campbell_diagram.pdf]]

# Bode Plots for different ratio wr/w0

The magnitude of the coupling terms are increasing with the rotation speed.


** Integral Force Feedback


** Direct Velocity Feedback


* Conclusion
<<sec:conclusion>>


* Acknowledgment


* Bibliography                                                       :ignore:
\bibliography{ref}
