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