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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?
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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
${2(D_{antenna}+D_{target})^2}\over\lambda$
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).

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

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