solution through the following equation[@Schmidt2008] $$\label{eq:TDTRSignal}
Z(T_{12}(\omega), t) = \sum\limits_{n = -\infty}^{\infty}H(T_{12}(\omega), \eta_0+n\eta_s)e^{in\eta_st},$$ where $\eta_0$ is the reference angular frequency of the periodic heating, $\eta_s$ is the angular sampling frequency set by the repetition rate of the laser pulses, and $H(T_{12}(\omega), \eta)$ is surface temperature response subject to a periodic heating at frequency $\eta$ given the transmission coefficients $T_{12}(\omega)$. Therefore, identifying $Z(T_{12}(\omega), t)$ requires computing $H(T_{12}(\omega), \eta)$, or the surface temperature frequency response to a periodic heating in a double-layer structure of a metal film on a substrate as shown in Fig. \[fig:Schematics\] (b).

Thermal transport in an isotropic crystal, assuming only cross-plane heat conduction, is described by the one-dimensional (1D) spectral Boltzmann transport equation (BTE) under the relaxation time approximation (RTA)[@Majumdar1993], $$\begin{aligned}
\label{eq:BTE}
\frac{\partial g_{\omega}}{\partial t} &+& \mu v_{\omega} \frac{\partial g_{\omega}}{\partial x} = -\frac{g_{\omega}+f_0(T_0)-f_0(T)}{\tau_{\omega}}+\frac{Q_{\omega}}{4\pi} \\
f_0(T) &=& \frac{1}{4\pi}\hbar \omega D(\omega) f_{BE}(T) \approx f_0(T_0)+\frac{1}{4\pi}C_{\omega}\Delta T,
\label{eq:BEDist_Linearized}\end{aligned}$$ where $g_{\omega} =\hbar\omega D(\omega)[f_{\omega}(x,t,\mu)-f_0(T_0)]$ is the deviational distribution function, $f_0 = f_0(x,t)$ is the equilibrium distribution function, $\mu = cos(\theta)$ is the directional cosine, $v_{\omega}$ is the phonon group velocity, $\tau_{\omega}$ is the phonon relaxation time, and $Q_{\omega}(x,t)$ is the spectral volumetric heat generation. Assuming a small temperature rise, $\Delta T = T - 