SCIENCE CHINA Information Sciences, Volume 61 , Issue 2 : 022302(2018) https://doi.org/10.1007/s11432-016-9042-8

Fractional full duplex cellular network: a stochastic geometry approach

Wenping BI 1,2,3, Limin XIAO 1,2,3, Xin SU 1,2,3, Shidong ZHOU 1,2,3,*
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  • ReceivedNov 13, 2016
  • AcceptedFeb 28, 2017
  • PublishedJul 28, 2017



This work was supported by National Basic Research Program of China (Grant No. 2012CB316002), National Natural Science Foundation of China (Grant No. 61631013), National High Technology Research and Development Program of China (863 Program) (Grant No. 2015AA01A706), National Natural Science Foundation of China (Grant No. 61321061), Tsinghua University Initiative Scientific Research Program (Grant No. 2015Z02-3), National S&T Major Project (Grant No. 2014ZX03001011), Key Project of International Science and Technology Innovation Cooperation Between the Government (Grant No. 2016YFE0122900), and Huawei Technologies.



Proof of Lemma sect. 3.1

Starting with the definition of laplace transform, we can get \begin{equation} \begin{aligned} {L_{{I_{{\rm h}q}}}}\left( s \right) \mathop = \limits^{\left( {\rm a} \right)} {{{\rm E}}_{{\Phi _{{\rm h}q}}}}\left\{ {\prod\limits_{z \in {\Phi _{{\rm h}q}}\backslash c_o} {{{{\rm E}}_{{h_{z,d_o}}}}\exp \left( { - s{h_{z,d_o}}D_{z,d_o}^{ - {\alpha _{\rm bu}}}} \right)} } \right\}\mathop = \limits^{\left( {\rm b} \right)} \exp \left\{ { - 2\pi {\lambda _{\rm s}}\int\nolimits_R^\infty {\left\{ {1 - {{{\rm E}}_{{\rm h}}}\left[ {{{\mathop{\rm e}\nolimits} ^{\left( { - s{h}v^{ - {\alpha _{\rm bu}}}} \right)}}} \right]} \right\}v{\rm d}v} } \right\}, \end{aligned} \tag{23}\end{equation} where $c_o$ and $d_o$ are $b_o$ ($u_o$) and $u_o$ ($b_o$) if $q=d(u)$, respectively. (a) follows the fact that $\Phi~_{{\rm~h}q}$ and $h_{z,d_o}$ are independent from each other. Therefore, the expectation order can be exchanged. (b) follows from the probability generating function of PPP. Carrying on the proof in (23), the followings can be obtained \begin{align} &\exp \left\{ { - 2\pi {\lambda _{\rm s}}\int\nolimits_R^\infty {\left\{ {1 - {{{\rm E}}_{\rm h}}\left[ {\exp \left( { - sh{v^{ - {\alpha _{\rm bu}}}}} \right)} \right]} \right\}v{\rm d}v} } \right\} \\ &\mathop = \limits^{\left( {\rm c} \right)} \exp \left\{ - \frac{2}{{{\alpha _{\rm bu}}}}\pi {\lambda _{\rm s}}{{{\rm E}}_{\rm h}}\left\{ - {{\left( {sh} \right)}^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}}} \right)- \frac{{{\alpha _{\rm bu}}}}{2}{R^2} + {{\left( {hs} \right)}^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}},hs{R^{ - {\alpha _{\rm bu}}}}} \right) \right\} \right\}, \tag{24} \end{align} where (c) is proofed in [17]. Based on the Rayleigh fading assumption, $h\sim~\exp(1)$, we can obtain the expectations as \begin{eqnarray}&\displaystyle{{{\rm E}}_{\rm h}}\left\{ {{{ h }^{\frac{2}{{{\alpha _{\rm bu}}}}}}} \right\} = \int\nolimits_0^\infty {{{ x }^{\frac{2}{{{\alpha _{\rm bu}}}}}}{{\rm e}^{ - x}} {\rm d}x} \mathop = \limits^{\left( {\rm d} \right)} \Gamma \left( {\frac{2}{{{\alpha _{\rm bu}}}} + 1} \right) , \tag{25} \\ &\displaystyle{ {\rm E}_{\rm h}}\left\{ {{{ h}^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}},hs{R^{ - {\alpha _{\rm bu}}}}} \right)} \right\}= \int\nolimits_0^\infty {{x^{\frac{2}{{{\alpha _{\rm bu}}}}}}\Gamma \left( { - \frac{2}{{{\alpha _{\rm bu}}}},xs{R^{ - {\alpha _{\rm bu}}}}} \right){{\rm e}^{ - x}}{\rm d}x}\mathop = \limits^{\left( {\rm e} \right)} J\left( {\frac{2}{{{\alpha _{\rm bu}}}} + 1,1, - \frac{2}{{{\alpha _{\rm bu}}}},s{R^{ - {\alpha _{\rm bu}}}}} \right), \tag{26} \end{eqnarray} where $J(\mu~,\beta~,\nu~,\gamma~)$ is defined in (11) and proofed in [17]. (d) and (e) can be obtained by Eqs. (3.478) and (3.381) 3), respectively. Substitute (25) and (26) into (24) and we can get the result of Lemma sect. 3.1.

