On the capacity of multiple antenna systems

Abstract

The significant role of multiple antenna techniques is vital to enable wireless systems to support the ever-rising demand for higher data rates and reliability. Thus, investigating these systems is continually important, and one of the essential aspects of this study is analyzing the capacity of such systems to gain insight into their performance. This paper presents several closed-form formulae to express the capacity of the multiple antenna system, by introducing newly derived finite and unconditionally valid solutions. It is also mathematically describing the outage probability of multiple antenna system in several scenarios. The numerical results show the tight fit between the obtained formulae and the Monte Carlo simulation outcomes.

Appendix

Here, we define the integral $\mathcal {I}(\mu ,\alpha )$ used in (10). An intractable solution of this integral can be found in ([21], $\S ~4.222.8^{12}$, p. 534 or in $\S ~4.337.5^{12}$, p. 576). Therefore, we decided to compute a more suitable solution of this integral as elaborated in the following steps

$$\begin{aligned} \mathcal {I}(\mu ,\alpha ) = \int _{0}^{\infty }\log (1+\alpha x)\textrm{e}^{-x}x^{\mu }dx \end{aligned}$$

(A1)

We use the integration by parts to compute $\mathcal {I}(\mu ,\alpha )=\int _{0}^{\infty }udv$. So, let’s assume $u = \log (1+\alpha x)$, then we have

$$\begin{aligned} du = \frac{\alpha }{1+\alpha x} dx \end{aligned}$$

(A2)

and let $dv =\textrm{e}^{-x}x^\mu$ and assume $\tau = \mu +1$, so, we have

$$\begin{aligned} dv = \textrm{e}^{-x}x^{\tau -1} \end{aligned}$$

(A3)

Then, using ([21], $\S ~2.321.2^{11}$, p. 106) we can find v

$$\begin{aligned} v= & {} -\textrm{e}^{-x}(\tau -1)!\sum _{l=1}^{\tau }\frac{x^{\tau -l}}{(\tau -l)!}\nonumber \\= & {} -\textrm{e}^{-x}\mu !~\sum _{l=1}^{\mu +1}\frac{x^{\mu -l+1}}{(\mu -l+1)!}\nonumber \\= & {} -\textrm{e}^{-x}\mu !~\sum _{l=0}^{\mu }\frac{x^{\mu -l}}{(\mu -l)!} \end{aligned}$$

(A4)

Thus

$$\begin{aligned} \mathcal {I}(\mu ,\alpha )= & {} \int _{0}^{\infty }u dv \nonumber \\= & {} \lim _{x\rightarrow \infty }(uv)-\lim _{x\rightarrow 0}(uv)-\int _{0}^{\infty }v du \end{aligned}$$

(A5)

The first two terms of (A5) are equal to zero. Therefore,

$$\begin{aligned} \mathcal {I}(\mu ,\alpha )= & {} \int _{0}^{\infty }\textrm{e}^{-x}\mu !~\sum _{l=0}^{\mu }\frac{x^{\mu -l}}{(\mu -l)!}\times \frac{\alpha }{1+\alpha x} dx\nonumber \\= & {} \mu !~\sum _{l=0}^{\mu }\frac{\alpha }{{(\mu -l)!}}\int _{0}^{\infty }\frac{\textrm{e}^{-x}x^{\mu -l}}{1+\alpha x} dx\nonumber \\= & {} \mu !~\textrm{e}^{1/\alpha }\sum _{l=0}^{\mu }\frac{1}{{(\mu -l)!}}\left( \frac{1}{\alpha }\right) ^{\mu -l}\nonumber \\{} & {} \times ~\Gamma (\mu -l)\Gamma \left[ 1-(\mu -l+1),1/\alpha \right] \nonumber \\= & {} \mu !~\textrm{e}^{1/\alpha }\!\sum _{l=0}^{\mu }\left( \frac{1}{\alpha }\right) ^{\mu -l}\Gamma \left[ 1\!-\!(\mu \!-\!l\!+\!1),\frac{1}{\alpha }\right] \end{aligned}$$

(A6)

Now, let $n =\mu -l+1$

$$\begin{aligned} \mathcal {I}(\mu ,\alpha ) = \mu !~\textrm{e}^{1/\alpha }\sum _{l=0}^{\mu }\left( \frac{1}{\alpha }\right) ^{n-1}\Gamma \left[ 1-n,1/\alpha \right] \end{aligned}$$

(A7)

and given that $E_n(z)=z^{n-1}\Gamma (1-n,z)$, we have

$$\begin{aligned} \mathcal {I}(\mu ,\alpha )= & {} \mu !~\textrm{e}^{1/\alpha }\sum _{l=0}^{\mu }E_{\mu -l+1}\left( \frac{1}{\alpha }\right) \nonumber \\= & {} \mu !~\textrm{e}^{1/\alpha }\sum _{l=0}^{\mu }E_{l+1}\left( \frac{1}{\alpha }\right) \nonumber \\= & {} \mu !~\textrm{e}^{1/\alpha }\sum _{l=1}^{\mu +1}E_{l}\left( \frac{1}{\alpha }\right) \end{aligned}$$

(A8)

But from (A3) we realize that v can be given in another form

$$\begin{aligned} v= & {} \ -\Gamma (\mu +1,x) \end{aligned}$$

(A9)

This enables us to express $\mathcal {I}(\mu , \alpha )$ in terms of Meijer G-function as follows

$$\begin{aligned} \mathcal {I}(\mu ,\alpha )= & {} -\int _{0}^{\infty }vdu \nonumber \\= & {} \int _{0}^{\infty }\Gamma (\mu +1,x)\frac{\alpha }{1+\alpha x}dx \nonumber \\= & {} -\alpha \cdot G_{2,3}^{3,1}\left( \frac{1}{\alpha } \Bigg |\begin{array} {c} 1,2 \\ 1,1,\mu +2 \\ \end{array} \right) \nonumber \\= & {} G_{2,3}^{3,1} \left( \frac{1}{\alpha } \Bigg |\begin{array} {c} 0,1 \\ 0,0,\mu +1 \\ \end{array} \right) \end{aligned}$$

(A10)