# Zero-phase filtering of a minimum-phase wavelet

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.16

Show that the result of passing a minimum-phase signal through a zero-phase filter is mixed phase.

### Background

Zero-phase signals are discussed in Sheriff and Geldart, 1995, section 15.5.6d, where it is shown that the spectrum of a zero-phase signal comprises products of pairs of factors of the form ${\displaystyle (az-1)(az^{-1}-1)=(1+a^{2})-a(z+z^{-1})=(1+a^{2})-2a{\rm {\;cos\;}}(\omega \Delta )}$, where ${\displaystyle a}$ can be complex. Because the imaginary part is zero, the phase is zero.

### Solution

Let ${\displaystyle \omega _{t}}$ be the minimum-phase signal and ${\displaystyle f_{t}}$ the zero-phase filter. Time-domain filtering is accomplished by convolution, ${\displaystyle \omega _{t}*f_{t}}$; in the frequency domain the result is ${\displaystyle W(z)F(z)}$. The factors of ${\displaystyle F(z)}$ occur in pairs of the form ${\displaystyle (az-1)(az^{-1}-1)}$, and each pair has roots ${\displaystyle z=a}$, ${\displaystyle 1/a}$. If ${\displaystyle |a|>1}$, then ${\displaystyle |1/a|<1}$. Thus, one member of each pair of roots is not minimum-phase and consequently the filtered signal is mixed-phase.