The expression presented in (11) is only valid for a fixed (time-invariant) channel, that is, the fading coefficients *K*_{
ij
}are fixed. The aim of this paper is to obtain an expression for the bit error probability when the fading coefficients vary according to a Rayleigh distribution.

### A. Bit Error Probability

The bit error probability associated with the signal from user 1, at user 2, for a fixed gain is described in (6). Now assuming a nonstatic situation, the average bit error probability can be computed averaging (6) with respect to a Rayleigh distribution

{\stackrel{\u0304}{P}}_{{e}_{12}}=E\left[{P}_{{e}_{12}}\right]=\frac{1}{2}\left(1-\sqrt{\frac{{\gamma}_{12}}{2+{\gamma}_{12}}}\right)

(12)

where γ_{12} is the average signal-to-noise ratio, defined as

{\gamma}_{12}=\frac{2{a}_{12}^{2}{\alpha}_{12}^{2}{N}_{c}}{{\sigma}_{2}^{2}}

(13)

From (11), we can define two random variables *U*_{1} and *U*_{2}, respectively, as

\frac{{v}_{\lambda}^{T}{v}_{1}}{\sqrt{{v}_{\lambda}^{T}{v}_{\lambda}}}=\sqrt{{U}_{1}}=\frac{\left({\left({K}_{10}{a}_{12}\right)}^{2}+\lambda {\left({K}_{10}{a}_{13}+{K}_{20}{a}_{23}\right)}^{2}\right)\sqrt{{N}_{c}}}{\left({\left({K}_{10}{a}_{12}\right)}^{2}+{\lambda}^{2}{\left({K}_{10}{a}_{13}+{K}_{20}{a}_{23}\right)}^{2}\right){\sigma}_{0}}

(14)

\frac{{v}_{\lambda}^{T}{v}_{2}}{\sqrt{{v}_{\lambda}^{T}{v}_{\lambda}}}=\sqrt{{U}_{2}}=\frac{\left({\left({K}_{10}{a}_{12}\right)}^{2}+\lambda \left({\left({K}_{10}{a}_{13}\right)}^{2}-{\left({K}_{20}{a}_{23}\right)}^{2}\right)\right)\sqrt{{N}_{c}}}{\left({\left({K}_{10}{a}_{12}\right)}^{2}+{\lambda}^{2}{\left({K}_{10}{a}_{13}+{K}_{20}{a}_{23}\right)}^{2}\right){\sigma}_{0}}

(15)

since *K*_{10} and *K*_{20} are Rayleigh distributed, the support of (14) will be always greater than zero. On the other hand, since we have negative values in the numerator of (15), its support will be all the real line. Taking this into account, we can rewrite (11) as

{P}_{{e}_{1}}=\left(1-{P}_{{e}_{12}}\right)Q\left(\sqrt{{U}_{1}}\right)+{P}_{{e}_{12}}Q\left({U}_{2}\right)

(16)

To obtain the error probability, we must average {P}_{{e}_{1}}, over the probability density function (PDF) of *U*_{1} and *U*_{2}[7]. Thus, we have to evaluate the integral

\begin{array}{c}{P}_{{e}_{f}}=\left(1-{\stackrel{\u0304}{P}}_{{e}_{12}}\right)\underset{0}{\overset{\infty}{\int}}Q\left(\sqrt{{u}_{1}}\right){f}_{{u}_{1}}\left({u}_{1}\right)d{u}_{1}+\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{\stackrel{\u0304}{P}}_{{e}_{12}}\underset{-\infty}{\overset{\infty}{\int}}Q\left({u}_{2}\right){{f}_{u}}_{{}_{2}}\left({u}_{2}\right)d{u}_{2}\end{array}

(17)

In order to calculate *P*_{
ef
}, we have to know the distribution of *U*_{1} and *U*_{2}, thus to facilitate the calculations, we assume an equal power allocation situation, where *a*_{12} = *a*_{13} = *a*_{23} = *a*. With this assumption the random variables *U*_{1} and *U*_{2} will be simplified to

{U}_{1}=\frac{{a}^{2}\left({K}_{10}^{2}+\lambda {\left({K}_{10}+{K}_{20}\right)}^{2}\right){N}_{c}}{\left({K}_{10}^{2}+{\lambda}^{2}{\left({K}_{10}+{K}_{20}\right)}^{2}\right){\sigma}_{0}^{2}}

(18)

