# Energy definition for Autocorrelation lag 0 and lag 1 for complex signals

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

• Just to be completely sure we're on the same page here: Your signal $x$ is a complex, multidimensional signal? – Marcus Müller Aug 31 '19 at 16:04
• 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. – Marcus Müller Aug 31 '19 at 16:06
• Doesn't the R auto-correlation matrix has the maximum peak at 0 i.e energy? x is complex signal 1*N – chaosmind Aug 31 '19 at 16:08
• 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. – Marcus Müller Aug 31 '19 at 16:14
• 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 – 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.

• Thank you for the detailed answer it is much appreciated! – chaosmind Sep 1 '19 at 9:24
• You're welcome! Glad it helped. – Florian Sep 1 '19 at 11:19
• 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 – chaosmind Sep 3 '19 at 7:36
• I had indeed made a typo that I just corrected. Your equation looks right, it should work. – Florian Sep 3 '19 at 8:07
• 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? – chaosmind Sep 4 '19 at 6:34