#+TITLE: Control in a rotating frame
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#+PROPERTY: header-args:matlab+ :output-dir Figures

* Introduction
The objective of this note it to highlight some control problems that arises when controlling the position of an object using actuators that are rotating with respect to a fixed reference frame.

In section [[sec:system]], a simple system composed of a spindle and a translation stage is defined and the equations of motion are written.
The rotation induces some coupling between the actuators and their displacement, and modifies the dynamics of the system.
This is studied using the equations, and some numerical computations are used to compare the use of voice coil and piezoelectric actuators.

Then, in section [[sec:control_strategies]], two different control approach are compared where:
- the measurement is made in the fixed frame
- the measurement is made in the rotating frame

In section [[sec:simscape]], the analytical study will be validated using a multi body model of the studied system.

Finally, in section [[sec:control]], the control strategies are implemented using Simulink and Simscape and compared.

* System
  :PROPERTIES:
  :HEADER-ARGS:matlab+: :tangle system_numerical_analysis.m
  :END:
  <<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 by the actuators of 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 sample with respect to a fixed reference frame.

#+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:
#+name: eq:inertial_frame_deriv
\begin{align*}
\frac{\partial L}{\partial x} & = -kx \\
\frac{\partial L}{\partial y} & = -ky \\
\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} & = m\ddot{x} \\
\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} & = m\ddot{y}
\end{align*}

The external forces applied to the mass are:
\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*}

By appling the Lagrangian equations, we obtain:
\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}

We then change coordinates from $(x, y)$ to $(d_x, d_y, \theta)$.
\begin{align*}
x & = d_u \cos{\theta} - d_v \sin{\theta}\\
y & = d_u \sin{\theta} + d_v \cos{\theta}
\end{align*}

We obtain:
\begin{align*}
\ddot{x} & = \ddot{d_u} \cos{\theta} - 2\dot{d_u}\dot{\theta}\sin{\theta} - d_u\ddot{\theta}\sin{\theta} - d_u\dot{\theta}^2 \cos{\theta}
           - \ddot{d_v} \sin{\theta} - 2\dot{d_v}\dot{\theta}\cos{\theta} - d_v\ddot{\theta}\cos{\theta} + d_v\dot{\theta}^2 \sin{\theta} \\
\ddot{y} & = \ddot{d_u} \sin{\theta} + 2\dot{d_u}\dot{\theta}\cos{\theta} + d_u\ddot{\theta}\cos{\theta} - d_u\dot{\theta}^2 \sin{\theta}
           + \ddot{d_v} \cos{\theta} - 2\dot{d_v}\dot{\theta}\sin{\theta} - d_v\ddot{\theta}\sin{\theta} - d_v\dot{\theta}^2 \cos{\theta} \\
\end{align*}

By injecting the previous result into the Lagrangian equation, we obtain:
\begin{align*}
 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
#+NAME: eq:du_coupled
\begin{equation}
 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}
\end{equation}
#+NAME: eq:dv_coupled
\begin{equation}
 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}
\end{equation}
#+end_important

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*).

The forces induced by the rotating reference frame are independent of the stiffness of the actuator.
The resulting effect of those forces should then be higher when using softer actuators.

** Numerical Values for the NASS
#+begin_src matlab :exports none :results silent :noweb yes
  <<matlab-init>>
#+end_src

Let's define the parameters for the NASS.
#+begin_src matlab :exports none :results silent
  mlight = 35; % Mass for light sample [kg]
  mheavy = 85; % Mass for heavy sample [kg]

  wlight = 2*pi; % Max rot. speed for light sample [rad/s]
  wheavy = 2*pi/60; % Max rot. speed for heavy sample [rad/s]

  kvc = 1e3; % Voice Coil Stiffness [N/m]
  kpz = 1e8; % Piezo Stiffness [N/m]

  wdot = 1; % Maximum rotation acceleration [rad/s2]

  d = 0.01; % Maximum excentricity from rotational axis [m]
  ddot = 0.2; % Maximum Horizontal speed [m/s]

