Average symbol error rate and ergodic capacity of switch-and-examine combining diversity receivers over the α-μ fading channel

Abstract

In this paper, we consider a switch-and-examine combining (SEC) diversity scheme operating over independent and identically distributed (i.i.d) branches assuming an α-μ fading channel. We derive expressions for the average symbol error rate (ASER) for a class of coherent modulation techniques considering this fading model as well as expressions for the ergodic capacity under the same assumptions. The results for the ASER and the ergodic capacity are shown to reduce to those previously reported in the literature for other channel models such as the Weibull model as a special case, which confirms the validity of the obtained expressions. Different aspects are studied including the effect of fading severity, the number of branches and the modulation scheme used. Also, insights on the optimal choice of the switching threshold are provided.

Appendices

Appendix I: Evaluating the integral $\int \text {erfc}(\sqrt {\psi \gamma }) ~\gamma ^{{\frac {\alpha \mu }{2}}-1}~e^{-\beta \gamma ^{\frac {\alpha }{2}}}d\gamma$

To evaluate the ASER expression in (7), one can notice that we are in need to solve an integral in the form of $\int \text {erfc}(\sqrt {\psi \gamma }) ~\gamma ^{{\frac {\alpha \mu }{2}}-1}~e^{-\beta \gamma ^{\frac {\alpha }{2}}}d\gamma$ with two different sets of limits of integration. We start evaluating this integral by performing integration by parts where $\gamma ^{{\frac {\alpha \mu }{2}}-1}~e^{-\beta \gamma ^{\frac {\alpha }{2}}}$ is integrated yielding $-\frac {2 \mathrm {\Gamma }(\mu ,\beta \gamma ^{\frac {\alpha }{2}})}{\alpha \beta ^{\mu }}$ [14, Eq. (2.33.11)] where Γ(s,x) is the upper incomplete Gamma function defined in [14, Eq. (8.350.2)] and $\text {erfc}(\sqrt {\psi \gamma })$ is differentiated yielding $\sqrt {\frac {\psi }{\pi \gamma }}~e^{-\psi \gamma }$. Hence, one can obtain:

$$\begin{array}{@{}rcl@{}} \int \text{erfc}(\sqrt{\psi\gamma}) ~\gamma^{{\frac{\alpha\mu}{2}}-1}~e^{-\beta\gamma^{\frac{\alpha}{2}}}d\gamma &= -\frac{2}{\alpha\beta^{\mu}}\text{erfc}(\sqrt{\psi\gamma})\mathrm{\Gamma}(\mu,\beta\gamma^{\frac{\alpha}{2}}) \\ &+ \frac{2}{\alpha\beta^{\mu}} \sqrt{\frac{\psi}{\pi}}\int\frac{e^{-\psi\gamma}\mathrm{\Gamma}(\mu,\beta\gamma^{\frac{\alpha}{2}})}{\sqrt{\gamma}}d\gamma \end{array} $$

(16)

The problem now is reduced to that of solving an integral of the form above. Towards that end, we start by using the following series expansion of the upper incomplete Gamma function given in [14, Eq. (8.352.4)]:

$$\begin{array}{*{20}l} \mathrm{\Gamma}(n,x)=(n-1)! e^{-x} \sum\limits_{m=0}^{n-1} \frac{x^{m}}{m!}. \end{array} $$

(17)

By substituting from (17) in (16), one can easily get

$$\begin{array}{*{20}l} \int\frac{e^{-\psi\gamma}\mathrm{\Gamma}(\mu,\beta\gamma^{\frac{\alpha}{2}})}{\sqrt{\gamma}}d\gamma &= \frac{\psi}{\pi}(\mu-1)!\sum\limits_{m=0}^{\mu-1} \frac{\beta^{m}}{m!} \\ &\quad \times \underbrace{\int \gamma^{\frac{\alpha m - 1}{2}} e^{-\psi\gamma} e^{-\beta\gamma^{\frac{\alpha}{2}}} d\gamma}_{I_{5}}. \end{array} $$

(18)

It is worth noting here that we need to solve the integral I ₅ in (18) using 2 sets of limits; one from 0 to ∞, while the other is from 0 to γ _th.

