Roger wrote:

In the last 24 hours, Roy posted a revised analysis that contains
results useful here. He presented a voltage example that resulted in a
steady state with steady state voltages 4 time initial value. Under
superposition, this would equate to 4 times the initial power residing
on the transmission line under the conditions presented. This concurs
with other authors who predict power on the transmission line may exceed
the delivered power due to reflected waves.

Your equation "Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)" is taken from
illumination and radiation theory to describe power existing at a point
in space near a reflecting surface. If we consider space to be a
transmission media, and the reflecting surface to be a discontinuity in
the transmission media, then we have a situation very similar to an
electrical transmission line near a line discontinuity.

It is entirely reasonable to consider that the reflection ratio between
the space transmission media and the reflective surface would result in
an storage factor equaling 4 times the peak power of the initial forward
wave. By storage factor, I simply mean the ratio of forward power to
total power on the transmission media under standing wave conditions.

Under open circuit conditions, a half wavelength transmission line will
have a storage factor of 2. Roy presented an example where the storage
factor was 4. Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

Power is neither stored nor conserved, so a power "storage factor" is
meaningless. Consider a very simple example. Let's charge a capacitor
with a constant current DC source. We'll apply 1 amp to a 1 farad
capacitor for 1 second. During that time, the power begins at zero,
since the capacitor voltage is zero, then it rises linearly to one watt
as the capacitor voltage rises to one volt at the end of the one second
period. So the average power over that period was 1/2 watt, and we put
1/2 joule of energy into the capacitor. (To confirm, the energy in a
capacitor is 1/2 * C * V^2 = 1/2 joule.) Was power "put into" or stored
in the capacitor?

Now we'll connect a 0.1 amp constant current load to the capacitor, in a
direction that discharges it. We can use an ideal current source for
this. The power measured at the capacitor or source terminals begins at
0.1 watt and drops linearly to zero as the capacitor discharges. The
average is 0.05 watt. Why are we getting less power out than we put in?

"Where did the power go?" is heard over and over, and let me assure you,
anyone taking care with his mathematics and logic is going to spend a
long time looking for it. So in this capacitor problem, where did the
power go?

It takes 10 seconds to discharge the capacitor, during which the load
receives the 1/2 joule of energy stored in the capacitor. Energy was
stored. Energy was conserved. Power was neither stored nor conserved.

Roy Lewallen, W7EL

Keith Dysart wrote:
You should really spend some time looking for a
reference to support your assertion that "It will
not be the impedance needed to calculate the
reflection coefficient seen by the reflected
waves."

You will not find one.

You have got to be kidding, Keith. Even some of the
people on your side will admit that the effective
reflection coefficient for a source supplying zero
power is |1.0| nowhere near the value you calculated.
I believe that is what Roy said.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
I assume that you have not provided a reference to support
this assertion because you have not been able to find one.

I provided the reference a number of times and you
chose to ignore it. The reference is the chapter on
interference in "Optics", by Hecht.

Unfortunately optics do not do well at explaining
transmission lines since they do not extend down
to DC.

On the contrary, most physics books on "Light" do
indeed extend down to DC. I'm surprised you don't
know that. "Light" and "visible light" are two
entirely different subjects. "Light" covers all
EM waves all the way down to DC.

I have yet to find anything about transmission lines
that needs constructive and destructive interference
for explanation.

Well, there's your entire problem in a nutshell. If
you don't ever look for something because you don't
"feel the need", you will never find it. Please don't
blame anyone else for your feelings. And you are not
alone. My mother never "felt the need" to understand
anything except God.

I am unsure why some are content to constrain
themselves to solution techniques and explanations
that only work on the special case of sinusoids.

Sinusoids are a test of your comprehension level.
Because we know if you cannot even comprehend the
most simple case, you have no hope of comprehending
anything more complicated. Don't feel bad. My
girlfriend cannot comprehend sinusoids either.
--
73, Cecil http://www.w5dxp.com

On Wed, 02 Jan 2008 13:22:42 -0600, Cecil Moore
wrote:

Keith Dysart wrote:
Perhaps you could make an attempt at writing a clear
description of the behaviour of such a system in terms
of charge flow and storage.

Since you are unable to understand the more simple example
using only one sine wave, what makes you think you are
capable of understanding the more complex step function?

For extra credit, try explaining why a traveling wave antenna has
standing waves on it! :-0

Keith Dysart wrote:

On Dec 27, 10:53 am, Cecil Moore wrote:

But I don't
comprehend the utility of the following:

The instantaneous value of voltage is 10 volts.
The instantaneous value of current is 1 amp.
The voltage and current are in phase.

