Distributed transmit-antenna selection in variable-gain relaying systems

Abstract

Recently, distributed transmit-antenna selection schemes have attracted great interest, since they capture the essential benefits of multi-antenna systems while reducing their cost, complexity, delay, and feedback overhead. In those distributed schemes, the antenna selection is based on local channel-state information, in contrast to their optimal centralized counterparts, which require knowing the channel state of all links. Herein, we design two such distributed schemes for a dual-hop variable-gain amplify-and-forward relaying system with one multi-antenna source, one single-antenna relay, and one single-antenna destination. The two schemes differ in the diversity method used at the destination, namely, selection combining or maximal-ratio combining, and in the selection rule accordingly. In addition to conceiving these new schemes, we analyze their outage performance. Since an exact analysis proves intractable, we tackle the outage probability in terms of lower-bound expressions and their asymptotes at high signal-to-noise ratio. Importantly, the derived bounds turn out to be almost indistinguishable from the true performance, assessed via simulation. Our results reveal that the proposed distributed schemes achieve the same diversity order of their optimal centralized counterparts and perform closely to these, specially when the relay is near the source or destination.

Appendices

Appendix 1: Proof of Lemma 1

By using (4) into the definition of $I_1$ in (8), a lower bound $I_{1}^{\text{LB}}$ can be derived as

$$\begin{aligned} I_{1}&\ge \Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\max \left[ Y_{\overline{i}},\min \left[ X_{\overline{i}},Z\right] \right]<\tau \right) \triangleq I_{1}^{\text{LB}} \\&= \Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\max \left[ Y_{\overline{i}}, X_{\overline{i}}\right]<\tau \right) \\&\overset{(a)}{=}\Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ \max \left[ Y_{i},X_{i}\right] \right\}<\tau \right) \\&=\underbrace{\Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ X_{i} \right\}<\tau \right) }_{\triangleq \rho } \Pr \left( Y_{i}<\tau \right) ^{N_t}, \end{aligned}$$

(24)

which coincides with the result in (9). In step (a) we have used the DAS/SC rule for $Z\ge \max _i \left\{ X_{i} \right\}$, given in (6).

Appendix 2: Proof of Lemma 2

By using (4) into the definition of $I_2$ in (8), a lower bound $I_{2}^{\text{LB}}$ can be derived as

$$\begin{aligned} I_{2}&\ge \Pr \left( Z< \underset{i}{\max }\left\{ X_{i} \right\} ,\max \left[ Y_{\underline{i}},\min \left[ X_{\underline{i}},Z\right] \right]<\tau \right) \triangleq I_{2}^{\text{LB}} \\&\overset{(a)}{=}\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ Y_{i}\right\}<\tau , \min \left[ X_{i},Z\right]<\tau \right) \\&=\,\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} ,\min \left[ X_{i},Z\right]<\tau \right) }_{\triangleq J} \Pr \left( Y_{i}<\tau \right) ^{N_t}, \end{aligned}$$

(25)

which coincides with the result in (10). In step (a) we have used the DAS/SC rule for $Z < \max _i \left\{ X_{i} \right\}$, given in (6).

Appendix 3: Proof of Proposition 1

Relying on basic principles of the probability theory, the term $\rho$ defined in (24) can be elaborated as

$$\begin{aligned} \rho&= \Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ X_{i} \right\}<\tau \right) \\&=\Pr \left( Z\ge \tau ,\underset{i}{\max }\left\{ X_{i} \right\}<\tau \right) \\&\quad +\Pr \left( Z<\tau ,\underset{i}{\max }\left\{ X_{i} \right\}<Z\right) \\&=\Pr \left( Z\ge \tau \right) \Pr \left( X_{i}<\tau \right) ^{N_t} \\&\quad +\int _0^\tau f_{Z}(z) \Pr \left( X_{i}<z\right) ^{N_t} dz. \end{aligned}$$

(26)

