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Problem: I am looking at an adaptive filtering application where the eigenvaluespread of the autocorrelation matrix $R$ is important for the convergence of the algorithm. For a single channel system the autocorrelation matrix $R$ for iterationstep $n$ can be calculated by $R=E\{ x(n) x^H(n)\}$ where $x(n)$ is the input signal of the adaptive filter at iterationstep $n$ consisting of a number of samples $N$ recorded over a timespan. The calculation of the eigenvalues is straight forward.

Question: What is the "multichannel equivalent" for $R$ in the case of e.g. an adaptive filtering multichannel application? Do I need to calculate some sort of autocorrelation tensor?

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You apply the same formula, but instead of using a scalar $x(n)$, you will have $x(n) \in \mathbb{C}^{M \times 1}$, where $M$ is the number of channels, and $x^{H}(n)$ is the $1 \times M$ matrix whose entries are the complex conjugate of the entries in $x(n)$. As a result $R \in \mathbb{C}^{M \times M}$ is an Hermitian matrix, and its eigenvalues are all real.

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  • $\begingroup$ But wouldnt $R \in {\mathbb{C}}^{M \times M \times 2N-1}$ ? $\endgroup$
    – Bulbasaur
    Commented Mar 9, 2021 at 8:50
  • $\begingroup$ It seemed too easy to be correct hehe. I interpreted E as the expected value,x(n) the covariance of one sample. You meant the matrix R, where each element $R_{i,j} = E\{x(n-i)x^H(n-j)\}$ for random vectors? By equalizing you mean to whiten the signal, to remove the correlation between different samples? $\endgroup$
    – Bob
    Commented Mar 9, 2021 at 9:55
  • $\begingroup$ I was wrong. You were right. Take my upvote. $\endgroup$
    – Bulbasaur
    Commented Apr 29, 2021 at 19:45
  • $\begingroup$ Thank you, now both of us are right :) $\endgroup$
    – Bob
    Commented Apr 30, 2021 at 4:48

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