Transmit Power Minimization for MIMO Systems of Exponential Average BER with Fixed Outage Probability

Abstract

This paper is concerned with a wireless multiple-antenna system operating in multiple-input multiple-output (MIMO) fading channels with channel state information being known at both transmitter and receiver. By spatiotemporal subchannel selection and power control, it aims to minimize the average transmit power (ATP) of the MIMO system while achieving an exponential type of average bit error rate (BER) for each data stream. Under the constraints on each subchannel that individual outage probability and average BER are given, based on a traditional upper bound and a dynamic upper bound of Q function, two closed-form ATP expressions are derived, respectively, which can result in two different power allocation schemes. Numerical results are provided to validate the theoretical analysis, and show that the power allocation scheme with the dynamic upper bound can achieve more power savings than the one with the traditional upper bound.

Appendix

The proof of Proposition 1

Since $\overline{P}_b(i) \le \frac{\xi _i}{2} \frac{\overline{\lambda }_{\mathrm{out}}(i)}{\overline{\lambda }_{\mathrm{mea}}(i)}$, then it follows from Lemma 2 that $\lambda _0(i)=\overline{\lambda }_{\mathrm{out}}(i)$. Furthermore, we can have

$$\begin{aligned} \hat{\rho }^{(i)} \left( \overline{P}_b(i),P_{\mathrm{out}}^{(i)}\right) =\int _{\overline{\lambda }_{\mathrm{out}}(i)}^{\infty }p_i f_i(\lambda _i)d\lambda _i. \end{aligned}$$

(60)

Substituting (29) into (60), we obtain

$$\begin{aligned} \hat{\rho }^{(i)} \left( \overline{P}_b(i),P_{\mathrm{out}}^{(i)}\right)= & {} \int _{\overline{\lambda }_{\mathrm{out}}(i)}^{\infty }\frac{\widehat{\hbox {SNR}}(i)}{\lambda _i} f_i(\lambda _i)d\lambda _i\nonumber \\&+\int _{\overline{\lambda }_{\mathrm{out}}(i)}^{\infty }\frac{2}{\beta _i\lambda _i} \ln \left( \frac{\lambda _i\Delta (i)}{\overline{\lambda }_{\mathrm{out}}(i)}\right) f_i(\lambda _i)d\lambda _i. \end{aligned}$$

(61)

With respect to (61), we define

$$\begin{aligned} \rho _s \left( \widehat{\hbox {SNR}}(i),\overline{\lambda }_{\mathrm{out}}(i)\right) =\int _{\overline{\lambda }_{\mathrm{out}}(i)}^{\infty }\frac{\widehat{\hbox {SNR}}(i)}{\lambda _i} f_i(\lambda _i)d\lambda _i \end{aligned}$$

(62)

and

$$\begin{aligned} \rho _{\Delta } \left( \overline{\lambda }_{\mathrm{out}}(i)\right) =\int _{\overline{\lambda }_{\mathrm{out}}(i)}^{\infty }\frac{2}{\beta _i\lambda _i} \ln \left( \frac{\lambda _i\Delta (i)}{\overline{\lambda }_{\mathrm{out}}(i)}\right) f_i(\lambda _i)d\lambda _i. \end{aligned}$$

(63)

In what follows, we consider to derive (62) and (63), respectively. $\square$

From Lemma 1 in [36], the marginal p.d.f. of the $i$-th largest eigenvalue $\lambda _i$ can be written as

$$\begin{aligned} f_i(\lambda _i)=\sum _{k=i}^m(-1)^{k-i}{k-1 \atopwithdelims ()i-1}{m \atopwithdelims ()k}f_{\mathrm{min:}k}(\lambda _i) \end{aligned}$$

(64)

where $f_{\mathrm{min:}k}(x)$ denotes the p.d.f. of the smallest random variable considered in a subset of k random variables over the set of all eigenvalues, and is given by

$$\begin{aligned} f_{\mathrm{min:}k}(x) &= \frac{kC}{m!}\sum _{{\mathcal {\alpha }}} \sum _{{\mathcal {\mu }}} \hbox {sgn}({\mathcal {\alpha }})\hbox {sgn} ({\mathcal {\mu }})A_{k}({\mathcal {\alpha }},{\mathcal {\mu }}) \nonumber \\&\times \,e^{-kx}{\mathbb {\sum }}_{{\mathcal {\tau }}} x^{\theta +\alpha _k+\mu _k-2+\sum _{\iota =1}^{k-1}\tau _{\iota }} \prod _{\iota =1}^{k-1}\frac{(\theta +\alpha _{\iota }+\mu _{\iota }-2)!}{\tau _{\iota }!}. \end{aligned}$$

(65)

With the help of the complementary incomplete gamma function $\Gamma (q,x)$, thus we can obtain the desired result (43) after a simple derivation.

The derivation of (63) is similar, but involves a process employing the following special function $\jmath _q(x)$ defined as [38]:

$$\begin{aligned} \jmath _q(x) &= \int _1^\infty t^{q-1}\ln t e^{-xt}dt\nonumber \\= & {} \frac{(q-1)!}{x^q}\sum _{k=0}^{q-1}\Gamma (k,x)/k!. \end{aligned}$$

(66)

Finally, again making use of (64), we can easily obtain the desired expression of $\rho _{\Delta }(\overline{\lambda }_{\mathrm{out}}(i))$ (44).