ДимитрийLEE

Transient response of mircoring modulators

2025-08-17T20:09:39.081Z

latex source code

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\begin{document}
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\title{Transient Response of Microring Modulators}
\author{Dingning Li} 
\affiliation{ECE, BU, Boston, MA}
% \author{Kai Zhang} 
% \email{kai.zhang@dukekunshan.edu.cn}
% \affiliation{Division of Natural and Applied Sciences, Duke Kunshan University, Kunshan, Jiangsu, 215300, China}

\date{\today}


\begin{abstract}

    We follow the CMT theory given by~\cite{little1997microring} and derive the transient switching behavior of a microring modulator. We found that when switching on a ring modulator, there would be ringing phenomenon showing up in the trough port optical intensity. While switching off a ring modulator would not induce any ringing.

\end{abstract}

\maketitle

\section{Introduction}

As shown in~\cite{little1997microring}, the transmission of a microring is 
\begin{equation}
    s_t = s_i-\jj\mu a,
\end{equation}
where $s_i(\omega)$ is the incident wave amplitude, $a(\omega)$ is the energy amplitude stored in the ring. 
\begin{center}\label{singlering}
\includegraphics[width=0.2\textwidth]{figs/singlering.png}
\captionof{figure}{A single ring modulator. In this letter we consider the case where there is no drop waveguide.}
\end{center}
From CMT, one can solve for the frequency domain transfer function
\begin{equation}
s_t(\omega) = 
\frac{ \jj(\omega - \omega_0) + \tfrac{1}{\tau} - \tfrac{2}{\tau_e} }
     { \jj(\omega - \omega_0) + \tfrac{1}{\tau} }\, s_i.
\end{equation}
And the energy stored in the ring is
\begin{equation}
a(\omega) = 
\frac{ -\jj \sqrt{\tfrac{2}{\tau_e}} }
     { \jj(\omega - \omega_0) + \tfrac{1}{\tau} }\, s_i.
\end{equation}
The coupling coefficient $\mu$ has relationship with $\tau_e$ given by
\begin{align}
\mu^2 &= \kappa^2 v_g (2\pi R)^{-1} \notag \\
      &= \tfrac{2}{\tau_e}.
\end{align}

However, the CMT only gives steady-state frequency domain responses. When one suddenly switches on/off a ring modulator, as shown in fig.~\ref{ringonoff}, we need to solve for the transient time domain response of the CMT equation
\begin{equation}
\frac{d}{dt} a = \left( j\omega_{0} - \frac{1}{\tau} \right) a \; - \; j\mu s_i.
\end{equation}

\begin{center}\label{ringonoff}
\includegraphics[width=0.45\textwidth]{figs/ringonoff.png}
\captionof{figure}{Switching off/on a microring modulator.}
\end{center}



\section{Math formulation}
Define $A(t)$ be the time domain function for $a(\omega=\omega_{\rm laser})$, then the linear ODE governing the evolving of the ring system becomes
\begin{equation}
\label{eq:main}
\frac{dA}{dt} + \Big(\jj\,\Delta\omega(t) + \gamma\Big) A = -\,\jj\,\mu s_i,
\qquad \gamma \equiv \frac{1}{\tau} > 0,
\end{equation}
with real parameters $\mu,s_i,\gamma,\omega$ and $\jj^2=-1$.  The detuning
$\Delta\omega(t)$ will be piecewise constant with a single step at $t=0$.
We assume the system is in steady state for $t<0$, so $A$ is continuous at the step:
$A(0^-)=A(0^+)$.

\subsection{General constant-coefficient solution (review)}
Consider the general linear ODE with constant complex coefficients
\begin{equation}
\label{eq:general}
\frac{dA}{dt} + C\,A = D , \qquad C,D \in \mathbb{C}.
\end{equation}
Multiplying by the integrating factor $e^{Ct}$ gives
\[
e^{Ct}\frac{dA}{dt} + C e^{Ct} A = D e^{Ct}
\;\; \Longrightarrow \;\;
\frac{d}{dt}\!\left(e^{Ct}A\right) = D e^{Ct}.
\]
Integrating from $t_0$ to $t$:
\[
e^{Ct}A(t) - e^{Ct_0}A(t_0) = \int_{t_0}^{t} D e^{Cs}\,ds
= \frac{D}{C}\Big(e^{Ct} - e^{Ct_0}\Big).
\]
Hence
\begin{equation}
\label{eq:general-solution}
A(t) = \frac{D}{C} + \Big(A(t_0) - \frac{D}{C}\Big)\,e^{-C\,(t-t_0)}.
\end{equation}
When $C$ is piecewise constant and changes at $t=0$ (no impulses), we solve
\eqref{eq:general} on each side and enforce continuity $A(0^-)=A(0^+)$.

Assume $\Re\{C\}>0$, and given the condition that the system is in steady state for $t<0$ so $t_0\rightarrow -\infty$, one can get 
\begin{equation}
    A(t=0)=\frac{D}{C}
\end{equation}

\subsection{Turning on: $\Delta\omega(t)$ steps from $0$ to $-\omega$ at $t=0$}
Here
\begin{align*}
&\Delta\omega(t) =\begin{cases}
0,& t<0,\\
-\omega,& t\ge 0,
\end{cases}
\qquad
\\&C_-=\gamma, \quad C_+=\gamma-\jj\omega, \quad 
\\&D=-\jj\,\mu s_i \equiv -\jj S \;\; (S\in\mathbb{R}).
\end{align*}


\paragraph{Pre-step steady state ($t<0$).}
Since $C_-=\gamma$ and $D=-\jj S$,
\[
A(0^-)=\frac{D}{C_-}=\frac{-\jj S}{\gamma}=-\jj S\,\tau.
\]

