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Outage Performance Analysis of Incremental Relay Selection for STBC AF Cooperative Networks

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Abstract

In this paper, selection and incremental schemes are investigated for multi-input multi-output (MIMO) cooperative systems in which multiple amplify-and-forward relays and space–time block coding (STBC) are applied. In order to make an efficient use of the degrees of freedom of channel and provide full diversity order in the multi-relay STBC network, the selection and incremental protocols are employed by exploiting limited feedback from destination. Maximum ratio combination technique is applied at the destination. Outage probabilities of the both cooperative schemes are analytically derived at high signal to noise ratio regime. It is shown the incremental selection STBC scheme leads to a higher performance compared to the STBC selection relaying scheme and conventional STBC MIMO channels; so that the outage capacity provided by the incremental selection scheme is twice of the selection scheme at the low outage probabilities.

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Correspondence to Sajad Poursajadi.

Appendix

Appendix

Claim

Let Z = f(UV) + f(UW) where f(xy) = xy/(x + y) and also U, V and W are independent random variables. In addition PDF of U is f U (u) = λ 2 ue λu, and V and W have exponential distribution of μ parameter. Letting h > 0, the probability Pr [Z < h] satisfies

$$ \mathop{\lim }\limits_{h \to 0} \frac{1}{{h^{2} }}\Pr \left[ {f(U,V) + f(U,W) < h} \right] = {{\left( {{{\lambda^{2} } \mathord{\left/ {\vphantom {{\lambda^{2} } 2}} \right. \kern-0pt} 2} + \mu^{2} } \right)} \mathord{\left/ {\vphantom {{\left( {{{\lambda^{2} } \mathord{\left/ {\vphantom {{\lambda^{2} } 2}} \right. \kern-0pt} 2} + \mu^{2} } \right)} 2}} \right. \kern-0pt} 2}. $$

Proof of Claim

It is clear that the solving of the math problem \( \mathop{\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\Pr (Z < z) = \mathop{\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}F_{Z} (z) \) is wanted where function F Z (z) indicates cumulative density function (CDF) of Z variable. So, \( \mathop{\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\Pr \left[ {Z \le z} \right] \) can be written as [24],

$$ \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\Pr \left[ {Z \le z} \right] = \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\mathop{\iiint}\limits_{D} {f_{UVW} (u,v,w)dudvdw} $$
(17)

where f UVW (uvw) indicates PDF, and region D should satisfy

$$ \frac{uv}{u + v} + \frac{uw}{u + w} < z $$
(18)

From (18) it is clear that for z → 0, v and w should be limited when u varies from 0 to infinite. And also, u should be limited when v and w vary from 0 to infinite. Hence, one can assume that D is made by two region as presented in Table 2. By this description, one can rewrite (17) as follows

$$ \begin{aligned} \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\Pr \left[ {Z \le z} \right] & = \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\mathop{\iiint}\limits_{{D_{1} }} {\,f_{UVW} (u,v,w)\,\,dudvdw} \\ & \quad + \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\,\mathop{\iiint}\limits_{{D_{2} }} {f_{UVW} (u,v,w)\,\,dudvdw} \\ & \quad - \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}\mathop{\iiint}\limits_{{D_{1} \cap D_{2} }} {f_{UVW} (u,v,w)\,\,dudvdw} \\ \end{aligned} $$
(19)

For the first term in the right hand side of (19) one has

$$ \Pr (Z < z,{\text{D}}_{ 1} ) = \int_{0}^{\infty } {\Pr \left[ {\frac{uV}{u + V} + \frac{uW}{u + W} < z} \right]} f_{U} (u)du $$
(20)

By taking \( X = \frac{uV}{u + V} \), \( Y = \frac{uW}{u + W} \) and using change of variable t = x/z, P[X + Y < z] satisfies

$$ P\left[ {X + Y < z} \right] = \int_{0}^{z} {P\left[ {Y < z - x} \right]} f_{X} (x)dx = z\int_{0}^{1} {P\left[ {Y < z(1 - t)} \right]} f_{X} (zt)dt $$
(21)

Since

$$ f_{Y} (y) = \frac{{\mu \,u^{2} }}{{(u - y)^{2} }}e^{{ - \frac{\mu y\,u}{u - y}}} $$

so it can be derived that \( \frac{\Pr (Y < y)}{y} \to \mu \) as y → 0. Using this fact and after some manipulations, (21) can be simplified as follows

$$ \begin{aligned} P\left[ {X + Y < z} \right] & = z^{2} \mu \int_{0}^{1} {(1 - t)} f_{X} (zt)dt \\ & = z^{2} \mu \left[ {(1 - t)\frac{{F_{X} (zt)}}{z}\left| {\begin{array}{*{20}c} 1 \\ 0 \\ \end{array} } \right. + \int_{0}^{1} {\frac{{F_{X} (zt)}}{z}dt} } \right] = \frac{{z^{2} \mu^{2} }}{2} \\ \end{aligned} $$
(22)

where in the last equality the fact that X variable has the same property as Y variable is applied. Finally one has

$$ \mathop {\lim }\limits_{z \to 0} \frac{{F_{Z} (z,{\text{D}}_{ 1} \,)}}{{z^{2} }} = \mu^{2} $$
(23)

