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 $B = R(0) + R(1)$.

  • $\begingroup$ Just to be completely sure we're on the same page here: Your signal $x$ is a complex, multidimensional signal? $\endgroup$ Aug 31 '19 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. $\endgroup$ Aug 31 '19 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 '19 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. $\endgroup$ Aug 31 '19 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 '19 at 16:17

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.

  • $\begingroup$ Thank you for the detailed answer it is much appreciated! $\endgroup$
    – chaosmind
    Sep 1 '19 at 9:24
  • $\begingroup$ You're welcome! Glad it helped. $\endgroup$
    – Florian
    Sep 1 '19 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 '19 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 '19 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 '19 at 6:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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