\paragraph{Post-step solution ($t\ge 0$).}
Using \eqref{eq:general-solution} with $t_0=0$ and $C_+=\gamma-\jj\omega$,
\begin{align*}
A(t)
&= \frac{D}{C_+} + \Big(A(0^+)-\frac{D}{C_+}\Big)e^{-C_+ t}
\\&= \frac{-\jj S}{\gamma-\jj\omega}
+ \left(\frac{-\jj S}{\gamma}-\frac{-\jj S}{\gamma-\jj\omega}\right)e^{-(\gamma-\jj\omega)t}.
\label{eq:A-case1}
\end{align*}
It is convenient to write
\begin{equation}
\begin{aligned}
    &A(t) = A_\infty + \Delta A\, e^{-\gamma t} e^{+\jj\omega t},
\quad
\\&A_\infty:=\frac{-\jj S}{\gamma-\jj\omega}, \quad
\\&\Delta A:=\frac{-\jj S}{\gamma}-\frac{-\jj S}{\gamma-\jj\omega}.
\end{aligned}
\end{equation}
A quick simplification gives
\begin{equation}
\label{eq:DeltaA1}
\Delta A
= -\,\frac{S\,\omega}{\gamma(\gamma-\jj\omega)}.
\end{equation}

\paragraph{Intensity \( |A(t)|^2 \).}
Compute
\begin{align*}
|A(t)|^2
&= \big|A_\infty + \Delta A\, e^{-\gamma t} e^{+\jj\omega t}\big|^2 \\
&= |A_\infty|^2 + |\Delta A|^2 e^{-2\gamma t}
   + 2 e^{-\gamma t}\,\Re\!\left\{A_\infty \overline{\Delta A}\, e^{-\jj\omega t}\right\}.
\end{align*}
Each term is elementary:
\[
|A_\infty|^2=\frac{S^2}{\gamma^2+\omega^2}, \qquad
|\Delta A|^2=\frac{S^2\,\omega^2}{\gamma^2(\gamma^2+\omega^2)} .
\]
For the cross term, note that
\[
A_\infty \overline{\Delta A}
= \frac{-\jj S}{\gamma-\jj\omega}\;\cdot\;
\frac{-S\,\omega}{\gamma(\gamma+\jj\omega)}
= \frac{\jj\,S^2\,\omega}{\gamma(\gamma^2+\omega^2)} .
\]
Thus
\[
\Re\!\left\{A_\infty \overline{\Delta A}\, e^{-\jj\omega t}\right\}
= \frac{S^2\,\omega}{\gamma(\gamma^2+\omega^2)}\;\sin(\omega t).
\]
Collecting all pieces,
\begin{equation}
\boxed{
|A(t)|^2
= \frac{S^2}{\gamma^2+\omega^2}\left[
1+\frac{\omega^2}{\gamma^2}e^{-2\gamma t}
+\frac{2\omega}{\gamma}\,e^{-\gamma t}\sin(\omega t)
\right], \ t\ge 0.}
\label{eq:case1-intensity}
\end{equation}
Checks: $|A(0)|^2=S^2/\gamma^2$ (matches pre-step steady value) and
$|A(\infty)|^2=S^2/(\gamma^2+\omega^2)$ (new steady state).




\subsection{Turning off: $\Delta\omega(t)$ steps from $-\omega$ to $0$ at $t=0$}
Now
\begin{align*}
    &\Delta\omega(t)=\begin{cases}
-\omega,& t<0,\\
0,& t\ge 0,
\end{cases}
\qquad
\\&C_-=\gamma-\jj\omega, \quad C_+=\gamma, \quad 
\\&D=-\jj S.
\end{align*}

\paragraph{Pre-step steady state ($t<0$).}
\[
A(0^-)=\frac{D}{C_-}=\frac{-\jj S}{\gamma-\jj\omega}.
\]

\paragraph{Post-step solution ($t\ge 0$).}
With $C_+=\gamma$ and $A(0^+)=A(0^-)$,
\begin{align*}
A(t)
&= \frac{D}{\gamma} + \Big(A(0^+)-\frac{D}{\gamma}\Big)e^{-\gamma t}
\\&= \frac{-\jj S}{\gamma}
+ \left(\frac{-\jj S}{\gamma-\jj\omega}-\frac{-\jj S}{\gamma}\right)e^{-\gamma t}.
\label{eq:A-case2}
\end{align*}
Define
\[
A_\infty:=\frac{-\jj S}{\gamma}, \quad
\Delta A:=\frac{-\jj S}{\gamma-\jj\omega}-\frac{-\jj S}{\gamma}
=\frac{S\,\omega}{\gamma(\gamma-\jj\omega)} .
\]

\paragraph{Intensity \( |A(t)|^2 \).}
Here the transient has no rotating phase (only $e^{-\gamma t}$), so
\begin{align*}
|A(t)|^2
&= |A_\infty|^2 + |\Delta A|^2 e^{-2\gamma t}
   + 2 e^{-\gamma t}\,\Re\!\left\{A_\infty \overline{\Delta A}\right\}.
\end{align*}
Compute
\[
|A_\infty|^2=\frac{S^2}{\gamma^2}, \qquad
|\Delta A|^2=\frac{S^2\,\omega^2}{\gamma^2(\gamma^2+\omega^2)} .
\]
For the cross term,
\begin{align*}
    A_\infty \overline{\Delta A}
&= \frac{-\jj S}{\gamma}\;\cdot\;\frac{S\,\omega}{\gamma(\gamma+\jj\omega)}
\\&= \frac{-\jj S^2 \omega}{\gamma^2(\gamma+\jj\omega)}
\\&= \frac{-\jj\gamma - \omega}{\gamma^2+\omega^2}\;\frac{S^2 \omega}{\gamma^2}.
\end{align*}
Hence
\[
\Re\!\left\{A_\infty \overline{\Delta A}\right\}
= -\,\frac{S^2\,\omega^2}{\gamma^2(\gamma^2+\omega^2)} .
\]
Putting everything together,
\begin{equation}
\boxed{
\begin{aligned}
    |A(t)|^2
&= \frac{S^2}{\gamma^{2}}\left[
1 - 2\alpha\,e^{-\gamma t} + \alpha\,e^{-2\gamma t}
\right],
\qquad
\\ \alpha&:=\frac{\omega^2}{\gamma^2+\omega^2}, \;\; t\ge 0.
\end{aligned}
}
\label{eq:case2-intensity}
\end{equation}
Checks: $|A(0)|^2= S^2/(\gamma^2+\omega^2)$ (pre-step steady), and
$|A(\infty)|^2=S^2/\gamma^2$ (new steady).




