# Nonminimum-delay wavelet

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Series Geophysical References Series Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing Enders A. Robinson and Sven Treitel 9 http://dx.doi.org/10.1190/1.9781560801610 9781560801481 SEG Online Store

What is the structure of a nonminimum-delay wavelet? Let the signature be the real-value, finite-length, nonminimum-delay wavelet ${\displaystyle s=\left(s_{0},{\rm {s}}_{1}{\rm {,\dots ,}}\ s_{\rm {N}}\right)}$. Let us call anything that has a flat magnitude spectrum white and anything that has a curved magnitude spectrum colored. The canonical representation is the key to signature deconvolution. The canonical representation states that a given nonminimum-delay wavelet s is equal to the convolution of (1) its minimum-delay counterpart b, which is a wavelet with the same color as the given nonminimum-delay wavelet s but with a minimum-phase spectrum, and (2) the all-pass wavelet p, which has a white (i.e., flat) magnitude spectrum and which carries the extra phase. The canonical representation of the signature is given by

 {\displaystyle {\begin{aligned}s=b*p.\end{aligned}}} (32)

The signature s is known. The problem is to find its two components.

Let ${\displaystyle S\left(\omega \right)}$ be the Fourier transform of the signature. The energy spectrum is defined as

 {\displaystyle {\begin{aligned}\Psi {\rm {\ }}\left(\omega \right)&{=|}S\left(\omega \right){\rm {|}}^{2}=\sum _{m=0}^{N}{s_{m}}e^{-i\omega m}\sum _{n=0}^{N}{s_{n}}e^{i\omega n}.\end{aligned}}} (33)

If we let ${\displaystyle m-n=k}$, we can rewrite this as

 {\displaystyle {\begin{aligned}\Psi (\omega )=\sum \limits _{m=0}^{N}{\sum \limits _{n=0}^{N}{s_{m}s_{n}e^{-i\omega (m-n)}=\sum \limits _{k=-N}^{N}{e^{-i\omega k}}\sum \limits _{n=0}^{N}{s_{n+k}s_{n}}.}}\end{aligned}}} (34)

The autocorrelation is defined as

 {\displaystyle {\begin{aligned}r_{k}=\sum _{n=0}^{N}{s_{n+k}}s_{n}.\end{aligned}}} (35)

The autocorrelation is symmetric; that is, ${\displaystyle r_{k}=r_{-k}}$ The energy spectrum can be written as

 {\displaystyle {\begin{aligned}\Psi \left(\omega \right)=\sum _{k=-N}^{N}{e^{-i\omega k}}r_{k}=r_{0}+2\sum _{i{=1}}^{N}{r_{k}}{\rm {\ cos\ }}\omega k\geq 0.\end{aligned}}} (36)

Thus, if we are given the signature s, we can compute its energy spectrum as well as its autocorrelation.

We now want to consider the inverse problem: Given the energy spectrum (or, equivalently, the autocorrelation), we want to find the wavelet that yields this energy spectrum. This inverse problem, as it stands, is not unique in that many wavelets have the given amplitude spectrum ${\displaystyle {\sqrt {\Psi \left(\omega \right)}}}$ Only one of these wavelets is minimum delay. This exceptional wavelet can be determined because it is possible to determine the minimum-phase spectrum from knowledge of its amplitude spectrum ${\displaystyle {\sqrt {\Psi \left(\omega \right)}}}$. Thus, we have enough information to compute the minimum-delay counterpart of the nonminimum-delay signature. This is how we do it:

We write the energy spectrum as the Z-transform

 {\displaystyle {\begin{aligned}\Psi \left(Z\right)=\sum _{k=-N}^{N}{Z^{k}}r_{k}.\end{aligned}}} (37)

We see that ${\displaystyle Z^{N}\Psi \left(Z\right)}$ is a polynomial of degree 2N. Because the autocorrelation is symmetric, it follows that ${\displaystyle Z^{N}\Psi \left(Z\right)=0}$ if and only if ${\displaystyle Z^{N}\Psi \left(Z^{-{1}}\right)=0}$. Thus, Z is a root of this polynomial if and only if ${\displaystyle Z^{-1}}$ is a root. Moreover, because the polynomial has real coefficients, it follows that for every complex root, the corresponding complex-conjugate root must occur. For example, suppose the autocorrelation is (2, 5, 2), where the center point 5 is at time index 0. The Z-transform of the autocorrelation is ${\displaystyle \Psi \left(Z\right){=2}Z^{-{1}}{+5+2}Z}$. The polynomial ${\displaystyle Z\Psi \left(Z\right){=2+5}Z+2Z^{2}}$ can be factored as ${\displaystyle \left(Z+2\right)\left(2Z{+\ 1}\right)}$. We see that -2 is a root and that ${\displaystyle -{\rm {1/2}}}$ is also a root.