By using this into (24), as well as the exponential PDFs and cumulative distribution functions (CDFs) of $X_i$, $Y_i$, and Z, $I_{1}^{\text{LB}}$ can be rewritten as

$$\begin{aligned} I_{1}^{\text{LB}}&=\left( 1-e^{-\frac{\tau }{\bar{Y}}}\right) ^{N_t}\left( e^{-\frac{\tau }{\bar{Z}}} \left( 1-e^{-\frac{\tau }{\bar{X}}}\right) ^{N_t} \right. \\&\qquad \left. +\underbrace{\int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}} \left( 1-e^{-\frac{z}{\bar{X}}}\right) ^{N_t} dz}_{\triangleq \sigma }\right) . \end{aligned}$$

(27)

Finally, by using the binomial theorem [19, Eq. (1.111)] to solve the integral term $\sigma$ defined in (27), a closed-form expression for $I_{1}^{\text{LB}}$ is then obtained as in (11).

Appendix 4: Proof of Proposition 2

Relying on basic principles of the probability theory, the term J defined in (25) can be elaborated as

$$\begin{aligned} J&=\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau \right) }_{\triangleq J_{1}} \\&\quad +\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , Z<\tau \right) }_{\triangleq J_{2}}\\ &\quad -\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau ,Z<\tau \right) }_{\triangleq J_{3}}. \end{aligned}$$

(28)

Below, the component terms $J_1$, $J_2$, and $J_2$ are analyzed. The term $J_1$ can be solved as

$$\begin{aligned} J_1&=\Pr \left( X_{i}<\tau \right) -\Pr \left( \underset{i}{\max }\left\{ X_{i} \right\}<Z, X_{i}<\tau \right) \\&=\Pr \left( X_{i}<\tau \right) -\Pr \left( X_{1}<Z,X_{2}<Z,\ldots ,X_{i}<Z,\ldots ,X_{N_{t}}<Z, X_{i}<\tau \right) \\&=\Pr \left( X_{i}<\tau \right) -\Pr \left( X_{1}<Z,X_{2}<Z,\ldots ,X_{i}<\min \left[ \tau ,Z\right] ,\ldots ,X_{N_{t}}<Z\right) \\&=\Pr \left( X_{i}<\tau \right) -\Pr \left( Z<\tau ,X_{1}<Z,X_{2}<Z,\ldots ,X_{i}<Z,\ldots ,X_{N_{t}}<Z\right) \\&\quad -\Pr \left( Z>\tau ,X_{1}<Z,X_{2}<Z,\ldots ,X_{i}<\tau ,\ldots ,X_{N_{t}}<Z\right) \\&=\Pr \left( X_{i}<\tau \right) -\Pr \left( Z<\tau ,\underset{i}{\max }\left\{ X_{i} \right\}<Z \right) \\&\quad -\Pr \left( Z>\tau , \underset{i}{\max }\left\{ X_{i} \right\}<Z,X_{i}<\tau \right) \\&\overset{(a)}{=}\left( 1-e^{-\frac{\tau }{\bar{X}}}\right) -\int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{a}{\bar{Z}}} \left( 1-e^{-\frac{a}{\bar{X}}}\right) ^{N_t} da \\ &\quad - \underbrace{\int _\tau ^\infty \frac{1}{\bar{Z}} e^{-\frac{b}{\bar{Z}}} \left( 1-e^{-\frac{b}{\bar{X}}}\right) ^{N_t-1}\left( 1-e^{-\frac{\tau }{\bar{X}}}\right) db}_{\triangleq \delta }, \end{aligned}$$

(29)

the term $J_2$ as

$$\begin{aligned} J_2&=\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , Z<\tau \right) \\&=\int _0^\tau f_{Z}(z) \Pr \left( z<\underset{i}{\max }\left\{ X_{i} \right\} \right) dz \\&\overset{(a)}{=}\int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}} \left( 1-\left( 1-e^{-\frac{z}{\bar{X}}}\right) ^{N_t}\right) dz, \end{aligned}$$