\section{Through port response calculation}

We can calculate the through-port field through
\begin{equation}
\label{eq:sthru-def}
s_t(t)=s_i-\jj\,\mu\,A(t).
\end{equation}
We will use \(S\equiv\mu s_i\) and \(\kappa\equiv\mu^2\).

Across the step at \(t=0\), \(A\) is continuous. We assume the pre-step is in steady state.

\subsection{Turning on: \(\ (\Delta\omega:0\to-\omega)\)}
In this case, the microring's resonance frequency is shifting away from the laser frequency.
Pre-step: \(C_-=\gamma\Rightarrow A(0^-)=D/C_-=-\jj S/\gamma\) with \(D=-\jj S\).
Post-step: \(C_+=\gamma-\jj\omega\). Using the constant-coefficient solution,
\begin{align}
A(t) &= \frac{-\jj S}{\gamma-\jj\omega}
      +\left(\frac{-\jj S}{\gamma}-\frac{-\jj S}{\gamma-\jj\omega}\right)e^{-(\gamma-\jj\omega)t}
\notag\\[-2pt]
     &=: A_\infty+\Delta A\,e^{-(\gamma-\jj\omega)t},
\label{eq:A-case1}
\end{align}
where
$$A_\infty=\frac{-\jj S}{\gamma-\jj\omega},$$
$$\Delta A=-\frac{S\,\omega}{\gamma(\gamma-\jj\omega)}.$$

\subsubsection{Through-port field \(s_t(t)\)}
Insert \eqref{eq:A-case1} into \eqref{eq:sthru-def}:
\begin{align}
-\jj\mu A_\infty &= -\jj\mu\cdot\frac{-\jj S}{\gamma-\jj\omega}
                   = -\frac{\kappa s_i}{\gamma-\jj\omega},\\
-\jj\mu\,\Delta A &= -\jj\mu\left(-\frac{S\omega}{\gamma(\gamma-\jj\omega)}\right)
                   = \frac{\jj\,\kappa s_i\,\omega}{\gamma(\gamma-\jj\omega)}.
\end{align}
Hence
\begin{equation}
\boxed{\,s_t(t)
= s_i\!\left[
1-\frac{\kappa}{\gamma-\jj\omega}
+\frac{\jj\,\kappa\,\omega}{\gamma(\gamma-\jj\omega)}\,e^{-(\gamma-\jj\omega)t}
\right],\quad t\ge 0.\,}
\label{eq:st-case1}
\end{equation}

\subsubsection{Through power \(|s_t(t)|^2\)}
Write \(s_t=s_i\,(T_\infty+B e^{-(\gamma-\jj\omega)t})\) with
\[
T_\infty=1-\frac{\kappa}{\gamma-\jj\omega}
=\frac{(\gamma-\kappa)-\jj\omega}{\gamma-\jj\omega},\quad
B=\frac{\jj\,\kappa\,\omega}{\gamma(\gamma-\jj\omega)}.
\]
Then
\begin{equation}
    \begin{aligned}
\frac{|s_t|^2}{s_i^2}
&= |T_\infty|^2+|B|^2 e^{-2\gamma t}
 +2e^{-\gamma t}\,\Re\!\big(T_\infty\,\overline{B}\,e^{+\jj\omega t}\big) \\
&= \frac{(\gamma-\kappa)^2+\omega^2}{\gamma^2+\omega^2}
 + \frac{\kappa^2\omega^2}{\gamma^2(\gamma^2+\omega^2)}\,e^{-2\gamma t} \\
&\quad + \frac{2\kappa\omega}{\gamma(\gamma^2+\omega^2)}\,e^{-\gamma t}
\Big[(\gamma-\kappa)\sin(\omega t)-\omega\cos(\omega t)\Big].
\label{eq:T-case1}
\end{aligned}
\end{equation}
% So
% \begin{equation}
% \boxed{%
% \begin{aligned}
% |s_t(t)|^2
% &= s_i^{\,2}\Bigg\{
% \frac{(\gamma-\kappa)^2+\omega^2}{\gamma^2+\omega^2} \\[6pt]
% &\quad + \frac{\kappa^2\omega^2}{\gamma^2(\gamma^2+\omega^2)}\,e^{-2\gamma t} \\[6pt]
% &\quad + \frac{2\kappa\omega}{\gamma(\gamma^2+\omega^2)}\,e^{-\gamma t}
% \big[(\gamma-\kappa)\sin\omega t-\omega\cos\omega t\big]
% \Bigg\}.
% \end{aligned}}
% \end{equation}


\subsection{Turning off: \(\ (\Delta\omega:-\omega\to 0)\)}
In this case, the microring's resonance frequency is shifting onto the laser frequency. Pre-step: \(C_-=\gamma-\jj\omega\Rightarrow A(0^-)=D/C_-=-\jj S/(\gamma-\jj\omega)\).
Post-step: \(C_+=\gamma\). Thus
\begin{equation}
    \begin{aligned}
A(t)&=\frac{-\jj S}{\gamma}
     +\left(\frac{-\jj S}{\gamma-\jj\omega}-\frac{-\jj S}{\gamma}\right)e^{-\gamma t}
     \\&=: A_\infty+\Delta A\,e^{-\gamma t},
\end{aligned}
\end{equation}
where
$$A_\infty=\frac{-\jj S}{\gamma},$$
$$\Delta A=\frac{S\,\omega}{\gamma(\gamma-\jj\omega)}.$$

\subsubsection{Through-port field \(s_t(t)\)}
\begin{align}
-\jj\mu A_\infty &= -\jj\mu\cdot\frac{-\jj S}{\gamma}=-\frac{\kappa s_i}{\gamma},\\
-\jj\mu\,\Delta A &= -\jj\mu\cdot\frac{S\omega}{\gamma(\gamma-\jj\omega)}
                  = -\frac{\jj\,\kappa s_i\,\omega}{\gamma(\gamma-\jj\omega)}.
\end{align}
Hence
\begin{equation}
\boxed{\,s_t(t)
= s_i\!\left[
1-\frac{\kappa}{\gamma}
-\frac{\jj\,\kappa\,\omega}{\gamma(\gamma-\jj\omega)}\,e^{-\gamma t}
\right],\qquad t\ge 0.\,}
\label{eq:st-case2}
\end{equation}

