
%===================================================================================== 
%\chapter{Determination of an ellipse's parameters with any 
%orientation in a tridimensionel cartesian placemark by knowledge 
%of its three components x, y, z in this placemark}
%=====================================================================================


%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

\cleardoublepage
\chapter{Method}\label{MET}

%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

\section{Goal}

A search-coil magnetometer provides 3 waveforms $B_{x}, B_{y}, B_{z}$, depending on time.
Main assumption in the following method is to consider the signal $B_{i} (t)$ as a sum of plane waves of various frequency,
each wave corresponding to a single $\vec{k}$ vector, i.e. an elliptic wave (see fig \ref{figure_1}).


 
\begin{figure}[!h]
\centering
\label{figure_1}
\epsfysize=5.0cm
\epsffile{./figures/figure_1.eps}
\caption{Elliptic plane wave for a single frequency, in a fixed measurement coordinate system. }
\end{figure}


The heart of the method is to do a Fourier analysis of the $B_{i}$(t) signals, assuming that each ray of the spectrum 
$\widetilde{B_{i}}(f_{k})$ corresponds to a plane wave.
So, the goal is to determine all the parameters of the ellipse corresponding to the ray $f_{k}$ from the $B_{i}(t)$ signals,
 whatever the initial coordinate system.
 \\
 \\
 Of course, for each frequency f corresponds an ellipse whose parameters are to be computed : \\
 \\
$ \bullet \hspace{1 mm} \vec{X_{i}}, \vec{Y_{i}}, \vec{Z_{i}}$ : the 3 axes of the ellipse, 
({\it i} = 1,3 i.e. 9 parameters) \\
$ \bullet \hspace{1 mm} (a, b, \varphi)$ : the magnitude of the major and minor axes, and the phase of the spining $\vec{B}$ vector at 
the given frequency f. (see fig \ref{figure_2})
 
\begin{figure}[!h]
\centering
\label{figure_2}
\epsfysize=4.0cm
\epsffile{./figures/figure_2.eps}
\caption{Ellipse of polarisation for a $\vec{B}$ spining vector. }
\end{figure}


 So a total of 12 parameters.\\
 \\
These 12 parametres can be found from the 3 input signals :
$ x_{i}(t) = \alpha_{i}\cos(\omega t+\Phi_{i})$  \hspace{2 mm} {\it i}=1,3\\
 assuming that a Fourier analysis provides the 6 coefficients $(\alpha_{i}, \Phi_{i})$
  for each frequency.\\
  
~\\


%===================================================================================== 
\section{Geometric considerations}
%===================================================================================== 

We begin with the equation of the ellipse reported on its axes (fig 2) : \\
%figure
 \begin{equation} \label{ellipse}
  \left\{
   \begin{array}{ll}
     X=a\cos(\omega t+\varphi) \\
   Y=b\sin(\omega t+\varphi) \\
   Z=0
\end{array}
  \right.
\end{equation}
   
$\varphi$ defines the initial position of the spining vector at angular frequency $\omega$, for $t=0$. \\

In the measurement coordinat system, the three axes $(\vec{X}, \vec{Y}, \vec{Z})$ of 
the ellipse are defined by their components :
\\
\begin{displaymath}
\begin{array}{ccc}
\vec{X} = \left(
        \begin{array}{c}
		X_{x}\\
		X_{y}\\
		X_{z}
	\end{array}\right)
&
\vec{Y} = \left(
	\begin{array}{c}
		Y_{x}\\
		Y_{y}\\
		Y_{z}
	\end{array}\right)
& 
\vec{Z} = \left(\begin{array}{c}
		Z_{x}\\
		Z_{y}\\
		Z_{z}
	\end{array}\right)
\end{array}
\end{displaymath}

~\\
The measured components $(x, y, z)$ at a given frequency can be deduced from $(X, Y, Z)$
by the following coordinate system transform :
\begin{center}
\begin{displaymath}
\left( \begin{array}{c}
                x(t)\\
		y(t)\\
		z(t)
	\end{array}
\right) = \left(\begin{array}{ccc}
X_{x} & Y_{x} & Z_{x}\\
X_{y} & Y_{y} & Z_{y}\\
X_{z} & Y_{z} & Z_{z} 
\end{array}\right)
\left(\begin{array}{c}
X(t)\\
Y(t)\\
Z(t)
\end{array}\right)
\end{displaymath}
\end{center}


  
~\\


%===================================================================================== 
\section{Problem to solve}
%===================================================================================== 



$X, Y, Z$ are sinusoidal signals defined in section \ref{ellipse}.
$x, y z$ are sinusoidals signals too,  which can be expressed as :
 \begin{equation} \label{sinu}
 \alpha_{i}\cos(\omega t + \Phi_{i})
 \end{equation}

So we have one matrix equation :


\begin{center}
\begin{Boxedminipage}[c]{10.5 cm}
 \begin{eqnarray} \label{matrix1}
~~\left( \begin{array}{c}
        \alpha_{x}\cos(\omega t + \Phi_{x}) \\
		\alpha_{y}\cos(\omega t + \Phi_{y}) \\
		\alpha_{z}\cos(\omega t + \Phi_{z}) 
	\end{array}
\right) = \left(\begin{array}{ccc}
X_{x} & Y_{x} & Z_{x}\\
X_{y} & Y_{y} & Z_{y}\\
X_{z} & Y_{z} & Z_{z} 
\end{array}\right)
\left(\begin{array}{c}
a\cos(\omega t + \varphi)\\
b\sin(\omega t + \varphi) \\
0
\end{array}\right)
\hspace{1.5 cm}
 \end{eqnarray}
\vspace{0.5 mm}
\end{Boxedminipage}\\
\end{center}