{U}_{2}=\frac{a\left({K}_{10}^{2}+\lambda \left({K}_{{}_{10}}^{2}+{K}_{{}_{20}}^{2}\right)\right)\sqrt{{N}_{c}}}{\left(\sqrt{{K}_{10}^{2}+{\lambda}^{2}{\left({K}_{10}+{K}_{20}\right)}^{2}}\right){\sigma}_{0}}

(19)

Since *U*_{1} depends on *K*_{10} and *K*_{20}, it is possible to write the cumulative distribution function (CDF) and the PDF of *U*_{1}, respectively, as

{F}_{{u}_{1}}\left({u}_{1}\right)=\int \underset{{k}_{10},{k}_{20}\in {D}_{{u}_{1}}}{\int}{f}_{{u}_{1}}\left({k}_{10},{k}_{20}\right)\mathsf{\text{d}}{k}_{10}\mathsf{\text{d}}k20

(20)

{f}_{{u}_{1}}\left({u}_{1}\right)=\frac{\mathsf{\text{d}}{F}_{{u}_{1}}\left({u}_{1}\right)}{d{u}_{1}}

(21)

In this case, {D}_{{u}_{1}} is the region of the *K*_{10} × *K*_{20} plane where

\frac{{a}^{2}\left({k}_{10}^{2}+\lambda {\left({k}_{10}+{k}_{20}\right)}^{2}\right){N}_{c}}{\left({k}_{10}^{2}+{\lambda}^{2}{\left({k}_{10}+{k}_{20}\right)}^{2}\right){\sigma}_{0}^{2}}\le {u}_{1}

(22)

Note that this region is very similar to a rotated ellipse but not exactly an ellipse.

Since *K*_{10} and *K*_{20} are independent Rayleigh distribution with parameters *α*_{10} and *α*_{20}, respectively, we have

{F}_{{u}_{1}}\left({u}_{1}\right)=\underset{{k}_{10}=0}{\overset{a\left({u}_{1}\right)}{\int}}\underset{{k}_{20}=0}{\overset{b\left({u}_{1}\right)}{\int}}{f}_{{k}_{10}{k}_{20}}\left({k}_{10,}{k}_{20}\right)\mathsf{\text{d}}{k}_{20}\mathsf{\text{d}}{k}_{10}

(23)

where

a\left({u}_{1}\right)=\frac{1}{{\lambda}_{1}}\sqrt{\frac{{u}_{1}{\lambda}_{2}}{{A}_{1}}}

(24)

b\left({u}_{1}\right)=\frac{\sqrt{2{A}_{1}\left({B}_{1}-\left(2{A}_{1}{k}_{10}^{2}-{u}_{1}\right)\lambda \right)}-2{A}_{1}{k}_{10}{\lambda}^{2}}{2{A}_{1}\lambda}

(25)

{A}_{1}=\frac{{a}^{2}{N}_{c}}{{\sigma}_{0}^{2}}

(26)

{B}_{1}=\lambda \sqrt{{u}_{1}\left({u}_{1}{\lambda}^{2}-4{A}_{1}{k}_{10}^{2}\left(\lambda -1\right)\right)}

(27)

now it is possible to derive the PDF of *U*_{1} easily as

{f}_{{u}_{1}}\left({u}_{1}\right)=\underset{{k}_{10}=0}{\overset{a\left({u}_{1}\right)}{\int}}\frac{\partial b\left({u}_{1}\right)}{\partial {u}_{1}}\frac{b\left({u}_{1}\right)}{{\alpha}_{20}^{2}}{\mathsf{\text{e}}}^{-\frac{b{\left({u}_{1}\right)}^{2}}{2{\alpha}_{20}^{2}}}\frac{{k}_{10}}{{\alpha}_{10}^{2}}{\mathsf{\text{e}}}^{-\frac{{k}_{10}^{2}}{2{\alpha}_{10}^{2}}}\mathsf{\text{d}}{k}_{10}

(28)

and unfortunately, it is not possible to evaluate (28) in a closed-form solution.

In order to validate the above formulation, Figure 1 shows the analytical and simulated PDF of *U*_{1}. Note the excellent agreement between them showing the correctness of our formulation.

