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Let $y[n]= h*x[n] + w[n]$, where $h$ is an unknown but deterministic parameter, $x[n]$ is a BPSK random variable with equal probability of +1 and -1, $w[n]$ are i.i.d. Gaussian with zero mean and unknown variance . Given $N$ observations find the estimate of $h$, and variance of noise.

$x[n]$ and $w[n]$ are independent.

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  • $\begingroup$ If we take average values of $y$, it will eliminate the effect of $w$ but will not help us estimate $h$ because $x$ is not known to you. If we consider average of squared values of $y$, $ P = \frac{1}{N}\sum_0^{N-1}y^2[n] = \frac{1}{N}\sum_0^{N-1}(h^2x_n^2 + w_n^2 + 2hx_nw_n) $ The expectation value of above estimator is $ E(P) = \frac{1}{N}\sum_0^{N-1}(h^2 \times 1 + \sigma^2 + 0) = h^2 + \sigma^2 $ Still you have 2 unknowns with only 1 estimator. $\endgroup$
    – jithin
    Mar 30 '20 at 17:02
  • $\begingroup$ Can we assume h is not time varying? $\endgroup$ Apr 29 '20 at 2:46
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Squaring Method

The input signal $x[n]$ can take two values: $+1$ or $-1$. After being multiplied by $h$, the signal becomes $|h|e^{j\angle h}$ (if $x[n]=+1$) or $|h|e^{j(\pi + \angle h)}$ (if $x[n]=-1$). By squaring this we get, $|h|^2e^{j2\angle h}$. Both points get mapped to this. Now we can pick out the magnitude and phase of $h$. In summary, the steps are:

  1. Square $y[n]$ to get $z[n]=y^2[n]$.
  2. Get the magnitude, $\hat{|h|}=\sqrt{\bigg|\frac{1}{N}\sum_n z[n]\bigg|}$. It was pointed out in the comments that this is a biased estimate with bias $\sigma^2$, so it is not favorable for a large $\sigma$. I describe another way to estimate $\sigma^2$ (without using $\hat{h}$) so that we can use this $\hat{\sigma}^2$ to remove the bias from $\hat{|h|}$.
  3. Get the phase, $\angle h=\frac{1}{2N} \sum_n \angle z[n]$
  4. Use $\hat{h}$ to get the noise variance, $\hat{\sigma}^2=\bigg(\frac{1}{N}\sum_n z[n]\bigg)-\hat{h}^2$. This comes from the fact that $E\big[z[n]\big]=h^2+\sigma^2$ (also see the clustering method is a better way to go about this as it does not depend on any other estimates).

Caveat: This method is valid only for phase offsets within $\pm \frac{\pi}{2}$. Anything beyond that, the received signal becomes indistinguishable and there are multiple phase shifts which could give the same signal.

Clustering Method

If $h$, can be correctly determined, then you can take hard decisions and calculate the noise variance using the distance of the received symbols from the decision you made.

If $h$ can't be determined right away, then this clustering based estimate of $|h|$ and $\sigma^2$ can be used: We know there should be two clusters ($\pm 1$) so we can use a simple method like k-means (https://en.wikipedia.org/wiki/K-means_clustering) with $k=2$. The algorithm is:

  1. Run k-means on the received samples and get a list of labels back labeling each sample to either cluster $1$ or $2$. Let $S_1$ be the set of sample indices in cluster $1$ and $S_2$ be the set of sample indices in cluster $2$.

  2. Now we are going to take all the points in each cluster, shift them be zero mean and take the variance (this will give us a noise variance estimate for each cluster). Then we will average over all of the cluster noise variance estimates to form the final estimate. $\hat{\sigma}^2=\frac{\bigg( \text{var}\big(y[n]-E[y[n]] \big)\big|_{n \in S_1} + \text{var}\big(y[n]-E[y[n]] \big)\big|_{n \in S_2} \bigg)}{2}$

  3. Now we can take the estimate from step $2$ and subtract off the bias term which we estimated here as $\hat{\sigma}^2$ to get the refined estimate $\hat{|h|}_0=\hat{|h|}-\hat{\sigma}^2$. You could also get $|h|$ using the magnitude of the cluster means, $|\hat{h}|=\frac{\text{E}\big(y[n]\big|_{n \in S_1}\big)+\text{E}\big(y[n]\big|_{n \in S_2}\big)}{2}$.

There is also a way get $\angle h$ from the clusters but it too has the criteria that $\angle h$ must be less than $\pm \frac{\pi}{2}$ so I will not detail it here.

Long story short: $|h|$ and $\sigma^2$ are doable, but $\angle h$ is only doable for some cases.

Code: https://github.com/B-William/DSPSE/blob/master/blindEstimationOfSignalParameterAndNoiseVariance.m

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  • $\begingroup$ And how would you determine the ML of x without knowing h, wouldn't the value of h appear in the ML of X $\endgroup$ Mar 31 '20 at 13:41
  • $\begingroup$ The above answer is being edited $\endgroup$
    – Engineer
    Mar 31 '20 at 13:42
  • $\begingroup$ But squaring the received samples will involve cross product terms between h*x(n) and w(n)... since there is no expectations taken in the answer then these do not disappear...I prefer to take modulus of y(n) , that works better I think $\endgroup$ Mar 31 '20 at 14:11
  • $\begingroup$ Also how do you get the variance of noise which is unknown $\endgroup$ Mar 31 '20 at 14:18
  • $\begingroup$ The cross terms average to zero since the noise is zero mean. I don't need the noise variance since it doesn't appear in my answer at all. All you need is the received samples (that includes the signal plus noise). $\endgroup$
    – Engineer
    Mar 31 '20 at 14:48

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