Performance analysis for 5G beamforming heterogeneous networks

Abstract

The heterogeneous network (HetNet) is an attractive solution to solve limited spectral efficiency, the increasing traffic demands, and crowded coverage. In 5G, massive multiple-input multiple-output and millimeter-wave technologies are considered to explore high array gain and alleviate high path loss with beamforming. In this paper, we propose a general tractable model for signal-to-interference-plus-noise ratio (SINR) analysis with cell expansion and beamforming for 5G. In this HetNet, each layer is characterized by particular parameters such as base station (BS) density, transmission power, beam gain, beamwidth, propagation loss, and bias factor. In our model, the expressions of outage probability for each layer and the whole network, which represent the SINR distributions, are derived through a typical user. We also analyze the rate performance by the minimum average user rates and the rate coverage probabilities. Furthermore, the spectrum reuse strategy, some layers are spectrum separation while others are spectrum sharing, are considered to consistent with the existing and future cellular network. From the simulations, system performance can be greatly improved by beamforming and affected by beamwidth, and the optimal bias changes with BS density and path loss. Furthermore, the proposed method provides practical deployment guidelines with some determined parameters.

Appendices

Appendix 1: Proof of Lemma 1

When $P_{r,i}>P_{r,j}$ for all the $j\in K , j\not =i$, the probability of the TPU associates with the BS in the ith layer can be given by

$$\begin{aligned} A_i&= E_{R_i}[{\mathbb {P}}[P_{r,i}(X_i)>\max \limits _{j,j\not =i}(P_{r,j})] ] \nonumber \\&=E_{R_i}\left[ \prod \limits _{j=1,j\not =i}^{K}{\mathbb {P}}[P_{r,i}(X_i)>P_{r,j}]\right] \nonumber \\&=E_{R_i}\left[ \prod \limits _{j=1,j\not =i}^{K}{\mathbb {P}}\left[ X_j>\Big (\frac{P_j}{P_i}\cdot \frac{B_j}{B_i}\cdot \frac{M_j}{M_i} \Big )^{1/{\alpha _{j}}} X_{i}^{{\alpha _i}/{\alpha _j}} \right] \right] \nonumber \\&=\int _0^\infty \prod \limits _{j=1,j\not =i}^{K} {\mathbb {P}}\left[ X_j> \widehat{V_j} ^{1/{\alpha _{j}}} r^{1/{\widehat{\alpha _j}}} \right] f_{X_i}(r)\mathrm {d}r, \end{aligned}$$

(21)

where $\widehat{V_j} =\widehat{P_j}\cdot \widehat{B_j} \cdot \widehat{M_j}$,

$$\begin{aligned}&\prod \limits _{j=1,j\not =i}^{K}{\mathbb {P}}\left[ X_j> \widehat{V_j} ^{1/{\alpha _{j}}} r^{1/{\widehat{\alpha _j}}} \right] \nonumber \\&\quad =\prod \limits _{j=1,j\not =i}^{K} e^{-\pi \lambda _j( \widehat{V_j} ^{2/{\alpha _{j}}} r^{2/{\widehat{\alpha _j}}})}, \end{aligned}$$

(22)

and

$$\begin{aligned} {f_{{X_i}}}(r) = {e^{ - \pi {\lambda _i}{r^2}}}2\pi {\lambda _i}r. \end{aligned}$$

(23)

Combining (21), (22), and (23), we have

(24)

If $\alpha _j=\alpha$, it follows that

$$\begin{aligned} A_i&= \frac{\lambda _i}{ \prod \nolimits _{j=1}^{K} \lambda _j {\widehat{V_j}}^{2/{\alpha }} }\nonumber \\&= \frac{\lambda _i}{{\mathop {\sum }\nolimits _{j = 1,j \ne i}^K} {\lambda _j}{{\widehat{V_j}}^{2/{\alpha }}} + {\lambda _i}}. \end{aligned}$$

(25)

Appendix 2: Proof of Lemma 3

$X_i>x$ is equal to $L_i>x$, the probability is

$$\begin{aligned} {\mathbb {P}}[X_i>x]&={\mathbb {P}}[L_i>x|n=i] \nonumber \\&= \frac{{\mathbb {P}}[L_i>x,n=i]}{{\mathbb {P}}[n=i]} =\frac{{\mathbb {P}}[L_i>x,n=i]}{A_i}, \end{aligned}$$

(26)

where

$$\begin{aligned}&{\mathbb {P}}[L_i>x,n=i] \nonumber \\&\quad ={\mathbb {P}}[{L_i}> x,{P_{r,i}}\left( {{L_i}} \right)> \mathop {\max }\limits _{j,j \ne i} ( {{P_{r,j}}} ) ] \nonumber \\&\quad =\int _0^\infty {\mathop \prod \limits _{j = 1,j \ne i}^K } {\mathbb {P}} [{P_{r,i}}( r )> {P_{r,j}}]{f_{{L_i}}}(r) \mathrm {d}r \nonumber \\&\quad = \int _{x}^{\infty } {\mathop \prod \limits _{j = 1,j \ne i}^K } {\mathbb {P}} [L_j > \widehat{V_j}^{1/{\alpha _j}}r^{1/\widehat{\alpha _j}}]f_{{L_i}}(r) \mathrm {d}r\nonumber \\&\quad = 2\pi {\lambda _i}\int _x^\infty r e^{ { - \pi \sum \limits _{j = 1}^K {\lambda _j}{{\widehat{V_j}}^{2/{\alpha _j}}}{r^{2/\widehat{{\alpha _j}}}}} }\mathrm {d}r. \end{aligned}$$

