Problem analysis
Due to fluctuation of ambient renewable energy, the relay nodes cannot harvest enough energy to maintain working status. So, the outage event will occur.
According to aforementioned descriptions, that is to say, if the direct link channel is good, or the relay link’s energy is more about E_{
r
}, and the relay link channel is good, the system can work well. We can express the overall successful communication events as
$$\begin{array}{*{20}l} &\{\text{direct link}\ (SD)\ \text{is good}\}\\ &\quad\bigcup_{\kappa}\left\{\vphantom{\bigcap}\{\kappa\text{th relay link}\ (SR_{k}D)\ \text{is good}\}\right.\\ &\quad\left. \bigcap\{\text{the energy at}\ R_{k}\ \text{is enough}\}\right\} \end{array} $$
(8)
Correspondingly, if the direct link channel is not good, at the same time, the relay link channel is not good, or the relay link node’s energy is small than E_{
r
}, the system outage will occur. The overall outage events can be expressed as
$$\begin{array}{*{20}l} &\{\text{direct link}\ (SD)\ \text{is bad}\}\\ &\quad\bigcap_{\kappa}\left\{\vphantom{\bigcup}\{\kappa \text{th relay link}\ (SR_{k}D)\ \text{is bad}\}\right.\\ &\quad\left. \bigcup\{\text{the energy at}\ R_{k}\ \text{is not enough}\}\right\} \end{array} $$
(9)
Assume the energyharvesting modules owned by the relay nodes are independent from each other. The overall outage probability in (9) following the transmission protocol can be given by
$$ p_{\text{total}}^{\mathrm{{(out)}}}=P_{sd}^{\mathrm{{(out)}}}\cdot\prod_{\kappa=1}^{K}{\left[p_{k}^{ex}+{\left(1p_{k}^{ex}\right)}\cdot P_{sr_{k}d}^{\mathrm{{(out)}}}\right]} $$
(10)
where \(P_{sd}^{\mathrm {{(out)}}}\) denotes the outage probability of the direct link can not work, \(P_{sr_{k}d}^{\mathrm {{(out)}}}\) denotes the outage probability of the relay link is not good or the energy of the relay node is small than E_{
r
}, and \(p_{k}^{ex}\) is the probability of relayaided link do not have enough energy to active itself to work, which have been discussed on the previous chapter.
In order to calculate ptotal(out), we have a detail analysis about \(P_{sd}^{\mathrm {{(out)}}}\) and \(P_{sr_{k}d}^{\mathrm {{(out)}}}\) in the next step.
Calculation of \(P_{sd}^{\mathrm {{(out)}}}\)
In direct transmission scenario, only the direct link between S and D is selected as a working link where all relay nodes do not work, and all the time slots in a signal block are used by the direct link.
The mutual information for the direction link of S−D achieved by zeromean circularly symmetric complex Gaussian input is defined as:
$$ {}{\begin{aligned}I_{sd}=\log_{2}\left(1+P_{s}\frac{h_{sd}^{2}}{N_{D}}\right)=\log_{2}\left(1+\frac{P_{s}}{N_{D}}h_{0}\right)=\log_{2}(1+\gamma h_{0}) \end{aligned}} $$
(11)
where \(\gamma =\frac {P_{s}}{N_{D}}\) denotes SNR, and for h_{0}=h_{