\subsubsection{Through power \(|s_t(t)|^2\)}
Here \(T_\infty=1-\kappa/\gamma\in\mathbb{R}\) and \(B=-\jj\kappa\omega/(\gamma(\gamma-\jj\omega))\).
Then
\begin{equation}
    \begin{aligned}
\frac{|s_t|^2}{s_i^2}
&= T_\infty^2+|B|^2 e^{-2\gamma t}+2e^{-\gamma t}\,T_\infty\,\Re(\overline{B}) \\
&= \left(1-\frac{\kappa}{\gamma}\right)^2
 + \frac{\kappa^2\omega^2}{\gamma^2(\gamma^2+\omega^2)}\,e^{-2\gamma t}
 \\&\quad+ \frac{2\kappa\omega^2}{\gamma(\gamma^2+\omega^2)}
   \left(1-\frac{\kappa}{\gamma}\right) e^{-\gamma t}.
\label{eq:T-case2}
\end{aligned}
\end{equation}
% So
% \begin{equation}
% \boxed{%
% \begin{aligned}
% |s_t(t)|^2
% &= s_i^{\,2}\Bigg\{ 
%     \left(1-\frac{\kappa}{\gamma}\right)^2 \\[6pt]
% &\quad + \frac{\kappa^2\omega^2}{\gamma^2(\gamma^2+\omega^2)}\,e^{-2\gamma t} \\[6pt]
% &\quad + \frac{2\kappa\omega^2}{\gamma(\gamma^2+\omega^2)}
%    \left(1-\frac{\kappa}{\gamma}\right) e^{-\gamma t}
% \Bigg\}.
% \end{aligned}}
% \end{equation}
No sinusoid appears because the post-step detuning is \(0\).

\subsection{Critical coupling simplifications \(\ (\kappa=\mu^2=\gamma)\)}
Under \(\kappa=\gamma\), we have
\begin{equation}
\boxed{\,\frac{|s_t(t)|^2}{s_i^2}
= \frac{\omega^2}{\gamma^2+\omega^2}
\Big[1+e^{-2\gamma t}-2e^{-\gamma t}\cos(\omega t)\Big]\,}
\end{equation}
for case I turning on, and
\begin{equation}
\boxed{\,\frac{|s_t(t)|^2}{s_i^2}
= \frac{\omega^2}{\gamma^2+\omega^2}\,e^{-2\gamma t}}
\end{equation}
for case II turning off.


Multiplying by \(s_i^2\) gives the absolute powers.

\paragraph{Sanity checks.}
At \(t=0\): Case 1 yields \(|s_t|^2=0\) (on-resonance critical coupling),  
Case 2 gives \(|s_t|^2=s_i^2\,\omega^2/(\gamma^2+\omega^2)\) (pre-step detuned steady state).
As \(t\to\infty\), Case 1 approaches the detuned steady state;
Case 2 decays to zero (critical coupling on resonance).





\section{Discussion}
In general, for $t\ge 0$ the solution has the form
\[
A(t)=A_\infty+\Delta A\,e^{-\gamma t}e^{-\jj\Delta\omega_+ t}.
\]
Then
\begin{align*}
    |A(t)|^2 = &|A_\infty|^2 + |\Delta A|^2 e^{-2\gamma t}
\\&+ 2 e^{-\gamma t}\,\Re\!\left\{A_\infty \overline{\Delta A}\,e^{-\jj\Delta\omega_+ t}\right\}.
\end{align*}

If the post-step detuning $\Delta\omega_+$ is nonzero (Case~1, $\Delta\omega_+=-\omega$),
the cross term contains $\sin(\omega t)$ and produces an exponentially damped oscillation.
If $\Delta\omega_+=0$ (Case~2), the cross term is time-independent and the intensity decays
monotonically with no sinusoid.
\begin{center}\label{onofftransient}
\includegraphics[width=0.45\textwidth]{figs/onofftransient.png}
\captionof{figure}{Transient response with critical coupling condition.}
\end{center}
Physical understanding of the transient picture would be developed below in detail.


\subsection{Transient time constant for two cases}
The post-step system has one pole at $-\gamma-j\Delta\omega_+$. Therefore the \emph{field} transient always decays as
\[
A_{\text{hom}}(t)\propto e^{-(\gamma+j\Delta\omega_+)t}\quad\Rightarrow\quad |A_{\text{hom}}(t)|\sim e^{-\gamma t},
\]
so the \textbf{field time constant is the same in both cases}: $\tau=1/\gamma$.

But why in Fig.~\ref{onofftransient}, the two cases show different decay rates? The through \emph{power} contains interference between the constant and the decaying field:
\[
\frac{|s_t(t)|^2}{s_i^2}
=|T_\infty|^2+\underbrace{|B|^2\,e^{-2\gamma t}}_{\text{pure transient}}
+\underbrace{2e^{-\gamma t}\,\Re\!\big(T_\infty\overline{B}\,e^{+j\Delta\omega_+ t}\big)}_{\text{cross (interference) term}}.
\]
Hence power traces generically have \emph{two} decays: an $e^{-2\gamma t}$ term (square of the field transient) and an $e^{-\gamma t}$ term (interference between steady and transient fields). Which one dominates---and whether it oscillates---depends on the \emph{post-step detuning} and the \emph{coupling}.

\subsubsection{Switching on: Step away from the laser {$(0\to-\omega)$}}
After the step $\Delta\omega_+=-\omega\neq 0$.
\begin{itemize}
  \item The transient \emph{rotates} at $\omega$: $e^{-(\gamma-j\omega)t}$.
  \item The cross term contains $\sin\omega t$ and $\cos\omega t$; power exhibits \textbf{damped oscillations} with an \textbf{envelope} $\sim e^{-\gamma t}$.
  \item The $e^{-2\gamma t}$ piece is also present (smaller at long times).
\end{itemize}
\emph{Interpretation:} The cavity initially stores energy (on resonance). After detuning, the leaked field from the ring rotates in phase relative to the direct waveguide path; their interference at the through port produces ringing at $\omega$, decaying with the cavity lifetime $\tau$.

