On the Achievable Rate for Wideband Channels with Estimated CSI

Abstract

We consider the achievable rate for frequency-selective fading channels when the channel state information (CSI) is to be estimated at the receiver. Since the estimated CSI is not perfect, the achievable rate must be degraded from that with perfect CSI. Using the rate-distortion theory, we study an upper bound on the achievable rate and investigate how the achievable rate can be maximized through an optimization problem by allocating the resources such as degrees of freedom (the transmission time in our work or transmission power) for the exploration of CSI (i.e., the channel estimation) using pilot symbols, and the exploitation of channels to transmit data symbols. Although our study is based on some ideal assumptions, the results could help develop flexible communication systems such as software defined radio (SDR) to achieve a best performance by optimizing radio resources for unknown channels.

Appendices

Appendix A: Proof of Theorem 4

The second order derivative of ${{\cal K}}(\alpha)$ is given by

$$ \frac{d^2 {{\cal K}}(\alpha)}{d \alpha^2} = - 2 \frac{d \psi\left(\bar \gamma(\alpha)\right)}{d \alpha} + (1-\alpha) \frac{d^2 \psi\left(\bar \gamma(\alpha)\right)}{d \alpha^2 }. $$

(52)

From this, we can see that if the following two conditions hold:

• (i) $\psi(\bar \gamma(\alpha))$ is nondecreasing;

• (ii) $\psi(\bar \gamma(\alpha))$ is concave,

then ${{\cal K}}(\alpha)$ is concave or $\frac{d^2 {{\cal K}}(\alpha)}{d \alpha^2} < 0$.

For the first condition, we need the following two results.

Lemma 2

$I_{\rm erg,d} (D) = \psi(\bar \gamma(D))$ is decreasing on the interval D ∈ [0, 1].

Proof

We can show that $\bar \gamma(D)$ is decreasing on D ∈ [0, 1]. Since ψ(x) is increasing, $I_{\rm erg,d} (D) = \psi(\bar \gamma(D))$ is decreasing. □

Lemma 3

If γ _a ≫ 1, D ^*(α) is nonincreasing on the interval α ∈ (0, 1).

Proof

If γ _a ≫ 1, D ^*(α) in Eq. 37 becomes

$$ D^*(\alpha) = \bar D^*(\alpha) { \buildrel \triangle \over = } \frac{ \sqrt{ \alpha^2 + 4 (1-\alpha) \alpha } - \alpha} {2 (1- \alpha )\alpha \gamma_{\rm a}}. $$

After some manipulations, we can show that $\frac{d}{d \alpha} \bar D^*(\alpha) \le 0$, α ∈ (0,1). □

From Lemmas 2 and 3, the first condition is satisfied if γ _a ≫ 1.

The following results are crucial to verify the second condition.

Lemma 4

$\psi(x) = E[\log (1+ x |h_l|^2)]$ is concave and nondecreasing.

Proof

Since log(x) is concave and a sum of concave functions are also concave [19], we can conlcude that $\psi(x) = E[\log (1+ x |h_l|^2)]$ is concave. Furthermore, since log(1 + x) is nondecreasing, ψ(x) is also nondecreasing. □

From Eq. 29, we have the following relation:

$$ \psi\left(\bar \gamma(\alpha)\right) = E\left[\log \left(1+ \bar \gamma(\alpha) |h_l|^2\right)\right]. $$

(53)

Lemma 5

If $\bar \gamma(\alpha)$ is concave, $\psi(\bar \gamma(\alpha))$ is concave.

Proof

According to [19], a composite function h(f(x)) is concave if h(x) is concave and nondecreasing and f(x) is concave. We can show that $\psi(\bar \gamma(\alpha)) = E[\log (1+ \bar \gamma(\alpha) |h_l|^2)]$ is concave because (1) ψ(x) is concave and nondecreasing from Lemma 4 and (2) $\bar \gamma(\alpha)$ is assumed to be concave. □

From Lemma 5, we now see that the second condition is satisfied if $\bar \gamma(\alpha)$ is concave.