  save('./mat/parameters.mat');
#+end_src

#+begin_src matlab :results table :exports results
  labels = {'Light sample mass [kg]', ...
            'Heavy sample mass [kg]', ...
            'Max rot. speed - light [rpm]', ...
            'Max rot. speed - heavy [rpm]', ...
            'Voice Coil Stiffness [N/m]', ...
            'Piezo Stiffness [N/m]', ...
            'Max rot. acceleration [rad/s2]', ...
            'Max mass excentricity [m]', ...
            'Max Horizontal speed [m/s]'};
  data = [mlight, mheavy, 60*wlight/2/pi, 60*wheavy/2/pi, kvc, kpz, wdot, d, ddot];
  data2orgtable(data', labels, {}, ' %.1e ')
#+end_src

#+RESULTS:
| Light sample mass [kg]         | 3.5e+01 |
| Heavy sample mass [kg]         | 8.5e+01 |
| Max rot. speed - light [rpm]   | 6.0e+01 |
| Max rot. speed - heavy [rpm]   | 1.0e+00 |
| Voice Coil Stiffness [N/m]     | 1.0e+03 |
| Piezo Stiffness [N/m]          | 1.0e+08 |
| Max rot. acceleration [rad/s2] | 1.0e+00 |
| Max mass excentricity [m]      | 1.0e-02 |
| Max Horizontal speed [m/s]     | 2.0e-01 |

** Euler and Coriolis forces - Numerical Result
First we will determine the value for Euler and Coriolis forces during regular experiment.
- *Euler forces*: $m d_v \ddot{\theta}$
- *Coriolis forces*: $2 m \dot{d_v} \dot{\theta}$

#+begin_src matlab :exports none :results silent
  Felight = mlight*d*wdot;
  Feheavy = mheavy*d*wdot;

  Fclight = 2*mlight*ddot*wlight;
  Fcheavy = 2*mheavy*ddot*wheavy;
#+end_src

The obtained values are displayed in table [[tab:euler_coriolis]].

#+begin_src matlab :results value table :exports results :post addhdr(*this*)
  data = [Fclight, Fcheavy ;
          Felight, Feheavy];
  data2orgtable(data, {'Coriolis', 'Euler'}, {'Light', 'Heavy'}, ' %.1fN ')
#+end_src

#+NAME: tab:euler_coriolis
#+CAPTION: Euler and Coriolis forces for the NASS
#+RESULTS:
|          | Light | Heavy |
|----------+-------+-------|
| Coriolis | 88.0N | 3.6N  |
| Euler    | 0.4N  | 0.8N  |

** Negative Spring Effect - Numerical Result
The negative stiffness due to the rotation is equal to $-m{\omega_0}^2$.

#+begin_src matlab :exports none :results silent
  Klight = mlight*wlight^2;
  Kheavy = mheavy*wheavy^2;
#+end_src

The values for the negative spring effect are displayed in table [[tab:negative_spring]].

This is definitely negligible when using piezoelectric actuators. It may not be the case when using voice coil actuators.

#+begin_src matlab :results value table :exports results :post addhdr(*this*)
  data = [Klight, Kheavy];
  data2orgtable(data, {'Neg. Spring'}, {'Light', 'Heavy'}, ' %.1f[N/m] ')
#+end_src

#+NAME: tab:negative_spring
#+CAPTION: Negative Spring effect
#+RESULTS:
|             | Light       | Heavy    |
|-------------+-------------+----------|
| Neg. Spring | 1381.7[N/m] | 0.9[N/m] |

** Limitations due to coupling
To simplify, we consider a constant rotating speed $\dot{\theta} = {\omega_0}$ and thus $\ddot{\theta} = 0$.