\subsubsection{Switching off: Step onto the laser {$(-\omega\to 0)$}}
After the step $\Delta\omega_+=0$.
\begin{itemize}
  \item The transient has \emph{no rotation}: $e^{-\gamma t}$.
  \item The cross term is \textbf{monotonic} ($\propto e^{-\gamma t}$) and its \emph{coefficient} is $\propto (1-\mu^2/\gamma)$.
  \item At \textbf{critical coupling} ($\mu^2=\gamma$), the cross term \emph{vanishes}, leaving a \textbf{pure} $e^{-2\gamma t}$ decay in power (apparent power time constant $\tau/2$).
\end{itemize}
\emph{Interpretation:} Once on resonance the leaked field’s phase is fixed relative to the input (no beating). The through power relaxes monotonically; exactly at critical coupling the direct and leaked fields cancel at steady state, removing the $e^{-\gamma t}$ interference path and exposing only the squared-transient $e^{-2\gamma t}$.

\subsection{Phasor and pole viewpoints (unified picture)}
\begin{itemize}
  \item \textbf{Pole view:} Post-step pole $-\gamma-j\Delta\omega_+$ has the same real part ($-\gamma$) in both cases $\Rightarrow$ same field lifetime $\tau$; imaginary part ($-\Delta\omega_+$) governs whether the transient \emph{rotates} (oscillations in power) or not.
  \item \textbf{Phasor view:} The through field is the vector sum of a constant (direct path) and a shrinking phasor (leakage). If the shrinking phasor \emph{spins} (detuned case), the sum’s length oscillates; if it \emph{slides} without spinning (on-resonance case), the sum changes monotonically.
\end{itemize}

\subsection{Practical takeaway}
\begin{itemize}
  \item The \textbf{intrinsic} cavity decay rate is always $\gamma$ ($\tau=1/\gamma$).
  \item Power traces can look faster or slower depending on whether the $e^{-\gamma t}$ cross term is present or suppressed:
    \begin{itemize}
      \item Away from laser: $e^{-\gamma t}$ envelope with oscillations at $\omega$ (always present).
      \item Onto the laser: $e^{-\gamma t}$ monotone cross term; at critical coupling it cancels, leaving only $e^{-2\gamma t}$.
    \end{itemize}
  \item Apparent differences in “decay time” between the two experiments reflect \emph{interference pathways}, not a change in the underlying photon lifetime.
\end{itemize}
\vspace{10pt}
\subsection{NRZ - not exactly on-resonance}

Some intuitive analysis tells us there would be oscillation in both cases because interference can happen also when switching off: there is still remaining energy in the through waveguide that would interfere with the energy leaked from the ring. The ringing frequency in this case would be the small deviation from $\Delta\omega=0$.

\bibliography{main.bib}

\end{document}

Generated with assistance of ChatGPT.

Lineshape function / Linewidth broadening -- through a RLC circuit example

2025-07-27T06:09:29.331Z

code for generating fig. 1:

import numpy as np
import matplotlib.pyplot as plt

# Frequency range
nu = np.linspace(0.9, 1.1, 1000)
nu0 = 1.0  # Resonance frequency

# Define Lorentzian function
def lorentzian(nu, nu0, Q):
    gamma = nu0 / (2 * Q)
    return 1 / ((nu - nu0)**2 + gamma**2)

# Two cases: small R (high Q) and large R (low Q)
Q_high = 100  # small R → high Q → narrow peak
Q_low = 10    # large R → low Q → broad peak

# Compute line shapes
f_highQ = lorentzian(nu, nu0, Q_high)
f_lowQ = lorentzian(nu, nu0, Q_low)

# Normalize the functions
f_highQ /= np.trapz(f_highQ, nu)
f_lowQ /= np.trapz(f_lowQ, nu)

# Plot
plt.figure(figsize=(8, 5))
plt.plot(nu, f_highQ, label='Small R (High Q)', linewidth=2)
plt.plot(nu, f_lowQ, label='Large R (Low Q)', linewidth=2)
plt.title('Line Shape Function vs. Resistance in RLC Circuit')
plt.xlabel('Frequency ν (arb. units)')
plt.ylabel('Normalized f(ν)')
plt.legend()
plt.grid(True)
plt.tight_layout()
plt.show()

LaTex code:

\documentclass[twocolumn,superscriptaddress,preprintnumbers,amsmath,amssymb]{revtex4}
%\usepackage{CJK}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{dcolumn}% Align table columns on decimal point
\usepackage{graphicx}
\usepackage{subfigure}
\usepackage[american]{circuitikz}
\usepackage{tikz}
\usepackage{bm}% bold math
\usepackage{epstopdf}
\usepackage{threeparttable}
\usepackage{float}
\usepackage{color}
\usepackage{mathtools} % add box around equation in align environment
\usepackage[font=small,labelfont=bf]{caption}


%\usepackage{times}
%\usepackage{appendix}
%\usepackage{fancyhdr}


\renewcommand\thesection{\arabic{section}}
\renewcommand\thesubsection{\thesection.\arabic{subsection}}
\renewcommand\thesubsubsection{\thesubsection.\arabic{subsubsection}}

%\renewcommand{\thefootnote}{\alph{footnote}}
%\renewcommand{\thefootnote}{\roman{footnote}}


\begin{document}
%\begin{CJK}{UTF8}{gbsn} % Use default fonts from CJK (see below)
\title{How Linewith Broadening Arise from Finite Lifetime}
\author{Dingning Li} 
\affiliation{ECE, BU, Boston, MA}
% \author{Kai Zhang} 
% \email{kai.zhang@dukekunshan.edu.cn}
% \affiliation{Division of Natural and Applied Sciences, Duke Kunshan University, Kunshan, Jiangsu, 215300, China}

\date{\today}


\begin{abstract}

    This is a derivation for the lineshape function $g(\nu)$. As pointed out by Yariv~\cite{yariv1976introduction}, the Lorentzian form for $g(\nu)$ can be revealed through a simple RLC circuit analogy. Below is the derivation of the power spectrum for a series RLC circuit when its driving source (with frequency $\omega_0$) suddenly turns off. This shows how lineshape functions (or broadening) arise from transient decaying systems with finite lifetime.