Following similar rationale, we now find the CDF and PDF of *U*_{2}. Note that in this case, the region of integration, {D}_{{u}_{2}}, will be given by

\frac{a\left({k}_{10}^{2}+\lambda \left({k}_{10}^{2}-{k}_{20}^{2}\right)\right)\sqrt{{N}_{c}}}{\left(\sqrt{{k}_{10}^{2}+{\lambda}^{2}{\left({k}_{10}+{k}_{20}\right)}^{2}}\right){\sigma}_{0}}\le {u}_{2}

(29)

leading to the following CDF and PDF, respectively, as

{F}_{{u}_{2}}\left({u}_{2}\right)=\left\{\begin{array}{c}\hfill \underset{{k}_{20}=\frac{\left|{u}_{{}_{2}}\right|}{{A}_{2}}}{\overset{\infty}{\int}}\underset{{k}_{10}=0}{\overset{\infty}{\int}}{f}_{{k}_{10}{k}_{20}}\left({k}_{10},{k}_{20}\right)\mathsf{\text{d}}{k}_{10}\mathsf{\text{d}}{k}_{20}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{u}_{2}<0,\hfill \\ \hfill \underset{{k}_{20}=0}{\overset{\infty}{\int}}\underset{{k}_{10}=0}{\overset{a\left({u}_{2}\right)}{\int}}{f}_{{k}_{10}{k}_{20}}\left({k}_{10},{k}_{20}\right)\mathsf{\text{d}}{k}_{10}\mathsf{\text{d}}{k}_{20}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{u}_{2}\ge 0.\hfill \end{array}\right.

(30)

and

{f}_{{u}_{2}}\left({u}_{2}\right)=\left\{\begin{array}{cc}\hfill \underset{\frac{\left|{u}_{2}\right|}{{A}_{2}}}{\overset{\infty}{\int}}\frac{\partial b\left({u}_{2}\right)}{\partial {u}_{2}}{f}_{{k}_{10}{k}_{20}}\left(b\left({u}_{2}\right),{k}_{20}\right)\mathsf{\text{d}}{k}_{20}\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}{u}_{2}<0,\hfill \\ \hfill \underset{0}{\overset{\infty}{\int}}\frac{\partial a\left({u}_{2}\right)}{\partial {u}_{2}}{f}_{{k}_{10}{k}_{20}}\left(a\left({u}_{2}\right),{k}_{20}\right)\mathsf{\text{d}}{k}_{20}\hfill & \hfill \mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}{u}_{2}\ge 0.\hfill \end{array}\right.

(31)

where

a\left({u}_{2}\right)=\Re \left(\frac{1}{2{A}_{2}{\lambda}_{1}\sqrt{3}}\left(\sqrt{{R}_{1}}+\sqrt{\frac{{R}_{2}}{2}}\right)\right)

(32)

b\left({u}_{2}\right)=\Re \left(\frac{1}{2{A}_{2}{\lambda}_{1}\sqrt{3}}\left(\sqrt{{R}_{1}}+\sqrt{\frac{{R}_{2}}{2}}\right)\right)

(33)

{A}_{2}=\frac{a\sqrt{{N}_{c}}}{{\sigma}_{0}}

(34)

where \Re \left(\cdot \right) denotes the real part of a number, and *R*_{1} and *R*_{2} are described in the Appendix.

In the same way as in the first case, (31) cannot be obtained in a closed-form solution. Figure 2 compares the analytical and simulated PDF of *U*_{2} in order to validate our formulation.

Once that the PDFs of *U*_{1} and *U*_{2} were exactly computed, it is possible to obtain the average bit error probability by simply substituting (28) and (31) into (17). Figure 3 shows the simulation result of the bit error probability and the result of our theoretical expression given in (17), where we can observe that both curves are almost coincident. In this figure, \mathsf{\text{SNR}}=\frac{P}{{\sigma}_{0}^{2}}. According to Section II-B, we consider three symbols periods, each period with an average power of *P*. Also, for simplicity, we consider that E\left[{K}_{10}^{2}\right],\phantom{\rule{2.77695pt}{0ex}}E\left[{K}_{20}^{2}\right] and E\left[{K}_{12}^{2}\right] are identical.

Although (17) presents the exact solution to the average bit error probability, in some cases, the complexity to compute this expression can be prohibitive. For this reason, we found a very accurate approximation for the bit error probability presented in the sequel.