~\\
Signals x y z are deduced from Fourier's analys which gives directly the coefficients
$(\alpha_{i},\Phi_{i})$ at the angular frequency $\omega=2\pi f_{k}$, for the ray $f_{k}$ considered.
~\\
~\\
So we have to determine the ellipse's parameters, ie $(a, b, \varphi)$ and the direction of its axes, 
thus the vector's coordinates 
$\vec{X}, \vec{Y}$ and  $\vec{Z}$ from equation (3), so a total of 12 unknown.
~\\
~\\
The previous matrix equation can be expressed as :
~\\
\begin{center}
\begin{Boxedminipage}[c]{8. cm}
\begin{center}
\begin{displaymath}
       ~~~\alpha_{i}\cos(\omega t + \Phi_{i})= X_{i} a \cos(\omega t + \varphi)  + Y_{i} b \sin(\omega t + \varphi)  \hspace{1 cm} i=x,y,z
\end{displaymath}
\end{center}

\vspace{0.1 cm}
\end{Boxedminipage}
\end{center}
~\\

Developping :
 $$\begin{array}{ll}
 \alpha_{i}[\cos(\omega t)\cos(\Phi_{i}) - \sin(\omega t)\sin(\Phi_{i})] = 
   X_{i} a [\cos(\omega t)\cos(\varphi) - \sin(\omega t)\sin(\varphi)]+Y_{i} b \sin(\omega t + \varphi)\\
   
\cos(\omega t) (\alpha_{i}\cos(\Phi_{i}) - X_{i} a \cos(\varphi) - Y_{i} b \sin(\varphi)) = 
\sin(\omega t) (-\alpha_{i} \sin(\Phi_{i}) + X_{i} a\sin(\varphi) - Y_{i}b \cos(\varphi)) 
\end{array}$$

And we obtain the 6 equations's system  (i=x, y, z) :\\

\begin{center}
\begin{Boxedminipage}[c]{7.7 cm}
	\begin{center}
		\begin{displaymath}
			\begin{array}{ll}
			&\alpha_{i}\cos(\Phi_{i})-X_{i}a\cos(\varphi)-Y_{i}b\sin(\varphi)=0\\
		       -&\alpha_{i}\sin(\Phi_{i})+X_{i}a\sin(\varphi)-Y_{i}b\cos(\varphi)=0 
			\end{array}
		\end{displaymath}  
	\end{center}
	\vspace{0.1 cm}
\end{Boxedminipage}
\end{center}
~\\
~\\
~\\
We can add the 6 orthonormalisation's relations of axes $\vec{X}, \vec{Y}, \vec{Z}$ (3 for orthogonal axes, and 3 for normalisation to 1)
.\\

