%\documentclass[aps,pre,preprint,a4paper,superscriptaddress,showpacs]{revtex4}
\documentclass[aps,a4paper,twocolumn]{revtex4}

\usepackage{amsmath}
\usepackage{times,mathptm}
%\usepackage{graphicx}% Include figure files
\usepackage[pdftex]{graphicx}   \usepackage{epstopdf}   %PDF Output
%\columnwidth9cm % \renewcommand{\baselinestretch}{1.2}

\newcommand{\mb}[2]{\begin{minipage}[t]{#1\columnwidth}\setlength{\parskip}{1.3ex plus 0.3ex minus 0.3ex} #2 \end{minipage}}
\newcommand{\mbc}[2]{\begin{minipage}[c]{#1\columnwidth}\setlength{\parskip}{1.3ex plus 0.3ex minus 0.3ex} #2 \end{minipage}}
\newcommand{\mbb}[2]{\mb{0.49}{#1}\hfill\mb{0.49}{#2}}

\newcommand{\infig}[1]{\includegraphics[width=\columnwidth]{./#1}} 
\newcommand{\ffig}[2]{\mb{#1}{~ file=./figs/#2}}
\newcommand{\fig}[2]{\mb{#1}{\infig{#2}}}
\newcommand{\figc}[2]{\mbc{#1}{\infig{#2}}}
\newcommand{\figb}[2]{\mbb{#1}{\infig{#2}}}

\setlength{\parskip}{1.3ex plus 0.1ex minus 0.1ex}
\setlength{\parindent}{0em}

\begin{document}

%Title of paper
\title{Eye opening by saturable absorber}

\author{ede}
% \affiliation{Humboldt-Universit\"at zu Berlin, Institut f\"ur Physik, Newtonstrasse 15, D-12489 Berlin, Germany}

\date{\today}

\maketitle
%

\section{Problem description}
%----------------------------

A short ($L \approx 100 \mu$m) unpumped SOA (open circuit) is considered. 

~ \hfill \fig{0.5}{device} \hfill ~

A pulse train $P(t)$ with a bad extinction ratio is injected. How does this device perform as an ''eye opener''? Which are the optimum operation point and design rules? I try to find formulae helping to answer these questions and to understand simulation and measurement. 
\\

As a measure for the effects I consider the
\begin{eqnarray}
\boxed{\text{\bf suppression} \hspace{2em} s(t)=-10\lg\left(\frac{P_\text{out}(t)}{T_0 P_\text{in}(t-t_\text{travel})}\right)}
\end{eqnarray}
$T_0=\exp(-\alpha _0L)$: CW transmittivity of the completely saturated device, $\alpha _0$: back ground absorption coefficient, $t_\text{travel}=L/c$: time of travel through the device, $c$: group velocity of light. 
\\

This quantitiy give the actual suppression in dB of the transmittivity due to incomplete saturation of the absorber. Its temporal behaviour is a direct measure for the influence of the device on the vertical opening of the eye.



\section{Equation of motion}
%------------------------

{\em Some assumptions:} \\
bit rate $f \ll 1/t_\text{travel} \approx $ THz, \\ spatially constant carrier density $N(t)$. \\

The suppression is determined by the actual carrier density as follows:
\begin{eqnarray}
s=g(N)L=\Gamma g'L \cdot (N-N_t).
\end{eqnarray}
$\Gamma=0.4 $: confinement factor, $N_t=10^{24}$m$^{-3}$: transparency density, $g'=5\cdot 10^{-20}$m$^2$: differential gain.
Thus, its evolution is governed by the carrier rate equation:
\begin{eqnarray}
\partial _t N = \frac{I}{eAL} -\frac{N}{\tau } - c g(N) S(t).
\end{eqnarray}
$I=0$: injection current (set to zero, but here I keep it general because perhaps better performance with $I\ne 0$), $A=5\cdot 10^{-13}$m$^2$: cross section area of AZ, $\tau =1 $ns: carrier life time,
\begin{eqnarray}
S(t) &=& \frac{\bar{P}(t)}{cA\hbar\omega } \hspace{3em} \text{photon density}
\end{eqnarray}
$\bar{P}(t)$: spatailly averaged power in the cavity, $\hbar\omega =0.8$eV: photon energy. 
\\

The resulting equation of motion for the suppression is
\begin{eqnarray}
&&\boxed{ \tau \,\partial _t s = s_0 - (1+p)s } \\
&& s_0= 10\lg(e) \Gamma g' L N_t (1-\frac{I}{I_t})=8.7\text{dB}\\
&&p=\frac{\bar{P}}{P_s}�, \hspace{1em} P_s=\frac{A \,\hbar\omega }{\tau \Gamma g'}=3.2\text{mW \footnotesize saturation power}
\end{eqnarray}
$s_0$ is the maximum possible suppression, i.e., an upper limit for the improvement of the eye opening.