Let us compute the roots of the polynomial. Any root of modulus one gives rise to an equal-delay component. Such situations are treated in Robinson (1967b)[1]. Here, we assume that there are no roots of modulus one. There are 2N roots in all. Of these roots, N of them, say, ${\displaystyle Z_{1},...,Z_{N}}$, will have modulus greater than one. The remaining N roots will have modulus less than one. Let us form the polynomial

 {\displaystyle {\begin{aligned}b_{N}\left(Z-Z_{1}\right)\left(Z-Z_{2}\right)\dots \left(Z-Z_{N}\right)=b_{0}+b_{1}Z+\dots +b_{N}Z^{N},\end{aligned}}} (38)

where the constant ${\displaystyle b_{N}}$ is determined by requiring that the wavelet b have the same energy as the signature s. In other words, the constant ${\displaystyle b_{N}}$ is determined by the requirement that

 {\displaystyle {\begin{aligned}b_{0}^{2}+b_{1}^{2}+\dots +b_{N}^{2}=s_{0}^{2}+s_{1}^{2}+\dots +s_{N}^{2}.\end{aligned}}} (39)

The wavelet b is thus the desired minimum-delay counterpart of the signature s.

The inverse ${\displaystyle b^{-{l}}=f=\left(f_{\rm {O}},f_{1}{\rm {\ ,\ }}f_{2},\dots \right)}$ of the minimum-delay counterpart b can be obtained by carrying out the polynomial division

 {\displaystyle {\begin{aligned}{\frac {1}{b_{0}+b_{1}Z+\dots +b_{N}Z^{N}}}=f_{0}+f_{1}Z+f_{2}Z^{2}+\dots .\end{aligned}}} (40)

Let us now describe a computing method to obtain the coefficients ${\displaystyle f_{i}}$. Write equation 40 as

 {\displaystyle {\begin{aligned}\sum _{m=0}^{N}{b_{m}}&Z^{m}\sum _{n=0}^{\infty }{f_{n}}Z^{n}{=1}.\end{aligned}}} (41)

If we let ${\displaystyle n=k-m}$, we can rearrange terms to give

 {\displaystyle {\begin{aligned}\sum _{k=0}^{\infty }&{\left(\sum _{m=0}^{N}{b_{m}}f_{k-m}\right)}Z^{k}{=1}.\end{aligned}}} (42)

For this equation to hold, we see that

 {\displaystyle {\begin{aligned}b_{0}f_{0}{=1}&\mathrm {\ and\ } \sum _{m=0}^{N}{b_{m}}f_{k-m}=0\mathrm {\ for\ } k{=1,2,3,}\dots .\end{aligned}}} (43)

Right away, we find the value ${\displaystyle f_{0}=b_{0}^{-{l}}}$. We shall put this value in the equation for ${\displaystyle k=1}$ and solve the equation for ${\displaystyle f_{1}}$. Then we put these values in the equation for ${\displaystyle k{=2}}$ and solve the equation for ${\displaystyle f_{2}}$. We continue in this way to solve the equations to obtain the inverse ${\displaystyle f=b^{-1}}$.

Next, we use the expression

 {\displaystyle {\begin{aligned}f*s=f*\left(b*p\right)=b^{-1}*\left(b*p\right)=p\end{aligned}}} (44)

to obtain an expression for the all-pass filter p. Once this all-pass operator has been found, its inverse follows without further computation by simply reversing the order of its coefficients. That is, the inverse of p is simply its reverse (with respect to the time index 0):

 {\displaystyle {\begin{aligned}p^{-1}=p^{R}.\end{aligned}}} (45)

Thus, we have obtained the components of the canonical representation ${\displaystyle s=b*p}$ of the signature, and the components of the signature inverse, namely

 {\displaystyle {\begin{aligned}s^{-1}={\left(b*p\right)}^{-1}=b^{-1}*p^{-1}=f*p^{R}.\end{aligned}}} (46)