So we have to resolve \underline {a system of 12 equations with 12 unknown} :\\
~\\
\begin{center}
\begin{Boxedminipage}[c]{8.5 cm}
	~\\
	\begin{equation} \label{equation1} 
		~~~\alpha_{i}\cos(\Phi_{i})-X_{i}a\cos(\varphi)-Y_{i}b\sin(\varphi) =0 
	\end{equation}
	 \begin{equation} \label{equation2}
	        -\alpha_{i}\sin(\Phi_{i})+X_{i}a\sin(\varphi)-Y_{i}b\cos(\varphi)=0 
 	\end{equation}
 	 \begin{equation} \label{equation3} 
 		\sum{X_{i}Y_{i}}=\sum{Y_{i}Z_{i}}=\sum{Z_{i}X_{i}} =0 
  	\end{equation}
  	\begin{equation} \label{equation4} 
  	  	\sum{X_{i}^{2}}=\sum{Y_{i}^{2}}=\sum{Z_{i}^{2}} =1
	\end{equation}
  	  	\tiny{~\\}
\end{Boxedminipage}
\end{center}
~\\
~\\
~\\

%=====================================================================================
\section{Analytical solution}
%=====================================================================================

%-------------------------------------------------------------------------------
\subsection{Principle}\label{Prin}

{\bf A)}
We can compute $X_{i}a$ and $Y_{i}b$, in fonction of $\varphi$, only from the equations \ref{equation1} and \ref{equation2}.
\\
%
\\
{\bf B)}
If $\sum{X_{i}Y_{i}}=0$ (see \ref{equation3})
we can write $\sum{(X_{i}a)(Y_{i}b)}=0$, which, considered to the result of {\bf A)}, allows the comparation of $\varphi$.
Thus we know at this moment $\varphi$, $(X_{i}a)$ and $(Y_{i}b)$.
\\
%
\\
{\bf C)}
%
Axes a and b can be obtained by the expression :
 ~\\
 $$ \left\{
  \begin{array}{c}
    a^{2}=\sum{(X_{i}a)^{2}}\\
   \! b^{2}=\sum{(Y_{i}a)^{2}}
 \end{array}
 \right. $$
%
\\
{\bf D)} 
%
Knowing $a$ and $b$, and $X_{i}a$ and $Y_{i}b$, we can compute $X_{i}$ and $Y_{i}$ \\
%
\\
{\bf E)}
%
At last $Z_{i}$ is calculated by $\vec{Z}=\vec{X}\times\vec{Y}$

~\\~\\
%=====================================================================================
\subsection{Calculations}
%=====================================================================================

%=====================================================================================
\subsubsection{A) Computation of {\large $X_{ia}$} and {\large $Y_{ia}$} }
%=====================================================================================

equation \ref{equation1} and \ref{equation2} are :

\bigskip
 $\left\{
 \begin{array}{ll}
      &\alpha_{i}\cos(\Phi_{i})-X_{i}a\cos(\varphi)-Y_{i}b\sin(\varphi) = 0\\
     -&\alpha_{i}\sin(\Phi_{i})+X_{i}a\sin(\varphi)-Y_{i}b\cos(\varphi) = 0
 \end{array}
\right.$
\hspace{1 cm} for {\it i}=x, y, z
~\\
~\\Multiply equation \ref{equation1} by $\cos\varphi$ and equation \ref{equation2} by $\sin\varphi$ :\\
~\\
 $ \left\{
 \begin{array}{ll}
    &\alpha_{i}\cos(\Phi_{i})\cos(\varphi) - X_{i}a\cos(\varphi)^{2} - Y_{i} b \sin(\varphi)\cos(\varphi) = 0\\
   -&\alpha_{i}\sin(\Phi_{i})\sin(\varphi) + X_{i}a\sin(\varphi)^{2} - Y_{i} b \cos(\varphi)\sin(\varphi) = 0
 \end{array}
\right.$
\hspace{1 cm} for {\it i}=x, y, z
~\\
\\
Adding the two equations  :\\

$\alpha_{i}(\cos(\Phi_{i})\cos(\varphi) + \sin(\Phi_{i})\sin(\varphi)) - X_{i}a = 0$
~\\
~\\
And we obtain :
 \begin{equation}\label{AddX} 
	\begin{Boxedminipage}[c]{4 cm}
	 \begin{center}
         $$ X_{i}a=\alpha_{i}\cos(\varphi-\Phi_{i})$$
 \end{center}
	  \vspace{1 mm}
	  \end{Boxedminipage}
          \end{equation}\\

Multiply \ref{equation1} by $\sin\varphi$ and \ref{equation2} by $\cos\varphi$ :\\
~\\
$\left\{
  \begin{array}{ll}
    &\alpha_{i}\cos(\Phi_{i})\sin(\varphi)-X_{i}a\cos(\varphi)\sin(\varphi)-Y_{i}b\sin(\varphi)^{2} = 0\\
   -&\alpha_{i}\sin(\Phi_{i})\cos(\varphi)+X_{i}a\sin(\varphi)\cos(\varphi)-Y_{i}b\cos(\varphi)^{2} = 0
 \end{array}
\right.$
~\\
~\\
Adding the two equations :\\
\\
$\alpha_{i}[\cos(\Phi_{i})\sin(\varphi) - \sin(\Phi_{i})\cos(\varphi)] - Y_{i}b = 0$\\

And we obtain :\\ 
 \begin{equation}\label{AddY} 
	\begin{Boxedminipage}[c]{4 cm}
	 \begin{center}
	$$Y_{i}b = \alpha_{i}\sin(\varphi-\Phi_{i})$$
	\end{center}
	  \vspace{1 mm}
	  \end{Boxedminipage}
          \end{equation}\\
 
%=================================================================
\subsubsection {B) Computation of {\large $\varphi$}} 
%=================================================================
 
From equation \ref{equation3} we have :  $\sum{X_{i}Y_{i}}=0$\\
~\\
We can also write : \\
~\\
$\sum{X_{i}aY_{i}b}=0$\\
~\\
Replacing $X_{i}a$ and $Y_{i}a$ by the results given by equation \ref{equation3} and equation \ref{equation4}, we obtain :\\
\\
$\sum{[\alpha_{i}^{2}sin(\varphi-\Phi_{i})\cos(\varphi-\Phi_{i})]}=0$\\
~\\or\\
\\$\frac{1}{2}\sum{\alpha_{i}^{2}\sin2(\varphi-\Phi_{i})}=0$
\\    
 ~\\
  By developping we found :
\\
\begin{equation*}
\begin{array}{l}
\alpha_{x}^{2}\sin(2\varphi-2\Phi_{x})+\alpha_{y}^{2}\sin(2\varphi-2\Phi_{y})+\alpha_{z}^{2}\sin(2\varphi-2\Phi_{z})=0\\
\\
