The short answer to your question is no, with practical impedances. It
turns out you'd need a very low impedance transmission line (on the
order of a few ohms, if I recall correctly) to track very well.

The easiest way to analyze a folded dipole is as two separate circuits,
common-mode (or "antenna") and differential-mode (or "transmission
line"). Superposition applies, so the two can be analyzed separately.

First consider the "antenna". Its contribution to the feedpoint
impedance is the same as a conventional monopole made from the two wires
in parallel, but multiplied by 4 due to the transforming action of the
"folding" process. (Other ratios are possible -- 4 is what you get if
the conductors are the same diameter.) This "antenna" does all the
radiating.

In parallel with the "antenna" at the feedpoint is the "transmission
line". This is the short-circuited transmission line made from the two
conductors, with transmission line velocity factor taken into account.
The "transmission line" part does no radiating.

Now, if you shorten the antenna, two things will happen. The reactance
of the "antenna" will become more negative, and its resistance will drop
some. This will show up at the feedpoint just like it would for an
unfolded monopole, but multiplied by 4. But you're also shortening the
transmission line, whose impedance also appears at the feedpoint, in
parallel with the transformed "antenna's". Assuming negligible loss in
the feedline, this will cause a change only in parallel reactance at the
feedpoint. Because of the relatively high impedance of the transmission
line, and the relative sharpness of the impedance change of the
"antenna" compared to the transmission line, there's very little
compensation in the case of most practical antennas. Remember that the
two impedances are in parallel, not series, so the high Z contribution
of the transmission line has little overall effect. As I recall from
doing an analysis some time ago, you get more broadening compared to a
conventional monopole from having a fatter equivalent "antenna"
conductor than you do from the transmission line stub.

The reactance of the "antenna" can be a bit tricky to calculate
accurately, but a number of modern programs (such as the free EZNEC
demo) do a good job of it. The transmission line part of the effect is
easy, either with a scientific calculator or one of the transmission
line programs which are readily available. So overall, it's not
difficult to find the actual impedances you'll see in practice.

Roy Lewallen, W7EL


The other John Smith wrote:
Good evening, Gentlemen.

A thought experiment:

Start with a regular 1/4-wave monopole ground plane. The literature says it
looks like half the value of a dipole, about 35 Ohms, when resonant. It
would be nice to have the resistance at the terminals be a bit higher, and I
very much value a grounded element anyway, so let's let it evolve into a
folded monopole. The literature says it should now have about 4 times the
terminal resistance of the original 1/4-wave we started with (about 140
Ohms). Huh. Now it's a bit high.

They tell me that shortening the antenna below resonance will lower the
resistance and introduce capacitance. But I think I have also seen in the
literature that the antenna can be viewed as a transmission line. A shorted
portion of parallel conductor transmission line (the folded monopole) less
than 1/4-wave long looks inductive. But wait! Which will win? Will the
shortness of the antenna look capacitive or will the transmission line
dominate and the antenna will look inductive?

Even better, is there some choice of the folded section wire diameters and
spacing that will give an inductance that will exactly offset the
capacitance due to shortness? So, then, is there a folded monopole of such
dimensions that the resistance is 50 Ohms (due to being shorter than 1/4
wave) with no terminal reactance (due to the inductive design of the
"transmission line" cancelled by the shortness of the antenna's
capacitance)?

Brain hurts.

John, KD5YI