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Start to rewrite all the equations

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Thomas Dehaeze
2020-06-25 12:18:44 +02:00
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@@ -1,4 +1,4 @@
#+TITLE: Active Damping of Rotating Positioning Platforms
#+TITLE: Decentralized Active Damping of Rotating Positioning Platforms
:DRAWER:
#+LATEX_CLASS: ISMA_USD2020
#+OPTIONS: toc:nil
@@ -77,6 +77,9 @@ Controller Poles are shown by black crosses (
** Describe the paper itself / the problem which is addressed :ignore:
Due to gyroscopic effects, the guaranteed robustness properties of Integral Force Feedback do not hold.
Either the control architecture can be slightly modfied or mechanical changes in the system can be performed.
** Introduce Each part of the paper :ignore:
This paper has been published
@@ -84,6 +87,9 @@ The Matlab code that was use to obtain the results are available in cite:dehaeze
* Dynamics of Rotating Positioning Platforms
** Studied Rotating Positioning Platform
# Introduce the fact that we need a simple system representing the rotating aspect.
# Simplest system where gyroscopic forces can be studied
Consider the rotating X-Y stage of Figure [[fig:system]].
@@ -112,64 +118,71 @@ Consider the rotating X-Y stage of Figure [[fig:system]].
# #+attr_latex: :width 0.5\linewidth
# [[file:figs/cedrat_xy25xs.jpg]]
** Equation of Motion
The system has two degrees of freedom and is thus fully described by the generalized coordinates $u$ and $v$ (describing the position of the mass in the rotating frame).
** Equations of Motion
The system has two degrees of freedom and is thus fully described by the generalized coordinates $[q_1\ q_2] = [d_u\ d_v]$ (describing the position of the mass in the rotating frame).
Let's express the kinetic energy $T$ and the potential energy $V$ of the mass $m$ (neglecting the rotational energy):
Dissipation function $R$
Kinetic energy $T$
Potential energy $V$
Let's express the kinetic energy $T$, the potential energy $V$ of the mass $m$ (neglecting the rotational energy) as well as the deceptive function $R$:
#+name: eq:energy_functions_lagrange
\begin{subequations}
\begin{align}
T & = \frac{1}{2} m \left( \left( \dot{u} - \Omega v \right)^2 + \left( \dot{v} + \Omega u \right)^2 \right) \\
R & = \frac{1}{2} c \left( \dot{u}^2 + \dot{v}^2 \right) \\
V & = \frac{1}{2} k \left( u^2 + v^2 \right)
T & = \frac{1}{2} m \left( \left( \dot{d}_u - \Omega d_v \right)^2 + \left( \dot{d}_v + \Omega d_u \right)^2 \right) \\
V & = \frac{1}{2} k \left( {d_u}^2 + {d_v}^2 \right) \\
R & = \frac{1}{2} c \left( \dot{d}_u{}^2 + \dot{d}_v{}^2 \right)
\end{align}
\end{subequations}
The Lagrangian is the kinetic energy minus the potential energy:
\begin{equation}
L = T - V
\end{equation}
From the Lagrange's equations of the second kind, the equation of motion is obtained ($q_1 = u$, $q_2 = v$).
The equations of motion are derived from the Lagrangian equation:
#+name: eq:lagrangian_equations
\begin{equation}
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) + \frac{\partial D}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = Q_i
\end{equation}
with $Q_i$ is the generalized force associated with the generalized variable $q_i$ ($Q_1 = F_u$ and $Q_2 = F_v$).
with $L = T - V$ is the Lagrangian and $Q_i$ is the generalized force associated with the generalized variable $q_i$ ($Q_1 = F_u$ and $Q_2 = F_v$).
#+name: eq:eom_coupled
\begin{subequations}
\begin{align}
m \ddot{u} + c \dot{u} + ( k - m \Omega ) u &= F_u + 2 m \Omega \dot{v} \\
m \ddot{v} + c \dot{v} + ( k \underbrace{-\,m \Omega}_{\text{Centrif.}} ) v &= F_v \underbrace{-\,2 m \Omega \dot{u}}_{\text{Coriolis}}
m \ddot{d}_u + c \dot{d}_u + ( k - m \Omega ) d_u &= F_u + 2 m \Omega \dot{d}_v \\
m \ddot{d}_v + c \dot{d}_v + ( k \underbrace{-\,m \Omega}_{\text{Centrif.}} ) d_v &= F_v \underbrace{-\,2 m \Omega \dot{d}_u}_{\text{Coriolis}}
\end{align}
\end{subequations}
# Explain Gyroscopic effects
The Gyroscopic effects can be seen from the two following terms:
- Coriolis Forces: coupling
- Centrifugal forces: negative stiffness
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*).
