The zeros of prediction filters

Because prediction filters are finite-impulse-response filters, they can be characterized by the zeros of their z-transform. From equation (2), we know that the zeros of the prediction filter $\hat{P}(\omega,s)$ are $\{e^{(\sigma_l+ip_l\omega)\Delta x};\ \ l=1,\ldots,L\}$.Therefore, we can express this filter as follows:

$$\hat{P}(\omega,s)=\prod^L_{l=1}(1-e^{(\sigma_l+ip_l\omega-s)\Delta \hat{x}}).$$ (10)

If we scan the amplitude spectrum of this filter over the s plane, we can find L notches at

 

$$s = \sigma_l+ip_l\omega,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ l=1,\ldots,L$$ (11)

that locate all the zeros of the z-transform of this filter. Similarly, we can express the prediction filter $P(\omega,s)$ as follows:

$$\begin{array} {lll} P(\omega,s) & = & \displaystyle{\prod^L_... ...{(\sigma_l+ip_l\omega-s)\Delta \hat{x}-i\phi_{m}})},\end{array}$$ (12)

where $\phi_{m}$ denotes the phases of the Mth order complex roots of the unity. Now, if we scan the amplitude spectrum of $P(\omega,s)$ over the s plane, we can find $M\times L$ notches at

 

$$s = \sigma_l+ip_l\omega+i{\phi_{m} \over \Delta \hat{x}},\ \... ... \ \ \ \ \ \ l=1,\ldots,L \ \ \ \hbox{and}\ \ \ m=1,\ldots,M,$$ (13)

M times as many notches as that of $\hat{P}(\omega,s)$.Comparing equation (13) with equation (11), it is apparent that these two equations become identical when $\phi_{m}$ is equal to zero. Thus, L out of $M\times L$ zeros of $P(\omega,s)$ are the zeros of $\hat{P}(\omega,s)$.Our goal is to identify these L zeros when $P(\omega,s)$ is known. If the component of data at frequency $\omega$ is not spatially aliased, then $P(\omega,s)$ has L zeros between two vertical lines $s=-i\pi/\Delta x$ and $s=+\pi/\Delta x$, which are L zeros of $\hat{P}(\omega,s)$. However, if the component of data at frequency $\omega$ is spatially aliased, the task of identifying the zeros becomes complicated and requires sophisticated algorithms.