Channel estimation for AF relaying using ML and MAP

Abstract

Channel state information is very important in cooperative communication for performance improvement or signal demodulation. Previous works on channel state information estimation mainly focus on least squares and minimum mean squared error estimators. In this work, several new maximum likelihood and maximum a posteriori estimators for the cooperative network are proposed. Two of them estimate the individual channel powers of different links using all pilots from the source node, while two of them estimate the individual channel gains of different links using pilots from the source node as well as the relay node. Numerical results show that the estimation error decreases when the signal-to-noise ratio increases or when the number of pilots increases. The estimators for the individual channel gains have normalized mean squared errors of less than 0.001 when the signal-to-noise ratio is 30 dB, and their bit error rate performances are very close to the perfect case. Importantly, channel estimation is performed at the destination only such that there is no extra complexity at the relay node.

Appendices

Appendix 1: derivation of (10) and (11)

The likelihood function can be written as

$$\begin{aligned} f(\bar{u},\bar{v}|g_R,g_I,h_R,h_I)=\frac{1}{[2\pi (g_R^2+g_I^2+1)\sigma ^2]^I}e^{-\frac{\sum _{i=1}^I[u_i-(g_Rh_R-g_Ih_I)]^2+\sum _{i=1}^I[v_i-(g_Rh_I+g_Ih_R)]^2}{2(g_R^2+g_I^2+1)\sigma ^2}} \end{aligned}$$

(23)

where $\bar{u}=[u_1 u_2 \cdots u_I]$ and $\bar{v}=[v_1 v_2 \cdots v_I]$. This gives the log-likelihood function as

$$\begin{aligned}&logf(\bar{u},\bar{v}|g_R,g_I,h_R,h_I) = -I\ln [2\pi (g_R^2+g_I^2+1)\sigma ^2] - \frac{\sum _{i=1}^I[u_i^2+v_i^2]}{2(g_R^2+g_I^2+1)\sigma ^2}\nonumber \\&-\,\frac{I(g_R^2+g_I^2)(h_R^2+h_I^2)}{2(g_R^2+g_I^2+1)\sigma ^2}+\frac{(g_Rh_R-g_Ih_I)\sum _{i=1}^Iu_i+(g_Rh_I+g_Ih_R)\sum _{i=1}^Iv_i}{(g_R^2+g_I^2+1)\sigma ^2}. \end{aligned}$$

(24)

By taking the derivatives of (24) with respect to $g_R$, $g_I$, $h_R$ and $h_I$ and setting the results to zero, one has

$$\begin{aligned}&-2Ig_R(g_R^2+g_I^2+1)+\frac{g_R}{\sigma ^2}\sum _{i=1}^I(u_i^2+v_i^2)-\frac{Ig_R}{\sigma ^2}(h_R^2+h_I^2) \nonumber \\&+ \,\frac{1+g_I^2-g_R^2}{\sigma ^2}(h_R\sum _{i=1}^Iu_i+h_I\sum _{i=1}^Iv_i)-\frac{2g_Rg_I}{\sigma ^2}(h_R\sum _{i=1}^Iv_i-h_I\sum _{i=1}^Iu_i) = 0, \end{aligned}$$

(25)

$$\begin{aligned}&-\,2Ig_I(g_R^2+g_I^2+1)+\frac{g_I}{\sigma ^2}\sum _{i=1}^I(u_i^2+v_i^2)-\frac{Ig_I}{\sigma ^2}(h_R^2+h_I^2) \nonumber \\&+\, \frac{1+g_R^2-g_I^2}{\sigma ^2}(h_R\sum _{i=1}^Iv_i-h_I\sum _{i=1}^Iu_i)-\frac{2g_Rg_I}{\sigma ^2}(h_R\sum _{i=1}^Iu_i+h_I\sum _{i=1}^Iv_i) = 0, \end{aligned}$$

(26)

$$\begin{aligned}&-I(g_R^2+g_I^2)h_R + g_R\sum _{i=1}^Iu_i + g_I\sum _{i=1}^Iv_i = 0, \end{aligned}$$

(27)

$$\begin{aligned}&-I(g_R^2+g_I^2)h_I + g_R\sum _{i=1}^Iv_i - g_I\sum _{i=1}^Iu_i = 0. \end{aligned}$$

(28)

Moreover, by solving (27) and (28) for $h_R$ and $h_I$, respectively, and substituting them in (25) or (26), one can derive (10). Also, using (27) and (28), one has (11).

