why scaling required in ADI-FDTD waveform?
I am doing simulation in ADI-FDTD and time step dt is 'n' times the maximum CFL limit of dt of normal FDTD. So, dt_adi=n*dt_fdtdmax. If I want to match the waveforms of ADI-FDTD and explicit FDTD the waveforms of ADI with dt above the CFL limit needs to be scaled up/ down by a factor of 'n' (this could be 'n^2' or 'sqrt(n)', I am not sure). When I multiply the ADI waveform with 'n' (i.e. scaled up by a factor of 'n') (also multiply by 'n' on time scale or x axis) only then my waveform matches with that of explicit FDTD.
Can anybody give the explanation why this scaling up is required? Also can you please confirm whether the scaling factor should be 'n', 'n^2' or 'sqrt(n)'? And whether it should be scaled down or up?
One hint could be in dt time, energy transferred by the EM wave is = constant*E^y*dt (E is the electric field).
I would highly appreciate if someone correct me in the above and give the (physical) explanation why it is required. Thank you very much.
In order to answer your question you have to give a few more details.
1) How do you insert your sources
2) Waveform refers to exactly what (field strength, intensity, ...)
Thank you for your reply
>In order to answer your question you have to give a few more details.
>1) How do you insert your sources
I simply considered a cubic space (say 100x100x100 cells), placed hard source (gaussian pulse at 3ghz) at the center (50,50,50). I am taking the observation at say 20 cells away (70,50,50). The whole space is filled with either air or any material (some value for epsilon, sigma). Whatever these assumptions are, I noticed the scaling is required.
>2) Waveform refers to exactly what (field strength, intensity, ...)
Waveform of electric field Ez
Hope these are sufficient
...
Not quite. The concrete code for the source or a mathematical description would be necessary.
In any case it is possible that the problem is not ADI-FDTD specific.
If you use the same type of sources with FDTD at 100% CFL and 50% CFL what kind
of scaling do you need?
Thank you.
I have found out that when CFL number=n is used, the source being used gets divided by 'n'. So the reason for the requirement to multiply any observed signal with 'n' became obvious. Thanks a lot for your support and advices.