This section presents the analysis of the performance of the system model in which closed-form expressions of the outage probability, throughput, ergodic rate and EE are determined in DTT and DLT modes.

### Outage performance

#### Outage probability at \(D_{1}\)

User \(D_1\) is not in outage when it can decode both signals \(x_1\) and \(x_2\) received from the BS. The outage probability at \(D_1\) is thus obtained by

$$\begin{aligned} P_{D_1} = 1 - \Pr \left( {{\gamma _{2,{D_1}}}> \gamma _{t{h_2}},{} {\gamma _{1,{D_1}}} > \gamma _{t{h_1}}} \right) , \end{aligned}$$

(12)

where, \(\gamma _{t{h_1}} = {2^{2{R_1}}} - 1\) and \(\gamma _{t{h_2}} = {2^{2{R_2}}} - 1\) represent the threshold SNRs at \(D_{1}\) for detecting signals \(x_1\) and \(x_2\), respectively.

### Theorem 1

*The outage probability at* \(D_{1}\) *is given by*

$$\begin{aligned} P_{D1} = 1 - {e^{ - \frac{{{\theta _1}}}{{{\Omega _1}}}}}, \end{aligned}$$

(13)

*where*, \({\theta _1} = \max ({\tau _1},{\nu _1}),{\tau _1} = \frac{{\gamma _{th2}}}{{\rho {\psi _I}({\alpha _2} - {\alpha _1}\gamma _{th2})}}\) *and* \({\nu _1} = \frac{{\gamma _{th1}}}{{{a_1}{\psi _I}\rho }}\) *with* \({\alpha _2} > {\alpha _1}\gamma _{th2}.\)

### Proof

From (12), the outage probability at \(D_{1}\) can be determined by

$$\begin{aligned} {P_{{D_1}}}&= {} 1\!-\!Pr\left( {\frac{{{\psi _I}{{\left| {{h_1}} \right| }^2}{\alpha _2}\rho }}{{{\psi _I}{{\left| {{h_1}} \right| }^2}{\alpha _1}\rho \!+\!1}}\!>\!{\gamma _{t{h_2}}},{} {} {} {} \psi _I{{\left| {{h_1}} \right| }^2}{\alpha _1}\rho \!>\! {\gamma _{t{h_1}}}} \right) \nonumber \\&= {} 1 - Pr\left( {{{\left| {{h_1}} \right| }^2}> \,\frac{{{\gamma _{t{h_2}}}}}{{\rho {\psi _I}\left( {{\alpha _2} - {\alpha _1}{\gamma _{t{h_2}}}} \right) }},\,\,{{\left| {{h_1}} \right| }^2} > \,\frac{{{\gamma _{t{h_1}}}}}{{{\alpha _1}\rho {\psi _I}}}} \right) \nonumber \\&= {} 1 - Pr({\left| {{h_1}} \right| ^2} \ge {\theta _1})\nonumber \\&= {} 1 - \int _{{\theta _1}}^\infty {{f_{{{\left| {{h_1}} \right| }^2}}}(x)dx} \end{aligned}$$

(14)

Applying the following equation

$$\begin{aligned} {f_{{h_i}}}(x) = \frac{1}{{{\Omega _i}}}\exp ( - \frac{x}{{{\Omega _i}}}),{} {} {} {} i \in \{ SD_{1},D_{1}D_{2}\mathrm{{\} }} \end{aligned}$$

Eq. (14) can be obtained as follows

$$\begin{aligned} {P_{{D_1}}}&= {} 1 - \int _{{\theta _{1}}}^\infty {\frac{1}{{{\Omega _1}}}{e^{\frac{{ - x}}{{{\Omega _1}}}}}dx}\nonumber \\&= {} 1 - {e^{ - \,\,\frac{{{\theta _{1}}}}{{{\Omega _1}}}}} \end{aligned}$$

(15)

The proof is completed. \(\square\)

### Corollary 1

*From* (15), *the outage probability at* \(D_{1}\) *for high SNR* \(\rho \rightarrow \infty\) *is expressed by*

$$\begin{aligned} P_{D_1}^\infty = \frac{{{\theta _1}}}{{{\Omega _1}}} \end{aligned}$$

(16)

