0
$\begingroup$

From the book Fundamentals of Massive MIMO, the section 2.3.4 shows that the expression of the received signal in fading channel with additive non-Gaussian noise and no CSI at receiver is dervied as: $$y = \sqrt{\rho}\operatorname{E}(g)x + \sqrt{\rho}(g-\operatorname{E}(g))x + w,$$

where $\rho$ is a constant, g stands the unknown channel and w represents the noise ~ N(0,1). As a result, the SINR of the signal is:

$$\text{SINR} = \frac{\rho |\operatorname{E}(g)|^2}{\rho \operatorname{Var}(g) + 1},$$

since $\operatorname{E}\{X-\operatorname{E}\{X\}\} = \operatorname{Var}\{X\}$.

However, I am still confused if I consider $g = \hat{g} + \widetilde{g}$, for $\hat{g}$ is the estimated channel and $\widetilde{g}$ is the estimation error, therefore the signal received at the receiver could be $y = \sqrt{\rho}\operatorname{E}(\hat{g})x + \sqrt{\rho}\operatorname{E}(\widetilde{g})x + w$, and the SINR is

$$\text{SINR} = \frac{\rho |\operatorname{E}\{\hat{g}\}|^2}{\rho |\operatorname{E}\{\widetilde{g}\}|^2 + 1}.$$

I cannot make sure that these two SINR expressions are equal, and I have no idea if the second SINR expression is correct or not. If there is any part which leads to my wrong understanding, please let me know. Thank you.

$\endgroup$

1 Answer 1

1
$\begingroup$

Using your notations, if $$g = \hat{g} + \widetilde{g}, \tag{1}$$ then $$\textrm{E}\{g\} = \text{E}\{\hat{g}\} + \text{E}\{\widetilde{g}\},\tag{2}$$

where $\text{E}\{\widetilde{g}\}$ is bias of the estimator.

The SINR of the book corresponds to the cases no bias, i.e. $\text{E}\{\widetilde{g}\} = 0$, and there is bias but it is known.

If the bias is unknown, what we get is $\sqrt{\rho}\text{E}(\hat{g})x$ and, therefore, $$y = \sqrt{\rho}gx+w =\\ \sqrt{\rho}\text{E}(\hat{g})x + \sqrt{\rho}(g-\text{E}(\hat{g}))x + w =\\ \sqrt{\rho}(\text{E}(g) - \text{E}\{\widetilde{g}\})x + \sqrt{\rho}(g-\text{E}(g) + \text{E}\{\widetilde{g}\})x + w, \tag{3}$$ and $$\text{SINR} = \frac{\rho |\text{E}(g) - \text{E}\{\widetilde{g}\}|^2}{\rho \text{Var}(g) + \rho |\text{E}\{\widetilde{g}\}|^2 +1} \tag{4}$$

$\endgroup$
7
  • $\begingroup$ Thank you. So it is just because I didn't notice that the original expression doesn't consider the bias which is the $\widetilde{g}$ in my notation? And one more question about format: how to show the '{' and '}' in equations as in your answer? I tried to use '\{' and '\}' as what I do in Overleaf Editing but it doesnot work ( such as to show 'E{g}' in an expression like (2) $\endgroup$ Jul 10, 2023 at 10:01
  • $\begingroup$ @LocoCitato No, the original expression is true even with bias if it is known. Because if the bias is known, we can easily subtract it in the estimate. The reasoning is that first, what do you mean by SINR? In this case, you want to inject it to the formula of channel capacity that is defined for AWGN channels. Therefore, you need to "convert" the received $y$ to the form of AWGN $Ax + n$, in which $A$ is known and $n$ is everything unknown (we assume that $n$ can be approximated as a normal random variable) ... $\endgroup$
    – AlexTP
    Jul 10, 2023 at 11:12
  • $\begingroup$ $A$ is the channel information obtained at the output of the channel estimator and, therefore, the SINR expression does depend on how you estimate the channel and use the estimate. The orignal expression assumed that the estimator gave us $\text{E}\{g\}$. If the estimator give us st biased, and we don't know that the estimate is biased, we will wrongly use the estimate as $\text{E}\{\hat{g}\}$ as $\text{E}\{g\}$, which leads to the Eq (4). $\endgroup$
    – AlexTP
    Jul 10, 2023 at 11:19
  • $\begingroup$ Of course, if you change your estimator, e.g. correctly estimate $\text{var}(g)$, or how to use the estimate (this is possible, but not for SINR that is used to evaluate channel capacity as we assume the best channel equalization without information loss). $\endgroup$
    – AlexTP
    Jul 10, 2023 at 11:23
  • $\begingroup$ About the formatting, I did use '\{' and '\}'. You can click "Edit" below my post to see how I formatted the Latex-like equations. $\endgroup$
    – AlexTP
    Jul 10, 2023 at 11:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.