I have to measure the S parameter for a rectangular sample with an antenna that works from 2 to 18 GHz. I need to measure a large band from 5 to 15 GHz and I have a problem with the known F.F. (Far Field) formula:

$$R = 2D^2/λ$$

Usually with $D$ is referred the aperture of the antenna, that in my case is 100 mm, while I found that for RCS the formula use as $D$ value the dimension of the sample (300 mm).

My questions are:

  • Why can the F.F. formula be referred to the antenna or to the sample and
  • when I need to use one or the other?

The far field formula is simply derived from a general idea that the phase error at the edge of the antenna aperture should be less than $22.5^\circ$ which means that $r'-r < {1\over16}\lambda$.
This figure is chosen because it gives gain errors of less than some small percentage, I can't remember the figure, it depends on the aperture field distribution too).

far field derivation

Now draw a new diagram with the large target object, and the source antenna, and work out the difference in path length between a ray from the centre of the object, to the centre of the antenna, and a ray between the edge of the object and the opposite edge of the antenna.

RCS measurement

The whole path difference $r'-r$ should be kept to below ${1\over16}\lambda$ for some level of accuracy.
You can see from this diagram (slide $r'$ up to make one triangle) that the "safe" far-field distance will be
This will guarantee that phase errors from any part of the antenna aperture, to any part of the reflecting target, are kept below $22.5^\circ$.

You might be able to relax this criterion by half if you're willing to tolerate a larger error, or know something about the aperture distribution of the antenna, or the reflection properties of the target. A sphere, for example, might not reflect from the whole surface, only the middle. Unfortunately a horn antenna will have a fairly uniform field.

$\lambda$ in all equations above needs to be calculated at the highest frequency you are measuring (but not the highest specified frequency of the horn).

  • 1
    $\begingroup$ @Allergenfree I've updated the formula... it's even further than I first thought, 16 m. $\endgroup$ – tomnexus May 14 '19 at 18:20
  • $\begingroup$ Thanks a lot for the very good answer, you completely solved my doubt $\endgroup$ – Allergenfree May 15 '19 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.