Impacts of imperfect SIC and imperfect hardware in performance analysis on AF non-orthogonal multiple access network

Abstract

In normal non-orthogonal multiple access (NOMA) network, perfect successive interference cancellation (SIC) and perfect hardware impairments are usually admitted at transceivers. However, imperfections of SIC and hardware exist in mathematical analysis of practical system, and such imperfections provide guideline to deploy NOMA in practical circumstances effectively. To overcome these disadvantages, this paper considers relaying NOMA as existence of hardware noise and interference of imperfect SIC operation and hence degraded performance under related factors can be addressed. We examine the system performance of the proposed system and develop a closed-form formulation of outage probabilities for each user. In high signal to noise ratio regime, we consider the corresponding asymptotic outage probability to provide meaningful insights in our proposed schemes. We also exhibit comparison study with other works related performance evaluation and with Decode and Forward based NOMA as useful benchmark. To further examine system performance of the proposed scheme, we derive formula and verify results in term of the throughput performance. Finally, numerical examples are performed to validate the effectiveness of the proposed scheme as practical challenges are raised, and compare these performance considerations with that in other system models with respect to outage behavior as varying related parameters.

Appendix

Proof of Proposition 1

To prove outage probability, we recall achieved SINR as presentation in previous section. In this case, the outage probability for $D_2$ can be computed by

$$\begin{aligned} \begin{aligned} {\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {\mathrm{I}} \right) } \buildrel \varDelta \over =&\Pr \left( \frac{{{\alpha }{\rho ^2}{X}{Z}}}{{{\rho ^2}{X}{{\left| g \right| }^2}{(1-\alpha )} + \rho {X} + \rho {Z} + 1}}< {\delta _{t{h_1}}},\right. \\&\left. \frac{{{(1-\alpha )}{\rho ^2}{X}{Z}}}{{\rho {X} + \rho {Z} + 1}}< {\delta _{t{h_2}}} \right) \\ =&\Pr \left( {{Z} < \frac{{\rho {X} + 1}}{{\rho \left( {\chi \rho {X} - 1} \right) }}} \right) \end{aligned}. \end{aligned}$$

(A.1)

In the case of $\chi \rho {\left| h \right| ^2} - 1 < 0$, it leads to following inequality ${\left| h \right| ^2} < {1 /{\rho \chi }}$, the outage probability in this case can be calculated as ${\mathrm{OP}}_{{\mathrm{D2,x2}}}^{\left( {\mathrm{I}} \right) } = 1$.

In the case of $\chi \rho {\left| h \right| ^2} - 1 > 0$, the outage probability is given by

$$\begin{aligned} {\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {\mathrm{I}} \right) }&= 1 - \int \limits _{\frac{1}{{\rho \chi }}}^\infty {\exp \left( { - \frac{{\rho z + 1}}{{\rho \left( {\chi \rho z - 1} \right) }} - z} \right) } dz\nonumber \\&= 1 - \frac{1}{{\rho \chi }} \times \exp \left( { - \frac{2}{{\rho \chi }}} \right) \nonumber \\&\times \int \limits _0^\infty {\exp \left( { - \frac{4}{{4\rho t}}\left( {\frac{1}{{\rho \chi }} + 1} \right) - \frac{t}{{\rho \chi }}} \right) } dt. \end{aligned}$$

(A.2)

After simple manipulation, it can be found final expression. This is end of the proof. $\square$

Proof of Proposition 2

The outage probability considered in previous section is rewritten as

$$\begin{aligned}&{\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{{\mathrm{ipsic}}} \nonumber \\&\quad = 1 - \int \limits _0^\infty {\exp \left( { - x} \right) \underbrace{\int \limits _{\frac{{{\delta _{t{h_2}}}}}{{\rho {(1-\alpha )}}}}^\infty {\exp \left( { - \frac{{{\delta _{t{h_2}}}\left( {\upsilon {\rho ^2}x + \rho z + 1} \right) }}{{\rho \left( {{(1-\alpha )}\rho z - {\delta _{t{h_2}}}} \right) }} - z} \right) } dz}_{{\varTheta _1}}dx}\nonumber \\ \end{aligned}$$

