# Noise power for decorrelating detector

I have the following setup: $$\mathbf{y}=\mathbf{R}\mathbf{s}+\mathbf{w}$$, where the details of each entity can be found at http://bard.ece.cornell.edu/downloads/sim_tools/spanc_me/report/node12.html. I'm not sure it is necessary information for my question though.

The decorrelating detector, performs the operation sign$$(\mathbf{R}^{-1}\mathbf{y})$$ to get the hard decision output. Multiplying by $$\mathbf{R}^{-1}$$ gives the following:

$$\mathbf{R}^{-1}\mathbf{y}=\mathbf{s}+\mathbf{R}^{-1}\mathbf{w}$$

I am interested in determining the noise power after I perform the decorrelating process. If $$\mathbf{w} \sim N(\mathbf{0},\sigma^2\mathbf{I})$$, then it can be shown that after the decorrelator, $$\mathbf{R}^{-1}\mathbf{w} \sim N(\mathbf{0},\mathbf{R}^{-1})$$.

My question is: what would be the "noise power" after decorrelating? That is, what is the power of $$\mathbf{R}^{-1}\mathbf{w}$$? For $$\mathbf{w}$$, the power is just the variance $$\sigma^2$$, but I am confused how to tell what is the power when the covariance matrix may have non-zero entries in the off-diagonals.

• If you interested in noise characteristics, why don't you simply compute the difference of $y - Rs$ and find the power of the noise signal? – Maxtron Jan 12 at 16:21
• If $R$ is hermitian matrix, then $\mathbf{z}\sim\mathcal{CN}\left(\mathbf{0},\,\sigma^2\mathbf{R}^{-2}\right)$. The noise power are the diagonal elements. For example, the first noise element has a power $\sigma^2[\mathbf{R}^{-2}]_{1,\,1}$, and so one. The off-diagonal elements are the cross-correlations between the different noise elements. But it's more useful to deal with the covariance matrix instead of the elements. What do you need the noise statistics for? Why did you multiply $\mathbf{y}$ by $\mathbf{R}^{-1}$ in the first place? what do you want to achieve by this? – BlackMath Jan 12 at 20:55