Complex-trace analysis

Problems in Exploration Seismology and their Solutions

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

Problem 9.26a

Given the wavelet $[10,\;8,\;0,\;-9,\;-11,\;-6,\;0,\;7,\;12,\;5,\;0,\;0]$, calculate the quadrature function, $g_{\bot }(t)$.

Background

A wave with continuously varying amplitude is of the form

${\begin{aligned}g(t)=A(t){\rm {\;cos\;}}\omega t,\end{aligned}}$

(9.26a)

where we assume that the rate of change of $A(t)$ is small compared with $\omega$ and the sampling frequency $1/\Delta$. If we take $A(t)$ fixed, the Hilbert transform (see Sheriff and Geldart, 1995, section 15.2.13) of $g(t)$, written $g_{\bot }(t)$, is

${\begin{aligned}g_{\bot }(t)=-A(t)\sin \omega t.\end{aligned}}$

(9.26b)

We note that $g(t)$ and $g_{\bot }(t)$ differ in phase by $-90^{\circ }$. We can combine the signals in equations (9.26a,b) to obtain the complex function $h(t)$:

${\begin{aligned}h(t)=g(t)+{\hbox{j}}g_{\bot }(t)=A(t)e^{-{\hbox{j}}\omega t}\end{aligned}}$

(9.26c)

(see Sheriff and Geldart, 1995, problem 15.12a for Euler’s formulas and Sheriff and Geldart, 1995, section 15.1.5 for a discussion of complex functions). The quantities $h(t)$ and $g_{\bot }(t)$ are known as the complex trace and the quadrature trace, respectively. $A(t)$ and its mirror image constitute the envelope of both $g(t)$ and $g_{\bot }(t)$.

${\begin{aligned}A(t)=[g^{2}(t)+g_{\bot }^{2}(t)]^{1/2}\end{aligned}}$

(9.26d)

We think of the complex trace as being traced by the tip of a vector of length $A(t)$ rotating in the complex plane as it moves perpendicular to the complex plane in the time direction (Figure 9.26a). The projection of the helical path generated by the tip of this vector onto the real plane is $g(t)$ and the projection onto the imaginary plane is $g_{\bot }$. The angle that this vector makes with the real plane is the instantaneous phase $\gamma (t)$ and the rotational speed is the instantaneous frequency $f_{i}=d\gamma /dt$.

${\begin{aligned}\gamma (t)&=\tan ^{-1}\left({\frac {g_{\bot }(t)}{g(t)}}\right).\end{aligned}}$

(9.26e)

${\begin{aligned}f(t)&=d\gamma (t)/dt={\frac {g(t){\frac {dg_{\bot }(t)}{dt)}}-g_{\bot }(t){\frac {dg(t)}{dt)}}}{g^{2}(t)+g_{\bot }^{2}(t)}}\approx {\frac {\Delta \gamma }{\Delta t}}={\frac {\gamma _{i+1}-\gamma _{i-1}}{2\Delta }}.\end{aligned}}$

(9.26f)

Figure 9.26a.  Graphs of $g_{t}$ (dashed line), $g_{\bot t}$ (solid line), and $\pm A$ (dotted lines).

Because $tan\theta ={\rm {\;tan\;}}(\theta +n\pi )$, $n$ integral, we generally add (or subtract) multiples of $\pi$ to $\gamma (t)$ to make it monotonically increasing (or decreasing). The derivative of the arc tangent in equation (9.26f) usually permits more stable calculation than taking the derivative of $\gamma (t)$ itself. However, the finite-difference expression is often sufficiently accurate.

To get $g_{\bot }(t)$ we find the Hilbert transform of $g(t)$ [see Sheriff and Geldart, 1995, equation (15.176)]:

${\begin{aligned}g_{\bot }(t)=g(t)*{\frac {-1}{\pi t}}.\end{aligned}}$

For a digital function $g_{t}$, this becomes [see Sheriff and Geldart, 1995, equation (9.107)]

${\begin{aligned}g_{\bot t}=g_{t}*q_{t}=\mathop {\sum } {_{n=-\infty }^{+\infty }}g_{t-n}(e^{{\hbox{j}}n\pi }-1)/\pi n,\end{aligned}}$

(9.26g)

where $q_{t}$ is the quadrature filter. Since $(e^{{\hbox{j}}n\pi }-1)=0$ or –2 according as $n$ is odd or even, the equation reduces to

${\begin{aligned}g_{\bot t}=\left({\frac {-2}{\pi }}\right)=\mathop {\sum } {_{n=-\infty }^{+\infty }}g_{t-n}/n,\quad \mathrm {n\ odd} .\end{aligned}}$

(9.26h)

Solution

We note that 11 samples constitute 5/4 cycles so, assuming 2-ms sampling, the period is 20(4/5) = 16 ms or $f$ = 62 Hz = dominant frequency.