The instantaneous power is 10 joules per 0 sec?


There is definitely a problem with that.

There is indeed a problem. The problem is that one amp does not equal
one coulomb per 0 sec.

73, ac6xg

People who are having trouble with the concept of a -1 voltage
reflection coefficient for a perfect voltage source might benefit from
the following exercise:

Look at my first analysis, where the perfect source was connected
directly to the transmission line. Make no assumptions about the
impedance or reflection coefficient it presents. Then, when the
reflection of the initial forward wave returns, calculate the value the
re-reflected wave must have in order to make the sum of the waves
present, which is the line input voltage, equal to the perfect source
voltage. The voltage of the perfect source can't change, by definition.
The ratio of the re-reflected wave to the returning wave is the voltage
reflection coefficient (since we're dealing with voltage waves).

I'll do it for you:

The forward wave was vf(t, x) = sin(wt - x)
The returning wave was vr(t, x) = sin(wt + x)

The returning wave will strike the input end of the line and create a
new forward wave with value vf2(t, 0) = Gs * vr(t, 0) at the input,
where Gs is the source voltage reflection coefficient.

Before the first forward wave returns, we have only vf(t, 0) = sin(wt)
at the input end of the line. This is of course the source voltage.

After the wave arrives and re-reflects, we have at the input end

vtot(t, 0) = vf(t, 0) + vr(t, 0) + vf2(t, 0)
= vf(t, 0) + vr(t, 0) * (1 + Gs)

This must equal the source voltage, which is the line input voltage, and
cannot change. So plugging in values:

sin(wt) = sin(wt) + sin(wt) * (1 + Gs)

Solving for Gs = Gs = -1

I have made no statement about the "impedance" of the perfect source.
The only thing I've required is that the voltage remains constant, which
is the very definition of a perfect source. You can do a similar
exercise to show that the voltage reflection coefficient of a perfect
current source is +1.

Roy Lewallen, W7EL

Keith Dysart wrote:
Cecil Moore rote:
But why don't you just say so clearly.

I explained why it only applies to coherent
waves. Every model has its limitations, even
yours unless you are presenting a theory of
everything.

Given that optical EM waves are only capable of
solving a subset of the uses of transmission lines,
it is not obvious why I should study them when
I can invest in learning approaches that will do
the whole job.

That's not a given and is in fact a falsehood.
RF waves are a subset of light waves. Perhaps
you are erroneously confusing "light waves" with
"visible light waves".

When a light wave is red-shifted to 10^12 times
its original wavelength, does it cease to be a
light wave? Feel free to answer yes or no.

Well, others more knowledgeable than I in optics
have disputed whether *your* approach accurately
represents those described in the textbooks.

The last resort is an argumentium ad verecundiam,
an appeal to reverence/authority. Who, in particular,
do you consider to be the High Priest of r.r.a.a?

The technical truth will win out in the long run.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Keith Dysart wrote:

Cecil Moore wrote:

The instantaneous power is 10 joules per 0 sec?


There is definitely a problem with that.

But an instantaneous value of 10 joules/sec; that
is useful.


But that instantaneous instant is NOT one second
long.

A rate of 10 joules/sec is valid at any time in which there is a 10
volt differential and a current flowing at a rate of one coulomb per
second (and a half coulomb in a half second, and a thousandth of
coulomb in a millisecond, and like that).

Exactly how long is that instantaneous
instant? 1 ms? 1 us? 1 ns? more? less?

That is a question more typically asked by someone who has never taken
a calculus class.

73, ac6xg

Roger wrote:
Perhaps the space transmission media also has a storage
factor of 4 under some conditions, as described by "Ptot = Ps + Pr +
2*SQRT(Ps*Pr)cos(A)".

Visualize visible interference rings between two equal
waves each of P magnitude, where the darkest of the rings
is completely black. If it is not obvious, the power in
the brightest of the rings would have to be 4P for the
energy to average out to the 2P in the source waves. That's
all there is to "black" destructive interference vs "bright"
constructive interference. (4P+0P)/2 = 2P = average power
It's a no-brainer for most folks.

On the source side of a Z0-match, the reflections are "black".
On the load side of a Z0-match with reflections, the forward
wave is "bright". What could be simpler?
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
I committed several hundred pages to fractals in the past, ...

Richard, why didn't you commit several hundred pages
to your premise that reflections from non-reflective
glass are brighter than the surface of the sun?
--
73, Cecil http://www.w5dxp.com