For a high γ _a, we have

$$\begin{array}{rll} \lim\limits_{\gamma_{\rm a} \rightarrow \infty} \frac{\bar \gamma(\alpha)}{\gamma_{\rm a}} &=& v (\alpha) \\ &{\buildrel \triangle \over = }& \frac{2 (1-\alpha)}{\sqrt{\frac{4-3 \alpha}{\alpha} } + 1}. \label{EQ:Va} \end{array}$$

(54)

After some manipulations, we can show that $\frac{d^2}{d \alpha^2} v(\alpha) \le 0$. This confirms that $\bar \gamma(\alpha)$ is concave when γ _a ≫ 1. This completes the proof.

Appendix B: Proof of Theorem 5

From Theorem 4, it is known that ${{\cal K}}(\alpha)$ is concave for γ _a ≫ 1. This implies that α ^* is the solution of the following equation:

$$ 0 = \frac{d}{d \alpha}{{\cal K}} (\alpha). \label{EQ:00} $$

(55)

As x increases, we can approximate ψ(x) as

$$ \psi(x) \simeq \log(x) + C, $$

where $C = E[\log|h_l|^2]$ is a constant. This allows a high SNR approximation of ${{\cal K}}(\alpha)$ as follows:

$$\begin{array}{rll}{{\cal K}}(\alpha) &\simeq& (1- \alpha) \log\left(\bar \gamma(\alpha)\right) + (1- \alpha) C \\ &\simeq& (1- \alpha) \left( \log(\gamma_a) + \log(v(\alpha) \right) + (1-\alpha) C . \label{EQ:acK} \end{array}$$

(56)

The second approximate results from Eq. 54. For convenience, logarithms are taken to the base e. Taking the derivate with respect to α, we have

$$ \frac{d}{d \alpha}{{\cal K}} (\alpha) = - \log_e \gamma_{\rm a} + \frac{1}{2 \alpha} - \frac{1}{4 \sqrt{\alpha}} + O(1). $$

(57)

As γ _a increases, α decreases and $\frac{1}{2 \alpha}$ becomes dominant. Thus, for a high SNR, the solution of Eq. 55 approaches

$$ \bar \alpha^* = \frac{1}{2 \log_e \gamma_{\rm a}}, \gamma_{\rm a} \gg 1. \label{EQ:ba*} $$

(58)

Appendix C: Proof of Theorem 6

From Eq. 56, we have

$$\begin{array}{rll}{\cal K}(\alpha^*) &\simeq& \left(1- \alpha^*\right) \left( \log(\gamma_a) + \log(v\left(\alpha^*\right) \right) + \left(1-\alpha^*\right) C , \gamma_{\rm a} \\ &\gg& 1. \label{EQ:AK} \end{array}$$

(59)

From Eq. 54, consider the Taylor series of v(α) at 0 as follows:

$$\begin{array}{rll} v (\alpha) &=& \frac{2 (1-\alpha)}{\sqrt{\frac{4-3 \alpha}{\alpha}} + 1} \\ &=& \sqrt{\alpha} - \frac{\alpha}{2} + O\left(\alpha^{3/2}\right). \end{array}$$

Since α ^* approaches 0 as γ _a increases, from Eq. 58, we have

$$ v\left(\alpha^*\right) \simeq \sqrt{\alpha^*} \simeq \frac{1}{2 \log_e \gamma_{\rm a}}, \gamma_{\rm a} \gg 1. $$

Thus,

$$ \log v\left(\alpha^*\right) \simeq - \frac{1}{2} \log \log \gamma_{\rm a} + \frac{1}{2} \log C_1, \label{EQ:va*} $$

(60)

where $C_1 = \frac{\log e}{2}$. Substituting Eq. 60 into Eq. 59, we have

$${\cal K}\left(\alpha^{*}\right) \simeq \log(\gamma_a) - \frac{1}{2} \log \log \gamma_{\rm a} + O(1), \gamma_{\rm a} \gg 1. \label{EQ:AK2} $$

(61)