Phase of composite wavelets

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

Using the wavelets

{\begin{aligned}W_{1}(z)=(2-z)^{2}(3-z)^{2};\quad W_{2}(z)=(4-z^{2})(9-z^{2}),\end{aligned}} calculate the composite wavelets:

{\begin{aligned}W_{1}(z)+W_{2}(z);\quad W_{1}(z)+zW_{2}(z);\quad zW_{1}(z)+W_{2}(z);\quad z^{-1}W_{1}(z)+W_{2}(z).\end{aligned}} Plot the composite wavelets in the time domain. All of these composite wavelets have the same frequency spectrum but different phase spectra because multiplication by $z$ shifts the phase. The results illustrate the effects of phase.

Background

It is shown in Sheriff and Geldart, 1995, section 15.1.5 that the phase of a complex quantity, $z=a+jb$ , is $\gamma ={\rm {tan}}^{-1}(b/a)$ , that is, ${\rm {\;tan\;}}\gamma =$ (imaginary part)/(real part). Since $z^{n}=e^{-{\hbox{j}}\omega n\Delta }={\rm {\;cos\;}}(\omega n\Delta )+j{\rm {\;sin\;}}(\omega n\Delta )$ , ${\rm {\;tan\;}}\gamma ={\rm {\;tan\;}}(\omega n\Delta )$ . If a wavelet has several elements, the imaginary part of the wavelet will be the sum of several sines and the real part will be the sum of several cosines. When the wavelet is multiplied by $z^{n}$ , both sums will change, hence the phase changes.

Solution

{\begin{aligned}W_{1}(z)=(2-z)^{2}(3-z)^{2}=(4-4z+z^{2})(9-6z+z^{2})\\=36-60z+37z^{2}-10z^{3}+z^{4}\leftrightarrow [36,\;-60,\;37.\;-10,\;1],\\W_{2}(z)=(4-z^{2})(9-z^{2})=36-13z^{2}+z^{4}\leftrightarrow [36,\;0,\;-13,\;0,\;1],\\W_{3}(z)=W_{1}(z)+W_{2}(z)=72-60z+24z^{2}-10z^{3}+2z^{4}\\\leftrightarrow [72,\;-60,\;24,\;-10,\;2],\\W_{4}(z)=W_{1}(z)+zW_{2}(z)=36-24z+37z^{2}-23z^{3}+z^{4}+z^{5}\\\leftrightarrow [36,\;-24,\;37,\;-23,\;1,\;1],\\W_{5}(z)=zW_{1}(z)+W_{2}(z)=36+36z-73z^{2}+37z^{3}-9z^{4}+z^{5}\\\leftrightarrow [36,\;36,\;-73,\;37,\;-9,\;1],\\W_{6}(z)=z^{-1}W_{1}(z)+W_{2}(z)=36z^{-1}-24+37z-23z^{2}+z^{3}+z^{4}\\\leftrightarrow [36,\;\mathop {24} \limits ^{\downarrow },\;-37,\;-23,\;-1,\;1].\\\end{aligned}} The convention is that (unless otherwise specified) the first nonzero element fixes the wavelet origin $t=0$ (see problem 9.12f). We start our plots in Figure 9.13a at $t=-1$ [except for $W_{6}(z)]$ in order to show the complete waveforms.

The two wavelets $W_{1}$ and $W_{2}$ are plotted first in Figure 9.13a and then the four sums of the wavelets. Note that two of the four factors that make up each of $W_{1}$ and $W_{2}$ are the same and the other two differ only in signs:

{\begin{aligned}W_{1}(z)=(2-z)(2-z)(3-z)(3-z);\\W_{2}(z)=(2-z)(2+z)(3-z)(3+z).\\\end{aligned}} However, the waveshapes differ significantly, especially in apparent frequency. Clearly relatively small changes in the equation of a wavelet can produce significant changes in the waveshape.

Because delaying the second wavelet or advancing the first produces the same waveshape, composite wavelets $W_{4}$ and $W_{6}$ are the same except for a time shift. Wavelets $W_{3}$ and $W_{5}$ have distinctly different waveshapes from those of $W_{4}$ . We conclude that a shift of one component relative to the other has a significant effect on the location of peaks and troughs and hence on the waveshape.