Case 1: Evaluating I ₅ from 0 to ∞:

We start by using the following expression given in [15, Eq. (2.3.2.13)]:

$$\begin{array}{*{20}l} \int\limits_{0}^{\infty} x^{\alpha-1} e^{-a x^{r}} e^{-sx} dx = U' \end{array} $$

(19)

where U ^′ is defined as:

$$ U\prime =\left\{ \begin{array}{ll} \sum\limits_{j = 0}^{q - 1} \frac{(-a)^{j}}{j!} \mathrm{\Gamma}(\alpha + r j) s^{-\alpha - r j} \times \\ {_{p+1}}F_{q} \left(1, {\Delta}(p,\alpha+rj); ({-1})^{q} z;\Delta(q,1+j)\right), \,\,\,\,\,0< r \leq 1 \\ \sum\limits_{h = 0}^{p-1} \frac{(-1)^{h}}{h!r} \mathrm{\Gamma}(\frac{\alpha+h}{r}) a^{-\frac{\alpha+h}{r}} s^{h} \times \\ {_{q+1}}F_{p} \left(1, {\Delta}(q,\frac{\alpha+h}{r}); ({-1})^{p} {z}^{-1}, {\Delta}(p,1+h)\right) , \,\,\,\,\,r > 1 \end{array}\right. $$

(20)

and _p F _q (a ₁,...,a _ℓ ;b ₁,...,b _k ;ζ) is the generalised hypergeometric function defined in [14, Eq. (9.14)]. Also, Δ(x,y)=y/x,(y+1)/x,...,(y+x−1)/x, and the argument z is defined as $z = \left (\frac {p}{s}\right )^{p} \left (\frac {a}{q}\right )^{q}$.

Case 2: Evaluating I ₅ from 0 to γ _th:

Using Taylor series expansion, one can replace $e^{-\beta \gamma ^{\frac {\alpha }{2}}}$ with $\sum \limits _{n = 0}^{\infty } \frac {(-1)^{n} \beta ^{n} \gamma ^{\frac {\alpha n}{2}}}{n!}$ and hence,

$$\begin{array}{*{20}l} \int\limits_{0}^{\gamma_{\text{th}}} \gamma^{\frac{\alpha m - 1}{2}} e^{-\psi\gamma} e^{-\beta\gamma^{\frac{\alpha}{2}}} d\gamma &= \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n} \beta^{n}}{n!} \\ &\quad \times\underbrace{\int\limits_{0}^{\gamma_{\text{th}}} \gamma^{(\frac{\alpha}{2}m + \frac{\alpha}{2}n - \frac{1}{2})} e^{-\psi\gamma} d\gamma}_{I_{6}}. \end{array} $$

(21)

I ₆ can be evaluated using the expression defined in [14, Eq. (3.351.1)] yielding

$$\begin{array}{*{20}l} I_{6} = \frac{\mathrm{\Upsilon} \left( \frac{1}{2}\left(\alpha m+\alpha n +1\right), \psi\gamma_{\text{th}} \right)}{ \psi^{\frac{1}{2}(\alpha m+\alpha n +1)}} \end{array} $$

(22)

where Υ(s,x) is the lower incomplete Gamma function defined in [14, Eq. (8.350.1)].

Substituting (22) into (21), one can get:

$$\begin{array}{*{20}l} \int\limits_{0}^{\gamma_{\text{th}}} \gamma^{\frac{\alpha m - 1}{2}} e^{-\psi\gamma} e^{-\beta\gamma^{\frac{\alpha}{2}}} d\gamma {}={} \sum\limits_{n = 0}^{\infty} \frac{(-1)^{n} \beta^{n}}{n!} \frac{\mathrm{\Upsilon} \left( \frac{1}{2}(\alpha m+\alpha n +1), \psi\gamma_{\text{th}} \right)}{ \psi^{\frac{1}{2}(\alpha m+\alpha n +1)}}. \end{array} $$

(23)

Substituting from (20) and (23) in the preceding equations, one can get the results presented in (8) and (10), respectively.

