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.