Filtros inversos

Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing

Series	Geophysical References Series
Title	Digital Imaging and Deconvolution: The ABCs of Seismic Exploration and Processing
Author	Enders A. Robinson and Sven Treitel
Chapter	5
DOI	http://dx.doi.org/10.1190/1.9781560801610
ISBN	ISBN 9781560801481
Store	SEG Online Store

Inverse operations constitute fundamental components of seismic inversion schemes (Hilterman, 1970, 1985; Robinson, 1981; Treitel and Lines, 1982, 2001; Bednar et al., 1983; Robinson et al., 1983; Hampson and Russell, 1984; Gersztenkorn et al., 1986; Lines and Treitel, 1984; Ulrych et al., 1990; Ulrych et al., 2001).

What is an inverse filter? Let us develop an expression for $y_{n}$ by looking backward in time. The easiest way is to carry out the polynomial division as given by the expression for the transfer function $H\left(Z\right)=+{1/}\left(1-Z^{T}\right)$. The result is

${\begin{array}{l}{\frac {}{1-cZ^{T}}}{\frac {1+cZ^{T}+c^{2}Z^{2T}+c^{3}Z^{3T}+c^{4}Z^{3T}+c^{4}Z^{4T}+c^{5}Z^{5T}+...}{1}}\\\;\;\;\;\;\;\;\;\;\;\;{\frac {1-c\;Z^{T}}{c\;Z^{T}}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\frac {cZ^{T}-c^{2}Z^{2T}}{c^{2}Z^{2T}}}\;\;\;\;\;\;\;\;\;\;\ .\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\frac {c^{2}Z^{2T}+c^{3}Z^{3T}}{-c^{3}Z^{3T}}}\;\;\\\end{array}}$

(40)

Thus, the transfer function has the infinitely long feedforward representation

${\begin{aligned}H\left(Z\right)=1+cZ^{T}+c^{2}Z^{2T}+c^{3}Z^{{3}T}+c^{4}Z^{{4}T}+c^{5}Z^{5T}+\end{aligned}}$

(41)

In addition, the required expression for the feedback filter’s output $y_{n}$ is

${\begin{aligned}y_{n}=x_{n}+cx_{n-T}+c^{2}x_{n-2T}+c^{3}x_{n-{3}T}+c^{4}x_{n-{4}T}+c^{5}y_{n-5T}+\dots .\end{aligned}}$

(42)

The result

${\begin{aligned}{\frac {1}{1-cZ^{T}}}=1+cZ^{T}+c^{2}Z^{2T}+c^{3}Z^{{3}T}+c^{4}Z^{{4}T}+c^{5}Z^{5T}+\dots \end{aligned}}$

(43)

shows that the filters $1+cZ^{T}+c^{2}Z^{2T}+c^{3}Z^{{3}T}+c^{4}Z^{{4}T}+c^{5}Z^{5T}+\dots$ and ${\text{1 – }}cZ^{T}$ are inverse to each other. That is, the two filters in tandem result in a filter whose impulse response is a unit spike. We can say that either filter removes the effect of the other.

Let us show that the feedback filter is stable for the minimum-delay case. When ${|}c{|<}1$ (which is a minimum-delay case), the impulse-response coefficients $(1,\;0,\;0,\;...,\;0,\;c_{1},\;0,\;0,\;...,\;0\;c^{2},\;0,\;0\;...,\;0,\;c^{3},\;0,\;0,\;...)$ tend to zero, thereby making this expression for the system output $y_{3}$ converge, and hence the feedback filter is stable. Suppose that c = 0.5. Then the system output at time index 3 is

${\begin{aligned}y_{3}=x_{3}-{0.5}x_{2}+0.25x_{1}-{0.125}x_{0}{+0.0625}x_{-1}-{0.03125}y_{-2}+\dots ,\end{aligned}}$

(44)

which is convergent.

We also can show that the feedback filter is unstable for the maximum-delay case. If ${|}c{|>}1$ (which is a maximum-delay case), then the constant box produces amplification, and the feedback effect grows unboundedly with respect to time. For example, by the same reasoning as before, the feedback filter’s output at time index 3 is given by the expression

${\begin{aligned}y_{3}=x_{3}-cx_{2}+c^{2}\left(x_{1}-cy_{0}\right)=x_{3}-cx_{2}+c^{2}x_{1}-c^{3}y_{0},\end{aligned}}$

(45)

and when ${|}c{|>}1$, the impulse-response coefficients

${\begin{aligned}{1,\ }-c,\ c^{2},-c^{3},\ c^{4},\ c^{5},\dots \end{aligned}}$

(46)

tend to infinity, thereby making the expression for the system output $y_{3}$ diverge. Hence, the feedback filter is unstable. As a numerical example, suppose that c = 2. Then the system output at time index 3 is given by the divergent summation

${\begin{aligned}y_{3}=x_{3}-2x_{2}+4x_{1}-{8}x_{0}+16x_{-1}-{32}y_{-2}+\dots \end{aligned}}$

(47)

How can a stable filter approximate an unstable filter? The foregoing stability analysis is fine and good, but what if we demand a feedback filter with feedback-loop amplification - that is, ${|}c{|>}1$? What weapon do we have to make such a filter stable? The only weapon we have is time delay. We must delay the filter’s output to obtain a semblance of stability in our system. Thus, by demanding feedback-loop amplification ${|}c{|>}1$, the factors of stability and output delay must be balanced against each other. With no delay in the output, the feedback filter is necessarily unstable. However, the longer we are willing to delay the filter’s output, the greater is the chance we have for constructing a stable filter that does approximately the same job as the unstable one (Treitel and Robinson, 1966).