(B.1)

in which,

$$\begin{aligned} {\varTheta _1}\left( x \right)&= \int \limits _{\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}\rho }}}^\infty {\exp \left( { - \frac{{{\delta _{t{h_2}}}\left( {\upsilon {\rho ^2}x + \rho z + 1} \right) }}{{\rho \left( {{(1-\alpha )}\rho z - {\delta _{t{h_2}}}} \right) }} - z} \right) } dz\nonumber \\&= \frac{1}{{{(1-\alpha )}\rho }}\exp \left( { - \frac{{2{\delta _{t{h_2}}}}}{{{(1-\alpha )}\rho }}} \right) \nonumber \\&\times \underbrace{\int \limits _0^\infty {\exp \left( { - \frac{{4{\delta _{t{h_2}}}}}{{4\rho t}}\left( {\upsilon {\rho ^2}x + \frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}}} + 1} \right) - \frac{t}{{{(1-\alpha )}\rho }}} \right) } dt}_{{\mu _1}} \end{aligned}$$

(B.2)

It is noted that $\mu _1$ in above expression can be further computed by

$$\begin{aligned} {\mu _1}&= 2\exp \left( { - \frac{{2{\delta _{t{h_2}}}}}{{{(1-\alpha )}\rho }}} \right) \times \int \limits _0^\infty \exp \left( { - x} \right) \nonumber \\&\quad \times \sqrt{\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}{\rho ^2}}}\left( {\upsilon {\rho ^2}x + \frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}}} + 1} \right) } \nonumber \\&\quad \times {\mathrm{K}_1}\left( {2\sqrt{\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}{\rho ^2}}}\left( {\upsilon {\rho ^2}x + \frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}}} + 1} \right) } } \right) dx \end{aligned}$$

(B.3)

Next, it can be expressed such outage by

$$\begin{aligned} {\mathrm{OP}}_{{\mathrm{D2,x2}}}^{{\mathrm{ipsic}}} = 1 - \exp \left( { - \frac{{2{\delta _{t{h_2}}}}}{{{(1-\alpha )}\rho }}} \right) \times I \end{aligned}$$

(B.4)

where, $I = \phi \left( {\frac{{\upsilon {\delta _{t{h_2}}}}}{{{(1-\alpha )}}},\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}{\rho ^2}}}\left( {\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}}} + 1} \right) } \right) = \phi \left( {a,b} \right)$, in which ${\frac{{\upsilon {\delta _{t{h_2}}}}}{{{(1-\alpha )}}}}=a$, ${\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}{\rho ^2}}}\left( {\frac{{{\delta _{t{h_2}}}}}{{{(1-\alpha )}}} + 1} \right) }=b$

Interestingly, it can be re-expressed $\phi \left( {a,b} \right)$ as below

$$\begin{aligned} \phi \left( {a,b} \right) \buildrel \varDelta \over =&2\int \limits _0^\infty {\exp \left( { - x} \right) \sqrt{ax + b} \times {\mathrm{K}_1}\left( {2\sqrt{ax + b} } \right) dx} \nonumber \\ =&2\exp \left( { - \frac{b}{a}} \right) \left\{ \underbrace{\sqrt{a} \int \limits _0^\infty {\sqrt{x} \times {\mathrm{K}_1}\left( {\sqrt{4a} \sqrt{x} } \right) \exp \left( { - x} \right) dx} }_{{\varDelta _1}}\right. \nonumber \\&\left. - \int \limits _0^b {\sqrt{y} \times {\mathrm{K}_1}\left( {2\sqrt{y} } \right) \exp \left( { - \frac{y}{a}} \right) dy} \right\} \end{aligned}$$

(B.5)

The integral formula in (B.5) can be obtained by using [32, vol. 4, eq. (1.1.2.3)]. The first term in above equation can be fulfilled by applying [32, vol. 4, eq. (3.16.2.4)]

$$\begin{aligned} {\varDelta _1}&= \sqrt{a} \times \int \limits _0^\infty {\sqrt{x} \times {\mathrm{K}_1}\left( {\sqrt{4a} \sqrt{x} } \right) \times \exp \left( { - x} \right) dx} \nonumber \\&= \sqrt{a} \times \frac{{\varGamma \left( 2 \right) \sqrt{4a} }}{4} \times \exp \left( {\frac{{4a}}{4}} \right) \times \varGamma \left( { - 1,\frac{{4a}}{4}} \right) \nonumber \\&= \frac{{a \times \varGamma \left( 2 \right) }}{2} \times \exp \left( a \right) \times \varGamma \left( { - 1,a} \right) . \end{aligned}$$

