Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
With two measurements you will obtain the correct
impedance.

With two measurements you will obtain an impedance.
It will not be the impedance needed to calculate
the reflection coefficient seen by the reflected
waves.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 9:24*am, Cecil Moore wrote:
Keith Dysart wrote:
With two measurements you will obtain the correct
impedance.

With two measurements you will obtain an impedance.
It will not be the impedance needed to calculate
the reflection coefficient seen by the reflected
waves.

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.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.

However, after the interference patterns have been
established, the reflected waves do not encounter that
source impedance. That is why the reflection coefficient
seen by the reflected waves is relatively unrelated to
the value of resistance in a Thevenin equivalent circuit.

You need to complete step 3 of the superposition process
to realize exactly what is happening. Reference the
irradiance equation from the field of EM wave optics
to ascertain the interference levels. What do you have
to lose by alleviating your ignorance?

Keith, until you take time to understand destructive and
constructive interference, you will never understand what
is happening inside a source and will be forever confused
by your blinders-on-come-hell-or-high-water method of
thinking. Optical physicists figured out a couple of
centuries ago exactly what you are wrestling with now.
Your present problem was already solved before your
grandfather was born.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Can you make this all work for a pulse, or a step
function?

Please reference a good book on optical EM waves
for a complete answer. It is *not me* making it work.
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.

How do you compute
Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
for a pulse or a step?

The above equation applies to coherent signals.
It is known not to work when the signals are not
coherent because the angle 'A' never reaches
a fixed steady-state value.

Or is your approach limited to sinusoids?

Again, it is not *my* approach and is described in any
textbook on "Optics" including Hecht and Born & Wolf.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 9:38*am, Cecil Moore wrote:
Keith Dysart wrote:
The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.

However, after the interference patterns have been
established, the reflected waves do not encounter that
source impedance. That is why the reflection coefficient
seen by the reflected waves is relatively unrelated to
the value of resistance in a Thevenin equivalent circuit.

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

You need to complete step 3 of the superposition process
to realize exactly what is happening. Reference the
irradiance equation from the field of EM wave optics
to ascertain the interference levels. What do you have
to lose by alleviating your ignorance?

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

Keith, until you take time to understand destructive and
constructive interference, you will never understand what
is happening inside a source and will be forever confused
by your blinders-on-come-hell-or-high-water method of
thinking. Optical physicists figured out a couple of
centuries ago exactly what you are wrestling with now.
Your present problem was already solved before your
grandfather was born.

I have yet to find anything about transmission lines
that needs constructive and destructive interference
for explanation. Volts, amps and superposition seem
to be able to do it all, and have the added benefit
of explaining the behaviour for step functions and
pulses. With the volts, amps and superposition,
sinusoids are just a special case of the general
one.

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

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Jan 2, 9:59*am, Cecil Moore wrote:
Keith Dysart wrote:
Can you make this all work for a pulse, or a step
function?

I accept your "NO" and agree that EM waves are
incapable of providing solutions for pulse or
step excitation.

But why don't you just say so clearly.

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

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.

It is *not me* making it work.

True. And as you have said, it does not work
for pulses or steps.

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.

Some who claim to have studied them thoroughly
seem to be constrained by their limitations. Is
that better?

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.

How do you compute
* Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
for a pulse or a step?

The above equation applies to coherent signals.
It is known not to work when the signals are not
coherent because the angle 'A' never reaches
a fixed steady-state value.

Or is your approach limited to sinusoids?

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

Standing-Wave Current vs Traveling-Wave Current
 

Roger Sparks wrote:
Well, I would like to think I could understand it, but maybe something is
wrong like Roy suggests. I think it is all falling into place, but our
readers are not all in agreement.

Anyone who will take the time to understand and is capable
of understanding, will understand. What I am reporting are
centuries-old proven scientific concepts. Readers need only
reference an optics textbook for irradiance, reflectance,
and transmittance. "Optics", by Hecht is easy reading.

Is this a Thevenin source? If so, what is the internal resistor set to in
terms of Z0?

The example was a Thevenin source driving a 1/2WL open
line through a source resistor equal to Z0, e.g. 50 ohms.

