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Power calc in 2D FDTD not a Poynting vector

时间:03-30 整理:3721RD 点击:
What is the relationship of the signal power calculation in the 2D FDTD to the Poynting vector?

I measure my signal power from a TE FDTD calculation with the formula :

Power in (watt/m^2) = 0.5 * sqrt((Ex*Hz)*(Ex*Hz) + (Ey*Hz)*(Ey*Hz)) with the wave propagating in the positive x direction.

Is this approach valid or should one not even contemplate this for 2D modelling ?

Hi bobilgner,

Are you looking for the time-averaged Poynting vector? Recall that this is derived from ExH*. As far as I know your approach is otherwise correct for FDTD, but I am not a computational electromagnetics expert.


I am looking to calculate the power of the signal in watt at a specific point. Imagine the signal propagating away from a vertical dipole antenna. Because of the symmetry in my layout I can use the 2D FDTD, but how can I determine the power in the signal at some y(vertical) and x(horizontal). I have used

Power in (watt/m^2) = 0.5 * sqrt((Ex*Hz)*(Ex*Hz) + (Ey*Hz)*(Ey*Hz))

but the results look unexpected. I would expect some drop-off proportional to 1/r^2, but do not get this.

You probably do want the time averaged Poynting vector then.

Are the individual fields fields correct (i.e. decay with distance)?

You were right of course. I calculated the quadrature (RMS) time average of the power and everything looked good. Decay was according to theory.

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