While $g_{t}$ is causal with elements $g_{0}$ to $g_{9}$, equation (9.26h) shows that $g_{\bot t}$ is not causal. We use equation (9.26h) to calculate $g_{\bot t}$ for $t=0$ to 9. Thus,

${\begin{aligned}g_{\bot 0}&=(-2/\pi )[g_{1}/(-1)+g_{3}/(-3)+g_{5}/(-5)+g_{7}/(-7)+g_{9}(-9)]\\&=0.64(8/1-9/3-6/5+7/7+5/9)=0.64\times 5.36=3.43;\\g_{\bot 1}&=0.64(g_{0}/1+g_{2}/(-1)+g_{4}/(-3)+g_{6}/(-5)+g_{8}/(-7)=0.64\\&=0.64[(0+0-11/(-3)+0+12/(-7)]=0.64\times 11.95=7.65.\end{aligned}}$

Continuing the calculation as far as $g_{\bot 9}$, we obtain for $g_{\bot t}$

${\begin{aligned}g_{\bot t}=\left[\mathop {-3.4} \limits ^{\downarrow },\;7.2,\;10.8,\;7.6,\;-2.3\;,\;-8.3\;,\;-10.3\;,\;-9.1\;,\;-0.4,\;7.0\right].\end{aligned}}$

Figure 9.26a shows $g_{t}$ and $g_{\bot t}$.

Problem 9.26b

What are values of the complex trace $h$($t$) and the amplitude of the envelope $A(t)$.

Solution

Using equations (9.26c,d), we get

${\begin{aligned}h_{0}=10-3.4j\qquad A_{0}=10.6\\h_{1}=8+7.2j\qquad A_{1}=10.8\\h_{2}=10.8j\qquad A_{2}=10.8\\h_{3}=-9+7.6j\qquad A_{3}=11.8\\h_{4}=-11-2.3j\qquad A_{4}=11.3\\h_{5}=-6-8.3j\qquad A_{5}=10.2\\h_{6}=-10.3j\qquad A_{6}=10.3\\h_{7}=7-9.1j\qquad A_{7}=11.4\\h_{8}=12-0.4j\qquad A_{8}=12.0\\h_{9}=5+7.0j\qquad A_{9}=8.6\end{aligned}}$

Figure 9.26b.  Graph of $h_{t}$.

Figure 9.26b is a graph of $h_{t};h_{0}$ is in the southeast quadrant, $h_{1}$ and $h_{2}$ in the northeast, $h_{3}$ in the northwest, $h_{4}$, $h_{5}$, and $h_{6}$ in the southwest, $h_{7}$ and $h_{8}$ in the southeast, and $h_{9}$ in the northeast. Thus the phase makes a little over one revolution, just as $h_{t}$ does.

The amplitude $A_{t}$ is shown dotted in figure 9.26a.

Problem 9.26c

Calculate the instantaneous phase $\gamma (t)$ and the instantaneous frequency $f(t)$ (adding to the phase multiples of $\pi$ as necessary to obtain a monotonically increasing function. Assume $\Delta =2$ ms.

Solution

We use equations (9.26e,f) to obtain values of $\gamma _{t}$ and $f_{t}$, $i=0$ to 9. To calculate $f_{t}$ using the derivative operator (problem 9.31) in equation (9.26f) for $i=0$ to 9, we need $\gamma _{-1}$ to $\gamma _{10}$. Since $g_{1}=0=g_{10}$, equation (9.26e) shows that both $\gamma _{-1}$ and $\gamma _{10}$ are $\pm 90^{\circ }$, depending on the signs of $g_{\bot (-1)}$ and $g_{\bot 10}$; to determine these signs we use equations (9.26h) to write

