An Analytical Framework for Multi-Antenna Wireless Networks

Abstract

This chapter presents a general analytical framework for large-scale multi-antenna wireless networks. We first introduce a general wireless network model, along with a brief survey of multi-antenna transmission techniques. Using tools from stochastic geometry, a unified framework is then presented for the tractable analysis of the multi-antenna wireless network model. To illustrate the effectiveness of this analytical framework, tractable expressions for the coverage analysis in both ad hoc and cellular networks are derived. It is shown that the presented framework makes the analysis of multi-antenna networks almost as tractable as single-antenna ones. Furthermore, it helps analytically gain key network design insights, such as revealing the impacts of the antenna size and network density.

Appendix

Proof of Corollary 3.1

Since $g\sim \mathrm {Gamma}(\kappa ,\beta )$, i.e., $f_g(u)=\frac{u^{\kappa -1}e^{-\frac{u}{\beta }}}{\beta ^\kappa \varGamma (\kappa )}$, according to (3.88), (3.89), and (3.92), we have

$$\begin{aligned} c_n&=\frac{(-s)^n}{n!}\frac{\mathrm {d}^n}{\mathrm {d}s^n}\left[ 1-\frac{\eta (s)}{\pi \lambda r_0^2}\right] \end{aligned}$$

(3.126)

$$\begin{aligned}&= -\frac{(-s)^n}{n!}\frac{\mathrm {d}^n}{\mathrm {d}s^n}\left\{ \delta (sr_0^{-\alpha })^\delta \mathbb {E}_g\left[ g^\delta \gamma (-\delta ,sr_0^{-\alpha }g)\right] \right\} \nonumber \\&=-\frac{(-s)^n}{n!}\frac{\mathrm {d}^n}{\mathrm {d}s^n}\int _1^\infty \mathbb {E}_g\left[ \exp \left( -sr_0^{-\alpha }v^{-\frac{\alpha }{2}}g\right) \right] \mathrm {d}v\\&= -\frac{(-s)^n}{n!}\int _1^\infty \left[ \frac{\mathrm {d}^n}{s^n}\frac{1}{\left( 1+\beta r_0^{-\alpha }v^{-\frac{\alpha }{2}}s\right) ^\kappa }\right] \mathrm {d}v\nonumber \\&= -\frac{\varGamma (\kappa +n)}{\varGamma (\kappa )\varGamma (n+1)}\left( \frac{\tau \beta }{\theta }\right) ^\frac{2}{\alpha }\int _{\left( \frac{\tau \beta }{\theta }\right) ^{-\frac{2}{\alpha }}}^\infty \frac{\left( v^{-\frac{\alpha }{2}}\right) ^n}{\left( 1+v^{-\frac{\alpha }{2}}\right) ^{\kappa +n}}\mathrm {d}v\nonumber \\&=\frac{\varGamma (\kappa +n)}{\varGamma (\kappa )\varGamma (n+1)}\frac{\delta }{\delta - n}\left( \frac{\tau \beta }{\theta }\right) ^n{}_2F_1\left( n+\kappa ,n-\delta ;n+1-\delta ;-\frac{\tau \beta }{\theta }\right) ,\nonumber \end{aligned}$$

(3.127)

where (3.127) follows from the definition of the lower incomplete gamma function $\gamma (s,x)$, and the last equality follows from the integral representation of the hypergeometric function [23, Sect. 9.14], which completes the proof.

Proof of Proposition 3.2

According to the two steps of applying Theorem 3.2 presented in Remark 4, first, we calculate the conditional Laplace transform, expressed as

$$\begin{aligned} \mathscr {L}(s)=\exp \left\{ -2\pi \lambda \int _0^\infty {\left( 1-\mathbb {E}_{g}[\exp (-sgv^{-\alpha })]\right) }v\mathrm {d}v\right\} . \end{aligned}$$

(3.128)