sd
}^{2}, h_{0} follows the Nakagamim distribution, whose PDF and outage probability is expressed as respectively
$$ p(h_{0})=\frac{{m_{sd}}^{m_{sd}}h_{0}^{m_{sd}1}}{\Omega_{sd}^{m_{sd}}\Gamma(m_{sd})}\exp\left(\frac{m_{sd}}{\Omega_{sd}}h_{0}\right) $$
(12)
Assume that the minimum acceptable rate equals to R_{0}, the outage probability \(P_{sd}^{\mathrm {{(out)}}}[R_{s,d}<R_{0}]\) under direct transmission protocol can be expressed as
$$\begin{array}{@{}rcl@{}} P_{sd}^{\mathrm{{(out)}}}&=&P[\log_{2}(1+\gamma h_{0})\leq R_{0}]=P[h_{0}\leq \gamma_{th}^{sd}]\\ &=&\int_{0}^{\gamma_{th}^{sd}}\frac{{m_{sd}}^{m_{sd}}h_{0}^{m_{sd}1}}{\Omega_{sd}^{m_{sd}}\Gamma(m_{sd})} \exp\left(\frac{m_{sd}}{\Omega_{sd}}h_{0}\right){dh}_{0}\\ &=&\frac{m_{sd}}{\Omega_{sd}\Gamma(m_{sd})}\int_{0}^{\gamma_{th}^{sd}} \left(\frac{m_{sd}}{\Omega_{sd}}h_{0}\right)^{m_{sd}1}\exp\left(\frac{m_{sd}}{\Omega_{sd}}h_{0}\right){dh}_{0}\\ &=&\frac{1}{\Gamma(m_{sd})}\int_{0}^{\frac{m_{sd}}{\Omega_{sd}}\gamma_{th}^{sd}}t^{m_{sd}1}e^{t}dt,t=\frac{m_{sd}}{\Omega_{sd}}h_{0}\\ &=&\frac{1}{\Gamma(m_{sd})}\Gamma\left(m_{sd},frac{m_{sd}}{\Omega_{sd}}\gamma_{th}^{sd}\right)\\ &=&1\exp{\left(\frac{m_{sd}}{\Omega_{sd}}\gamma_{th}^{sd}\right)}\sum_{n=0}^{m_{sd}1} \left(\frac{m_{sd}\gamma_{th}^{sd}}{\Omega_{sd}}\right)^{n}\frac{1}{n!} \end{array} $$
(13)
where \(\gamma _{th}^{sd}=\frac {2^{R_{0}}1}{\gamma }\).
Calculation of \(P_{sr_{k}d}^{\mathrm {{(out)}}}\)
For cooperative relay system [24, 25], if someone relay aided a link with enough energy among K’s relay, it will be selected as a work node on the condition that the direct link cannot provide information transmission because of deep fading. The cooperative relay works as

Step 1:
the system prefers to consider the direct link to transmit information with. If h_{s,d}^{2} is bigger than the threshold value, the direct link shown in (1) will be selected to work for transmitting information. And the relay link will not need to be set up. Otherwise, go to Step 2.

Step 2:
some relay link among K’s, as shown in (2) and (3), will be selected and activated to work instead of a direct link. If it also fails, then there will be an outage event occurred.
Since the relay node R_{
k
} is deployed to amplifyandforward relay scheme. Considering the relationship between received and signal \(y_{s,r_{k}}(t)\) and retransmitted signal \(x_{r_{k}}(t)\) at R_{
k
}, as shown in (2) and (3), we can get
$$ x_{r_{k}}(t)=G\cdot y_{s,r_{k}}(t) $$
(14)
Correspondingly, the destination D’s received instantaneous SNR [21] can be express as
$$\begin{array}{@{}rcl@{}} \gamma_{eq}&=&\frac{h_{s{r_{k}}}^{2}h_{{r_{k}}d}^{2}G^{2}P_{s}}{h_{r_{k}d}^{2}G^{2}N_{R}+N_{D}} =\frac{\frac{h_{s{r_{k}}}^{2}P_{s}}{N_{R}}\frac{h_{{r_{k}}d}^{2}P_{r_{k}}}{N_{D}}}{\frac{h_{s{r_{k}}}^{2}P_{s}}{N_{R}} +\frac{h_{{r_{k}}d}^{2}P_{r_{k}}}{N_{D}}+1}\\ &=&\frac{\gamma_{sr_{k}}\gamma_{r_{k}d}}{\gamma_{sr_{k}}+\gamma_{r_{k}d}+1} \end{array} $$
