0
\$\begingroup\$

The working principle and fields of an antenna can often be shown by considering a parallel LC circuit. The plates can be moved to opposite ends and the inductor can be stretched out, both of which decreases the capacitance and inductance of the circuit and increases the resonant frequency.

enter image description here

In the limiting case of a half-wave dipole, the antenna must be center-fed (if fed by a current source) and connected to the transmission line. To accomplish this, the inductor in the middle needs to be cut and there is no current flow directly through the inductor:

enter image description here As can be seen in the picture above, the center is insulated, while this point in the LC circuit corresponds to the center of the inductor, which is conducting.

While in a regular LC circuit a current flows through the inductor, it now moves through the whole transmission line at each cycle. It seems to me that these circuits are not equivalent anymore, since that introduces all the impedance of the transmission line into the circuit.

It's not so clear to me what changes if the inductor is still connected (apart from presenting a short circuit to the transmission line), and does this "cutting" change the inductance of the antenna itself?

\$\endgroup\$
6
  • \$\begingroup\$ I have a little experience with antennas, but I have no idea what you are talking about. What exactly means: "...the inductor in the middle has to be cut" for a half-wave dipole? Why has the inductor to be cut? Can you provide a picture of what you mean? \$\endgroup\$ Commented Oct 17, 2018 at 12:02
  • \$\begingroup\$ @StefanWyss I added a picture of a dipole where it is clear what I meant by "cutting". If you take a half-wave length of wire and attach the terminals of the transmission line to it in the center without insulating the pieces at the center point, it will just be a short circuit. On the other hand, there are things like delta matches that use exactly such a "whole" antenna - what exactly is happening here? How is the behavior of the antenna affected by the resistance at the center point (insulator)? \$\endgroup\$
    – DK2AX
    Commented Oct 17, 2018 at 12:09
  • \$\begingroup\$ I don't think they meant physically cut, I think they meant to replace it with 2 inductors each having inductance L/2. \$\endgroup\$ Commented Oct 17, 2018 at 12:18
  • \$\begingroup\$ @CristobolPolychronopolis Sure, let me put it like this: if I spread apart the plates of the capacitor in a LC circuit, I can pass a current between one plate and the other (through the inductor). In the case of the half-wave dipole, the circuit is broken by the insulator: a current needs to flow through the transmission line instead of only through the antenna. \$\endgroup\$
    – DK2AX
    Commented Oct 17, 2018 at 12:22
  • \$\begingroup\$ In the first picture, you have the ground at the bottom of the antenna, while the feed is at the top. This is undesireable for mechanical reasons. \$\endgroup\$
    – Janka
    Commented Oct 17, 2018 at 12:27

1 Answer 1

2
\$\begingroup\$

Consider the wire antenna itself, before you cut it to insert the feed line. It has a certain resonant frequency based on its length. When operated at this resonant frequency, there will be a standing wave on the antenna, with a certain voltage distribution (zero at the middle and high at the ends) and current distribution (zero at the ends and high in the middle).

You can think of the middle of the antenna as the "inductor" and the ends as the "capacitor", but these effects are really both now distributed along the entire length of the antenna.

The antenna also has a characteristic impedance that varies along its length that you can get by dividing the voltage by the current. Near the ends of the antenna, the impedance is high because the voltage swing is very high while the current is very tiny. At the center, the impedance is lowest because the voltage is much lower and the current much higher.1

The point is, if you cut the antenna at any point and connect a feedline of the same impedance at that point, it does not fundamentally change the operation of the antenna at all. It just provides a means of coupling electrical energy directly into or out of the antenna. Both the "effective inductance" and the "effective capacitance" are not changed by doing this.


1 The impedance never goes all the way to zero or infinity because of the way that the fields of the antenna interact with the free space around it.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.