For condition number 2 below, is this a Thevenin equivalent resistance?

Yes, all the rules of superposition and Thevenin equivalency
are being followed.

1. To add the total power of each wave and then also add the power
reduction of the phase relationship, would be creating power unless the
cos(A) is negative. Something seems wrong here, probably my understanding.

Yep, I haven't given the other corresponding equation. The
pair of equations that satisfy the conservation of energy
requirements are (in Ramo&Whinnery Poynting vector notation):

|Pz+| = P1 + P2 + 2*SQRT(P1*P2)cos(A) eq.1

|Pz-| = P3 + P4 + 2*SQRT(P3*P4)cos(-A) eq.2

In a transmission line, the Poynting vector for Pz+ and the
Poynting vector for Pz- are pointed in opposite directions.

The destructive interference, as you observed, carries a
negative sign. This negative sign does NOT create energy.
It means that some excess amount of destructive interference
energy must go somewhere. It naturally goes into constructive
interference somewhere else. Please see:

http://www.mellesgriot.com/products/optics/oc_2_1.htm

http://micro.magnet.fsu.edu/primer/j...ons/index.html

2. The two voltages should be equal, therfore the power delivered by each
to the source resister would be equal.

Under each of the two superposition steps, yes. However, when
the signals are combined, interference results, and the power
in the source resistor is reduced to zero. That "excess" energy
must go somewhere and it appears as constructive interference
redistributed toward the load. See the web pages above.

3. The power delivered to the source resistor will arrive a different times
due to the phase difference between the two waves.

We are talking average power. Please don't get bogged down in
instantaneous power which would be the same when integrated.

4. If the reflected power returns at the same time as the delivered power
(to the source resistor), no power will flow. This because the resistor in
a Thevenin is a series resistor. Equal voltages will be applied to each side
of the resitor. The voltage difference will be zero.

Yes, the result is destructive interference.

5. If the reflected power returns 180 degrees out of phase with the applied
voltage (to the source resistor), the voltage across the resistor will
double with each cycle, resulting in an ever increasing current.(and power)
into the reflected wave.

Good time to switch over to a Norton equivalent.

This is correct for this condition. The problem with the equation comes
when cos(90) which can easily happen.

That's no problem at all. It is just constructive interference
which implies destructive interference somewhere else. Destructive
interference must always equal constructive interference to avoid
violation of the conservation of energy principle.

Light, especially solar energy, is a collection of waves of all magnitudes
and frequencies. We should see only about 1/2 of the power that actually
arrives due to superposition. The reflection from a surface should create
standing waves in LOCAL space that will contain double the power of open
space. I think your equation Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A) would be
correct for a collection of waves, but incorrect for the example.

A collection of waves is not likely to be coherent so the equation
would not work. That equation works for any coherent EM wave. CW RF
waves are coherent.

I tried to read your article a couple of weeks ago, but I found myself not
understanding. I have learned a lot since then thanks to you, Roy, and
others. I will try to read it again soon.

You might want to pick up a copy of "Optics", by Hecht
available from http://www.abebooks.com Please pay close
attention to the chapters on interference and superposition.

So what do you think Cecil?

Also try HP's AN 95-1, an s-parameter app-note available on the
web. Pay close attention to what they say about power and what
happens when you square the s-parameter equations. Surprise!
You get the power interference equation above.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
The resistor in the Thevenin/Norton equivalent source is selected with
some criteria in mind. What I would like to do is to design a Thevenin
source to provide 1v across a 50 ohm transmission line INFINATELY long,
ignoring ohmic resistance. I would like the source resistor to absorb as
much power as the line, so that if power ever returns under reflected
wave conditions, it can all be absorbed by the resistor. I think the
size of such a resistor will be 50 ohms.

If you allow reflected energy to flow through the
source resistor, destructive interference is often
the result and the resistor therefore will not
dissipate the reflected power.

The only way to get the source resistor to dissipate
all of the reflected power is to cause total constructive
interference within the source. That requires all of the
energy from the load side of the system.

When the source rejects reflected energy due to destructive
interference, that energy has no choice except to reverse
direction and flow back toward the load as constructive
interference energy.

