# Energy Harvesting over Rician Fading Channel:
## 3 System Performance  
### Outage Probability  
Since the random variable (RV) \(X_{i}\) is Rician distributed for \(i=1,2\), the probability density function (PDF) of \(X_{i}\) can be found as  
\[f_{X_{i}}(x)=\frac{(K+1)e^{-K}}{\lambda_{i}}e^{-\frac{(K+1)x}{\lambda_{i}}}I_{ 0}\left(2\sqrt{\frac{K(K+1)x}{\lambda_{i}}}\right), \tag{10}\]  
where \(\lambda_{i}\) is the mean value of the RV \(X_{i},i=1,2\), \(K\) is the Rician K-factor defined as the ratio of the power of the line-of-sight (LOS) component to the scattered components and \(I_{0}\left(\cdot\right)\) is the zero-th order modified Bessel function of the first kind.  
Using the equality \(I_{0}(x)=\sum\limits_{l=0}^{\infty}\frac{x^{2l}}{2^{2l}(l)^{2}}\)[32], Equation (10) can be rewritten as  
\[f_{X_{i}}(x)=\zeta_{i}e^{-K}\sum\limits_{l=0}^{\infty}\frac{\left(\zeta_{i}K \right)^{l}}{\left(l!\right)^{2}}x^{l}e^{-\zeta_{i}x}, \tag{11}\]  
where \(\zeta_{i}=\frac{K+1}{\lambda_{i}}\).  
The cumulative density function (CDF) of the RV \(X_{i}\) (\(i=1,2\)) can be derived as in [33]:  
\[F_{X_{i}}(x)=\int\limits_{0}^{x}f_{X_{i}}(t)dt=1-e^{-K}\sum\limits_{l=0}^{ \infty}\sum\limits_{m=0}^{l}\frac{K^{l}\zeta_{i}^{m}}{l!m!}x^{m}e^{-\zeta_{i}x}. \tag{12}\]  
The outage probability of this system is defined as the probability that the end-to-end SNR falls below a desired threshold--let us say, \(\gamma_{th}\). In this case, it can be derived as  
\[P_{out\_1}=F_{\gamma_{th}}(\gamma_{th})=\Pr\left(\gamma_{1}<\gamma_{th}\right) =\Pr\left(\frac{X_{1}X_{2}}{X_{1}X_{2}a+X_{1}^{2}a+X_{1}b+c}<\gamma_{th}\right), \tag{13}\]  
where \(\gamma_{th}=2^{2R}-1\), \(R\) is the source transmission rate:  
\[\begin{split}\Pr\left(\frac{X_{1}X_{2}}{X_{1}X_{2}a+X_{1}^{2}a+X_ {1}b+c}<\gamma_{th}\right)=\Pr\left\{X_{2}(X_{1}-\gamma_{th}X_{1}a)<\gamma_{th }X_{1}^{2}a+\gamma_{th}X_{1}b+\gamma_{th}c\right\},\\ =\begin{cases}\Pr\left\{X_{2}<\frac{\gamma_{th}X_{1}^{2}a+\gamma_ {th}X_{1}b+\gamma_{th}c}{X_{1}-\gamma_{th}X_{1}a}\right\},&\text{if }1-\gamma_{th}a \geq 0,\\ 1,&\text{if }1-\gamma_{th}a<0.\end{cases}\end{split} \tag{14}\]  
In this subsection, we assume that \(1-\gamma_{th}a\) is positive, which is reasonable in practice. From Equation (14), we can derive the exact-form expression for the outage probability of the system. This is stated by the following theorem.  
**Theorem 1** (Outage probability--Exact form).: _The exact outage probability for node \(S_{1}\) of the proposed half-duplex bidirectional wireless sensor network using time-switching energy harvesting strategy over a Rician fading channel can be expressed as_  
\[\begin{split} P_{out\_1}=1-2\zeta_{1}e^{-2K}e^{-b_{1}\zeta_{2} \hat{\tau}_{th}}&\sum\limits_{l=0}^{\infty}\sum\limits_{k=0}^{ \infty}\sum\limits_{m=0}^{l}\sum\limits_{n=0}^{m}\sum\limits_{p=0}^{n}\frac{K ^{l+k}\xi_{1}^{k}(\zeta_{2}\hat{\tau}_{th})^{\frac{3m-2n+p+k+1}{2}}a^{m-n}b^{p} c^{\frac{m-p+k+1}{2}}}{l!p!(n-p)!(m-n)!(k!)^{2}(\zeta_{1}+a\zeta_{2}\hat{\tau}_{ th})^{\frac{m-2n+p+k+1}{2}}}\\ &\times K_{m-2n+p+k+1}\left(2\sqrt{(\zeta_{1}+a\zeta_{2}\hat{\tau }_{th})\zeta_{2}\hat{\tau}_{th}c}\right),\end{split} \tag{15}\]  
_where \(a=\kappa^{2}+\kappa_{r}^{2}(1+\kappa^{2}),b=\frac{(1+\kappa_{r}^{2})}{P/\sigma _{r}^{2}},c=\frac{(1+\kappa^{2})}{\Psi^{2}/\sigma_{r}^{2}},\zeta_{i}=\frac{K+ 1}{\lambda_{i}}\), and \(\hat{\tau}_{th}=\frac{\gamma_{th}}{1-a\gamma_{th}}\)._  
Proof of Theorem 1.: See Appendix A.  
**Remark 1**.: _The outage probability at node \(S_{2}\) can also be obtained by exchanging the roles of \(\zeta_{1}\) and \(\zeta_{2}\) in Equation (15)._