Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

I wonder why we can not get the same results?

If you mean for the calculation of the voltage at 100 ns and 20
degrees down the line, it's because of my error. It should be sin(36 +
20 degrees) ~ 0.83.

From the equation vf(t,x) = sin(wt-x), I am getting
vf(36 degrees, 20 degrees) = sin(36-20) = sin(16) = 0.276v.

I should quit before I get farther behind! In my haste I made a second
error by adding rather than subtracting the 20 degrees. Again, you're
correct and I apologize for the error.

I'll take extra care to try and avoid goofing up again.

Could we look at five points on the example? (The example has frequency
of 1 MHz, entered the transmission line 100 ns prior to the time of
interest, and traveled 36 degrees into the line)

Ok.

Using zero voltage on
the leading edge as a reference point on the sine wave, and the input
point on the transmission line as the second reference point, find the
voltage at (36, -20), (36, 0), (36, 20), (36, 36)and (36, 40). All
points are defined in degrees.

I'm a little confused by your dual reference points, both of which seem
to refer to physical positions. In my analysis, I give an equation for a
voltage as a function of both time and physical position. The time (t)
reference point is the time at which the source is connected. At that
instant, the sine wave source voltage is zero. The reference point for
position (x) is the input end of the line. The voltage at any time and
any position can be determined from the equation, provided that the wave
being referred to has reached the position on the line at or before the
time being evaluated (and provided that I don't make a stupid arithmetic
error -- or two). You'll note that in my analysis, I usually give a time
at which the equation is valid -- this is to insure that the wave has
reached any point of interest on the line.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.

The equation vf = sin(wt - x) isn't valid until the wave has reached
point x on the line. It's also not valid at any point not on the line.
So it can't be used under those conditions.

At line input, at 0 degrees, sin(36+0) = 0.59v.

100 ns after turn-on, the wave has propagated 36 degrees down the line,
so it's present at the input end (x = 0) and the equation is valid. The
voltage at the line input is as you calculated, 0.59 v. at that time.

On the line, at +20 degrees, sin(36-20) = 0.276v.

Correct.

On the line, at +36 degrees, sin(36-36) = 0v.

Correct.

On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

Correct.

Each of us must be using a different reference point because we are
getting different results.

Without my careless errors, we get the same result except for the first
case.

As I mentioned, time domain analysis of transmission lines is
relatively rare. But there's a very good treatment in Johnk,
_Engineering Electromagnetic Fields and Waves_. Another good reference
is Johnson, _Electric Transmission Lines_. Near the end of Chapter 14,
he actually has an example with a zero-impedance source: "Let us
assume that the generator is without impedance, so that any wave
arriving at the transmitting end of the line is totally reflected with
reversal of voltage; the reflection factor at the sending end is thus
-1." And he goes through a brief version of essentially the same thing
I did to arrive at the steady state.

I wonder if it is possible that Johnson, _Electric Transmission Lines_,
presented an incorrect example? That occasionally happens, but rarely.

Certainly it happens, but both he and I get the same, correct result
using the same method. And it seems unlikely that both he and Johnk made
the same error.

Or maybe some subtle condition assumption is different making the
example unrelated to our experiment?

I'm not sure why you see this as necessary. Do you have an analysis
which correctly predicts the voltage at all times on the line after
startup but which uses some different interaction of waves returning to
the source? If so, please present it. I've presented mine and shown that
I arrived at the correct result.

clip.......
The "perfect voltage source" controls or better, overwhelms any
effect that might be caused by the reflected wave. It completely
defeats any argument or description about reflected waves.

If you'd like, I can pretty easily modify my analysis to include a
finite source resistance of your choice. It will do is modify the
source reflection coefficient and allow the system to converge to
steady state. Would you like me to? There's no reason the choice of a
perfect voltage source should interfere with the understanding of
what's happening -- none of the phenomena require it.

Just for conversation, we could place a 50 ohm resistor in parallel with
the full wave 50 ohm transmission line, which is open ended. At
startup, the "perfect voltage source" would see a load of 50/2 = 25
ohms.

No, at startup, the source always sees Z0, regardless of the load
termination. The load termination has no effect at the source end until
the reflected wave returns. And if the line is terminated in Z0, then
there is no reflected wave and the source sees Z0 forever. To the load,
a terminated line looks exactly like a plain resistor of value Z0. (I'm
assuming a lossless line as we have been all along.)

At steady state, the "perfect voltage source" would see a load of
50 ohms in parallel with "something" from the transmission line. The
power output from the "perfect voltage source" would be reduced below
the startup output. We would arrive at that conclusion using traveling
waves, tracing the waves as they move toward stability.

With the line terminated in 50 ohms, steady state is reached as soon as
the initial forward wave reaches the load.

If the termination is some value other than Z0, the steady state
impedance seen by the transmission line will be the load impedance (for
our example one wavelength line or any line an integral number of half
wavelengths long). When starting, the initial impedance is always Z0 for
one round trip, for any line length. Then it will jump or down. With
each successive round trip, the impedance will either monotonically step
in the direction of the final value, or oscillate around it but with the
excursion decreasing each time. Which behavior you'll see and the rate
at which it converges are dictated by the relationship among source,
load, and characteristic impedances.

Would it be acceptable to use a perfect CURRENT source, along with a
parallel resistor. Then CURRENT would remain constant, but voltage
would vary. Again, power into the test circuit would vary.

The transmission line will react to a perfect current source in parallel
with a resistance exactly the same as for a perfect voltage source with
a series resistance. Put them in boxes and there's no test you can
devise which can distinguish them by tests of the terminal
characteristics. So yes, that's fine, or just a black box about which
the contents are completely unknown except for any two of the open
circuit voltage, short circuit current, or impedance. (It's assumed that
the box contains some linear circuit that doesn't change during the
analysis.) The results will be identical for any of the three choices.

When we define both the source voltage and the source impedance, we also
define the source power. Two of the three variables in the power
equation are defined, so power is defined.

We define the maximum amount of power which can be extracted from the
box, but not the power that it's delivering. If we open circuit or short
circuit the box or terminate it with a pure reactance, it's delivering
no power at all. If we terminate it with the complex conjugate of its
impedance, it's delivering V^2 / (2 * Rp) where Rp is the parallel
equivalent load resistance and V is the open circuit voltage. With any
other termination it's delivering less than that but more than zero.

