# Control in a rotating frame

## Table of Contents

- 1. Introduction
- 2. System
  - 2.1. System description
  - 2.2. Equations
  - 2.3. Numerical Values for the NASS
  - 2.4. Euler and Coriolis forces - Numerical Result
  - 2.5. Negative Spring Effect - Numerical Result
  - 2.6. Limitations due to coupling
    - 2.6.1. Numerical Analysis
  - 2.7. Limitations due to negative stiffness effect
- 3. Control Strategies
  - 3.1. Measurement in the fixed reference frame
  - 3.2. Measurement in the rotating frame
- 4. Multi Body Model - Simscape
  - 4.1. Parameter for the Simscape simulations
  - 4.2. Identification in the rotating referenced frame
    - 4.2.1. Low rotation speed and High rotation speed
  - 4.3. Identification in the fixed frame
  - 4.4. Identification from actuator forces to displacement in the fixed frame
  - 4.5. Effect of the rotating Speed
    - 4.5.1. TODO Use realistic parameters for the mass of the sample and stiffness of the X-Y stage
    - 4.5.2. TODO Check if the plant is changing a lot when we are not turning to when we are turning at the maximum speed (60rpm)
  - 4.6. Effect of the X-Y stage stiffness
    - 4.6.1. TODO At full speed, check how the coupling changes with the stiffness of the actuators
- 5. Control Implementation
  - 5.1. Measurement in the fixed reference frame

## 1 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 2, 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 3, 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 4, the analytical study will be validated using a multi body model of the studied system.

Finally, in section 5, the control strategies are implemented using Simulink and Simscape and compared.

## 2 System

### 2.1 System description

The system consists of one 2 degree of freedom translation stage on top of a spindle (figure 1).

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.

Figure 1: Schematic of the mecanical system

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 \(\vec{i}_v\)

### 2.2 Equations

Based on the figure 1, 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 \(m\):

\begin{align} \label{org93a4d45} 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.

\begin{equation} \label{org19136da} 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:

\begin{align*} \label{org4fc9f2b} \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{equation} \label{orgf3ca0ca} 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} \begin{equation} \label{org5e2eb96} 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}

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.

### 2.3 Numerical Values for the NASS

Let's define the parameters for the NASS.

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

### 2.4 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: \(2 m \dot{d_v} \dot{\theta}\)

The obtained values are displayed in table 1.

Table 1: Euler and Coriolis forces for the NASS
| Light | Heavy
--- | --- | ---
Coriolis | 88.0N | 3.6N
Euler | 0.4N | 0.8N

### 2.5 Negative Spring Effect - Numerical Result

The negative stiffness due to the rotation is equal to \(-m{\omega_0}^2\).

The values for the negative spring effect are displayed in table 2.

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

Table 2: Negative Spring effect
| Light | Heavy
--- | --- | ---
Neg. Spring | 1381.7[N/m] | 0.9[N/m]

### 2.6 Limitations due to coupling

To simplify, we consider a constant rotating speed \(\dot{\theta} = {\omega_0}\) and thus \(\ddot{\theta} = 0\).

From equations \eqref{orgf3ca0ca} and \eqref{org5e2eb96}, 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{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}

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

#### 2.6.1 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 2).
- with the heavy sample (figure 3).

Figure 2: Relative Coupling for light mass and high rotation speed

Figure 3: Relative Coupling for heavy mass and low rotation speed

Coupling is higher for actuators with small stiffness.

### 2.7 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 3.

Table 3: Maximum rotation speed at which negative stiffness is negligible (\(0.1\sqrt{\frac{k}{m}}\))
| 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.

Negative stiffness effect has very important effect when using soft actuators.

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 4)

Table 4: Minimum possible stiffness
| Light | Heavy
--- | --- | ---
k min [N/m] | 2199 | 89

## 3 Control Strategies

### 3.1 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 4.

Figure 4: Control with a measure from fixed frame

The loop gain is then \(L = G(\theta) K J(\theta)\).