From equations [[eq:du_coupled]] and [[eq:dv_coupled]], we obtain:
 \begin{align*}
 (m s^2 + (k - m{\omega_0}^2)) d_u &= F_u + 2 m {\omega_0} s d_v \\
 (m s^2 + (k - m{\omega_0}^2)) d_v &= F_v - 2 m {\omega_0} s d_u \\
\end{align*}

From second equation:
\[ d_v = \frac{1}{m s^2 + (k - m{\omega_0}^2)} F_v - \frac{2 m {\omega_0} s}{m s^2 + (k - m{\omega_0}^2)} d_u \]

And we re-inject $d_v$ into the first equation:
\begin{equation*}
(m s^2 + (k - m{\omega_0}^2)) d_u = F_u + \frac{2 m {\omega_0} s}{m s^2 + (k - m{\omega_0}^2)} F_v - \frac{(2 m {\omega_0} s)^2}{m s^2 + (k - m{\omega_0}^2)} d_u
\end{equation*}

\begin{equation*}
\frac{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2}{m s^2 + (k - m{\omega_0}^2)} d_u = F_u + \frac{2 m {\omega_0} s}{m s^2 + (k - m{\omega_0}^2)} F_v
\end{equation*}

Finally we obtain $d_u$ function of $F_u$ and $F_v$.
\[ d_u = \frac{m s^2 + (k - m{\omega_0}^2)}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2} F_u + \frac{2 m {\omega_0} s}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2} F_v \]

Similarly we can obtain $d_v$ function of $F_u$ and $F_v$:
\[ d_v = \frac{m s^2 + (k - m{\omega_0}^2)}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2} F_v - \frac{2 m {\omega_0} s}{(m s^2 + (k - m{\omega_0}^2))^2 + (2 m {\omega_0} s)^2} F_u \]

The two previous equations can be written in a matrix form:
#+begin_important
\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}
#+end_important

Then, coupling is negligible if $|-m \omega^2 + (k - m{\omega_0}^2)| \gg |2 m {\omega_0} \omega|$.

*** Numerical Analysis
We plot on the same graph $\frac{|-m \omega^2 + (k - m {\omega_0}^2)|}{|2 m \omega_0 \omega|}$ for the voice coil and the piezo:
- with the light sample (figure [[fig:coupling_light]]).
- with the heavy sample (figure [[fig:coupling_heavy]]).

#+HEADER: :exports none :results silent
#+begin_src matlab
  f = logspace(-1, 2, 1000);

  figure;
  hold on;
  plot(f, abs(-mlight*(2*pi*f).^2 + kvc - mlight * wlight^2)./abs(2*mlight*wlight*2*pi*f), 'DisplayName', 'Voice Coil')
  plot(f, abs(-mlight*(2*pi*f).^2 + kpz - mlight * wlight^2)./abs(2*mlight*wlight*2*pi*f), 'DisplayName', 'Piezo')
  plot(f, ones(1, length(f)), 'k--', 'HandleVisibility', 'off')
  set(gca, 'xscale', 'log'); set(gca, 'yscale', 'log');
  xlabel('Frequency [Hz]');
  legend('Location', 'northeast');
  hold off;
#+end_src

#+HEADER: :tangle no :exports results :results file :noweb yes
#+HEADER: :var filepath="Figures/coupling_light.png" :var figsize="normal-normal"
#+begin_src matlab
  <<plt-matlab>>
#+end_src

#+NAME: fig:coupling_light
#+CAPTION: Relative Coupling for light mass and high rotation speed
#+RESULTS:
[[file:Figures/coupling_light.png]]

#+HEADER: :exports none :results silent
#+begin_src matlab
  f = logspace(-1, 2, 1000);

  figure;
  hold on;
  plot(f, abs(-mheavy*(2*pi*f).^2 + kvc - mheavy * wheavy^2)./abs(2*mheavy*wheavy*2*pi*f), 'DisplayName', 'Voice Coil')
  plot(f, abs(-mheavy*(2*pi*f).^2 + kpz - mheavy * wheavy^2)./abs(2*mheavy*wheavy*2*pi*f), 'DisplayName', 'Piezo')
  plot(f, ones(1, length(f)), 'k--', 'HandleVisibility', 'off')
  set(gca, 'xscale', 'log'); set(gca, 'yscale', 'log');
  xlabel('Frequency [Hz]');
  legend('Location', 'northeast');
  hold off;
#+end_src

#+HEADER: :tangle no :exports results :results file :noweb yes
#+HEADER: :var filepath="Figures/coupling_heavy.png" :var figsize="normal-normal"
#+begin_src matlab
  <<plt-matlab>>
#+end_src

#+NAME: fig:coupling_heavy
#+CAPTION: Relative Coupling for heavy mass and low rotation speed
#+RESULTS:
[[file:Figures/coupling_heavy.png]]

#+begin_important
Coupling is higher for actuators with small stiffness.
#+end_important

** Limitations due to negative stiffness effect
If $\max{\dot{\theta}} \ll \sqrt{\frac{k}{m}}$, then the negative spring effect is negligible and $k - m\dot{\theta}^2 \approx k$.