** Transfer Functions in the Laplace domain
# Laplace Domain
Using the Laplace transformation on the equations of motion eqref:eq:eom_coupled, the transfer functions from $[F_u,\ F_v]$ to $[d_u,\ d_v]$ are obtained:
#+name: eq:oem_laplace_domain
\begin{subequations}
\begin{align}
u &= \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v \\
v &= \frac{-2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v
d_u &= \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v \\
d_v &= \frac{-2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v
\end{align}
\end{subequations}
Without rotation $\Omega = 0$ and the system corresponds to two uncoupled one degree of freedom mass-spring-damper systems:
#+name: eq:oem_no_rotation
\begin{subequations}
\begin{align}
d_u &= \frac{1}{m s^2 + cs + k} F_u \\
d_v &= \frac{1}{m s^2 + cs + k} F_v
\end{align}
\end{subequations}
** Change of Variables / Parameters for the study
# Change of variables
In order this study is more independent on the system parameters, the following change of variable is performed:
- $\omega_0 = \sqrt{\frac{k}{m}}$: Natural frequency of the mass-spring system in $\si{\radian/\s}$
- $\xi = \frac{c}{2 \sqrt{k m}}$: Damping ratio
#+name: eq:tf_d
\begin{equation}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
\bm{G}_d
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} = \bm{G}_d \begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
Where $\bm{G}_d$ is a $2 \times 2$ transfer function matrix.
@@ -189,69 +202,18 @@ With:
\end{align}
\end{subequations}
- $\omega_0 = \sqrt{\frac{k}{m}}$: Natural frequency of the mass-spring system in $\si{\radian/\s}$
- $\xi$ damping ratio
$G_{dp}$ describes to poles of the system, $G_{dz}$ the zeros of the diagonal terms and $G_{dc}$ the coupling.
# Parameters
- $k = \SI{1}{N/m}$, $m = \SI{1}{kg}$, $c = \SI{0.05}{\newton\per\meter\second}$
- $\omega_0 = \SI{1}{\radian\per\second}$, $\xi = 0.025$
** Constant Rotational Speed
To simplify, let's consider a constant rotational 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 : Evolution of the poles as a function of the rotational speed $\Omega$
#+attr_latex: :environment subfigure :width 0.4\linewidth :align c
| file:figs/campbell_diagram_real.pdf | file:figs/campbell_diagram_imag.pdf |
| <<fig:campbell_diagram_real>> Real Part | <<fig:campbell_diagram_imag>> Imaginary Part |
# #+name: fig:campbell_diagram
# #+caption: Campbell Diagram
# #+attr_latex: :scale 1
# [[file:figs/campbell_diagram.pdf]]
** System Dynamics and Campbell Diagram
# Bode Plots for different ratio wr/w0
The magnitude of the coupling terms are increasing with the rotation speed.
The bode plot of $\bm{G}_d$ is shown in Figure [[fig:plant_compare_rotating_speed]].
# Describe the dynamics
#+name: fig:plant_compare_rotating_speed
#+caption: Bode Plots for $\bm{G}_d$
@@ -259,12 +221,34 @@ The magnitude of the coupling terms are increasing with the rotation speed.
| file:figs/plant_compare_rotating_speed_direct.pdf | file:figs/plant_compare_rotating_speed_coupling.pdf |
| <<fig:plant_compare_rotating_speed_direct>> Direct Terms $d_u/F_u$, $d_v/F_v$ | <<fig:plant_compare_rotating_speed_coupling>> Coupling Terms $d_v/F_u$, $d_u/F_v$ |
# #+name: fig:plant_compare_rotating_speed
# #+caption: Caption
# #+attr_latex: :scale 1
# [[file:figs/plant_compare_rotating_speed.pdf]]
* Integral Force Feedback
# Campbell Diagram
The poles are the roots of $G_{dp}$.
Two pairs of complex conjugate poles (supposing small damping $\xi \approx 0$):
\begin{subequations}
\begin{align}
p_1 &= \pm j (\omega_0 - \Omega) \\
p_2 &= \pm j (\omega_0 + \Omega)
\end{align}
\end{subequations}
When the rotation speed in non-null, the resonance frequency is split 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.
# The system goes unstable at some frequency w0
When the rotational speed $\Omega$ reaches $\omega_0$, the real part of one pair of complex conjugate becomes position meaning is system is unstable.
The stiffness of the X-Y stage is too small to hold to rotating payload hence the instability.
Stiff positioning platforms should be used if high rotational speeds or heavy payloads are used.
#+name: fig:campbell_diagram
#+caption: Campbell Diagram : Evolution of the poles as a function of the rotational speed $\Omega$
#+attr_latex: :environment subfigure :width 0.4\linewidth :align c
| file:figs/campbell_diagram_real.pdf | file:figs/campbell_diagram_imag.pdf |
| <<fig:campbell_diagram_real>> Real Part | <<fig:campbell_diagram_imag>> Imaginary Part |
* Decentralized Integral Force Feedback
** Control Schematic
Force Sensors are added in series with the actuators as shown in Figure [[fig:system_iff]].
