Moment Generating Function Based Performance Analysis of Maximal-Ratio Combining Diversity Receivers in the Generalized-K Fading Channels

Abstract

In this paper, we have analyzed the performance of maximal ratio combing (MRC) diversity receiver of the wireless communication systems over the composite fading environment, which is modelled by using the generalized-K distribution. However, this distribution has been considered as a versatile distribution for the precise modelling of a great variety of the short-term fading in conjunction with the long-term fading (shadow fading) channel conditions. In this proposed analysis, we have derived the mathematical expression for the moment generating function (MGF) of the generalized-K fading channel model that is used to evaluate a novel closed-form expression of the average bit error rate for (BER) the binary phase-shift keying /binary frequency-shift keying and average symbol error rate (SER) for the rectangular quadrature amplitude modulation scheme. We have also derived the mathematical expressions for the outage probability as well as the channel capacity for the generalized-K fading channel model.

Appendices

Appendix: MGF of Generalized-K Fading Channel

Equation (4) can be obtained by substituting the value of $f_\gamma (\gamma )$ from Eq. (2) into Eq. (3) and the MGF of generalized K-fading channel can be written as:

$$\begin{aligned} M_\gamma (s)=\frac{2\left( { \Xi } \right) ^{{(\alpha +1)}/2} {}{}}{\Gamma (mL)\Gamma (k)}\int \limits _0^\infty {\exp (-s\gamma ){}{}} {}{}(\gamma )^{{(\alpha -1)}/2}K_\beta \left[ {2\sqrt{ \Xi \gamma }} \right] d\gamma \end{aligned}$$

(20)

By putting $K_\beta \left( {2\sqrt{\Xi \gamma }\cdot } \right)$ as $\frac{1}{2}G{\small \begin{array}{ll} 2&{} 0 \\ 0&{} 2 \\ \end{array} }\left[ { {\Xi \gamma } \bigg |{\small \begin{array}{l} {\small {\begin{array}{ll} &{} \\ \end{array} }} \\ {\small {\begin{array}{cc} {\beta /2}&{} {{-\beta }/2} \\ \end{array} }} \\ \end{array} }} \right]$ [16, Equation (8.4.23.1)] in Eq. (20), we get:

$$\begin{aligned} M_\gamma (s)=\frac{\left( \Xi \right) ^{(\alpha +1)/2}}{\Gamma (mL)\Gamma (k)}\int \limits _0^\infty {\gamma ^{(\alpha -1)/2}e^{-s\gamma }} G{\begin{array}{ll} 2&{} 0 \\ 0&{} 2 \\ \end{array} }\left[ { {\Xi \gamma } \bigg |{\begin{array}{ll} {{\begin{array}{ll} &{} \\ \end{array} }} \\ {{\begin{array}{cc} {\beta /2}&{} {{-\beta }/2} \\ \end{array} }} \\ \end{array} }} \right] {}{}d\gamma \end{aligned}$$

(21)

By putting $\int _0^\infty {x^{s-1}} e^{-\sigma {}x}G{\small \begin{array}{ll} m&{} n \\ p&{} q \\ \end{array} }\left[ {\omega {}x\left| {\small {\begin{array}{c} {\left( {a_p } \right) } \\ {\left( {b_q } \right) } \\ \end{array} }} \right. } \right] dx=\sigma ^{-s}G{\small \begin{array}{cc} m&{} {n+1} \\ {p+1}&{} q \\ \end{array} }\left[ { {\frac{\omega }{\sigma } } \bigg |{\small \begin{array}{ll} {\small {\begin{array}{cc} {1-s}&{} {\left( {a_p } \right) } \\ \end{array} }} \\ {\small {\begin{array}{cc} {\left( {b_q } \right) }&{} \\ \end{array} }} \\ \end{array} }} \right]$ form [16, Equation (2.24.3.1)] into Eq. (21) provides equation (4).