(30)

and the term $J_3$ as

$$\begin{aligned} J_3&=\text{Pr}\left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau ,Z<\tau \right) \\&=\text{Pr}\left( X_{i}<\tau ,Z<\tau \right) \\&\quad -\text{Pr}\left( Z>\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau ,Z<\tau \right) \\&=\text{Pr}\left( X_{i}<\tau ,Z<\tau \right) \\&\quad -\int _0^\tau f_Z\left( z\right) \text{Pr}\left( z>\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau \right) dz \\&\overset{(a)}{=}\,\left( 1-e^{-\frac{\tau }{\bar{X}}}\right) \left( 1-e^{-\frac{\tau }{\bar{Z}}}\right) \\&\quad -\int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}}\left( 1-e^{-\frac{z}{\bar{X}}}\right) ^{N_t}dz, \end{aligned}$$

(31)

where in step (a) of the above expressions we have used the exponential PDFs and CDFs of $X_i$ and Z. Finally, by substituting (29)–(31) into (28) and then into (25), with use of the exponential CDF of $Y_i$, a single-fold integral form expression for $I_{2}^{\text{LB}}$ is then obtained as in (12).

Appendix 5: Proof of Proposition 3

By using the MacLaurin series of the exponential function [19, Eq. (1.211.1)] into the term $\sigma$ defined in (27), we obtain

$$\begin{aligned} \sigma&\simeq \int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}} \left( \frac{z}{\bar{X}}\right) ^{N_t} dz \\&=\frac{1}{\bar{Z}} \left( \frac{1}{\bar{X}}\right) ^{N_t} \sum _{n=0}^{\infty } \left( \frac{1}{\bar{Z}}\right) ^{n} \frac{1}{n!} \frac{\tau ^{N_t+n+1}}{N_t+n+1}. \end{aligned}$$

(32)

Then, by substituting (32) into (27), and by using again the MacLaurin series of the exponential function, $I_1^{\text{LB}}$ can be asymptotically expressed as

$$\begin{aligned} I_{1}^{\text{LB}}\simeq \left( \frac{\tau }{\bar{Y}}\right) ^{N_t}\left( \left( 1-\frac{\tau }{\bar{Z}} \right) \left( \frac{\tau }{\bar{X}}\right) ^{N_t} +\frac{1}{\bar{Z}} \left( \frac{1}{\bar{X}}\right) ^{N_t}\frac{\tau ^{N_t+1}}{N_t+1}\right) . \end{aligned}$$

(33)

Finally, by preserving only the lowest-order terms in (33), it reduces to $\tilde{I}_1^{\text{LB}}$ as in (13).

Appendix 6: Proof of Proposition 4

By using the MacLaurin series of the exponential function into the term $\delta$ defined in (29), we obtain

$$\begin{aligned} \delta&= \left( 1-e^{-\frac{\tau }{\bar{X}}}\right) \int _\tau ^\infty \frac{1}{\bar{Z}} e^{-\frac{b}{\bar{Z}}} \left( 1-e^{-\frac{b}{\bar{X}}}\right) ^{N_t-1} db \\&= \left( 1-e^{-\frac{\tau }{\bar{X}}}\right) \left\{ \int _0^\infty \frac{1}{\bar{Z}} e^{-\frac{b}{\bar{Z}}} \left( 1-e^{-\frac{b}{\bar{X}}}\right) ^{N_t-1} db\right. \\&\quad \left. -\int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{b}{\bar{Z}}} \left( 1-e^{-\frac{b}{\bar{X}}}\right) ^{N_t-1} db \right\} \\&\simeq \frac{\tau }{\bar{Z}} B\left( \frac{\bar{X}}{\bar{Z}},N_t\right) \\&\quad -\frac{\tau }{\bar{Z}} \left( \frac{1}{\bar{X}}\right) ^{N_t+1} \sum _{n=0}^{\infty } \left( \frac{1}{\bar{Z}}\right) ^{n} \frac{1}{n!} \frac{\tau ^{N_t+n}}{N_t+n}. \end{aligned}$$