### Proof

From (12), when \(\rho \rightarrow \infty\), the outage probability at \(D_{1}\) with \(1 - {e^{-x}} \approx x\) is given by

$$\begin{aligned} P_{D_1}^{\infty }&= {} 1 - {P_r}\left( {{{\left| {{h_1}} \right| }^2} \ge \theta _1} \right) \nonumber \\&= {} 1 - {e^{ - \frac{{\theta _1}}{{\Omega {\,_1}}}}}\nonumber \\&= {} \frac{{{\theta _1}}}{{{\Omega _1}}} \end{aligned}$$

(17)

The proof is completed. \(\square\)

Based on (15) and \({\alpha _2} > {\alpha _1}\gamma _{th2}\), \(P_{D_1}\) depends on \(\tau _{1}\) and the random variable \(\Omega _{1}\) (\(|h_1|^2\)). The closer the *d*, the lower the \(P_{D_1}\). This means that a better transmission quality can be achieved, and vice versa.

#### Outage probability at \(D_{2}\) for no direct link

Since \(D_1\) can not detect \(x_2\) as well as \(D_2\) can not recover the forwarded information from \(D_1\), the \(D_2\) is in outage. Hence, the outage probability at \(D_2\) is derived as (see (18)). By calculating \(J_2\) and \(J_3\), the outage probability for no direct link is determined by

$$\begin{aligned} \begin{array}{l} P_{{D_2},nodir} = \underbrace{\Pr \left( {{\gamma _{2,{D_1}}}< \gamma _{t{h_2}}^{HD}} \right) }_{{J_2}} + \underbrace{\Pr \left( {{\gamma _{2,{D_2}}} < \gamma _{t{h_2}}^{HD},{\gamma _{2,{D_1}}} > \gamma _{t{h_2}}^{HD}} \right) }_{{J_3}}, \end{array} \end{aligned}$$

(18)

### Theorem 2

*The outage probability at*
\(D_{2}\)
*can be obtained by*

$$\begin{aligned} P_{{D_2},nodir} = 1 - {e^{ - \frac{{{\tau _1}}}{{{\Omega _1}}}}}+ \int \limits _{{\tau _1}}^\infty {\left( {1 - {{\mathop {e}\nolimits } ^{ - \frac{{\gamma _{t{h_2}}}}{{x{\psi _E}\rho {\Omega _2}}}}}} \right) \frac{1}{{{\Omega _1}}}\exp \left( {\frac{{- x}}{{{\Omega _1}}}} \right) } dx. \end{aligned}$$

(19)

### Proof

Considering the Rayleigh fading channel, \(J_2\) can be given by

$$\begin{aligned} {J_2} = 1 - \exp \left( {\frac{{ - {\tau _1}}}{{{\Omega _1}}}} \right) . \end{aligned}$$

(20)

and \(J_3\) can be expressed as (see(21)).

$$\begin{aligned} \begin{array}{l} {J_3}=\Pr \left( {{{\left| {{h_2}} \right| }^2}{{\left| {{h_1}} \right| }^2}{\psi _E}\rho<\gamma _{t{h_2}},\frac{{{{\left| {{h_1}} \right| }^2}{\psi _I}{\alpha _2}\rho }}{{{\psi _I}{{\left| {{h_1}} \right| }^2}{\alpha _1}\rho +1}}>\gamma _{t{h_2}}} \right) \\ \\ \quad \,=\left\{ \begin{array}{l} \Pr \left( {{{\left| {{h_2}} \right| }^2}<\frac{{\gamma _{t{h_2}}}}{{{{\left| {{h_1}} \right| }^2}{\psi _E}\rho }}, {{\left| {{h_1}} \right| }^2}>\frac{{\gamma _{t{h_2}}}}{{{\psi _I}\rho \left( {{\alpha _2}-{\alpha _1}\gamma _{t{h_2}}} \right) }}} \right) ,{\alpha _2}> {\alpha _1}\gamma _{t{h_2}}\\ \\ 0,\,\,{\alpha _2} \le {a_1}\gamma _{t{h_2}} \end{array} \right. \\ \\ \quad \,= \int \limits _{\frac{{\gamma _{t{h_2}}}}{{{\psi _I}\rho \left( {{\alpha _2} - {\alpha _1}\gamma _{t{h_2}}} \right) }}}^\infty {\int \limits _0^{\frac{{\gamma _{t{h_2}}}}{{x{\psi _E}\rho }}} {{f_{{{\left| {{h_1}} \right| }^2}}}(x){f_{{{\left| {{h_2}} \right| }^2}}}(y)dxdy} }= \int \limits _{{\tau _1}}^\infty {\frac{1}{{{\Omega _{1}}}}\left[ {1 - \exp \left( {\frac{{ - \gamma _{t{h_2}}}}{{x{\psi _E}\rho {\Omega _{2}}}}} \right) } \right] \exp \left( {\frac{{ - x}}{{{\Omega _{1}}}}} \right) } dx. \end{array} \end{aligned}$$

