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Consider an original stipulated model, not measured material data: thermal diffusivity $\alpha =10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, period $P=100\mathrm{s}$, driven temperature amplitude ${\Theta }_0=4\mathrm{K}$, and thermometers at $x_1=0.020\mathrm{m}$ and $x_2=0.050\mathrm{m}$. Without side losses, $a=b=17.7245{\mathrm{m}}^{-1}$ and thermal penetration depth $\delta =0.05642\mathrm{m}$. ThusThe two no-loss estimates and the combined estimate all return $10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Now retain the same thermal diffusivity and drive but introduce $\lambda =0.010{\mathrm{s}}^{-1}$. The exact complex wavenumber gives $a=19.1863{\mathrm{m}}^{-1}$ and $b=16.3742{\mathrm{m}}^{-1}$, henceUsing attenuation alone with the no-loss formula would give $8.5343\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$; using lag alone would give $1.17174\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. Their disagreement is informative. If an independent volumetric heat capacity measurement gives $\rho c_p=2.0\times 10^6\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, the combined estimate implies thermal conductivity $k=200\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H=2.0\times 10^4\mathrm{W}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$.

The exact complex amplitudes can be checked independently by discretizing $\alpha {\hat{\theta }}^{\prime \prime }-(\lambda +i\omega )\hat{\theta }=0$ and solving the resulting linear system with a tridiagonal matrix. Halving the spatial step quarters the discrepancy, as expected for centered finite differences. Separately fitting the real time traces to a constant plus sine and cosine terms recovers both amplitudes and their lag; this also checks the sign convention used in the combined estimate.

Write ${\delta }_j=\sqrt{2K/{\omega }_j}$ for each component's thermal penetration depth. Its contribution at depth isThus its amplitude is multiplied by $e^{-x/{\delta }_j}$ and its wave phase lags the surface by $x/{\delta }_j$ radians. The corresponding unwrapped time lag is $x/({\delta }_j{\omega }_j)$; a phase measured from a single periodic signal is observable only modulo $2\pi$.

Higher-frequency variations have smaller penetration depths. In particular,The daily oscillation is therefore strongly attenuated at depths where a substantial annual oscillation survives. This is a thermal diffusion wave, whose phase lag is not the arrival time of a sharp thermal front. The displayed solution is the established periodic response; an additional transient would depend on initial subsurface data not needed for this interpretation.

The thermal penetration depth of a harmonic ground temperature disturbance scales as the inverse square root of its angular frequency. For yearly and daily forcing, a 365-day year gives ${\delta }_Y/{\delta }_D=\sqrt{365}\simeq 19.1$. At ten daily penetration depths the daily component is attenuated by $e^{-10}$, whereas the annual one retains about $59.2\%$ of its surface amplitude. At ten annual penetration depths, even the annual component is attenuated by $e^{-10}$. These are consequences of the linear thermal diffusion wave model, with homogeneous constant thermal diffusivity.

For the no-side-loss thermal diffusion wave, the thermal penetration depth $\delta =1/a=\sqrt{2\alpha /\omega }=\sqrt{\alpha P/\pi }$ is the distance over which the amplitude falls by $e^{-1}$ and the wave phase lags by one radian. It is a periodic counterpart of the thermal diffusion length, with its numerical factor fixed by this definition. A shorter period probes a shallower region; a longer period increases sensitivity to the far-end boundary condition.