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