281 lines
12 KiB
Org Mode
281 lines
12 KiB
Org Mode
#+TITLE: Control in a rotating frame
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#+HTML_HEAD: <link rel="stylesheet" type="text/css" href="css/htmlize.css"/>
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#+HTML_HEAD: <script type="text/javascript" src="js/readtheorg.js"></script>
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#+LATEX_CLASS: cleanreport
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#+LaTeX_CLASS_OPTIONS: [tocnp, secbreak, minted]
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#+STARTUP: overview
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#+LaTeX_HEADER: \usepackage{svg}
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#+LaTeX_HEADER: \newcommand{\authorFirstName}{Thomas}
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#+LaTeX_HEADER: \newcommand{\authorLastName}{Dehaeze}
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#+LaTeX_HEADER: \newcommand{\authorEmail}{dehaeze.thomas@gmail.com}
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#+PROPERTY: header-args:matlab :session *MATLAB*
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#+PROPERTY: header-args:matlab+ :tangle rotating_frame.m
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#+PROPERTY: header-args:matlab+ :comments org
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#+PROPERTY: header-args:matlab+ :exports both
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#+PROPERTY: header-args:matlab+ :eval no-export
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#+PROPERTY: header-args:matlab+ :output-dir Figures
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* Goal of this note
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The control objective is to stabilize the position of a rotating object with respect to a non-rotating frame.
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The actuators are also rotating with the object.
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We want to compare the two different approach:
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- the measurement is made in the fixed frame
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- the measurement is made in the rotating frame
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* System
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<<sec:system>>
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** System description
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The system consists of one 2 degree of freedom translation stage on top of a spindle (figure [[fig:rotating_frame_2dof]]).
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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 $v$.
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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.
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#+name: fig:rotating_frame_2dof
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#+caption: Schematic of the mecanical system
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[[./Figures/rotating_frame_2dof.png]]
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In the following block diagram:
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- $G$ is the transfer function from the forces applied in the actuators to the measurement
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- $K$ is the controller to design
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- $J$ is a Jacobian matrix usually used to change the reference frame
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Indices $x$ and $y$ corresponds to signals in the fixed reference frame (along $\vec{i}_x$ and $\vec{i}_y$):
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- $D_x$ is the measured position of the sample
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- $r_x$ is the reference signal which corresponds to the wanted $D_x$
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- $\epsilon_x$ is the position error
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Indices $u$ and $v$ corresponds to signals in the rotating reference frame ($\vec{i}_u$ and $\vec{i}_v$):
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- $F_u$ and $F_v$ are forces applied by the actuators
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- $\epsilon_u$ and $\epsilon_v$ are position error of the sample along $\vec{i}_u$ and $\vec{i}_v$
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** Equations
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<<sec:equations>>
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Based on the figure [[fig:rotating_frame_2dof]], we can write the equations of motion of the system.
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Let's express the kinetic energy $T$ and the potential energy $V$ of the mass $m$:
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#+name: eq:energy_inertial_frame
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\begin{align}
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T & = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) \\
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V & = \frac{1}{2} k \left( x^2 + y^2 \right)
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\end{align}
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The Lagrangian is the kinetic energy minus the potential energy.
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#+name: eq:lagrangian_inertial_frame
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\begin{equation}
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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)
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\end{equation}
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The partial derivatives of the Lagrangian with respect to the variables $(x, y)$ are:
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#+name: eq:inertial_frame_deriv
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\begin{align*}
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\frac{\partial L}{\partial x} & = -kx \\
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\frac{\partial L}{\partial y} & = -ky \\
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\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} & = m\ddot{x} \\
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\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} & = m\ddot{y}
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\end{align*}
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The external forces applied to the mass are:
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\begin{align*}
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F_{\text{ext}, x} &= F_u \cos{\theta} - F_v \sin{\theta}\\
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F_{\text{ext}, y} &= F_u \sin{\theta} + F_v \cos{\theta}
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\end{align*}
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By appling the Lagrangian equations, we obtain:
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\begin{align}
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m\ddot{x} + kx = F_u \cos{\theta} - F_v \sin{\theta}\\
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m\ddot{y} + ky = F_u \sin{\theta} + F_v \cos{\theta}
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\end{align}
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We then change coordinates from $(x, y)$ to $(d_x, d_y, \theta)$.