Appendix: Average BER for BPSK/BFSK Modulation Scheme

For the proof of Eq. (6), we assume, $I_1 =\int _0^{\pi /2} {M_\gamma } \left( {\frac{g}{\sin ^{2}\theta }} \right) d\theta$ in Eq. (5). By using Eq. (4), $I_{1}$ can be expressed as:

$$\begin{aligned} I_1 =\frac{\left( { \Xi } \right) ^{\frac{\alpha +1}{2}}1}{\pi \Gamma (mL)\Gamma (k)}\int \limits _0^{\frac{\pi }{2}} { \left( {\frac{g}{\sin ^{2}\theta }} \right) ^{-\frac{(\alpha +1)}{2}}G{\begin{array}{ll} 2&{} 1 \\ 1&{} 2 \\ \end{array} }\left[ {\left. {\frac{ \Xi \sin ^{2}\theta }{g}} \right| {\begin{array}{cc} {{(1-\alpha )}/2}&{} \\ {\beta /2}&{} {-\beta /2} \\ \end{array} }} \right] } d\theta \end{aligned}$$

(22)

By changing the variable $t=\sin ^{2}\theta$ and after some mathematical manipulation, Eq. (22) can be expressed as:

$$\begin{aligned} I_1 =\frac{\left( { \Xi /g} \right) ^{\frac{\alpha +1}{2}}}{2\pi \Gamma (mL)\Gamma (k)}\int \limits _0^1 { \left( t \right) ^{\frac{(\alpha )}{2}}\left( {1-t} \right) ^{{-1}/2}G{\begin{array}{l@{\quad }l} 2&{} 1 \\ 1&{} 2 \\ \end{array} }\left[ {\left. {\frac{ \Xi t}{g}} \right| {\begin{array}{cc} {{(1-\alpha )}/2}&{} \\ {\beta /2}&{} {-\beta /2} \\ \end{array} }} \right] } dt \end{aligned}$$

(23)

By using $\int _0^1 {x^{s-1}} \left( {1-x} \right) ^{t-1}G{\small \begin{array}{ll} m&{} n \\ p&{} q \\ \end{array} }\left[ {\omega {}x\left| {{\small \begin{array}{cc} {\left( {a_p } \right) } \\ {\left( {b_q } \right) } \\ \end{array} }} \right. } \right] dx\!=\!\Gamma (t)G{\small \begin{array}{cc} m&{} {n\!+\!1} \\ {p\!+\!1}&{} {q\!+\!1} \\ \end{array} }\left[ {\omega \left| {{\small \begin{array}{cc} {{\small \begin{array}{cc} {1\!-\!s}&{} {\left( {a_p } \right) } \\ \end{array} }} \\ {{\small \begin{array}{cc} {\left( {b_q } \right) }&{} {1\!-\!s\!-\!t} \\ \end{array} }} \\ \end{array} }} \right. } \right]$ from [16, Equation (2.24.2)], the Eq. (23) can be expressed as:

$$\begin{aligned} I_1 =\frac{ \left( {\Xi /g} \right) ^{\frac{(1+\alpha )}{2}}\sqrt{\pi }}{2 \Gamma (mL)\Gamma (k)}G{\begin{array}{l@{\quad }l} 2&{} 2 \\ 2&{} 3 \\ \end{array} }\left[ {\left. {\frac{ \Xi }{g}} \right| {\begin{array}{c@{\quad }c} {{-{}{}{}\alpha }/2}&{} {{(1-\alpha )}/2} \\ {\beta /2}&{} {{\begin{array}{c@{\quad }c} {-\beta /2}&{} {-{(\alpha +1)}/2} \\ \end{array} }} \\ \end{array} }} \right] \end{aligned}$$

(24)

By using Eqs. (5) and (24), we can obtain Eq. (6).

