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True position of the source
One source localisation problem
Initial inverse solution
Measured self-potential
4
−50
−100
−100
3
−150
−150
−200
−200
2
−250
−250
−300
−300
1
−350
−350
−400
0
−400
−450
−450
Observed voltage
Observed voltage + noise
Predicted data
−1
−500
−500
−550
−550
100
150
200
250
300
350
400
450
0.01
0.02
500
0.03
0.04
0.05
0.06
0.07
0.08
0.09
Self-potential (mV)
X
(m)
(a)
(b)
Ninth iteration (using compactness)
Measured self-potential
−50
100
−100
−100
90
−150
−150
80
−200
−200
70
−250
−250
True source current 800 µA/m 3
60
−300
−300
50
−350
−350
40
−400
−400
30
−450
−450
20
Observed voltage
Observed voltage + noise
Predicted data
−500
−500
10
−550
0
−550
100
150
200
250
300
350
400
450
500
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
X
(m)
Self-potential (mV)
(c)
(d)
Figure 4.19 Test of the electrical source localization for a single source localized in an inhomogeneous medium. The resistivity
distribution is given in Figure 4.15a (see Table 4.4). A Gaussian noise with a standard deviation equal to 10% of the computed
data mean has been added to the data. The sources are analyzed in terms of volumetric source current distributions. a) Source
current distribution at the first iteration. b) Comparison between the true self-potential distribution and the one determined
from the inverted source current density distribution. c) Source current distribution using compactness. d) Comparison between
the true self-potential distribution and the one determined from the inverted source current density distribution using
compactness. ( See insert for color representation of the figure .)
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