@@ -276,53 +260,85 @@ Force Sensors are added in series with the actuators as shown in Figure [[fig:sy
#+attr_latex: :scale 1
[[file:figs/system_iff.pdf]]
** Equations
The sensed forces are equal to:
** Plant Dynamics
The forces measured by the force sensors are equal to:
#+name: eq:measured_force
\begin{equation}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\begin{bmatrix} F_u \\ F_v \end{bmatrix} - (c s + k)
\begin{bmatrix} d_u \\ d_v \end{bmatrix}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\begin{bmatrix} F_u \\ F_v \end{bmatrix} - (c s + k)
\begin{bmatrix} d_u \\ d_v \end{bmatrix}
\end{equation}
Which then gives:
Re-injecting eqref:eq:tf_d into eqref:eq:measured_force yields:
#+name: eq:tf_f
\begin{equation}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\bm{G}_{f}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} = \bm{G}_{f} \begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
Where $\bm{G}_f$ is a $2 \times 2$ transfer function matrix.
\begin{equation}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\bm{G}_f =
\frac{1}{G_{fp}}
\begin{bmatrix}
G_{fz} & -G_{fc} \\
G_{fc} & G_{fz}
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
with:
\begin{align}
G_{fp} &= \left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right)^2 + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2 \\
G_{fz} &= \left( \frac{s^2}{{\omega_0}^2} - \frac{\Omega^2}{{\omega_0}^2} \right) \left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right) + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2 \\
G_{fc} &= \left( 2 \xi \frac{s}{\omega_0} + 1 \right) \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)
\end{align}
** Plant Dynamics
# Explain the two real zeros => change of gain but not of phase
# The alternating poles and zeros properties of collocated IFF holds
# but additional real zeros are added
The zeros of the diagonal terms are the roots of $G_{fz}$ (supposing small damping):
\begin{subequations}
\begin{align}
z_1 &= \pm j \omega_0 \sqrt{\frac{1}{2} \sqrt{8 \frac{\Omega^2}{{\omega_0}^2} + 1} + \frac{\Omega^2}{{\omega_0}^2} + \frac{1}{2} } \\
z_2 &= \pm \omega_0 \sqrt{\frac{1}{2} \sqrt{8 \frac{\Omega^2}{{\omega_0}^2} + 1} - \frac{\Omega^2}{{\omega_0}^2} - \frac{1}{2} }
\end{align}
\end{subequations}
The frequency of the two complex conjugate zeros $z_1$ is between the frequency of the two pairs of complex conjugate poles $p_1$ and $p_2$.
This is the expected behavior of a collocated pair of actuator and sensor.
However, the two real zeros $z_2$ induces an increase of +2 of the slope without change of phase (Figure [[fig:plant_iff_compare_rotating_speed]]).
This represents non-minimum phase behavior.
# Explain physically why the real zeros
# Show that the low frequency gain is no longer zero
The low frequency gain, for $\Omega < \omega_0$, is no longer zero:
#+name: low_freq_gain_iff_plan
\begin{equation}
\bm{G}_{f0} = \lim_{\omega \to 0} \left| \bm{G}_f (j\omega) \right| = \begin{bmatrix}
\frac{- \Omega^2}{{\omega_0}^2 - \Omega^2} & 0 \\
0 & \frac{- \Omega^2}{{\omega_0}^2 - \Omega^2}
\end{bmatrix}
\end{equation}
It increase with the rotational speed $\Omega$.
#+name: fig:plant_iff_compare_rotating_speed
#+caption: Bode plot of $\bm{G}_f$ for several rotational speeds $\Omega$
#+attr_latex: :scale 1
[[file:figs/plant_iff_compare_rotating_speed.pdf]]
# Show that the low frequency gain is no longer zero
** Decentralized Integral Force Feedback
# Write the analytical value of the low frequency gain
\begin{equation}
\bm{K}_F(s) = g \cdot \frac{1}{s}
\end{equation}
# Explain the two real zeros => change of gain but not of phase
# Explain physically why
** Integral Force Feedback
# Problem of zero with a positive real part
Also, as one zero has a positive real part, the *IFF control is no more unconditionally stable*.
This is due to the fact that the zeros of the plant are the poles of the closed loop system with an infinite gain.
Thus, for some finite IFF gain, one pole will have a positive real part and the system will be unstable.
# General explanation for the Root Locus Plot
@@ -335,6 +351,9 @@ Which then gives:
#+attr_latex: :scale 1
[[file:figs/root_locus_pure_iff.pdf]]
# IFF is usually known for its guaranteed stability (add reference)
# This is not the case anymore due to gyroscopic effects
# Physical Interpretation
At low frequency, the gain is very large and thus no force is transmitted between the payload and the rotating stage.
@@ -350,12 +369,16 @@ This means that at low frequency, the system is decoupled (the force sensor remo
# Equation with the new control law
\begin{equation}
\bm{K}_{F}(s) = g \cdot \frac{1}{s} \cdot \underbrace{\frac{s/\omega_i}{1 + s/\omega_i}}_{\text{HPF}} = g \cdot \frac{1}{s + \omega_i}
\end{equation}
# Explain why it is usually done and why it is done here: the problem is the high gain at low frequency => high pass filter
** Feedback Analysis
# Explain what do we mean for Loop Gain (loop gain for the decentralized loop)
# Explain that now the low frequency loop gain does not reach a gain more than 1 (if g not so high)
@@ -365,7 +388,10 @@ This means that at low frequency, the system is decoupled (the force sensor remo
[[file:figs/loop_gain_modified_iff.pdf]]
