antenna-theory.com

[Howard Edwards]

Time and again, I read in books relating to antennas, electromagnetic theory, propagation etc, that field strength is expressed in "volts per metre" (British spelling !). Then they leave it at that. Just what does this mean, and how is it measured ? If they are talking about "metre" in free space, in which direction is it taken? Is it in the same direction as the polarity of the electric wave component of the EM wave ? Thus if an antenna puts out a vertically polarized signal, is the field strength measured in the vertical plain ? If this is indeed so, I still don't quite understand what is actually going on, particularly in that we are dealing with a signal whose instantaneous value is constantly changing. If the wavelength of the signal is two metres, then it looks to me as if you measure the potential over a distance of one metre, then you will get a zero reading as the instantaneous value at both points will be the same, as there is no difference in potential.
I am obviously and seriously missing something here.
It would be helpful if books were to show diagrams at this point which show the waves and how / why the field strength is expressed in "volts per metre" ! I suppose I could take a field strength meter out, tune it in to a particular frequency and the dial would show such and such microvolts per metre, but I would hardly be the wiser in terms of understanding what is actually going on !
So that I can get a decent night's sleep, can anyone help, or can someone refer me to the "Ultimate guide in EM propagation" - where "Ultimate" means 'Brilliant explanations' ? !!
Thank you
Howard

[Schubert]

Here's some notes:

1) Yes the electric field is continuously changing (for non d.c. fields), and hence the voltage is changing as well. Don't make that more complicated than it is. Just think about things one moment at a time. The voltage follows the Electric Field.

2) You make the point that if the wavelength is 2 meters, and you measure the potential over a distance of one meter, then you will get a zero reading. Answer: No. Think about an x-y-z coordinate system. Suppose a plane wave is propagating in the +z-direction, and that the electric field is polarized in the x-direction. Then for any instant of time, the field is constant in the entire x-y plane, for any value of z. The electric field only varies in the z-direction. So the voltage difference at 1 meter in the x-direction, you will see a voltage difference. The voltage difference across 1 meter in the z-direction will be zero. Try to understand plane waves and this should make sense.

Does this clear some of it up?

[R. Fry]

As commonly used, "field strength" in volts/meter is a measure of the electric field vector of an e-m wave, and refers to the potential difference existing between two points in space separated by a linear distance of 1 meter.

The greatest free-space E-field intensity will exist the same planes(s) as the maximum E-field(s) radiated by the antenna, and define its polarization (not polarity).

The polarization of an electromagnetic wave is defined by the physical orientation of its electric field. For linear radiators such as a dipole and vertical monopole, the direction of polarization is that of the physical orientation of the radiator.

The polarity of an e-m wave is determined by its electric field vector, which reverses every 180 degrees along the propagated wave, regardless of the polarization of the wave.

The applet at http://www.amanogawa.com/archive/Polarization2/Polarization2-2.html is useful to visualize this. To see vertical polarization, first set the Ey field to zero, and start the animation (top center of the page). Then set the Ex field to zero and the Ey field to one to see horizontal polarization.

In this applet if you set the Ey and Ex field equal (say at 1 each), and their phase relationship to -90 degrees using the slider below the Ey and Ex sliders, then the resulting e-m field is perfect, right-hand circular polarization. The animation shows a field vector of constant magnitude rotating once per wavelength.

[Howard Edwards]

Very grateful for your comments - I shall study them and get back to you on this matter. Do you agree that it is often poorly put over in books ?
H