Spitz for variable covariances

Since signal and noise are uncorrelated, the spectrum of data is the spectrum of the signal plus that of the noise. An equation for this idea is  

$$\sigma_d^2 \eq \sigma_s^2 \ + \ \sigma_n^2$$ (15)

This says resonances in the signal and resonances in the noise will both be found in the data. When we are given $\sigma_d^2$ and $\sigma_n^2$ it seems a simple matter to subtract to get $\sigma_s^2$.Actually it can be very tricky. We are never given $\sigma_d^2$ and $\sigma_n^2$;we must estimate them. Further, they can be a function of frequency, wave number, or dip, and these can be changing during measurements. We could easily find ourselves with a negative estimate for $\sigma_s^2$ which would ruin any attempt to segregate signal from noise. An idea of Simon Spitz can help here.

Let us reexpress equation (15) with prediction-error filters.

$${1\over \bar A_d A_d} \eq {1\over \bar A_s A_s} \ + \ {1\ov... ...\ {\bar A_n A_n} \over ( {\bar A_s A_s} ) ( {\bar A_n A_n}) }$$ (16)

Inverting

$${ \bar A_d A_d} \eq { ( {\bar A_s A_s} ) \ ( {\bar A_n A_n}) \over {\bar A_s A_s} \ +\ {\bar A_n A_n} }$$ (17)

The essential feature of a PEF is its zeros. Where a PEF approaches zero, its inverse is large and resonating. When we are concerned with the zeros of a mathematical function we tend to focus on numerators and ignore denominators. The zeros in ${\bar A_s A_s}$compound with the zeros in ${\bar A_n A_n}$to make the zeros in ${\bar A_d A_d}$.This motivates the ``Spitz Approximation.''

$${ \bar A_d A_d} \eq ( {\bar A_s A_s} )\ ( {\bar A_n A_n})$$ (18)

It usually happens that we can find a patch of data where no signal is present. That's a good place to estimate the noise PEF A_n. It is usually much harder to find a patch of data where no noise is present. This motivates the Spitz approximation which by saying A_d = A_s A_n tells us that the hard-to-estimate A_s is the ratio A_s = A_d / A_n of two easy-to-estimate PEFs.

It would be computationally convenient if we had A_s expressed not as a ratio. For this, form the signal $\bold u = \bold A_n \bold d$by applying the noise PEF A_n to the data $\bold d$.The spectral relation is

$$\sigma_u^2 \eq \sigma_d^2 / \sigma_n^2$$ (19)

Inverting this expression and using the Spitz approximation we see that a PEF estimate on $\bold u$ is the required A_s in numerator form because

$$A_u \eq A_d / A_n \eq A_s$$ (20)