Keith Dysart wrote:
In a stub driven with a step function, where is the
energy stored?

Depends upon which valid model one is using.

1. Reflection Model - the energy is stored in the
forward and reflected traveling waves.

2. The LCLCLC transmission line model - the energy
is alternately stored in the L's and C's.

3. The Sloshing Model - I'll let Roy handle that one.
--
73, Cecil http://www.w5dxp.com

On Wed, 02 Jan 2008 14:42:01 -0600, Cecil Moore
wrote:

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?

Can't get past the trauma of your sunburn, can you? :-)

What do you see when you look in a conjugate mirror? (have to provide
two questions in the hope you can score 50% - MENSA standards).

Roy Lewallen wrote:
"Where did the power go?"

Or more correctly, where did the energy go?
Was it destroyed or created? (Rhetorical)
--
73, Cecil http://www.w5dxp.com

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

Ideal traveling-wave antennas have no standing waves.
You are simply pointing out the obvious difference
between the ideal and the real world. So what new?
--
73, Cecil http://www.w5dxp.com

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

The answer is typically asserted by someone who doesn't
know the difference between joules and joules per second.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
For future reference, however, just remember: Fields first, then power
or energy. That's the way superposition really works.

Way back before optical physicists could measure light
wave fields, they were dealing with reflectance,
transmittance, and irradiance - all involving power
or energy. They are still using those concepts today
proven valid over the past centuries. Optical physicists
calculate the fields *AFTER* measuring the power density
and they get correct consistent answers.

"Way back" is irrelevant. One only needs to open a serious text book on
Optics, such as Born and Wolf, to see how optical physicists perform
analysis today.


Quoting HP AN 95-1: "The previous four equations show that
s-parameters are simply related to power gain and mismatch
loss, quantities which are often of more interest than the
corresponding voltage functions."

I agree with this statement completely (surprised??). S-parameter
analysis is very useful. However, the "corresponding voltage functions"
are equally valid, even if not as "interesting". What you might also
notice in AN 95-1 is that there is no mention of incident and reflected
waves on a transmission line, each carrying energy (or power or whatever
you prefer), and passing like ships in the night.

You like to talk about conservation of energy, implying that your
"powerful" reflected wave model is essential to meeting the conservation
of energy requirement. In fact, your model is a poster child for the
violation of energy conservation. Electromagnetic energy, like any
energy, is a scalar quantity, and it is only positive. It is not
possible to "net" the non-zero energy contributed from your
counter-traveling waves to zero. The direction of the wave propagation
does not change the sign of the energy. Be careful here; energy is *not*
the same as the energy flux or Poynting vector. Don't mix terms that
have totally different units. What *can* be assigned negative values are
the fields. (Voltage and current are not exactly "fields", but they will
work for these transmission line examples.) A "net" of zero volts or
current is exactly what happens at the standing wave nodes resulting
from the counter-traveling waves. After you have done the superposition
correctly, using fields, not energy or power, then you can easily
determine the energy and power state as needed. Conservation of energy
will be automatically satisfied, assuming no mathematical blunders. The
Maxwell equations would be pretty useless if they did not provide
conservation of energy.

For future reference, just remember: Fields first, then power
or energy. That's the way superposition really works.

73,
Gene
W4SZ

Keith Dysart wrote:

On Dec 29, 2:31 pm, Cecil Moore wrote:

Roger wrote:

Are there reflections at point "+"? Traveling waves going in opposite
directions must pass here, therefore they must either pass through one
another, or reflect off one another.

In the absence of a real physical impedance discontinuity,
they cannot "reflect off one another". In a constant Z0 transmission
line, reflections can only occur at the ends of the line and only
then at an impedance discontinuity.


Roger: an astute observation. And Cecil thinks he has
the ONLY answer. Allow me to provide an alternative.

Many years ago, when I first encountered this news group
and started really learning about transmission lines, I
found it useful to consider not only sinusoidallly
excited transmission lines, but also pulse excitation.
It sometimes helps remove some of the confusion and
clarify the thinking. So for this example, I will use
pulses.

Consider a 50 ohm transmission line that is 4 seconds
long with a pulse generator at one end and a 50 ohm
resistor at the other.

The pulse generator generates a single 1 second pulse
of 50 volts into the line. Before and after the pulse
its output voltage is 0. While generating the pulse,
1 amp (1 coulomb/s) is being put into the line, so
the generator is providing 50 watts to the line.

After one second the pulse is completely in the line.
The pulse is one second long, contains 1 coulomb of
charge and 50 joules of energy. It is 50 volts with
1 amp: 50 watts.

Let's examine the midpoint (2 second) on the line.
At two seconds the leading edge of the pulse arrives
at the midpoint. The voltage rises to 50 volts and
the current becomes 1 amp. One second later, the
voltage drops back to 0, as does the current. The
charge and the energy have completely passed the
midpoint.

When the pulse reaches the end of the line, 50
joules are dissipated in the terminating resistor.

Notice a key point about this description. It is
completely in terms of charge. There is not a single
mention of EM waves, travelling or otherwise.

Now we expand the experiment by placing a pulse
generator at each end of the line and triggering
them to each generate a 50V one second pulse at
the same time. So after one second a pulse has
completely entered each end of the line and these
pulse are racing towards each other at the speed
of light (in the line). In another second these
pulses will collide at the middle of the line.