${\begin{aligned}g_{\bot (-1)}&={\frac {-2}{\pi }}\mathop {\sum } \nolimits _{-\infty }^{+\infty }{\frac {g_{-1-n}}{n}}\\&={\frac {2}{\pi }}\mathop {\sum } \limits _{}^{}{\frac {g_{-1-n}}{-n}},\;n=-1,\;-3,\;-5,\;-7,\;-9\\&=0.64(g_{0}/1+g_{2}/3+g_{4}/5+g_{6}/7+g_{8}/9)\\&=0.64(10/1-11/5+12/9)=\mathrm {positive} ,\\g_{\bot 10}&=-0.64(g_{9}/1+g_{7}/3+g_{5}/5+g_{3}/7+g_{1}/9)\\&=-0.64(5/1+7/3-6/5-9/7+8/9)=\mathrm {negative} .\end{aligned}}$

Thus, ${\rm {\;tan\;}}\gamma _{-1}=+\infty$, $\gamma _{-1}=90^{\circ }$, ${\rm {\;tan\;}}\gamma _{10}=-\infty$ , $\gamma _{10}=-90^{\circ }$.

We now calculate the values of $\gamma _{i}$ and $f_{i}$, $i=0$ to 9.

${\begin{aligned}\gamma _{0}&={\tan }^{-1}\left({\frac {g_{\bot 0}}{g_{0}}}\right)={\tan }^{-1}(-3.4/10)&f_{0}=\{\gamma _{1}-\gamma _{-1})/2\times 0.002\times 360^{*}\\&=-19^{\circ };+180^{\circ }=161^{\circ }&=132/1.44=92\mathrm {Hz} \\\gamma _{1}&={\tan }^{-1}[(7.2/8]&f_{1}=109/1.44=76\ \mathrm {Hz} \\&=42^{\circ }+180^{\circ }=222^{\circ }&\\\gamma _{2}&=90^{\circ }+180^{\circ }=270^{\circ }&f_{2}=98/1.44=68\mathrm {Hz} \\\gamma _{3}&={\tan }^{-1}[7.6/(-9)]&f_{3}=102/1.44=71\mathrm {Hz} \\&=-40^{\circ }+360^{\circ }=320^{\circ }&\\\gamma _{4}&={\tan }^{-1}[-2.3/(-11)]&f_{4}=94/1.44=65\mathrm {Hz} \\&=12^{\circ }+360^{\circ }=372^{\circ }&\\\gamma _{5}&={\tan }^{-1}[-8.3/-6]&f_{5}=78/1.44=54\mathrm {Hz} \\&=54^{\circ }+360^{\circ }=414^{\circ }&\\\gamma _{6}&=90^{\circ }+360^{\circ }=450^{\circ }&f_{6}=74/1.44=51\mathrm {Hz} \\\gamma _{7}&={\tan }^{-1}[-9.1/7]&f_{7}=88/1.44=61\mathrm {Hz} \\&=-52^{\circ }+540^{\circ }=488^{\circ }&\\\gamma _{8}&={\tan }^{-1}[-0.4/12]&f_{8}=106/1.44=74\mathrm {Hz} \\&=-2^{\circ }+540^{\circ }=538^{\circ }&\\\gamma _{9}&={\tan }^{-1}[7.0/5]=54^{\circ }+540^{\circ }=594^{\circ }&f_{9}=92/1.44=64\mathrm {Hz} \end{aligned}}$

$*$Note: Division by 360 gives revolutions/second of the vector $A(t)$, equivalent to hertz.

The average instantaneous frequency is 68 Hz, the standard deviation being 11 Hz.

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Also in this chapter

• Fourier series
• Space-domain convolution and vibroseis acquisition
• Fourier transforms of the unit impulse and boxcar
• Extension of the sampling theorem
• Alias filters
• The convolutional model
• Water reverberation filter
• Calculating crosscorrelation and autocorrelation
• Digital calculations
• Semblance
• Convolution and correlation calculations
• Properties of minimum-phase wavelets
• Phase of composite wavelets
• Tuning and waveshape
• Making a wavelet minimum-phase
• Zero-phase filtering of a minimum-phase wavelet
• Deconvolution methods
• Calculation of inverse filters
• Inverse filter to remove ghosting and recursive filtering
• Ghosting as a notch filter
• Autocorrelation
• Wiener (least-squares) inverse filters
• Interpreting stacking velocity
• Effect of local high-velocity body
• Apparent-velocity filtering
• Complex-trace analysis
• Kirchhoff migration
• Using an upward-traveling coordinate system
• Finite-difference migration
• Effect of migration on fault interpretation
• Derivative and integral operators
• Effects of normal-moveout (NMO) removal
• Weighted least-squares

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