In article
,
RHF wrote:

On Dec 29, 12:37*pm, Telamon
wrote:
In article
,



*RHF wrote:
On Dec 28, 8:36*pm, Telamon
wrote:
In article ,
*John Smith wrote:

SNIP

I never even commented on where the placement of the matchbox
would be, and, as everyone knows, anywhere along the line you
can place it. *The best place would be between the coax
(feedline) and the antenna-

Right. That's because you are to stupid to understand a concept
until someone rubes your nose in it. This would not even occur
to you until someone else brought it up.

-that is, taking for granted that the match from your rig to
the feedline is perfect.

SNIP

You are really worried about the match of 50 ohm *coax to your
radios 50 ohm output? Now that's funny.

IIRC - The Characteristically 50 Ohm Impedance Coax Cable is
'only' 50 Ohms nominal . . . Until you attach something to it.

SNIP

Nope. The cable itself has a characteristic impedance of some
design value. The spacing and size of the conductors along with the
dielectric constant of the insulator between them dictates the
impedance of the coax.

You are confusing the characteristic impedance of the coax with its
ability to be an effective transmission line. The coax only behaves
as an effective transmission line when both ends of it are
terminated at its characteristic impedance.


Telamon,

OK -restatement- The "Measured" {by You} 50 Ohm Impedance Coax Cable
is 'only' 50 Ohms nominal* * Until you attach something to it.

-IF- You then attach a Transmitters Output that is a Nominal 50 Ohms
to one end of the Coax Cable and then the 'other' end will still
"Measure*" about 50 Ohms. * This is what the Antenna will see.

However -if- You attach an Unknown "Z" Antenna and Ground to one end
of the Coax Cable; then the 'other' end may "Measure*" near or far
from 50 Ohms. * This is what the Transmitter will see.

Unknown "Z" Antenna = Random Wire Antenna

as always . . . i may be 'w-r-o-n-g' - iane ~ RHF

You are just confusing a few things. You need to understand that at RF
all parts of a circuit are not "seen" by the RF energy "at the same
time." The energy has to propagate through the circuit. This is
different from DC where the whole circuit "is seen" by the energy
source at once. I'm sure that at DC you are familiar with adding up
resistor networks or loads into a total load resistance where you can
figure out what the total current would be if you applied a certain
voltage. This is also know as a lump sum circuit.

At RF since it takes time for the energy to propagate through parts of
the circuit so they are not seen at the same time and you have to use
vector math that has magnitude and phase components, instead of just
magnitude, to describe the current that results from an applied
voltage. Now with this vector math representing the circuit impedance
as opposed to just magnitude resistance you can make transformations
similar to an equivalent DC total load resistance for RF current
calculations but you have to keep in mind that these are time or phase
dependent. A complication of this is some of the energy can even go
backward depending on the circuit so these time dependent voltages and
currents need to be summed as vectors with magnitude and phase, which
represent a voltage or current at a spot in the circuit.

The practical upshot of this is that RF paths like coax have to be
viewed as transmission lines where the RF energy only "sees" a part of
the coax at any one moment in time and after a short period of time the
energy "sees" the next part of the coax and so on until it reaches the
end. With this understanding it is the "environment" of the coax that
causes it to represent an "impedance" or how it reacts to the RF energy
as a complex resistance to its flow. This reactive environment is
created by the size, spacing, and DC resistance of the two conductors
along with the dielectric value of the insulator between them.

Now with that said you can still make the R total type of RF impedance
calculation with a resistor or a reactive load on the far end of a coax
cable as a total transformed circuit impedance value but it does not
change in any way the intrinsic impedance of the coax itself, which is
fully dependent on its physical construction.

Now, if you are still with me, the coax will only function properly as a
transmission line when the source impedance at one end and the load
impedance at the other end are the same value. In this state all the RF
energy is internal to the coax and if the load or source impedance is
not the same currents start to flow on the outside of the coax so it
will not shield properly. All the energy will not even enter the coax at
the source end and energy will be reflected at the load end creating
standing waves of energy on the coax, which is the sum of the forward
and reverse waves at any one point. Where these waves sum the voltages
can be many times higher than the source causing excessive heating or
even breakdown of the dielectric at that point or arcing at the
connectors. Under these conditions the coax specifications will not be
met. The coax will not meet its isolation, insertion loss, or VSWR
specifications even though there is nothing wrong with it.

--
Telamon
Ventura, California