## 4 Methods for Mitigating Jamming
### ABER Calculation  
#### 4.3.1 Analysis of the ABER for SIMO in Scenario 1  
The PDF of the received signal can be derived from Equations (8) and (30) as follows  
\[f_{y_{D}}(y_{D})=\left\{ \begin{array}{l}\frac{1}{\sqrt{p_{I}/ \rho}}\exp\bigg{(}-\frac{\sigma_{AAA}^{2}}{2\sigma_{AAA}^{2}}\bigg{)}\delta \bigg{(}\frac{y_{D}}{\sqrt{p_{I}/\rho}}\bigg{)}\\ +\frac{1}{\sqrt{p_{I}/\rho}}\bigg{[}1-\exp\bigg{(}-\frac{\sigma_{AAA}^{2}}{2 \sigma_{AAA}^{2}}\bigg{)}\bigg{]}\frac{a_{I}\beta_{I}\xi_{I}^{2}}{A_{0}\Gamma (a_{I})\Gamma(\beta_{I})}\\ \times G_{1,3}^{3,0}\bigg{(}\frac{a_{I}\beta_{I}y_{D}}{A_{0}\sqrt{p_{I}/\rho}} \bigg{|}\begin{array}{l}\xi_{I}^{2}\\ \zeta_{J}^{2}-1,\alpha_{J}-1,\beta_{J}-1\end{array}\bigg{)},\quad\text{for $s=0$}\\ \\ \frac{1}{\sqrt{p_{I}/\rho}}\exp\bigg{(}-\frac{\sigma_{AAA}^{2}}{2\sigma_{AAA}^{2 }}\bigg{)}\delta\bigg{(}\frac{y_{D}-\sqrt{P_{s}}s\sum\limits_{n=1}^{N}a_{in}h _{in}}{\sqrt{p_{I}/\rho}}\bigg{)}\\ +\frac{1}{\sqrt{p_{I}/\rho}}\bigg{[}1-\exp\bigg{(}-\frac{\sigma_{AAA}^{2}}{2 \sigma_{AA}^{2}}\bigg{)}\bigg{]}\frac{a_{I}\beta_{I}\xi_{I}^{2}}{A_{0}\Gamma (\alpha_{J})\Gamma(\beta_{I})}\\ \times G_{1,3}^{3,0}\bigg{(}\frac{a_{I}\beta_{I}\bigg{(}y_{D}-\sqrt{P_{s}}s \sum\limits_{n=1}^{N}a_{in}h_{in}\bigg{)}}{A_{0}\sqrt{p_{I}/\rho}}\bigg{|} \begin{array}{l}\xi_{I}^{2}\\ \zeta_{J}^{2}-1,\alpha_{J}-1,\beta_{J}-1\end{array}\bigg{)},\ \ \text{for $s=1$}\\ \end{array} \right. \tag{36}\]  
Let \(h_{sum}=\sum\limits_{n=1}^{N}a_{n}h_{in}\), then by substituting (18) and (36), we obtain  
\[ \begin{array}{l}P_{e1a}(h_{i})=\frac{\rho}{2}\bigg{[}1-\exp\bigg{(}-\frac{ \sigma_{AAA}^{2}}{2\sigma_{AAA}^{2}}\bigg{)}\frac{a_{I}\beta_{I}\xi_{I}^{2}}{A _{0}\Gamma(\alpha_{I})\Gamma(\beta_{I})}\\ \times G_{2,4}^{4,0}\bigg{(}\frac{a_{I}\beta_{I}\sqrt{\rho_{I}}h_{sum}}{A_{0} \int}\bigg{|}\begin{array}{l}\xi_{I}^{2},0\\ -1,\xi_{I}^{2}-1,\alpha_{J}-1,\beta_{J}-1\end{array}\bigg{)}\end{array}  \tag{37}\]  
Then, we apply the Mellin transform to (43) using 07.34.22.0004.01 [29], which results in  
\[\begin{array}{l}\varphi_{h_{in}}(z)=\frac{a_{I}\beta_{I}\xi_{I}^{2}}{A_{0} \Gamma(a_{I})\Gamma(\beta_{I})}\bigg{[}1-\exp\bigg{(}-\frac{\sigma_{AAA}^{2}}{2 \sigma_{AAA}^{2}}\bigg{)}\bigg{]}\frac{a_{I}\beta_{I}\xi_{I}^{2}}{A_{0}\Gamma (\alpha_{I})\Gamma(\beta_{I})}\bigg{(}\frac{a_{I}\beta_{I}\alpha_{I}\beta_{I}}{ A_{0}}\bigg{)}^{-z}\\ \times\frac{\Gamma(\xi_{I}^{2}-1+z)\Gamma(a_{I}-1+z)\Gamma(\beta_{I}-1+z) \Gamma(\xi_{I}^{2}-1+z)\Gamma(a_{I}-1+z)\Gamma(\beta_{I}-1+z)}{\Gamma(\xi_{I }^{2}+z)\Gamma(\xi_{I}^{2}+z)} \tag{38}\]  
Considering the independent channels, the Mellin transform of \(h_{sum}\) can be expressed as  
\[\varphi_{h_{sum}}(z)=\prod\limits_{n=1}^{N}\varphi_{h_{in}}(z)=\big{(}\varphi_ {h_{in}}(z)\big{)}^{N} \tag{39}\]By substituting (38) into (39) and performing the Mellin inverse transform, the PDF of \(h_{sum}\) can be obtained as follows  
\[ \begin{split} f_{h_{sum}}(x)=\left(\frac{\alpha_{p}\beta_{p}\xi_{p}^{ 2}}{\Lambda_{0p}\Gamma(x_{p})\Gamma(\beta_{p})}\frac{\alpha_{q}\beta_{q}\xi_{q} ^{2}}{\Lambda_{0p}\Gamma(x_{q})\Gamma(\beta_{q})}\right)^{N}\\ \times\xi_{2N,\delta N}^{6N,0}\left(\left(\frac{\alpha_{p}\beta_{ p}\alpha_{q}\beta_{q}}{\Lambda_{0p}\Lambda_{0q}}\right)^{N}\times\left|\begin{array}{c} \mathbf{S_{1}}\\ \mathbf{S_{2}},\mathbf{S_{3}},\mathbf{S_{4}}\end{array}\right.\right)\end{split}  \tag{40}\]  
\[\text{where}\quad\mathbf{S_{1}}=\underbrace{\sum_{p^{\prime}}^{N}\cdots,\xi_{p^ {\prime}}^{2}}_{N}\underbrace{\sum_{q^{\prime}}^{N}\cdots,\xi_{q^{\prime}}^{2} }_{N},\quad\mathbf{S_{2}}=\underbrace{\xi_{p}^{2}-1,\cdots,\xi_{p}^{2}-1}_{N},\underbrace{\alpha_{p}-1,\cdots,\alpha_{p}-1}_{N},\]  
\[\mathbf{S_{3}}=\underbrace{\beta_{p}-1,\cdots,\beta_{p}-1}_{N},\underbrace{ \xi_{q}^{2}-1,\cdots,\xi_{q}^{2}-1}_{N},\text{and}\ \mathbf{S_{4}}=\underbrace{\alpha_{q}-1,\cdots,\alpha_{q}-1}_{N}, \underbrace{\beta_{q}-1,\cdots,\beta_{q}-1}_{N}.\]  
Now, using (37), (40), and 07.34.21.0011.01 [29], after some rigorous mathematics, the final closed-form expression of the ABER during jamming active is given by Appendix A (A1).