\alpha_{x}^{2}[\sin2\varphi\cos2\Phi_{x}-\sin2\Phi_{x}\cos2\varphi]+\\
\alpha_{y}^{2}[\sin2\varphi\cos2\Phi_{y}-\sin2\Phi_{y}\cos2\varphi]+\\
\alpha_{z}^{2}[\sin2\varphi\cos2\Phi_{z}-\sin2\Phi_{z}\cos2\varphi]=0\\
\\
\sin2\varphi[\alpha_{x}^{2}\cos2\Phi_{x}+\alpha_{y}^{2}\cos2\Phi_{y}+
	     \alpha_{z}^{2}\cos2\Phi_{z}]-\cos2\varphi[\alpha_{x}^{2}\sin2\Phi_{x}+
	     \alpha_{y}^{2}\sin2\Phi_{y}+\alpha_{z}^{2}\sin2\Phi_{z}]=0\\
	     \\
\sin2\varphi\sum{\alpha_{i}^{2}\cos2\Phi_{i}}=\cos2\varphi\sum{\alpha_{i}^{2}\sin2\Phi_{i}} 
\end{array}
 \end{equation*}
 ~\\
Thus $\varphi$ is given by :
\begin{equation} \label{1}
 \begin{Boxedminipage}[c]{4.5 cm}
   $$~~\tan2\varphi = \frac{\sum_{i}\alpha_{i}^{2}\sin2\Phi_{i}}{\sum_{i} \alpha_{i}^{2}\cos2\Phi_{i}} \hspace{2 cm} i=x, y, z\\$$
    \tiny{~\\}
    \end{Boxedminipage}\\
 \end{equation} 

\bigskip
%=================================================================
\subsubsection{C) Computation of major and minor axes}
%=================================================================

 Equation \ref{equation4} give : $\sum{X_{i}^{2}} = 1$ and $\sum{Y_{i}^{2}}=1$ \\
~\\
   So we can write :  \hspace{3.5 mm}  $\left\{
	\begin{array}{ll} 
   		 \sum{(X_{i}a)^{2}}&=a^{2}\\
     		 \sum{(Y_{i}b)^{2}}&=b^{2}
   		 \end{array}
    	 \right.$
  ~\\
~\\
 But from equation \ref{equation3} and \ref{equation4} we have : \hspace{2mm} $\left\{
         \begin{array}{ll} 
       X_{i}a&=\alpha_{i}\cos(\varphi-\Phi_{i})  \hspace{1 cm} from ~equation~ \ref{equation4} \\
       Y_{i}b&=\alpha_{i}\sin(\varphi-\Phi_{i}) \hspace {1 cm} from ~equation~ \ref{equation3}
     \end{array}
    \right.
    $

~\\
And we obtain the length of the axes :


 \begin{equation} \label{axe}
 \begin{Boxedminipage}[c]{4.5 cm}
      	\begin{center}
     	  \begin{displaymath}
     	  \begin{array}{ll}
              a^{2}=\sum{\alpha_{i}^{2}\cos^{2}(\varphi-\Phi_{i})}\\
              ~\\
              b^{2}=\sum{\alpha_{i}^{2}\sin^{2}(\varphi-\Phi_{i})} 
       	\end{array}
      \end{displaymath}
     	\end{center}
 		\tiny{~\\}
     \end{Boxedminipage}
     \end{equation}
~\\
%================================================================
\subsubsection{D) Computation of the axes' direction :}
%================================================================
~\\
Knowing a and b, and using equation \ref{AddX} and \ref{AddY} we deduce directly :\\
      \begin{equation} \label{deduction}
      \begin{Boxedminipage}[c]{4 cm}
      	\begin{center}
     	  \begin{displaymath}
     	  \begin{array}{ll}
       \displaystyle  X_{i}=\frac{1}{a}\alpha_{i}\cos(\varphi-\Phi_{i})\\
       ~\\
       \displaystyle ~Y_{i}=\frac{1}{b}\alpha_{i}\sin(\varphi-\Phi_{i})
     		\end{array}
     		 \end{displaymath}
     		\tiny{ ~\\}
     	\end{center}
  %   	\vspace{1 mm}
     \end{Boxedminipage}
     \end{equation}
     
%=============================================================
\subsubsection{E) Computation of  $\vec{k}$ direction}~\\
%=============================================================
  We have :   $\vec{k}=\vec{Z}=\vec{X}\times\vec{Y}$ ~\\
       So : 
    	 \begin{displaymath}
    		 \begin{array}{c}
    		  Z_{x}=X_{y}Y{z} - X_{z}Y{y}\\
    		  Z_{y}=X_{z}Y{x} - X_{x}Y{z}\\
    		  Z_{z}=X_{y}Y{z} - X_{z}Y{y}
  		  \end{array}
  	\end{displaymath}
   ~\\
   It's possible to developp :~\\
 
  \begin{math}
	\begin{array}{lll}
   		\displaystyle X_{i}Y_{j} = \frac{1}{ab}\alpha_{i}\alpha{j}\cos(\varphi-\Phi_{i})\sin(\varphi-\Phi_{j})~\\
   	       \displaystyle  X_{i}Y_{j}-X_{j}Y_{i}=\frac{1}{ab}(\alpha_{i}\alpha_{j}\cos(\varphi-\Phi_{i})\sin(\varphi-\Phi_{j})-
\alpha_{j}\alpha_{i}\cos(\varphi-\Phi_{j})\sin(\varphi-\Phi_{i}))~\\
   	      \\
   	        \displaystyle X_{i}Y_{j}-X_{j}Y_{i}=\frac{\alpha_{i}\alpha_{j}}{ab}\sin(\Phi_{i}-\Phi_{j})\\
   		\end{array}
   	   \end{math}
      
~\\
\\
Thus $\vec{k}$ is given by : \\
\begin{equation} \label{developp}
\begin{Boxedminipage}[c]{4. cm}
     \begin{center}
     	\begin{displaymath}
       	  ~~Z_{k}=\frac{\alpha_{i}\alpha_{j}}{ab}\sin(\Phi_{i}-\Phi_{j})\hspace{2 cm} i, j, k = x, y, z
         \end{displaymath}
      \end{center}
      \vspace{1 mm}
 \end{Boxedminipage}
 \end{equation}
 
 ~\\
%===================================================================================== 
   \subsection{Verification}
%=====================================================================================    
~\\
If 
$\alpha_{x}=a
     \alpha_{y}=b
    \alpha_{z}=0$
   \begin{displaymath}
  ~~~~~ \left\{
     \begin{array}{ll} 
    \Phi_{x}=\varphi\\
    \Phi_{y}= \varphi - \frac{\pi}{2}
      \end{array}
    \right.
    \end{displaymath}
  We must find :   
 \begin{displaymath}
   \left\{
     \begin{array}{ll} 
     X_{x}=1\\
     X_{y}=0\\
     X_{z}=0
 \end{array}
    \right.
    \end{displaymath}
  