Second term in the right hand side of (19) can be calculated as

$$ \begin{aligned} \Pr \left[ {Z \le z,{\text{D}}_{ 2 1} } \right] & = \int_{0}^{\infty } {\int_{0}^{\infty } {\Pr \left[ {\frac{Uv}{U + v + \delta } + \frac{Uw}{U + w + \delta } < z} \right]} } f_{V} (v)f_{W} (w)dvdw \\ & = \int_{0}^{\infty } {\int_{0}^{\infty } {\int_{0}^{{u_{z} }} {f_{U} (u)f_{V} (v)f_{W} (w)dvdw} } } du \\ \end{aligned} $$
(24)

To derive exact integration bounds along u, one should solve (18) as

$$ (v + w - z)u^{2} + (2vw - z(v + w))u - zvw < 0 $$
(25)

Clearly u should be in the region [u 0 , u 1 ] for z → 0, where

$$ \begin{aligned} u_{0} (z) & = \frac{z(v + w) - 2vw - \sqrt \Delta }{2(v + w - z)} \\ u_{1} (z) & = \frac{z(v + w) - 2vw + \sqrt \Delta }{2(v + w - z)} \\ \end{aligned} $$

and Δ = 4v 2 w 2 + z 2(v − w)2. Since u 0 < 0 for z → 0, therefore one has u ∊ [0, u 1(z)] in (24). Now it can be derived that

$$ \mathop {\lim }\limits_{z \to 0} \frac{1}{z}F_{Z} (z,{\text{D}}_{ 2} ) = \frac{\partial }{\partial z}F_{Z} (z,{\text{D}}_{ 2} )\left| {_{{_{z = 0} }} } \right. = \int_{0}^{\infty } {\int_{0}^{\infty } {\left( {\frac{\partial }{\partial z}u_{1} (z)\left| {_{{_{z = 0} }} } \right.} \right)} } f_{U} \left( {u_{1} (z)\left| {_{{_{z = 0} }} } \right.} \right))f_{V} (v)f_{W} (w)dvdw $$
(26)

By taking again differentiation from (26) one derive that

$$ \begin{aligned} & \left. {\mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}F_{Z} (z,{\text{D}}_{ 2} ) = \frac{{\partial^{2} }}{{\partial z^{2} }}F_{Z} (z,{\text{D}}_{ 2} )} \right|_{z = 0} \hfill \\ & \quad = \int_{0}^{\infty } {\int_{0}^{\infty } {\left. {\left[ {\frac{{\partial^{2} }}{{\partial z^{2} }}u_{1} (z)f_{U} (u_{1} (z)) + \left( {\frac{\partial }{\partial z}u_{1} (z)} \right)^{2} \frac{\partial }{\partial z}f_{U} (u_{1} (z))} \right]} \right|_{z = 0} } } f_{V} (v)f_{W} (w)dvdw \hfill \\ \end{aligned} $$
(27)

It can be derived that u 1(0) = 0, \( \frac{\partial }{\partial z}u_{1} (z)\left| {_{{_{z = 0} }} } \right. = 1/2 \), f U (0) = 0 and \( \frac{\partial }{\partial z}f_{U} (z)\left| {_{{_{z = 0} }} } \right. = \lambda^{2} \). Therefore, one has

$$ \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}F_{Z} (z\,,\,\,{\text{D}}_{ 2} \,) = {{\lambda^{2} } \mathord{\left/ {\vphantom {{\lambda^{2} } 4}} \right. \kern-0pt} 4}. $$
(28)

To solve the third term of (19), it should be note that all of the limits of variables are upper bounded by z, so one has

$$ F_{Z} (z,{\text{D}}_{ 1} \cap {\text{D}}_{ 2} ) = \mathop{\iiint}\limits_{{D_{1} \cap D_{2} }} {f_{UVW} (u,v,w)dudvdw} \le \int_{0}^{z} {\int_{0}^{z} {\int_{0}^{z} {f_{UVW} (u,v,w)dudvdw} } } \, \le \, z^{3} $$
(29)

where last inequality in above benefits from f UVW (uvw) ≤ 1. Therefore, clearly it can be derived that

$$ \mathop {\lim }\limits_{z \to 0} \frac{1}{{z^{2} }}F_{Z} (z,{\text{D}}_{ 1} \cap {\text{D}}_{ 2} ) = 0. $$
(30)

Finally by applying (23), (28) and (30) in (19), claim is proved.

Table 2 Distinction the regions of variables u, v and w

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Poursajadi, S., Madani, M.H. Outage Performance Analysis of Incremental Relay Selection for STBC AF Cooperative Networks. Wireless Pers Commun 83, 2317–2331 (2015). https://doi.org/10.1007/s11277-015-2523-y

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