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

### 3.2 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 5

Figure 5: Control with a measure from rotating frame

The loop gain is \(L = G K\).

## 4 Multi Body Model - Simscape

### 4.1 Parameter for the Simscape simulations

    w = 2*pi; % Rotation speed [rad/s]

    theta_e = 0; % Static measurement error on the angle theta [rad]

    m = 5; % mass of the sample [kg]

    mTuv = 30;% Mass of the moving part of the Tuv stage [kg]
    kTuv = 1e8; % Stiffness of the Tuv stage [N/m]
    cTuv = 0; % Damping of the Tuv stage [N/(m/s)]

    mlight = 5; % Mass for light sample [kg]
    mheavy = 55; % 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]

    d = 0.01; % Maximum excentricity from rotational axis [m]

### 4.2 Identification in the rotating referenced frame

We initialize the inputs and outputs of the system to identify.

    %% 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');

We start we identify the transfer functions at high speed with the light sample.

    w = wlight; % Rotation speed [rad/s]
    m = mlight; % mass of the sample [kg]

    kTuv = kpz;
    Gpz_light = linearize(mdl, io, 0.1);
    Gpz_light.InputName  = {'Fu', 'Fv'};
    Gpz_light.OutputName = {'Du', 'Dv'};

    kTuv = kvc;
    Gvc_light = linearize(mdl, io, 0.1);
    Gvc_light.InputName  = {'Fu', 'Fv'};
    Gvc_light.OutputName = {'Du', 'Dv'};

Then we identify the system with an heavy mass and low speed.

    w = wheavy; % Rotation speed [rad/s]
    m = mheavy; % mass of the sample [kg]

    kTuv = kpz;
    Gpz_heavy = linearize(mdl, io, 0.1);
    Gpz_heavy.InputName  = {'Fu', 'Fv'};
    Gpz_heavy.OutputName = {'Du', 'Dv'};

    kTuv = kvc;
    Gvc_heavy = linearize(mdl, io, 0.1);
    Gvc_heavy.InputName  = {'Fu', 'Fv'};
    Gvc_heavy.OutputName = {'Du', 'Dv'};

Finally, we plot the coupling ratio in both case (figure 6). We obtain the same result than the analytical case (figures 2 and 3).

And then with the heavy sample.

    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'};

Plot the ratio between the main transfer function and the coupling term:

#### 4.2.1 Low rotation speed and High rotation speed

    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'};

    figure;
    bode(G_low, G_high);

### 4.3 Identification in the fixed frame

Let's define some options as well as the inputs and outputs for linearization.

    %% 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');

We then define the error estimation of the error and the rotational speed.

    %% Run the linearization
    angle_e = 0;
    rot_speed = 0;

Finally, we run the linearization.

    G = linearize(mdl, io, 0);

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

    %% 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'};

    %% 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'};

    figure;
    bode(G);
    % exportFig('G_x_y', 'wide-tall');

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

### 4.4 Identification from actuator forces to displacement in the fixed frame

    %% 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');

    rot_speed = 2*pi;
    angle_e = 0;
    G = linearize(mdl, io, 0.0);

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

    rot_speed = 2*pi;
    angle_e = 0;
    G1 = linearize(mdl, io, 0.4);

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

    rot_speed = 2*pi;
    angle_e = 0;
    G2 = linearize(mdl, io, 0.8);

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

    figure;
    bode(G, G1, G2);
    exportFig('G_u_v_to_x_y', 'wide-tall');

### 4.5 Effect of the rotating Speed

#### 4.5.1 TODO Use realistic parameters for the mass of the sample and stiffness of the X-Y stage

#### 4.5.2 TODO Check if the plant is changing a lot when we are not turning to when we are turning at the maximum speed (60rpm)

### 4.6 Effect of the X-Y stage stiffness

#### 4.6.1 TODO At full speed, check how the coupling changes with the stiffness of the actuators

## 5 Control Implementation

### 5.1 Measurement in the fixed reference frame

Author: Thomas Dehaeze

Created: 2019-01-23 mer. 15:21

Validate