### B. Approximate Bit error Probability

The main problem in order to obtain a simpler expression for the bit error probability is to simplify the PDFs of *U*_{1} and *U*_{2} given, respectively, in (14) and (15). In order to obtain an approximation, the expressions (14) and (15) can be reduced when λ = 1, *σ*_{0} = 1 and *a*_{12} = *a*_{13} = *a*_{23} = 1. Therefore, the new random variables are given by

{U}_{1}^{\prime}={N}_{c}\left({K}_{10}^{2}+{\left({K}_{10}+{K}_{20}\right)}^{2}\right)

(35)

{U}_{2}^{\prime}=\frac{\sqrt{{N}_{c}}\left(2{K}_{10}^{2}-{K}_{20}^{2}\right)}{\sqrt{{K}_{10}^{2}+{\left({K}_{10}+{K}_{20}\right)}^{2}}}

(36)

Considering {D}_{{u}^{\prime}1} as the region of the plane *K*_{10} × *K*_{20} where {N}_{c}\left({k}_{10}^{2}+{\left({k}_{10}+{k}_{20}\right)}^{2}\right)\le {u}_{1}^{\prime}, it can be seen that {D}_{{u}^{\prime}1} corresponds to the area of an ellipse whose center is in the origin (0, 0). Unfortunately, the evaluation of the integral (20) is rather complex for the domain {D}_{{u}^{\prime}1}. For this reason, we consider a simplified version of {D}_{{u}^{\prime}1}, as being the area of a circle expressed as {k}_{10}^{2}+{k}_{20}^{2}\le {u}_{1}^{\prime}. This simplification can be applied since a circle corresponds to a particular case of the general ellipse. Hence

{F}_{{{u}^{\prime}}_{1}}\left({{u}^{\prime}}_{1}\right)=\underset{{k}_{20}=-\sqrt{{{u}^{\prime}}_{1}}}{\overset{\sqrt{{{u}^{\prime}}_{1}}}{\int}}\underset{{k}_{10}=-\sqrt{{{u}^{\prime}}_{1}-{k}_{20}^{2}}}{\overset{\sqrt{{{u}^{\prime}}_{1}-{k}_{20}^{2}}}{\int}}{f}_{{k}_{10}{k}_{20}}\left({k}_{10},{k}_{20}\right)\mathsf{\text{d}}{k}_{10}\mathsf{\text{d}}{k}_{20}

(37)

This gives

\begin{array}{c}{f}_{{{u}^{\prime}}_{1}}\left({{u}^{\prime}}_{1}\right)=\\ \underset{{k}_{20}=-\sqrt{{{u}^{\prime}}_{1}}}{\overset{\sqrt{{{u}^{\prime}}_{1}}}{\int}}\frac{1}{2\sqrt{{{u}^{\prime}}_{1}-{k}_{20}^{2}}}\left\{{f}_{{k}_{10}{k}_{20}}\right.\left(\sqrt{{{u}^{\prime}}_{1}-{k}_{20}^{2}},{k}_{20}\right)+\\ \left(\right)close="\}">\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{f}_{{k}_{10}{k}_{20}}\left(-\sqrt{{{u}^{\prime}}_{1}-{k}_{20}^{2}},{k}_{20}\right)& \mathsf{\text{d}}{k}_{20}\end{array}\n

(38)

Since *K*_{10} and *K*_{20} are independent Rayleigh distributed with parameters *α*_{1} and *α*_{2}, respectively, the PDF of *U'*_{
1
}given in (38) will result in a chi-square probability distribution with four degrees of freedom [8]. Therefore, our approximation of *U*_{1} will be given by

{f}_{{u}_{1}}\left({u}_{1}\right)\approx \frac{4{u}_{1}}{{\gamma}_{1}^{2}}{\mathsf{\text{e}}}^{-2{u}_{1}\u2215{\gamma}_{1}}

(39)

where *γ*_{1} is the mean of *U*_{1} given in (14)

{\gamma}_{1}=E\left[{U}_{1}\right]

(40)

Figure 4 shows the comparison between our approximate PDF given in (39) and the computer simulation for the PDF of *U*_{1} given in (14) for two different values of λ keeping the same values for *a*_{12} = 1, *a*_{13} = 2, and *a*_{23} = 3. We observe that the curves are very close for both values of λ. Although only these two cases are presented here, many other cases were compared and the approximation still remains very good.

