# 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 Rayleigh Fading Channels  
Substituting (5) into (8) and solving the integral gives the CDF for the Rayleigh fading channel. By substituting the obtained solution for (8) and (5) in (7), the PDF is obtained at the output of the SC combiner for the case of the Rayleigh fading channel:  
\[f_{SC,Ray}(\gamma)=\frac{2}{\gamma_{0}}e^{-2\frac{\gamma}{\gamma_{0}}}+\frac{2} {\gamma_{0}}e^{-\frac{\gamma}{\gamma_{0}}}. \tag{9}\]  
The BER of the WSN transceiver based on ZigBee technology operating in an AWGN environment and with a signal under impact of Rayleigh fading, previously passed through an SC combiner with two branches, is obtained when (9) and (2) are included in (4):  
\[ \begin{array}{c}BER_{Zigbene,SC,Ray}=\int\limits_{0}^{\infty}BER_{AWGN}( \gamma)f_{SC,Ray}(\gamma)d\gamma\\ =\int\limits_{0}^{\infty}\frac{1}{\gamma_{0}}erfc\big{(}\sqrt{T\gamma}\big{)}e^ {-2\frac{\gamma}{\gamma_{0}}}d\gamma+\int\limits_{1}^{\infty}\frac{1}{\gamma_ {0}}erfc\big{(}\sqrt{T\gamma}\big{)}e^{-\frac{\gamma}{\gamma_{0}}}d\gamma\\ -\int\limits_{0}^{\infty}\frac{1}{\gamma_{0}}erfc^{2}\big{(}\sqrt{T\gamma} \big{)}e^{-2\frac{\gamma}{\gamma_{0}}}d\gamma-\int\limits_{0}^{\infty}\frac{1 }{\gamma_{0}}erfc^{2}\big{(}\sqrt{T\gamma}\big{)}e^{-\frac{\gamma}{\gamma_{0} }}d\gamma\\ =I_{10.1}+I_{10.2}-I_{10.3}-I_{10.4}.\end{array}  \tag{10}\]  
By use of the transformation for the erfc(\(\cdot\)) function [22] Eq. (06.27.26.0006.01) and the transformation for exponential function [23] Eq. (8.4.3), integral \(I_{10.1}\) reduces to the following form:  
\[I_{10.1}=\int\limits_{0}^{\infty}\frac{1}{\gamma_{0}}erfc\Big{(}\sqrt{T\gamma }\Big{)}e^{-2\frac{\gamma}{\gamma_{0}}}d\gamma=\frac{1}{\gamma_{0}}\int \limits_{0}^{\infty}\frac{1}{\sqrt{\pi}}G_{1,2}^{2,0}\bigg{(}T\gamma\bigg{|} \begin{matrix}1\\ 0,\frac{1}{2}\end{matrix} \bigg{)}G_{0,1}^{1,0}\bigg{(}2\frac{\gamma}{\gamma_ {0}}\bigg{|} \begin{matrix}-\\ 0\end{matrix} \bigg{)}d\gamma, \tag{11}\]  
where G(\(\cdot\)) is the Maijer G-function. By using [24] Eq. (07.34.21.0011.01), integral \(I_{10.1}\) obtains the following solution:  
\[I_{10.1}=\frac{1}{\sqrt{\pi}\gamma_{0}T}G_{2,2}^{1,2}\bigg{(}\frac{2}{T\gamma _{0}}\bigg{|} \begin{matrix}0,-\frac{1}{2}\\ 0,-1\end{matrix} \bigg{)}. \tag{12}\]  
Figure 1: SC combiner with two inputs. By using the same transformations as for solving the integral \(I_{10.1}\), the solution of the integral \(I_{10.2}\) is reduced to:  
\[I_{10.2}=\frac{1}{\sqrt{\pi}\tau_{0}T}G_{2,2}^{1,2}\bigg{(}\frac{1}{T\gamma_{0}} \bigg{|} \begin{array}{c}0,-\frac{1}{2}\\ 0,-1\end{array} \bigg{)}. \tag{13}\]  
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), integral \(I_{10.3}\) reduces to the following form:  
\[ \begin{array}{c}I_{10.3}=\int\limits_{0}^{\infty}\frac{1}{4\tau_{0}}erfc^{2}( \sqrt{T\gamma})e^{-2\frac{\gamma}{\gamma_{0}}}d\gamma\\ =\frac{1}{4\pi\gamma_{0}}\sum\limits_{n=0}^{\infty}\frac{1}{n!}\left(\frac{-2 }{\gamma_{0}}\right)^{n}\int\limits_{0}^{\infty}\gamma^{n}G_{1,2}^{2,0} \bigg{(}T\gamma\bigg{|}\begin{array}{c}1\\ 0,\frac{1}{2}\end{array}\bigg{)}G_{1,2}^{2,0}\bigg{(}T\gamma\bigg{|}\begin{array} []{c}1\\ 0,\frac{1}{2}\end{array}\bigg{)}d\gamma.\end{array}  \tag{14}\]  
By using [24] Eq. (07.34.21.0011.01), integral \(I_{10.3}\) obtains the following solution:  
\[I_{10.3}=\frac{1}{4\pi\gamma_{0}T}\sum\limits_{n=0}^{\infty}\frac{1}{n! }\left( \frac{-2}{T\gamma_{0}}\right)^{n}G_{3,3}^{2,2}\bigg{(}1\bigg{|} \begin{array}[] {c}-n,-n-\frac{1}{2},1\\ 0,\frac{1}{2},-n-1\end{array} \bigg{)}. \tag{15}\]  
By using the same transformations as for solving the integral \(I_{10.3}\), the solution of the integral \(I_{10.4}\) is reduced to:  
\[I_{10.4}=\frac{1}{4\pi\gamma_{0}T}\sum\limits_{n=0}^{\infty}\frac{1}{n! }\left( \frac{-1}{T\gamma_{0}}\right)^{n}G_{3,3}^{2,2}\bigg{(}1\bigg{|} \begin{array}[] {c}-n,-n-\frac{1}{2},1\\ 0,\frac{1}{2},-n-1\end{array} \bigg{)}.