(15)
$$\begin{array}{@{}rcl@{}} G^{2}&=&\frac{P_{r_{k}}}{h_{s{r_{k}}}^{2}P_{s}+N_{R}} \end{array} $$
(16)
Then we can get the mutual information for the S−R_{
k
}−D link as
$$ I_{sr_{k}d}=\frac{1}{2}W\log_{2}(1+\gamma_{eq}) $$
(17)
Given γ_{
eq
} in (15), which is the random variable denoting the received SNR at destination node D, then we can derive exact expressions for the CDF of γ_{
eq
}. And the CCDF of γ_{(eq)} is \(\bar {F}_{\gamma _{eq}}(\gamma)\) [26, 27]
$$\begin{array}{@{}rcl@{}} \bar{F}_{\gamma_{eq}}(\gamma\!)\!\!&=&\!2\exp{\left[\left(\frac{m_{sr_{k}}}{\Omega_{sr_{k}}}+ \frac{m_{r_{k}d}}{\Omega_{r_{k}d}}\right)\gamma\right]}\\ \!&\cdot&\!\sum_{n=0}^{m_{sr_{k}}1}\sum_{k=0}^{m_{r_{k}d}1}\sum_{m=0}^{k}\left\{ \frac{{\frac{\Omega_{sr_{k}}}{m_{sr_{k}}}}^{\frac{nm+12m_{sr_{k}}}{2}}{\frac{\Omega_{r_{k}d}}{m_{r_{k}d}}}^{\frac{mn12k}{2}} }{m!(km)!n!(m_{sr_{k}}n1)!}\right.\\ \!&\cdot&\!\left. K_{nm+1}\!\left(\!2\sqrt{\frac{m_{sr_{k}}m_{r_{k}d}\gamma(\gamma+1)}{\Omega_{sr_{k}} \Omega_{r_{k}d}}}\right)\!{\left(\!1\,+\,\frac{1}{\gamma}\!\right)}\!^{\frac{n+m+1}{2}}\gamma^{m_{sr_{k}}+k}\vphantom{\frac{{\frac{\Omega_{sr_{k}}}{m_{sr_{k}}}}^{\frac{nm+12m_{sr_{k}}}{2}}{\frac{\Omega_{r_{k}d}}{m_{r_{k}d}}}^{\frac{mn12k}{2}} }{m!(km)!n!(m_{sr_{k}}n1)!}}\right\}\\ \end{array} $$
(18)
So, CDF of γ_{(eq)} is \(F_{\gamma _{eq}}(\gamma)\), which is given by
$$ P_{sr_{k}d}^{\mathrm{(out)}}=P_{r}(\gamma_{eq}<\gamma_{th})=F_{\gamma_{eq}}(\gamma)=1\bar{F}_{\gamma_{eq}}(\gamma) $$
(19)
System overall outage probability
Substituting (13) and (19) into (10), the explicit closedform expression of outage probability for the relayaided cooperative transmission protocol by the harvested energy can be explained as follows
$$\begin{array}{@{}rcl@{}} P_{\text{total}^{(out)}}\!\!&=&\!\!\left\{1\exp{\left(\frac{m_{sd}}{\Omega_{sd}}\gamma_{th}^{sd}\right)} \sum_{n=0}^{m_{sd}1}\left(\frac{m_{sd}\gamma_{th}^{sd}}{\Omega_{sd}n!}\right)^{n}\right\}\\ &\cdot&\!\!\prod_{k=1}^{K}\left\{p_{k}^{ex}+{(1p_{k}^{ex})}\left\{12\exp{\left[\left(\frac{m_{sr_{k}}}{ \Omega_{sr_{k}}}+\frac{m_{r_{k}d}}{\Omega_{r_{k}d}}\right)\gamma\right]}\right.\right.\\ &\cdot&\!\!\sum_{n=0}^{m_{sr_{k}}1}\sum_{k=0}^{m_{r_{k}d}1}\sum_{m=0}^{k} \frac{{\frac{\Omega_{sr_{k}}}{m_{sr_{k}}}}^{\frac{nm+12m_{sr_{k}}}{2}}{\frac{\Omega_{r_{k}d}}{m_{r_{k}d}}}^{\frac{mn12k}{2}} }{m!(km)!n!(m_{sr_{k}}n1)!} \\ &\cdot&\!\! \left.\left.K_{nm+1}\!{\left(\!2\sqrt{\frac{m_{sr_{k}}m_{r_{k}d}\gamma(\gamma+1)}{\Omega_{sr_{k}}\Omega_{r_{k}d}}}\right)} {\left(\!1\,+\,\frac{1}{\gamma}\right)}^{\frac{n+m+1}{2}}\gamma^{m_{sr_{k}}+k}\right\}\right\}\\ \end{array} $$
(20)