Where is TIme Zero Position in gprMax?

[Zacharias]

Hello there,

I have created a model by means of gprMax, the input file of which contains the following commands:

/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#title: B-scan from an iron rod of 1cm radius buried in sand in 30 centimeter depth

#domain: 5 3 0.005
#dx_dy_dz: 0.005 0.005 0.005
#time_window: 30e-9

#material: 3 0.01 1 0 sand

#waveform: ricker 1 6e8 my_ricker
#hertzian_dipole: z 0.1 2.5 0 my_ricker
#rx: 0.2 2.5 0
#src_steps: 0.025 0 0
#rx_steps: 0.025 0 0

#box: 0 0 0 5 2.5 0.005 sand
#cylinder: 2.5 2.19 0 2.5 2.19 0.005 0.01 pec

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

You will also find attached a picture of the Ascan when transceiver is exactly above the cylinder shaped target and a picture of Bscan. The index of x-coordinate of this position is 95 if we start counting from 1. 

I have used the hyperbola fitting technique in order to estimate the propagation velocity in the medium, in this case in sand.The intention of this is to compare the estimated velocity from hyperbola fitting with the theoretical one that derives from the relation v = c/sqrt(er). In order to do this, I need the coordinates of hyperbola in (x, t), where x is the horizontal axis of Bscan (GPR's moving direction) in meters (m) and t is the vertical axis of Bscan in seconds (s). The problem is that I have to pick a time from the reflected pulse with respect to a time zero reference. That could be t=0 or that time position that refers to the reflection of the ground surface. I do not know which one could be the correct.

Another problem is that If the first pulse we see in the picture corresponds to the direct wave travels through the air from transmitter to receiver, where is the reflection from ground surface? In the paper "Where is True Time Zero?" the first pulse is addressed as the ground surface reflection, so the process following that accounts for this assumption. Namely, if you choose as time zero reference the time corresponds to the maximum positive value of the first pulse, then, if the polarity of the reflected pulse of target remains the same with the first pulse, you as well have to choose the time of the maximum positive peak of the pulse. Likewise, if the polarity is reversed, you have to choose the time of the maximum negative peak of the pulse.

What is the case in the gprMax model I have created?

I have tried 2 versions of time zero reference:

1. I assumed that the first pulse concerns the ground surface reflection, so I chose as time zero position the time of maximum negative peak of the first pulse. Subsequently I picked from the target reflected pulse the time of maximum positive peak because the polarity has changed. With these given, the hyperbola fitting technique for a known cylinder's radius (r = 0.01m) yielded an estimated velocity of v_est = 1.895e8 m/s , when the theoretical is calculated as v_ther = c/sqrt(er) = 3e8/sqrt(3) =  1.732e8 m/s. The time coordinate of the apex of hyperbola was estimated as t_apex = 3.10164 ns. The corresponding depth of the cylinder was calculated through the relation: depth = t*v_est/2 = 0.29388039 m , which is very close to the real value of depth which is 0.3 m or 30 cm.

2. I assumed as time zero position the t = 0 s. The velocity by means of hyperbola fitting was estimated as v_est = 1.731 m/s, very close to the theoretical one (v_ther = 1.732e8 m/s). The time of hyperbola's apex was estimated as t_apex = 4.09228 ns . This leads to a depth equal to 0.354186834 m which incorporates a greater error than the previous version. What is happening? Why this divergence?

What would your choice be for time zero position and what time position would you pick for the hyperbola's apex of this exact model?     

Sorry for size of the message, but I found it difficult to explain myself as clearly as possible.


Best Regards,

Zacharias

[Craig Warren]

Hi Zacharias,

Sorry for the delay in responding, I just haven't had time to look into your issue. Your explanation is very clear. Here are a few initial thoughts:

- The first part of the response you see in the A-scan is a combination of the direct wave (from transmitter to receiver) and the reflected wave from the ground. You can (in the model) raise the source and receiver off the surface of the ground which will allow you to see (and separate) these two waves.

- As you have read in the paper by Richard Yelf, there are many different approaches to choosing time zero, and not one correct method.

- I think the first approach you have mentioned seems reasonable.

Kind regards,

Craig

[Zacharias]

Hello again,

thanks for the response!

- Regarding the first part of the response:  I run some models with different gpr-ground offsets and I verify the fact that the first pulse is a combination of the direct and ground wave, and I also realized that the ground wave's influence to the field strength of the compound wave is small in comparison with the direct wave's.

- The first approach of time reference seems to lead to an estimated velocity (by means of hyperbola fitting technique) that diverges quite enough from the theoretical one, while the second leads to an estimated velocity very close to the theoretical one. I still can not figure out which is the right one, but I slant towards the second one.

[Antonis Giannopoulos]

Hi Zacharias,

The obvious time zero in a numerical simulation is the beginning of the simulation at t=0. This is a lot less ambiguous than the actual time zero at a real GPR system which although can be  similarly defined it is harder to establish and record. The problem arises when you need to pick responses for estimating time of arrival and the shape of hyperbolae. If you use the actual time zero then you cannot pick as arrival time the one that corresponds to the maximum of the waveform but you must pick the time of the onset of the response when its amplitude is really very small. This is a difficult process even in a simulation and extremely tricky and error prone on real GPR data. It is clear to see why this is the case as the gprMax model response is rising very slowly from a low value towards its peak. In a model you can try and correct for this by calculating the time delay to the first maximum of the feeding waveform and use this in your calculation to adjust the time zero. I think this will help. However, in real GPR you do not have this ability. If you move your t_apex in your second option to the correct value then both velocity and depth will be OK. 

Time zero is a rather interesting topic for real data and a model can help illustrate some pitfalls but it will be better using a proper antenna model as all other near field effects are included. Obviously, the assumptions inherent in hyperbola fitting etc. do not take into account the presence of near field effects that are especially important for very shallow depth targets. So, in essence this approach will work but not perfectly when the targets get too close. The further away they are the better.

Hope this helps you to understand a bit better the problem. The key is to experiment and be critical of analysing your responses and you will get interesting and hopefully useful results.

Regards

Antonis

[Anadyr]

Just to understand how one could deal with the time zero problem.

Would it make sense to run the same script for simulation again removing the cylinder?

So you get 2 <file>.out results.

result_normal.out
result_air.out

Loading the data from those files and subtracting the air simulation from the normal simulation, like:

electro_field = electro_normal - electro_air

and then I would work with the electro_field only. To me this makes sense because I can remove the direct coupling Tx -> Rx and the first reflection wave coming from the first surface.

Are there any arguments this speak against this method?

Thanks a lot for your replies.

[Anadyr]

Sorry guys,

I have made I mistake: I wrote air but what I meant is that the simulation enviroment is made of air and box. Only the cylinder is removed.

Sorry for the misunderstanding