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AN INTERACTIVE COMPUTER/GRAPHIC-DISPLAY-TERMINALSYSTEM FOR INTERPRETATION OF RESISTIVITY
SOUNDINGS *
BY
H. K. JOHANSEN **
ABSTRACT
JOHANSEN, H. K., 1975, An Interactive Com puter/Graphic-Display-Terminal Sys temfor Interpretation of Res istivity Soundings, Ge ophysical Prospecting 23,449-458.
A fast computer-procedure giving the apparent resistivity curve as well as the partialderivatives with respect to the layer-parameters is presented. It is based on the linearfilter method developed by D. P. Ghosh in1971.
The sampling frequency isIO points per decade, and 3 decades are covered. The maxi-mu m relative error is less than10-3, and in mo st cas es orders o f magnitude sma ller.The computation time on a CD C 6400 for one curve given in 30 points ranges linearlyfrom .17s for a two-layer case to .36s for a ten-layer cas e.
The procedure is used to plot mas ter curve s interactively on a graphic display terminal(Tektronix 4010) connected to the CD C 6400. By trial-and-error adjustme nts a set of
layer-parameters is found, giving essen tially the measured c urve.
INTRODUCTION
The linear filter method developed by Ghosh (1971, 1,971”) has opened anew range of possibilities for the interpretation of resistivity soundings. Itis well suited for computerization. Because of the small number of operationsneeded the execution time is almost negligible. One can afford to calculatemaster curves based on purely guessed layer-sequences and just throw themaway if they do not fit the data.
In his paper Ghosh points out that a sampling frequency of three pointsper decade is a good compromise when the tools are a desk calculator andlogarithmic paper, because a relative accuracy better than .02 is achieved usingonly nine filter coefficients. However, when working with an electronic computerthere is no reason not to use about hundred coefficients, if this proves to beadvantageous.
The direct interpretation method is to calculate the resistivity transform Tfrom the field data and then derive the layer-parameters (resistivities pj and
* Presented at the 35th meeting of the European Assoc iation of Exploration Geo-phy sicists , Brighton, June1973.
** Laboratory of Geop hysics, Finlandsgade 8, DK-8200 Arhus N, Denm ark.
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INTERACTIVE TERMINAL SYS TEM FOR RESISTIVITY INTERPRETATION 451
We note the following property of the convolution integral (3) :
pas(4 = i T(Y + S)b(x - Y - WY (5)-m
/ which holds for any shift S.
Now let F, G and H denote the Fourier-transforms of pas, T and b respect-ively. Then
F(f) = G(f) . H(f). (6)
Since all information about the layer sequence is contained in T aloneH(f) = F(fMf) is the same function for all cases (pj, &). The violent oscilla-
tions of b(x) for large x imply that H exists only in a generalized sense. Hencethe following indirect approach (Ghosh 1971) to the determination of H ischosen :
The particular pair of functions
P(X) = exp (3X)/(1 + exp (2x))5’2
T(y) = Q ~XP (- Y) =P (- exp (- ~1)satisfy (3).
(7)
Their Fourier transforms are found by numerical integration, and themodulus Qdf) and phase O(f) of the quotient F/G determine H.
For a given sampling interval Ax the Nyquist frequency isfp = ~/(zAx),and the corresponding sine-response function is given by
c(x) = I/fN 's"Qcf,m (a(f) + zxfx)df. (8)0
As observed by Koefoed (1972) it is convenient to sample C(X) at positionsshifted to the left by the amount
s = (A+) @(fN) (9)because it makes the sampling points coincide asymptotically with the nodesof C(X), thus reducing the length of the filter.
The apparent resistivity pas in the sampling points x( = iAx is then givenapproximately by
mazpas(iAx) - ’C T((i -j)Ax + S)C(jAx - S) (10)
j =Gnh
where jnzgn and j nta5 are chosen so that the filter coefficients with smaller orlarger indices can be neglected.
Geophysical Prospecting, Vo l. 23 29
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4.52 H. K. JOHANSEN
ALGORITHM FOR CALCULATION OF T
The resistiv ity transform T(p, . . . . ., pi; dr, . . . . ., dnr-1; h) for a N-layerstratified earth (see fig. I) can be calculated recursively (Koefoed 1970).