(34)

Then, by substituting (32) and (34) into (12), and by using again the MacLaurin series of the exponential function, $I_2^{\text{LB}}$ can be asymptotically expressed as

$$\begin{aligned} I_{2}^{\text{LB}}&\simeq \left( \frac{\tau }{\bar{Y}}\right) ^{N_t}\left( \frac{\tau }{\bar{X}}- \frac{\tau }{\bar{Z}} B\left( \frac{\bar{X}}{\bar{Z}},N_t\right) \right. \\&\quad -\frac{1}{\bar{Z}} \left( \frac{1}{\bar{X}}\right) ^{N_t+1} \frac{\tau ^{N_t+1}}{N_t}+\frac{\tau }{\bar{Z}}\\ &\quad \left. -\frac{1}{\bar{Z}} \left( \frac{1}{\bar{X}}\right) ^{N_t} \frac{\tau ^{N_t+1}}{N_t+1}-\frac{\tau ^2}{\bar{X}\bar{Z}}\right) . \end{aligned}$$

(35)

Finally, by preserving only the lowest-order terms in (35), it reduces to $\tilde{I}_2^{\text{LB}}$ as in (14).

Appendix 7: Proof of Proposition 5

An asymptotic expression for the outage probability is obtained by preserving only the lowest-order terms in the sum $\tilde{I}_1^{\text{LB}}+\tilde{I}_2^{\text{LB}}$. From (13) and (14), it becomes apparent that the individual diversity orders of $\tilde{I}_1^{\text{LB}}$ and $\tilde{I}_2^{\text{LB}}$ are $2 N_t$ and $N_t+1$, respectively. Therefore, only for the particular case of $N_t = 1$ these two diversity orders coincide, so that the both terms must be preserved in the asymptotic outage bound. Otherwise, when $N_t \ge 2$, the diversity order of $\tilde{I}_1^{\text{LB}}$ is higher than that of $\tilde{I}_2^{\text{LB}}$, and thus $\tilde{I}_1^{\text{LB}}$ can be ignored. Taking all this into account, we arrive at (15).

Appendix 8: Proof of Lemma 3

By using (5) into the definition of $L_1$ in (16), a lower bound $L_{1}^{\text{LB}}$ can be derived as

$$\begin{aligned} L_{1}&\ge \Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,Y_{\overline{i}}+\min \left[ X_{\overline{i}},Z\right]<\tau \right) \triangleq I_{1}^{\text{LB}} \\&= \Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,Y_{\overline{i}}+ X_{\overline{i}}<\tau \right) \\&\overset{(a)}{=} \Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ Y_{i}+X_{i}\right\} <\tau \right) , \end{aligned}$$

(36)

which coincides with the result in (17). In step (a) we have used the DAS/MRC rule for $Z\ge \max _i \left\{ X_{i} \right\}$, given in (7).

Appendix 9: Proof of Lemma 4

By using (5) into the definition of $L_2$ in (16), a lower bound $L_{2}^{\text{LB}}$ can be derived as

$$\begin{aligned} L_{2}&\ge \Pr \left( Z< \underset{i}{\max }\left\{ X_{i} \right\} ,Y_{\underline{i}}+\min \left[ X_{\underline{i}},Z\right]<\tau \right) \triangleq L_{2}^{\text{LB}} \\&\overset{(a)}{=} \Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ Y_{i}\right\} + \min \left[ X_{i},Z\right] <\tau \right) , \end{aligned}$$

(37)

which coincides with the result in (18). In step (a) we have used the DAS/MRC rule for $Z < \max _i \left\{ X_{i} \right\}$, given in (7).

