Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

A wave is defined as a progressive vibrational disturbance propagated
through a medium, such as air, without progress or advance of the parts
or particles themselves, as in the transmission of sound, light, and an
electromagnetic field. Light, for example, is also calld luminous or
radiant energy. Sound and radio waves are also examples of energy in
motion.

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

Best regards, Richard Harrison, KB5WZI

On Sun, 23 Dec 2007 13:34:49 -0600, (Richard Harrison) wrote:

Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

A wave is defined as a progressive vibrational disturbance propagated
through a medium, such as air, without progress or advance of the parts
or particles themselves, as in the transmission of sound, light, and an
electromagnetic field. Light, for example, is also calld luminous or
radiant energy. Sound and radio waves are also examples of energy in
motion.

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

Best regards, Richard Harrison, KB5WZI

It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

In addition to another comment above that implies that reflected power is reactive power, this is not
true--reflected power is as real as forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in phase in the forward direction, they are
traveling 180° out of phase in the rearward direction. Multiplying voltage and current while 180° different in
phase results in the same power as when they are in phase.

Walt, W2DU

Hi Walt,

I'm a little confused here. I hope you can straighten me out.

Walter Maxwell wrote:

It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves. But you're saying there are currents and voltages "in"
the standing wave. Are you referring to the total current and voltage at
any point along the line? If so, why are they always in quadrature?
Certainly, the total V and I are in quadrature if the line is terminated
by an open, short, or purely reactive load. But not in any other case.
Or do you regard a line as having a "standing wave" with its own voltage
and current which are different from the total V and I? If so, how do
you define a "standing wave"? Are there separate equations for "standing
wave" V and I that are different than for total V and I?

. . .

Roy Lewallen, W7EL

Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves.

If you have "Fields and Waves ..." by Ramo and Whinnery,
take a look at the equation for standing wave voltage
and current on page 285 of the 2nd edition.

Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz)

Roy, that is NOT the equation for an envelope.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves.

If you have "Fields and Waves ..." by Ramo and Whinnery,
take a look at the equation for standing wave voltage
and current on page 285 of the 2nd edition.

Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz)

Roy, that is NOT the equation for an envelope.

It's too bad you don't have the foggiest notion as to just
what it is the equation of. Much of the thousands of posts
could have been avoided if that were the case.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Cecil Moore wrote:
Roy Lewallen wrote:
I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves.

If you have "Fields and Waves ..." by Ramo and Whinnery,
take a look at the equation for standing wave voltage
and current on page 285 of the 2nd edition.

Ex = E*e^j(wt-Bz) + E'*e^j(wt+Bz)

Roy, that is NOT the equation for an envelope.

It's too bad you don't have the foggiest notion as to just
what it is the equation of. Much of the thousands of posts
could have been avoided if that were the case.

The technical content of your posting is noted. Here is what
it is the equation of:

http://www.chemmybear.com/standing.html
--
73, Cecil http://www.w5dxp.com

Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection. Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay. Terman shows the vector diagrams of
incident and reflected waves combined to produce a voltage distribution
on an almost lossless transmission line (Zo=R) for an open circuit case
and for a resistive load case where the load is Zo in Fig. 4-3 on
page 91 of his 1955 opus. Indeed, the angle between the incident and
reflected voltages is 90-degrees in either case.

In Fig. 4-4 on page 92, Terman shows voltage and current distributions
produced on low-loss transmission lines by different load impedances and
in every case volts and amps are in quadrature.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection.

Yes.

Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay.

?? Which waves? Forward voltage and reverse current? Forward and reverse
voltage?

Terman shows the vector diagrams of
incident and reflected waves combined to produce a voltage distribution
on an almost lossless transmission line (Zo=R) for an open circuit case
and for a resistive load case where the load is Zo in Fig. 4-3 on
page 91 of his 1955 opus. Indeed, the angle between the incident and
reflected voltages is 90-degrees in either case.

Please look carefully at those diagrams. The horizontal axis is the
distance along the line. The diagrams are showing the relationship
between voltage and current envelopes as a function of position. The
graphs aren't showing the time phase of V and I, which is the matter
under discussion.

In Fig. 4-4 on page 92, Terman shows voltage and current distributions
produced on low-loss transmission lines by different load impedances and
in every case volts and amps are in quadrature.

I don't have that diagram in my 1947 Third Edition, but I'm sure that if
you'll study the diagram and accompanying text you'll find that it's
also a graph of peak amplitude vs position, not V or I as a function of
time.

Let me pose a very simple problem. Suppose you have a quarter wavelength
of 50 ohm transmission line terminated in 25 ohms and driven with a 100
volt RMS sine wave source. Consider the phase of the voltage source to
be the reference of zero phase angle.

1. What is the current at the line input? [Answer: 1 ampere, at 0 degree
phase]
2. What is the ratio of V to I at the line input? [Answer: 100 at an
angle of zero divided by 1 at an angle of zero = 100 + j0 ohms]
3. How do you resolve this with the graphs in Terman and your
explanation of the voltage and current being in quadrature everywhere
along the line?

Feel free to repeat this with any other line length. I'll wait very
patiently for the length which produces V and I in quadrature at the
input. A very interesting result of that would be that no power would be
consumed from the source, so if any reaches the load then we've created
power.

I'd be glad to post the equation relating Z (the ratio of V to I) at the
line input or any point along the line to the load and characteristic
impedances. But I'm afraid it would be wasted effort, since there's a
great reluctance here to actually work an equation or understand its
meaning. But good and accurate graphs of what Richard has claimed the
Terman graphs show (but don't) can be found in the _ARRL Antenna Book_.
In the 20th and 21st Editions, the graphs are Fig. 12 on p. 24-9. In
other editions, they're probably also Fig. 12 in the Transmission Lines
chapter. In the graphs, the angle between the I and E vectors is the
relative phase angle between the two, and also the angle of the
impedance of the point where the vectors are shown.

Roy Lewallen, W7EL

Roy Lewallen wrote:
3. How do you resolve this with the graphs in Terman and your
explanation of the voltage and current being in quadrature everywhere
along the line?

The standing wave voltage is *ALWAYS* in quadrature with
the standing wave current at all points and at all times.
If the total voltage is not in quadrature with the total
current, there is a traveling wave in the mixture.
--
73, Cecil http://www.w5dxp.com

Richard Harrison wrote:
Roy Lewallen, W7EL wrote:
"Certainly, the total V and I are in quadrature if the line is
terminated by an open, short, or purely reactive load. But not in any
other case."

Something else is at work. The reflection reverses direction of the wave
producing a 180-degree phase shift in either voltage or current, but not
both, if there is a reflection. Because the waves are traveling at the
sane speed in approaching each other, they produce a phase reversal in a
distance of only 90-degrees instead of 180-degrees. This places the
waves in quadrature to stay.

Seems you two are arguing about two different things.

If Z0 is purely resistive: Pure standing waves are *ALWAYS*
in quadrature, i.e. the sine of the angle between V and
I is always 1.0. Pure traveling waves are are *ALWAYS* in
phase or 180 degrees out of phase, i.e. the cosine of the
angle between V and I is always 1.0.

In a mixed environment of standing waves and traveling
waves, the angle between V and I can assume any value.
--
73, Cecil http://www.w5dxp.com