(21)

The outage probability at \(D_2\) is given by

$$\begin{aligned} P_{{D_2},nodir} = \,{J_2}\, + \,{J_3}.\, \end{aligned}$$

(22)

\(\square\)

### Corollary 2

*The outage probability at* \(D_{2}\) *for high SNR can be determined as* (*see*(23)), *where* \({K_1}(.)\) *is the first order modified Bessel function of the second kind* [55, *Eq*.(3.324.1)].

$$\begin{aligned} P_{{D_2},nodir}^{\infty }&= {} \Pr \left( {\frac{{{\alpha _2}}}{{{\alpha _1}}}< \gamma _{t{h_2}}} \right) + \Pr \left( {{{\left| {{h_2}} \right| }^2}< \frac{{\gamma _{t{h_2}}}}{{{\psi _E}\rho {{\left| {{h_1}} \right| }^2}}},\frac{{{\alpha _2}}}{{{\alpha _1}}}> \gamma _{t{h_2}}} \right) \nonumber \\&= {} \Pr \left( {{{\left| {{h_2}} \right| }^2} < \frac{{\gamma _{t{h_2}}}}{{{\psi _E}\rho {{\left| {{h_1}} \right| }^2}}},\frac{{{\alpha _2}}}{{{\alpha _1}}} > \gamma _{t{h_2}}} \right) \nonumber \\&= {} \int \limits _0^\infty {\left[ {1 - \exp \left( {\frac{{ - \gamma _{t{h_2}}}}{{{\psi _E}\rho {\Omega _2}x}}} \right) } \right] } \frac{1}{{{\Omega _1}}}\exp \left( {\frac{{ - x}}{{{\Omega _1}}}} \right) dx \nonumber \\&= {} 1 - 2\sqrt{\frac{{\gamma _{t{h_2}}}}{{{\psi _E}\rho {\Omega _1}{\Omega _2}}}} {K_1}\left( {2\sqrt{\frac{{\gamma _{t{h_2}}}}{{{\psi _E}\rho {\Omega _1}{\Omega _2}}}} } \right) . \end{aligned}$$

(23)

#### Outage probability at \(D_{2}\) for User Relaying with Direct Link

When \(x_{2}\) can be detected at \(D_{1}\) but the SINR is smaller than the target SNR after MRC or both \(D_{1}\) and \(D_{2}\) can not detect \(x_{2}\), the outage probability will occur at \(D_{2}\) and is given by (see(24))

$$\begin{aligned} {P_{{D_2},dir}} = \underbrace{{P_r} (\gamma _{{D_2}}^{MRC}< {\gamma _{th{{}_2}}})}_{{J_4}}\underbrace{{P_r}\left( {{\gamma _{2,{D_1}}} > {\gamma _{th{{}_2}}}} \right) }_{{J_5}} + {} \underbrace{{P_r}({\gamma _{2,{D_1}}}< {\gamma _{th{{}_2}}},{\gamma _{1,{D_2}}} < {\gamma _{th{{}_2}}})}_{{J_6}}, \end{aligned}$$

(24)

### Theorem 3

*The outage probability at* \(D_{2}\) *can be given by* (*see*(25))

$$\begin{aligned} P_{{D_2},dir}&= {} \int _0^\infty {\int _0^{\psi _{I}\,{\tau _1}} {\frac{1}{{{\Omega _0}{\Omega _1}}}\left( {1-{e^{-\frac{{\gamma _{t{h_2}}}}{{x\psi _{E}\rho {\Omega _2}}}+\frac{{y{\alpha _2}}}{{x\psi _{E}{\Omega _2}\left( {y{\alpha _1}\rho +1} \right) }}}}} \right) {e^{-\frac{x}{{\Omega {_1}}}\!-\!\frac{y}{{{\Omega _0}}}}}dxdy}}\times {e^{ - \frac{{{\tau _1}}}{{{\Omega _1}}}}}\,\nonumber \\&+ \left( {1 - {e^{ - \frac{{\tau _1}}{{{\Omega _1}}}}}} \right) \left( {1 - {e^{ - \,\,\frac{{{\tau _1}\,\psi _{I}}}{{{\Omega _0}}}}}} \right) \end{aligned}$$