Appendix 10: Proof of Proposition 6

Relying on basic principles of the probability theory, the lower bound $L_{1}^{\text{LB}}$ in (36) can be elaborated as

$$\begin{aligned} L_{1}^{\text{LB}}&=\Pr \left( Z\ge \underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ Y_{i}+X_{i}\right\}<\tau \right) \\&=\int _0^\infty f_{Z}(z) \Pr \left( \underset{i}{\max }\left\{ X_{i} \right\} \le z,\underset{i}{\max }\left\{ Y_{i}+X_{i}\right\}<\tau \right) dz \\&=\int _0^\infty f_{Z}(z) \Pr \left( \underset{i}{\max }\left\{ X_{i} \right\} \le \min \left[ z,\tau \right] ,\underset{i}{\max }\left\{ Y_{i}+X_{i}\right\}<\tau \right) dz \\&=\int _0^\tau f_{Z}(z) \Pr \left( \underset{i}{\max }\left\{ X_{i} \right\} \le z ,\underset{i}{\max }\left\{ Y_{i}+X_{i}\right\}<\tau \right) dz \\&\quad +\int _\tau ^\infty f_{Z}(z) \Pr \left( \underset{i}{\max }\left\{ X_{i} \right\} \le \tau ,\underset{i}{\max }\left\{ Y_{i}+X_{i}\right\}<\tau \right) dz \\&=\int _0^\tau f_{Z}(z) {\left( \int _0^z f_{X_i}(x) \Pr \left( Y_{i}+x<\tau \right) dx\right) } ^{N_t} dz \\&\quad +\int _\tau ^\infty f_{Z}(z) {\left( \int _0^\tau f_{X_i}(x) \Pr \left( Y_{i}+x<\tau \right) dx\right) }^{N_t} dz \\&\overset{(a)}{=}\int _0^\tau f_{Z}(z) {\underbrace{\left( 1- e^{-\frac{z}{\overline{X}}}-\frac{1}{\overline{X}}e^{-\frac{\tau }{\overline{Y}}} \int _0^z e^{-x\left( \frac{1}{\overline{X}}-\frac{1}{\overline{Y}}\right) }dx\right) }_{\triangleq \alpha _1}} ^{N_t} dz \\&\quad +\int _\tau ^\infty f_{Z}(z) {\underbrace{\left( 1- e^{-\frac{\tau }{\overline{X}}}-\frac{1}{\overline{X}}e^{-\frac{\tau }{\overline{Y}}} \int _0^\tau e^{-x\left( \frac{1}{\overline{X}}-\frac{1}{\overline{Y}}\right) }dx\right) }_{\triangleq \alpha _2}}^{N_t} dz, \end{aligned}$$

(38)

where in step (a) we have used the exponential PDFs and CDFs of $X_i$ and $Y_i$. Finally, by using the exponential PDF of Z into (38), a single-fold integral-form expression for $L_{1}^{\text{LB}}$ is then obtained as in (19).

Appendix 11: Proof of Proposition 7

Relying on basic principles of the probability theory, the lower bound $L_{2}^{\text{LB}}$ in (37) can be elaborated as

$$\begin{aligned} L_{2}^{\text{LB}}&=\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} ,\underset{i}{\max }\left\{ Y_{i}\right\} + \min \left[ X_{i},Z\right]<\tau \right) \\&=\int _0^\tau f_{Y_i}(y) \underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} ,\min \left[ X_{i},Z\right] <\tau -y\right) }_{\triangleq K} dy, \end{aligned}$$

(39)

where the term K defined above can be split into three components as

$$\begin{aligned} K&=\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau -y\right) }_{\triangleq K_{1}} \\&\quad +\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , Z<\tau -y\right) }_{\triangleq K_{2}} \\&\quad -\underbrace{\Pr \left( Z<\underset{i}{\max }\left\{ X_{i} \right\} , X_{i}<\tau -y,Z<\tau -y\right) }_{\triangleq K_{3}}. \end{aligned}$$

(40)

By comparing (40) and (28), it becomes apparent that the definitions of $K_1$, $K_2$, and $K_3$ coincide with those of $J_1$, $J_2$, and $J_3$, respectively, except that $\tau$ (in the latter) is replaced by $\tau -y$ (in the former). Therefore, expressions for $K_1$, $K_2$, and $K_3$ can be respectively obtained as in (29), (30), and (31), by substituting $\tau -y$ for $\tau$. Finally, by applying those expressions into (40) and then into (39), with use of the exponential PDF of $Y_i$, a two-fold integral form expression for $L_{2}^{\text{LB}}$ is then obtained as in (20).

