Don't forget, as always happens, that when Zo is assumed to be purely
resistive, as is always done, it automatically forces wire resistance loss
to equal shunt conductance loss. You've lost a degree of freedom. Which
has a considerable bearing on your arguments.
----
Reg

Reg Edwards wrote:
Don't forget, as always happens, that when Zo is assumed to be purely
resistive, as is always done, it automatically forces wire resistance loss
to equal shunt conductance loss. You've lost a degree of freedom. Which
has a considerable bearing on your arguments.

Reg, seems I learned back in the dark ages that if Z0 is assumed to
be purely resistive, it forces wire resistance loss to equal shunt
conductance loss - AND BOTH OF THEM ARE EQUAL TO ZERO, i.e. the
line is lossless.

Actually, the argument is not about lossless lines, but about lines
with an attenuation factor term. I'm assuming that when Z0 is not
purely resistive, the ratio of voltage to current still equals a
constant Z0. If the E-field is attenuated by series I^2*R losses,
the H-field will supply energy to the E-field in an amount that will
maintain the Z0 constant ratio. If the H-field is attenuated by shunt
I^2*R losses, the E-field will supply energy to the H-field in an
amount that will maintain the Z0 constant ratio. That's why the
attenuation factors are identical - there's simply no other alternative.
--
73, Cecil, W5DXP

"Cecil Moore" wrote in message
...
Reg Edwards wrote:
Don't forget, as always happens, that when Zo is assumed to be purely
resistive, as is always done, it automatically forces wire resistance
loss
to equal shunt conductance loss. You've lost a degree of freedom.
Which
has a considerable bearing on your arguments.

Reg, seems I learned back in the dark ages that if Z0 is assumed to
be purely resistive, it forces wire resistance loss to equal shunt
conductance loss - AND BOTH OF THEM ARE EQUAL TO ZERO, i.e. the
line is lossless.

Actually, the argument is not about lossless lines, but about lines
with an attenuation factor term. I'm assuming that when Z0 is not
purely resistive, the ratio of voltage to current still equals a
constant Z0. If the E-field is attenuated by series I^2*R losses,
the H-field will supply energy to the E-field in an amount that will
maintain the Z0 constant ratio. If the H-field is attenuated by shunt
I^2*R losses, the E-field will supply energy to the H-field in an
amount that will maintain the Z0 constant ratio. That's why the
attenuation factors are identical - there's simply no other alternative.
--
73, Cecil, W5DXP

===============================

Dear Cec,

You ought to have more sense than try to argue with ME about transmission
lines. You had better return to the dark-ages before Heaviside and start
again from square one.

If I say, when Zo is made purely resistive, that Series Resistance and Shunt
Conductance losses automatically become equal to each other even when
NEITHER is zero, then I really do mean "When Zo is made purely resistive,
Series Resistance and Shunt Conductance losses automatically become equal to
each other even when NEITHER is zero."

So your argument, whatever it is, falls as flat as a pancake on Good Friday.
;o) ;o)
----
Reg

Reg Edwards wrote:
So your argument, whatever it is, falls as flat as a pancake on Good Friday.

I am assuming lossy lines, where R G, as is typical of transmission
lines used at HF frequencies. So exactly where does my argument fall flat?

Actually, it seems that it is your argument that is falling flat as R is
rarely, if ever, equal to G at HF frequencies. Where would we ever obtain
such a terrible dielectric at HF that G would be equal to R? Maybe 9913
filled with water?
--
73, Cecil, W5DXP

In article , Cecil Moore
wrote:

I am assuming lossy lines, where R G, as is typical of transmission
lines used at HF frequencies. So exactly where does my argument fall flat?

If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

David

--
David Ryeburn

To send e-mail, use "ca" instead of "caz".

David Ryeburn wrote:
If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

Sorry, I should have said (effect of R) (effect of G)
on the losses. Or 0.1 ohms/meter in series has a greater
effect than 1/(0.9 micromhos/meter) in parallel. That's
about 7 magnitudes difference based on an example in
"Transmission Lines" by Chipman.
--
73, Cecil, W5DXP

David Ryeburn wrote:

Cecil Moore wrote:
I am assuming lossy lines, where R G, as is typical of transmission
lines used at HF frequencies. So exactly where does my argument fall flat?

If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

Sorry, should have said, "... where R/Z0 G*Z0, as is typical of
transmission lines used at HF frequencies."
--
73, Cecil, W5DXP

On Fri, 19 Nov 2004 23:48:22 -0600, Cecil Moore
wrote:
I am assuming lossy lines, where R G, as is typical of transmission
how can you say one of them is much larger than the other?
Sorry, should have said, "... where R/Z0 G*Z0
just a matter of Z0² ... no big deal
perhaps it shoulda been R· Z0² G ;-)

If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

============================

Its easy for Cecil. He doesn't have the slightest trouble.

Reg Edwards wrote:
If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

Its easy for Cecil. He doesn't have the slightest trouble.

Just forgot to render them both dimensionless with the Z0
term. I suspect you knew what I meant anyway. :-)
--
73, Cecil, W5DXP