User Capacity Scaling Laws of Multi-user Fading Channels in the Presence of MMSE Channel Estimators

Abstract

Considering development of delay-sensitive applications in wireless communications and cellular networks along with lack of system resources and environmental factors including fading and noise all together make it impossible for all users to be active and to achieve quality of service requirements simultaneously. To analyze wireless systems performance, user capacity is defined as the maximum number of users that can be activated at the same time. In order to obtain user capacity scaling laws, channel distribution information is required and channel gains can be estimated by different channel estimators such as minimum mean square error estimators. In this paper, estimated channel gains are substituted for true channel gains and effects of imperfect channel estimation errors on the user capacity scaling laws are analyzed. It is shown that the user capacity scales double logarithmically and there is a gap depending on channel estimators accuracy between the lower and upper bounds. Moreover, assuming estimated channels for different users are independent and identically distributed, it is shown that the user capacity scaling laws are asymptotically tight.

Appendices

Appendix 1: Rayleigh Fading

Consider the multi-user channel, with the independent gains $\hat{h}_i\sim {\mathcal {C}}{\mathcal {N}}\left( 0,\hat{\sigma }^2_{h,i}\right)$ for $i=1,\ldots ,n$. Then, $|\hat{h}_i|\sim { Rayleigh}\left( \sqrt{\hat{\sigma }^2_{h,i}/2}\right)$ and $|\hat{h}_i|^2\sim {\Gamma }\left( 1,\hat{\sigma }^2_{h,i}\right)$ for $i=1,\ldots ,n$. Hence, the cumulative distribution function of $|\hat{h}_i|^2$ according to the Gamma distribution is given by

$$\begin{aligned} {\mathbf {F}}_{Ry}(x;\hat{\sigma }^2_{h,i})= \frac{\gamma \left( 1,x/\hat{\sigma }^2_{h,i}\right) }{\Gamma (1)} =\gamma \left( 1,x/\hat{\sigma }^2_{h,i}\right) \end{aligned}$$

(29)

where $\Gamma (\cdot ) , \Gamma (\cdot ,\cdot )$ and $\gamma (\cdot ,\cdot )$ are the gamma, the upper incomplete gamma and the lower incomplete gamma functions respectively (see [18, Eqs. (13)–(15) and (23)]).

Appendix 2: Rician Fading

Consider the multi-user channel with independent gains $\hat{h}_i\sim {\mathcal {C}}{\mathcal {N}}\left( \hat{\mu }_i,\hat{\sigma }^2_{h,i}\right)$, for $i=1,\ldots ,n$; if $\tilde{h}_i\sim {\mathcal {C}}{\mathcal {N}}\left( \hat{\mu }_i,2\right)$, according to Lemma 9.1, $|\hat{h}_i|=|\sqrt{\hat{\sigma }^2_h/2}\tilde{h}_i|\sim Rice\left( \sqrt{\hat{\sigma }^2_{h,i}/2},\sqrt{\hat{\sigma }^2_{h,i}/2}\hat{\mu }_i\right)$ and $|\tilde{h}_i|^2\sim {\mathcal {N}}{\mathcal {C}} \chi _2^2\left( \hat{\mu }_i^2\right)$ (i.e. non-central Chi-square distribution with two degrees of freedom) with the cumulative distribution function

$$\begin{aligned} {\mathbf {F}}_{{\mathcal {N}}{\mathcal {C}} \chi _2^2}(x;2,\hat{\mu }_i^2) = \sum _{j=0}^{\infty } e^{-\hat{\mu }_i^2/2} \frac{\left( \hat{\mu }_i^2/2\right) ^j}{j!}\, \frac{\gamma (j+1,x/2)}{\Gamma (j+1)} \end{aligned}$$

As $\hat{h}=\sqrt{\hat{\sigma }^2_h/2}\cdot \tilde{h}$, the distribution function of $|\hat{h}|^2$ equals to

$$\begin{aligned} {\mathbf {F}}_{Rc}(x;\hat{\sigma }^2_{h,i},\hat{\mu }_i)&= {\mathbf {F}}_{{\mathcal {N}}{\mathcal {C}} \chi _2^2}\left( 2 x/\hat{\sigma }^2_{h,i};2,\hat{\mu }_i^2\right) \nonumber \\&= \sum _{j=0}^{\infty } e^{-\hat{\mu }_i^2/2} \frac{\left( \hat{\mu }_i^2/2\right) ^j}{j!}\, \frac{\gamma \left( j+1, x/\hat{\sigma }^2_{h,i}\right) }{\Gamma (j+1)}. \end{aligned}$$

(30)

Lemma 9.1

Consider a Rician random variable $X\sim Rice(\xi ,\kappa )$. For any positive $r$ , if $Y=rX,\,Y$ is distributed as $Rice(r\xi ,r\kappa )$.

Proof

$$\begin{aligned} f_Y(y)&= \frac{1}{|r|}f_X(y/r)=\frac{1}{r}f_X(y/r) \\&= \frac{y/r}{r\kappa ^2}\exp \left( - \dfrac{y^2/r^2+\xi ^2}{2\kappa ^2}\right) I_0 \left( \dfrac{y/r\xi }{\kappa ^2} \right) \\&= \frac{y}{(r\kappa )^2}\exp \left( - \dfrac{y^2+(r\xi )^2}{2(r\kappa )^2}\right) I_0 \left( \dfrac{y(r\xi )}{(r\kappa )^2} \right) \end{aligned}$$

$\square$

Appendix 3: Nakagami Fading

Consider the multi-user channel with independent absolute gains $|\hat{h}_i| \sim Nakagami({\hat{\Omega }}_i,\hat{m}_i)$ for $i=1,\ldots ,n$. Then, $|\hat{h}_i|^2 \sim \Gamma (\hat{m}_i, {\hat{\Omega }}_i/\hat{m}_i)$ with the cumulative distribution function

$$\begin{aligned} {\mathbf {F}}_{Nk}(x;\hat{m}_i,{\hat{\Omega }}_i) = \frac{\gamma \left( \hat{m}_i,\frac{\hat{m}_i }{{\hat{\Omega }}_i}x\right) }{\Gamma (\hat{m}_i)}. \end{aligned}$$

(31)