# Not the system can be stable for small values of g
# Actually, the system becomes unstable for g > ...
# Actually, the system becomes unstable for g > ... => it has been verified
\begin{equation}
g_\text{max} = \omega_i \left( \frac{{\omega_0}^2}{\Omega^2} - 1 \right) \label{eq:iff_gmax}
\end{equation}
#+name: fig:root_locus_modified_iff
#+caption: Root Locus for IFF with and without the HPF
@@ -414,16 +440,50 @@ This means that at low frequency, the system is decoupled (the force sensor remo
# Maybe add the fact that this is equivalent to amplified piezo for instance
** Plant Dynamics
# Equations: sensed force
\begin{equation}
\begin{bmatrix} f_u \\ f_v \end{bmatrix} =
\bm{G}_k
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
\begin{equation}
\begin{bmatrix} f_u \\ f_v \end{bmatrix} =
\frac{1}{G_{kp}}
\begin{bmatrix}
G_{kz} & -G_{kc} \\
G_{kc} & G_{kz}
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
With:
\begin{align}
G_{kp} &= \left( \frac{s^2}{{\omega_0^\prime}^2} + 2\xi^\prime \frac{s}{{\omega_0^\prime}^2} + 1 - \frac{\Omega^2}{{\omega_0^\prime}^2} \right)^2 + \left( 2 \frac{\Omega}{\omega_0^\prime}\frac{s}{\omega_0^\prime} \right)^2 \\
G_{kz} &= \left( \frac{s^2}{{\omega_0^\prime}^2} + \frac{k_p}{k + k_p} - \frac{\Omega^2}{{\omega_0^\prime}^2} \right) \left( \frac{s^2}{{\omega_0^\prime}^2} + 2\xi^\prime \frac{s}{{\omega_0^\prime}^2} + 1 - \frac{\Omega^2}{{\omega_0^\prime}^2} \right) + \left( 2 \frac{\Omega}{\omega_0^\prime}\frac{s}{\omega_0^\prime} \right)^2 \\
G_{kc} &= \left( 2 \xi^\prime \frac{s}{\omega_0^\prime} + \frac{k}{k + k_p} \right) \left( 2 \frac{\Omega}{\omega_0^\prime}\frac{s}{\omega_0^\prime} \right)
\end{align}
# New parameters
where:
- $\omega_0^\prime = \frac{k + k_p}{m}$
- $\xi^\prime = \frac{c}{2 \sqrt{(k + k_p) m}}$
** Effect of the Parallel Stiffness on the Plant Dynamics
# Negative Stiffness due to rotation => the stiffness should be larger than that
# TODO: Verify that
# For kp < negative stiffness => real zeros
# For kp > negative stiffness => complex conjugate zeros
\begin{equation}
\frac{k_p}{k + k_p} - \frac{\Omega^2}{{\omega_0^\prime}^2} > 0
\end{equation}
Which is equivalent to
\begin{equation}
k_p > m \Omega^2
\end{equation}
#+name: fig:plant_iff_kp
#+caption: Bode Plot of $f_u/F_u$ without parallel spring, with parallel springs with stiffness $k_p < m \Omega^2$ and $k_p > m \Omega^2$
@@ -432,6 +492,8 @@ This means that at low frequency, the system is decoupled (the force sensor remo
# Location of the zeros as a function of kp
# Try to show that we don't have anymore real zeros that was making the system non-minimum phase
# Show that it is the case on the root locus
#+name: fig:root_locus_iff_kp
@@ -445,6 +507,10 @@ This means that at low frequency, the system is decoupled (the force sensor remo
# Explain that we have k = ka + kp constant in order to have the same resonance
# Attainable damping generally proportional to the distance between the poles and zeros (add reference, probably preumont)
# The zero is the poles of the system without the force sensors => w =
# Thus, small kp is wanted: kp close to m Omega^2 should give the optimal damping but is not acceptable for robustness reasons
# Large Stiffness decreases the attainable damping
@@ -482,6 +548,27 @@ This means that at low frequency, the system is decoupled (the force sensor remo
# Write the equations
\begin{equation}
\begin{bmatrix} v_u \\ v_v \end{bmatrix} =
\bm{G}_v
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
\begin{equation}
\begin{bmatrix} v_u \\ v_v \end{bmatrix} =
\frac{s}{k} \frac{1}{G_{vp}}
\begin{bmatrix}
G_{vz} & G_{vc} \\
-G_{vc} & G_{vz}
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
With:
\begin{align}
G_{vp} &= \left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right)^2 + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2 \\
G_{vz} &= \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \\
G_{vc} &= 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0}
\end{align}
# Show that the rotation have somehow less impact on the plant than for IFF

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@@ -1,4 +1,4 @@
% Created 2020-06-24 mer. 16:28
% Created 2020-06-25 jeu. 10:07
% Intended LaTeX compiler: pdflatex
\documentclass{ISMA_USD2020}
\usepackage[utf8]{inputenc}
@@ -36,7 +36,14 @@
\usepackage{tikz}
\usetikzlibrary{shapes.misc}
\date{}
\title{Active Damping of Rotating Positioning Platforms}
\title{Decentralized Active Damping of Rotating Positioning Platforms}
\hypersetup{
pdfauthor={},
pdftitle={Decentralized Active Damping of Rotating Positioning Platforms},
pdfkeywords={},
pdfsubject={},
pdfcreator={Emacs 27.0.91 (Org mode 9.4)},
pdflang={English}}
\begin{document}
\maketitle
@@ -46,7 +53,7 @@
}
\section{Introduction}
\label{sec:org3cbd2ff}
\label{sec:org5780a8f}
\label{sec:introduction}
Controller Poles are shown by black crosses (
\begin{tikzpicture} \node[cross out, draw=black, minimum size=1ex, line width=2pt, inner sep=0pt, outer sep=0pt] at (0, 0){}; \end{tikzpicture}
@@ -55,9 +62,9 @@ This paper has been published
The Matlab code that was use to obtain the results are available in \cite{dehaeze20_activ_dampin_rotat_posit_platf}.