Now if the we use such a source, a reflection would bring additional
power back to the input.

This presumes that there are power waves bouncing around on the line.

We would need to begin the analysis all over
again as if we were restarting the experiment, this time with two
voltages applied (the source and reflected voltages). Then we would
have the question: Should the two voltages should be added in series, or
in parallel?

That's not a question I can answer, since it's a consequence of a
premise I don't believe. If your premise has merit, you should be able
to express it mathematically and arrive at the answer to the question.
Feel free to do an analysis using multiple sources and traveling waves
of power. If you get the correct answer for how a line actually behaves,
we'll try it out with a number of different conditions and see if it
holds up.

Your answer has been to use a reflection factor of -1, which would be to
reverse the polarity. This presents a dilemma because when the
reflected voltage is equal to the forward voltage, the sum of either the
parallel or series addition is zero.

That's correct but no dilemma.

You can see what that does to our
analysis. Power just disappears so long as the reflective wave is
returning, as if we are turning off the experiment during the time the
reflective wave returns.

By "our" analysis do you mean my analysis or the one you propose?
There's no rule saying power can't disappear -- power is not conserved.
In fact, look at any point along the open circuited transmission line
and you'll find that the power "disappears" -- goes to zero -- twice
each V or I cycle. When you first turn on the source, the source
produces average power. In steady state, the average power is zero. It
"disappeared", without any dissipative elements in the circuit. Energy
*is* conserved, however, and there is no energy disappearing in my
analysis. Feel free to try and conserve power in your analysis if you
want. But energy had better be conserved in the process.

Would a "perfect CURRENT source" without any restrictions about
impedance work as an initial point for you? That would be a source that
supplied one amp but rate of power delivery could vary.

Sure. The only difference to the analysis procedure would be to change
the source reflection coefficient from -1 to +1, since a perfect current
source has an infinite impedance. I don't know how it would affect the
final outcome without going through the steps. Of course, the reflection
coefficient for the current wave would go from +1 to -1, so I imagine
you'd run into the same conceptual problem if you tried doing an
analysis of the current waves.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Keith Dysart wrote:
On Dec 30, 9:51 am, Cecil Moore wrote:
Roger wrote:
Roy Lewallen wrote:
Yes, that is correct. The impedance of the source (a perfect voltage
source) is zero, so the reflection coefficient seen by the reverse
traveling wave is -1.
The logic of this assumption eludes me. In fact, it seems completely
illogical and counter to the concept of how voltage and current waves
are observed to move on a transmission line.

The reflection coefficient for a short is -1, is it not?

One way of viewing a short is that it is a perfect
voltage source (i.e. 0 output impedance) set to 0
volts.

Good point. The power disappears here, but energy transfer is evident
(we have current). We both have power and do not have power.

At the source, when the reflect wave returns and re-reflects, we have 2v
from the reflected wave matched with 1v from the source. We have both
1v and 2v.

OK.

Setting it to some other voltage (or function describing
the voltage) does not alter its output impedance. It
there fore creates the same reflection of the travelling
wave, regardless of the voltage function it is generating.

A real world voltage source has an output impedance. Use
this impedance to compute the reflection coefficient.
The reflection will be the same regardless of the
voltage function being generated by the source.

...Keith

Thanks for pointing out the short circuit voltage view. I had not
thought of that.

What would you think of using a perfect current source for these thought
experiments? My suggested perfect current source would supply only as
much current as could be absorbed by the load, so no power would be used
by the source, and current would be limited by the load.

73, Roger, W7WKB
continuing and expanding.

Maybe we should be considering a perfect POWER source, which could only
emit power, never absorb it. The power out would be defined by load
impedance, just as it is for the perfect voltage source.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:

I predict that the pulse arriving at the left end will
have the same voltage, current and energy profile as
the pulse launched at the right end and the pulse
arriving at the right end will be similar to the
one launched at the left.

They will appear exactly AS IF they had passed
through each other.

The difficulty with saying THE pulses passed
through each other arises with the energy. The
energy profile of the pulse arriving at the left
will look exactly like that of the one launched
from the right so it will seem that the energy
travelled all the way down the line for delivery
at the far end. And yet, from the experiment above,
when the pulses arriving from each end have the
same shape, no energy crosses the middle of the
line.

So it would seem that the energy that actually
crosses the middle during the collision is
exacly the amount of energy that is needed to
reconstruct the pulses on each side after the
collision.

If all the energy that is launched at one end
does not travel to the other end, then I am
not comfortable saying that THE pulse travelled
from one end to the other.

But I have no problem saying that the system
behaves AS IF the pulses travelled from one
end to the other.

You said that:

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.

Yet it sounds like you are saying that despite this charge repulsion and
bouncing of waves off each other, each wave appears to be completely
unaltered by the other? It seems to me that surely we would detect some
trace of this profound effect.

. . .

Is there any test you can conceive of which would produce different
measurable results if the pulses were repelling and bouncing off each
other or just passing by without noticing the other?

There are equations describing system behavior on the assumption of no
wave interaction, and these equations accurately predict all measurable
aspects of line behavior without exception. Have you developed equations
based on this charge interaction which predict line behavior with
equal accuracy and universal applicability?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 5:30*pm, Roger wrote:
Cecil Moore wrote:
Roger wrote:
[snip]
However, I can see the dilemma faced by a purist who sees 2v from a
reflected wave (because the reflected wave has returned to the source
and reflected as if it were an open end) and the 1v from the source at
exactly the same location. *Something must be wrong.

A perfect voltage source has an output impedance of 0,
the same impedance as a short. The reflection
coefficient is -1.

The returning wave has a voltage Vref(t) which reflects
as -Vref(t).

So total V(t) at the output is
V(t) = Vsig(t) + Vref(t) - Vref(t)
= Vsig(t)

as expected.

I don't recall any examples using perfect CURRENT sources. *I think a
perfect current source would supply a signal that could respond to
changing impedances correctly. *It should solve the dilemma caused by
the rise in voltage which occurs when when a traveling wave doubles
voltage upon encountering an open circuit, or reversing at the source.

What do you think?

