# Complex-trace analysis

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 9.26a

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

### Background

A wave with continuously varying amplitude is of the form

 {\displaystyle {\begin{aligned}g(t)=A(t){\rm {\;cos\;}}\omega t,\end{aligned}}} (9.26a)

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

 {\displaystyle {\begin{aligned}g_{\bot }(t)=-A(t)\sin \omega t.\end{aligned}}} (9.26b)

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

 {\displaystyle {\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 ${\displaystyle h(t)}$ and ${\displaystyle g_{\bot }(t)}$ are known as the complex trace and the quadrature trace, respectively. ${\displaystyle A(t)}$ and its mirror image constitute the envelope of both ${\displaystyle g(t)}$ and ${\displaystyle g_{\bot }(t)}$.

 {\displaystyle {\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 ${\displaystyle A(t)}$ 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 ${\displaystyle g(t)}$ and the projection onto the imaginary plane is ${\displaystyle g_{\bot }}$. The angle that this vector makes with the real plane is the instantaneous phase ${\displaystyle \gamma (t)}$ and the rotational speed is the instantaneous frequency ${\displaystyle f_{i}=d\gamma /dt}$.

 {\displaystyle {\begin{aligned}\gamma (t)&=\tan ^{-1}\left({\frac {g_{\bot }(t)}{g(t)}}\right).\end{aligned}}} (9.26e)

 {\displaystyle {\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 ${\displaystyle g_{t}}$ (dashed line), ${\displaystyle g_{\bot t}}$ (solid line), and ${\displaystyle \pm A}$ (dotted lines).

Because ${\displaystyle tan\theta ={\rm {\;tan\;}}(\theta +n\pi )}$, ${\displaystyle n}$ integral, we generally add (or subtract) multiples of ${\displaystyle \pi }$ to ${\displaystyle \gamma (t)}$ 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 ${\displaystyle \gamma (t)}$ itself. However, the finite-difference expression is often sufficiently accurate.

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

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

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

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

 {\displaystyle {\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 ${\displaystyle f}$ = 62 Hz = dominant frequency.

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

{\displaystyle {\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 ${\displaystyle g_{\bot 9}}$, we obtain for ${\displaystyle g_{\bot t}}$

{\displaystyle {\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 ${\displaystyle g_{t}}$ and ${\displaystyle g_{\bot t}}$.

## Problem 9.26b

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

### Solution

Using equations (9.26c,d), we get

{\displaystyle {\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 ${\displaystyle h_{t}}$.

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

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

## Problem 9.26c

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

### Solution

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

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

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

{\displaystyle {\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}}}

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

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