\section{Dynamics of Rotating Positioning Platforms}
\label{sec:org3cf58d1}
\label{sec:orga8db619}
\subsection{Studied Rotating Positioning Platform}
\label{sec:orgf321431}
\label{sec:org70ddefe}
Consider the rotating X-Y stage of Figure \ref{fig:system}.
\begin{itemize}
@@ -74,66 +81,75 @@ Consider the rotating X-Y stage of Figure \ref{fig:system}.
\caption{\label{fig:system}Schematic of the studied System}
\end{figure}
\subsection{Equations of Motion}
\label{sec:org647b64d}
The system has two degrees of freedom and is thus fully described by the generalized coordinates \([q_1\ q_2] = [d_u\ d_v]\) (describing the position of the mass in the rotating frame).
\subsection{Equation of Motion}
\label{sec:org9612ace}
The system has two degrees of freedom and is thus fully described by the generalized coordinates \(u\) and \(v\) (describing the position of the mass in the rotating frame).
Let's express the kinetic energy \(T\) and the potential energy \(V\) of the mass \(m\) (neglecting the rotational energy):
Dissipation function \(R\)
Kinetic energy \(T\)
Potential energy \(V\)
Let's express the kinetic energy \(T\), the potential energy \(V\) of the mass \(m\) (neglecting the rotational energy) as well as the deceptive function \(R\):
\begin{subequations}
\label{eq:energy_functions_lagrange}
\begin{align}
T & = \frac{1}{2} m \left( \left( \dot{u} - \Omega v \right)^2 + \left( \dot{v} + \Omega u \right)^2 \right) \\
R & = \frac{1}{2} c \left( \dot{u}^2 + \dot{v}^2 \right) \\
V & = \frac{1}{2} k \left( u^2 + v^2 \right)
T & = \frac{1}{2} m \left( \left( \dot{d}_u - \Omega d_v \right)^2 + \left( \dot{d}_v + \Omega d_u \right)^2 \right) \\
V & = \frac{1}{2} k \left( {d_u}^2 + {d_v}^2 \right) \\
R & = \frac{1}{2} c \left( \dot{d}_u{}^2 + \dot{d}_v{}^2 \right)
\end{align}
\end{subequations}
The Lagrangian is the kinetic energy minus the potential energy:
\begin{equation}
L = T - V
\end{equation}
From the Lagrange's equations of the second kind, the equation of motion is obtained (\(q_1 = u\), \(q_2 = v\)).
The equations of motion are derived from the Lagrangian equation:
\begin{equation}
\label{eq:lagrangian_equations}
\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) + \frac{\partial D}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = Q_i
\end{equation}
with \(Q_i\) is the generalized force associated with the generalized variable \(q_i\) (\(Q_1 = F_u\) and \(Q_2 = F_v\)).
with \(L = T - V\) is the Lagrangian and \(Q_i\) is the generalized force associated with the generalized variable \(q_i\) (\(Q_1 = F_u\) and \(Q_2 = F_v\)).
\begin{subequations}
\label{eq:eom_coupled}
\begin{align}
m \ddot{u} + c \dot{u} + ( k - m \Omega ) u &= F_u + 2 m \Omega \dot{v} \\
m \ddot{v} + c \dot{v} + ( k \underbrace{-\,m \Omega}_{\text{Centrif.}} ) v &= F_v \underbrace{-\,2 m \Omega \dot{u}}_{\text{Coriolis}}
m \ddot{d}_u + c \dot{d}_u + ( k - m \Omega ) d_u &= F_u + 2 m \Omega \dot{d}_v \\
m \ddot{d}_v + c \dot{d}_v + ( k \underbrace{-\,m \Omega}_{\text{Centrif.}} ) d_v &= F_v \underbrace{-\,2 m \Omega \dot{d}_u}_{\text{Coriolis}}
\end{align}
\end{subequations}
The Gyroscopic effects can be seen from the two following terms:
\begin{itemize}
\item Coriolis Forces: coupling
\item Centrifugal forces: negative stiffness
\end{itemize}
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 \textbf{centrifugal forces}).
