## 3 Analysis of BER for SISO System
### Scenario 1
#### 3.1.1 BER during Jamming Active  
Let us define a random variable, \(T\triangleq\sqrt{N_{J}}h_{J}=\sqrt{P_{j}/\rho}h_{J}\). Using (8), the PDF of T is given as \[ \begin{split} f_{T}(T)&=\frac{1}{\sqrt{p_{f}/\rho}}\exp \left(-\frac{\sigma_{20 aA}^{2}}{2\sigma_{aA}^{2}}\right)\delta\left(\frac{T}{ \sqrt{p_{f}/\rho}}\right)+\frac{1}{\sqrt{p_{f}/\rho}}\bigg{[}1-\exp\left(- \frac{\sigma_{20 aA}^{2}}{2\sigma_{aAaA}^{2}}\right)\bigg{]}\\ &\times\frac{\alpha_{j}\beta_{j}\zeta_{j}^{2}}{A_{0_{f}}\Gamma( \alpha_{j})\Gamma(\beta_{j})}G_{1,3}^{3,0}\left(\frac{\alpha_{j}\beta_{j}T}{A_ {0_{f}}\sqrt{p_{f}/\rho}}\bigg{|}\begin{array}{c}\xi_{j}^{2}\\ \zeta_{j}^{2}-1,\alpha_{j}-1,\beta_{j}-1\end{array}\right)\end{split}  \tag{14}\]  
When jamming is present, the energy of the jamming significantly outweighs that of Gaussian white noise, which allows us to disregard the Gaussian white noise. Therefore, we can derive that \(T=y_{1}-\sqrt{P_{g}}h_{i}\)s. By using (14), we can further deduce that  
\[ \begin{split} f_{y_{1}}(y_{1})&=\frac{1}{\sqrt{p_{f}/ \rho}}\exp\left(-\frac{\sigma_{20 aA}^{2}}{2\sigma_{aAaA}^{2}}\right)\delta \left(\frac{y_{1}-\sqrt{P_{g}}h_{i}}{\sqrt{p_{f}/\rho}}\right)+\frac{1}{\sqrt{ p_{f}/\rho}}\bigg{[}1-\exp\left(-\frac{\sigma_{20 aA}^{2}}{2\sigma_{aAaA}^{2}} \right)\bigg{]}\\ &\times\frac{\alpha_{j}\beta_{j}\zeta_{j}^{2}}{A_{0_{f}}\Gamma( \alpha_{j})\Gamma(\beta_{j})}G_{1,3}^{3,0}\left(\frac{\alpha_{j}\beta_{j}(y_{ 1}-\sqrt{P_{g}}h_{i})}{A_{0_{f}}\sqrt{p_{f}/\rho}}\bigg{|}\begin{array}{c} \xi_{j}^{2}\\ \zeta_{j}^{2}-1,\alpha_{j}-1,\beta_{j}-1\end{array}\right)\end{split}  \tag{15}\]  
(15) can be expanded based on the value of the symbol \(s\). \[ \begin{split} f_{y_{1}}(y_{1})=\left\{\begin{array}{ll}\frac{1} {\sqrt{p_{f}/\rho}}\exp\left(-\frac{\sigma_{20 aA}^{2}}{2\sigma_{aAaA}^{2}} \right)\delta\left(\frac{y_{1}}{\sqrt{p_{f}/\rho}}\right)+\frac{1}{\sqrt{p_{f }/\rho}}\bigg{[}1-\exp\left(-\frac{\sigma_{20 aA}^{2}}{2\sigma_{aAaA}^{2}} \right)\bigg{]}\\ \times\frac{\alpha_{j}\beta_{j}\zeta_{j}^{2}}{A_{0_{f}}\Gamma( \alpha_{j})\Gamma(\beta_{j})}G_{1,3}^{3,0}\left(\frac{\alpha_{j}\beta_{j}(y_{ 1}-\sqrt{P_{g}}h_{i})}{A_{0_{f}}\sqrt{p_{f}/\rho}}\bigg{|}\begin{array}{c} \xi_{j}^{2}\\ \zeta_{j}^{2}-1,\alpha_{j}-1,\beta_{j}-1\end{array}\right),&\text{for $s=0$}\\ \frac{1}{\sqrt{p_{f}/\rho}}\exp\left(-\frac{\sigma_{20 aA}^{2}}{2\sigma_{aAaA}^ {2}}\right)\delta\left(\frac{y_{1}-\sqrt{P_{g}}h_{i}}{\sqrt{p_{f}/\rho}} \right)+\frac{1}{\sqrt{p_{f}/\rho}}\bigg{[}1-\exp\left(-\frac{\sigma_{20 aA}^{2 }}{2\sigma_{aAaA}^{2}}\right)\bigg{]}\\ \times\frac{\alpha_{j}\beta_{j}\zeta_{j}^{2}}{A_{0_{f}}\Gamma( \alpha_{j})\Gamma(\beta_{j})}G_{1,3}^{3,0}\left(\frac{\alpha_{j}\beta_{j}(y_{ 1}-\sqrt{P_{g}}h_{i})}{A_{0_{f}}\sqrt{p_{f}/\rho}}\bigg{|}\begin{array}{c} \xi_{j}^{2}\\ \zeta_{j}^{2}-1,\alpha_{j}-1,\beta_{j}-1\end{array}\right),&\text{for $s=1$}\end{array}\right.\end{split}  \tag{16}\]  
ABER can be expressed based on the occurrence of jamming states. \[ \begin{split} P_{e1}=P(\Lambda=1)[P(x=1)P(v|x=1)+P(x=0)P(v|x=0) ]+P(\Lambda=0)P_{ei}\end{split}  \tag{17}\]  
where \(P_{ei}\) represents BER during jamming idle states and \(v\) denotes the occurrence of a bit error event. Here, we assume \(P(x=1)=P(x=0)=0.5\), indicating that the symbols '\(1\)' and '\(0\)' are transmitted with equal probabilities. When jamming is active and the jamming noise significantly exceeds the additive white Gaussian noise, the additive white Gaussian noise can be disregarded when calculating BER. As a result, the expression for calculating the BER under jamming active, derived from (17), is as follows  
\[ \begin{split} P_{e1a}=\rho\bigg{[}0.5\int_{th}^{\infty}f_{y}(y|s= 0)dy+0.5\int_{0}^{th}f_{y}(y|s=1)dy\bigg{]}\end{split}  \tag{18}\]  
According to [25], \(th=\sqrt{P_{g}}h_{i}\). Substituting (19) into (21) and integrating using 07.34.21.0084.01 and 07.34.21.0085.01 is as  
\[