Appendix 12: Proof of Proposition 8

By using the MacLaurin series of the exponential function into the term $\alpha _1$ defined in (38), we obtain

$$\begin{aligned} \alpha _1&=1- e^{-\frac{z}{\bar{X}}}-\frac{1}{\bar{X}}e^{-\frac{\tau }{\bar{Y}}} \int _0^z e^{-x\left( \frac{1}{\bar{X}}-\frac{1}{\bar{Y}}\right) }dx \\&\simeq 1-\left( 1- \frac{z}{\bar{X}}+\frac{z^2}{2\bar{X}^2}\right) \\&\quad -\frac{1}{\bar{X}}e^{-\frac{\tau }{\bar{Y}}} \int _0^z \left( 1-x\left( \frac{1}{\bar{X}}-\frac{1}{\bar{Y}}\right) \right) dx \\&\simeq \frac{1}{2} \frac{z^2}{\bar{X}\bar{Y}}. \end{aligned}$$

(41)

The term $\alpha _2$, also defined in (38), can be obtained in the same way, by substituting $\tau$ for z. Then, by using (41) into (38), $L_1^{\text{LB}}$ can be asymptotically expressed as

$$\begin{aligned} L_{1}^{\text{LB}}&\simeq \int _0^\tau \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}} \left( \frac{1}{2} \frac{z^2}{\bar{X}\bar{Y}}\right) ^{N_t} dz \\&\quad +\int _\tau ^\infty \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}} \left( \frac{1}{2} \frac{\tau ^2}{\bar{X}\bar{Y}}\right) ^{N_t} dz. \end{aligned}$$

(42)

Finally, by solving the integrals in (42) with use of the MacLaurin series of the exponential function, and preserving only the lowest-order terms, it reduces to $\tilde{L}_1^{\text{LB}}$ as in (21).

Appendix 13: Proof of Proposition 9

To begin with, let us rewrite the lower bound $L_2^{\text{LB}}$ in (20) as

$$\begin{aligned} L_{2}^{\text{LB}}&=\underbrace{\int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( 1-e^{-\frac{y}{\bar{Y}}}\right) ^{N_t-1}\left( 1-e^{-\frac{\tau -y}{\bar{X}}}\right) dy}_{\triangleq \epsilon _1} \\&\quad +\underbrace{\int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( 1-e^{-\frac{y}{\bar{Y}}}\right) ^{N_t-1} \left( 1-e^{-\frac{\tau -y}{\bar{Z}}}\right) dy}_{\triangleq \epsilon _2} \\&\quad -\underbrace{\int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( 1-e^{-\frac{y}{\bar{Y}}}\right) ^{N_t-1}\int _{\tau -y}^\infty \frac{1}{\bar{Z}} e^{-\frac{b}{\bar{Z}}} \left( 1-e^{-\frac{b}{\bar{X}}}\right) ^{N_t-1}\left( 1-e^{-\frac{\tau -y}{\bar{X}}}\right) db dy}_{\triangleq \epsilon _3} \\&\quad -\underbrace{ \int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( 1-e^{-\frac{y}{\bar{Y}}}\right) ^{N_t-1}\int _0^{\tau -y} \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}}\left( 1-e^{-\frac{z}{\bar{X}}}\right) ^{N_t} dz dy}_{\triangleq \epsilon _4} \\&\quad -\underbrace{\int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( 1-e^{-\frac{y}{\bar{Y}}}\right) ^{N_t-1} \left( 1-e^{-\frac{\tau -y}{\bar{X}}}\right) \left( 1-e^{-\frac{\tau -y}{\bar{Z}}}\right) dy}_{\triangleq \epsilon _5}. \end{aligned}$$