\subsection{Transfer Functions in the Laplace domain}
\label{sec:org1590670}
\label{sec:org55c9228}
Using the Laplace transformation on the equations of motion \eqref{eq:eom_coupled}, the transfer functions from \([F_u,\ F_v]\) to \([d_u,\ d_v]\) are obtained:
\begin{subequations}
\label{eq:oem_laplace_domain}
\begin{align}
u &= \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v \\
v &= \frac{-2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v
d_u &= \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v \\
d_v &= \frac{-2 m \Omega s}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_u + \frac{ms^2 + cs + k - m \Omega^2}{\left( m s^2 + cs + k - m \Omega^2 \right)^2 + \left( 2 m \Omega s \right)^2} F_v
\end{align}
\end{subequations}
Without rotation \(\Omega = 0\) and the system corresponds to two uncoupled one degree of freedom mass-spring-damper systems:
\begin{subequations}
\label{eq:oem_no_rotation}
\begin{align}
d_u &= \frac{1}{m s^2 + cs + k} F_u \\
d_v &= \frac{1}{m s^2 + cs + k} F_v
\end{align}
\end{subequations}
\subsection{Change of Variables / Parameters for the study}
\label{sec:orgb7d090c}
In order this study is more independent on the system parameters, the following change of variable is performed:
\begin{itemize}
\item \(\omega_0 = \sqrt{\frac{k}{m}}\): Natural frequency of the mass-spring system in \(\si{\radian/\s}\)
\item \(\xi = \frac{c}{2 \sqrt{k m}}\): Damping ratio
\end{itemize}
\begin{equation}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} =
\bm{G}_d
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\label{eq:tf_d}
\begin{bmatrix} d_u \\ d_v \end{bmatrix} = \bm{G}_d \begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
Where \(\bm{G}_d\) is a \(2 \times 2\) transfer function matrix.
@@ -153,68 +169,11 @@ With:
\end{align}
\end{subequations}
\begin{itemize}
\item \(\omega_0 = \sqrt{\frac{k}{m}}\): Natural frequency of the mass-spring system in \(\si{\radian/\s}\)
\item \(\xi\) damping ratio
\end{itemize}
\(G_{dp}\) describes to poles of the system, \(G_{dz}\) the zeros of the diagonal terms and \(G_{dc}\) the coupling.
\subsection{Constant Rotational Speed}
\label{sec:orgd9375df}
To simplify, let's consider a constant rotational speed \(\dot{\theta} = \Omega\) and thus \(\ddot{\theta} = 0\).
\begin{equation}
\label{eq:coupledplant}
\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}
\begin{equation}
\label{eq:coupled_plant}
\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.
\begin{equation}
\label{eq:coupled_plant_no_rot}
\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}
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 \ref{fig:campbell_diagram}).
\begin{figure}[htbp]
\begin{subfigure}[c]{0.4\linewidth}
\includegraphics[width=\linewidth]{figs/campbell_diagram_real.pdf}
\caption{\label{fig:campbell_diagram_real} Real Part}
\end{subfigure}
\begin{subfigure}[c]{0.4\linewidth}
\includegraphics[width=\linewidth]{figs/campbell_diagram_imag.pdf}
\caption{\label{fig:campbell_diagram_imag} Imaginary Part}
\end{subfigure}
\caption{\label{fig:campbell_diagram}Campbell Diagram : Evolution of the poles as a function of the rotational speed \(\Omega\)}
\centering
\end{figure}
The magnitude of the coupling terms are increasing with the rotation speed.
\subsection{System Dynamics and Campbell Diagram}
\label{sec:org24f5f5f}
The bode plot of \(\bm{G}_d\) is shown in Figure \ref{fig:plant_compare_rotating_speed}.
\begin{figure}[htbp]
\begin{subfigure}[c]{0.45\linewidth}
@@ -230,10 +189,33 @@ The magnitude of the coupling terms are increasing with the rotation speed.
\end{figure}
\section{Integral Force Feedback}
\label{sec:org95f47e8}
When the rotation speed in non-null, the resonance frequency is split 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.
When the rotational speed \(\Omega\) reaches \(\omega_0\), the real part of one pair of complex conjugate becomes position meaning is system is unstable.
The stiffness of the X-Y stage is too small to hold to rotating payload hence the instability.
Stiff positioning platforms should be used if high rotational speeds or heavy payloads are used.
\begin{figure}[htbp]
\begin{subfigure}[c]{0.4\linewidth}
\includegraphics[width=\linewidth]{figs/campbell_diagram_real.pdf}
\caption{\label{fig:campbell_diagram_real} Real Part}
\end{subfigure}
\begin{subfigure}[c]{0.4\linewidth}
\includegraphics[width=\linewidth]{figs/campbell_diagram_imag.pdf}
\caption{\label{fig:campbell_diagram_imag} Imaginary Part}
\end{subfigure}
\caption{\label{fig:campbell_diagram}Campbell Diagram : Evolution of the poles as a function of the rotational speed \(\Omega\)}
\centering
\end{figure}
\section{Decentralized Integral Force Feedback}
\label{sec:orgd957fd6}
\subsection{Control Schematic}
\label{sec:org8bb26ea}
\label{sec:orgc01d8cf}
Force Sensors are added in series with the actuators as shown in Figure \ref{fig:system_iff}.