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\begin{align*}
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x & = d_u \cos{\theta} - d_v \sin{\theta}\\
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y & = d_u \sin{\theta} + d_v \cos{\theta}
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\end{align*}
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We obtain:
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\begin{align*}
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\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}
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- \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} \\
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\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}
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+ \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} \\
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\end{align*}
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By injecting the previous result into the Lagrangian equation [[eq:lagrangian_eq_inertial]], we obtain:
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\begin{align*}
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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}
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-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}
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+ k d_u \cos{\theta} - k d_v \sin{\theta} = F_u \cos{\theta} - F_v \sin{\theta} \\
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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}
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+ 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}
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+ k d_u \sin{\theta} + k d_v \cos{\theta} = F_u \sin{\theta} + F_v \cos{\theta}
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\end{align*}
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Which is equivalent to:
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\begin{align*}
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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
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-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}}
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+ k d_u - k d_v \frac{\sin{\theta}}{\cos{\theta}} = F_u - F_v \frac{\sin{\theta}}{\cos{\theta}} \\
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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
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+ 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}}
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+ k d_u + k d_v \frac{\cos{\theta}}{\sin{\theta}} = F_u + F_v \frac{\cos{\theta}}{\sin{\theta}}
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\end{align*}
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We can then subtract and add the previous equations to obtain the following equations:
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#+begin_important
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\begin{align*}
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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} \\
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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} \\
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\end{align*}
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#+end_important
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** Analysis
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We obtain two differential equations that are coupled through:
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- *Euler forces*: $m d_v \ddot{\theta}$
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- *Coriolis forces*: $2 m \dot{d_v} \dot{\theta}$
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Without the coupling terms, each equation is the equation of a one degree of freedom mass-spring system with mass $m$ and stiffness $k-d_u m\dot{\theta}^2$.
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Thus, the term $-d_u m\dot{\theta}^2$ acts like a negative stiffness (due to *centrifugal forces*).
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* Analytical Computation of forces for the NASS
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For the NASS, the Euler forces should be less of a problem as $\ddot{\theta}$ should be very small when conducting an experiment.
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First we will determine the value for Euler and Coriolis forces during regular experiment.
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** Euler and Coriolis forces
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Let's define the parameters for the NASS.
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#+begin_src matlab :exports code :results silent
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mlight = 35; % [kg]
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mheavy = 85; % [kg]
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wlight = 2*pi; % [rad/s]
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wheavy = 2*pi/60; % [rad/s]
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wdot = 1; % [rad/s2]
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d = 0.1; % [m]
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ddot = 0.2; % [m/s]
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#+end_src
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#+begin_src matlab :exports none :results silent
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Felight = mlight*d*wdot;
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Feheavy = mheavy*d*wdot;
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Fclight = 2*mlight*d*wlight;
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Fcheavy = 2*mheavy*d*wheavy;
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#+end_src
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We then compute the corresponding values of the Coriolis and Euler forces, and the obtained values are displayed in table [[tab:euler_coriolis]].
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#+begin_src matlab :results value table :exports results :post addhdr(*this*)
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ans = sprintf(' | Light | Heavy | \n Coriolis | %.1f N | %.1f N | \n Euler | %.1f N | %.1f N', Fclight, Fcheavy, Felight, Feheavy)
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#+end_src
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#+NAME: tab:euler_coriolis
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#+CAPTION: Euler and Coriolis forces for the NASS
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#+RESULTS:
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| | Light | Heavy |
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|----------+--------+-------|
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| Coriolis | 44.0 N | 1.8 N |
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| Euler | 3.5 N | 8.5 N |
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** Spring Softening Effect
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#+begin_src matlab :exports none :results silent
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Klight = mlight*d*wdot^2;
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Kheavy = mheavy*d*wdot^2;
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#+end_src
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The values for the spring softening effect are displayed in table [[tab:spring_softening]].