\end{abstract}

 \maketitle

\section*{1. Circuit Model and Transient Response}

Consider a series RLC circuit excited at its resonant frequency and then disconnected at time \( t = 0 \). The circuit exhibits a damped oscillation governed by the differential equation:
\begin{equation}
L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = 0.
\end{equation}
This is a standard second-order linear ODE with characteristic equation:
\begin{equation}
\lambda^2 + \frac{R}{L} \lambda + \frac{1}{LC} = 0.
\end{equation}
Define the natural angular frequency and damping coefficient:
\begin{equation}
\omega_0 = \frac{1}{\sqrt{LC}}, \qquad \gamma = \frac{R}{2L}.
\end{equation}
In the underdamped regime (\( \gamma < \omega_0 \)), the solution is:
\begin{equation}
q(t) = A e^{-\gamma t} \cos(\omega_d t + \phi), \qquad \omega_d = \sqrt{\omega_0^2 - \gamma^2}.
\end{equation}
The current is the derivative of charge:
\begin{equation}
I(t) = \frac{dq}{dt} = A e^{-\gamma t} [-\gamma \cos(\omega_d t + \phi) - \omega_d \sin(\omega_d t + \phi)].
\end{equation}
This can be written in cosine form as:
\begin{equation}
I(t) = I_0 e^{-\gamma t} \cos(\omega_d t + \phi').
\end{equation}

\section*{2. Fourier Transform and Spectrum}

We are interested in the frequency-domain representation of this transient response:
\begin{equation}
x(t) = A e^{-\gamma t} \cos(\omega_d t), \qquad t \geq 0.
\end{equation}
We compute its Fourier transform:
\begin{equation}
X(\nu) = \int_0^\infty A e^{-\gamma t} \cos(\omega_d t) e^{-2\pi i \nu t} \, dt.
\end{equation}
Use Euler's formula:
\begin{equation}
\cos(\omega_d t) = \frac{1}{2} \left( e^{i \omega_d t} + e^{-i \omega_d t} \right),
\end{equation}
so
\begin{equation}
X(\nu) = \frac{A}{2} \int_0^\infty \left[ e^{-\gamma t} e^{i (\omega_d - 2\pi \nu) t} + e^{-\gamma t} e^{-i (\omega_d + 2\pi \nu) t} \right] dt.
\end{equation}
Define:
\begin{equation}
a_+ = \gamma - i(\omega_d - 2\pi \nu), \qquad
a_- = \gamma + i(\omega_d + 2\pi \nu).
\end{equation}
Then
\begin{equation}
X(\nu) = \frac{A}{2} \left( \int_0^\infty e^{-a_+ t} dt + \int_0^\infty e^{-a_- t} dt \right)
= \frac{A}{2} \left( \frac{1}{a_+} + \frac{1}{a_-} \right).
\end{equation}

\section*{3. Near-Resonance Approximation}

Let \( \nu_0 = \omega_d / (2\pi) \), and define \( \Delta = \nu - \nu_0 \). Then:
\begin{equation}
a_+ = \gamma - i (2\pi \Delta), \qquad
a_- \approx \gamma + i (2\pi (2\nu_0 + \Delta)).
\end{equation}
Since \( \nu_0 \gg \gamma \), the second term \( 1/a_- \) contributes little. Therefore, near resonance:
\begin{equation}
X(\nu) \approx \frac{A}{2} \cdot \frac{1}{\gamma - i 2\pi (\nu - \nu_0)}.
\end{equation}
Taking the squared magnitude (power spectrum):
\begin{equation}
|X(\nu)|^2 \propto \frac{1}{\gamma^2 + [2\pi (\nu - \nu_0)]^2}.
\end{equation}

\section*{4. Lorentzian Form and Quality Factor}

Define the FWHM of the spectrum as:
\begin{equation}
\Delta \nu = \frac{\gamma}{\pi}.
\end{equation}
Then the spectrum becomes:
\begin{equation}
f(\nu) = \frac{1}{(\nu - \nu_0)^2 + (\Delta \nu / 2)^2}.
\end{equation}
We define the quality factor as:
\begin{equation}
Q = \frac{\nu_0}{\Delta \nu} = \frac{\pi \nu_0}{\gamma}.
\end{equation}
Thus the spectrum becomes:
\begin{equation}
f(\nu) = \frac{1}{(\nu - \nu_0)^2 + (\nu_0 / 2Q)^2}.
\end{equation}
This is the standard Lorentzian line shape.

\section*{5. Conclusion}

From the derivation above, we have seen that finite lifetime of a decaying process will lead to linewidth broadening. In the case for RLC circuit, lifetime is determined by the $R$ value. When $R$ is small, $Q$ would be large and the circuit has longer lifetime, leading to narrower spectral width. When $R$ is large, $Q$ would be small and the circuit has shorter lifetime, leading to broader lineshape. 

\begin{center}
\includegraphics[width=0.35\textwidth]{figs/output-4.png}
\captionof{figure}{Two different cases of lineshape function $f(\nu)$.}
\label{fig:ingaasp_absorp}
\end{center}

\section*{6. Notes on Circuit Topology and Q Definitions}

The exact expression for \( Q \) depends on whether the RLC circuit is series or parallel:

\begin{itemize}
  \item \textbf{Series RLC circuit:}
  \begin{equation}
  Q_\text{series} = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{\omega_0 L}{R}, \qquad \gamma = \frac{R}{2L}
  \end{equation}
  \item \textbf{Parallel RLC circuit:}
  \begin{equation}
  Q_\text{parallel} = \omega_0 R C = 2\pi \nu_0 R C, \qquad \gamma = \frac{1}{2RC}
  \end{equation}
\end{itemize}

Regardless of topology, the frequency-domain spectrum from the transient decay is always Lorentzian in form:
\begin{equation}
f(\nu) = \frac{1}{(\nu - \nu_0)^2 + (\nu_0 / 2Q)^2}.
\end{equation}

\bibliography{main.bib}

\end{document}

Generated with assistance of ChatGPT.