Let's estimate what is the maximum rotation speed for which the negative stiffness effect is still negligible ($\omega_\text{max} = 0.1 \sqrt{\frac{k}{m}}$). Results are shown table [[tab:negative_stiffness]].
#+begin_src matlab :results table :exports results :post addhdr(*this*)
  data = 0.1*60*(1/2/pi)*[sqrt(kvc/mlight), sqrt(kpz/mlight);
                       sqrt(kvc/mheavy), sqrt(kpz/mheavy)];
  data2orgtable(data, {'Light', 'Heavy'}, {'Voice Coil', 'Piezo'}, ' %.0f[rpm] ')
#+end_src

#+NAME: tab:negative_stiffness
#+CAPTION: Maximum rotation speed at which negative stiffness is negligible ($0.1\sqrt{\frac{k}{m}}$)
#+RESULTS:
|       | Voice Coil | Piezo     |
|-------+------------+-----------|
| Light | 5[rpm]     | 1614[rpm] |
| Heavy | 3[rpm]     | 1036[rpm] |

The negative spring effect is proportional to the rotational speed $\omega$.
The system dynamics will be much more affected when using soft actuator.

#+begin_important
  Negative stiffness effect has very important effect when using soft actuators.
#+end_important

The system can even goes unstable when $m \omega^2 > k$, that is when the centrifugal forces are higher than the forces due to stiffness.

From this analysis, we can determine the lowest practical stiffness that is possible to use: $k_\text{min} = 10 m \omega^2$ (table [[tab:min_k]])

#+begin_src matlab :results table :exports results :post addhdr(*this*)
  data = 10*[mlight*2*pi, mheavy*2*pi/60]
  data2orgtable(data, {'k min [N/m]'}, {'Light', 'Heavy'}, ' %.0f ')
#+end_src

#+NAME: tab:min_k
#+CAPTION: Minimum possible stiffness
#+RESULTS:
|             | Light | Heavy |
|-------------+-------+-------|
| k min [N/m] |  2199 |    89 |

* Control Strategies
  <<sec:control_strategies>>
** Measurement in the fixed reference frame
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 $L = G(\theta) K J(\theta)$.

One question we wish to answer is: is $G(\theta) J(\theta) = G(\theta_0) J(\theta_0)$?

** Measurement in the rotating frame
Let's consider that the measurement is made in the rotating reference frame.

The corresponding block diagram is shown figure [[fig:control_measure_rotating_2dof]]

#+name: fig:control_measure_rotating_2dof
#+caption: Control with a measure from rotating frame
[[./Figures/control_measure_rotating_2dof.png]]

The loop gain is $L = G K$.

* Multi Body Model - Simscape
  :PROPERTIES:
  :HEADER-ARGS:matlab+: :tangle simscape_analysis.m
  :END:
  <<sec:simscape>>

#+begin_src matlab :exports none :results silent :noweb yes
  <<matlab-init>>
  load('./mat/parameters.mat');
#+end_src

#+begin_src matlab :exports none :results silent
  open rotating_frame.slx
#+end_src

** Identification in the rotating referenced frame
We initialize the inputs and outputs of the system to identify.
#+begin_src matlab :exports code :results silent
  %% Options for Linearized
  options = linearizeOptions;
  options.SampleTime = 0;

  %% Name of the Simulink File
  mdl = 'rotating_frame';

  %% Input/Output definition
  io(1) = linio([mdl, '/fu'], 1, 'input');
  io(2) = linio([mdl, '/fv'], 1, 'input');

  io(3) = linio([mdl, '/du'], 1, 'output');
  io(4) = linio([mdl, '/dv'], 1, 'output');
#+end_src