Let us now illustrate this point, although for the moment we shall speak metaphorically. We live in a world in which we know only a one-way flow of time. Everything we do is subject to this unidirectional flow. Now let us look at the two-faced Roman god Janus (after whom the month of January is named). Janus can see both forward and backward in time. We can see only backward in time - that is, we can see only the history of a time series up to the present time n, and we cannot see the future development of the time series, no matter how hard we try. However, Janus can see that future just as well as he can see the past. To him, time is not a unidirectional flow. He has a vantage point denied to us. How can we reach that vantage point? We can reach it only by waiting - that is, by delaying - until a finite amount of time has passed, during which period some or most of the information required for stability has become past history. Thus, to reach for stability in our feedback filter, we must delay our output. The greater the delay is, the greater are our chances.

Let us develop an expression for the output by looking forward in time. The equation for the basic feedback filter is

${\begin{aligned}y_{n}=x_{n}-cy_{n-1}.\end{aligned}}$

(48)

Let us now examine this equation by looking forward in time instead of looking backward in time as we did previously. We rewrite the equation as

${\begin{array}{l}{\frac {}{cZ+1)}}{\frac {\;\;\;\;(c\;Z)^{-1}-(c\;Z)^{-2}+(c\;Z)^{-3}-(c\;Z)^{-4}+(c\;Z)^{-5}+...}{1}}\\\;\;\;\;\;\;\;\;\;\;\;{\frac {1-(c\;Z)^{-1}}{(c\;Z)^{-1}}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\frac {(c\;Z)^{-1}-(c\;Z)^{-2}}{(c\;Z)^{-2}}}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\frac {(c\;Z)^{-2}+(c\;Z)^{-3}}{(c\;Z)^{-3}}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.\\\end{array}}$

(49)

Thus, if c = 2 the expression for the feedback filter’s output $y_{3}$ at time index 3 is

${\begin{aligned}y_{3}={\frac {1}{c}}x_{4}-{\frac {1}{c^{2}}}x_{5}+{\frac {1}{c^{3}}}x_{6}-{\frac {1}{c^{4}}}x_{7}+{\frac {1}{c^{5}}}x_{8}-\dots \end{aligned}}$

(50)

Here, $y_{3}$ is expressed in terms of the future values $x_{4},x_{5},x_{6},x_{7},x_{8},\dots .$, of the input. When ${|}c{|>1}$, the terms of the sequence

${\begin{aligned}{\frac {1}{c}},\ -{\frac {1}{c^{2}}},\ {\frac {1}{c^{3}}},\ -{\frac {1}{c^{4}}},\ {\frac {1}{c^{5}}},\dots \end{aligned}}$

(51)

tend to zero, thereby making the expression for the output $y_{3}$ converge and hence the closed-loop feedback system stable in terms of future time. As a numerical example, suppose that c = 2. Then the system output at time index 3 is

${\begin{aligned}y_{3}=0.5x_{4}-{0.25}x_{5}+0.l25x_{6}-{0.0625}x_{7}+0.03l25x_{8}-\dots ,\end{aligned}}$

(52)

which is convergent.

Let us show that a time delay serves to approximate the desired stability. We have only one problem. At time index 3, the future values $x_{4},x_{5},x_{6},x_{7},x_{8},\dots$ that are needed for the foregoing expression for $y_{3}$ are not available. Therefore, to make our expression for $y_{3}$ meaningful, we must wait until enough of these future values are available. As an approximation, then, let us truncate our expression for $y_{3}$ so that we use the expression

${\begin{aligned}{0.5}x_{4}-0{0.25}x_{5}+0.l25x_{6}-{0.0625}x_{7}+0.03l25x_{8}.\end{aligned}}$

(53)

This truncated expression serves as an approximation for the desired output $y_{3}$. Rearranging the order of the terms in this expression, we obtain

${\begin{aligned}{0.03125}x_{8}{0.0625}x_{7}+0.125x_{6}-{0.25}x_{5}+0.5x_{4}.\end{aligned}}$

(54)

Suppose that we then build a causal FIR filter with the five-length memory function (0.03125, –0.0625, 0.125, –0.25, 0.5). Because this filter is an FIR filter, it necessarily is stable. If we let the time series $x_{t}$ be the input to this filter, the output at time index n = 8 will be an approximation to the desired $y_{3}$. In other words, to approximate the desired output $y_{3}$, we have introduced a time delay of 8 – 3 = 5 time units, thereby obtaining a causal FIR filter with a five-length memory function. By increasing this time delay, we can obtain causal FIR filters that yield better approximations. A perfect approximation is reached only by introducing an infinite time delay (Treitel and Robinson, 1969).

Now let us summarize the stabilization of feedback filters. The stability of a feedback filter is determined by its delay properties. In particular, the feedback filter is stable if and only if its transfer function is a minimum-delay function. Feedback filters that are unstable can be stabilized approximately by the introduction of output delay. The approximation becomes increasingly better as this delay becomes larger.

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