# Explanation for: A monopole antenna must contain a resistor (or equivalent) and therefore must have 2 terminals?

Could someone explain why this sentence makes sense:

A monopole antenna transfers energy from electrical domain to the electromagnetic domain, hence must contain (equivalently) a resistor, hence must have 2 or more terminals.

Why must it contain a resistor and how can you conclude that you need 2 terminals because of the resistor?

Also,

A monopole is half of a dipole with groundplane in its symmetry plane and the drive is between antenna-feedpoint and ground(plane)

What is the symmetry plane? What does drive mean here? The transmitter?

• It is good form to indicate whom you are citing. Apr 5, 2015 at 17:48

Why must it contain a resistor

The author is most likely referring to the radiation resistance. A circuit delivers electrical energy to a resistor where it is converted to heat, i.e., the energy is not stored but is lost to the environment.

Analogously, a circuit delivers electrical energy to an antenna where it is converted to electromagnetic radiation which propagates away at the speed of light.

Thus, in this sense, the antenna is a 'radiation resistor'. It is a two-terminal system where one terminal is the antenna proper and the other terminal is ground.

What is the symmetry plane?

Essentially, there is an 'image' antenna on the other side of the ground plane. From the linked article:

The radio waves from an antenna element that reflect off a ground plane appear to come from a mirror image of the antenna located on the other side of the ground plane. In a monopole antenna, the radiation pattern of the monopole plus the virtual "image antenna" make it appear as a two element center-fed dipole antenna. So a monopole mounted over an ideal ground plane has a radiation pattern identical to a dipole antenna.

What does drive mean here? The transmitter?

Yes. Or, more precisely, the output terminals of the transmitter.

• "the antenna is a 'radiation resistor'" So do you mean that the 'radiotion resistor' is a part of the antenna or the entire antenna is just a resistor? "one terminal is the antenna proper" What does 'antenna proper' mean in this context? One terminal is the output of the transmitter and the other is the ground(plane)? Apr 5, 2015 at 18:48
• PeterH the antenna can be easily compared to a transmission line en.wikipedia.org/wiki/Transmission_line#/media/… . Yes, antennas are passive components (therefore have a resistance), but if you'd want to make a comparison, they would appear not only as a resistor, but a combination of everything you can see in the picture. Jul 14, 2015 at 11:29

The above information is totally false. Notice no authoritative sources are cited.

This statement violates the laws of physics:

" circuit delivers electrical energy to an antenna where it is converted to electromagnetic radiation which propagates away at the speed of light."

The first clue that it is gross error is the obviously false statement that RF energy propagates at "the speed of light" NOTHING except light does that, any EM field is bound to propagate at LESS than c, depending on the velocity of propagation.

Antenna do not radiate electro-magnetic energy. The E field is maintained around and collapses back into the antenna, to then produce a M field which then radiates at a distance. (assuming an antenna intended to radiate a long distance)

This statement is easy to observe as false:

"Thus, in this sense, the antenna is a 'radiation resistor'. It is a two-terminal system where one terminal is the antenna proper and the other terminal is ground."

An end (bottom) fed monopole vertical does not have two terminals, but still it is an antenna which radiates. It is not a resistor, for radiation 'resistance' is not resistance, it is only crudely assumed as that as a function of feed point current and assumed, falsely, to be resistive as it is a loss from the antenna. There is no resistance involved in radiation. The R component is due to skin effect, not radiation.

It is false that the ground supplies half the antenna. Nothing in the Earth is 1/4 wavelength in any dimension, the Earth is infinitely large. How can a ground connection at 10 MHz, be a quarter wave replacement, and also be so at 20 MHz? Impossible.

If it were true that the "ground replaces part of the antenna" then vertical antennas mounted at more than a half (or one) wavelength above ground would not have that 'ground effect' and would not function.

The ground myth comes from the problem of an open ended feedline, not the antenna or Earth.

See: http://my.ece.msstate.edu/faculty/donohoe/ece4990notes1.pdf p.3 "open circuit transmission line..." This is the basis for all the myths of grounds, ground planes, 1/4 wave images (which are all proven false by vertical antennas mounted high above Earth. The ground (Earth) either attenuates/absorbs RF energy (a total loss) or reflects a portion. Conecting the shield of a feedline coaxial cable to "ground" constitutes a circuit that takes the open ended feedline to a closed circuit, there is no magic in "ground", it is EM fields being coupled between the antenna and ground, thus across the end of the feedline.

INstead of internet rumour, read The Self and Mutual INductances of Linear Conductors by Edward Rosa, 1908, especially about the antenna return circuit, and works by "Grover". Study "inductance of monopole antennas."