A perfect current source has an output impedance of
infinity, just like an open circuit. The reflection
coefficient is 1.

Similar to the reflected voltage for the perfect
voltage source, the reflected current cancels leaving
just the current from the perfect current source.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

Maybe we should be considering a perfect POWER source, which could only
emit power, never absorb it. The power out would be defined by load
impedance, just as it is for the perfect voltage source.

Why do you consider that to be necessary?

A very serious problem with this is that you've now got a nonlinear
element, and we can no longer use linear circuit analysis and would have
to resort to the basic time domain differential equations describing
voltage and current relationships. Feel free to do the analysis; I won't
do that one.

If your concepts of how a transmission line works can't handle simple
cases like a perfect voltage source in series with a resistance, you
might consider reexamining your concepts.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 5:53*pm, Roger wrote:
Keith Dysart wrote:
On Dec 30, 9:51 am, Cecil Moore wrote:
Roger wrote:
Roy Lewallen wrote:
Yes, that is correct. The impedance of the source (a perfect voltage
source) is zero, so the reflection coefficient seen by the reverse
traveling wave is -1.
The logic of this assumption eludes me. *In fact, it seems completely
illogical and counter to the concept of how voltage and current waves
are observed to move on a transmission line.

The reflection coefficient for a short is -1, is it not?

One way of viewing a short is that it is a perfect
voltage source (i.e. 0 output impedance) set to 0
volts.

Good point. *The power disappears here, but energy transfer is evident
(we have current). *We both have power and do not have power.

At the source, when the reflect wave returns and re-reflects, we have 2v
from the reflected wave matched with 1v from the source. *We have both
1v and 2v.

OK.



Setting it to some other voltage (or function describing
the voltage) does not alter its output impedance. It
there fore creates the same reflection of the travelling
wave, regardless of the voltage function it is generating.

A real world voltage source has an output impedance. Use
this impedance to compute the reflection coefficient.
The reflection will be the same regardless of the
voltage function being generated by the source.

...Keith

Thanks for pointing out the short circuit voltage view. *I had not
thought of that.

What would you think of using a perfect current source for these thought
experiments? *My suggested perfect current source would supply only as
much current as could be absorbed by the load, so no power would be used
by the source, and current would be limited by the load.

A perfect current source always provides the current
that it is set to regardless of the load impedance.
It does so by changing its output voltage.

(A perfect voltage source always provides the voltage
that it is set to regardless of the load impedance.
It does so by changing its output current.)

Doing the thought experiments with both voltage and
current sources is quite useful. The results are
always complementary, but I find that is solidifies
the learnings.

Real sources can be modelled as Thevenin sources;
that is, an ideal voltage source in series with
a resistor equal to the output impedance.

They can similarly be modelled as Norton sources;
that is, an ideal current source in parallel with
a resistor equal to the output impedance.

And a Thevenin source can be transformed into an
equivalent Norton source. Example...

A 100 V ideal voltage source in series with 50 ohms
has exactly the same output characteristics as a
2 amp ideal current source in parallel with a 50 ohm
resistor. Test using any load impedance: open, short,
any resistance value.

Either one is matched to a 50 ohm line (because
the output impedance is 50 ohms) and any reverse
wave will not be re-reflected but will disappear
into the generator.

...Keith

PS: The above descriptions generalize for impedances
that are not pure resistances but also contain
reactance.

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 30, 5:13*pm, Cecil Moore wrote:
Keith Dysart wrote:
My model requires nothing of EM waves, in-so-far-as
it is completely described in terms of charge. Just
the basics for current flow.

Yes, I agree. Your model doesn't even require the
EM waves to obey the laws of physics. It's a lot
like my mother's model which requires only God.

Are you saying that charge can not be used to
understand circuit behaviour? Are all the books
wrong? Please expand.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

So far I've seen a number of comments that claim my analysis is wrong
for one reason or the other, in spite of the fact that it correctly
predicts the line behavior as verified by SPICE. The main thrust of the
arguments is that my analysis fails to agree with preconceived notions
of traveling power waves and the invalidity of superposition.

Yet I've seen no alternative analysis or equations which correctly
predict the startup line behavior by using other methods which
incorporate these notions.

Why?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:
Roger wrote:
Roy Lewallen wrote:

I wonder why we can not get the same results?

If you mean for the calculation of the voltage at 100 ns and 20
degrees down the line, it's because of my error. It should be sin(36
+ 20 degrees) ~ 0.83.

From the equation vf(t,x) = sin(wt-x), I am getting
vf(36 degrees, 20 degrees) = sin(36-20) = sin(16) = 0.276v.

I should quit before I get farther behind! In my haste I made a second
error by adding rather than subtracting the 20 degrees. Again, you're
correct and I apologize for the error.

I'll take extra care to try and avoid goofing up again.

Could we look at five points on the example? (The example has
frequency of 1 MHz, entered the transmission line 100 ns prior to the
time of interest, and traveled 36 degrees into the line)

Ok.

Using zero voltage on the leading edge as a reference point on the
sine wave, and the input
point on the transmission line as the second reference point, find the
voltage at (36, -20), (36, 0), (36, 20), (36, 36)and (36, 40). All
points are defined in degrees.

I'm a little confused by your dual reference points, both of which seem
to refer to physical positions. In my analysis, I give an equation for a
voltage as a function of both time and physical position. The time (t)
reference point is the time at which the source is connected. At that
instant, the sine wave source voltage is zero. The reference point for
position (x) is the input end of the line. The voltage at any time and
any position can be determined from the equation, provided that the wave
being referred to has reached the position on the line at or before the
time being evaluated (and provided that I don't make a stupid arithmetic
error -- or two). You'll note that in my analysis, I usually give a time
at which the equation is valid -- this is to insure that the wave has
reached any point of interest on the line.

Not on the line yet, at -20 degrees, sin(36+20) = 0.83.

The equation vf = sin(wt - x) isn't valid until the wave has reached
point x on the line. It's also not valid at any point not on the line.
So it can't be used under those conditions.

At line input, at 0 degrees, sin(36+0) = 0.59v.

100 ns after turn-on, the wave has propagated 36 degrees down the line,
so it's present at the input end (x = 0) and the equation is valid. The
voltage at the line input is as you calculated, 0.59 v. at that time.