Gradshteyn I, Ryzhik I. Table of Integrals, Series, and Products. Manhattan: Academic Press, 2014. 346–370.

Proof of Lemma 3.2

Starting with the definition of laplace transform, we can get \begin{align} &{L_{{I_{{\rm f}q}}}}\left( s \right) \mathop = \limits^{\left({\rm f}\right)}{{{\rm E}}_{D,h,{\Phi _{\rm fd}}}}\left\{ {\prod\limits_{z \in {\Phi _{{\rm fd}\backslash b_o}}} {\exp \left( { - s{P_{\rm fd}}{h_{z,d_o}}{{k}_{z,d_o}}D_{z,d_o}^{ - {\alpha _{\rm d}}}} \right)} } \right\}{{{\rm E}}_{D,h,{\Phi _{\rm fu}}}}\left\{ {\prod\limits_{x \in {\Phi _{\rm fu}}\backslash c_o} {\exp \left( { - s{P_{\rm fu}}{h_{x,d_o}}{{k}_{x,d_o}}D_{x,d_o}^{ - {\alpha _{\rm u}}}} \right)} } \right\} \\ & \mathop = \limits^{\left({\rm g}\right)} \exp \left\{ { - 2\pi {\lambda _{\rm s}}{{\rm E}_{\rm h}}\int\nolimits_{R_{\rm d}}^\infty {\left\{ {1 - \left[ {\exp \left( { - {s_{\rm d}}h{v^{ - {\alpha _{\rm d}}}}} \right)} \right]} \right\}v{\rm d}v} } \right\} \exp \left\{ { - 2\pi {\lambda _{\rm s}}{{\rm E}_{\rm h}}\int\nolimits_{R_{\rm u}} ^\infty {\left\{ {1 - \left[ {\exp \left( { - {s_{\rm u}}h{v^{ - {\alpha _{\rm u}}}}} \right)} \right]} \right\}v{\rm d}v} } \right\}, \tag{27} \end{align} where $c_o$ and $d_o$ are $\emptyset$ ($u_o$) and $u_o$ ($b_o$) if $q=d({\rm~u})$, respectively. (f) follows the fact that $\Phi~_{\rm~fu}$ and $\Phi~_{\rm~fd}$ are assumed to be independent. In Step (g), when we focus on the downlink, the inter-user interference distance can be arbitrarily close, therefore the integration lower bound $R_{\rm~u}$ is $\eta~_{\rm~uu}R$ when taking the interference cancelation strategy into consideration. Based on the association principle, the interfering distance between focused downlink user and neighbor BSs is at least $R$, so $R_{\rm~d}~=~R$. While when considering uplink transmission, the inter-BS interference distance between focused BS and the neighbor BSs can also be arbitrarily close, as a consequence $R_{\rm~d}$ is $\eta~_{\rm~bb}R$ when SIC is applied. In the same way, the distance from the neighbor uplink users to the focused BS is at least $R$, which gives $R_{\rm~u}=~R$. Then following the proof in sect. 1, we can get the result in (13).

Proof of Lemma 3.3

Before solving joint LT of $I_{\rm~d}$ and $\hat~I_{\rm~d}$, some properties of the sum of two exponential distribution random variables are analysed.