    \begin{displaymath}
   \left\{
     \begin{array}{ll} 
	Y_{x}=0\\
	Y_{y}=1\\
	Y_{z}=0
	 \end{array}
    \right.
    \end{displaymath}
   
   \begin{displaymath}
   \left\{
     \begin{array}{ll} 
	Z_{x}=0\\
	Z_{y}=0 \\
	Z_{z}=1
	 \end{array}
	 \right.
    \end{displaymath}
    ~\\
    

%===================================================================================== 

\underline{Verification of $\varphi$} : \\
%===================================================================================== 
~\\~\\
$\displaystyle
 \tan(2\varphi)=\frac{\sum{(\alpha_{i})^{2}\sin(2\Phi_{i})}}{\sum{(\alpha_{i})^{2} \cos(2\Phi_{i})}} 
 =\frac{a^{2}\sin(2\varphi)+b^{2}\sin(2\varphi-\pi)}{a^{2}\cos(2\varphi)+b^{2}\cos(2\varphi-\pi)}$
~\\
~\\

$\displaystyle
 \tan(2\varphi)=\frac{a^{2}\sin(2\varphi)-b^{2}\sin(2\varphi)}{a^{2}\sin(2\varphi)-b^{2}\sin(2\varphi)}
  =\frac{(a^{2}-b^{2})\sin(2\varphi)}{(a^{2}-b^{2})\cos(2\varphi)}=\tan(2\varphi)$ \\
~\\~\\


%===================================================================================== 

 \underline{Verification of $a^{2}$ and $b^{2}$} : \\
 %===================================================================================== 

\begin{displaymath}
	\left\{
	\begin{array}{ll}
	a^{2}=\sum{\alpha_{i}^{2}\cos^{2}(\varphi-\Phi_{i})} = a^{2} + b^{2}\cos^{2}(\varphi - (\varphi+ \frac{\pi}{2})) = a^{2} \\
	b^{2}=\sum{\alpha_{i}^{2}\sin^{2}(\varphi-\Phi_{i})} = b^{2}\sin^{2}(\varphi - (\varphi+ \frac{\pi}{2})) = b^{2}
	\end{array}
	\right.
\end{displaymath}
~\\
%===================================================================================== 
 
\underline{Verification of $X_{i}, Y_{i}, Z_{i}$} : \\
%===================================================================================== 
~\\~\\
$X_{i} = \frac{1}{a} \alpha_{i} \cos(\varphi - \Phi_{i} )$

   \begin{displaymath}
     ~~~\left\{
     \begin{array}{ll} 
 \displaystyle X_{x} = \frac{1}{a} a =1 \\
 \\
 \displaystyle X_{y} = \frac{1}{a} b \cos (\frac{-\pi}{2}) = 0\\
 \\
 \displaystyle X_{z} = 0 
      \end{array}
    \right.
    \end{displaymath}
\\
$\displaystyle 
Y_{i} = \frac{1}{a} \alpha_{i} \sin(\varphi - \Phi_{i} )$
   \begin{displaymath}
   \displaystyle
    \left\{
     \begin{array}{ll} 
\displaystyle Y_{x} = \frac{1}{a} a\sin(0) =0 \\
\\
\displaystyle Y_{y} = \frac{1}{b} b \sin (\frac{\pi}{2}) = 1\\
\\
\displaystyle Y_{z} = 0 
      \end{array}
    \right.
    \end{displaymath}
\\
$\displaystyle 
Z_{k}= \frac{\alpha_{i}\alpha_{j}}{ab}\sin(\Phi_{i}-\Phi_{j})$
   \begin{displaymath}
	~~~~~~~~~~~\left\{
		\begin{array}{ll} 
		\displaystyle Z_{x} = \frac{\alpha_{y}\alpha_{z}}{ab}\sin(\Phi_{y}-\Phi_{z}) =0 \\
		\\
		\displaystyle Z_{y} = \frac{\alpha_{z}\alpha_{x}}{ab}\sin(\Phi_{z}-\Phi_{x}) =0 \\
		\\
		\displaystyle Z_{z} = \frac{\alpha_{x}\alpha_{y}}{ab}\sin(\Phi_{x}-\Phi_{y}) =1
     		 \end{array}
        \right.
\end{displaymath}

~\\
~\\
%===================================================================================== 
\section{Practical application}
%===================================================================================== 
~\\
We have a fixed coordinate's system x y z in which we record signals x(t), y(t), z(t).\\
By complex FFT  we obtain coefficients $\bf{C_{\eta}^{x}}, \bf{C_{\eta}^{y}}, \bf{C_{\eta}^{z}}$ 
~\\
\begin{displaymath}
  \left\{
    \begin{array}{c}
	\bf{C_{\eta}^{x}}=\rho_{x}e^{i\theta_{x}} \\
	\bf{C_{\eta}^{y}}=\rho_{y}e^{i\theta_{y}} \\
	\bf{C_{\eta}^{z}}=\rho_{z}e^{i\theta_{z}} 
   \end{array}
  \right.
\end{displaymath}
~\\ With $(\bf{C_{-\eta}^{x}})=(\bf{C_{\eta}^{x}})^*$ because x(t) is a real signal.
Idem for y and z.\\


Signals x(t), y(t), z(t) are then decomposed into a sum of monochromatic's signals : ~\\
\begin{displaymath}
\left\{
	\begin{array}{ll}
		x_{\eta\omega_{0}}(t) = 2\rho_{x} \cos(\eta\omega_{0}t + \Phi_{x}) \Longrightarrow 
  			\left\{
     				\begin{array}{ll} 
					\alpha_{x} = 2\rho_{x}\\
					\Phi_{x} = \Theta_{x}
     				\end{array}
    			\right.
\\
		y_{\eta\omega_{0}}(t) = 2\rho_{y} \cos(\eta\omega_{0}t + \Phi_{y}) \Longrightarrow
   			 \left\{
    		 		\begin{array}{ll} 
					\alpha_{y} = 2\rho_{y}\\
					\Phi_{y} = \Theta_{y}
     		 		\end{array}
    			\right.
\\
		z_{\eta\omega_{0}}(t) = 2\rho_{y}\cos(\eta\omega_{0}t + \Phi_{z}) \Longrightarrow 
   			\left\{
     				\begin{array}{ll} 
					\alpha_{z} = 2\rho_{z}\\
					\Phi_{z} = \Theta_{z}
      				\end{array}
   			 \right.
\\
      \end{array}
\right.
\end{displaymath}
~\\

For each frequency $\displaystyle f = \frac{\omega}{2\pi} = \frac{\eta\omega_{0}}{2\pi}$ we determine completely an ellipse's parameters, 
with frequency $ \frac{\omega}{2\pi}$, and a projection three axes x, y, z would give the three monochromatic's signals 
$x_{\omega}(t), y_{\omega}(t), z_{\omega}(t)  $ exactly similar to those recorded. \\
(Nevertheless we have to prove this decomposition is single). \\
~\\
~\\
\underline {So ellipse parameters are} : \\
~\\
1) $(a, b, \varphi) $ = major axe, minor axe, ellipse's phase centered in the system $XY$ \\
~\\
2) ($\vec{X}, \vec{Y}, \vec{Z}$) = coordinates of the main axis of the ellipse in the (x, y, z) system :
$\vec{X}$=major axe,
$\vec{Y}$=minor axe,
$\vec{Z}$=normal to the wave plane.



~\\
~\\
~\\

%===================================================================================== 
\section{Summary of useful formulae}\label{summary}
%===================================================================================== 
~\\
We begin with a mesure's system xyz in which by a Fourier's analysis, we know the 3 components of monochromatic signal. 
\begin{center}
\begin{displaymath}
\left( \begin{array}{c}
                x\\
		y\\
		z
	\end{array}
\right) = \left(\begin{array}{ccc}
\alpha_{x} \cos(\omega t + \Phi_{x})\\
\alpha_{y}\cos(\omega t + \Phi_{y})\\
\alpha_{z}\cos(\omega t + \Phi_{z}) 
\end{array}\right)
\end{displaymath}
\end{center}
So we know $\alpha_{i}$ and $\Phi_{i}$\\
\\
This 3 components correspond to the projection on xyz of an ellipse, which reported to its main axes XYZ written :
\begin{center}
	\begin{displaymath}
		\left( \begin{array}{c}
                X\\
		Y\\
		Z
	\end{array}
		\right) =
			 \left(\begin{array}{ccc}
			a \cos(\omega t + \varphi)\\
			b\sin(\omega t + \varphi)\\
			0 
			\end{array}\right)
	\end{displaymath}
\end{center}
~\\


 The previous computations allows us to determine :\\
~\\
\begin{center}
\bf{1)	{The absolute phase $\varphi$} :}\\
~\\
     \begin{Boxedminipage}[c]{5 cm}
        ~\\
     	\begin{displaymath}
    	\tan(2\varphi) = \frac{\sum_{i}\alpha_{i}^{2}\sin(2\Phi_{i})}{\sum_{i}\alpha_{i}^{2}\cos(2\Phi_{i})}   
    	\end{displaymath} 
    	~
     \end{Boxedminipage}
\end{center}

~\\
~\\
    \begin{center}
\bf{2)  {Major and minor axes' length} : }\\
~\\
     \begin{Boxedminipage}[c]{9 cm}
       \begin{displaymath}
 	a^{2}=\sum_{i}\alpha_{i}^{2}\cos(\varphi - \Phi_{i})     \\
	~\\
	~~~~~b^{2}=\sum_{i}\alpha_{i}^{2}\sin(\varphi - \Phi_{i})     
	\end{displaymath}
	~
      \end{Boxedminipage}\\
   \end{center}

