On Fri, 05 Nov 2004 07:45:16 GMT, Richard Clark
wrote:

Good day Richard,

Sorry this isn't too timely, but I've been off visiting a dear friend
and soon to be silent key.

|On Thu, 04 Nov 2004 20:38:35 -0700, Wes Stewart
|wrote:
|
|Or, model a short lossless monopole over perfect ground and determine
|the feedpoint R. In this case, R is totally due to radiation loss,
|i.e. "radiation resistance." Add a lossless loading inductance
|somewhere in the middle and see what happens to R.
|
|Hi Wes,
|
|The difference between the two (perfect/real) insofar as Z is hardly
|remarkable.

True.
|
|First I will start with a conventionally sized quarterwave and by
|iteration approach the short antenna and observe effects. I am using
|the model VERT1.EZ that is in the EZNEC distribution and modifying it
|by turns. For instance, I immediately turn on the wire loss.
|
|40mm thick radiator 10.3 meters tall:
| Impedance = 36.68 + J 2.999 ohms
|which lends every appearance to expectation of Rr that could be
|expected from a lossless perfect grounded world.
|Best gain is
| -0.03dBi
|
|next iteration:
|
|cut that sucker in half:
| Impedance = 6.867 - J 301 ohms
|which, again, conforms to most authorities on the basis of Rr.
|best gain
| 0.16dBi
|How about that! More gain than for the quarterwave (but hardly
|remarkable). This makes me wonder why any futzing is required except
|for the tender requirements of the SWR fearing transmitter (which, by
|the way, could be as easily taken care of with a tuner).
|
|next iteration:
|
|load that sucker for grins and giggles:
| load = 605 Ohms Xl up 55%
| Impedance = 13.43 + J 0.1587 ohms
|Did I double Rr? (Only my hairdresser knows.)
|best gain
| 0.13dBi
|Hmm, losing ground for our effort, it makes a pretty picture of
|current distribution that conforms to all the descriptions here (sans
|the balderdash of curve fitting to a sine wave). I am sure someone
|will rescue this situation from my ineptitude by a better load
|placement, so I will leave that unfinished work to the adept
|practitioners.
|
|next iteration:
|
|cut that sucker down half again (and remove the load):
| Impedance = 1.59 - J 624.6 ohms
|Something tells me that this isn't off the scale of the perfect
|comparison.
|best gain:
| 0.25dBi
|Hmm, the trend seems to go counter to intuition.
|
|next iteration:
|
|-sigh- what charms could loading bring us?
| load = 1220 Ohms Xl up 55%
| Impedance = 3.791 + J 1.232 ohms
|more than doubled the Rr?
|best gain:
| 0.23dBi
|
|Now, all of this is for a source that is a constant current generator;
|we've monkeyed with the current distribution and put more resistance
|(Rr?) into the equation with loading; and each time loading craps in
|the punch bowl.
|
|So much for theories of Rr being modified by loading. I would
|appreciate other effort in kind to correct any oversights I've made
|(not just the usual palaver of tedious "explanations" - especially
|those sophmoric studies of current-in/current-out).

Sure. Your calculations are impeccable so far (although it would be
better to use more than 10 segments IMHO).

I believe it is your contention that loading to resonance with an
arbitrarily positioned inductor, or not loading at all, does not
affect the gain, and the radiation resistance is not the same as the
changing feedpoint resistance.

I am in the other camp, along with Hansen, Devoldere, et. al. who say
that the current distribution does affect the radiation resistance
(and in the real world, the gain/efficiency).

I hope you would agree that the normalized gain would be a good proxy
for efficiency.

For example if we use the lossless 1/4 wavelength monopole over
perfect ground as a reference, then gain with respect to that (5.15
dBi) would be an indicator of efficiency.

I believe that you will agree that the efficiency can be determined
by:

Rr
eta = ------------- Eq.1
Rr + Rg + Rl

where Rr = radiation resistance
Rg = ground resistance
Rl = all other resistances (conductor, etc)

I think you would also agree that for the full-sized monopole over
perfect ground the feedpoint resistance of ~36 Ohm = radiation
resistance.

As an old (sorry [g]) metrologist, you're very familiar with
substitution, so let's set Rl = 0 (lossless case) and eta to 0.5 (-3
dB). Per Eq. 1, Rg = Rr.

So in our model, if I add a simulated ground resistance, Rg, that
reduces the gain by 3 dB, I have by substitution, determined the
radiation resistance.

Sure enough, if I add a 36 Ohm load at the bottom of the perfect 1/4
wave monopole, the gain drops to 2.14 dBi, and the feedpoint
resistance doubles.

I will let you try this with the other cases. I trust you will find
that the radiation resistance does decrease with shorter radiators
and/or lower loading points.

I too I would appreciate other effort in kind to correct any
oversights I've made.

Wes