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I am studying the role of an auto-correlation matrix for random signals and the difference of energy between a lag 0 and lag 1 matrix.

Consider a complex input signal $x(k)=[x1,x2]^T$ and $x(k-1)=[x0,x1]^T$, as column vectors with auto-correlation matrix of lag 0, $R(0)$, with diagonal $r(0)$ and auto-correlation matrix of lag 1 $R(1)$, with diagonal $r(1)$.

where $R(0)=E[x(k)x^H(k)]$ and $R(1)=E[x(k)x^H(k-1)]$

  • what is the relationship between the energy resulting from $R(0)$ and $R(1)$? can we thus say that eigenvalue max of $R(0)$ is greater then that of $R(1)$
  • and the difference in diagonal entries $r(0)$ and $r(1)$?

given that $r(0)$ is always positive real and $r(1)$ can be complex.

  • What can we comment of the real and imaginary part of the $r(1)$ entry?
  • and where does the energy resides i.e. real portion or imaginary?

  • Additionally to a wide sense stationary signal $x(k)$ can we assume that $r(1)$ is approximately equal to $r(0)$?

Thus if we have $B = R(0) + R(1)$, we know that $R(0)$ is positive definite, symmetric and hermitian however $R(1)$ is not. Can we say that eigenvalue max of $B$ is eigenvalue max of $R(0)$ + the norm of that of $R(1)$ assuming its complex?

My apologies for the successive questions however I am trying to understand the differences between the auto-correlation matrix at lag 0 and lag 1 in terms of eigenvalues and energy because I have to work with a matrix $B = R(0) + R(1)$.

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  • $\begingroup$ Just to be completely sure we're on the same page here: Your signal $x$ is a complex, multidimensional signal? $\endgroup$ – Marcus Müller Aug 31 at 16:04
  • $\begingroup$ What is "energy resulting from (a matrix) R(0)"? I've not heard energy resulting from matrices so far, so I think we'll need you to exactly define that. $\endgroup$ – Marcus Müller Aug 31 at 16:06
  • $\begingroup$ Doesn't the R auto-correlation matrix has the maximum peak at 0 i.e energy? x is complex signal 1*N $\endgroup$ – chaosmind Aug 31 at 16:08
  • $\begingroup$ A matrix doesn't have a peak. And if your signal is onedimensional, then there's no two autocorrelation matrices. I don't really know how to interpret any of what you say – edit your question to include your mathematical definition of what $R(0)$ is, in formalized writing. $\endgroup$ – Marcus Müller Aug 31 at 16:14
  • $\begingroup$ If the matrix x is a 1*N vector doesn't R(0) = E[x(k)x^H(k)] where E is the expectation and H is the hermitian i.e. conjugate transpose. thus i get an N*N R(0) and for the R(1) is can be defined as E[x(k)x^H(k-1)] for stationary signals it is k-k+1 hence R(1). My apologies I will check the help for the formatting $\endgroup$ – chaosmind Aug 31 at 16:17
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I had some problems understanding your questions, but from what you wrote I'm assuming you meant $\mathbf x(k) = [x(k), x(k+1)]^T$ so that $\mathbf x(k-1) = [x(k-1), x(k)]^T$ (using bold faced letters to distinguish scalars from vectors). Is that right?

In this case, expanding your equations, we get the following: $$ R(0) = \begin{bmatrix} r(k,k) & r(k,k+1) \\ r(k+1,k) & r(k+1,k+1) \end{bmatrix}, \quad R(1) = \begin{bmatrix} r(k,k-1) & r(k,k) \\ r(k+1,k-1) & r(k+1,k) \end{bmatrix}.$$

What we can see is that $R(0)$ is Hermitian symmetric (since $r(k,k+1) = r(k+1,k)^*$) and positive definite. Otherwise, unless further assumptions are made, we cannot argue much.

If you assume that your sequence is white sense stationary, we have $r(k,\ell) = r_{\ell-k}$ and we obtain $$ R(0) = \begin{bmatrix} r_0 & r_1 \\ r_1^* & r_0 \end{bmatrix}, \quad R(1) = \begin{bmatrix} r_1^* & r_0 \\ r_2^* & r_1^* \end{bmatrix}.$$ Moreover $r_0$ represents the signal's variance ("energy") and you have $|r_1|\leq r_0$ and $|r_2|\leq r_0$ from Schwartz' inequality. In particular, you can say that $r_1 = r_0 \rho_1$ where $\rho_1$ is the normalized (Pearson) correlation coefficient between adjacent lags of your sequence. Its magnitude is in $[0,1]$ (0 for an uncorrelated sequence, 1 for a constant [or alternating] sequence) and its phase can be anything. This answers why $r_1$ is complex and how to interpret real and imaginary part. It's only close to $r_0$ if the temporal correlation is high. For uncorrelated sequences its zero and hence nowhere close to $r_0$.

Now, $R(0)$ will have eigenvalues $r_0 \pm | r_1|$, so the max eigenvalue is between $r_0$ (for an uncorrelated sequence) and $2r_0$ (for a constant sequence). The eigenvalues of $R(1)$ will be $\left(r_1 \pm \sqrt{r_0 r_2}\right)^*$. From this, one can see that the dominant eigenvalue of $R(0)$ will be greater than or equal to the dominant eigenvalue of $R(1)$ with equality only for a constant sequence.

Finally, the eigenvalues of $R(0)+R(1)$ can be shown to be $ r_0 + r_1^* \pm \sqrt{(r_1^* + r_2^*) (r_1 + r_0)}$, so it's not just the dominant eigenvalue of $R(0)$ plus something.

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  • $\begingroup$ Thank you for the detailed answer it is much appreciated! $\endgroup$ – chaosmind Sep 1 at 9:24
  • $\begingroup$ You're welcome! Glad it helped. $\endgroup$ – Florian Sep 1 at 11:19
  • $\begingroup$ I just have a question regarding the eigenvalues of $R(0)+R(1)$, which formula did you use to get this form? I wasn't able to reach the same results, I used $lambda^2-Tr(B)*lambda+det(B)$ where $Tr$ is trace and $det$ is the determinant $\endgroup$ – chaosmind Sep 3 at 7:36
  • $\begingroup$ I had indeed made a typo that I just corrected. Your equation looks right, it should work. $\endgroup$ – Florian Sep 3 at 8:07
  • $\begingroup$ Thank you! Can we say that $||r(0)+r^*(1)||$ i.e. the diagonal entry of matrix $B$ is the energy? or it only applies for the auto-correlation matrix with symmetric properties? $\endgroup$ – chaosmind Sep 4 at 6:34

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