I know that I can find the autocorrelation matrix of a series of finite length sequences via rank-1 updates using the relation:

$\mathbf{R}[k] = \frac{1}{k}\mathbf{R}[k-1] + \frac{1}{k+1}\mathbf{r}[k]\mathbf{r}^{H}[k]$

Where $k$ is the discrete time index and $\mathbf{r}$ is the autocorrelation of the current signal sequence. The true autocorrelation can then be found via the first column of the matrix (see https://en.wikipedia.org/wiki/Autocorrelation_matrix)

So I imagine I can do the same for the PSD, correct? Namely:

$\mathbf{S}[k] = \frac{1}{k}\mathbf{S}[k-1] + \frac{1}{k+1}\mathbf{s}[k]\mathbf{s}^{H}[k]$

Where $\mathbf{s}$ is the fourier spectrum of the current frame. The PSD will then be found as the first column of $\mathbf{S}$. Correct?

Is there a way I can transform $\mu: \mathbf{R} \rightarrow \mathbf{S}$ directly, that is, without having to compute the PSD of every sequence frame?


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