(27)

Put (27) into (26), we have

$$\begin{aligned} {\mathbb {P}}[X_i>x] = \frac{{2\pi {\lambda _i}}}{{{A_i}}}\int _x^\infty r e^{ { - \pi \sum \limits _{j = 1}^K {{\lambda _j}} {{\widehat{V_j}}^{2/{\alpha _j}}}{r^{2/\widehat{{\alpha _j}}}}} }\mathrm {d}r. \end{aligned}$$

(28)

The CDF of $X_i$ is $F_{{X_i}}(x)=1-{\mathbb {P}}[X_k>x]$, and PDF is

$$\begin{aligned} {f_{{X_i}}}(x)&= \frac{{\mathrm {d}{F_{{X_i}}}(x)}}{{\mathrm {d}x}} \nonumber \\&= \frac{{2\pi {\lambda _i}}}{{{A_i}}}x e^{ { - \pi \sum \limits _{j = 1}^K {{\lambda _j}} {{\widehat{V_j}}^{2/{\alpha _j}}}{x^{2/\widehat{{\alpha _j}}}}} }. \end{aligned}$$

(29)

Appendix 3: Proof of Theorem 1

From (12), we can get the outage probability of the ith layer,

$$\begin{aligned} {\mathcal {O}}_i&= 1 - \int _{0}^\infty {\mathbb {P}}{[{\text{ SINR }}_i(x)> \tau ]}{f_{{X_i}}}(x) \mathrm {d}x\nonumber \\&= 1 - \frac{{2\pi {\lambda _i}}}{{{A_i}}}\int _{0}^\infty {\mathbb {P}}{[{\text{ SINR }}_i(x) > \tau ]} \cdot x \cdot \nonumber \\&\quad \exp \left\{ - \pi \sum \limits _{j = 1}^K {\lambda _j} {{\widehat{V_j}}^{2/{\alpha _j}}}{x^{2/\widehat{{\alpha _j}}}} \right\} \mathrm {d}x, \end{aligned}$$

(30)

where $f_{{X_i}}$ is given in Lemma 3. We can rewrite the SINR in (3) as $\gamma _i (x) = {\text {SINR}}_{i}(x) = \frac{{{g_{i,0}}}}{{{x^{{\alpha _i}}}{{({P_i}{M_i}m)}^{ - 1}}Q}}$, where $Q = \sum \nolimits _{j = 1}^K \omega _{ij} {{I_j} + W}$, $I_j = \sum \nolimits _{k \in {\varPhi _j}\backslash {B_{i0}}} {{D_{jk}^*}{P_j}{h_{jk}}{{\left| {{Y_{jk}}} \right| }^{ - {\alpha _j}}}}$.

The complementary CDF of SINR of the TPU with a distance x from the associated BS in ith layer will be

$$\begin{aligned}&{\mathbb {P}}[\gamma _i (x)>\tau ] \nonumber \\&\quad ={\mathbb {P}}[{g_{i,0}} > {x^{{\alpha _i}}}{({P_i}{M_i}m)^{ - 1}}Q\tau ] \nonumber \\&\quad = \int _0^\infty \exp \left\{ { - {x^{{\alpha _i}}}{{({P_i}{M_i}m)}^{ - 1}}\tau q} \right\} {f_Q}(q) \mathrm {d}q \nonumber \\&\quad = \mathbb {E}_Q [\exp \left\{ { - {x^{{\alpha _i}}}{{({P_i}{M_i}m)}^{ - 1}}\tau q} \right\} ] \nonumber \\&\quad = \exp \left\{ \frac{-\tau }{\text {SNR}}\right\} \cdot \prod _{j=1}^K ({\mathcal{L}}_{I_j}(x^{\alpha _i}{{({P_i}{M_i}m)}^{ - 1}}\tau )) ^{\omega _{ij}}, \end{aligned}$$