We define $G$ as the sum of two exponential distribution random variables as follows: \begin{equation} G = {s_2}{h_1} + {s_2}{h_2}, \tag{28}\end{equation} where both $h_1$ and $h_2$ follow exponential distribution, i.e, $h_1~~\text{and}~~h_2\sim~{\rm~exp}(1)$. Then, when $s_1=s_2=s$, $G$ is the sum of two exponential random variables with same rate. As presented in [17], $G$ follows Erlang distribution. While, if $s_1~\ne~s_2$, $G$ follows the hypo-exponential distribution.

Based on the PDF of $G$, we can have \begin{equation} {{{\rm E}}_G}\left( {{G^\delta }} \right) = \left \{ \begin{aligned} &{s^\delta }\Gamma \left( {\delta + 2} \right), \text{if $s_1=s_2=s$}, \\ &\frac{1}{{{s_2} - {s_1}}}\Gamma \left( {\delta + 1} \right)\left( {{s_2}^{\delta + 1} - {s_1}^{\delta + 1}} \right), \text{otherwise}, \end{aligned} \right. \tag{29}\end{equation} where (29) can be obtained by Eq. (3.478)$^{3)}$. Furthermore, according to Eq. (3.381)$^{3)}$, we can also obtain \begin{equation} \begin{array}{l} {{\rm E}_G}\left( {{G^{\mu - 1}}\Gamma \left( {\nu ,\gamma G} \right)} \right) = \left\{ \begin{array}{l} {s}^2J\left( {\mu {\rm{ + }}1,{s}^{ - 1},\nu ,\gamma } \right), {\rm{ if }} {{\rm{s}}_{\rm{1}}}{\rm{ = }}{{\rm{s}}_{\rm{2}}}{\rm{ = s }}, \\ \frac{1}{{{s_2} - {s_1}}}\left( {J\left( {\mu ,\frac{1}{{{s_2}}},\nu ,\gamma } \right) - J\left( {\mu ,\frac{1}{{{s_1}}},\nu ,\gamma } \right)} \right), {\rm{ otherwise}}, \end{array} \right. \end{array} \tag{30}\end{equation} where $J(~{\cdot,~\cdot,~\cdot,~\cdot~})$ is defined in Eq. (11). We also start with the definition of joint LT. \begin{align}&{\mathcal{L}_{{{\hat I}_{\rm d}},{I_{\rm d}}}}\left( {{s_{\rm h}},{s_{\rm f}}} \right) = {{{\rm E}}_{{{\hat I}_{\rm d}},{I_{\rm d}}}}\left( {{{\rm e}^{ - {s_{\rm h}}{{\hat I}_{\rm d}} - {s_{\rm f}}{I_{\rm d}}}}} \right) \\ &\mathop = \limits^{\left(k \right)}{\rm E}\left( \exp \left( - {s_{\rm hd}}\left( {\sum\limits_{z \in {\Phi _{\rm hd}}\backslash b_o} {{h_{z,u_o}}{D_{z,u_o}}^{ - {\alpha _{\rm bu}}}} } \right)- \left( {\sum\limits_{z \in {\Phi _{{\rm fd}\backslash b_o}}} {{s_{\rm fd}}{h_{z,u_o}}{{ {{D_{z,u_o}}} }^{ - {\alpha _{\rm bu}}}}} + \sum\limits_{x \in {\Phi _{\rm fu}}} {{s_{\rm fu}}{h_{x,u_o}}{{ {{D_{x,u_o}}} }^{ - {\alpha _{\rm uu}}}}} } \right) \right) \right) \\ &\mathop = \limits^{\left( l \right)} {\rm E}\left( \exp \left( - \left( {\sum\limits _{z \in {\Phi _{\rm fd}}\backslash b_o} {\left( {{s_{\rm hd}}{h_{z,u_o}} + {s_{\rm fd}}{h_{z,u_o}}} \right){D_{z,u_o}}^{ - {\alpha _{\rm bu}}}} } \right) - {s_{\rm fu}}\left( {\sum\limits_{x \in {\Phi _{\rm fu}}} {{h_{x,u_o}}{{\left( {{D_{x,u_o}}} \right)}^{ - {\alpha _{\rm uu}}}}} } \right) \right) \right) \\ &\mathop = \limits^{\left( m \right)} \exp \left\{ { - 2\pi {\lambda _{\rm s}}{\rm E}\int\nolimits_R^\infty {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - \left( {{{s}_{\rm fd}}{h_{\rm fd}} + {{s}_{\rm hd}}{h_{\rm hd}}} \right){v^{ - {\alpha _{\rm bu}}}}} \right)}}v{\rm d}v} } \right\}\exp \left\{ { - 2\pi {\lambda _{\rm s}}{\rm E}\int\nolimits_{\eta _{\rm uu}R} ^\infty {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {{s}_{\rm fu}}{h_{\rm fu}}{v^{ - {\alpha _{\rm uu}}}}} \right)}}v{\rm d}v} } \right\} \\ &\mathop = \limits^{\left( n \right)}\exp \left\{ - 2\pi {\lambda _{\rm s}}{{\rm E}_{{G_{\rm u}}}}\left\{ - \frac{1}{{{\alpha _{\rm bu}}}}{G_{\rm u}}^{\frac{2}{{{\alpha _{\rm bu}}}}}\Gamma \left( - \frac{2}{{{\alpha _{\rm bu}}}}\right) - \frac{1}{2}{R^2}+ \frac{1}{{{\alpha _{\rm bu}}}}{G_{\rm u}}^{\frac{2}{{{\alpha _{\rm bu}}}}}\Gamma \left( - \frac{2}{{{\alpha _{\rm bu}}}},\frac{{{G_{\rm u}}}}{{{R^\alpha }}}\right) \right\} \right\}\omega \left( {{\alpha _{\rm uu}},s_{\rm fu}\eta _{\rm uu}R } \right), \tag{31} \end{align} where ${s}_{\rm~fd}={{s_{\rm~f}}}{{P_{\rm~fd}}}{{k}_{\rm~fd}},~~{s}_{\rm~hd}={{s_{\rm~h}}}{{P_{\rm~hd}}}{{k}_{\rm~hd}}$ and${s_{\rm~fu}}={{s_{\rm~f}}}{{P_{\rm~fu}}}{{k}_{\rm~fu}}$ in Step (k). In Step (l), the interference BS sets in HD RBs and FD RBs are the same, therefore, the interfering distances between the focused downlink user and the BSs after classifying the user as CEU is the same as the distances when the focused user is regarded as cell CCU. Hence, we can write them together as the first sum item of Step (l). $G_{\rm~u}={{{s}_{\rm~fd}}{h_{\rm~fd}}~+~{{s}_{\rm~hd}}{h_{\rm~hd}}}$ which is the same as (28) and its properties have already been given by (29) and (30). Then the conclusion of Lemma 3.3 is obtained.