Carrier Lifetime in Semiconductor Optical Amplifiers (SOA)

2025-07-20T06:02:52.610Z

Code for generating Fig. 1:

import numpy as np
import matplotlib.pyplot as plt

# Parameters
R_inj = 1e27      # carriers/cm^3/s
R_st = 0.3e27     # stimulated recombination rate
tau = 1e-9        # carrier lifetime in seconds
t0 = 3e-9         # time at which R_st turns off

# Time array (in seconds)
t = np.linspace(0, 8e-9, 500)

# Compute N(t)
N = np.zeros_like(t)

# For t < t0
mask1 = t < t0
N[mask1] = (R_inj - R_st) * tau * (1 - np.exp(-t[mask1] / tau))

# For t >= t0
N_t0 = (R_inj - R_st) * tau * (1 - np.exp(-t0 / tau))  # N(t0)
N_inf = R_inj * tau
mask2 = t >= t0
N[mask2] = N_inf * (1 - np.exp(-(t[mask2] - t0) / tau)) + N_t0 * np.exp(-(t[mask2] - t0) / tau)

# Plotting
plt.figure(figsize=(8, 4))
plt.plot(t * 1e9, N / 1e18, label="Carrier Density $N(t)$")
plt.axvline(t0 * 1e9, color='gray', linestyle='--', label="$t_0$")
plt.xlabel("Time (ns)")
plt.ylabel("Carrier Density ($10^{18}$ cm$^{-3}$)")
plt.title("Carrier Density Evolution with Step Change in $R_{st}$ at $t_0$")
plt.legend()
plt.grid(True)
plt.tight_layout()
plt.show()

Code for Latex

\documentclass[twocolumn,superscriptaddress,preprintnumbers,amsmath,amssymb]{revtex4}
%\usepackage{CJK}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{dcolumn}% Align table columns on decimal point
\usepackage{graphicx}
\usepackage{subfigure}
\usepackage[american]{circuitikz}
\usepackage{tikz}
\usepackage{bm}% bold math
\usepackage{epstopdf}
\usepackage{threeparttable}
\usepackage{float}
\usepackage{color}
\usepackage{mathtools} % add box around equation in align environment
\usepackage[font=small,labelfont=bf]{caption}


%\usepackage{times}
%\usepackage{appendix}
%\usepackage{fancyhdr}


\renewcommand\thesection{\arabic{section}}
\renewcommand\thesubsection{\thesection.\arabic{subsection}}
\renewcommand\thesubsubsection{\thesubsection.\arabic{subsubsection}}

%\renewcommand{\thefootnote}{\alph{footnote}}
%\renewcommand{\thefootnote}{\roman{footnote}}


\begin{document}
%\begin{CJK}{UTF8}{gbsn} % Use default fonts from CJK (see below)
\title{Material Properties for InGaAsP}
\author{Dingning Li} 
\affiliation{ECE, BU, Boston, MA}
% \author{Kai Zhang} 
% \email{kai.zhang@dukekunshan.edu.cn}
% \affiliation{Division of Natural and Applied Sciences, Duke Kunshan University, Kunshan, Jiangsu, 215300, China}

\date{\today}


\begin{abstract}

    Physical meaning of carrier lifetime is discussed in this letter. It has been shown that carrier lifetime does not only represents how fast carriers decay, but also represents how fast the carriers can be generated, which is important for SOA applications. The slow gain recovery~\cite{lal2007monolithic,lal2006thesis}, can be understood through taking carrier lifetime into consideration.

\end{abstract}

 \maketitle

\section*{1. Case 1: Turning On Current Injection}

When a constant current is turned on at $t = 0$, the carrier density $N(t)$ evolves according to the rate equation:
\[
\frac{dN}{dt} = R_{\text{inj}} - \frac{N}{\tau}
\]
where:
\begin{itemize}
  \item $R_{\text{inj}}$ is the constant carrier injection rate,
  \item $\tau$ is the carrier lifetime,
  \item $\frac{N}{\tau}$ represents spontaneous and nonradiative recombination.
\end{itemize}

This is a first-order linear ODE. The solution is obtained using the integrating factor method:
\[
\mu(t) = e^{t/\tau}
\]
Multiplying both sides:
\[
e^{t/\tau} \frac{dN}{dt} + \frac{1}{\tau} e^{t/\tau} N = R_{\text{inj}} e^{t/\tau}
\]
\[
\frac{d}{dt}\left(N e^{t/\tau}\right) = R_{\text{inj}} e^{t/\tau}
\]
\[
N(t) = R_{\text{inj}} \tau \left(1 - e^{-t/\tau}\right)
\]
This shows an exponential rise to the steady-state carrier density $N_\infty = R_{\text{inj}} \tau$ with time constant $\tau$.

\section*{2. Case 2: Stimulated Recombination Turns Off at $t = t_0$}

Initially, an optical signal causes stimulated recombination at rate $R_s$, and the rate equation becomes:
\[
\frac{dN}{dt} = R_{\text{inj}} - R_s - \frac{N}{\tau}
\]

\subsection*{For $t < t_0$:}

The solution is again a first-order linear ODE:
\[
\frac{dN}{dt} + \frac{N}{\tau} = R_{\text{inj}} - R_s
\]
\[
N(t) = (R_{\text{inj}} - R_s)\tau \left(1 - e^{-t/\tau} \right)
\]

\subsection*{For $t \geq t_0$:}

Stimulated recombination is turned off, so:
\[
\frac{dN}{dt} + \frac{N}{\tau} = R_{\text{inj}}
\]
Let $N(t_0)$ be the carrier density at the moment of switching:
\[
N(t_0) = (R_{\text{inj}} - R_s)\tau \left(1 - e^{-t_0/\tau} \right)
\]

The solution for $t \ge t_0$ is:
\[
N(t) = R_{\text{inj}}\tau \left(1 - e^{-(t - t_0)/\tau} \right) + N(t_0) e^{-(t - t_0)/\tau}
\]

This shows exponential recovery to a higher steady-state $N_\infty = R_{\text{inj}} \tau$, governed by the same carrier lifetime $\tau$.

\begin{center}
\includegraphics[width=0.35\textwidth]{figs/carrier_dynamics_plot.png}
\captionof{figure}{Carrier density $N(t)$ as a function of time, with $R_{st}$ turning off at $t_0$. }
\label{fig:ingaasp_absorp}
\end{center}


\bibliography{main.bib}

\end{document}

Generated with assistance of ChatGPT.