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This is definitely negligible when using piezoelectric actuators. It may not be the case when using voice coil actuators.
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#+begin_src matlab :results value table :exports results :post addhdr(*this*)
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ans = sprintf(' | Light | Heavy | \n Spring Soft. | %.1f N/m | %.1f N/m', Klight, Kheavy)
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#+end_src
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#+NAME: tab:spring_softening
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#+CAPTION: Spring Softening effect
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#+RESULTS:
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| | Light | Heavy |
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|--------------+---------+---------|
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| Spring Soft. | 3.5 N/m | 8.5 N/m |
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* Control Strategies
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<<sec:control_strategies>>
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** Measurement in the fixed reference frame
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First, let's consider a measurement in the fixed referenced frame.
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The transfer function from actuator $[F_u, F_v]$ to sensor $[D_x, D_y]$ is then $G(\theta)$.
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Then the measurement is subtracted to the reference signal $[r_x, r_y]$ to obtain the position error in the fixed reference frame $[\epsilon_x, \epsilon_y]$.
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The position error $[\epsilon_x, \epsilon_y]$ is then express in the rotating frame corresponding to the actuators $[\epsilon_u, \epsilon_v]$.
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Finally, the control low $K$ links the position errors $[\epsilon_u, \epsilon_v]$ to the actuator forces $[F_u, F_v]$.
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The block diagram is shown on figure [[fig:control_measure_fixed_2dof]].
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#+name: fig:control_measure_fixed_2dof
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#+caption: Control with a measure from fixed frame
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[[./Figures/control_measure_fixed_2dof.png]]
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The loop gain is then $L = G(\theta) K J(\theta)$.
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*** QUESTION Is the loop gain is changing with the angle ?
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Is \[ G(\theta) J(\theta) = G(\theta_0) J(\theta_0) \] ?
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** Measurement in the rotating frame
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Let's consider that the measurement is in the rotating reference frame.
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The corresponding block diagram is shown figure [[fig:control_measure_rotating_2dof]]
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#+name: fig:control_measure_rotating_2dof
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#+caption: Control with a measure from rotating frame
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[[./Figures/control_measure_rotating_2dof.png]]
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The loop gain is $L = G K$.
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* Effect of the rotating Speed
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<<sec:effect_rot_speed>>
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** TODO Use realistic parameters for the mass of the sample and stiffness of the X-Y stage
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** TODO Check if the plant is changing a lot when we are not turning to when we are turning at the maximum speed (60rpm)
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* Effect of the X-Y stage stiffness
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<<sec:effect_stiffness>>
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** TODO At full speed, check how the coupling changes with the stiffness of the actuators
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* Noweb Snippets :noexport:
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:PROPERTIES:
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:HEADER-ARGS:matlab+: :results none :exports none :tangle no
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:HEADER-ARGS:emacs-lisp+: :tangle no :eval no-export
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:END:
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** Matlab Init
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#+NAME: matlab-init
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#+BEGIN_SRC matlab :results none :exports none
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clear; close all; clc;
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addpath('./src/');
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ans = 0;
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#+END_SRC
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** Matlab Export Figure
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#+NAME: matlab-export-figure
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#+BEGIN_SRC matlab :var fn="figure" ft="png" size="normal-normal" :exports none
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exportFig(fn, size);
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ans = [fn, ".", ft];
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#+END_SRC
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** Add hline to org table
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#+name: addhdr
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#+begin_src emacs-lisp :var tbl=""
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(cons (car tbl) (cons 'hline (cdr tbl)))
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#+end_src
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