To obtain a coverage probability expression for arbitrarily distributed interferers’ power gains, we propose to swap the order of the integral and the expectation. In this way, part of the exponent is given by

$$\begin{aligned} \begin{aligned}&{=\,\,\,}2\mathbb {E}_{g}\left\{ \int _0^\infty {\left[ 1-\exp (-sgv^{-\alpha })\right] }v\mathrm {d}v\right\} \\&=\mathbb {E}_g\left\{ (sg)^\frac{2}{\alpha }\frac{2}{\alpha }\int _0^1\frac{v}{1-v}\left[ -\ln (1-v)\right] ^{-\frac{2}{\alpha }-1}\mathrm {d}v\right\} \\&=\mathbb {E}_g\left\{ (sg)^\delta \varGamma (1-\delta )\right\} . \end{aligned} \end{aligned}$$

(3.129)

Therefore, the log-Laplace transform can be written as

$$\begin{aligned} \eta (s)=-\pi \lambda \varGamma (1-\delta )s^\delta \mathbb {E}_g\left[ g^\delta \right] , \end{aligned}$$

(3.130)

and the nonzero entries of $\mathbf {A}_M$ are determined by

$$\begin{aligned} \begin{aligned} a_n&=\frac{(-s)^n}{n!}\eta ^{(n)}(s)\\&=-\frac{(-1)^n}{n!} \pi \lambda r_0^2\varGamma (1-\delta )(\delta )_n\left( \frac{\tau }{\theta }\right) ^\delta \mathbb {E}_g\left[ g^\delta \right] . \end{aligned} \end{aligned}$$

(3.131)

Since there is no need to take an expectation over $r_0$ in the ad hoc network model, the derivation steps similar to (3.97) and (3.98) are unnecessary, and the proof is complete.

Proof of Corollary 3.3

According to (3.106), the Laplace transform of noise and interference is

$$\begin{aligned} \mathscr {L}(s)=p_0=e^{\eta (s)}=\exp \left( -s\sigma _\mathrm {n}^2-\pi \lambda \varGamma (1-\delta )s^\delta \mathbb {E}_g\left[ g^\delta \right] \right) . \end{aligned}$$

(3.132)

Note that $\varGamma (1-\delta )$ is a positive term due to the fact that $0<\delta <1$. Hence, the Laplace transform $p_0$ is a convex and monotonically decreasing function with respect to the transmitter density $\lambda$.

Furthermore, according to (3.107), the signs of $\{a_n\}_{n=1}^{M-1}$ are critical, i.e.,

$$\begin{aligned} a_n= -\frac{(-1)^n}{n!}(\delta )_n\pi \lambda \varGamma (1-\delta )s^\delta \mathbb {E}_g\left[ g^\delta \right] +s\sigma _\mathrm {n}^2\mathbbm {1}(n=1). \end{aligned}$$

(3.133)

Since $(-1)^n(\delta )_n=(-\delta )^{(n)}<0$ with $(x)^{(n)}$ denoting the rising factorial, we have $a_n>0$ for $1\le n\le M$. Recall that the recursive relations between $\{p_n\}_{n=1}^{M-1}$ are

$$\begin{aligned} p_n=\sum _{i=0}^{n-1}\frac{n-i}{n}a_{n-i}p_i. \end{aligned}$$

(3.134)

Since the term $\frac{n-i}{n}a_{n-i}$ is positive, it turns out that all $\{p_n\}_{n=1}^{M-1}$ have the same monotonicity and convexity with respect to $\lambda$. Recalling that $p_\mathrm {c}(\tau )=\sum _{n=0}^{M-1}p_n$, the monotonicity and concavity in Corollary 4.3 have been proved. Next, we prove the expression (3.111).

We first write $\mathbf {A}_M^\prime$ in the form

$$\begin{aligned} \mathbf {A}_M^\prime =a_0^\prime \mathbf {I}_M+(\mathbf {A}_M^\prime -a_0^\prime \mathbf {I}_M). \end{aligned}$$

(3.135)

Since $\mathbf {A}_M^\prime$ is a lower triangular Toeplitz matrix, the second part is a nilpotent matrix, i.e., $(\mathbf {A}_M^\prime -a_0^\prime \mathbf {I}_M)^n=\mathbf {0}$ for $n\ge M$. Hence, according to the properties of matrix exponential, we have

$$\begin{aligned} e^{\mathbf {A}_M}=e^{\lambda \mathbf {A}_M^\prime }=e^{a_0^\prime \lambda }\cdot \sum _{n=0}^{M-1}\frac{1}{n!}\left[ \lambda \left( \mathbf {A}_M^\prime -a_0^\prime \mathbf {I}_M\right) \right] ^n. \end{aligned}$$

