Wayne wrote:


"gareth" wrote in message ...

Consider 100W at 3.6MHz propagating along some 50 ohm
coax, which terminates suddenly but with 1/4 inch of the central
conductor protruding.

Now there's no difficulty in feeding all that power into that
1/4 inch because it is so short compared to a wavelength
that there is a uniformity of voltage and current along it,
and it will be essentially the same as that existing in the last
gnat's cock of the coax.

Attach a hi-impedance scope probe to the end of that
1/4 inch and all the power being delivered through the
coax will be detectable right at the tip of that 1/4 inch.

Now, will that 1/4 inch antenna radiate all the power that
is being successfully fed to it at 3.6MHz, or will the
configuration behave merely as an open-circuit with all
the power being reflected back down the coax?

A number of contributors to this NG claim that the 1/4 inch
stub antenna will radiate the full 100W at 3.6MHz,
snip

In this example, the transmitter delivers 100w to the coax, but only a small
part of that is delivered to the 1/4 inch "antenna".

The 1/4 inch will radiate the power delivered to it just as well as a to a
full sized antenna.

Matching and radiation are two different subjects.

Correct but it is much easier to see if one looks at it in terms of
voltage and current and calculates the power at any given point.

The equivelant circuit to a 100 W, 50 Ohms transmitter is a 141.4 V
voltage source with a 50 Ohm resistor.

The impedance of this antenna is a very small value, assume 0.001 Ohms
as a reasonable value for analysis.

There is 141.4 V applied to 50.001 Ohms, so the current is 141.4/50.001.

The power in the antenna is I^2R, which is (141.4/50.001)^2*0.001,
which my calculator says is 0.008 W.

Where is the other 99.992 W?

It is dissipated in the internal resistance of the transmitter and
given real world coax, some of it is dissipated as heat in the coax
because of the huge VSWR.


--
Jim Pennino