(25)

### Proof

From (24), the outage probability at \(D_{2}\) is determined by (see(26)), (see(27)), (see(28))

$$\begin{aligned} {J_4}&= {} \Pr \left( {{{\left| {{h_2}} \right| }^2}< \frac{{\gamma _{t{h_2}}}}{{{{\left| {{h_1}} \right| }^2}\psi _{E}\rho }} - \frac{{{{\left| {{h_0}} \right| }^2}{\alpha _2}}}{{{{\left| {{h_1}} \right| }^2}\psi _{E}\left( {{{\left| {{h_0}} \right| }^2}{\alpha _1}\rho + 1} \right) }},{{\left| {{h_0}} \right| }^2} < {\tau _1}\,\psi _{I}} \right) \nonumber \\&= {} \int _0^\infty {\int _0^{\psi _{I}\,{\tau _1}} {\int \limits _0^{\frac{{\gamma _{t{h_2}}}}{{x\,\psi _{E}\rho }}\,\, - \,\,\frac{{y{\alpha _2}}}{{x\,\psi _{E}\left( {y\,{\alpha _1}\rho + 1} \right) }}} {{f_{{{\left| {h{\,_1}} \right| }^2}}}\left( x \right) {f_{{{\left| {{h_0}} \right| }^2}}}\left( y \right) {f_{{{\left| {h{\,_2}} \right| }^2}}}\left( z \right) dxdydz} } }\nonumber \\&= {} \int _0^\infty {\int _0^{\psi _{I}\,{\tau _1}} {\frac{1}{{{\Omega _0}{\Omega _1}}}\left( {1 - {e^{ - \,\,\frac{{\gamma _{t{h_2}}}}{{x\,\psi _{E}\rho \,\Omega {\,_2}}}\,\, + \,\,\frac{{y{\alpha _2}}}{{x\,\psi _{E}\Omega {\,_2}\left( {y\,{\alpha _1}\rho + 1} \right) }}}}} \right) {e^{ - \,\,\frac{x}{{\Omega {\,_1}}}\,\, - \,\,\frac{y}{{{\Omega _0}}}}}dxdy} } \end{aligned}$$

(26)

$$\begin{aligned} {J_5}&= {} {\left| {{h_1}} \right| ^2} > {\tau _1}=\int _{\tau _1}^\infty {{f_{{{\left| {h{\,_1}} \right| }^2}}}\left( x \right) dx}= {e^{ - \,\,\frac{{\tau _1}}{{\Omega {\,_1}}}}} \end{aligned}$$

(27)

$$\begin{aligned} {J_6}&= {} \Pr \left( {{{\left| {{h_0}} \right| }^2}< \psi _{I}{} {} {\tau _1},{{\left| {{h_1}} \right| }^2} < {\tau _1}} \right) = \int _0^{{\tau _1}} {\int _0^{\psi _{I}{} {\tau _1}} {{f_{{{\left| {h{{}_1}} \right| }^2}}}\left( x \right) } } {f_{{{\left| {{h_{{} 0}}} \right| }^2}}}\left( y \right) dxdy\;\;\;{} {} {} {} {} {}\nonumber \\&= {} \left( {1 - {e^{ - \frac{{\psi _{I}{} {\tau _1}}}{{{\Omega _0}}}}}} \right) \left( {1 - {e^{ - \frac{{{\tau _1}}}{{{\Omega _1}}}}}} \right) \end{aligned}$$

(28)

\(\square\)

### Throughput for DLT mode

#### User relaying without direct link

With a given constant *R*, the transmitted information of the source node depends on the outage probability performance due to wireless fading channels. Therefore, the throughput of the system is determined by

$$\begin{aligned} \tau _{t,nodir} = \left( {1 - P_{{D_1}}} \right) {R_1} + \left( {1 - P_{{D_2},nodir}} \right) {R_2}, \end{aligned}$$

(29)

where \(P_{D_1}\) and \(P_{{D_2},nodir}\) can be achieved from (15) and (19), respectively.