On the line, at +20 degrees, sin(36-20) = 0.276v.

Correct.

On the line, at +36 degrees, sin(36-36) = 0v.

Correct.

On the line, at +40 degrees, sin(undefined, -40) = (wave has not arrived)

Correct.

Each of us must be using a different reference point because we are
getting different results.

Without my careless errors, we get the same result except for the first
case.

I make errors as well, so no criticism from here. I am just glad that
we are on the same page on how we use the equation.

As for the (36, -20) point, I was indeed specifying a point that had not
been supplied to the transmission line. It was a prediction, as if the
wave existed physically and was moving toward our experiment. We could
not actually observe it until time = 56 degrees.

As I mentioned, time domain analysis of transmission lines is
relatively rare. But there's a very good treatment in Johnk,
_Engineering Electromagnetic Fields and Waves_. Another good
reference is Johnson, _Electric Transmission Lines_. Near the end of
Chapter 14, he actually has an example with a zero-impedance source:
"Let us assume that the generator is without impedance, so that any
wave arriving at the transmitting end of the line is totally
reflected with reversal of voltage; the reflection factor at the
sending end is thus -1." And he goes through a brief version of
essentially the same thing I did to arrive at the steady state.

I wonder if it is possible that Johnson, _Electric Transmission
Lines_, presented an incorrect example? That occasionally happens,
but rarely.

Certainly it happens, but both he and I get the same, correct result
using the same method. And it seems unlikely that both he and Johnk made
the same error.

Or maybe some subtle condition assumption is different making the
example unrelated to our experiment?

I'm not sure why you see this as necessary. Do you have an analysis
which correctly predicts the voltage at all times on the line after
startup but which uses some different interaction of waves returning to
the source? If so, please present it. I've presented mine and shown that
I arrived at the correct result.

clip.......
The "perfect voltage source" controls or better, overwhelms any
effect that might be caused by the reflected wave. It completely
defeats any argument or description about reflected waves.

If you'd like, I can pretty easily modify my analysis to include a
finite source resistance of your choice. It will do is modify the
source reflection coefficient and allow the system to converge to
steady state. Would you like me to? There's no reason the choice of a
perfect voltage source should interfere with the understanding of
what's happening -- none of the phenomena require it.

Just for conversation, we could place a 50 ohm resistor in parallel
with the full wave 50 ohm transmission line, which is open ended. At
startup, the "perfect voltage source" would see a load of 50/2 = 25 ohms.

No, at startup, the source always sees Z0, regardless of the load
termination. The load termination has no effect at the source end until
the reflected wave returns. And if the line is terminated in Z0, then
there is no reflected wave and the source sees Z0 forever. To the load,
a terminated line looks exactly like a plain resistor of value Z0. (I'm
assuming a lossless line as we have been all along.)

At steady state, the "perfect voltage source" would see a load of 50
ohms in parallel with "something" from the transmission line. The
power output from the "perfect voltage source" would be reduced below
the startup output. We would arrive at that conclusion using
traveling waves, tracing the waves as they move toward stability.

With the line terminated in 50 ohms, steady state is reached as soon as
the initial forward wave reaches the load.

If the termination is some value other than Z0, the steady state
impedance seen by the transmission line will be the load impedance (for
our example one wavelength line or any line an integral number of half
wavelengths long). When starting, the initial impedance is always Z0 for
one round trip, for any line length. Then it will jump or down. With
each successive round trip, the impedance will either monotonically step
in the direction of the final value, or oscillate around it but with the
excursion decreasing each time. Which behavior you'll see and the rate
at which it converges are dictated by the relationship among source,
load, and characteristic impedances.

Would it be acceptable to use a perfect CURRENT source, along with a
parallel resistor. Then CURRENT would remain constant, but voltage
would vary. Again, power into the test circuit would vary.

The transmission line will react to a perfect current source in parallel
with a resistance exactly the same as for a perfect voltage source with
a series resistance. Put them in boxes and there's no test you can
devise which can distinguish them by tests of the terminal
characteristics. So yes, that's fine, or just a black box about which
the contents are completely unknown except for any two of the open
circuit voltage, short circuit current, or impedance. (It's assumed that
the box contains some linear circuit that doesn't change during the
analysis.) The results will be identical for any of the three choices.

When we define both the source voltage and the source impedance, we
also define the source power. Two of the three variables in the power
equation are defined, so power is defined.

We define the maximum amount of power which can be extracted from the
box, but not the power that it's delivering. If we open circuit or short
circuit the box or terminate it with a pure reactance, it's delivering
no power at all. If we terminate it with the complex conjugate of its
impedance, it's delivering V^2 / (2 * Rp) where Rp is the parallel
equivalent load resistance and V is the open circuit voltage. With any
other termination it's delivering less than that but more than zero.

Now if the we use such a source, a reflection would bring additional
power back to the input.

This presumes that there are power waves bouncing around on the line.

We would need to begin the analysis all over
again as if we were restarting the experiment, this time with two
voltages applied (the source and reflected voltages). Then we would
have the question: Should the two voltages should be added in series,
or in parallel?

That's not a question I can answer, since it's a consequence of a
premise I don't believe. If your premise has merit, you should be able
to express it mathematically and arrive at the answer to the question.
Feel free to do an analysis using multiple sources and traveling waves
of power. If you get the correct answer for how a line actually behaves,
we'll try it out with a number of different conditions and see if it
holds up.

Your answer has been to use a reflection factor of -1, which would be
to reverse the polarity. This presents a dilemma because when the
reflected voltage is equal to the forward voltage, the sum of either
the parallel or series addition is zero.

That's correct but no dilemma.

You can see what that does to our analysis. Power just disappears so
long as the reflective wave is returning, as if we are turning off the
experiment during the time the reflective wave returns.

By "our" analysis do you mean my analysis or the one you propose?
There's no rule saying power can't disappear -- power is not conserved.
In fact, look at any point along the open circuited transmission line
and you'll find that the power "disappears" -- goes to zero -- twice
each V or I cycle. When you first turn on the source, the source
produces average power. In steady state, the average power is zero. It
"disappeared", without any dissipative elements in the circuit. Energy
*is* conserved, however, and there is no energy disappearing in my
analysis. Feel free to try and conserve power in your analysis if you
want. But energy had better be conserved in the process.