(43)

Now, high-SNR asymptotic expressions for the terms $\epsilon _1$, $\epsilon _2$, $\epsilon _3$, $\epsilon _4$, and $\epsilon _5$ defined above can be obtained by appropriately using the MacLaurin series of the exponential function into each definition. For $\epsilon _1$, we obtain

$$\begin{aligned} \epsilon _1&\simeq \int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( \frac{y}{\bar{Y}}\right) ^{N_t-1}\left( 1-e^{-\frac{\tau -y}{\bar{X}}}\right) dy \\&=N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t}\int _0^\tau \left( e^{-\frac{y}{\bar{Y}}} \left( y\right) ^{N_t-1}\right. \\&\quad \left. -e^{-\frac{y}{\bar{Y}}} \left( y\right) ^{N_t-1}e^{-\frac{\tau }{\bar{X}}}e^{\frac{y}{\bar{X}}}\right) dy \\&=N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t}\int _0^\tau e^{-\frac{y}{\bar{Y}}} \left( y\right) ^{N_t-1}dy \\&\quad -N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t} e^{-\frac{\tau }{\bar{X}}} \int _0^\tau e^{-\frac{y}{\bar{Y}}} \left( y\right) ^{N_t-1}e^{\frac{y}{\bar{X}}}dy \\&=N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t}\sum _{n=0}^{\infty } \left( -\frac{1}{\bar{Y}}\right) ^{n} \frac{1}{n!} \frac{\tau ^{N_t+n}}{N_t+n} \\&\quad - N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t} \sum _{n=0}^{\infty } \left( -\frac{1}{\bar{Y}}+\frac{1}{\bar{X}}\right) ^{n} \frac{1}{n!} \frac{\tau ^{N_t+n}}{N_t+n} \\&\quad +N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t} \frac{\tau }{\bar{X}} \sum _{n=0}^{\infty } \left( -\frac{1}{\bar{Y}}+\frac{1}{\bar{X}}\right) ^{n} \frac{1}{n!} \frac{\tau ^{N_t+n}}{N_t+n} \\&\simeq \left( \frac{1}{\bar{Y}}\right) ^{N_t} \frac{1}{\bar{X}} \left( \frac{\tau ^{N_t+1}}{N_t+1} \right) . \end{aligned}$$

(44)

From (43), note that an expression for $\epsilon _2$ can be obtained by replacing $\bar{X}$ with $\bar{Z}$ into the expression for $\epsilon _1$, which gives

$$\begin{aligned} \epsilon _2\simeq \left( \frac{1}{\bar{Y}}\right) ^{N_t} \frac{1}{\bar{Z}} \left( \frac{\tau ^{N_t+1}}{N_t+1} \right) . \end{aligned}$$

(45)

In addition, expressions for $\epsilon _3$, $\epsilon _4$, and $\epsilon _5$ are obtained as follows:

$$\begin{aligned} \epsilon _3&\simeq \int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( 1-e^{-\frac{y}{\bar{Y}}}\right) ^{N_t-1} \\&\quad \int _{\tau -y}^\infty \frac{1}{\bar{Z}} e^{-\frac{b}{\bar{Z}}} \left( 1-e^{-\frac{b}{\bar{X}}}\right) ^{N_t-1}\left( 1-e^{-\frac{\tau -y}{\bar{X}}}\right) db dy \\&\simeq \int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( \frac{y}{\bar{Y}}\right) ^{N_t-1} \frac{\left( \tau -y\right) }{\bar{Z}} B\left( \frac{\bar{X}}{\bar{Z}},N_t\right) dy \\&\simeq \frac{1}{\bar{Z}}\left( \frac{1}{\bar{Y}}\right) ^{N_t} \left( \frac{\tau ^{N_{t}+1}}{N_t+1}\right) B\left( \frac{\bar{X}}{\bar{Z}},N_t\right) \end{aligned}$$

