July 27
Lineshape function / Linewidth broadening -- through a RLC circuit example
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()
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\begin{document}
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\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}