Richard,

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

You can reread what I said, if you care to understand, rather than pick
a sentence out of context.

73,
Gene
W4SZ



Richard Harrison wrote:
Gene, W4SZ wrote:
"Standing waves are not static."

Incredible!

My "American College dictionary" defines "standing wave": "a
distribution of wave displacements , such that the distribution in space
is periodic, with fixed maximum and minimum points, with the maxima
occuring everywhere at the same time, as in vibration of strings,
electric potentials, acoustic pressures, etc."

Note the word "fixed" in the definition. That`s a synonym for "static".

For how this applies to antennas and transmission lines, see page 177 of
Kraus` "Antennas", third edition, Figure 6-7. Notice that current
reverses 1/2-wavelength back from the antenna`s open-circuit endjust as
it does in the case of the open-circuit transmission-line, as shown by
Terman on page 92 of "Electronic and Radio Engineering", 1955 edition,
and on page 94 in FiG. 4-5 (a). This all starts at the reflection point
and progresses the same regardless of the length of the antenna or
transmission-line. It is due to superposition of the forward and
reflected waves, just as Cecil maintains.

Advice: Never argue with Kraus and Terman.

Best regards, Richard Harrison, KB5WZI

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.

Question is: Why do you promote W8JI's stuff when it is quite obvious
that he is NOT familiar with standing waves. If he were familiar with
standing waves, he wouldn't be asserting that net standing wave current
flows into the bottom of the loading coil and out the top of the loading
coil. Are you absolutely sure that you want to promote the alleged
"information" on W8JI's web page as absolute fact? If he is so right
and so capable of defending his assertions, why isn't he here right
now? (Trying to get W8JI to follow me down the Primrose Path :-) as
he did when he asserted that "differential" effects are "completely
unrelated" to "common mode" effects.)
--
73, Cecil, W5DXP

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