(46)

$$\begin{aligned} \epsilon _4&\simeq \int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( \frac{y}{\bar{Y}}\right) ^{N_t-1} \\&\quad \int _0^{\tau -y} \frac{1}{\bar{Z}} e^{-\frac{z}{\bar{Z}}}\left( \frac{z}{\bar{X}}\right) ^{N_t} dz dy \\&=\left( \frac{1}{\bar{Y}}\right) ^{N_t} N_t \int _0^\tau e^{-\frac{y}{\bar{Y}}} \left( y\right) ^{N_t-1} \frac{1}{\bar{Z}} \left( \frac{1}{\bar{X}}\right) ^{N_t} \\&\quad \sum _{n=0}^{\infty } \left( \frac{1}{\bar{Z}}\right) ^{n} \frac{1}{n!} \frac{\left( \tau -y\right) ^{N_t+n+1}}{N_t+n+1} dy \\&\simeq \left( \frac{1}{\bar{Y} \bar{X}}\right) ^{N_t} \frac{1}{\bar{Z}}\frac{N_t}{(N_t+1)} \\&\quad \int _0^\tau e^{-\frac{y}{\bar{Y}}} \left( y\right) ^{N_t-1}\left( \tau -y\right) ^{N_t+1} dy \\&=\left( \frac{1}{\bar{Y} \bar{X}}\right) ^{N_t} \frac{1}{\bar{Z}}\frac{N_t}{(N_t+1)} \\&\quad \sum _{k=0}^{\infty } \left( {\begin{array}{c}N_{t}+1\\ k\end{array}}\right) (-1)^k\tau ^{N_t+1-k} \int _0^\tau e^{-\frac{y}{\bar{Y}}} y^{N_t-1+k} dy \\&\simeq \left( \frac{1}{\bar{Y} \bar{X}}\right) ^{N_t} \frac{1}{\bar{Z}}\frac{ \tau ^{N_t+1} }{(N_t+1)} \sum _{k=0}^{\infty } \left( {\begin{array}{c}N_{t}+1\\ k\end{array}}\right) (-1)^k \frac{N_t}{Nt+k} \\&=\left( \frac{1}{\bar{Y} \bar{X}}\right) ^{N_t} \frac{1}{\bar{Z}}\frac{ \tau ^{N_t+1} }{(N_t+1)} \frac{\Gamma (Nt+1) \Gamma (N_t+2)}{\Gamma (2Nt+2)} \end{aligned}$$

(47)

$$\begin{aligned} \epsilon _5&\simeq \int _0^\tau N_t \frac{1}{\bar{Y}} e^{-\frac{y}{\bar{Y}}} \left( \frac{y}{\bar{Y}}\right) ^{N_t-1} \left( \frac{\tau -y}{\bar{X}}\right) \left( \frac{\tau -y}{\bar{Z}}\right) dy \\&=N_t \left( \frac{1}{\bar{Y}}\right) ^{N_t} \frac{1}{\bar{X}}\frac{1}{\bar{Z}}\int _0^\tau e^{-\frac{y}{\bar{Y}}}\left( \tau ^2 y{N_t-1} -2\tau y{N_t}+y{N_t+1} \right) dy \\&\simeq \left( \frac{1}{\bar{Y}}\right) ^{N_t} \frac{1}{\bar{X}\bar{Z}}\frac{2 \tau ^{N_t+2}}{(N_t+1)(N_t+2)}. \end{aligned}$$

(48)

Finally, by using (44)–(48) into (43), and preserving only the lowest-order terms, it reduces to $\tilde{L}_2^{\text{LB}}$ as in (22).