Maybe you should reconsider the circulator+load-resistor?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Follow up and more corrections.
Roger wrote:
Correction. Please note the change below. I apologize for the error.

Roger Sparks wrote:
"Cecil Moore" wrote in message
...
Roger wrote:
The principles of superposition are mathematically usable, not too
hard, and I think very revealing. Yes, if we use part of the model,
we must use it all the way. To do otherwise would be error, or worse.

clip some

Roger, I have explained all of this before. If you are
capable of understanding it, I take my hat off to you.
Here's what happens in that ideal voltage source using
the rules of superposition.


Well, I would like to think I could understand it, but maybe something
is wrong like Roy suggests. I think it is all falling into place, but
our readers are not all in agreement.

Could I ask a couple of questions to make sure I am understanding your
preconditions?

Is this a Thevenin source? If so, what is the internal resistor set
to in terms of Z0?

1. With the reflected voltage set to zero, the source
current is calculated which results in power in the
source resistor. Psource = Isource^2*Rsource


For condition number 2 below, is this a Thevenin equivalent resistance?

2. With the source voltage set to zero, the reflected
current is calculated which results in power in the
source resistor. Pref = Iref^2*Rsource

Now superpose the two events. The resultant power
equation is: Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(A)
where 'A' is the angle between the source current
and the reflected current. Note this is NOT power
superposition. It is the common irradiance (power
density) equation from the field of optics which
has been in use in optical physics for a couple of
centuries so it has withstood the test of time.

OK. A few reactions.

1. To add the total power of each wave and then also add the power
reduction of the phase relationship, would be creating power unless
the cos(A) is negative. Something seems wrong here, probably my
understanding.

2. The two voltages should be equal, therfore the power delivered by
each to the source resister would be equal.

3. The power delivered to the source resistor will arrive a different
times due to the phase difference between the two waves.

4. If the reflected power returns at the same time as the delivered
power (to the source resistor), no power will flow. This because the
resistor in a Thevenin is a series resistor. Equal voltages will be
applied to each side of the resitor. The voltage difference will be
zero.

5. If the reflected power returns 180 degrees out of phase with the
applied voltage (to the source resistor), the voltage across the
resistor will double with each cycle, resulting in an ever increasing
current.(and power) into the reflected wave.

For the voltage source looking into a 1/2WL open
circuit stub, angle 'A' is 180 degrees. Therefore
the total power in the source resistor is:

Ptot = Ps + Pr + 2*SQRT(Ps*Pr)cos(180) or

Ptot = Ps + Pr - 2*SQRT(Ps*Pr) = 0

This is correct for this condition. The problem with the equation
comes when cos(90) which can easily happen.

I would suggest the equation Ptot = Ps+ Pr*(sine(A) where angle 'A' is
the angle between positive peaks of the two waves. This angle will
rotate twice as fast as the signal frequency due to the relative
velocity between the waves..
No, this angle is fixed by system design. If the system changes, this
angle will rotate twice as fast as the angle change of a single wave.
The equation should be OK.

The equation can NOT be OK for anything except a single interaction
point. It can not be used to plot the interaction between waves because
the wave location information is not present.

Ptot in the source resistor is zero watts but we
already knew that. What is important is that the
last term in that equation above is known as the
"interference term". When it is negative, it indicates
destructive interference. When Ptot = 0, we have
"total destructive interference" as defined by
Hecht in "Optics", 4th edition, page 388.

Whatever the magnitude of the destructive interference
term, an equal magnitude of constructive interference
must occur in a different direction in order to
satisfy the conservation of energy principle. Thus
we can say with certainty that the energy in the
reflected wave has been 100% re-reflected by the
source. The actual reflection coefficient of the
source in the example is |1.0| even when the source
resistor equals the Z0 of the transmission line.
I have been explaining all of this for about 5 years
now. Instead of attempting to understand these relatively
simple laws of physics from the field of optics, Roy
ploinked me. Hopefully, you will understand.