~\\
~\\
\begin{center}
\bf{3)	{Ellipse's axes directions} : }\\
~\\
   \begin{Boxedminipage}[c]{12.5 cm}
     \begin{displaymath}
	X_{i}=\frac{1}{a}\alpha_{i}\cos(\varphi-\Phi_{i})\\
	~\\
	~~~Y_{i}=\frac{1}{b}\alpha_{i}\sin(\varphi-\Phi_{i})\\
	~\\
	~~~ Z_{i}=\frac{1}{ab}\alpha_{j}\alpha_{k}\sin(\Phi_{j}-\Phi_{k})  
     \end{displaymath}
     ~  
   \end{Boxedminipage}\\
\end{center}

~\\~\\~\\
The implementation of these formulas is made in the next chapter \ref{Cod}, 
in the form of codes written in FORTRAN 77, FORTRAN 90, C and IDL.
Chapter \ref{Ex} show various examples of wave polarisation observed by 
the CLUSTER/STAFF-SC experiment which valid the method.

%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

\cleardoublepage
\chapter{Code source to compute polarisation parameters}\label{Cod}

%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

%==========================================================================
\section{Fortran 77 code}
%==========================================================================

\normalsize \mdseries
The Fortran code was initially written in F77, as given below. 
Conversion in F90 has been done on next section,
with a minimum of modifications, only to be compiled as a polarlib.f90 code.

\begin{changemargin}{1.5cm}{3.5cm}
{\tt\footnotesize\baselineskip=8.1pt
\listinginput[5]{1}{../polarlib.f}
}
\end{changemargin}

~\\
%==========================================================================
\section{Fortran 90 code}
%==========================================================================


\begin{changemargin}{1.5cm}{3.5cm}
{\tt\footnotesize\baselineskip=8.2pt
\listinginput[5]{1}{../polarlib.f90}
}
\end{changemargin}

~\\
%==========================================================================
\section{C code}
%==========================================================================

The following C code has been obtained from the f77 code and the f2C tool. 
Then, aesthetics of the C code has been improved, and code itself checked 
by using C compiler on Ubuntu Linux machine.
\bigskip

\begin{changemargin}{1.5cm}{3.5cm}
{\tt\footnotesize\baselineskip=8.2pt
\listinginput[5]{1}{../polarlib.c}
}
\end{changemargin}

\newpage
%==========================================================================
\section{IDL code}
%==========================================================================

IDL code has been deduced from F77 version in May 2015. 
\bigskip

\begin{changemargin}{1.5cm}{3.5cm}
{\tt\footnotesize\baselineskip=8.1pt
\listinginput[5]{1}{../polarlib.pro}
}
\end{changemargin}