(3.136)

Since it has been shown that $a_n^\prime >0$ for $n\ge 1$, $\mathbf {A}_M^\prime -a_0^\prime \mathbf {I}_M$ is a strictly lower triangular Toeplitz matrix with all positive entries, and so are the matrices $(\mathbf {A}_M^\prime -a_0^\prime \mathbf {I}_M)^n$. Therefore,

$$\begin{aligned} \left\| e^{\lambda \mathbf {A}_M^\prime }\right\| _1= e^{a_0^\prime \lambda }\cdot \sum _{n=0}^{M-1}\frac{1}{n!}\left[ \lambda ^n\left\| \left( \mathbf {A}_M^\prime -a_0^\prime \mathbf {I}_M\right) ^n\right\| _1\right] , \end{aligned}$$

(3.137)

which completes the proof of Corollary 4.3.

Proof of Proposition 3.4

According to Theorem 3.1, the outage probability is $p_\mathrm {o}(\tau )=1-\sum _{n=0}^{M-1}\bar{p}_n$, and then we have

$$\begin{aligned} \begin{aligned} \underset{M\rightarrow \infty }{\lim }\frac{p_\mathrm {o}(M+1)}{p_\mathrm {o}(M)}&=1-\underset{M\rightarrow \infty }{\lim }\frac{\bar{p}_M}{1-\sum _{n=0}^{M-1}\bar{p}_n}\\&=1-\underset{M\rightarrow \infty }{\lim }\frac{1}{1-\sum _{n=M}^\infty \frac{\bar{p}_n}{\bar{p}_M}}. \end{aligned} \end{aligned}$$

(3.138)

Since $r_\mathrm {c}$ is the radius of convergence of the power series $\bar{P}(z)$, i.e., $r_\mathrm {c}=\underset{n\rightarrow \infty }{\lim }\frac{\bar{p}_n}{\bar{p}_{n+1}}$, the above equation can be further simplified as

$$\begin{aligned} \underset{M\rightarrow \infty }{\lim }\frac{p_\mathrm {o}(M+1)}{p_\mathrm {o}(M)}=1-\underset{M\rightarrow \infty }{\lim }\frac{1}{\sum _{n=0}^\infty \left( \frac{1}{r_\mathrm {c}}\right) ^n}=\frac{1}{r_\mathrm {c}}. \end{aligned}$$

(3.139)

According to (3.126), the coefficients in the power series C(z) are given by

$$\begin{aligned} c_n=\frac{(-s)^n}{n!}c_0^{(n)}(s), \end{aligned}$$

(3.140)

where $c_0(s)=-\delta (sr_0^{-\alpha })^\delta \mathbb {E}_g\left[ g^\delta \gamma (-\delta ,sr_0^{-\alpha }g)\right]$. By reversely applying the Taylor expansion, the power series C(z) can be written as

$$\begin{aligned} C(z)=\sum _{n=0}^\infty c_n z^n=\sum _{n=0}^\infty \frac{(-sz)^n}{n!}c_0^{(n)}(s)=c_0((1-z)s). \end{aligned}$$

(3.141)

Recalling that in (3.97) we proved that $\bar{P}(z)=\frac{1}{C(z)}$, thus the radius of convergence of $\bar{P}(z)$ is the solution of the equation $C(r_\mathrm {c})=c_0((1-r_\mathrm {c})s)=0$, which is equivalent to (3.120).

Next, we prove that the solution $r_\mathrm {c}$ to equation (3.120) is larger than 1. The left-hand side of (3.120) can be rewritten as

$$\begin{aligned} \begin{aligned}\mathbb {E}_g\left[ {}_1F_1\left( -\delta ;1-\delta ;\frac{(r_\mathrm {c}-1)\tau }{\theta }g\right) \right] 1+\delta \mathbb {E}_g\left[ \int _0^1\frac{1-e^{\frac{(r_\mathrm {c}-1)\tau }{\theta }gv}}{v^{1+\delta }}\mathrm {d}v\right] . \end{aligned} \end{aligned}$$

(3.142)

Since $0<\delta <1$, $\tau >0$, $\beta >0$, and g is assumed as a nonnegative random variable with arbitrary distributions, it is seen from (3.142) that $C(r_\mathrm {c})$ is a monotonically decreasing function of $r_\mathrm {c}$. Furthermore, it is easy to check that, when $r_\mathrm {c}=1$, we have $C(1)=1$. Following the monotonicity of $C(r_\mathrm {c})$ and the fact that $C(1)>0$, we conclude that there exists only one solution of (3.120) that is larger than 1.