Would a "perfect CURRENT source" without any restrictions about
impedance work as an initial point for you? That would be a source
that supplied one amp but rate of power delivery could vary.

Sure. The only difference to the analysis procedure would be to change
the source reflection coefficient from -1 to +1, since a perfect current
source has an infinite impedance. I don't know how it would affect the
final outcome without going through the steps. Of course, the reflection
coefficient for the current wave would go from +1 to -1, so I imagine
you'd run into the same conceptual problem if you tried doing an
analysis of the current waves.

Roy Lewallen, W7EL

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance.
Mathematically, the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed by
the source.

Just as for both perfect voltage and perfect current sources, the actual
power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being
zero.

Would the "perfect power source" be acceptable to you?

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 30, 5:53 pm, Roger wrote:
Keith Dysart wrote:
On Dec 30, 9:51 am, Cecil Moore wrote:
Roger wrote:
Roy Lewallen wrote:
Yes, that is correct. The impedance of the source (a perfect voltage
source) is zero, so the reflection coefficient seen by the reverse
traveling wave is -1.
The logic of this assumption eludes me. In fact, it seems completely
illogical and counter to the concept of how voltage and current waves
are observed to move on a transmission line.
The reflection coefficient for a short is -1, is it not?
One way of viewing a short is that it is a perfect
voltage source (i.e. 0 output impedance) set to 0
volts.
Good point. The power disappears here, but energy transfer is evident
(we have current). We both have power and do not have power.

At the source, when the reflect wave returns and re-reflects, we have 2v
from the reflected wave matched with 1v from the source. We have both
1v and 2v.

OK.



Setting it to some other voltage (or function describing
the voltage) does not alter its output impedance. It
there fore creates the same reflection of the travelling
wave, regardless of the voltage function it is generating.
A real world voltage source has an output impedance. Use
this impedance to compute the reflection coefficient.
The reflection will be the same regardless of the
voltage function being generated by the source.
...Keith
Thanks for pointing out the short circuit voltage view. I had not
thought of that.

What would you think of using a perfect current source for these thought
experiments? My suggested perfect current source would supply only as
much current as could be absorbed by the load, so no power would be used
by the source, and current would be limited by the load.

A perfect current source always provides the current
that it is set to regardless of the load impedance.
It does so by changing its output voltage.

(A perfect voltage source always provides the voltage
that it is set to regardless of the load impedance.
It does so by changing its output current.)

Doing the thought experiments with both voltage and
current sources is quite useful. The results are
always complementary, but I find that is solidifies
the learnings.

Real sources can be modelled as Thevenin sources;
that is, an ideal voltage source in series with
a resistor equal to the output impedance.

They can similarly be modelled as Norton sources;
that is, an ideal current source in parallel with
a resistor equal to the output impedance.

And a Thevenin source can be transformed into an
equivalent Norton source. Example...

A 100 V ideal voltage source in series with 50 ohms
has exactly the same output characteristics as a
2 amp ideal current source in parallel with a 50 ohm
resistor. Test using any load impedance: open, short,
any resistance value.

Either one is matched to a 50 ohm line (because
the output impedance is 50 ohms) and any reverse
wave will not be re-reflected but will disappear
into the generator.

...Keith

PS: The above descriptions generalize for impedances
that are not pure resistances but also contain
reactance.

Thanks for this summary Keith. Discussion about traveling waves keeps
getting trapped with reflected waves being inverted, absorbed, or
disappear because of source problems. We end up discussing the source,
not the traveling waves.

In a posting to Roy, I propose using a perfect power source, which would
be a combination of perfect voltage and perfect current. The advantage
is that output power would follow the load, not be permanently defined
by the load like most consider to be the case for a perfect voltage or
current source.

I won't repeat that posting here, but please comment.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance. Mathematically,
the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed by
the source.

Just as for both perfect voltage and perfect current sources, the actual
power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being zero.

Would the "perfect power source" be acceptable to you?

No. As I mentioned on another thread, this is a nonlinear device, which
would not permit using the linear circuit analysis I used. It would
require reverting to fundamental differential equations, which I'm not
willing to do. I'll be willing to use absolutely any linear source
(which doesn't change during the analysis period) you can devise. Any
linear source can be reduced to a Thevenin or Norton equivalent to
produce identical results.

Can't your concept of transmission lines deal with linear sources? If
not, why not?

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance.
Mathematically, the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed
by the source.

Just as for both perfect voltage and perfect current sources, the
actual power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being zero.

Would the "perfect power source" be acceptable to you?

No. As I mentioned on another thread, this is a nonlinear device, which
would not permit using the linear circuit analysis I used. It would
require reverting to fundamental differential equations, which I'm not
willing to do. I'll be willing to use absolutely any linear source
(which doesn't change during the analysis period) you can devise. Any
linear source can be reduced to a Thevenin or Norton equivalent to
produce identical results.

Can't your concept of transmission lines deal with linear sources? If
not, why not?

Roy Lewallen, W7EL

No, my concept of transmission lines deals fine with linear sources, it
is the non-linear constant voltage source and constant current source
that can not handle a second source of power arriving at a time later
than the original pulse.

Thank you for much thoughtful discussion. I have learned a lot, and
sharpened my skills. I won't drop the topic, because it can bear great
fruit, but let's you and I drop it for a while.

You may be right about differential equations needed for an perfect
POWER source, but so far, I don't think so. I am being honest that I
think your analysis applies up to the point of using the -1 reflection
factor at the source. In fact, I will probably use it in the future,
with an attribute to you.

Following Keith's post discussing zero voltage as a current source, I
can see why you might insist on using the -1 reflection factor. As time
passes, I will try to improve the concept of the perfect POWER source to
see if that can bridge the conceptual differences, but retain the
relative simplicity of sine waves adding.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
Roger wrote:

I agree that we would have the same problem with a perfect current
source which had an infinite impedance.

How about using a perfect POWER source, that could not absorb power.
Output power would be limited by the external impedance.
Mathematically, the perfect POWER source would be described by

Ps = (Vp^2)/Zvp = Zip * Ip^2 where Ps = maximum power output, Vp =
voltage from perfect voltage source, Zvp = 0, Zip = infinity, and Ip =
current from a perfect current source.