All of this is explained in my three year old Worldradio
energy article on my web page.

http://www.w5dxp.com/energy.htm
--
73, Cecil http://www.w5dxp.com

Light, especially solar energy, is a collection of waves of all
magnitudes and frequencies. We should see only about 1/2 of the power
that actually arrives due to superposition. The reflection from a
surface should create standing waves in LOCAL space that will contain
double the power of open space. I think your equation Ptot = Ps + Pr
+ 2*SQRT(Ps*Pr)cos(A) would be correct for a collection of waves, but
incorrect for the example.

I tried to read your article a couple of weeks ago, but I found myself
not understanding. I have learned a lot since then thanks to you,
Roy, and others. I will try to read it again soon.

So what do you think Cecil?

73, Roger, W7WkB


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)".

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Jan 2, 1:41 am, Roger wrote:
Keith Dysart wrote:
On Jan 1, 6:00 pm, Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Could you better describe how you determine that the source has a Z0
equal to the line Z0? I can guess that you use a Thévenin equivalent
circuit and set the series resistor to Z0.
Probably the simplest way is to put the entire source circuitry into a
black box. Measure the terminal voltage with the box terminals open
circuited, and the current with the terminals short circuited. The ratio
of these is the source impedance. If you replace the box with a Thevenin
or Norton equivalent, this will be the value of the equivalent circuit's
impedance component (a resistor for most of our examples).
If the driving circuitry consists of a perfect voltage source in series
with a resistance, the source Z will be the resistance; if it consists
of a perfect current source in parallel with a resistance, the source Z
will be the resistance. You can readily see that the open circuit V
divided by the short circuit I of these two simple circuits equals the
value of the resistance.
The power output of the Thévenin equivalent circuit follows the load.
Sorry, I don't understand this. Can you express it as an equation?
There seems to be some confusion as to the terms "Thévenin equivalent
circuit", "ideal voltage source", and how impedance follows these
sources.
Two sources we all have access to are these links:
Voltage source:
http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem
I don't disagree with anything I read there.
But you may not quite have the concept of impedance
correct.
The impedance of the Thevenin/Norton equivalent source
is not V/I but rather the slope of the line representing
the relationship of the voltage to the current.
When there is a source present, this
line does not pass through the origin. Only for
passive components does this line pass through
the origin in which case it becomes V/I.
Because the voltage to current plot for an ideal
voltage source is horizontal, the slope is 0 and hence
so is the impedance. For an ideal current source
the slope is vertical and the impedance is infinite.
Hoping this helps clarify....
...Keith
Yes, I can understand that there is no change in voltage no matter what
the current load is, so there can be no resistance or reactive component
in the source. The ideal voltage link also said that the ideal source
could maintain voltage no matter what current was applied. Presumably a
negative current through the source would result in the same voltage as
a positive current, which is logical if the source has zero impedance.

This is true.

During the transition from negative current passing through the source
to positive current passing through the source, the current at some time
must be zero.

This would occur, for example, when the output terminals
are open circuited. Or connected to something that had
the same open-circuit voltage as your source.

How is the impedance of the perfect source defined at
this zero current point?

This is not a different question than: How is the
resistance of a resistor defined when it has no
current flowing in it?

It continues to satisfy the relation V = I * R,
though one can not compute R from the measured
V and I.

How is the impedance of the attached system
defined?

It is unknown, but has a voltage equal to the
voltage of the source.

The resistor in the Thevenin/Norton equivalent source is selected with
some criteria in mind. What I would like to do is to design a Thevenin
source to provide 1v across a 50 ohm transmission line INFINATELY
long, ignoring ohmic resistance. I would like the source resistor to
absorb as much power as the line, so that if power ever returns under
reflected wave conditions, it can all be absorbed by the resistor. I
think the size of such a resistor will be 50 ohms.

That is the correct value to not have any reflections
at the source.

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.

...Keith

Very interesting! It makes sense to me. I must be gone most of the day
today so will be QRT for a while.

I am gaining an appreciation for the time boundaries here. To properly
account for the power, we would need to integrate the source power over
the entire time from switch on to switch off plus the time required for
the reflected wave to completely return and die out.

If we consider the extended time period, within that period the power
delivered by the source should equal the power dissipated by the
resistor. The forward and reflected waves should only serve to be
storage for the power during portions of the extended time period.

73, Roger, W7WKB