~\\~\\
\bigskip
%==========================================================================
\section{IDL Example of use }
%==========================================================================

\bigskip

\subsection{IDL Test Procedure}
%------------------------------

In section  \ref{summary} we saw that, after a Fourier transform of the 3 waveforms, we know
the amplitude $\alpha_{i}$ and phase  $\Phi_{i}$
of the 3 components of monochromatic signal, for each frequency {\it\large f}, by : 
{\baselineskip=6.pt
\begin{center}
\begin{displaymath}
\left( \begin{array}{c}
        x_{f}(t)\\
		y_{f}(t)\\
		z_{f}(t)
	\end{array}
\right) = \left(\begin{array}{ccc}
\alpha_{x}\cos(2\pi f t + \Phi_{x})\\
\alpha_{y}\cos(2\pi f t + \Phi_{y})\\
\alpha_{z}\cos(2\pi f t + \Phi_{z}) 
\end{array}\right)
\end{displaymath}
\end{center}
}
\bigskip
So we know $\alpha_{k}$ and $\Phi_{k}$ (with k= x, y, z) by the complex coefficient of the Fourier transform at the {\it\large f} frequency,  ie : 

\begin{center}
$X_k = \alpha_{k} e^{i \Phi_{k}}$ 
\end{center} 

The input complex coefficient $X_k$ are named `{\tt Sx, Sy, Sz}' in the IDL test procedure below.
~\\

\begin{changemargin}{1.5cm}{3.5cm}
{\tt\footnotesize\baselineskip=8.6pt
\listinginput[5]{1}{../test.pro}
}
\end{changemargin}


\newpage

\subsection{Result}
%--------------------------

By lauching IDL and running the previous test procedure, after compilation of polarlib.pro library, 
we obtained the following results :
~\\

\begin{changemargin}{1.cm}{0.5cm}
{\tt\footnotesize\baselineskip=8.4pt
\listinginput[5]{1}{../test.resu}
}
\end{changemargin}


We can see that we find the elliptical polarization defined as input, with $\vec{k} // \vec{Z}$, and the main characteristics:

\begin{center}
   $
   \begin{array}{lll}
    a = & 9.96810 &\simeq 10. \\ 
    b = & 4.99641 &\simeq 5. \\
    e = & 0.865308 &\simeq 0.866025 \\
    \varphi = & 44.8574 &\simeq 45. \\
    & & \\
    Kz = & 1.0 ~&~ \\
    Az = &0.999998 &\simeq 1.0 \\
   \end{array}
   $
\end{center}

To have a complete spectrum, the {\tt cpolar} routine must be called for each frequency of the complex arrays {\tt Sx, Sy, Sz}. Note that the first $N/2$ points of the arrays {\tt Sx(N), Sy(N), Sz(N)}, correspond to the frequency $f_i = (i-1)\delta f $ for $i=1,N/2$.

~\\
The successive spectra are then plotted in the form of a time-frequency
spectrogram, as we can see in the following chapter \ref{Ex}.

%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

\cleardoublepage
\chapter{Examples of experimental results}\label{Ex}.

%XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

%==========================================================================
\section{Description of plotted parameters}
%==========================================================================

The following examples show the results of the visualisation program, which plot the main polarisation parameters in several pannels described below.

\subsection{Total power}

This is a classical spectrogram of the total power, in time-frequency diagram, given by the 
three input signal Sx(f), Sy(f) and Sz(f).
Color scale is logarithmic, and corresponding values are given by the upper line of the color 
palet (top-right).

\subsection{Exentricity}

The corresponding spectrogram gives the value of the exentricity of the polarisation ellipse.
Color code is dark-blue for e=0 (circular wave) and red for e=1 (degenerated ellipse 
along a line).

\subsection {K direction}\label{pol}

The K direction is in fact the normal to the plane of the ellipse, and is determined by the two polar angles $\theta$ and $\phi$.
Convention is following :
{\center
~~$\theta < 90$ : right handed mode\\
  $\theta > 90$ : left  handed mode\\
}
Color code for $\theta$ and $\phi$ are given in color palet top-right.

\subsection{Major axis direction}

As K vector, this direction is given by the two polar angles $\theta$ and $\phi$.

\subsection{Remarks}

$\bullet$ {If we have exentricity=0 (circular wave) the major axis direction is undetermined.}

$\bullet$ {If we have exentricity=1 (linear polarisation) K could be undetermined, but major axis
 direction is measured.}

$\bullet$ {If we have $\theta$=0, $\phi$ is undetermined.}



\newpage
%==========================================================================
\section{Circular polarisation with K // Bo}
%==========================================================================

Very clear example showing a circular right handed polarisation (exentricity in blue, 
ie e=0, then circular wave, and $\theta_K$ in blue, ie $\theta_K =0$, then right handed
polarisation (see \ref{pol}). 
As $\theta_K =0$, $\phi_K$ is undetermined.
As the circular polarization is slightly flattened, the direction of the major axis can nevertheless be determined ($\theta =90$\textdegree).

\begin{figure}[!h]
\centering
\label{visupolar_2_20040130_164700_128_HBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_2_20040130_164700_128_HBR.ps}
\caption{Polarisation parameters for a Whistler recorded in HBR mode on CLUSTER/STAFF-SC. }
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_2_20040130_164700_128_HBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_2_20040130_164700_128_HBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_2_20040130_164700_128_HBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_2_20040130_164700_128_HBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Linear polarisation with K $//$ or $\perp$ to Bo}
%==========================================================================

First event at $\sim$ 06:58 is a large band event, with linear polarisation
in the plane orthogonal to $B_0$, and a K direction parallel or antiparallel to $B_0$. 

Second event at $\sim$ 07:10 is a linear monochromatic wave at $\sim$ 2 Hz, of same characteristics.

