On Jun 18, 11:26 pm, Keith Dysart wrote:
On Jun 18, 7:04 am, K1TTT wrote:

On Jun 18, 12:00 am, Keith Dysart wrote:
except i can measure If and Ir separately and can see they are both
zero.

How would you measure them?

you said you had a directional wattmeter, so do i. mine reads zero,
what does yours read?

That would most likely be because your wattmeter is AC coupled.
Working
with DC, you need a DC-coupled meter, in which case it would read as
I predicted.

mine is dc coupled... still nothing. please provide the schematic or
model number for the one you use to get such a reading.


A highly instructive exercise is to examine the schematic for your
directional wattmeter and work out how it implements the expressions
Vf(t)=(V(t)+R0*I(t))/2
Vr(t)=(V(t)-R0*I(t))/2

Pf=avg(Vf(t)*Vf(t)/R0)
Pr=avg(Vr(t)*Vr(t)/R0)
or some equivalent variant using If and Ir.

You will find a voltage sampler, a current sampler, some scaling,
and an addition (or subtraction). If your meter is analog, the
squaring function is usually implemented in the non-linear meter
scale. And the averaging function might be peak-reading or rely
on meter inertia.

You can, though, have an arbitrary current as an intermediate
result in solving a problem. With the two battery example, reduce
the resistor to 0 and an infinitely large current is computed in
both directions, though we know in reality that no current is
actually flowing.

no it isn't. in the limit at R-0 0V/R still looks like 0 to me.

When computing the intermediate results using superposition, you
have the full battery voltage in to 0 ohms. First the battery on
the left, then the one on the right, and then you add the two
currents to get the total current. 0 is always the total, but
the intermediate results can go to infinity if you connect
the two batteries directly in parallel.

there is no battery on the right, unless you have changed the case
again. i thought we were doing the open circuit coax with the dc
source on one end.

but even if you have the case above and connect another 100v battery
on the open end of the line there will be no current flowing back as
there is no voltage difference to drive it.



I certainly agree that there is no real current flowing, but the
fictititious If and Ir are indeed present (or pick a better word
if you like).

now you say they are fictitious, yet you have formulas for them and
can measure them... until you come to your senses and admit that
reflections are real there isn't much we can do for you so the rest of
your post is based on an incorrect premise.

Then what are the equations for computing forward and reflected waves
and 'power'? That is where I started. Is that the premise you consider
to be incorrect?

yep, your 'waves' don't include the proper term to make them
propagating waves that satisfy maxwell's equations... so they can't be
real... so your initial premise is incorrect.


Do follow through the arithmetic in the previously provided examples
and see if you can find any faults. Point me to any errors and I will
gladly reconsider my position.

If not, perhaps it is time for you to start questioning your
assumptions.

...Keith

my assumptions are perfectly valid... it is yours that are flawed from
the very beginning.