## 5 Conclusions
Based on central-limit theorem, the random phase \(\phi\) follows the standard normal distribution, so \(M_{\phi}(\omega)\) is given by Equation (A4). \[M_{\phi}(\omega)=\exp\left(-\frac{\sigma_{\phi}^{2}\omega^{2}}{2}\right)\] (A4)  
\(\sigma_{\phi}^{2}\) is the variance of phase \(\phi\). When the wavefront is expanded by the infinite Zernike terms, and the first \(J\) terms are corrected, \(\sigma_{\phi}^{2}\) is equal to \(\Delta_{J}\), shown as Equation (6)  
After massive algebraic calculus, Equations (A2) and (A3) is expressed as Equations (A5) and (A6). \[ \begin{cases}\overline{\chi_{r}}=S\exp\bigl{(}\overline{\Lambda_{k}}\bigr{)} \overline{\exp\bigl{(}\Lambda_{k}-\overline{\Lambda_{k}}\bigr{)}}\exp\biggl{(} -\frac{\sigma_{\phi}^{2}}{2}\biggr{)}\\ \overline{\chi_{i}}=0\end{cases} \] (A5)  
\[ \begin{cases}\frac{\sigma_{r}^{2}}{S}=\frac{\exp\bigl{(}2\overline{\Lambda_{k }}\bigr{)}\overline{\exp\bigl{(}\Lambda_{k}-\overline{\Lambda_{k}}\bigr{)}} }{2N}\Bigl{[}1+\exp\Bigl{(}-2\sigma_{\phi}^{2}\Bigr{)}\Bigr{]}-\frac{\exp \bigl{(}2\overline{\Lambda_{k}}\bigr{)}\overline{\bigl{[}\exp\bigl{(}\Lambda_{k }-\overline{\Lambda_{k}}\bigr{)}\bigr{]}^{2}}}{N}\exp\Bigl{(}-\sigma_{\phi}^{2} \Bigr{)}\\ \frac{\sigma_{r}^{2}}{S}=\frac{\exp\bigl{(}2\overline{\Lambda_{k}}\bigr{)}\exp \bigl{(}2\Lambda_{k}-\overline{\Lambda_{k}}\bigr{)}}{2N}\Bigl{[}1-\exp\Bigl{(} -2\sigma_{\phi}^{2}\Bigr{)}\Bigr{]}\end{cases} \] (A6)  
Considering the atmospheric channel, the area that the signal light covers at the optical pupil of the receiving telescope is much larger than the aperture, so we assume that the speckles spread equally at this small area and follow a uniform distribution. Thus, \(r_{k}\) follows a 2-dimentional uniform distribution, as shown in Equation (A7). \[p_{R_{k}}(r_{k})=\left\{ \begin{array}{ll}\frac{1}{\pi R^{2}},&|r_{k}|\leq R \\ 0,&|r_{k}|\geq R\end{array} \right.\] (A7)As a result, the expressions of \(\overline{\Lambda_{k}}\), \(\overline{\exp\left(\Lambda_{k}-\overline{\Lambda_{k}}\right)}\) and \(\overline{\exp 2\left(\Lambda_{k}-\overline{\Lambda_{k}}\right)}\) are given by Equations (A8)-(A10), respectively. \[\overline{\Lambda_{k}}=2\pi\int_{0}^{R}\left(-\frac{r_{k}^{2}}{\omega_{a}^{2} }\right)\cdot\frac{1}{\pi R^{2}}\cdot r_{k}\mathrm{d}r_{k}=-\frac{R^{2}}{2 \omega_{a}^{2}}\] (A8)  
\[ \begin{array}{cc}\overline{\exp\left(\Lambda_{k}-\overline{\Lambda_{k}} \right)}&=\frac{\exp\left(\Lambda_{k}\right)}{\exp\left(\overline{\Lambda_{k} }\right)}=\frac{2\pi}{\exp\left(\Lambda_{k}\right)}\int_{0}^{R}\exp\left(- \frac{r_{k}^{2}}{\omega_{a}^{2}}\right)\cdot\frac{1}{\pi R^{2}}\cdot r_{k}dr_ {k}\\ &\qquad=\frac{\omega_{a}^{2}}{R^{2}}\left[1-\exp\left(\frac{R^{2}}{\omega_{a}^ {2}}\right)\right]/\exp\left(-\frac{R^{2}}{2\omega_{a}^{2}}\right)\\ \overline{\exp 2\left(\Lambda_{k}-\overline{\Lambda_{k}}\right)}&=\frac{\exp \left(2\Lambda_{k}\right)}{\exp\left(2\Lambda_{k}\right)}=\frac{2\pi}{\exp \left(2\Lambda_{k}\right)}\int_{0}^{R}\exp\left(-\frac{2r_{k}^{2}}{\omega_{a}^ {2}}\right)\cdot\frac{1}{\pi R^{2}}\cdot r_{k}\mathrm{d}r_{k}\\ &\qquad=\frac{\omega_{a}^{2}}{2R^{2}}\left[1-\exp\left(\frac{2R^{2}}{\omega_{a} ^{2}}\right)\right]/\exp\left(-\frac{R^{2}}{\omega_{a}^{2}}\right)\end{array} \] (A9)  
Thus, the value of \(\overline{\chi_{r}}\), \(\overline{\chi_{i}}\), \(\sigma_{r}^{2}\) and \(\sigma_{i}^{2}\) are shown by Equations (A11) and (A12). \[