The power output could be infinite, but power could never be absorbed
by the source.

Just as for both perfect voltage and perfect current sources, the
actual power output would be limited by external loads.

The impedance of the perfect POWER source would be Vp/Ip, both
controlled by external loads. To me that means that the output power
from the perfect POWER source would follow the impedance presented by
the load, but power going into the source would be defined as being zero.

Would the "perfect power source" be acceptable to you?

No. As I mentioned on another thread, this is a nonlinear device, which
would not permit using the linear circuit analysis I used. It would
require reverting to fundamental differential equations, which I'm not
willing to do. I'll be willing to use absolutely any linear source
(which doesn't change during the analysis period) you can devise. Any
linear source can be reduced to a Thevenin or Norton equivalent to
produce identical results.

Can't your concept of transmission lines deal with linear sources? If
not, why not?

Roy Lewallen, W7EL

No, my concept of transmission lines deals fine with linear sources, it
is the non-linear constant voltage source and constant current source
that can not handle a second source of power arriving at a time later
than the original pulse.

Thank you for much thoughtful discussion. I have learned a lot, and
sharpened my skills. I won't drop the topic, because it can bear great
fruit, but let's you and I drop it for a while.

You may be right about differential equations needed for an perfect
POWER source, but so far, I don't think so. I am being honest that I
think your analysis applies up to the point of using the -1 reflection
factor at the source. In fact, I will probably use it in the future,
with an attribute to you.

Following Keith's post discussing zero voltage as a current source, I
can see why you might insist on using the -1 reflection factor. As time
passes, I will try to improve the concept of the perfect POWER source to
see if that can bridge the conceptual differences, but retain the
relative simplicity of sine waves adding.

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

I was afraid that a mathematical treatment wouldn't be appropriate for
this forum -- it is, after all, an amateur newsgroup -- and it looks
like I was right. It's taken a considerable amount of time to develop
and present the analysis and answer questions as completely as I can,
but it appears from the responses that the time was largely wasted. It's
been my hope that at least a few "lurkers" have gained some insight from
it. If so, please drop me an email letting me know one way or the other
-- don't worry, I'll keep your email confidential. I'd been considering
making up an analytical and quantitative analysis of where the energy
goes in a transmission line. But it's certainly pointless if my analysis
of even the most basic development of line voltage can't be understood
or believed, as seems to be the case.

Roy Lewallen, W7EL

Standing-Wave Current vs Traveling-Wave Current
 

On Sun, 30 Dec 2007 19:08:44 -0800, Roger wrote:

No, my concept of transmission lines deals fine with linear sources, it
is the non-linear constant voltage source and constant current source
that can not handle a second source of power arriving at a time later
than the original pulse.

Hi Roger,

That is not the problem of the sources, it is your problem alone in
coming to terms with the network. Any reference that reaches deep
into the fundamentals of lines and networks begins with the Thevenin
source (and at least one off of my shelf delves into two Thevenin
sources facing each other).
"There are waves of identical frequency traveling
in both directions on the line, but their amplitudes
and phases are independently variable,
and neither can be called "incident" or "reflected"
waves."

It is notable that the author expressly offers only one set of
equations for the distributed voltage and current along a line and
distinctly says:
"It is worth noting that (8.1) and (8.2) [those equations]
are also applicable to Fig. 8-2 below [showing the dual
source configuration with deliberate matches], a
distinctly different transmission line circuit."

What makes it "a distinctly different transmission line circuit" is
that it is in fact one source feeding both ends. It settles the hash
about frequency, phase, coherence, amplitude, matching, and all the
other folderol that attends many of Cecil's jejune postings:
"Here a single source supplies signals to both ends
of a transmission line section, through networks that terminate
the section in its characteristic impedance at each end."

As time
passes, I will try to improve the concept of the perfect POWER source to
see if that can bridge the conceptual differences, but retain the
relative simplicity of sine waves adding.

The two equations, needing no reference to a constant power source:
V(z) = V1· e^-y·z + V2· e^+y·z
I(z) = (V1· e^-y·z - V2· e^+y·z) / Zo
where
z: the distance along a line
y = a + jB
a: nepers per unit length of line
B = 2 · pi / wavelength
V1 & V2 are arbitrary voltage phasors to be determined by
boundary conditions at the ends of the line.

It took very little effort to then proceed to the obvious:
"Whenever two waves of identical frequency travel
in opposite directions on a transmission line ...
the fundamental phenomenon of interference or
'standing waves' occurs. ... exhibits periodic maxima
and minima ... in its most striking form when
the two oppositely directed waves have equal
amplitude and the transmission line system has
zero attenuation."

Elaborations of stepped pulses of energy had better resolve to
identical analysis using the math above, or the strain of elaboration
has led to sterile inventions.

73's
Richard Clark, KB7QHC

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:
They engaged in typical author-speak.

I think not.

You are welcome to your opinion. Authors always
couch their assertions in probabilities by
never uttering an absolute lest they be proved
wrong by one esoteric example.

Well, there is no energy flowing through the '+' points.

I proved that energy is flowing through the '+' points
before the line is cut. Superposition proves that there
is energy flowing through those points.

And I have no issue if you wish to claim that there
are reflections at these points, though I might use
'bouncing' to differentiate from reflections occuring
at points with non-zero reflection coefficients.

So bouncing is what happens at points with zero reflection
coefficients. You seem to have invented a new religion.

No energy is flowing (q.v. IEEE definition of
instantaneous power), and yet you want energy
to be flowing.

Lots of energy is flowing in both directions.
Only the *NET* energy flow is zero.

Although many have tried to prove that the output (source)
impedance is the impedance encountered by the reflected waves,
all of those numerous experiments have failed.

You, Cecil, are the only one who believes this. Any good
book on transmission lines will tell you otherwise.

I am not surprised that you are ignorant of the raging
arguments that have been going on primarily between
Bruene and Maxwell and their respective supporters.
I believe it continued to rage in the 2007 letters to
the QEX editors.

Web references and Spice models which agree that "the
output (source) impedance is the impedance encountered
by the reflected waves" have been previously provided,
but you refused to explore them.