Proof of Proposition 3.5

According to (3.131), we have

$$\begin{aligned} \begin{aligned} A(z)&=\sum _{n=0}^\infty a_n z^n =\sum _{n=0}^\infty \frac{(-sz)^n}{n!}\eta ^{(n)}(s)=\eta ((1-z)s)\\&=-\pi \lambda r_0^2\varGamma (1-\delta )\mathbb {E}_g\left[ g^\delta \right] (1-z)^\delta =-\mu (1-z)^\delta . \end{aligned} \end{aligned}$$

(3.143)

Therefore, with the formulas (3.95) and (3.97), we have the closed-form expression

$$\begin{aligned} \bar{P}(z)=P(z)=e^{{A(z)}}=e^{-\mu (1-z)^\delta }. \end{aligned}$$

(3.144)

When $\alpha =4$, i.e., $\delta =1/2$, the power series $\bar{P}(z)$ is given by $\bar{P}(z)=\sum _{n=0}^\infty \bar{p}_nz^n=e^{-\mu \sqrt{1-z}}$. According to the definition of the modified Bessel function of the second kind $K_n(x)$ [45, Ch. 9.6], we have

$$\begin{aligned} \bar{p}_n=\sqrt{\frac{2\mu }{\pi }}\frac{(\mu /2)^n}{n!}K_{n-\frac{1}{2}}(\mu ). \end{aligned}$$

(3.145)

Then, define the ratio to test the monotonicity as

$$\begin{aligned} \frac{\bar{p}_{n+1}}{\bar{p}_n}&=\frac{\mu }{2(n+1)}\frac{K_{n+\frac{1}{2}}(\mu )}{K_{n-\frac{1}{2}}(\mu )}\nonumber \\&\le \frac{n+\sqrt{n^2+\mu ^2}}{2(n+1)}, \end{aligned}$$

(3.146)

where the inequality adopted in (3.146) comes from [46, Th. 1]. Finally, it can be checked that $\frac{n+\sqrt{n^2+\mu ^2}}{2(n+1)}<1$ when $n>\frac{\mu ^2}{4}-1$, which completes the proof.

Proof of Proposition 3.6

By performing coefficient extraction to (3.144), we have

$$\begin{aligned} \begin{aligned} \bar{P}(z)&=e^{\mu (1-z)^\delta }=\sum _{k=0}^\infty \frac{\mu ^k(1-z)^{\delta k}}{k!}\\&=\sum _{k=0}^\infty \frac{\mu ^k}{k!}\sum _{n=0}^\infty \frac{(-1)^n}{n!}(\delta k)_nz^n \sum _{n=0}^\infty \left[ \frac{(-1)^n}{n!}\sum _{k=0}^\infty \frac{\mu ^k}{k!}(\delta k)_n\right] z^n, \end{aligned} \end{aligned}$$

(3.147)

and it follows that

$$\begin{aligned} \bar{p}_n&=\frac{(-1)^n}{n!}\sum _{k=0}^\infty \frac{(-\mu )^k}{k!}(\delta k)_n \nonumber \\&=\frac{(-1)^n}{n!}\sum _{k=0}^\infty \frac{(-\mu )^k}{k!} \sum _{p=0}^n\rho (n,p)(\delta k)^p \end{aligned}$$

(3.148)

$$\begin{aligned}&=\frac{(-1)^n}{n!}\sum _{p=0}^n\rho (n,p)\delta ^p\sum _{k=0}^\infty \frac{(-\mu )^k}{k!} k^p \nonumber \\&=\frac{(-1)^ne^{-\mu }}{n!}\sum _{k=1}^n\rho (n,k)T_k(-\mu )\delta ^k, \end{aligned}$$

(3.149)

where steps (3.148) and (3.149) reversely apply the definition of the Stirling numbers of the first kind and the Touchard polynomial, respectively.