changing the yee cell sizes in my scan changed the reults of the scan in an unexpected way??

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אופק פרסלר

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Dec 10, 2025, 7:06:38 AM (12 days ago) Dec 10
to gprMax-users
hello, Ive been looking into the time it took for a simulation to run VS some change in scan variables and came across a weird result.
Below are a-scans of a simulation I ran four times with different scan resolutions (dx,dy,dz were all equal and changed between scans) the file names were res"x" where "x" was the value I chose for dx,dy,dz.
examples of .in file I ran: 
1.
#domain: 0.8 2 0.8
#dx_dy_dz: 0.02 0.02 0.02
#time_window: 2e-8

#material: 5 0 1 0 half_space
#waveform: ricker 1 6e8 my_ricker

#hertzian_dipole: z 0.44 1.7 0.4 my_ricker
#rx: 0.38 1.7 0.4
#box: 0 0 0 0.4 0.95 0.4 half_space
#cylinder: 0.4 0.3 0.4 0.4 0.32 0.4 0.12 pec

2.
#domain: 0.8 2 0.8
#dx_dy_dz: 0.01 0.01 0.01
#time_window: 2e-8

#material: 5 0 1 0 half_space
#waveform: ricker 1 6e8 my_ricker

#hertzian_dipole: z 0.44 1.7 0.4 my_ricker
#rx: 0.38 1.7 0.4
#box: 0 0 0 0.4 0.95 0.4 half_space
#cylinder: 0.4 0.3 0.4 0.4 0.32 0.4 0.12 pec

The output files are also added below, but I also put the Ez component of each a-scan in a single graphFigure2_resVSsig.png
and also a zoom photo of the target and air/ground signature that are hard to see:
Figure2_resVSsig-NoAir.png
res0.015.out
res0.005.out
res0.01.out
res0.02.out

Antonis Giannopoulos

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Dec 11, 2025, 10:32:52 PM (11 days ago) Dec 11
to gprMax-users
The result is not weird. It is what is expected. The waveform command specifies the current (I) in Amps of the Hertzian dipole  but what determines the strength of the field is the actual current density. Look at the definition of J at


So, the current density that is used at the source point is different for different spatial steps. You can adjust your amplitude value for (I)  and the results will look the same. If your model is well resolved and the results are accurate for different resolutions normalising should produce nearly identical results. if you have big discrepancies when you do this then the results are either numerically not accurate or then there is a problem. A Hertzian dipole is a theoretical source. If you model a real antenna and use proper excitation difference resolutions will give similar results in terms of amplitudes when converted to voltages at the receiver. 

Hope this helps

Antonis
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