For a layer (pi-I, dN-1) on top of a substratum (pN) the transform is
TN-I(~) = pN-11 - kN- 1 eX p (- 2dN- oh)
I + kN.e I exp (- ZdN- lh) (11)where
kN-1 = (PN-1 - PN)/(~N-1 + PN). (12)
layer 1 9 1 dl
/--T’(h)
layeri “i rTi+l (A)
layer N-l
j -TN-1 (A)
t
QN-1 dN-1
A
layer N
(infinite substratum)QN
Fig. I. A N-layer horizontally stratified earth with definition of symbols used in theformulas.
The transform Tj for layer (pj, dj) on top of the sequence (p~+l, . . . . . , pN;d,+l, . . . . . dN-1) with transform TJ+I is given by
w,(h) + Tj+$) .T’(A) = I + Wj(A)Tj+,(A)/p; ’
j=N-z,N--3 ,..., 2,1; (13)
where
W,(V = PjI - exp (- zd+)
I + exp (- zd+) *
Recursive application of (13) and (14) yield T = Tl(A).
(14)
We find easily thatASYMPTOTIC EXPRESSIONS
(15)
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INTERACTIVE TERMINAL SYSTEM FOR RESISTIVITY INTERPRETATION453
and hence
Tj(h) --f pjh-+co
Tj +1(A) A --f o . (16)
The way in which these limits are approached depends on the specific setof layer-parameters considered. Let us look at the simple two-layer case:
-
For large A we have
TN-I(~) Mp~-1(1 - zkN-1 exp (-;zd~-IA)). (17)
The relative departure from the asymptotic value
A = I TN-&) - plv-1 I / plv-I (1s)
is smaller than some small number E if onlyh > - ln(a/z)/(zdlv- 1). (19)
Together withA$ = exp (- iAx) (20)
this gives an upper bound on i. If E = 10-2, &,-I = I and Ax = (In IO)/IO,we get i < - 5.
Something similar holds in the general case for T = TI and pl, althoughthe expansion of T is more complicated than (17). Experience has shown thatwe are on the safe side if i c - 9.
For small A we have
There are two extreme cases corresponding to kN-1 w -l I :
TN-I(~) M PN(I - (~N/~N-I}~~N-I) for PN > PN-1
and
(22)
TN-I(~) M ~N(I +{p~-I/~N}~~N-I) forp~<p~-l. (23)Since the resistivity ratios can take on all positive values, we have to com-
promise.Let us consider (23) say. If PN-~/PN 5 10~ and dii-1 < Io3, then A 5 IOs2
if only A 5 10-l’.This happens if i > IOO.Experience shows that i > IOO is sufficient also in multi-layer cases.
CUTOFF AND ACCURACYApart from the sampling frequency there are two factors affecting the
accuracy: the tails of the filter must be cut off at some point, and the coeffi-cients can only be calculated with a limited precision.
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454 H. K. JOHANSEN
The problem is how to compromise on these two aspects when we are aimingat a relative accuracy better than 10-3.
Let us first consider the cutoff problem.
For convenience we write TtL-j) and C(J) for T((i -j)Ax + S) and C(jAx-S)in (10).
We split the sum into three parts:
p,s(Q = { ; + ‘*i’ + i } T(“-5) C(j); i = 0, I, 2, . . . . (24)j--m I +1 i = a
Now let us choose j, and jz so that T can be approximated by its asymptoticvalues pi and pl in the first and third term of (24) in all cases.
Then
where
(25)
C$~I) = i C(f) and C$~S)= i C(5) (26)I--m 1=1%
The analysis of the asymptotic behaviour indicates that j, = -100 andjz = 40 is what we need. C$-loo) and Cp”) have been found by applying theEuler transformation to the tails of the filter.
To what precision should the coefficients be known if we want to handleresistivity ratios up to p~/pl N10~ ? Consider the part of the curve where
PM N pl. There are terms T * C in the sum (25) with T w pN =10~ Q m.The round-off error in such a term should be no more than N I o/o0 of pl.Since the order of magnitude of the largest coefficients is I, all coefficientsmust be given with N 8 decimals. It is difficult to judge the absolute precisionof a two-fold numerical integration, and there is no guarantee that the decimals
given are accurate. However, we carry them all in our calculations, just toreduce round-off errors.
