Integrating power in far field projections

The section explains how to integrate the far fields on the default hemispherical surface. This is typically done to calculate the amount of far field power within some range of angles.

Note: The descriptions and examples of the far field projection calculation on the following pages are primarily intended for users of FDTD.   For users interested in calculating far field projections with MODE, these descriptions are basically still correct, although some subtle differences do exist.

Power Integrals

In general, we want to integrate power over a given solid angle in the far field. There are 2 ways this can be done

• We integrate the fraction of total electric field intensity (|E|2) over the solid angle that we are interested in, and multiply by the normalized power transmission through the monitor in the near field.

$$\text{far field fraction}=\frac{\int_{cone}\mid\mathbf{E}\mid^2}{\int_{hemisphere}\mid\mathbf{E}\mid^2}\\ \text{Power norm}=\text{far field fraction}*\text{near field power transmission}$$

• We calculate the Poynting vector in the far field and integrate the power over a given solid angle. We then normalize to the original source power. In the far field, it is easy to calculate the normal component of the Poynting vector directly from the Electric fields by using the plane wave relationships between E and H.  We know the tangential components the Poynting vector are zero for a plane wave.

$$\text{Poynting}_\bot=n\sqrt{\frac{\epsilon_0}{\mu_0}}\mid\mathbf{E}\mid^2\\Power=\frac{1}{2}\int_{cone}Re(\text{Poynting}_\bot)\\ \text{Power norm}=\frac{\text{Power}}{\text{SourcePower}}$$

Both methods will give the same result.

The script file will calculate the normalized power in a given cone around the z-axis defined by a half-angle. It will do the calculation by both methods described above. The desired half angle can be set in the first executed line of the script file

# choose the half angle over which we will integrate
half_angle = 30; #in degrees

We expect that using a half angle of 30 degrees will account for 50% of the power. When we run with 30 degrees, we find have the following output

> solver_far_field2;
The half angle is: 30 degrees at (theta,phi)=(0,0)
  The normalized transmission by Method 1 is: 45.6613 %
  The normalized transmission by Method 2 is: 44.5107 %

If we try several angles we find that

As expected, we find that half the power is radiated within a half-angle cone of 30 degrees around the z-axis.