*** Piezo and Voice coil
We start we identify the transfer functions at high speed with the light sample.
#+begin_src matlab :exports code :results silent
  rot_speed = wlight;
  angle_e = 0;
  m = mlight;

  k = kpz;
  c = 1e3;
  Gpz_light = linearize(mdl, io, 0.1);

  k = kvc;
  c = 1e3;
  Gvc_light = linearize(mdl, io, 0.1);

  Gpz_light.InputName  = {'Fu', 'Fv'};
  Gpz_light.OutputName = {'Du', 'Dv'};
  Gvc_light.InputName  = {'Fu', 'Fv'};
  Gvc_light.OutputName = {'Du', 'Dv'};
#+end_src

#+begin_src matlab :exports none :results silent
  figure;
  bode(Gpz_light, Gvc_light);
#+end_src

And then with the heavy sample.
#+begin_src matlab :exports code :results silent
  rot_speed = wheavy;
  angle_e = 0;
  m = mheavy;

  k = kpz;
  c = 1e3;
  Gpz_heavy = linearize(mdl, io, 0.1);

  k = kvc;
  c = 1e3;
  Gvc_heavy = linearize(mdl, io, 0.1);

  Gpz_heavy.InputName  = {'Fu', 'Fv'};
  Gpz_heavy.OutputName = {'Du', 'Dv'};
  Gvc_heavy.InputName  = {'Fu', 'Fv'};
  Gvc_heavy.OutputName = {'Du', 'Dv'};
#+end_src

#+begin_src matlab :exports none :results silent
  figure;
  bode(Gpz_heavy, Gvc_heavy);
#+end_src

Plot the ratio between the main transfer function and the coupling term:
#+begin_src matlab :results silent :exports none
  freqs = logspace(-2, 3, 1000);

  figure;
  hold on;
  plot(freqs, abs(squeeze(freqresp(Gvc_light('Du', 'Fu'), freqs, 'Hz'))))./abs(squeeze(freqresp(Gvc_light('Dv', 'Fu'), freqs, 'Hz')));
  plot(freqs, abs(squeeze(freqresp(Gpz_light('Du', 'Fu'), freqs, 'Hz'))))./abs(squeeze(freqresp(Gpz_light('Dv', 'Fu'), freqs, 'Hz')));
  hold off;
  set(gca, 'XScale', 'log'); set(gca, 'YScale', 'log');
  xlabel('Frequency [Hz]'); ylabel('Coupling ratio');
  legend({'Voice Coil', 'Piezoelectric'})
#+end_src

#+begin_src matlab :results silent :exports none
  freqs = logspace(-2, 3, 1000);

  figure;
  hold on;
  plot(freqs, abs(squeeze(freqresp(Gvc_heavy('Du', 'Fu'), freqs, 'Hz'))))./abs(squeeze(freqresp(Gvc_heavy('Dv', 'Fu'), freqs, 'Hz')));
  plot(freqs, abs(squeeze(freqresp(Gpz_heavy('Du', 'Fu'), freqs, 'Hz'))))./abs(squeeze(freqresp(Gpz_heavy('Dv', 'Fu'), freqs, 'Hz')));
  hold off;
  set(gca, 'XScale', 'log'); set(gca, 'YScale', 'log');
  xlabel('Frequency [Hz]'); ylabel('Coupling ratio');
  legend({'Voice Coil', 'Piezoelectric'})
#+end_src

*** Low rotation speed and High rotation speed
#+begin_src matlab :exports code :results silent
  rot_speed = 2*pi/60; angle_e = 0;
  G_low = linearize(mdl, io, 0.1);

  rot_speed = 2*pi; angle_e = 0;
  G_high = linearize(mdl, io, 0.1);

  G_low.InputName  = {'Fu', 'Fv'};
  G_low.OutputName = {'Du', 'Dv'};
  G_high.InputName  = {'Fu', 'Fv'};
  G_high.OutputName = {'Du', 'Dv'};
#+end_src

#+begin_src matlab :results silent
  figure;
  bode(G_low, G_high);
#+end_src

** Identification in the fixed frame
Let's define some options as well as the inputs and outputs for linearization.
#+begin_src matlab :exports code :results silent
  %% Options for Linearized
  options = linearizeOptions;
  options.SampleTime = 0;