(31)

the Laplace transform of $I_j$ is

$$\begin{aligned}&{\mathcal{L}}_{I_j}(x^{\alpha _i}{{({P_i}{M_i}m)}^{ - 1}}\tau ) \nonumber \\&\quad =\mathbb {E}_{I_j} \left[ e^{-x^{\alpha _i}{{({P_i}{M_i}m)}^{ - 1}}\tau I_j} \right] \nonumber \\&\quad =\mathbb {E}_{\varPhi _j} \left[ e^{ - {x^{{\alpha _i}}}\widehat{{P_j}}{({M_j}m)^{ - 1}}\tau \sum \limits _{k \in {\varPhi _j}} {{D_{jk}^*}{h_{jk}}{{\left| {{Y_{jk}}} \right| }^{ - {\alpha _j}}}} } \right] \nonumber \\&\quad =\mathbb {E}_{\varPhi _j} \left[ \prod \limits _{k \in {\varPhi _j}} {\exp \left\{ { - {x^{{\alpha _i}}}\widehat{{P_j}}\tau \frac{{{D_{jk}^*}}}{{{M_j}m}}{h_{jk}}{{\left| {{Y_{jk}}} \right| }^{ - {\alpha _j}}}} \right\} } \right] , \end{aligned}$$

(32)

From (32), the formula can be further deduced by

$$\begin{aligned}&{\mathcal{L}}_{I_j}(x^{\alpha _i}{{({P_i}{M_i}m)}^{ - 1}}\tau ) \nonumber \\&\quad \overset{(a)}{=} \exp \left\{ - 2\pi {\lambda _j} \int _{{z_j}}^\infty {(1 - {{\mathcal{L}}_{{h_j}}}({x^{{\alpha _i}}}\widehat{P_j}\tau \frac{{{D_{j}^*}}}{{M_j}m}{y^{ - {\alpha _j}}})y \mathrm {d}y)} \right\} \nonumber \\&\quad \overset{(b)}{=} \exp \left\{ { - 2\pi {\lambda _j} \int _{{z_j}}^\infty {(1 - \frac{1}{{1 + {x^{{\alpha _i}}}\widehat{{P_j}}\tau \frac{{{D_{j}^*}}}{{{M_j}m}}{y^{ - {\alpha _j}}}}}y \mathrm {d}y)} } \right\} \nonumber \\&\quad = \exp \left\{ { - 2\pi {\lambda _j} \int _{{z_j}}^\infty {\frac{y}{{1 + {{\left( {{x^{{\alpha _i}}}\widehat{{P_j}}\tau \frac{{{D_{j}^*}}}{{{M_j}m}}} \right) }^{ - 1}}{y^{{\alpha _j}}}}}\mathrm {d}y} } \right\} , \end{aligned}$$

(33)

where (a) comes from

$E[\prod \limits _{x \in \varPhi } {f(x)} ] = \exp \left\{ { - \lambda \int _{{R^2}} {[1 - f(x)] \mathrm {d}x} } \right\}$, and (b) follows ${h_j}\sim \exp (1)$. The interference signals are from the radius $z_j$ of the circle to the infinity, where ${z_j} = {\widehat{V_j}^{1/{\alpha _j}}}{x^{1/\widehat{{\alpha _j}}}}$. Let $u = {\left( {{x^{{\alpha _i}}}\widehat{{P_j}}\tau \frac{{{D_{j}^*}}}{{{M_j}m}}} \right) ^{ - 2/{\alpha _j}}}{y^2}$, then

$$\begin{aligned}&{\mathcal{L}}_{I_j}(x^{\alpha _i}{{({P_i}{M_i}m)}^{ - 1}}\tau ) \nonumber \\&\quad =\exp \left\{ { - \pi {\lambda _j}{{\left( {\widehat{{P_j}}\tau \frac{{{D_{j}^*}}}{{{M_j}m}}} \right) }^{2/{\alpha _j}}} {x^{2/\widehat{{\alpha _j}}}} \int _{m'}^\infty {\frac{1}{{1 + {u^{{\alpha _j}/2}}}}du} } \right\} \nonumber \\&\quad = \exp \left\{ - \pi {\lambda _j}{\left( {\widehat{P_j}\frac{{{D_{j}^*}}}{{{M_j}m}}} \right) ^{2/{\alpha _j}}}{x^{2/\widehat{\alpha _j}}}Z(\tau ,{\alpha _j},\widehat{B_j}) \right\} , \end{aligned}$$

(34)

where

$$\begin{aligned} Z(\tau ,{\alpha _j},\widehat{{B_j}}) = {\tau ^{2/{\alpha _j}}}\int _{m'}^\infty {\frac{1}{{1 + {u^{{\alpha _j}/2}}}}du}. \end{aligned}$$

(35)

Taking (34) into (31), we have

$$\begin{aligned}&{\mathbb {P}} [{\gamma _i}(x) > \tau ] \nonumber \\&\quad = \exp \left\{ { \frac{- \tau }{\text{ SNR }} - \pi \sum \limits _{j = 1}^K \omega _{ij} {{\lambda _j}} {{\left( {\widehat{{P_j}}\frac{{{D_{j}^*}}}{{{M_j}m}}} \right) } ^{2/{\alpha _j}}} {x^{2/\widehat{{\alpha _j}}}}Z(\tau ,{\alpha _j},\widehat{{B_j}})} \right\} . \end{aligned}$$

(36)

Combining (36) and (30), the outage probability of a layer can be obtained from (13). Moreover, by taking (13) into (11), the outage probability of the network can be got from (14).