# Impacts of Diversity Technique Application in Order to Reduce the Fading in IEEE 802.15.4 Networks
## 3 BER in the Case of an SC Combiner
### BER in the Case of SC Combiners for Nakagami-\(m\) Fading Channels  
By substituting (6) into (8) and solving the integral by using [25] Eq. (3.351.1) the CDF for Nakagami-\(m\) fading channels is obtained:  
\[F_{Naka}(\gamma)=\frac{(m-1)!}{\Gamma(m)}\Bigg{(}1-\sum\limits_{k=0}^{m-1} \frac{1}{k!}\bigg{(}\frac{m\gamma}{\gamma_{0}}\bigg{)}^{k}e^{-\frac{m\gamma}{ \gamma_{0}}}\bigg{)}. \tag{18}\]  
Substituting (6) and (18) into (7), the PDF is obtained at the output of the SC combiner for the case of the Nakagami-\(m\) fading channel:  
\[f_{SC,Naka}(\gamma)=\frac{2m^{m}(m-1)!}{\Gamma^{2}(m)\gamma_{0}^{m}}\Bigg{(} \gamma^{m-1}e^{-\frac{m\gamma}{\gamma_{0}}}-\sum\limits_{k=0}^{m-1}\frac{1}{k! }\bigg{(}\frac{m}{\gamma_{0}}\bigg{)}^{k}\gamma^{k+m-1}e^{-\frac{2m\gamma}{ \gamma_{0}}}\Bigg{)}. \tag{19}\]  
The BER of the WSN transceiver based on ZigBee technology operating in an AWGN environment and with a signal under the impact of Nakagami-\(m\) fading, previously passed through an SC combiner with two branches, is obtained when (19) and (2) are included in (4):\[BER_{Zigber,SC,Naka}=\int\limits_{0}^{\infty}BER_{AWGN}(\gamma)f_{SC,Naka}(\gamma)d \gamma \tag{20}\] \[=\frac{2m^{m}(m-1)!}{\Gamma^{2}(m)\gamma_{0}^{m}}\left[\int\limits_ {0}^{\infty}\int\limits_{0}^{\infty}\frac{1}{2}\gamma^{m-1}e^{-\frac{m\gamma}{ \gamma_{0}}}erfc(\sqrt{T\gamma})d\gamma-\int\limits_{0}^{\infty}\frac{1}{8} \gamma^{m-1}e^{-\frac{m\gamma}{\gamma_{0}}}erfc^{2}(\sqrt{T\gamma})d\gamma\right.\] \[\left.-\sum\limits_{k=0}^{m-1}\frac{1}{2k}\left(\frac{m}{\gamma_ {0}}\right)\int\limits_{0}^{\infty}\gamma^{k+m-1}e^{-\frac{2m\gamma}{\gamma_{0 }}}erfc(\sqrt{T\gamma})d\gamma-\sum\limits_{k=0}^{m-1}\frac{1}{8k^{2}}\left( \frac{m}{\gamma_{0}}\right)\int\limits_{0}^{\infty}\gamma^{k+m-1}e^{-\frac{2m \gamma}{\gamma_{0}}}erfc^{2}(\sqrt{T\gamma})d\gamma\right]\] \[=\frac{2m^{m}(m-1)!}{\Gamma^{2}(m)\gamma_{0}^{m}}[I_{20.1}-I_{20. 2}-I_{20.3}-I_{20.4}].\]  
By using the transformation for the erfc(\(\cdot\)) function [22] Eq. (06.27.26.0006.01) and the transformation for the exponential function [23] Eq. (8.4.3), as well as by using [24] Eq. (07.34.21.0011.01), integral \(I_{20.2}\) obtains the following solution:  
\[I_{20.1}=\frac{1}{2\sqrt{\pi}T^{m}}\sum\limits_{n=0}^{\infty}\frac{1}{n!} \left(-\frac{m}{T\gamma_{0}}\right)^{n}G_{3,3}^{2,2}\left(1\left| \begin{array} []{cc}1-n-m,\frac{1}{2}-n-m,1\\ 0,\ \frac{1}{2},\ -n-m\end{array} \right.\right). \tag{21}\]  
By using the transformation for the erfc(\(\cdot\)) function [22] Eq. (06.27.26.0006.01) and the transformation for the exponential function [23] Eq. (8.4.3), as well as by using [24] Eq. (07.34.21.0011.01), integral \(I_{20.3}\) obtains the following solution:  
\[I_{20.3}=\frac{1}{2\sqrt{\pi}T^{m}}\sum\limits_{k=0}^{m-1}\frac{1}{k! }\left( \frac{m}{T\gamma_{0}}\right)^{k}G_{2,2}^{1,2}\left(\frac{2m}{T\gamma_{0}} \right| \begin{array}{cc}1-k-m,&\frac{1}{2}-k-m\\ 0,&-k-m\end{array} \right). \tag{22}\]  
By using the transformation for the erfc(\(\cdot\)) function [22] Eq. (06.27.26.0006.01) and the transformation for the exponential function [23] Eq. (8.4.3), as well as by using [24] Eq. (07.34.21.0011.01), integral \(I_{20.3}\) obtains the following solution:  
\[I_{20.3}=\frac{1}{2\sqrt{\pi}T^{m}}\sum\limits_{k=0}^{m-1}\frac{1}{k! }\left( \frac{m}{T\gamma_{0}}\right)^{k}G_{2,2}^{1,2}\left(\frac{2m}{T\gamma_{0}} \right| \begin{array}{cc}1-k-m,&\frac{1}{2}-k-m\\ 0,&-k-m\end{array} \right). \tag{23}\]  
By using the transformation for the erfc(\(\cdot\)) function [22] Eq. (06.27.26.0006.01) and the transformation for the exponential function [25] Eq. (1.211.1), as well as by using [24] Eq. (07.34.21.0011.01), integral \(I_{20.4}\) obtains the following solution:  
\[I_{20.4}=\frac{1}{8\pi T^{m}}\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{m-1} \frac{(-2)^{n}}{n!k! }\left(\frac{m}{T\gamma_{0}}\right)^{n+k}G_{3,3}^{2,2} \left(1\left| \begin{array}{cc}1-n-k-m,\frac{1}{2}-n-k-m,1\\ 0,&\frac{1}{2},\ -n-k-m\end{array} \right.\right).