Herztian dipole + radiation patterns

[JamesGoh]

Currently reading over some lecture notes on Hertzian dipoles and I want to confirm my understanding on why the E-field pattern is donut shaped.

1) The Herztzian (aka. 1/2 wave ) diploe has two terminals attached to a transmission line feeder. The feeder sends RF current into the terminals which in turn produce EM waves.

2) Since the transmission line feeder itself consists of current going from one side to another, it is natural that the E-field component of the waves travel from one terminal to another (producing a circular pattern, similar to charges in the Earth's magnetic field)

3) Viewing this pattern from above, one can see the "donut" shaped pattern of the E-field

Thoughts, comments ??

Please feel free to correct me if I am wrong in any concept or theory

 

[Xezlec]

I don't know about the phrase "similar to charges in the Earth's magnetic field", but other than that I think you're on the right track. At any point in time, there will be electric field lines "spraying out" of the positive side of the dipole, wrapping around kind of like a donut and crunching together to squeeze into the negative side of the dipole. The polarity (i.e. the direction of the field, forward or backward, along those lines) will be oscillating back and forth.

It is also helpful not to forget about the magnetic component of the field. Since you have a current in a wire (the dipole), you also have circles of magnetic field pointing around that wire, approximately like a cylinder with the wire as its axis. Since the current is oscillating back and forth, the magnetic field is oscillating between the two different directions around that cylinder. Since a changing magnetic field produces a changing electric field in circles around it, you can now visualize a changing electric field looping around and through the cylinder of magnetic field, like if you wrap a string through a straw and back around to the same end again. That changing electric field will be a bit like the surface of a donut, where the magnetic field is kind of like rings of cream filling inside.

Oh, God, I'm so hungry now...

 

[Antenna Guy]

1) The Herztzian (aka. 1/2 wave ) diploe has two terminals attached to a transmission line feeder. The feeder sends RF current into the terminals which in turn produce EM waves.

Hertzian dipoles are much shorter than a half-wave - but as far as radiation patterns go, they have roughly the same pattern.

2) Since the transmission line feeder itself consists of current going from one side to another, it is natural that the E-field component of the waves travel from one terminal to another

That's one way of looking at it, but there is a way that can be applied more generally.

Are you familiar with finite element modeling?

In lieu of that, imagine a spherical coordinate system centered on the dipole such that the x-axis is aligned with the elements, and the z-axis is orthogonal to the elements. Define \theta=0 as the +z direction, and \phi=0 as the x-z plane (\theta>0 on the +x side).

If we restrict ourselves to \phi=0, the currents in the dipole have no projection perpendicular to the plane, so the fields of this plane would be considered aligned with theta (theta-polarized). Similarly, if we restrict ourselves to \phi=90 (the y-z plane), the currents in the dipole have no projection in the plane, so the fields of this plane would be considered aligned with phi (phi-polarized).

3) Viewing this pattern from above, one can see the "donut" shaped pattern of the E-field

What you are calling "above", I would call "a distant point on the x-axis" (defined above). The fields are zero in that direction because the effective aperture is zero.

I'm sure you're aware that the 3D "picture" you describe has an arbitrary size in 3D space. The three dimensions of the pattern you refer to are two spherical angles, and one magnitude (represented as r). It is assumed that the pattern is at constant radius in 3D space, and beyond \frac{2D^2}{\lambda}=\frac{\lambda}{2} the pattern does not change significantly with increasing r.

Care to venture a guess where r=0 with respect to the dipole?

Regards,

Bill

 

[JamesGoh]

What you are calling "above", I would call "a distant point on the x-axis" (defined above). The fields are zero in that direction because the effective aperture is zero.

"Above" refers to the H-plane (or viewing the circular ends of the 2 rods).

Please feel free to correct me

 

[Antenna Guy]

"Above" refers to the H-plane (or viewing the circular ends of the 2 rods).

Please feel free to correct me

Well - a plane isn't a direction, but I know what you mean.

Consider that "above" is (typically) a different direction when one considers a dipole "above" a ground plane.

Regards,

Bill

[addendum: The ends of the dipole fall in the E-plane, not the H-plane]