#### User relaying with direct link

The throughput of system is given by

$$\begin{aligned} \tau _{t,dir} = \left( {1 - P_{{D_1}}} \right) {R_1} + \left( {1 - P_{{D_2},dir}} \right) {R_2}, \end{aligned}$$

(30)

where \(P_{{D_1}}\) and \(P_{{D_2},dir}\) can be achieved from (15) and (25), respectively.

### Ergodic rate for DTT mode

#### Ergodic rate at \(D_{1}\)

The achievable rate at \(D_{1}\) where \(D_{1}\) can detect \(x_{2}\) is given by

$$\begin{aligned} {R_{{D_1}}} = \frac{1}{2}{\log _2}\left( {1 + {\gamma _{{D_1}}}} \right) . \end{aligned}$$

(31)

### Theorem 4

*The ergodic rate at* \(D_{1}\) *is determined by*

$$\begin{aligned} R_{{D_1}} = \frac{{ - \exp \left( {\frac{1}{{{\psi _I}{\alpha _1}\rho {\Omega _1}}}} \right) }}{{2\ln 2}}Ei\left( {\frac{{ - 1}}{{{\psi _I}{\alpha _1}\rho {\Omega _1}}}} \right) , \end{aligned}$$

(32)

*where* *Ei*(.) *indicates the exponential integral function* [55, *Eq*.(3.354.4)].

### Proof

See Appendix 1. \(\square\)

#### Ergodic rate at \(D_{2}\) for User Relaying Without Direct Link

Since \(x_2\) needs to be detected at both \(D_1\) and \(D_2\), the achievable rate at \(D_{2}\) is given by

$$\begin{aligned} {R_{{D_2},nodir}}{} = {} \frac{1}{2}{\log _2}\left( {1 + \min \left( {{\gamma _{2,{D_1}}},{\gamma _{2,{D_2}}}} \right) } \right) . \end{aligned}$$

(33)

### Theorem 5

*The ergodic rate at*
\(D_{2}\)
*is given by*

$$\begin{aligned} R_{{D_2},nodir} = \frac{1}{{2\ln 2}}\int \limits _0^{\frac{{{\alpha _2}}}{{{\alpha _1}}}} {\left[ {\frac{{{e^{ - \frac{x}{{{\psi _I}\rho \left( {{\alpha _2} - {\alpha _1}x} \right) {\Omega _1}}}}}}}{{1 + x}}} \right. } \left. {+ \frac{{\int _{\frac{x}{{{\psi _I}\rho \left( {{\alpha _2} - {\alpha _1}x} \right) }}}^\infty {\frac{1}{{{\Omega _1}}}\left( {1 - {e^{ - \frac{x}{{y\rho {\psi _E}{\Omega _2}}}}}} \right) {e^{ - \frac{y}{{{\Omega _1}}}}}dy} }}{{1 + x}}} \right] dx. \end{aligned}$$

(34)

### Proof

See Appendix 2 . \(\square\)

### Remark 1

The ergodic rate in the asymptotic expression at \(D_{2}\) for high SNR region \(\rho \rightarrow \infty\) is obtained by

$$\begin{aligned} R_{{D_2},nodir}^{\infty }=\frac{1}{{2\ln 2}}\int \limits _0^\infty {\frac{{1 - {F_X}(x)}}{{1 + x}}dx}. \end{aligned}$$

(35)

From the analytical result in (35), this expression can be deployed by

$$\begin{aligned} R_{{D_2},nodir}^{\infty }\!=\!\frac{1}{{2\ln 2}}\!\int \limits _0^{\frac{{{\alpha _2}}}{{{\alpha _1}}}} {\frac{{\!2\sqrt{\!\frac{x}{{{\psi _E}\rho \,{\Omega _1}{\Omega _2}}}} \,{K_1}\left( {\!2\sqrt{\!\frac{x}{{{\psi _E}\rho \,{\Omega _1}{\Omega _2}}}} } \right) }}{{1 + x}}dx}. \end{aligned}$$

(36)

### Proof

See Appendix 3. \(\square\)