The actual precision of the filter was tested in the same way as suggested-y Ghosh: The two functions
e3X
and
d’) = P1 + b2 - d (I + e2x)3/2 (27)
T(x) = PI + (~2 - PI) exp (-- exp (- 4)
form a transform pair resembling a two-layer case.
(28)
Applying the filter to T yields a peonv to be compared with p. In fig.2 theresult is shown for pl = I and p2 = IO. The error Ap = pconv - p is seen tobe very small, indeed.
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1 80.00, 0.01 0.1 1 10 100 1000 m
Fig. 2. Accurac y test . peonv is the convolution of the resistivity transform given inequation (28) in the text, with pi =I and pi = IO. The other curve shows the differencebetween peonv and the exac t function pesae t given in equation (27). The relative error is
in this case less than one part in106.
TABLE I
The filter coefficients corresponding to a san@ling frequency ofIO points perdecade. Ax = (ln IO)/IO , j, = -100, jz =40, S = --I.py,g458, C -loo) =
6.174~~ - 5, C ““) = 6.0goglo - 4
i C(j) . 108 j C(1) . 108 j ccn . 108 j C(l) . 10s
-99
-98
-97-96
-95
-94
-93-92
-91
-90
-89-88
.-87
-86-45-84-83-82-81-80-79-78-77
-76
-75-74
-73-72
-71-70
-69-68
-67-66
-1248412726
-1297513231
-1349413765
-14043
14330-14625
14930-15244
15567-15901
16246-I 6602
16971
--I7352
17746-18154
18577-19015
19469
-1994120429
-20936
21463-22oog
22577-23166
23779-24416
25079-25768
26487
,
-64 28016
-63 -28830
-62 29680-61 -30568
-60 31496
-59 -32467
-58 33484
Y-57 -34549-56 35666
-55 -36838
-54 38069
-53 -39363
-52 40724
-51 -42156-50 43666
-49 -45259-48 46940
-47 -48717
-46 50596
-45 -52587
-44 54697
-43 --56936
-42 59314-41 -61845
-40 64540-39 -67414
-38 70484
-37 -73767-36 77284-35 -81057
-34 85111
-33 -89475
-32 94183
-31 -99267
-29-28
-27-26
-25
-24
-23-22-21
-20
---I9-18
-17
-16---I5
-14-13-12
-11
-10
1;
1:
-5-4
-3-2-1
0
I
2
3
4
-110741
117248
-124303132085
-140461149959
-159826171917
-182946199955
-209469239052
-234543
304916-234124
453990
-106745899282
550573
2442523
3250077
79266751302334525610307
4115074164zy8og
72803988
36118538-100406442-242172543
20052460
444506381-489348908
wG m398
6
i9
IO
II
12
13
1415161718
1920
21
22
23
242526
2728
29
3031
32
33
34
3536
37
38
39
61285163--29362w
15817356-9504597
6226174-435i505
3198475
-24414931920840
-1548505
1273595-1065148
903512
-775750673079
-589375520264
-462558413891
-372478
336951-306251
279543-256168
235594-217394201216
---I86773I73826
-162176151657
-142126
133463-125568
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456 H. K. JOHANSEN
TABLE z
Exewtion times vewus mumber of layers for the CDC 6400.
no. of layers 2 3 4 5 6 IO
CP time (set) 0.174 0.191 0.217 0.234 0.271 0.364
When p2 is increased Ap grows proportional to pi. Thus when the resistivityratio is p2/p1 = 104, the maximum relative error is sti ll smaller than IO-~,
which is satisfactory.A comparison with the results of J. C. van Dam (1965) showed complete
agreement within the limits indicated above. The filter coefficients are givenin table I . The computation time of the present method depends only on thenumber of layers. Characteristic values for the CDC 6400 are given in table 2.They are seen to be almost negligible.