@@ -244,29 +226,28 @@ Force Sensors are added in series with the actuators as shown in Figure \ref{fig
\end{figure}
\subsection{Equations}
\label{sec:orgbd9ebe0}
The sensed forces are equal to:
\label{sec:orge5896ec}
The forces measured by the force sensors are equal to:
\begin{equation}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\begin{bmatrix} F_u \\ F_v \end{bmatrix} - (c s + k)
\begin{bmatrix} d_u \\ d_v \end{bmatrix}
\label{eq:measured_force}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\begin{bmatrix} F_u \\ F_v \end{bmatrix} - (c s + k)
\begin{bmatrix} d_u \\ d_v \end{bmatrix}
\end{equation}
Which then gives:
Reinjecting \eqref{eq:tf_d} into \eqref{eq:measured_force} yields:
\begin{equation}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\bm{G}_{f}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\label{eq:tf_f}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} = \bm{G}_{f} \begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
\begin{equation}
\begin{bmatrix} f_{u} \\ f_{v} \end{bmatrix} =
\bm{G}_f =
\frac{1}{G_{fp}}
\begin{bmatrix}
G_{fz} & -G_{fc} \\
G_{fc} & G_{fz}
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
\begin{align}
@@ -275,8 +256,17 @@ Which then gives:
G_{fc} &= \left( 2 \xi \frac{s}{\omega_0} + 1 \right) \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)
\end{align}
\begin{equation}
\bm{G}_f =
\frac{1}{\left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right)^2 + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2}
\begin{bmatrix}
\left( \frac{s^2}{{\omega_0}^2} - \frac{\Omega^2}{{\omega_0}^2} \right) \left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right) + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2 & -G_{fc} \\
G_{fc} & \left( \frac{s^2}{{\omega_0}^2} - \frac{\Omega^2}{{\omega_0}^2} \right) \left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right) + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2
\end{bmatrix}
\end{equation}
\subsection{Plant Dynamics}
\label{sec:org392809f}
\label{sec:org0a22a10}
\begin{figure}[htbp]
\centering
@@ -284,9 +274,8 @@ Which then gives:
\caption{\label{fig:plant_iff_compare_rotating_speed}Bode plot of \(\bm{G}_f\) for several rotational speeds \(\Omega\)}
\end{figure}
\subsection{Integral Force Feedback}
\label{sec:org049877c}
\subsection{Problems with Integral Force Feedback}
\label{sec:orgd432439}
\begin{figure}[htbp]
\centering
@@ -297,15 +286,17 @@ Which then gives:
At low frequency, the gain is very large and thus no force is transmitted between the payload and the rotating stage.
This means that at low frequency, the system is decoupled (the force sensor removed) and thus the system is unstable.
\section{Integral Force Feedback with High Pass Filters}
\label{sec:org54452db}
\label{sec:org2e1883a}
\subsection{Modification of the Control Low}
\label{sec:org325cdd4}
\label{sec:org218110f}
\begin{equation}
\bm{K}_{F}(s) = \frac{1}{s} \underbrace{\frac{s/\omega_i}{1 + s/\omega_i}}_{\text{HPF}} = \frac{1}{s + \omega_i}
\end{equation}
\subsection{Feedback Analysis}
\label{sec:org5efee77}
\label{sec:org03090fc}
\begin{figure}[htbp]
\centering
@@ -320,7 +311,7 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\end{figure}
\subsection{Optimal Cut-Off Frequency}
\label{sec:orgd5828e4}
\label{sec:org6ba4e55}
\begin{figure}[htbp]
\centering
@@ -335,9 +326,9 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\end{figure}
\section{Integral Force Feedback with Parallel Springs}
\label{sec:org22884d6}
\label{sec:org90ec20f}
\subsection{Stiffness in Parallel with the Force Sensor}
\label{sec:orgb871bfd}
\label{sec:org60d6640}
\begin{figure}[htbp]
\centering
@@ -345,25 +336,59 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\caption{\label{fig:system_parallel_springs}System with added springs \(k_p\) in parallel with the actuators}
\end{figure}
\begin{equation}
\begin{bmatrix} f_u \\ f_v \end{bmatrix} =
\bm{G}_k
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
\begin{equation}
\begin{bmatrix} f_u \\ f_v \end{bmatrix} =
\frac{1}{G_{kp}}
\begin{bmatrix}
G_{kz} & -G_{kc} \\
G_{kc} & G_{kz}
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
With:
\begin{align}
G_{kp} &= \left( \frac{s^2}{{\omega_0^\prime}^2} + 2\xi^\prime \frac{s}{{\omega_0^\prime}^2} + 1 - \frac{\Omega^2}{{\omega_0^\prime}^2} \right)^2 + \left( 2 \frac{\Omega}{\omega_0^\prime}\frac{s}{\omega_0^\prime} \right)^2 \\
