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Given measurements $$\begin{align} Z_1 &= x + N_1\\ Z_2 &= x + N_2 \end{align}$$ where $N_1$ and $N_2$ are independent zero-mean Gaussian random variables with variances $\sigma_1^2$ and $\sigma_2^2$ respectively, it can be shown that the minimum-mean-square-error (MMSE) estimate of $x$ in terms of $Z_1$ and $Z_2$ is a linear function of $Z_1$ and $...


4

Generally, the frequency estimation problem decouples from the amplitude + phase estimation problem. As I said in the comments, you could just do: $$ \hat{A} = \left|\sum_{n=0}^{N-1} x_n e^{-j 2\pi\hat{f} n} \right| $$ where $\hat{f}$ is your frequency estimate. Regarding your issues: because the number of points isn't high enough to hit the right ...


3

A PRN sequence is a Pseudo-Random Noise sequence, often generated by using an Linear Feedback Shift Register (LFSR) with the feedback taps done by using a primitive irreducible polynomial in GF{2}, which is the Golois Field of 2 elements. When a primitive and irreducible polynomial in GF{2} is used, the LFSR will produce a "maximum length sequence", meaning ...


2

An approach that might be worthwhile investigating is some sort of on-line method. Because of the highly non-linear nature of your signal model, the standard Kalman filter is out, but an Extended Kalman filter (or, perhaps, a particle filter) might be worth looking at. The idea is that you set up your signal model state to be: $$ x_t = \left[ \alpha\ \...


2

The basic idea behind maximum entropy models is that you want to make the least assumption about the data. This is considered equivalent to retaining as much unpredictability as possible, as quantified by entropy. For more information I refer you to this Wikipedia article. Information and unpredictability are closely related. If you think of the signal as a ...


2

I'm not sure what's you model is. Let's say it is something like: $$ y = H x + n $$ Now, using the Least Squares model is optimal (In the MSE sense) when $ n $ is AWGN (It is the linear optimal estimator if the noise is white). So unless the noise in your model is colored, no gain by filtering the data before applying the Least Squares method. Now, what ...


1

You could simply run your signal through some sort of high pass filter or shelf and compare the energy before and after the filter. If the ratio is high, you have a lot of high frequencies in your signal and should probably do less smoothing. If the ratio is low, there are not a lot of high frequencies and you can do more aggressive smoothing. The ...


1

In the case $a$ is a random variable (r.v.) we distinguish between two types of unbiased: (hereinafter I'm switching to $\theta$ and $\hat{\theta}$ instead of $a$ and $\hat{a}$ respectively): Unbiased in the strong sense: given by $E\left \{ \left ( \hat{\theta} - \theta \right )| \theta\right \} = 0, \quad \forall \theta$. Most often tha n not it is hard ...


1

Fristly, check my answer again, I edited it because the first paragraph wasn´t extrictly right, I hope you notice the difference. (1) YES, in the example of the sample mean, its variance it is also the CRLB, so if N goes to infinity, the CRLB tends to zero. It means that if you want fewer deviation from the expectation of the estimator, you nedd larger ...


1

You may want an estimator with minimun variance because each time you calculate it, it is likely to be closer to its expectation, and in the case that the estimator is unbiased, closer to the true parameter. An estimator is inconsisten if increasing the number of samples of the dataset to calculate it doesn't reduces its variance. The sample mean estimator ...


1

Since $ M \in \mathbb{S}^{N}_{++} $ (In other convention $ M \succ 0 $) by Cholesky Decomposition there is a Triangular Matrix $ R \in \mathbb{R}^{N \times N} $ such that $ M = {R}^{T} R $. Using this fact one could prove $ 1 \iff 2 $ as following: $$\begin{align*} \arg \min_{\hat{x}} \mathbb{E} \left[ {\left( \hat{x} - x \right)}^{T} M \left( \hat{x} - x \...


1

Use the Cholesky decomposition of $M$ as a change of coordinates. $tr(ABC) = tr(CAB)$ - trace is invariant under cyclic permutations, and $tr(scalar)=scalar)$. Thus, $tr(M E[ x x^T]) = tr(E[M x x^T]) = E[tr(M x x^T) ] = E[tr( x^T M x ) ] = E[x^T M x]$. Now, replace $x$ with $\hat{x}-x$.


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