Proof of Theorem sect. 4.1

The proof begins with the definition of the CEU and coverage probability, then we can obtain \begin{align} &F_{{\rm edge},q}\left( {T} \right) = \mathbb{P}\left( {{{{\rm SINR}}} > T|{\rm SINR} < {{\gamma}_q}} \right) \\ &\mathop = \limits^{\left({\rm h}\right)} \frac{{{\rm E}\left( {\mathbb{P}\left( {{{\rm SINR}} > T,{\rm SINR} < {{\gamma}_q}|R,{I_{{\rm f}q}},{I_{{\rm h}q}}} \right)} \right)}}{{{\rm E}\left( {\mathbb{P}\left( {{\rm SINR} < {{\gamma}_q}|R,{I_{{\rm f}q}}} \right)} \right)}}\mathop = \limits^{\left(I\right)} \frac{{{\rm E}\left( {{{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm h}}\left( {{\sigma ^2} + {{\hat I}_{\rm e}}} \right)} \right)}}\left( {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}}\left( {{\sigma ^2} + {I_{\rm e}} + {\delta }{I_{\rm SI}}} \right)} \right)}}} \right)} \right)}}{{{\rm E}\left( {\left( {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}}\left( {{\sigma ^2} + {I_{\rm e}} + {\delta }{I_{\rm SI}}} \right)} \right)}}} \right)} \right)}}, \tag{32} \end{align} where SINR and $I_{\rm~e}$ denote the received SINR and interference when the user is regarded as CCU. While SINR$~and~$hat I_rm e$~represent~the~received~SINR~and~interference~after~the~user~is~classified~as~CEU.~SINR~is~defined~as~(\ref{equation2}).~Step~(h)~follows~Bayes~theorem.~Step~(I)~is~obtained~based~on~two~facts.~one~is~the~independency~of~the~small~scale~fading~over~FD~and~HD~RBs,~the~other~fact~is~that~the~small~scale~fading~is~assumed~to~be~Rayleigh,~i.e,~$h∼ rm exp1)$,~so~$mathbbPh<t)=1-exp (-t)$.~Then~carry~on~the~proof~in~Step~(I)~and~we~have \begin{align} \label{equation_coverage_edge} &F_{{\rm edge},q}\left( {T} \right) = \mathbb{P}\left( {{{\rm SINR}} > T|{\rm SINR} < {{\gamma}_q}} \right)= \frac{{\int\nolimits_{R = 0}^\infty {{\rm E}\left( {{{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm h}}{\sigma ^2}} \right)}}\left( {{{\mathop{\rm e}\nolimits} ^{ - {s_{\rm h}}{{\hat I}_{\rm e}}}} - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}} {{{\sigma }^2} + {s_{\rm h}}{\hat I_{\rm e}} + {s_{\rm f}}{I_{\rm e}}} } \right)}}} \right)} \right){f_R}\left( R \right){\rm d}R} }}{{\int\nolimits_{R = 0}^\infty {{\rm E}\left( {1 - {{\mathop{\rm e}\nolimits} ^{\left( { - {s_{\rm f}}\left( {{{\sigma }^2} + {I_{\rm e}}} \right)} \right)}}} \right){f_R}\left( R \right){\rm d}r} }}\nonumber\\ &\mathop = \limits^{\left({\rm j}\right)} \frac{{\int\nolimits_{R = 0}^\infty {\left( {{{\mathop{\rm e}\nolimits} ^{ - {s_{\rm h}}{\sigma ^2}}}\left( {{L_{{I_{{\rm h}q}}}}\left( {{s_{\rm h}}} \right) - {{\mathop{\rm e}\nolimits} ^{ - {s_{\rm f}}\sigma {^2}}}{L_U}\left( {{s_{\rm h}},{s_{\rm f}}} \right)} \right)} \right){f_R}\left( R \right){\rm d}R} }}{{\int\nolimits_{R = 0}^\infty {\left( {1 - {{\mathop{\rm e}\nolimits} ^{ - {s_{\rm f}}\sigma {^2}}}{L_{{I_{{\rm f}q}}}}\left( {{s_{\rm f}}} \right)} \right){f_R}\left( R \right){\rm d}R} }}, \end{align} where~$f_R(R)$~is~defined~in~(\ref{distribution_r}).~We~can~obtain~Step~(j)~according~to~the~definition~of~LT.~$L_I_m$~is~the~LT~of~interference~$I_m$~($m ∈ łeft rm hq,rm fq right$~)~and~they~are~defined~in~(\ref{laplace_hd})~and~(\ref{laplace_fq}).~The~joint~LT~$L_U$~is~different~in~the~downlink~and~uplink~transmission.~In~downlink~transmission,~the~interfering~BSs~are~the~same~before~and~after~allocating~HD~RBs~to~the~CEUs,~which~results~in~$hat I_rm e$~and~$I_rm e$~are~correlated~with~each~other.~As~discussed~in~Lemma~\ref{Theorm_laplace_jiont},~$L_U$~is~defined~in~\ref{L_U}.~While~in~the~uplink,~the~uplink~interfering~user~sets~have~already~changed~when~the~serving~RBs~of~CEU~switch~from~FD~RBs~to~HD~RBs.~Therefore,~in~uplink~$L_U$~can~be~written~as~$L_rm hułeft(s_rm hright)L_rm fułeft(s_rm fright). Then the conclusion of Theorem sect. 4.1 is obtained.


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  • Figure 1

    An exemplary interference scenario of HD/FD cellular network. (a) Synchronous TDD cellular network; protectłinebreak (b) FD cellular network.

  • Figure 2

    Resource blocks allocation.

  • Figure 9

    SE vs. successive interference cancellation capability.

  • Table 1   Simulation parameters
    RB band width Number of RB Thermal noise $P_b$ $P_{\rm~u}$ $\eta~$ $\lambda_{\rm~s}$ $\lambda_{\rm~u}$ $\gamma_{\rm~d}$ ${\left|~{{\varepsilon~}}~\right|^2}$ Path loss [25]
    1 MHz 128 $-$174 dBm$\cdot$Hz$^{-1}$ 0.1 W 0.1 W 0 ${10^{{\rm{~-~}}3}}$ m$^{-2}$ ${1}$ m$^{-2}$ 0 dB $-110$ dB 140.7 + 36.7lg$R$
    ($R$ in km)

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