The control objective is to stabilize the position of a rotating object with respect to a non-rotating frame.
The actuators are also rotating with the object.
We want to compare the two different approach:
- the measurement is made in the fixed frame
- the measurement is made in the rotating frame
* System
<<sec:system>>
** System description
The system consists of one 2 degree of freedom translation stage on top of a spindle (figure [[fig:rotating_frame_2dof]]).
The control inputs are the forces applied in the translation stage ($F_u$ and $F_v$). As the translation stage is rotating around the Z axis due to the spindle, the forces are applied along $u$ and $v$.
The measurement is either the $x-y$ displacement of the object located on top of the translation stage or the $u-v$ displacement of the actuators.
#+name: fig:rotating_frame_2dof
#+caption: Schematic of the mecanical system
[[./Figures/rotating_frame_2dof.png]]
In the following block diagram:
- $G$ is the transfer function from the forces applied in the actuators to the measurement
- $K$ is the controller to design
- $J$ is a Jacobian matrix usually used to change the reference frame
Indices $x$ and $y$ corresponds to signals in the fixed reference frame (along $\vec{i}_x$ and $\vec{i}_y$):
- $D_x$ is the measured position of the sample
- $r_x$ is the reference signal which corresponds to the wanted $D_x$
- $\epsilon_x$ is the position error
Indices $u$ and $v$ corresponds to signals in the rotating reference frame ($\vec{i}_u$ and $\vec{i}_v$):
- $F_u$ and $F_v$ are forces applied by the actuators
- $\epsilon_u$ and $\epsilon_v$ are position error of the sample along $\vec{i}_u$ and $\vec{i}_v$
** Equations
<<sec:equations>>
Based on the figure [[fig:rotating_frame_2dof]], we can write the equations of motion of the system.
Let's express the kinetic energy $T$ and the potential energy $V$ of the mass $m$:
#+name: eq:energy_inertial_frame
\begin{align}
T & = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \\
V & = \frac{1}{2} k \left( x^2 + y^2 \right)
\end{align}
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 partial derivatives of the Lagrangian with respect to the variables $(x, y)$ are:
m \ddot{d_u} \cos{\theta} - 2m\dot{d_u}\dot{\theta}\sin{\theta} - m d_u\ddot{\theta}\sin{\theta} - m d_u\dot{\theta}^2 \cos{\theta}
-m \ddot{d_v} \sin{\theta} - 2m\dot{d_v}\dot{\theta}\cos{\theta} - m d_v\ddot{\theta}\cos{\theta} + m d_v\dot{\theta}^2 \sin{\theta}
+ k d_u \cos{\theta} - k d_v \sin{\theta} = F_u \cos{\theta} - F_v \sin{\theta} \\
m \ddot{d_u} \sin{\theta} + 2m\dot{d_u}\dot{\theta}\cos{\theta} + m d_u\ddot{\theta}\cos{\theta} - m d_u\dot{\theta}^2 \sin{\theta}
+ m \ddot{d_v} \cos{\theta} - 2m\dot{d_v}\dot{\theta}\sin{\theta} - m d_v\ddot{\theta}\sin{\theta} - m d_v\dot{\theta}^2 \cos{\theta}
+ k d_u \sin{\theta} + k d_v \cos{\theta} = F_u \sin{\theta} + F_v \cos{\theta}
\end{align*}
Which is equivalent to:
\begin{align*}
m \ddot{d_u} - 2m\dot{d_u}\dot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - m d_u\ddot{\theta}\frac{\sin{\theta}}{\cos{\theta}} - m d_u\dot{\theta}^2
-m \ddot{d_v} \frac{\sin{\theta}}{\cos{\theta}} - 2m\dot{d_v}\dot{\theta} - m d_v\ddot{\theta} + m d_v\dot{\theta}^2 \frac{\sin{\theta}}{\cos{\theta}}
+ k d_u - k d_v \frac{\sin{\theta}}{\cos{\theta}} = F_u - F_v \frac{\sin{\theta}}{\cos{\theta}} \\
m \ddot{d_u} + 2m\dot{d_u}\dot{\theta}\frac{\cos{\theta}}{\sin{\theta}} + m d_u\ddot{\theta}\frac{\cos{\theta}}{\sin{\theta}} - m d_u\dot{\theta}^2
+ m \ddot{d_v} \frac{\cos{\theta}}{\sin{\theta}} - 2m\dot{d_v}\dot{\theta} - m d_v\ddot{\theta} - m d_v\dot{\theta}^2 \frac{\cos{\theta}}{\sin{\theta}}
+ k d_u + k d_v \frac{\cos{\theta}}{\sin{\theta}} = F_u + F_v \frac{\cos{\theta}}{\sin{\theta}}
\end{align*}
We can then subtract and add the previous equations to obtain the following equations:
#+begin_important
\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} \\
m \ddot{d_v} + (k - m\dot{\theta}^2) d_v &= F_v - 2 m\dot{d_u}\dot{\theta} - m d_u\ddot{\theta} \\
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 $k-d_u m\dot{\theta}^2$.
Thus, the term $-d_u m\dot{\theta}^2$ acts like a negative stiffness (due to *centrifugal forces*).
First, let's consider a measurement in the fixed referenced frame.
The transfer function from actuator $[F_u, F_v]$ to sensor $[D_x, D_y]$ is then $G(\theta)$.
Then the measurement is subtracted to the reference signal $[r_x, r_y]$ to obtain the position error in the fixed reference frame $[\epsilon_x, \epsilon_y]$.
The position error $[\epsilon_x, \epsilon_y]$ is then express in the rotating frame corresponding to the actuators $[\epsilon_u, \epsilon_v]$.
Finally, the control low $K$ links the position errors $[\epsilon_u, \epsilon_v]$ to the actuator forces $[F_u, F_v]$.
The block diagram is shown on figure [[fig:control_measure_fixed_2dof]].
#+name: fig:control_measure_fixed_2dof
#+caption: Control with a measure from fixed frame
[[./Figures/control_measure_fixed_2dof.png]]
The loop gain is then $L = G(\theta) K J(\theta)$.