#### Ergodic rate at \(D_{2}\) for user relaying with direct link

The ergodic rate at \(D_{2}\) is given by

$$\begin{aligned} {R_{{D_2},dir}} = E\left[ {\frac{1}{2}{{\log }_2}\left( {1 + \min \left( {{\gamma _{2,{D_1}}},\gamma _{{D_2}}^{MRC}} \right) } \right) } \right] . \end{aligned}$$

(37)

### Theorem 6

*From* (37), *the ergodic rate at* \(D_{2}\) *can be computed by* (*see*(38))

$$\begin{aligned} R_{{D_2},dir}\!=\!\frac{1}{{2\ln 2}}\int \limits _0^{\frac{{{\alpha _2}}}{{{\alpha _1}}}} {\left[ {\frac{{{e^{\!-\!\frac{x}{{\psi _{I}\rho \left( {{\alpha _2} - {\alpha _1}x} \right) {\Omega _1}}}}}}}{{1\!+\!x}}} \right. } \left. {\!-\!\frac{{\int _0^\infty {\int _{\frac{x}{{\psi _{I}\rho \left( {{\alpha _2}\!-\!{\alpha _1}x} \right) }}}^\infty {\frac{1}{{{\Omega _1}{\Omega _0}}}\left( {1\!-\!{e^{\!-\!\frac{{x\left( {y{\alpha _1}\rho \!+\!1} \right) \!+\!y{\alpha _2}\rho }}{{z\rho \psi _{E}{\Omega _2}\left( {y{\alpha _1}\rho \!+\!1} \right) }}}}} \right) {e^{\!-\!\frac{y}{{{\Omega _0}}}\!-\!\frac{z}{{{\Omega _1}}}}}dydz} } }}{{1\!+\!x}}} \right] dx. \end{aligned}$$

(38)

### Proof

See Appendix 4. \(\square\)

### Remark 2

The ergodic rate in the asymptotic expression at \(D_{2}\) for high SNR region \(\rho \rightarrow \infty\) is given by

$$\begin{aligned} R_{{D_2},dir}^{\infty } = \frac{1}{{2\ln 2}}\int \limits _0^\infty {\frac{{1 - {F_X}\left( x \right) }}{{1 + x}}dx} \end{aligned}$$

(39)

From (39), this expression can be deployed by

$$\begin{aligned} R_{{D_2},dir}^{\infty } = \frac{1}{{2\ln 2}}\int \limits _0^{\frac{{{\alpha _2}}}{{{\alpha _1}}}} {\frac{{2\sqrt{\frac{x}{{{\psi _E}\rho \,{\Omega _1}{\Omega _2}}}} \,\!{K_1}\!\left( {2\sqrt{\frac{x}{{{\psi _E}\rho \,{\Omega _1}{\Omega _2}}}} } \right) }}{{1 + x}}dx} \end{aligned}$$

(40)

### Proof

See Appendix 5. \(\square\)

#### Ergodic rate of the system for user relaying without direct link

The ergodic rate of system is determined by

$$\begin{aligned} \tau _{r,nodir} = R_{{D_1}} + R_{{D_2},nodir}, \end{aligned}$$

(41)

where \(R_{{D_1}}\) and \(R_{{D_2},nodir}\) can be obtained from (32) and (34), respectively.

#### Ergodic rate of the system for user relaying with direct link

The ergodic rate of system is thus expressed by

$$\begin{aligned} \tau _{r,dir} = R_{{D_1}} + R_{{D_2},dir}, \end{aligned}$$

(42)

where \(R_{{D_1}}\) and \(R_{{D_2},dir}\) can be obtained from (32) and (38), respectively.

### Energy efficiency

The EE can be determined as the ratio of the total data rate over the total consumed power in entire network, which is given by \(\mathrm{{EE}} \buildrel \Delta \over = \frac{R}{{{P_S} + {P_r}}}\). The energy efficiency of user relaying systems can be given as

$$\begin{aligned} E{E_\phi } = \frac{{2\tau _\phi ^{HD}}}{{\rho \left( {1 + {\psi _E}{\Omega _1}} \right) }}, \end{aligned}$$

(43)

where \(\phi \in \left( {t,r} \right)\), denotes the system energy efficiency in DLT mode and DTT mode, respectively.