[Physics] Special Relativity I

2020-03-25T10:02:12.297Z

This is learning note about special relativity. Your comments and criticisms are always welcome.

!!! Attention !!!

The material in this post does not guarantee accuracy and correctness. If any fault is found OR any improvement can be made, PLEASE LEAVE COMMENT for future improvement and correction

[Math Tutorial Series] Vector Field and its Properties

2020-03-02T15:31:20.751Z

This post mainly foces on basic introduction to vector fields and its properties. Your comments and criticisms are always welcome.

!!! Attention !!!

The material in this post does not guarantee accuracy and correctness. If any fault is found OR any improvement can be made, PLEASE LEAVE COMMENT for future improvement and correct.

[Physics] Statistical mechanics

2020-02-14T09:41:07.815Z

This post is a note from Mehran Kardar's book. Your comments and criticisms are always welcome.

Remark: the book seems very confusing, though after understanding the whole thing in 1.1, 1.2, 1.3, the logic makes a lot of sense. yet now i have been in trouble in keep going. A transfer into another textbook may happen soon.

!!! Attention !!!

The material in this post does not guarantee accuracy and correctness. If any fault is found OR any improvement can be made, PLEASE LEAVE COMMENT for future improvement and correct.

[Math Tutorial Series] Complex exponential function I

2020-02-09T10:29:28.965Z

This post mainly foces on basic introduction to complex exponential functions and imaginary numbers. Your comments and criticisms are always welcome.

This is the first post in Math Tutorial Series. This post mainly foces on basic introduction to complex exponential functions and imaginary numbers. Following posts about this topic will come up soon, covering applications in Physics and going deeper into its math principles.

!!! Attention !!!

The material in this post does not guarantee accuracy and correctness. If any fault is found OR any improvement can be made, PLEASE LEAVE COMMENT for future improvement and correct.

[# ERROR FIXING] LAMMPS writing movie failure & undefined reference to `png_create_write_struct'

2020-01-29T08:05:20.746Z

/*Remark: This is the first ERROR FIXING record of this blog.*/

platform information: raspberrypi 4.

Error report:

ERROR on proc 0: Support for writing movies not included (../dump_movie.cpp:52)

/usr/bin/ld: image.o: in function `LAMMPS_NS::Image::write_PNG(_IO_FILE*)':
/home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1032: undefined reference to `png_create_write_struct'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1036: undefined reference to `png_create_info_struct'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1042: undefined reference to `png_set_longjmp_fn'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1043: undefined reference to `png_destroy_write_struct'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1038: undefined reference to `png_destroy_write_struct'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1047: undefined reference to `png_init_io'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1048: undefined reference to `png_set_compression_level'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1049: undefined reference to `png_set_IHDR'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1066: undefined reference to `png_set_text'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1067: undefined reference to `png_write_info'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1073: undefined reference to `png_write_image'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1074: undefined reference to `png_write_end'
/usr/bin/ld: /home/pi/lammps/lammps-7Aug19/src/Obj_serial/../image.cpp:1076: undefined reference to `png_destroy_write_struct'
collect2: error: ld returned 1 exit status
make[1]: *** [Makefile:89: ../lmp_serial] Error 1
make[1]: Leaving directory '/home/pi/lammps/lammps-7Aug19/src/Obj_serial'
make: *** [Makefile:217: serial] Error 2

As seen from above, there are two errors. Actually, the second error occur while trying to fix the first error.

ERROR DESCRIPTION

1. According to MANUAL (Aug 06, 2019, p.365), the first error:

Support for writing images in JPEG format not included LAMMPS was not built with the -DLAMMPS_JPEG switch in the Makefile.
Support for writing images in PNG format not included LAMMPS was not built with the -DLAMMPS_PNG switch in the Makefile.
Support for writing movies not included LAMMPS was not built with the -DLAMMPS_FFMPEG switch in the Makefile

2. According to Internet resources, the second error:

https://blog.csdn.net/yaoyefengchen/article/details/20012723

https://www.linuxidc.com/Linux/2014-02/97344.htm

ERROR FIXING

The fixing process of the first error should follow MANUAL (Aug 06, 2019, p.38).

open the corresponding Makefile document, under /src/MAKE. I only edited the makefile for lmp_serial.
find the string 'LMP_INC ='.
add the switch: '-DLAMMPS_FFMPEG' to the end of that expression.
save and exist.

[Remark]: since here I didn't turn on the switch for generating JPEG and PNG images, some other configurations are omitted (as pointed out on p.38, writing paths of libraries for images etc.). Though only part of the task had finished and presented here, I believe the rest (lmp_mpi, image) should be easy.

The second error looks tricky. We should notice that, the real error message is not the ones directly shown at the end:

collect2: error: ld returned 1 exit status
make[1]: *** [Makefile:89: ../lmp_serial] Error 1
make[1]: Leaving directory '/home/pi/lammps/lammps-7Aug19/src/Obj_serial'
make: *** [Makefile:217: serial] Error 2

These are not real error messages, which fooled me into finding out reasons on the Internet and took a long time. Albeit some interesting posts are found online:

https://stackoverflow.com/questions/15273904/make-error-make1-directories-error-1

The real solution should follow from the links mentioned in the previous section. However, I would reform it in my way:

find where is the library for png: '$locate libpng'. The first returned result should be the package we want. In my case '/lib/arm-linux-gnueabihf/libpng12.so.0'.
open the corresponding Makefile. Add the package found in step 1 & ' -lpng' to the 'LIB = '. In this case of LAMMPS, the final result should be 'LIB = /lib/arm-linux-gnueabihf/libpng12.so.0 -lpng.
save and exist.

[Remark]: it is slightly different from the links. I think the links would also work.

Reflection

The two errors here are quite strange, which haven't occured on the Cluster and Lab-Desktop. Moreover, the reason for the second error is opaque to me. It is very weird that an error that never happened before suddenly appear, under very similiar conditions. Whether there is any connections between the two is also unknown.