On Wed, 28 Feb 2007 20:35:27 GMT, Owen Duffy wrote:

Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

The questions a

Is there any internal inconsistency in the scenario characterisation, if
so, identify / explain?

Hi Owen,

After a fashion, yes. You don't have enough information. The lesson
of past correspondence reveals the source resistance is not a benign
element that can be discarded in the examination of these problems.

Explain? I thought I had by example, so to continue by simple
analysis we have to ask: "What is the source R?"

If it is 70 Ohms, it sees a 70 Ohm load and an Impedance Match. In
effect there is no other way for a transmitter that is capable of 100W
being able to source 100W to a load, is there?

If it is 50 Ohms, it immediately sees a mismatch at the connector -
you did do your lumped equivalent of the transmission line system,
didn't you? Hence a transmitter limited to supplying 100W could not
supply 100W to a mismatched load.

If it is 600 Ohms, we again see a massive mismatch, further discussion
is unnecessary as being repetitive.

What is the heat dissipated in the transmitter (and why)?

I will skip that exercise as obvious pending discussion below.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

Presuming the 70 Ohm source as it is the only one capable of
supporting the 100W application of power to the 70 Ohm load (by your
definition and limitations) we shall proceed to the determination of
those pesky powers with absolute finality.

The load voltage is 84 volts (to sufficient accuracy). The voltage at
the source port is 84 volts (to sufficient accuracy). The line has
been defined to be lossless. The same potential at the same angle is
returned to the source through 360 degrees of travel and no loss
(analysis allows us to treat the load as a generator). That signal
applied to the transmitter finds there is no potential difference, and
thus there is no current flow. This variously describes either a
short or an open; and either way it represents an infinite mismatch -
ALL the reflected power is re-reflected to the load.

Voila, another statistical curiosity!

Tuners do this more generally and we use tuners to re-reflect the
power, and to not suffer the transmitter the additional heat burden of
reflected power arriving at the finals.

If we repeat this exercise with the other source resistances, we stand
to appreciate a paradox in the initial condition of:
the DC input power to the transmitter is 200W.
forcing us to resolve the additional heat burden (roughly 50W)
exceeding the reflected power's ability to contribute that much
(especially when all the reflected power is re-reflected); or worse,
the source going into clipping. A non-linear source is not a pleasant
thing to behold.

73's
Richard Clark, KB7QHC