Appendix: Integration I$_{3}$

To derive Eq. (11), we proceed by using [15, Equation (6.643.3)] and Eq. (2), the MGF of the generalized K-fading distribution can be expressed as:

$$\begin{aligned} M_\gamma (s)= \left( {\frac{ \Xi }{s}} \right) ^{\alpha /2}\exp \left( {\frac{\Xi }{2s}} \right) W_{-\alpha /2,\beta /2} \left( {\frac{\Xi }{s}} \right) \end{aligned}$$

(25)

where $W_{-\mu ,\nu } \left( \cdot \right)$ is the Whittaker function as discussed in [15, Equation (9.220.4)]. From Eq. (10) and (25), the integral $I_{3}$ can be expressed as:

$$\begin{aligned} I_3 = \int \limits _0^{\frac{\pi }{4}} {\left( {\frac{ \Xi \sin ^{2}\theta }{g_{QAM} }} \right) ^{\alpha /2}\exp \left( {\frac{\Xi \sin ^{2}\theta }{2g_{QAM} }} \right) W_{-\alpha /2,\beta /2} \left( {\frac{\Xi \sin ^{2}\theta }{g_{QAM} }} \right) d\theta } \end{aligned}$$

(26)

By using the transformation $t = 2 \sin ^{2}\theta$ and after some mathematical manipulation, the Eq. (26) can be written as:

$$\begin{aligned} I_3 \!= \!\frac{1}{2\sqrt{2}} \left( {\frac{ \Xi }{2g_{QAM} }} \right) ^{\alpha /2}\int \limits _0^1 {t^{{(\alpha -1)}/2}\left( {1\!-\!t/2} \right) ^{{-1}/2}\exp \left( {\frac{\Xi t}{4g_{QAM} }} \right) W_{-\alpha /2,\beta /2} \left( {\frac{\Xi t}{2g_{QAM} }} \right) dt}\nonumber \\ \end{aligned}$$

(27)

With the help of [15, Equation (9.220.4)], [15, Equation (9.220.3)] and [15, Equation (9.220.2)] Eq. (27) can be expressed as:

$$\begin{aligned} I_3&= \frac{1}{2\sqrt{2}} \left( {\frac{ \Xi }{2g_{QAM} }} \right) ^{\alpha /2}\int \limits _0^1 {t^{{(\alpha -1)}/2}\left( {1-t/2} \right) ^{{-1}/2} \left[ {\frac{\Gamma (-\beta )}{\Gamma (1/2-\beta /2+\alpha /2)}} \right. } \left( {\frac{\Xi t}{2g_{QAM} }} \right) ^{{(\beta +1)}/2}\nonumber \\&\quad \times 1F1\left( {\frac{\alpha +\beta +1}{2},\beta +1,\frac{\Xi t}{2g_{QAM} }} \right) +\frac{\Gamma (\beta )}{\Gamma (1/2+\beta /2+\alpha /2)}\left( {\frac{\Xi t}{2g_{QAM} }} \right) ^{{(-\beta +1)}/2}\nonumber \\&\quad \left. {\times 1F1\left( {\frac{\alpha -\beta +1}{2},\beta +1,\frac{\Xi t}{2g_{QAM} }} \right) } \right] dt \end{aligned}$$

(28)

where, 1F$_{1}$ is the confluent hpergeometric function as discussed in [15, Equation (9.210.1)]. By using [15, Equation (9.210.1)] and [15, Equation (9.111)], Eq. (28) can be written as Eq. (11).