Third event, at $\sim$ 07:20-07:30 is a pseudo monochromatic wave, 
at about 10 Hz, always linear polarised
(e$\sim$1) but with K perpendicular to $B_0$ ($\theta_K =90$\textdegree).

\begin{figure}[!h]
\centering
\label{visupolar_1_20040504_065000_512_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_1_20040504_065000_512_NBR.ps}
\caption{Polarisation parameters for 3 events recorded in NBR mode on CLUSTER/STAFF-SC. }
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_1_20040504_065000_512_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_1_20040504_065000_512_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_1_20040504_065000_512_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_1_20040504_065000_512_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Linear monochromatic wave with harmonics with K~$\perp$ to Bo}
%==========================================================================

Nice case of monochomatic wave, with linear polarisation (e$\sim$1) and a K vector 
perpendicular to $B_0$  ($\theta_K =90$\textdegree).
Note that this value of $\theta_K$ is confirmed by the direction of major axis
along $B_0$ ($\theta =0$ or $180$).

\begin{figure}[!h]
\centering
\label{visupolar_1_20040520_121000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_1_20040520_121000_256_NBR.ps}
\caption{Polarisation parameters for a linear monochromatic wave along Bo, with harmonics.}
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_1_20040520_121000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_1_20040520_121000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_1_20040520_121000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_1_20040520_121000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Same, but unpolarised waves before event}
%==========================================================================
Same kind of wave than previous event, but with a slight frequency variation
(from 09:30 to 09:50) and harmonics. 

Before, and during the begining of the wave, unpolarised event appears 
at low frequency ($\sim$2 Hz) .

\begin{figure}[!h]
\centering
\label{visupolar_1_20010923_092000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_1_20010923_092000_256_NBR.ps}
\caption{Polarisation parameters for a linear monochromatic wave along Bo.}
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_1_20010923_092000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_1_20010923_092000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_1_20010923_092000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_1_20010923_092000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Linear monochromatic wave with numberous harmonics}
%==========================================================================

Same kind of wave than previous events (linear polarisation along $B_0$),  
with a slight frequency variation and numberous harmonics. 

\begin{figure}[!h]
\centering
\label{visupolar_1_20090116_120000_512_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_1_20090116_120000_512_NBR.ps}
\caption{Polarisation parameters for a linear monochromatic wave along Bo, with numberous harmonics. }
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_1_20090116_120000_512_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_1_20090116_120000_512_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_1_20090116_120000_512_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_1_20090116_120000_512_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}

\newpage
%==========================================================================
\section{Perfect linear waves}
%==========================================================================

Very clear example of linear polarisation along $B_0$
(exentricity=1, $\theta_K =90$\textdegree).


\begin{figure}[!h]
\centering
\label{visupolar_2_20031105_164000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_2_20031105_164000_256_NBR.ps}
\caption{Polarisation parameters for perfect linear monochromatic wave along Bo. }
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_2_20031105_164000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_2_20031105_164000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_2_20031105_164000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_2_20031105_164000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Same with time dependent frequency}
%==========================================================================
Same kind of wave than previous example, but with a time variation of the main frequency.


\begin{figure}[!h]
\centering
\label{visupolar_2_20031124_172000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_2_20031124_172000_256_NBR.ps}
\caption{Polarisation parameters for a linear monochromatic wave along Bo with frequency variation.}
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_2_20031124_172000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_2_20031124_172000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_2_20031124_172000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_2_20031124_172000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Large band circular polarisation, with k $//$ Bo}
%==========================================================================
Example of large frequency band event, with circular polarisation (exentricity $\sim$ 0)
and K$// B_0$ ($\theta_K =0$).


\begin{figure}[!h]
\centering
\label{visupolar_2_20080724_183000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_2_20080724_183000_256_NBR.ps}
\caption{Polarisation parameters for large band circular polarisation, with k $//$ Bo. }
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_2_20080724_183000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_2_20080724_183000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_2_20080724_183000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_2_20080724_183000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}

\newpage
%==========================================================================
\section{Other stable linear polarisation with constant frequency}
%==========================================================================
Another example of linear polarisation along $B_0$
(exentricity=1, $\theta_K =90$\textdegree) with constant frequency and harmonics.


\begin{figure}[!h]
\centering
\label{visupolar_3_20010613_014000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_3_20010613_014000_256_NBR.ps}
\caption{Polarisation parameters for stable linear polarisation with constant frequency.}
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_3_20010613_014000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_3_20010613_014000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_3_20010613_014000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_3_20010613_014000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}


\newpage
%==========================================================================
\section{Perfect large band circular waves}
%==========================================================================
Perfect example of large frequency band circular wave, with K // $B_0$.
Exentricity=0,  $\theta_K =0$ (then right handed polarisation).


As the circular polarization is slightly flattened, the direction of the major axis can nevertheless be determined ($\theta =90$\textdegree).


\begin{figure}[!h]
\centering
\label{visupolar_3_20040715_140000_256_NBR}
\epsfysize=18.0cm
\epsffile{./figures/visupolar_3_20040715_140000_256_NBR.ps}
\caption{Polarisation parameters for perfect large band circular waves with K // $B_0$.}
\end{figure}

\newpage
\begin{figure}[!h]
\centering
\label{visuspectro_3_20040715_140000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectro_3_20040715_140000_256_NBR.ps}
\caption{Classical spectrogram from the same event, in Left-Right-Z components. }
\end{figure}

\begin{figure}[!h]
\centering
\label{visuspectra_3_20040715_140000_256_NBR}
\epsfysize=10.0cm
\epsffile{./figures/visuspectra_3_20040715_140000_256_NBR.ps}
\caption{Average spectra of the above spectrogram, in Left-Right-Z components. }
\end{figure}