What will happen? Recall one of the basics about
charge: like charge repel. So it is no surprise
that these two pulses of charge bounce off each
and head back from where they came. At the center
of the line, for one second the voltage is 100 V
(50 V from each pulse), while the current is
always zero. No charge crossed the mid-point. No
energy crossed the mid-point (how could it if
the current is always zero (i.e. no charge
moves) at the mid-point.

It is a minor extension to have this model deal
with sinusoidal excitation.

What happens when these pulses arrive back at the
generator? This depends on generator output
impedance. If it is 50 ohms (i.e. equal to Z0),
then there is no reflection and 1 joule is
dissipated in each generator. Other values
of impedance result in more complicated
behaviour.

So do the travelling waves "reflect" off each
other? Save the term "reflect" for those cases
where there is an impedance discontinuity and
use "bounce" for those cases where no energy
is crossing a point and even Cecil may be
happy. But bounce it does.

...Keith

It's fairly safe to make this argument when both pulses are identical.
I challenge you to obtain this result when they are not. :-)

73, Jim AC6XG

Keith Dysart wrote:
. . .
But do not expect the power dissipated in the resistor
to increase by the same amount as the "reflected power".
In general, it will not. This is what calls into question
whether the reflected wave actually contains energy.

Do some simple examples with step functions. The math
is simpler than with sinusoids and the results do not
depend on the phase of the returning wave, but simply
on when the reflected step arrives bach at the source.

Examine the system with the following terminations on
the line: open, shorted, impedance greater than Z0,
and impedance less than Z0.

Because excitation with a step function settles to
the DC values, the final steady state condition is
easy to compute. Just ignore the transmission line
and assume the termination is connected directly
to the Thevenin generator. When the line is present,
it takes longer to settle, but the final state will
be the same with the line having a constant voltage
equal to the voltage output of the generator which
will be the same as the voltage applied to the load.

Then do the same again, but use a Norton source. You
will find that conditions which increase the dissipation
in the resistor of the Thevenin equivalent circuit
reduce the dissipation in the resistor of the Norton
equivalent circuit and vice versa.

This again calls into question the concept of power
in a reflected wave, since there is no accounting
for where that "power" goes.

I heartily second Keith's recommendations.

For some simple illustrations of one problem with chasing "power waves"
around, see http://eznec.com/misc/Food_for_thought.pdf, particularly the
"Forward and Reverse Power" section beginning on p. 6 and the table on
p. 8. This was originally written and posted more than five years ago
and, to my knowledge, the problems it raises with the concept of "power
waves" still haven't been addressed in the thousands of postings on the
topic in the intervening time.

Roy Lewallen, W7EL

Keith Dysart wrote:
On Jan 2, 9:59 am, Cecil Moore wrote:


Please reference a good book on optical EM waves
for a complete answer.



It is a body of physics knowledge that has existed
since long before you were born. It should have
been covered in your Physics 201 class. That you
are apparently unaware of such is a display of
basic ignorance of the science of EM waves.



The basic theory applies specifically to coherent
waves (which are the only EM waves capable of truly
interfering). CW RF waves are close enough to ideal
coherency that the theory works well. It would no
doubt work for a coherent Fourier series as well
but I don't want to spend the time necessary
to prove that assertion.



Again, it is not *my* approach and is described in any
textbook on "Optics" including Hecht and Born & Wolf.

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

In any case, being applicable only to sinusoids
limits the general applicability to transmission
lines which happily work at DC.

...Keith


It is sadly amusing that Cecil takes so much comfort in optics. The
electromagnetic theory for optics (e.g. somewhere in the vicinity of
visible light) is of course identical to the electromagnetic theory for
HF. The preferred applications and shortcuts are sometimes a bit
different, but that is simply a matter of convenience and of no
importance here.

I have a couple of editions of Born and Wolf, which is a high level
reference and often considered the standard for optics. I have been
unable to find even one mention of "constructive" or "destructive"
interference in their writing. Of course they delve into the topic of
interference in excruciating detail. They don't, however, ascribe any
particular mysticism or magic to interference. It is simply what happens
when the wave fields are superposed.

The more popular accounts, such as the FSU Java applet on interference,
the Melles-Griot web site, and apparently the text by Hecht, stay a bit
further from rigorous analysis. Therefore they resort to handwaving
requirements such as destructive must be balanced by constructive, blah,
blah, blah.

Adding the voltages in the manner you and Roy have done is precisely the
same operation as Cecil's interference method, without the emotional
baggage.

73,
Gene
W4SZ

On Wed, 02 Jan 2008 21:19:53 GMT, Cecil Moore
wrote:
For extra credit, try explaining why a traveling wave antenna has
standing waves on it! :-0

Ideal traveling-wave antennas have no standing waves.

Ideal? You mean your dream of desires. Schoolgirl stuff for diaries
under pillows, not technical discussion (not that I'm surprised, I
actually enjoy your tendency toward heart-throb writing).

You are simply pointing out the obvious difference
between the ideal and the real world. So what new?

What new? What new shows us you don't get any extra credit, that what
new! You can try again, if you wish. That is, if you can resuscitate
the corpse you strangled when your "purpose" went off the rails.

+ +
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