Newsgroups: rec.radio.amateur.misc

> How does current flow in an antenna that's not a folded dipole?  There's no
> return path for electrons to move in.  Also, someone mentioned a signal
> starting out in one element, coming back in another, and then going back
> out in a third one.  How does it know which element to start out on, and 
> which one to come back on (I'm assuming that all the elements are 
> connected to the same feed line).
> 
> Dube                    AB5AP                <dube@cpdvax.csc.ti.com>

The best way to visualize the operation of a simple dipole is to first
become familiar with the behavior of an open-ended transmission line.
In such a line, a "forward" travelling signal will be completely reflected 
from the open end, as the reflection coefficient is 1.  This will set up
a "standing wave" pattern on the line, with a voltage maximum and current
minimum at the open end, and then repeating every half-wavelength back to
the source (which, for simplicity's sake, we'd better assume is matched!).
In between these points (i.e., one-quarter wave in either direction from
a voltage maximum), are points at which the *current* is at a maximum, and
the voltage is at a minimum.  So every quarter-wave back from the open end,
we have "something special" happening - either a voltage or a current
maximum.

Now comes the tricky part - while maintaining this image in your mind,
visualize the line "opening up" to a point a quarter-wave back from the
original open end.  In other words, if you're visualizing a twinlead line,
pull one conductor in one direction, and the other in the opposite direction,
until you've separated one-quarter-wavelength's worth of the original line.
You now have a dipole!  What's more, the midpoint of the dipole - a point 
which was 1/4-wave back from the original end - is a current maximum and
voltage minimum.  In short, this is a low-impedance point.  The original
"standing wave" is basically maintained on the "opened-up" line, and so
we would also expect the end points of the dipole to be points of high voltage,
but low current.  And so they are.  To see how an electromagentic wave is
"launched" from this antenna, you simply need to recall that there will
be an electric field between these endpoints (since there is a high potential
difference between them), and the high current at the midpoint means that
there will be a magnetic field *around* the antenna - oriented 90 degrees
out from the E-field, and also 90 degres off in time!  This is all that's
needed for an EM wave, and sure enough, the thing radiates.

This is an over-simplified explanation; for one thing, the act of opening
the line up introduces what is basically "new" inductance into the picture,
and a resonant dipole must therefore be somewhat shorter than a full
half-wavelength.  Also, realize that there are voltage and current
*distributions* along the antenna, and so EVERY part is contributing to the
radiation, not just the endpoint and the middle.  The feedpoint impedance
of such a structure, at resonance, is right around 73 ohms, which is not
too bad a match to 50-ohm line.  (And in reality, the impedance may be
slightly lower anyway, due to factors which don't show up in the "pure"
theoretical analysis.)


Bob Myers  KC0EW   Hewlett-Packard Co.      |Opinions expressed here are not
                   User Interface Tech. Div.|those of my employer or any other
myers@fc.hp.com    Fort Collins, Colorado   |sentient life-form on this planet.