Appendix II: Evaluating the integral $\int \ln \left (1+\gamma \right ) \gamma ^{\frac {\alpha \mu }{2} -1} e^{-\beta \gamma ^{\frac {\alpha }{2}}} d\gamma$

We first consider the integral with the limits 0 and ∞. We will represent each function within this integral in terms of the Meijer’s G function. We then use [16, Eq. (21)] to obtain:

$$\begin{array}{*{20}l} &\int\limits_{0}^{\infty} \ln\left(1+\gamma\right) \gamma^{\frac{\alpha \mu}{2} -1} e^{-\beta\gamma^{\frac{\alpha}{2}}} d\gamma = \\ &\frac{\ell^{-1}\left(k\right)^{\frac{1}{2}}}{\left(2\pi\right)^{(\ell-1)+\frac{1}{2}(k-1)}} \mathbf{G}_{2\ell,k+2\ell}^{k+2\ell,\ell}\left({} \left(\frac{\beta}{k}\right)^{k}{}\Bigg\arrowvert{}\begin{array}{c} \Delta(\ell,\frac{-\alpha\mu}{2}),\Delta(\ell,1+\frac{\alpha\mu}{2}) \\ \Delta(k,0),\Delta(\ell,\frac{-\alpha\mu}{2}),\Delta(\ell,\frac{-\alpha\mu}{2})\\ \end{array}{}\right). \end{array} $$

(24)

We next consider the same integral but with the limits 0 and γ _th. First, we start executing the integral by parts to get :

$$\begin{array}{*{20}l} \int\limits_{0}^{\gamma_{\text{th}}} \mathrm{\ln}\left(1+\gamma\right) \gamma^{\frac{\alpha \mu}{2} -1} e^{-\beta\gamma^{\frac{\alpha}{2}}} d\gamma = &- \frac{\mathrm{\Gamma}\left(\mu, \beta \gamma_{\text{th}}^{\frac{\alpha}{2}}\right)}{\frac{\alpha}{2} \beta^{\mu}} \text{ln}(1 + \gamma_{\text{th}})\\& + \frac{1}{\frac{\alpha}{2} \beta^{\mu}} \underbrace{\int\limits_{0}^{\gamma_{\text{th}}} \frac{\mathrm{\Gamma}(\mu,\beta \gamma^{\frac{\alpha}{2}})}{1+\gamma} d\gamma}_{I_{7}} \end{array} $$

(25)

To solve I ₇, we start by using the series expansion of the upper incomplete Gamma function given in (17) along with the Taylor series expansion for $e^{-\beta \gamma ^{\frac {\alpha }{2}}}$ to get

$$\begin{array}{*{20}l} I_{7} = \mathrm{\Gamma}(\mu) \sum\limits_{m = 0}^{\mu - 1}\sum\limits_{n = 0}^{\infty} \frac{(-1)^{n} \beta^{m+n}}{m! n!} \underbrace{\int\limits_{0}^{\gamma_{\text{th}}} \frac{\gamma^{\frac{\alpha}{2}(m+n)}}{1+\gamma} d\gamma}_{I_{8}}. \end{array} $$

(26)

Using [14, Eq. (3.194.4)], which states that

$$\begin{array}{*{20}l} {\int\limits_{0}^{U}} \frac{x^{\zeta-1}}{\left(1+\rho x\right)^{\nu}}dx = \frac{U^{\zeta}}{\zeta} {_{2}}F_{1}\left(\nu,\zeta;1+\zeta;-\rho U\right), \end{array} $$

(27)

one can solve I ₈ easily. By substituting from the preceding equations, (14) and (15) are straightforwardly obtained.