Appendix: Outage Prabability

To derive Eq. (16), we proceed by substituting the value of $f_\gamma (\gamma )$ from Eq. (2) into Eq. (15), $P_{out}$ can be expressed as:

$$\begin{aligned} P_{out} =1-\int \limits _{\gamma _{th} }^\infty {\frac{2\left( \gamma \right) ^{{(\alpha -1)}/2}}{\Gamma (mL)\Gamma (k)}\left( \Xi \right) ^{{\left( {\alpha +1} \right) }/2}} K_\beta \left[ {2\sqrt{\Xi \gamma }} \right] d\gamma \end{aligned}$$

(29)

By replacing $K_\beta \left( \cdot \right)$ as $\frac{1}{2}G{\begin{array}{ll} 2&{} 0 \\ 0&{} 2 \\ \end{array} }\left[ { {\Xi \gamma } \bigg |{\begin{array}{cc} {{\begin{array}{cc} &{} \\ \end{array} }} \\ {{\begin{array}{cc} {\beta /2}&{} {{-\beta }/2} \\ \end{array} }} \\ \end{array} }} \right]$ from [16, Equation (8.4.23.1)] in Eq. (29), we get:

$$\begin{aligned} P_{out} =1-\frac{\left( \Xi \right) ^{{\left( {\alpha +1} \right) }/2}}{\Gamma (mL)\Gamma (k)}\int \limits _{\gamma _{th} }^\infty {\left( \gamma \right) ^{{(\alpha -1)}/2}G{\begin{array}{ll} 2&{} 0 \\ 0&{} 2 \\ \end{array} }\left[ {\left. {\Xi \gamma } \right| {\begin{array}{cc} {{\begin{array}{cc} &{} \\ \end{array} }} \\ {{\begin{array}{cc} {\beta /2}&{} {{-\beta }/2} \\ \end{array} }} \\ \end{array} }} \right] } d\gamma \end{aligned}$$

(30)

Now, with the help of $\int _u^\infty {x^{s-1}} (x-u)^{t-1}G{\small \begin{array}{ll} m&{} n \\ p&{} q \\ \end{array} }\left[ {\omega {}x\left| {{\small \begin{array}{cc} {\left( {a_p } \right) } \\ {\left( {b_q } \right) } \\ \end{array} }} \right. } \right] dx=\Gamma (t) G{\small \begin{array}{cc} {m+1}&{} n \\ {p+1}&{} {q+1} \\ \end{array} } \left[ {\omega {}\left| {{\small \begin{array}{cc} {{\begin{array}{cc} {\left( {a_p } \right) }&{} {1-s} \\ \end{array} }} \\ {{\small \begin{array}{cc} {1-s-t}&{} {\left( {b_q } \right) } \\ \end{array} }} \\ \end{array} }} \right. } \right]$ from [16, Equation (2.24.2.3)], the Eq. (30) can be expressed as Eq. (16).

Appendix: Channel Capacity by Channel Inversion with Fixed Rate

To derive Eq. (19), we proceed by sssuming, $I_6 =\int _0^\infty {M_\gamma } \left( s \right) ds$ in Eq. (18). Using Eq. (4) $I_{6}$ can be expressed as:

$$\begin{aligned} I_6 =\int \limits _0^\infty {\frac{\left( {\Xi /s} \right) ^{\frac{\alpha +1}{2}}}{\Gamma (mL)\Gamma (k)}} G{\begin{array}{ll} 2&{} 1 \\ 1&{} 2 \\ \end{array} }\left[ {\left. {\frac{\Xi }{s}} \right| {\begin{array}{cc} {{\left( {1-\alpha } \right) }/2} \\ {{\begin{array}{cc} {\beta /2}&{} {{-\beta }/2} \\ \end{array} }} \\ \end{array} }} \right] ds \end{aligned}$$

(31)

By putting 1/s = t in Eq. (31) and using [15, Equation (7.811.4)], Eq. (31) can be written as:

$$\begin{aligned} I_6 =\frac{\Xi \Gamma \left( {{(\alpha +\beta -1)}/2} \right) \Gamma \left( {{(\alpha -\beta -1)}/2} \right) }{\Gamma (mL){}{}\Gamma (k)} \end{aligned}$$

(32)

If we substitute value of $I_{6}$ from Eq. (32) to Eq. (18) it results Eq. (19).