Gene Fuller wrote:
I am quite familiar with standing waves, thank you. I have no
disagreements with Terman, Kraus, Balanis, or any other legitimate experts.

What I am still not understanding, is since the exponential equations
for voltage and current in a transmission line are identical except
for the Z0 term, how can something happen to the current without
the same thing happening to the voltage at the same time? How can
something happen to the voltage without also happening to the current
at the same time? In a matched system, the voltage and current arrives
at the load at exactly the same time attenuated by exactly the same
amount. But that voltage didn't flow and that current didn't drop???
--
73, Cecil, W5DXP

On Thu, 11 Nov 2004 18:46:18 -0600, Cecil Moore
wrote:
the voltage and current arrives at the load at exactly the same time
Only if you skip a battery off someone's skull.

What I am still not understanding, is . . . .


To clear away all misconceptions and confusion, try returning to square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V

----
Reg

Reg Edwards wrote:
To clear away all misconceptions and confusion, try returning to square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V

Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?
--
73, Cecil, W5DXP

Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V


Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?

Depends on whether 'a' is in series or in shunt.

73, ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?

Depends on whether 'a' is in series or in shunt.

It would be too much of a coincidence for 'a', the attenuation
factor, to be the same whether in series or in shunt. Most of
the attenuation at HF is due to I^2*R losses, a series event.

Transmission lines are distributed networks involved with EM wave
energy transmission. I^2*R losses can cause a decrease in current
just as it can cause a decrease in voltage. The sequence of events
is obvious.

1. The RF voltage drops because of I^2*R losses.
2. The proportional E-field decreases because of the voltage drop.
3. Since the E-field to H-field ratio is fixed by Z0, the H-field
decreases as does the ExH power in the wave.
4. Since the RF current is proportional to the H-field, the current
decreases by the same percentage as the voltage.

The chain of cause and effect is obvious. The current decreases
because of I^2*R losses in the transmission line which is a
distributed network, not a circuit.
--
73, Cecil, W5DXP

Jim Kelley wrote:



Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V



Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?


Depends on whether 'a' is in series or in shunt.

73, ac6xg

Apparently I should have added that distributed resistive losses in
series create distributed voltage drops with a common current through
all, but distributed resistive losses in shunt create a distribution of
Norton current sources, where the shunt current on the transmission line
or radiator at a given point is the sum of all equivalent current
sources between that point and the end of the transmission line/radiator.

73, Jim AC6XG

Jim Kelley wrote:
Apparently I should have added that distributed resistive losses in
series create distributed voltage drops with a common current through
all, but distributed resistive losses in shunt create a distribution of
Norton current sources, where the shunt current on the transmission line
or radiator at a given point is the sum of all equivalent current
sources between that point and the end of the transmission line/radiator.

Distributed shunt resistive losses would imply dielectric losses
which certainly exist but are minimum at HF. We could even assume
a worst case open-wire transmission line made from resistance wire
and located in the vacuum of free space. There doesn't seem to be
any valid way to justify asserting that shunt losses exactly equal
series losses in every possible transmission line at every possible
frequency under every possible conditions. Asserting such is just
an admission that one it trying to force reality to match the math
model rather than vice versa.

In a flat transmission line without reflections, if the E-field
drops, the characteristic impedance of the transmission line
forces energy to migrate from the H-field to the E-field, such
that the constant V/I ratio remains equal to Z0. Thus, the H-field
supplies energy to compensate for the losses in the E-field.
--
73, Cecil, W5DXP

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

Cecil,

I am curious how you separate "reality" from the "math model". I have
never directly measured the properties of open-wire transmission lines
located in the vacuum of free space. Have you?

How do you know that "reality" is correct and the "math model" is wrong?

The second paragraph is even more curious. Do you have a reference for
this migration of energy from the H-field to supply the suffering
E-field? I must have missed that day in class.

73,
Gene
W4SZ

Cecil Moore wrote:


Distributed shunt resistive losses would imply dielectric losses
which certainly exist but are minimum at HF. We could even assume
a worst case open-wire transmission line made from resistance wire
and located in the vacuum of free space. There doesn't seem to be
any valid way to justify asserting that shunt losses exactly equal
series losses in every possible transmission line at every possible
frequency under every possible conditions. Asserting such is just
an admission that one it trying to force reality to match the math
model rather than vice versa.

In a flat transmission line without reflections, if the E-field
drops, the characteristic impedance of the transmission line
forces energy to migrate from the H-field to the E-field, such
that the constant V/I ratio remains equal to Z0. Thus, the H-field
supplies energy to compensate for the losses in the E-field.
--
73, Cecil, W5DXP