A similar rationale can be applied in order to find a good approximation for *U*_{2}. The region of the *K*_{10} × *K*_{20} plane where {U}_{2}^{\prime}\le {u}_{2}^{\prime} is similar to (29). Note that the range of {{U}_{2}}^{\prime} varies from -\infty \le {{u}_{2}}^{\prime}\le \infty, discarding all the distributions with positive support. In order to observe the behavior of the PDF of {{U}_{2}}^{\prime}, a large number of simulations were performed, and the Gaussian distribution proves to fit extremely well in all the cases. Therefore, assuming a Gaussian distribution, the following can be written

{P}_{{e}_{f}}\approx \frac{1}{4}\left(1+\sqrt{\frac{{\gamma}_{12}}{2+{\gamma}_{12}}}\right)\left(1-\frac{\sqrt{{\gamma}_{1}}\left({\gamma}_{1}+6\right)}{{\left({\gamma}_{1}+4\right)}^{3\u22152}}\right)+\frac{1}{2}\left(1-\sqrt{\frac{{\gamma}_{12}}{2+{\gamma}_{12}}}\right)Q\left(\frac{{\gamma}_{2}}{\sqrt{1+{v}^{2}}}\right)

(41)

{f}_{{u}_{2}}\left({u}_{2}\right)\approx \frac{1}{\sqrt{2\pi {\nu}^{2}}}{\mathsf{\text{e}}}^{-\frac{{\left({u}_{{}_{2}}-{\gamma}_{{}_{2}}\right)}^{2}}{2{\nu}^{2}}}

(42)

where

{\gamma}_{2}=E\left[{U}_{2}\right]

(43)

{\nu}^{2}=\mathsf{\text{var}}\phantom{\rule{1em}{0ex}}\left({U}_{2}\right)

(44)

Figure 5 shows the comparison between the approximate PDF given in (42) and the computer simulation for the PDF of *U*_{2} given in (15), for two different values of λ. Note that the approximation is less accurate for small values of λ, but this inaccuracy does not have a significant influence in the bit error probability. In all the cases, the approximation fits very well the exact PDF of *U*_{2}.

Using (39) and (42) into (17), it is possible to obtain a very accurate approximate bit error probability. Fortunately, both integrals can be found in a closed-form solution as

\underset{0}{\overset{\infty}{\int}}Q\left(\sqrt{{u}_{1}}\right)\frac{4{u}_{1}}{{\gamma}_{1}^{2}}{\mathsf{\text{e}}}^{-2{u}_{1}\u2215{\gamma}_{1{{\mathsf{\text{d}}}_{u}}_{{}_{1}}}}=\frac{1}{2}\left(1-\frac{\sqrt{{\gamma}_{1}}\left({\gamma}_{1}+6\right)}{{\left({\gamma}_{1}+4\right)}^{3\u22152}}\right)

(45)

and

\underset{-\infty}{\overset{\infty}{\int}}Q\left({u}_{2}\right)\frac{1}{\sqrt{2\pi {\nu}^{2}}}{\mathsf{\text{e}}}^{-\frac{{\left({u}_{2}-{\gamma}_{2}\right)}^{2}}{2{\nu}^{2}}}\mathsf{\text{d}}{u}_{2}=Q\left(\frac{{\gamma}_{2}}{\sqrt{1+{\nu}^{2}}}\right)

(46)

All these calculations lead to the approximate bit error probability for the λ-MRC detector as shown in (41), where *γ*_{12} is given in (13), *γ*_{1} is given in (40), *γ*_{2} is given in (43), and *ν*_{
2
}is given in (44).

Assuming an equal power allocation scheme (*a*_{12} = *a*_{13} = *a*_{23} = *a*), Figure 6 shows the comparison between the theoretical bit error probability presented in (17) using the exact PDFs (28) and (31) and our approximation given in (41). We can observe that both curves are almost the same, validating our approximation.

Our results are quite exact for a different power allocation scheme as well. This can be seen in Figure 7, where a comparison between the exact simulated bit error probability and our approximation given in (41) was performed. In this figure, the following parameters were used *α*_{10} = *α*_{20} = 1 and *α*_{12} = 0.8.

The final approximate expression allows us to determine the optimal value for λ in each case. As stated in [1], when the BS believes that the inter-user channel is "perfect", then λ = 1 and the optimal detector turns out to be the maximal ratio combining [7]. As the inter-user channel becomes more unreliable, i.e., as {P}_{{e}_{12}} increases, the value of the best λ decreases toward to zero. In order to demonstrate this behavior, Figure 8 shows the optimized λ* versus the inter-user channel parameter α_{12}. This curve was obtained using computational optimization techniques that minimizes our approximate bit error probability (41) with respect to λ for each value of the inter-user channel parameter, *α*_{12}. The direct channel parameters *α*_{10} = *α*_{20} = 1 were kept constant, and the equal power allocation scheme (*a*_{12} = *a*_{13} = *a*_{23} = 1) was adopted.