JLD said:

How can an AC current exist in a wire that is connected to some

circuitry at one end but is unconnected, open, floating at the other

end?

I've tried to explain this in the past to various people, with

varying degrees of success, and here's the best I've been able

to come up with so far.

Imagine a circuit consisting of a 1V source - a battery, let's say,

and a 1 ohm resistor, connected by whatever length of wire you

find convenient and with a switch at the battery. You close the

switch, and what happens? Well, by Ohm's Law, we have

1V/1 ohms, and we see 1A of current flowing, right?

But what happens if the resistor is one mile away from the battery,

or ten miles, or a hundred miles, or...? Ignoring the resistance of

the wire for the moment, we've still got a problem, because we

know that nothing, not even electricity, travels at truly infinite

speed. It will take longer for the electricty to "get to" the resistor

if it's a hundred miles away than if it's ten feet away, right? Not

a whole lot longer, on a human time scale, but still definitely long

enough to measure. (Work it out - even things travelling at the

speed of light take about a nanosecond to go one foot, so for

100 miles...) But if that's true, then how much current is flowing

when we first close the switch? (There should be SOME - it

has to get to the resistor eventually, right? - but the battery clearly

can't just "know" that there's a 1 ohm resistor and magically send

out 1A of current right from the start, so what DOES happen?)

This is where we get the idea of "characteristic impedance." Even

before the source "sees" the final load, there IS an impedance

presented by the wire itself (even if it has no resistance at all, as we

are imagining here), and it comes from the physical configuration of

the wires - it is due to the capacitance and inductance presented by

the conductive path itself, the "distributed" capacitance and inductance

that is the same, for a given configuration, all along the path (and so

is given in terms of units like "picofarads per meter" or "microhenries

per foot" or some such). Without getting into the math of how it's

derived, there IS a "load" presented by this impedance to the source,

immediately upon the closure of the switch, and that set the voltage

and current relationship.

So now we've got the electricity going toward the resistor - what

happens when it gets there? What if the value of the resistance does

NOT match the value of the characteristic impedance of the line?

The proper resistance between voltage and current, per Ohm's Law,

MUST always be maintained, and this is what leads to "reflections"

occuring on the line. Reflected voltage and current signals are

created such that (1) the proper relationship exists at the load, and

(2) the relationship of the reflected voltage and current signals are

per the characteristic impedance of the line.

At this point, you're probably wondering what this has to do with

antennas. But now think about what happens if the source isn't

a battery (a DC source) but instead is AC - something that, in

effect, is "turning on and off" every so often, on a regular basis.

This results in a reflected AC signal at the load, if a mismatch

exists - and the combination of the "forward" and "reflected"

signals means that there will be points on the line where the

peaks coincide (and reinforce) and where the two waves

interfere with each other - this results in a pattern of maximum

and minimum amplitudes along the line, the so-called "standing

wave" pattern. Just two more steps until we're at an antenna.

Now, imagine what happens if the wires happen to be some

multiple of a quarter-wavelength long at the frequency of the source.

This means that there will be one of the peaks or minimums of this

"standing wave" out at the load end of the line. If the wave is exactly

a quarter-wave long, in fact, there's a voltage peak (and a current

minimum) right where the load is. In fact, the current can zero at

this point - if you simply remove the load resistor and leave the line

open! This means that we have a situation where the reflected wave

carries all of the energy of the forward wave.

Last step - since you've already taken off the resistor, there's no real

need to leave the conductors in parallel as you're probably envisioning

them. "Open up" this quarter-wavelength of line, so that you wind up

with a half-wavelength long structure - two conductors, leading out

from the source in opposite directions. (This change in the geometry

does change the characteristic impedance, etc., but the basic idea,

that you've got a voltage maximum out at the ends of the line, and

conversely a current maximum right at the source, still holds true.)

But now look at what we've got - you have these wires sticking "up

and down" from the source, with voltage peaks (of opposite

polarity) at either end, and a current peak in the middle. Currents

make magnetic fields, and a voltages across a space create electric

fields - and here we are with a structure that's arranged to create

these in the right relationship so as to create an EM wave. And that's

exactly what happens - the energy, rather than being reflected back

into the source, gets "launched" as an electromagnetic wave.

This is, of course, from the "transmitting" point of view, but the same

argument works in reverse for receiving. Incoming EM waves

induce voltages and currents in the antenna in the proper relationship

so as to create electrical signals that can be accessed at the center

point (so instead of the source being there, we now see a load there).

And, of course, what we've described is the classic "dipole" antenna,

but this is the same basic idea behind a simple vertical whip. In the

whip antenna, or more formally the quarter-wave vertical, only one of

the two elements is actually there, and is driven, and it "works against"

the ground plane below it. (The analysis of this structure shows that

it works as if the "missing" element were actually mirrored "in" the

ground plane.)

This is a somewhat simplified, and clearly non-mathematical, treatment

of this question, but hopefully you now at least sort of get the idea.

Bob M.