1
$\begingroup$

enter image description here

I read in some text books that

  1. E-plane is formed by taking constant azimuth angle and scanning elevation angle from -90:90.
  2. H-plane is formed by taking a plane perpendicular to E-plane.

What is the meaning of taking a plane perpendicular to E-plane?? Is it to take constant elevation angle and scan azimuth angle from -90:90?

$\endgroup$
  • 1
    $\begingroup$ Please, share with us the text book where you found this information. $\endgroup$ – Brian K1LI Dec 3 at 12:54
  • 1
    $\begingroup$ Harry L. Van Trees - Optimum Array Processing (Detection, Estimation, and Modulation Theory, Part IV) (2002).PageNo:242. $\endgroup$ – kartheek Dec 4 at 11:42
  • $\begingroup$ Hello, and welcome to this site! That seems to be a well-respected, scholarly book. I can't seem to find the book where he says that, but could it be that he says something along the lines of what Marcus said? $\endgroup$ – Mike Waters Dec 4 at 19:43
  • $\begingroup$ I searched other editions of this book (since the 2002 edition is not available to view on books.google.com, and I couldn't find E-plane. Could you please edit your question and post a photo of that page? Thanks! $\endgroup$ – Mike Waters Dec 4 at 20:05
  • 1
    $\begingroup$ Sorry, this page seems to have nothing to do with E- or H-Planes. Could you explain, @kartheek? $\endgroup$ – Marcus Müller Dec 5 at 19:25
4
$\begingroup$

Those definition are false.

The E-plane is defined as the plane in which the E-field varies over time. The H-plane is the plane in which the H-field varies over time.

There's nothing more to it.

Logically, the definition of planes only makes sense for linearly polarized antennas.

In isotropic (meaning: behaving the same from every angle) media (e.g., air), the H-plane is always perpendicular to the E-plane. Perpendicular is when two planes are at a right angle to each other.

Whether the E-plane is in azimuth or elevation or somewhere in between depends on the direction of polarization of the antenna.

$\endgroup$
  • $\begingroup$ @hobbs-KC2G ooops, yes. $\endgroup$ – Marcus Müller Dec 4 at 8:25
  • $\begingroup$ He has included the page in question. Any thoughts? $\endgroup$ – Mike Waters Dec 5 at 16:44
  • 1
    $\begingroup$ @MikeWaters not really – that page nowhere mentions an E- or H-Plane. $\endgroup$ – Marcus Müller Dec 5 at 19:26

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.