## 3 BER Calculation in Frequency Selective Fading MIMO Orientations
Considering two RVs \(X\) and \(Y\), then the cumulative distribution function (cdf) of the ratio \(Z=Y/X\) can be derived using the formula below [19] (\(SINR_{k,d}\geq 0\)). \[F_{Z}(z)=P(Y\leq zX,X>0)=\int\limits_{0}^{\infty}\left|\int\limits_{\infty}^{ \infty}f_{Y}(y)dy\right|f_{X}(x)dx \tag{14}\]  
The differentiation of \(F_{Z}(z)\) leads to the pdf of \(Z\). \[f_{Z}(z)=\frac{d}{dz}F_{Z}(z)=\int_{0}^{\infty}[xf_{Y}(xz)]f_{X}(x)dx \tag{15}\]  
Throughout the rest of this manuscript, all variables will be indexed by the corresponding transmission dimension (i.e., \(d\)). Therefore, if \(Y_{d}\) denotes the square of \(\gamma_{k,d}\), which is normally distributed with mean/standard deviation equal to \(\mu_{Y_{d}}/\sigma_{Y_{d}}\), respectively, and \(X_{d}\) is the denominator of Equation (13), then the following is true. \[ \begin{split} f_{Z_{d}}(z)=\int\limits_{0}^{\infty}\frac{x}{2 \sqrt{xz}}\frac{1}{\alpha_{Y_{d}}\sqrt{2}}e^{-\frac{(\sqrt{xz}-\mu_{Y_{d}})^{2 }}{2\sigma_{Y_{d}}^{2}}}\frac{x^{\alpha_{d}-1}e^{-\frac{\kappa}{\theta_{d}}}} {\Gamma(\alpha_{d})\theta_{d}^{\alpha_{d}}dx}=\\ \frac{e^{-\frac{\mu_{Y_{d}}}{2\sigma_{Y_{d}}^{2}}}}{\alpha_{Y_{d} }\Gamma(\alpha_{d})\theta_{d}^{\alpha_{d}}\sqrt{2 nz}}\int\limits_{0}^{\infty} \frac{x}{2\sqrt{xz}}e^{-\frac{x}{2\sigma_{Y_{d}}^{2}}}e^{\frac{\sqrt{xz}\mu_{ Y_{d}}}{\sigma_{Y_{d}}^{2}}}x^{\alpha_{d}-1}e^{-\frac{x}{\theta_{d}}}dx \end{split}  \tag{16}\]  
where \(\Gamma(\text{x})\) is the Gamma function and \(\alpha_{d}/\theta_{d}\) denote the shape/scale parameter of the Gamma distribution. If \(u=\sqrt{x}\) then \(du=\frac{1}{2\sqrt{x}}dx\). Hence, the equation below is found. \[ \begin{split} f_{Z_{d}}(z)=\frac{e^{-\frac{\mu_{Y_{d}}}{2\sigma_ {Y_{d}}^{2}}}}{\Gamma(\alpha_{d})\alpha_{Y_{d}}\theta_{d}^{\alpha_{d}}\sqrt{2 nz }}\int\limits_{0}^{\infty}e^{-u^{2}(\frac{z}{2\omega_{Y_{d}}^{2}}+\frac{1}{ \theta_{d}})}e^{\frac{u\sqrt{xz}\mu_{Y_{d}}}{\sigma_{Y_{d}}^{2}}}\frac{u^{2a_{ d}}du}{\sigma_{Y_{d}}^{2}}u^{2a_{d}}du\end{split}  \tag{17}\]  
For large number of users and minimum correlation, it can be assumed that \(a_{d}\approx[a_{d}]\), where \([\text{x}]\) is the integer part of x. Starting from the known integral, Equation (18) is found, which is shown below. \[\int\limits_{0}^{\infty}e^{-gx^{2}+px}dx=e^{\frac{p^{2}}{4g}}\sqrt{\frac{\pi} {4}}\left\{g^{-1/2}+\text{erf}\left(\frac{p}{2\sqrt{g}}\right)\right\} \tag{18}\]  
and considering the Taylor series expansion for erf(x), then Equation (19) is uncovered. \[\int\limits_{0}^{\infty}e^{-gx^{2}+px}dx=e^{\frac{p^{2}}{4g}}\Bigg{\{}\sqrt{\frac{ \pi}{4}}g^{-1/2}+\sum\limits_{n=0}^{\infty}\frac{{{{(-1)}^{n}}p^{2n+1}}}{n!(2n+1 )2^{2n+1}g^{n+1}}\Bigg{\}} \tag{19}\]  
Using the \(a_{d}\)th derivative of Equation (19) with respect to \(g\), \(f_{Z_{d}}(z)\) can be written as:  
\[f_{Z_{d}}(z)=\frac{\frac{{{z_{d}^{2}}_{Y_{d}}}}{{{(-1)}^{a_{d}}}{{e^{(2\sigma_{ Y_{d}^{2}++\frac{{{v_{d}}}}{{d_{d}}})}}}}{{{(2\sigma_{Y_{d}}^{2}+\frac{{{v_{d}}}}{{d_{ d}}})}}}{{{(2\sigma_{Y_{d}}^{2}+\frac{{{v_{d}}}}{{d_{d}}})}}}}{\sum\limits_{n=0}^{ \infty}\frac{{{{(-1)}^{n}}\left({\frac{{{v_{d}}}{{\varepsilon_{nY_{d}}}}}}{{2 \sigma_{Y_{d}}^{2}}}}\right)^{2n+1}\prod\limits_{j=1}^{a_{d}}(-n-j)}{\sqrt{n!(2n+1)\left({\frac{{{z_{d}}}}{{2\sigma_{Y_{d}}^{2}}}}+\frac{1}{{d_{d}}} \right)^{n+1}}}} \tag{20}\]  
For the derivation of the Gamma pdf, the Maximum Likelihood Parameter Estimation (MLPE) method in Reference [20] is taken into account. Lastly, for the BPSK modulation equivalent, BER can be expressed as Reference [21] (\(Q\) function is considered in the integral). \[BER=(1/dim)\sum\limits_{d=1}^{dim}\int\limits_{0}^{\infty}Q\Big{(}\sqrt{2z} \Big{)}f_{Z_{d}}(z)dz \tag{21}\]  
Note that for the derivation of Equations (17) and (20), it has been assumed that \(X_{d}\) and \(Y_{d}\) are uncorrelated. In the opposite case, the following transformation can be applied. \[