Appendix 2: derivation of (12) and (13)

The likelihood function using (3) and (4) is derived as

$$\begin{aligned} f(\bar{u},\bar{v}|g_R,g_I,h_R,h_I)=\frac{e^{-\frac{g_R^2+g_I^2+h_R^2+h_I^2}{2\alpha ^2}}}{[2\pi (g_R^2+g_I^2+1)\sigma ^2]^I2\pi \alpha ^2}e^{-\frac{\sum _{i=1}^I[u_i-(g_Rh_R-g_Ih_I)]^2+\sum _{i=1}^I[v_i-(g_Rh_I+g_Ih_R)]^2}{2(g_R^2+g_I^2+1)\sigma ^2}} \end{aligned}$$

(29)

where the extra exponential function comes from the distribution of the channel gain. From (29), the log-likelihood function is

$$\begin{aligned}&logf(\bar{u},\bar{v}|g_R,g_I,h_R,h_I) = -I\ln [2\pi (g_R^2+g_I^2+1)\sigma ^2] - \frac{\sum _{i=1}^I[u_i^2+v_i^2]}{2(g_R^2+g_I^2+1)\sigma ^2}\nonumber \\&-\,\frac{I(g_R^2+g_I^2)(h_R^2+h_I^2)}{2(g_R^2+g_I^2+1)\sigma ^2}+\frac{(g_Rh_R-g_Ih_I)\sum _{i=1}^Iu_i+(g_Rh_I+g_Ih_R)\sum _{i=1}^Iv_i}{(g_R^2+g_I^2+1)\sigma ^2}\nonumber \\&-\,\frac{g_R^2+g_I^2+h_R^2+h_I^2}{2\alpha ^2}-\ln [2\pi \alpha ^2]. \end{aligned}$$

(30)

Then, by taking derivatives of (30) with respect to $g_R$, $g_I$, $h_R$ and $h_I$ and setting them to zero, one has

$$\begin{aligned}&-2Ig_R(g_R^2+g_I^2+1)+\frac{g_R}{\sigma ^2}\sum _{i=1}^I(u_i^2+v_i^2)-\frac{Ig_R}{\sigma ^2}(h_R^2+h_I^2)-\frac{2g_Rg_I}{\sigma ^2}(h_R\sum _{i=1}^Iv_i-h_I\sum _{i=1}^Iu_i) \nonumber \\&+\, \frac{1+g_I^2-g_R^2}{\sigma ^2}(h_R\sum _{i=1}^Iu_i+h_I\sum _{i=1}^Iv_i) -\frac{g_R}{\alpha ^2}(g_R^2+g_I^2+1)^2= 0, \end{aligned}$$

(31)

$$\begin{aligned}&-2Ig_I(g_R^2+g_I^2+1)+\frac{g_I}{\sigma ^2}\sum _{i=1}^I(u_i^2+v_i^2)-\frac{Ig_I}{\sigma ^2}(h_R^2+h_I^2)-\frac{2g_Rg_I}{\sigma ^2}(h_R\sum _{i=1}^Iu_i+h_I\sum _{i=1}^Iv_i) \nonumber \\&+\, \frac{1+g_R^2-g_I^2}{\sigma ^2}(h_R\sum _{i=1}^Iv_i-h_I\sum _{i=1}^Iu_i) -\frac{g_I}{\alpha ^2}(g_R^2+g_I^2+1)^2= 0, \end{aligned}$$

(32)

$$\begin{aligned}&-I(g_R^2+g_I^2)h_R + g_R\sum _{i=1}^Iu_i + g_I\sum _{i=1}^Iv_i -\frac{h_R}{\alpha ^2}(g_R^2+g_I^2+1)\sigma ^2= 0, \end{aligned}$$

(33)

$$\begin{aligned}&-I(g_R^2+g_I^2)h_I + g_R\sum _{i=1}^Iv_i - g_I\sum _{i=1}^Iu_i -\frac{h_I}{\alpha ^2}(g_R^2+g_I^2+1)\sigma ^2= 0. \end{aligned}$$

(34)

By solving (33) and (34) for $h_R$ and $h_I$, respectively, and using them in (31) or (32), one has an equation for $x=g_R^2+g_I^2$ as

$$\begin{aligned} -(x+1)+\frac{R_2}{2\sigma ^2}-\frac{x}{2\sigma ^2}\frac{|R_1|^2}{[(x+1)\sigma ^2/(I\alpha ^2)+x]^2}+\frac{(1-x)|R_1|^2}{2\sigma ^2[(x+1)\sigma ^2/(I\alpha ^2)+x]}-\frac{(x+1)^2}{2I\alpha ^2}=0. \end{aligned}$$

(35)

This nonlinear equation does not have a closed-form expression for the solution but can be easily solved using numerical computations. Assume that the solution to (35) is $x_0$. Thus, one has (12). Using (33), (34) and (12), the MAP estimator for the channel power $|h|^2$ can be derived as (13).