August 17
Transient response of mircoring modulators
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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}