(B.6)

The second term in (B.5) is difficult to solve in closed-form because of the Bessel function and exponential function. This is end of proof. $\square$

Proof of Proposition 3

Extracting from obtained SNDR as findings in previous section, we formula outage probability at $D_2$ as

$$\begin{aligned} {\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {{\mathrm{II}}} \right) }&= \Pr \left( {{\mathrm{SND}}{{\mathrm{R}}_{1 \rightarrow 2}}< {\delta _{t{h_1}}},{\mathrm{SND}}{{\mathrm{R}}_2} < {\delta _{t{h_2}}}} \right) \nonumber \\&= 1 - \Pr \left( {{\mathrm{SND}}{{\mathrm{R}}_{1 \rightarrow 2}} \ge {\delta _{t{h_1}}},{\mathrm{SND}}{{\mathrm{R}}_2} \ge {\delta _{t{h_2}}}} \right) \end{aligned}$$

(C.1)

In particular, after replacing specific SNDR expressions, we have following formula

$$\begin{aligned} {\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {{\mathrm{II}}} \right) }&= \Pr \left( \min \left( {\frac{{{\alpha }}}{{{\delta _{t{h_1}}}}} - \left( {{(1-\alpha )} + {d_0}} \right) ,\frac{{{(1-\alpha )}}}{{{\delta _{t{h_2}}}}}} \right) \right. \nonumber \\&\left. {\rho ^2}{X}{Z}< {d_1}\rho {X} + {d_2}\rho {Z} + 1 \right) \nonumber \\&= \Pr \left( {\chi {\rho ^2}{X}{Z} < {d_1}\rho {X} + {d_2}\rho {Z} + 1} \right) \end{aligned}$$

(C.2)

It is further expressed by

$$\begin{aligned} {\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {{\mathrm{II}}} \right) } = \Pr \left( {{Z} < \frac{{{d_1}\rho {X} + 1}}{{\rho \left( {\chi \rho {X} - {d_2}} \right) }}} \right) . \end{aligned}$$

(C.3)

It is noted as $\chi \buildrel \varDelta \over = \min \left( {\frac{{{\alpha }}}{{{\delta _{t{h_1}}}}} - \left( {{(1-\alpha )} + {d_0}} \right) ,\frac{{{(1-\alpha )}}}{{{\delta _{t{h_2}}}}}} \right)$.

We obtain ${\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {{\mathrm{II}}} \right) } = 1$ as $\chi \rho {\left| h \right| ^2} - {d_2} < 0$.

In case of $\chi \rho {\left| h \right| ^2} - {d_2} \ge 0$ the related outage probability is given by

$$\begin{aligned} {\mathrm{OP}}_{{\mathrm{D_2,x_2}}}^{\left( {{\mathrm{II}}} \right) }&= 1 - \frac{1}{{{\varOmega _1}{\varOmega _2}\rho \chi }}\int \limits _0^\infty \exp \left( - \frac{{{d_1}}}{{{\varOmega _1}\rho \chi }} \right. \nonumber \\&\quad \left. - \frac{1}{{\rho t}}\left( {\frac{{{d_1}{d_2}}}{\chi } + 1} \right) - \frac{t}{{{\varOmega _1}{\varOmega _2}\rho \chi }} - \frac{{{d_2}}}{{{\varOmega _2}\rho \chi }} \right) dt\nonumber \\&= 1 - \frac{1}{{{\varOmega _1}{\varOmega _2}\rho \chi }}\exp \left( { - \frac{{\left( {{d_1}{\varOmega _2} + {d_2}{\varOmega _1}} \right) }}{{{\varOmega _1}{\varOmega _2}\rho \chi }}} \right) \nonumber \\&\times \int \limits _0^\infty {\exp \left( { - \frac{4}{{4\rho t}}\left( {\frac{{{d_1}{d_2}}}{\chi } + 1} \right) - \frac{t}{{{\varOmega _1}{\varOmega _2}\rho \chi }}} \right) } dt. \end{aligned}$$

(C.4)

Finally, a closed-form expression of the outage probability for $D_2$ in case of existence of hardware impairment is written as in Proposition 3 after performing simple computation. It completes the proof. $\square$