  %% Name of the Simulink File
  mdl = 'rotating_frame';

  %% Input/Output definition
  io(1) = linio([mdl, '/fx'], 1, 'input');
  io(2) = linio([mdl, '/fy'], 1, 'input');

  io(3) = linio([mdl, '/dx'], 1, 'output');
  io(4) = linio([mdl, '/dy'], 1, 'output');
#+end_src

We then define the error estimation of the error and the rotational speed.
#+begin_src matlab :exports code :results silent
  %% Run the linearization
  angle_e = 0;
  rot_speed = 0;
#+end_src

Finally, we run the linearization.
#+begin_src matlab :exports code :results silent
  G = linearize(mdl, io, 0);

  %% Input/Output names
  G.InputName  = {'Fx', 'Fy'};
  G.OutputName = {'Dx', 'Dy'};
#+end_src

#+begin_src matlab :exports code :results silent
  %% Run the linearization
  angle_e = 0;
  rot_speed = 2*pi;
  Gr = linearize(mdl, io, 0);

  %% Input/Output names
  Gr.InputName  = {'Fx', 'Fy'};
  Gr.OutputName = {'Dx', 'Dy'};
#+end_src

#+begin_src matlab :exports code :results silent
  %% Run the linearization
  angle_e = 1*2*pi/180;
  rot_speed = 2*pi;
  Ge = linearize(mdl, io, 0);

  %% Input/Output names
  Ge.InputName  = {'Fx', 'Fy'};
  Ge.OutputName = {'Dx', 'Dy'};
#+end_src

#+begin_src matlab :exports code :results silent
  figure;
  bode(G);
  % exportFig('G_x_y', 'wide-tall');

  figure;
  bode(Ge);
  % exportFig('G_x_y_e', 'normal-normal');
#+end_src

** Identification from actuator forces to displacement in the fixed frame
#+begin_src matlab :exports code :results silent
  %% Options for Linearized
  options = linearizeOptions;
  options.SampleTime = 0;

  %% Name of the Simulink File
  mdl = 'rotating_frame';

  %% Input/Output definition
  io(1) = linio([mdl, '/fu'], 1, 'input');
  io(2) = linio([mdl, '/fv'], 1, 'input');

  io(3) = linio([mdl, '/dx'], 1, 'output');
  io(4) = linio([mdl, '/dy'], 1, 'output');
#+end_src

#+begin_src matlab :exports code :results silent
  rot_speed = 2*pi;
  angle_e = 0;
  G = linearize(mdl, io, 0.0);

  G.InputName  = {'Fu', 'Fv'};
  G.OutputName = {'Dx', 'Dy'};
#+end_src

#+begin_src matlab :exports code :results silent
  rot_speed = 2*pi;
  angle_e = 0;
  G1 = linearize(mdl, io, 0.4);

  G1.InputName  = {'Fu', 'Fv'};
  G1.OutputName = {'Dx', 'Dy'};
#+end_src

#+begin_src matlab :exports code :results silent
  rot_speed = 2*pi;
  angle_e = 0;
  G2 = linearize(mdl, io, 0.8);

  G2.InputName  = {'Fu', 'Fv'};
  G2.OutputName = {'Dx', 'Dy'};
#+end_src

#+begin_src matlab :exports code :results silent
  figure;
  bode(G, G1, G2);
  exportFig('G_u_v_to_x_y', 'wide-tall');
#+end_src

** Effect of the rotating Speed
  <<sec:effect_rot_speed>>

#+begin_src matlab :exports none :results silent :noweb yes
  <<matlab-init>>
#+end_src

*** TODO Use realistic parameters for the mass of the sample and stiffness of the X-Y stage
*** TODO Check if the plant is changing a lot when we are not turning to when we are turning at the maximum speed (60rpm)
** Effect of the X-Y stage stiffness
  <<sec:effect_stiffness>>
*** TODO At full speed, check how the coupling changes with the stiffness of the actuators
* Control Implementation
  <<sec:control>>
** Measurement in the fixed reference frame