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\begin{document}

\sf

\include{020625}
\include{020702}

%\Large $z/L_c+\eta (x_0)$ 

{\bf 9. 7. 02}
\section{Stationary solution of (\ref{equ_p0byp1}) with gain difference}
%-----------------------------------------------------------

Up to now it was assumed that the two modes have identical gain. In praxis this can be adjusted only approximately. Therefore, I look now on the influence of a gain difference $\Delta g$ between the two modes. 

In this case, the only change of Equ. (\ref{equ_p0byp1}) is the additive term $\Delta g$ on the right hand side. Keeping the 
\underline{Assumptions} $\partial _t =0$,  $P_0+P_1=$const.$=P_{sat}$, $g\approx \alpha $, the solution changes to 
\begin{eqnarray} 
&&\boxed{~ \frac{z}{L_c}=\eta(x)-\eta(x_0)
~~~~ \text{ with the function } ~~
\eta(x)=\frac{\ln\left(\frac{|\delta +x|}{1+x}\right)}{1-\delta }
                       +\frac{\ln\left(\frac{|\delta +x|}{1-x}\right)}{1+\delta }~}
\\ \nonumber                       
&& x=\frac{P_0-P_1}{P_0+P_1}, ~~~~~
\delta = \Delta g L_c, ~~~~~
L_c =\frac{1}{\alpha \varepsilon P_{sat}}.
\end{eqnarray}
\figc{0.45}{statneu} \hfill \mbc{0.5}{
This solution is shown in the Figure for $\Delta g=0$ (thick black) and $\Delta g L_c=0.5$.  
The only influence of the initial power ratio parameter $x_0=x(z=0)$ is the starting point on this curve. For $x_0 > - \Delta_g L_c$, the solution moves up and approaches 1 for $\eta \rightarrow \infty$. Otherwise, it moves down approaching -1. In case without gain difference, this means suppression of the smaller initial power. With gain difference, the critical ratio $x_0$ for suppression is given by $- \Delta_g L_c$. 
}

For $\Delta g \ne 0$, the suppression becomes less pronounced in that branche in the Fig. belonging to the mode with the lower gain. 

{\bf No suppression effect for $|\Delta g| L_c > 1 $ !!!}

\section{Estimate of the characteristic length $L_c $}
%-----------------------------------------------------

First we write $L_c = \frac{1}{\alpha }~ \frac{P_\varepsilon }{P_{sat}}$. Here is $P_\varepsilon =\hbar\omega v_g w d /\varepsilon \approx 1$W the nonlinear gain saturation power that should not be confused with the saturation power $P_{sat}$ of the SOA. The latter one is typically 100mW, yielding the estimate $L_c\approx 10/\alpha \approx 5$ mm if assuming $\alpha = 20 $cm$^{-1}$. 

Shorter $L_c$ require higher losses and/or lower SOA saturation power and/or lower $P_\varepsilon $. However, SOA saturation depends on $\alpha $. 

Therefore a second, different estimate. From rate equations, it can be derived (derivation will be addad later)
\begin{eqnarray}
L_c = \frac{1}{\alpha } + L_\varepsilon 
~~~~ \text{with} ~~
L_\varepsilon = \frac{P_\varepsilon }{\hbar\omega } ~\frac{e}{(I-I_{tr})/L} \approx \frac{1 \text{ A/mm}}{(I-I_{tr})/L}\times 1 \text{ mm}
\end{eqnarray}
The first term contributes about 0.5 mm, the last one depends on how big the current can be made. With 1 A/mm we would arrive at altogether $L_c=1.5$ mm. 

Taking more realistically 5 mm, we must have $\Delta g \ll 2$ cm$^{-1}$. This is quite small. 
\end{document}

%
{\bf Saturation condition:} 
Temporal average $\langle [\dots ] \rangle$ of the braket = 0.

\begin{eqnarray}
g=\text{const.} \stackrel{(\ref{satcond})}{\Rightarrow} g=\frac{\alpha}{1-\varepsilon \langle |E|^2 \rangle}. 
\end{eqnarray}


