4.3. Water cloud model

As a first-order radiative transfer solution, the WCM model expresses the total backscattering signals as the summation of surface and volume scattering components, σ<sup>0</sup> total <sup>¼</sup> <sup>Γ</sup><sup>2</sup> σ0 surfaceþ σ0 volume. The surface scattering can be modeled using the bare soil moisture model such as the previous IEM and Oh models. The vegetation two-way attenuation on the surface scattering power is modeled by <sup>Γ</sup><sup>2</sup> <sup>¼</sup> exp ð Þ �2τ<sup>=</sup> cos <sup>θ</sup> .

The vegetation layer is assumed to be comprised of homogenous water particles with a uniform distribution, and volume scattering component can be expressed from vegetation scattering albedo and optical depth such as σ<sup>0</sup> volume <sup>¼</sup> <sup>0</sup>:75<sup>ω</sup> <sup>1</sup> � <sup>Γ</sup><sup>2</sup> cos <sup>θ</sup>. Accounting the polarization leads to the following empirical volume power [8]:

$$\begin{aligned} \sigma\_{\text{volume}}^{IV-pda} &= \sigma\_{\text{volume}}^{H\text{H}-pda} = 0.74\omega \left[ 1 + 0.54\omega \tau - 0.24(\omega \tau)^2 \right] \left[ 1 - \exp\left( -2.12\tau/\cos\theta \right) \right] \cos\theta\\ \sigma\_{\text{volume}}^{HV-pda} &= \omega \left[ 0.044\omega \tau - 0.018(\omega \tau)^2 + 0.006(\omega \tau)^3 \right] \left[ 1 - \exp\left( -11.7\tau/\cos\theta \right) \right] \cos\theta \end{aligned} \tag{33}$$

At the moderate or high frequency such as C- and X-bands, the dihedral scattering is negligible. However, at low frequency such as L-band, the dihedral scattering component must be accounted, which can be quantified as [8]

$$
\sigma\_{dhedral}^{H-pld} = 1.9\omega \left[1 + 0.9\omega \tau + 0.4(\omega \tau)^2\right] \left[1 - \exp\left(-1.93\tau/\cos\theta\right)\right] \exp\left(-1.37\tau^{1.12}/\cos\theta\right)
$$

$$
\exp\left(-0.84(\text{ks})^2 \cos\theta\right) |\mathcal{R}\_{Hl}|^2 \cos\theta
$$

$$
\sigma\_{dhedral}^{HV-pld} = 0.013\omega \left[1 + 7.85\omega\tau + 7.9(\omega \tau)^2\right] \left[1 - \exp\left(-9.62\tau/\cos\theta\right)\right] \exp\left(-1.02\tau^{1.38}/\cos\theta\right)
$$

$$
\exp\left(-2.9(\text{ks})^2 \cos\theta\right) \left(|\mathcal{R}\_{HH}|^2 + |\mathcal{R}\_{VV}|^2\right) 0.5 \cos\theta
$$

$$
\sigma\_{dhedral}^{VV-pld} = 0\tag{34}
$$