\section{General solution}
%-------------------------
The general solution of the equation of motion is
\begin{eqnarray}
s(t)&=&s_0\int_{-\infty}^t \frac{dt'}{\tau } \exp\left(-\int_{t'}^t \frac{dt''}{\tau }(1+p(t''))\right)
\\ \label{solution}
&=& s_0 \int_0^\infty \frac{dt'}{\tau } ~ \exp\left(\!-\int_{0}^{t'} \frac{dt''}{\tau }(1+p(t-t''))\!\right)
\end{eqnarray}
{\em Qualitative discussion:} the integral is always $\le 1$. It becomes $=1$ only in the limit $p(t)\rightarrow 0$ (maximum suppression for very small signals). The bigger $p$ becomes, the faster decays the integrand (the exp-function) to zero, reducing $s$. $s(t)$ does not depend on the actual power $p(t)$ only, but it also remembers earlier powers. The memory time is of the order of that $t'$ for which the exponent becomes -1, i.e. roughly
\begin{eqnarray}
\text{memory time:} \hspace{1em} t_\text{mem}\approx \frac{\tau }{1+\langle p \rangle},
\end{eqnarray}
$\langle p \rangle$: average of $p$ over the memory time. Signals weaker than $P_s=3$mW: $t_\text{mem}\approx \tau =1$ns, i.e. $s$ averages over about 40 periods of a 40 GHz signal and shows only a small modulation at this frequency. It responds much more to slower changes of the pulse train as accumulations of many 1-bits or many 0-bits, which causes the unwanted pattern effects. 
\\

To get the variations of $s$ due to patterns smaller than those at the operation frequency $f=40$ GHz, one must achieve $f t_{mem}\le 1$, independent of the value of $s_0$. 
\\

Thus, $t_{mem}$ has to be decreased to the order of 25ps or less.  One possibility is using true saturable absorbers with a short $\tau $. The same effect is achieved with increasing $\langle P\rangle /P_s$ up to about 40. With our parameters this required mean powers of about $40\cdot 3$mW $\approx 100$ mW. In order to find out which parameters have to be changed to reduce this very high value, I rewrite the memory time in the high power limit:
\begin{eqnarray}
t_\text{mem}\approx \frac{\hbar \omega  }{\langle P \rangle}~ \frac{A}{\Gamma g'} \hspace{2em} \text{if } \langle P \rangle \gg P_s.
\end{eqnarray}
Note: $\tau $ has canceled out in this limit. The only possibility to reduce the memory time is to increase the effective differential gain $\Gamma g' /A$.

In the following I support this discussion by considering some special cases. 

\section{Special cases}
%----------------------

\subsection{CW injection ($p(t)=$const.)}
%----------------------------------------

Integral (\ref{solution})  is easily calculated:
\begin{eqnarray}
s=\frac{s_0}{1+p}.
\end{eqnarray}
This gives the CW characterisitics

~ \hfill \fig{0.6}{kennl} \hfill ~

Additionally I have plotted here the derivative parameter
\begin{eqnarray}
s'= \frac{d (s/s_0)}{d (10\lg(P/P_s)} =\frac{1}{10\lg e} ~ \frac{p}{(1+p)^2}.
\end{eqnarray}
It shows the differential change of the relative suppression $s/s_0$ per dB change of the input power. Its maximum at $p=P/P_s=1$ is here about 0.06, i.e., the maximum differential change of $s$ is about $0.06 s_0 = 0.5$dB per dB change of the input power. 
\\

If e.g. we have an extinction of 5dB at input, the static extinction at output is about $5 + 0.5 * 5 = 7.5$dB. 

\subsection{Periodic pulse sequence}
%-----------------------------------

Hier muss ich erst mal aufgeben. Keine Zeit mehr. 


\newpage

\end{document}