APPLICATIONS
At our Laboratory we have at our disposal a graphic display terminal(Tektronix 4010) connected to the Aarhus University computing center’sCDC 6400 via telephone. We are using this facility in the following inter-
active way:I) The measured resistivity values are displayed with error bars indicating
the estimated accuracy.z) A (skilled) guess of the layer-parameters is entered by the interpreter.3) The corresponding p app-curve is computed and displayed.4) Trial-and-error adjustments of the layer-parameters are made until the
theoretical papp-curve agrees with the measured data.
1 IO 100 AU2 1000 mFig. 3. An example of the Tektronix 4010 graphical display illustrating the trial-and-error procedure. The inclined bars indicate the measured values i 5 . The layer para-
mete rs for the three consec utive guesse s are given in table 3.
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INTERACTIVE TERMINAL SYSTEM FOR RESISTIVITY INTERPRET ATION457
An example of this procedure is shown in fig. 3. The parameters are listedin table 3.
It is sometimes desirable to let the computer carry out the tedious
step 4 by an iterative least squares procedure. To this end the partial derivativesof papp with respect to the layer-parameters are needed. Recursive formulaefor them are given in the appendix.
TABLE 3Layer parameters for the three cwves shown in fig. 3.
curve I 2 3
layerno.
I
2 20 I5 27 15 27 I.53 160 40 160 40 160 21
4 3.5 3.5 3.5
APPENDIX
THE PARTIAL DERIVATIVES OFT
In order to get a recursive relationship for the partial derivatives we dif-ferentiate (13) : For convenience we will refer to the layer-parameters as pk,meaning either pk or dk, in cases where it does not matter which one we aretalking about.
Consider first eq. (II). We put a = exp (- zhd~-I), b = kN-1 a, andR = 4p~-1 *b/(1 + b)2.
Then
~TN-I3PJJ
- R *pit-l/(pN-1 + pN121
~TN-1~ = TN-JPN-I-R * PN/(PN-1 -~PIv)~~+N-1
(29)
(31)
Since the pk for k = N, N - I, . .: . .j + I enter TJ only through T~+I,we have
JTj 3Tf ~JT~+I
apk - 3Tj+1 aj’kk=N,N-I, . . . . . j +I. (32)
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458 JOHANSEN, INTERACTIVE TERMINAL
Also
aTj 3T3 4P3a---3d3 - 3w3 (I +a+
where
a = exp (- zd31).
Putting I + W3T3+,/p~ = c, we have
SYSTEM
(33)
JT3 3T3 w3-=--3P3
a w3 p3 + 2(w3 + T3 +I) W3 T3 +I p3 - 3/c2
and
3T33T3+1
= (I - w,“/p ;yc2
iJTi__ = (I - T;+Jp;)/c2aw3
(34)
These expressions hold for j = N - 2, N - 3, . . . . . , 2, I; so that recursiveapplication gives the partial derivatives of T = TI.
Since the function C(X) in (IO) is independent of the layer-sequence wehave simply :
(37)
where Tea-3) = T((i - j)Ax + S) and C(3) =C(jAx - S).
REFERENCES
GH OS H, D. P., 1971, The Application of Linear Filter Theory to the Direct Interpretationof Geoelectrical Res istivity Sounding Measu remen ts, Geophysical Prospecting rg,
192-217GH OS H, D. P., rg7ra, Inverse Filter Coefficients for the Com putation of Apparent Res ist-ivity Standard Curve s for a Horizontally Stratified Earth, Geophysical Prospecting19, 769-775.
KOEP OED , O., 1968, The Application of the Kernel Function in Interpreting GeoelectricalMea surem ents, Geoexploration Monographs, Series I, No . 2, Gebriider Borntraeger,Berlin-Stuttgart.
KOE FOE D, O., 1970, A Fa st Method for Determining the Layer Distribution from theRaised Kernel Function, Geophysical Prospecting 18, 564-570.
KOE FOE D, O., Igp, A Note on the Linear Filter Method of Interpreting Res istivitySounding Data , Geophysical Prospecting 20, 403-405.
VAN DA M, J. C. , 1965, A Simple Method for the Calculation of Standard Graphs to beUse d in Geo-electrical Prospecting, Geophys ical Prospecting 13, 37-65.