No, I asked you to measure the reflection coefficients
and report the results. You refused to do so. If the issue
had ever been resolved, it would be common knowledge and
we wouldn't be arguing about it.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:

The impedance of a perfect voltage source IS zero.

Please measure it and report your results.

The fact that 0 can not be achieved does not detract

If the goal is to confuse, confound, and promote
ignorance, be my guest.

A voltage source has two sides? Explain!

All of my references illustrate that voltage source
with two leads. Please pass a TV signal through it
to prove it is a zero impedance.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
However, I can see the dilemma faced by a purist who sees 2v from a
reflected wave (because the reflected wave has returned to the source
and reflected as if it were an open end) and the 1v from the source at
exactly the same location. Something must be wrong.

It can happen at a Z0-match point where destructive
interference exists on the source side and constructive
interference exists on the load side.

Here's an example:

Source-+---1/2WL 300 ohm line---50 ohm load

Assume the source voltage is 1 volt
The reflected voltage is ~2.5v
The forward voltage is ~3.5v
The load voltage is 1 volt

I don't recall any examples using perfect CURRENT sources.

A Norton equivalent circuit is a perfect current source.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Richard Clark wrote:
Given this equivalency, the forced power presumption collapses. Power
in the black box (if in fact that is the intent of this coy
"perfection") cannot be known.

Every reference I have on Thevenin and Norton equivalent
circuits say essentially that same thing.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
My suggested perfect current source would supply only as
much current as could be absorbed by the load, so no power would be used
by the source, and current would be limited by the load.

That would be a Norton equivalent source with the
parallel source resistor set to infinity.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
Maybe we should be considering a perfect POWER source, which could only
emit power, never absorb it. The power out would be defined by load
impedance, just as it is for the perfect voltage source.

I have been using such a source as an example for years.
I call it the SGCL, signal generator equipped with a
circulator and load resistor. Here's the diagram.


100wSG---1---2----
\ /
3
|
R

No reflections incident upon the source because they
are all dissipated in the resistor R. It's a great
way to simplify examples.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Similar to the reflected voltage for the perfect
voltage source, the reflected current cancels leaving
just the current from the perfect current source.

Zero power dissipation in a Thevenin equivalent
source equates to maximum power dissipation in
a Norton equivalent source. That sure shoots
down any argument about dissipation in the source.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
And a Thevenin source can be transformed into an
equivalent Norton source. Example...

A 100 V ideal voltage source in series with 50 ohms
has exactly the same output characteristics as a
2 amp ideal current source in parallel with a 50 ohm
resistor. Test using any load impedance: open, short,
any resistance value.

Indeed, take a look at the internal power dissipation
for an open and short and tell us again why those
sources are identical.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Cecil Moore wrote:
Your model doesn't even require the
EM waves to obey the laws of physics. It's a lot
like my mother's model which requires only God.

Are you saying that charge can not be used to
understand circuit behaviour? Are all the books
wrong? Please expand.

Your EM waves reflect when there is no reflective media.
That is a violation of the laws of physics. Please
demonstrate two light waves reflecting off of each other
in free space.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
So far I've seen a number of comments that claim my analysis is wrong
for one reason or the other, in spite of the fact that it correctly
predicts the line behavior as verified by SPICE.

Roy, you wrote down the equation for V(x,t) for a
standing wave voltage. Now write down the equation
for I(x,t) for a standing wave current and tell us
again how to use that standing wave current with
its constant phase at different points up and down
the line, to measure the delay through the line.

Since Roy has ploinked me, he will ignore this posting.
Would some kind soul please reply to this posting so
Roy will see it? Thanks.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
So far I've seen a number of comments that claim my analysis is wrong
for one reason or the other, ...

Would you mind sharing those reasons with us?
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
I won't repeat that posting here, but please comment.

Will this source satisfy your needs? It is linear
and it prohibits reflected energy from reaching
the source making tracking energy rather simple.
I usually define it as a 50 ohm device but other
impedances could be chosen.

100w--1---2--
\ /
3
|
R
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roy Lewallen wrote:
I'd been considering
making up an analytical and quantitative analysis of where the energy
goes in a transmission line.

It would certainly be informative to see how you get
those reflected voltages and reflected currents to
exist without energy.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 31, 12:43*am, Cecil Moore wrote:
Keith Dysart wrote:
A voltage source has two sides? Explain!

All of my references illustrate that voltage source
with two leads. Please pass a TV signal through it
to prove it is a zero impedance.

Oh that is what you meant. So an amplifier has 4 sides?

Anyway, examples of what you are looking for exist
in most amplifiers. They are often called blocking
capacitors. They charge to the DC voltage but pass
the signal. They are not particularly good voltage
sources, but that is their role in the circuit, and
they do it admirably well.

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 31, 1:32*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
Your model doesn't even require the
EM waves to obey the laws of physics. It's a lot
like my mother's model which requires only God.

Are you saying that charge can not be used to
understand circuit behaviour? Are all the books
wrong? Please expand.

Your EM waves reflect when there is no reflective media.
That is a violation of the laws of physics. Please
demonstrate two light waves reflecting off of each other
in free space.

Do you ever answer a direct question?

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 31, 12:36*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
They engaged in typical author-speak.

I think not.

You are welcome to your opinion. Authors always
couch their assertions in probabilities by
never uttering an absolute lest they be proved
wrong by one esoteric example.

Of course. That is writing with precision, as I said.

Well, there is no energy flowing through the '+' points.

I proved that energy is flowing through the '+' points
before the line is cut.

Unfortunately, this has not being proved as yet, but
merely assumed.

Superposition proves that there
is energy flowing through those points.

I thought that you had previously agreed that it was
not appropriate to superpose power. This can be
revisited if you have changed your mind.

And I have no issue if you wish to claim that there
are reflections at these points, though I might use
'bouncing' to differentiate from reflections occuring
at points with non-zero reflection coefficients.

So bouncing is what happens at points with zero reflection
coefficients. You seem to have invented a new religion.

No. I was merely offering an alternate term tht you might
use when claiming that they reflect, since reflect causes
you such grief.

No energy is flowing (q.v. IEEE definition of
instantaneous power), and yet you want energy
to be flowing.