G_{kz} &= \left( \frac{s^2}{{\omega_0^\prime}^2} + \frac{k_p}{k + k_p} - \frac{\Omega^2}{{\omega_0^\prime}^2} \right) \left( \frac{s^2}{{\omega_0^\prime}^2} + 2\xi^\prime \frac{s}{{\omega_0^\prime}^2} + 1 - \frac{\Omega^2}{{\omega_0^\prime}^2} \right) + \left( 2 \frac{\Omega}{\omega_0^\prime}\frac{s}{\omega_0^\prime} \right)^2 \\
G_{kc} &= \left( 2 \xi^\prime \frac{s}{\omega_0^\prime} + \frac{k}{k + k_p} \right) \left( 2 \frac{\Omega}{\omega_0^\prime}\frac{s}{\omega_0^\prime} \right)
\end{align}
where:
\begin{itemize}
\item \(\omega_0^\prime = \frac{k + k_p}{m}\)
\item \(\xi^\prime = \frac{c}{2 \sqrt{(k + k_p) m}}\)
\end{itemize}
\subsection{Effect of the Parallel Stiffness on the Plant Dynamics}
\label{sec:org4d37cce}
\label{sec:org3ec34fe}
\begin{equation}
\frac{k_p}{k + k_p} - \frac{\Omega^2}{{\omega_0^\prime}^2} > 0
\end{equation}
Which is equivalent to
\begin{equation}
k_p > m \Omega^2
\end{equation}
\begin{figure}[htbp]
\centering
\includegraphics[scale=1]{figs/plant_iff_kp.pdf}
\caption{\label{fig:plant_iff_kp}Bode Plot of \(f_u/F_u\) without any parallel stiffness, with a parallel stiffness \(k_p < m \Omega^2\) and with \(k_p > m \Omega^2\)}
\caption{\label{fig:plant_iff_kp}Bode Plot of \(f_u/F_u\) without parallel spring, with parallel springs with stiffness \(k_p < m \Omega^2\) and \(k_p > m \Omega^2\)}
\end{figure}
\begin{figure}[htbp]
\centering
\includegraphics[scale=1]{figs/root_locus_iff_kp.pdf}
\caption{\label{fig:root_locus_iff_kp}Root Locus for IFF without any parallel stiffness, with a parallel stiffness \(k_p < m \Omega^2\) and with \(k_p > m \Omega^2\)}
\caption{\label{fig:root_locus_iff_kp}Root Locus for IFF without parallel spring, with parallel springs with stiffness \(k_p < m \Omega^2\) and \(k_p > m \Omega^2\)}
\end{figure}
\subsection{Optimal Parallel Stiffness}
\label{sec:orgd19b212}
\label{sec:org9c47159}
\begin{figure}[htbp]
\centering
@@ -379,9 +404,9 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\end{figure}
\section{Direct Velocity Feedback}
\label{sec:org6904969}
\label{sec:org5cb3076}
\subsection{Control Schematic}
\label{sec:org103e18b}
\label{sec:orgaaa522f}
\begin{figure}[htbp]
\centering
@@ -389,14 +414,34 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\caption{\label{fig:system_dvf}System with relative velocity sensors and with decentralized controllers \(K_V\)}
\end{figure}
\subsection{Equations}
\label{sec:org793c22d}
\label{sec:orge0a4555}
\begin{equation}
\begin{bmatrix} v_u \\ v_v \end{bmatrix} =
\bm{G}_v
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
\begin{equation}
\begin{bmatrix} v_u \\ v_v \end{bmatrix} =
\frac{s}{k} \frac{1}{G_{vp}}
\begin{bmatrix}
G_{vz} & G_{vc} \\
-G_{vc} & G_{vz}
\end{bmatrix}
\begin{bmatrix} F_u \\ F_v \end{bmatrix}
\end{equation}
With:
\begin{align}
G_{vp} &= \left( \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \right)^2 + \left( 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0} \right)^2 \\
G_{vz} &= \frac{s^2}{{\omega_0}^2} + 2 \xi \frac{s}{\omega_0} + 1 - \frac{{\Omega}^2}{{\omega_0}^2} \\
G_{vc} &= 2 \frac{\Omega}{\omega_0} \frac{s}{\omega_0}
\end{align}
\subsection{Relative Direct Velocity Feedback}
\label{sec:orgc28d518}
\label{sec:org5401110}
\begin{figure}[htbp]
\centering
@@ -405,14 +450,14 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\end{figure}
\section{Comparison of the Proposed Active Damping Techniques for Rotating Positioning Stages}
\label{sec:org6af1fdb}
\label{sec:org4cbf163}
\subsection{Physical Comparison}
\label{sec:orgdff3aa2}
\label{sec:org5eba275}
\subsection{Attainable Damping}
\label{sec:org22c8f42}
\label{sec:org44635e0}
\begin{figure}[htbp]
\centering
@@ -422,7 +467,7 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\subsection{Transmissibility and Compliance}
\label{sec:org3e2cf56}
\label{sec:org58e9594}
\begin{figure}[htbp]
@@ -438,13 +483,12 @@ This means that at low frequency, the system is decoupled (the force sensor remo
\centering
\end{figure}
\section{Conclusion}
\label{sec:orge292803}
\label{sec:org292b448}
\label{sec:conclusion}
\section*{Acknowledgment}
\label{sec:orgaf681fb}
\label{sec:orgff7af07}
\bibliography{ref.bib}
\end{document}

13
paper/ref.bib
View File

@@ -56,3 +56,16 @@
year = {2018},
publisher = {Springer},
}
@inproceedings{preumont91_activ,
author = {Andre Preumont and Jean-Paul Dufour and Christian Malekian},
title = {Active damping by a local force feedback with piezoelectric
actuators},
booktitle = {32nd Structures, Structural Dynamics, and Materials
Conference},
year = 1991,
doi = {10.2514/6.1991-989},
url = {https://doi.org/10.2514/6.1991-989},
month = {apr},
publisher = {American Institute of Aeronautics and Astronautics},
}