Lots of energy is flowing in both directions.
Only the *NET* energy flow is zero.

Back to superposing power.

Although many have tried to prove that the output (source)
impedance is the impedance encountered by the reflected waves,
all of those numerous experiments have failed.

You, Cecil, are the only one who believes this. Any good
book on transmission lines will tell you otherwise.

I am not surprised that you are ignorant of the raging
arguments that have been going on primarily between
Bruene and Maxwell and their respective supporters.
I believe it continued to rage in the 2007 letters to
the QEX editors.

That argument was more about whether the output impedance
of an amateur transmitter was well defined. In my encounters
with the arguments I don't recall any claim that if it
was well defined and equal to the line impedance, then
there would be a reflection. If this claim was made,
then someone needed to revisit the books.

Web references and Spice models which agree that "the
output (source) impedance is the impedance encountered
by the reflected waves" have been previously provided,
but you refused to explore them.

No, I asked you to measure the reflection coefficients
and report the results. You refused to do so.

Exactly. You refused to check your textbooks. You refused to
review the web references. You refused to examine the
spice models. Instead you ask me to measure something and
expect me to believe that some measurement I make will
convince you. Just another way to delay.

If the issue
had ever been resolved, it would be common knowledge and
we wouldn't be arguing about it.

There is only one place that this is being argued. If you
would review your textbooks, if you would look at the web
references, or if you would examine the spice models, you
would learn that the argument was settled long ago.
(Actually, there probably never was an argument, except
on r.r.a.a.)

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

On Dec 31, 1:27*am, Cecil Moore wrote:
Keith Dysart wrote:
And a Thevenin source can be transformed into an
equivalent Norton source. Example...

A 100 V ideal voltage source in series with 50 ohms
has exactly the same output characteristics as a
2 amp ideal current source in parallel with a 50 ohm
resistor. Test using any load impedance: open, short,
any resistance value.

Indeed, take a look at the internal power dissipation
for an open and short and tell us again why those
sources are identical.

Are you saying that there output characteristics are
different because of differing internal power dissipations?

And are you not the lad who mentioned that one should
never examine the internal power dissipation of a Norton
or Thevenin equivalent circuit because, in general, it
bears no relation to the dissipation in the real
circuit? Or have you changed your view?

...Keith

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
I won't repeat that posting here, but please comment.

Will this source satisfy your needs? It is linear
and it prohibits reflected energy from reaching
the source making tracking energy rather simple.
I usually define it as a 50 ohm device but other
impedances could be chosen.

100w--1---2--
\ /
3
|
R

Thanks for alerting me to this device, which is real.

Thanks also for pointing out (in another posting) that the Norton source
is a constant current source.

But no, this will not do. The problem is that "We want to investigate a
1/2 wave length of transmission line, excited at one end. How soon is
stability reached?"

No doubt, past postings have not so clearly stated the scope of
investigation.

Unstated, and perhaps confusing, is the the power level must be limited
to some real value.

Also unstated, but reasonably assumed, is the understanding that the
exciting source would not be part of the circuit except as needed to
provide the excitation.

This last assumption was never achieved because the responders all
insisted that the source must be part of the investigation. It was
further insisted that the source invert the reflected waves with a -1
reflection factor. Had a +1 factor been acceptable, which is equivalent
to the reflection factor of an open circuit, the assumption would have
been realized.

Several responders discussed the Norton or Thévenin equivalent circuits.
These circuits assume steady state conditions, and are intended to be
replacements for groups of components. A very useful concept, but not
appropriate for startup excitation of a transmission line unless the
intent was to investigate the source AND transmission line as a system.

More later. I as still thinking about the "perfect power source".

73, Roger, W7WKB

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Anyway, examples of what you are looking for exist
in most amplifiers. They are often called blocking
capacitors. They charge to the DC voltage but pass
the signal. They are not particularly good voltage
sources, but that is their role in the circuit, and
they do it admirably well.

Sorry, no modifications allowed. I simply want you
to prove that the voltage source impedance minus
the series resistor is zero ohms as you say.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
Do you ever answer a direct question?

No answer needed or expected for irrelevant rhetorical
questions.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Roger wrote:
The problem is that "We want to investigate a
1/2 wave length of transmission line, excited at one end. How soon is
stability reached?"

I guess the answer depends upon your definition of
"stability" above.

You might start with a loaded version:

http://www.w5dxp.com/1secsgat.gif
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
No. I was merely offering an alternate term tht you might
use when claiming that they reflect, since reflect causes
you such grief.

When your alternative involves something supernatural
happening when the reflection coefficient is zero,
I think I will pass.

Back to superposing power.

Power can certainly be added using the power density
equations but scalars cannot be superposed.

... the argument was settled long ago.

Absolutely false since the argument is still raging.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Keith Dysart wrote:
On Dec 31, 1:27 am, Cecil Moore wrote:
Indeed, take a look at the internal power dissipation
for an open and short and tell us again why those
sources are identical.

Are you saying that there output characteristics are
different because of differing internal power dissipations?

No, I'm saying what all the references say - that power
dissipation inside an equivalent source is irrelevant.
Yet you seem convinced that you know the internal power
reflection coefficient. You are contradicting the
references.
--
73, Cecil http://www.w5dxp.com

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:


Lots of energy is flowing in both directions.
Only the *NET* energy flow is zero.


Cecil,

I guess you still have not gone back to the books to try to understand
what electromagnetic energy is all about. A good review of the Poynting
theorem would help to minimize the sort of nonsense you spouted above.

73,
Gene
W4SZ

Standing-Wave Current vs Traveling-Wave Current
 

Cecil Moore wrote:
Roger wrote:
Maybe we should be considering a perfect POWER source, which could
only emit power, never absorb it. The power out would be defined by
load impedance, just as it is for the perfect voltage source.

I have been using such a source as an example for years.
I call it the SGCL, signal generator equipped with a
circulator and load resistor. Here's the diagram.


100wSG---1---2----
\ /
3
|
R

No reflections incident upon the source because they
are all dissipated in the resistor R. It's a great
way to simplify examples.


Cecil,

It is interesting that you ridicule zero impedance ideal sources, and
then you provide an example with a circulator. Do you really think a
circulator is more ideal than a good voltage source?

73,
Gene
W4SZ